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From: SourceForge.net <noreply@so...>  20050430 11:10:09

Bugs item #1192935, was opened at 20050430 04:10 Message generated for change (Tracker Item Submitted) made by Item Submitter You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1192935&group_id=4933 Category: Xmaxima Group: Fix for 5.9.2 Status: Open Resolution: None Priority: 5 Submitted By: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: termsubstitution in function sum is too late Initial Comment: (%i1) "Lower RiemannSum, Intervall [0,1], 10 parts"$ (%i2) /*Function*/ f(x):=x^2$ (%i3) /*area width*/ b: 1/10$ (%i4) /*area depending on k=0,...,9*/ Ak: b*f(k*b)$ (%i5) Ak; 2 k (%o5)  1000 (%i6) /* wrong */ sum(Ak,k,0,9); 2 k (%o6)  100 obviously, we first have the summation and then the symbolsubstitution (%i7) sum(''Ak,k,0,9); 57 (%o7)  200 finally, this is ok, but the usage of the doublequote is hard to explain to beginners. it gets even worse, if you use a function A(k):=k*f(k*b)$ for a sum with n partitions: (%i8) "Lower RiemannSum, Intervall [0,1], n parts"$ (%i9) /*area width*/ b:1/n$ (%i10) /*area depending on k=0,...,n1*/ A(k):=k*f(k*b)$ (%i11) A(k); 3 k (%o11)  2 n (%i12) sum(A(k),k,0,n1); n  1 ==== (%o12) > A(k) / ==== k = 0 in that case you need ''(A(k)), a bracket and the doublequote. is it possible to patch the sumfunction in that way, that we first have the term or symbolsubstitution and second the summation?  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1192935&group_id=4933 
From: SourceForge.net <noreply@so...>  20050501 12:09:41

Bugs item #1192935, was opened at 20050430 06:10 Message generated for change (Comment added) made by willisbl You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1192935&group_id=4933 Category: Xmaxima Group: Fix for 5.9.2 Status: Open Resolution: None Priority: 5 Submitted By: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: termsubstitution in function sum is too late Initial Comment: (%i1) "Lower RiemannSum, Intervall [0,1], 10 parts"$ (%i2) /*Function*/ f(x):=x^2$ (%i3) /*area width*/ b: 1/10$ (%i4) /*area depending on k=0,...,9*/ Ak: b*f(k*b)$ (%i5) Ak; 2 k (%o5)  1000 (%i6) /* wrong */ sum(Ak,k,0,9); 2 k (%o6)  100 obviously, we first have the summation and then the symbolsubstitution (%i7) sum(''Ak,k,0,9); 57 (%o7)  200 finally, this is ok, but the usage of the doublequote is hard to explain to beginners. it gets even worse, if you use a function A(k):=k*f(k*b)$ for a sum with n partitions: (%i8) "Lower RiemannSum, Intervall [0,1], n parts"$ (%i9) /*area width*/ b:1/n$ (%i10) /*area depending on k=0,...,n1*/ A(k):=k*f(k*b)$ (%i11) A(k); 3 k (%o11)  2 n (%i12) sum(A(k),k,0,n1); n  1 ==== (%o12) > A(k) / ==== k = 0 in that case you need ''(A(k)), a bracket and the doublequote. is it possible to patch the sumfunction in that way, that we first have the term or symbolsubstitution and second the summation?  >Comment By: Barton Willis (willisbl) Date: 20050501 07:09 Message: Logged In: YES user_id=895922 We've discussed this before; see http://www.math.utexas.edu/pipermail/maxima/2003/004869.html http://www.math.utexas.edu/pipermail/maxima/2003/004870.html Maybe a simple workaround such as mysum(f,v,lo,hi):=block([acc:0], if integerp(lo) and integerp(hi) then for i from lo thru hi do acc:acc+substitute(i,v,f) else acc:funmake('mysum,[f,v,lo,hi]), acc) or mysum(f,lo,hi):= block([acc:0], if integerp(lo) and integerp(hi) then for i : lo thru hi do acc : acc + apply(f,[i]) else acc : funmake('mysum,[f,lo,hi]), acc); might work for you, Barton  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1192935&group_id=4933 
From: SourceForge.net <noreply@so...>  20050530 15:11:25

Bugs item #1192935, was opened at 20050430 06:10 Message generated for change (Settings changed) made by amundson You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1192935&group_id=4933 Please note that this message will contain a full copy of the comment thread, including the initial issue submission, for this request, not just the latest update. >Category: Lisp Core >Group: None Status: Open Resolution: None Priority: 5 Submitted By: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: termsubstitution in function sum is too late Initial Comment: (%i1) "Lower RiemannSum, Intervall [0,1], 10 parts"$ (%i2) /*Function*/ f(x):=x^2$ (%i3) /*area width*/ b: 1/10$ (%i4) /*area depending on k=0,...,9*/ Ak: b*f(k*b)$ (%i5) Ak; 2 k (%o5)  1000 (%i6) /* wrong */ sum(Ak,k,0,9); 2 k (%o6)  100 obviously, we first have the summation and then the symbolsubstitution (%i7) sum(''Ak,k,0,9); 57 (%o7)  200 finally, this is ok, but the usage of the doublequote is hard to explain to beginners. it gets even worse, if you use a function A(k):=k*f(k*b)$ for a sum with n partitions: (%i8) "Lower RiemannSum, Intervall [0,1], n parts"$ (%i9) /*area width*/ b:1/n$ (%i10) /*area depending on k=0,...,n1*/ A(k):=k*f(k*b)$ (%i11) A(k); 3 k (%o11)  2 n (%i12) sum(A(k),k,0,n1); n  1 ==== (%o12) > A(k) / ==== k = 0 in that case you need ''(A(k)), a bracket and the doublequote. is it possible to patch the sumfunction in that way, that we first have the term or symbolsubstitution and second the summation?  Comment By: Barton Willis (willisbl) Date: 20050501 07:09 Message: Logged In: YES user_id=895922 We've discussed this before; see http://www.math.utexas.edu/pipermail/maxima/2003/004869.html http://www.math.utexas.edu/pipermail/maxima/2003/004870.html Maybe a simple workaround such as mysum(f,v,lo,hi):=block([acc:0], if integerp(lo) and integerp(hi) then for i from lo thru hi do acc:acc+substitute(i,v,f) else acc:funmake('mysum,[f,v,lo,hi]), acc) or mysum(f,lo,hi):= block([acc:0], if integerp(lo) and integerp(hi) then for i : lo thru hi do acc : acc + apply(f,[i]) else acc : funmake('mysum,[f,lo,hi]), acc); might work for you, Barton  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1192935&group_id=4933 
From: SourceForge.net <noreply@so...>  20050613 02:34:22

Bugs item #1192935, was opened at 20050430 05:10 Message generated for change (Comment added) made by robert_dodier You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1192935&group_id=4933 Please note that this message will contain a full copy of the comment thread, including the initial issue submission, for this request, not just the latest update. Category: Lisp Core Group: None Status: Open Resolution: None Priority: 5 Submitted By: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: termsubstitution in function sum is too late Initial Comment: (%i1) "Lower RiemannSum, Intervall [0,1], 10 parts"$ (%i2) /*Function*/ f(x):=x^2$ (%i3) /*area width*/ b: 1/10$ (%i4) /*area depending on k=0,...,9*/ Ak: b*f(k*b)$ (%i5) Ak; 2 k (%o5)  1000 (%i6) /* wrong */ sum(Ak,k,0,9); 2 k (%o6)  100 obviously, we first have the summation and then the symbolsubstitution (%i7) sum(''Ak,k,0,9); 57 (%o7)  200 finally, this is ok, but the usage of the doublequote is hard to explain to beginners. it gets even worse, if you use a function A(k):=k*f(k*b)$ for a sum with n partitions: (%i8) "Lower RiemannSum, Intervall [0,1], n parts"$ (%i9) /*area width*/ b:1/n$ (%i10) /*area depending on k=0,...,n1*/ A(k):=k*f(k*b)$ (%i11) A(k); 3 k (%o11)  2 n (%i12) sum(A(k),k,0,n1); n  1 ==== (%o12) > A(k) / ==== k = 0 in that case you need ''(A(k)), a bracket and the doublequote. is it possible to patch the sumfunction in that way, that we first have the term or symbolsubstitution and second the summation?  >Comment By: Robert Dodier (robert_dodier) Date: 20050612 20:34 Message: Logged In: YES user_id=501686 OK, for the record, here is a definition of $sum which implements one of the ideas that has been proposed, namely this policy: evaluate the summand after binding the summation variable to itself. (defmspec $sum (l) (setq l (cdr l)) (if (= (length l) 4) (progv (list (cadr l)) (list (cadr l)) (dosum (meval (car l)) (cadr l) (meval (caddr l)) (meval (cadddr l)) t)) (wnaerr '$sum))) This version yields the results expected by the person who originated this bug report, and yields the results predicted by S Macrakis in the email referenced below (http://www.math.utexas.edu/pipermail/maxima/2003/004869.html), and run_testsuite() runs to completion with no errors. I'm in favor of making this change, but I'm just recording this code snippet here for future reference; no changes planned at the moment. For comparison here is the current version of $sum: (defmspec $sum (l) (setq l (cdr l)) (if (= (length l) 4) (dosum (car l) (cadr l) (meval (caddr l)) (meval (cadddr l)) t) (wnaerr '$sum)))  Comment By: Barton Willis (willisbl) Date: 20050501 06:09 Message: Logged In: YES user_id=895922 We've discussed this before; see http://www.math.utexas.edu/pipermail/maxima/2003/004869.html http://www.math.utexas.edu/pipermail/maxima/2003/004870.html Maybe a simple workaround such as mysum(f,v,lo,hi):=block([acc:0], if integerp(lo) and integerp(hi) then for i from lo thru hi do acc:acc+substitute(i,v,f) else acc:funmake('mysum,[f,v,lo,hi]), acc) or mysum(f,lo,hi):= block([acc:0], if integerp(lo) and integerp(hi) then for i : lo thru hi do acc : acc + apply(f,[i]) else acc : funmake('mysum,[f,lo,hi]), acc); might work for you, Barton  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1192935&group_id=4933 
From: SourceForge.net <noreply@so...>  20050613 11:31:00

Bugs item #1192935, was opened at 20050430 06:10 Message generated for change (Comment added) made by willisbl You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1192935&group_id=4933 Please note that this message will contain a full copy of the comment thread, including the initial issue submission, for this request, not just the latest update. Category: Lisp Core Group: None Status: Open Resolution: None Priority: 5 Submitted By: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: termsubstitution in function sum is too late Initial Comment: (%i1) "Lower RiemannSum, Intervall [0,1], 10 parts"$ (%i2) /*Function*/ f(x):=x^2$ (%i3) /*area width*/ b: 1/10$ (%i4) /*area depending on k=0,...,9*/ Ak: b*f(k*b)$ (%i5) Ak; 2 k (%o5)  1000 (%i6) /* wrong */ sum(Ak,k,0,9); 2 k (%o6)  100 obviously, we first have the summation and then the symbolsubstitution (%i7) sum(''Ak,k,0,9); 57 (%o7)  200 finally, this is ok, but the usage of the doublequote is hard to explain to beginners. it gets even worse, if you use a function A(k):=k*f(k*b)$ for a sum with n partitions: (%i8) "Lower RiemannSum, Intervall [0,1], n parts"$ (%i9) /*area width*/ b:1/n$ (%i10) /*area depending on k=0,...,n1*/ A(k):=k*f(k*b)$ (%i11) A(k); 3 k (%o11)  2 n (%i12) sum(A(k),k,0,n1); n  1 ==== (%o12) > A(k) / ==== k = 0 in that case you need ''(A(k)), a bracket and the doublequote. is it possible to patch the sumfunction in that way, that we first have the term or symbolsubstitution and second the summation?  >Comment By: Barton Willis (willisbl) Date: 20050613 06:30 Message: Logged In: YES user_id=895922 If we decide to use Robert's $sum function, the translate property for sum will need to be changed. Consider: (1) Redefine $sum as Robert suggested. (2) Try this: (%i1) ak : k^2$ (%i2) g(a,n) := sum(a,k,1,n)$ (%i3) g(ak,5); (%o3) 55 < correct for Robert's $sum function (%i4) translate(g); (%o4) [g] (%i5) g(ak,5); (%o5) 5*k^2 < yeech There might be other things that need fixing: (%i8) properties(sum); (%o8) [Special Evaluation Form,OUTATIVE,NOUN,RULE] Barton  Comment By: Robert Dodier (robert_dodier) Date: 20050612 21:34 Message: Logged In: YES user_id=501686 OK, for the record, here is a definition of $sum which implements one of the ideas that has been proposed, namely this policy: evaluate the summand after binding the summation variable to itself. (defmspec $sum (l) (setq l (cdr l)) (if (= (length l) 4) (progv (list (cadr l)) (list (cadr l)) (dosum (meval (car l)) (cadr l) (meval (caddr l)) (meval (cadddr l)) t)) (wnaerr '$sum))) This version yields the results expected by the person who originated this bug report, and yields the results predicted by S Macrakis in the email referenced below (http://www.math.utexas.edu/pipermail/maxima/2003/004869.html), and run_testsuite() runs to completion with no errors. I'm in favor of making this change, but I'm just recording this code snippet here for future reference; no changes planned at the moment. For comparison here is the current version of $sum: (defmspec $sum (l) (setq l (cdr l)) (if (= (length l) 4) (dosum (car l) (cadr l) (meval (caddr l)) (meval (cadddr l)) t) (wnaerr '$sum)))  Comment By: Barton Willis (willisbl) Date: 20050501 07:09 Message: Logged In: YES user_id=895922 We've discussed this before; see http://www.math.utexas.edu/pipermail/maxima/2003/004869.html http://www.math.utexas.edu/pipermail/maxima/2003/004870.html Maybe a simple workaround such as mysum(f,v,lo,hi):=block([acc:0], if integerp(lo) and integerp(hi) then for i from lo thru hi do acc:acc+substitute(i,v,f) else acc:funmake('mysum,[f,v,lo,hi]), acc) or mysum(f,lo,hi):= block([acc:0], if integerp(lo) and integerp(hi) then for i : lo thru hi do acc : acc + apply(f,[i]) else acc : funmake('mysum,[f,lo,hi]), acc); might work for you, Barton  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1192935&group_id=4933 
From: SourceForge.net <noreply@so...>  20050614 05:00:44

Bugs item #1192935, was opened at 20050430 05:10 Message generated for change (Comment added) made by robert_dodier You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1192935&group_id=4933 Please note that this message will contain a full copy of the comment thread, including the initial issue submission, for this request, not just the latest update. Category: Lisp Core Group: None Status: Open Resolution: None Priority: 5 Submitted By: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: termsubstitution in function sum is too late Initial Comment: (%i1) "Lower RiemannSum, Intervall [0,1], 10 parts"$ (%i2) /*Function*/ f(x):=x^2$ (%i3) /*area width*/ b: 1/10$ (%i4) /*area depending on k=0,...,9*/ Ak: b*f(k*b)$ (%i5) Ak; 2 k (%o5)  1000 (%i6) /* wrong */ sum(Ak,k,0,9); 2 k (%o6)  100 obviously, we first have the summation and then the symbolsubstitution (%i7) sum(''Ak,k,0,9); 57 (%o7)  200 finally, this is ok, but the usage of the doublequote is hard to explain to beginners. it gets even worse, if you use a function A(k):=k*f(k*b)$ for a sum with n partitions: (%i8) "Lower RiemannSum, Intervall [0,1], n parts"$ (%i9) /*area width*/ b:1/n$ (%i10) /*area depending on k=0,...,n1*/ A(k):=k*f(k*b)$ (%i11) A(k); 3 k (%o11)  2 n (%i12) sum(A(k),k,0,n1); n  1 ==== (%o12) > A(k) / ==== k = 0 in that case you need ''(A(k)), a bracket and the doublequote. is it possible to patch the sumfunction in that way, that we first have the term or symbolsubstitution and second the summation?  >Comment By: Robert Dodier (robert_dodier) Date: 20050613 23:00 Message: Logged In: YES user_id=501686 Actually, after looking at the code some more (specifically DOSUM in src/asum.lisp) it looks to me like the intent is indeed to evaluate the summand after binding the summation index to itself. In addition, there is code to assume the index is between its lower and upper limits (although assume/forget is handled incorrectly, see 851765). I believe that the observed behavior is a bug, in the narrow sense that the observed behavior is different from what the code author intended. See also my comments on SF bug # 740134. Here is a different patch. I like this one better than the previous one. In DOSUM: ;; ORIGINAL: MEVAL ONCE ;; (mbinding (lind l*i) (meval exp)) (mbinding (lind l*i) (meval (meval exp))) In MEVALSUMARG: ;; ORIGINAL: MEVALATOMS ;; (resimplify (mevalatoms (if (and (not (atom exp)) (resimplify (meval (if (and (not (atom exp)) With these changes, the results are as expected for the examples cited by the original poster, and also the example about the translated function produces the correct result, and some examples adapted from another bug report (740134) yield correct results. ex 1: f(x):=x^2$ b: 1/10$ Ak: b*f(k*b)$ sum(Ak,k,0,9); => 57/200 ex 2: A(k):=k*f(k*b)$ b:1/n$ A(k) => k^3/n^2 sum(A(k),k,0,n1) => ('sum(k^3,k,0,n1))/n^2 sum(A(k),k,0,n1), simpsum => ((n1)^4 + 2*(n1)^3 + (n1)^2) / (4*n^2) ex 3: ak : k^2$ g(a,n) := sum(a,k,1,n)$ g(ak,5) => 55 translate (g)$ g(ak,5) => 55 some other examples, adapted from bug # 740134: ex 4: sum (print (i), i, 1, 3) => prints 1, 2, 3 then returns 6 ex 5: sum (integrate (x^i ,x), i, 0, 2) => x^3/3 + x^2/2 + x ex 6: sum (integrate (1/(x^i + 1), x), i, 0, 1) => log(x+1) + x/2 ex 7: f[i](x):=x^i$ g[i](x):=x^i$ h[i](x):=x^i$ /* reference f[i] and g[i]  see 740134 for the effect this has on previous defn of sum */ f[i]$ g[i](t)$ sum (f[i](x), i, 0, n) => 'sum (x^i, i, 0, n) sum (g[i](x), i, 0, n) => 'sum (x^i, i, 0, n) sum (h[i](x), i, 0, n) => 'sum (x^i, i, 0, n) ex 8: sum (integrate (x^i, x), i, 0, n) => 'sum (x^(i+1) / (i+1), i, 0, n)  Comment By: Barton Willis (willisbl) Date: 20050613 05:30 Message: Logged In: YES user_id=895922 If we decide to use Robert's $sum function, the translate property for sum will need to be changed. Consider: (1) Redefine $sum as Robert suggested. (2) Try this: (%i1) ak : k^2$ (%i2) g(a,n) := sum(a,k,1,n)$ (%i3) g(ak,5); (%o3) 55 < correct for Robert's $sum function (%i4) translate(g); (%o4) [g] (%i5) g(ak,5); (%o5) 5*k^2 < yeech There might be other things that need fixing: (%i8) properties(sum); (%o8) [Special Evaluation Form,OUTATIVE,NOUN,RULE] Barton  Comment By: Robert Dodier (robert_dodier) Date: 20050612 20:34 Message: Logged In: YES user_id=501686 OK, for the record, here is a definition of $sum which implements one of the ideas that has been proposed, namely this policy: evaluate the summand after binding the summation variable to itself. (defmspec $sum (l) (setq l (cdr l)) (if (= (length l) 4) (progv (list (cadr l)) (list (cadr l)) (dosum (meval (car l)) (cadr l) (meval (caddr l)) (meval (cadddr l)) t)) (wnaerr '$sum))) This version yields the results expected by the person who originated this bug report, and yields the results predicted by S Macrakis in the email referenced below (http://www.math.utexas.edu/pipermail/maxima/2003/004869.html), and run_testsuite() runs to completion with no errors. I'm in favor of making this change, but I'm just recording this code snippet here for future reference; no changes planned at the moment. For comparison here is the current version of $sum: (defmspec $sum (l) (setq l (cdr l)) (if (= (length l) 4) (dosum (car l) (cadr l) (meval (caddr l)) (meval (cadddr l)) t) (wnaerr '$sum)))  Comment By: Barton Willis (willisbl) Date: 20050501 06:09 Message: Logged In: YES user_id=895922 We've discussed this before; see http://www.math.utexas.edu/pipermail/maxima/2003/004869.html http://www.math.utexas.edu/pipermail/maxima/2003/004870.html Maybe a simple workaround such as mysum(f,v,lo,hi):=block([acc:0], if integerp(lo) and integerp(hi) then for i from lo thru hi do acc:acc+substitute(i,v,f) else acc:funmake('mysum,[f,v,lo,hi]), acc) or mysum(f,lo,hi):= block([acc:0], if integerp(lo) and integerp(hi) then for i : lo thru hi do acc : acc + apply(f,[i]) else acc : funmake('mysum,[f,lo,hi]), acc); might work for you, Barton  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1192935&group_id=4933 
From: SourceForge.net <noreply@so...>  20050614 14:42:15

Bugs item #1192935, was opened at 20050430 05:10 Message generated for change (Comment added) made by robert_dodier You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1192935&group_id=4933 Please note that this message will contain a full copy of the comment thread, including the initial issue submission, for this request, not just the latest update. Category: Lisp Core Group: None Status: Open Resolution: None Priority: 5 Submitted By: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: termsubstitution in function sum is too late Initial Comment: (%i1) "Lower RiemannSum, Intervall [0,1], 10 parts"$ (%i2) /*Function*/ f(x):=x^2$ (%i3) /*area width*/ b: 1/10$ (%i4) /*area depending on k=0,...,9*/ Ak: b*f(k*b)$ (%i5) Ak; 2 k (%o5)  1000 (%i6) /* wrong */ sum(Ak,k,0,9); 2 k (%o6)  100 obviously, we first have the summation and then the symbolsubstitution (%i7) sum(''Ak,k,0,9); 57 (%o7)  200 finally, this is ok, but the usage of the doublequote is hard to explain to beginners. it gets even worse, if you use a function A(k):=k*f(k*b)$ for a sum with n partitions: (%i8) "Lower RiemannSum, Intervall [0,1], n parts"$ (%i9) /*area width*/ b:1/n$ (%i10) /*area depending on k=0,...,n1*/ A(k):=k*f(k*b)$ (%i11) A(k); 3 k (%o11)  2 n (%i12) sum(A(k),k,0,n1); n  1 ==== (%o12) > A(k) / ==== k = 0 in that case you need ''(A(k)), a bracket and the doublequote. is it possible to patch the sumfunction in that way, that we first have the term or symbolsubstitution and second the summation?  >Comment By: Robert Dodier (robert_dodier) Date: 20050614 08:42 Message: Logged In: YES user_id=501686 OK, here is another attempt. I think I now understand this comment by Stavros in bug report # 740134  "First evaluate foo with i bound to itself, and check if that result is free of i. If so, return the product. If not, *substitute* (don't evaluate) i=lowerlimit, i=lowerlimit+1, etc." ;; ORIGINAL: MEVAL ONCE ;; (mbinding (lind l*i) (meval exp)) ;; TRIED THIS: MEVAL TWICE ;; (mbinding (lind l*i) (meval (meval exp))) ;; THIRD TIME IS A CHARM ?? ($substitute *i ind (mbinding (lind l*i) (meval exp))) This version now yields this result  ex 9: f(x) := sum (x, i, 1, 3); f(x) => 3 x as expected, as well as the same results for ex 1 through ex 8.  Comment By: Robert Dodier (robert_dodier) Date: 20050613 23:00 Message: Logged In: YES user_id=501686 Actually, after looking at the code some more (specifically DOSUM in src/asum.lisp) it looks to me like the intent is indeed to evaluate the summand after binding the summation index to itself. In addition, there is code to assume the index is between its lower and upper limits (although assume/forget is handled incorrectly, see 851765). I believe that the observed behavior is a bug, in the narrow sense that the observed behavior is different from what the code author intended. See also my comments on SF bug # 740134. Here is a different patch. I like this one better than the previous one. In DOSUM: ;; ORIGINAL: MEVAL ONCE ;; (mbinding (lind l*i) (meval exp)) (mbinding (lind l*i) (meval (meval exp))) In MEVALSUMARG: ;; ORIGINAL: MEVALATOMS ;; (resimplify (mevalatoms (if (and (not (atom exp)) (resimplify (meval (if (and (not (atom exp)) With these changes, the results are as expected for the examples cited by the original poster, and also the example about the translated function produces the correct result, and some examples adapted from another bug report (740134) yield correct results. ex 1: f(x):=x^2$ b: 1/10$ Ak: b*f(k*b)$ sum(Ak,k,0,9); => 57/200 ex 2: A(k):=k*f(k*b)$ b:1/n$ A(k) => k^3/n^2 sum(A(k),k,0,n1) => ('sum(k^3,k,0,n1))/n^2 sum(A(k),k,0,n1), simpsum => ((n1)^4 + 2*(n1)^3 + (n1)^2) / (4*n^2) ex 3: ak : k^2$ g(a,n) := sum(a,k,1,n)$ g(ak,5) => 55 translate (g)$ g(ak,5) => 55 some other examples, adapted from bug # 740134: ex 4: sum (print (i), i, 1, 3) => prints 1, 2, 3 then returns 6 ex 5: sum (integrate (x^i ,x), i, 0, 2) => x^3/3 + x^2/2 + x ex 6: sum (integrate (1/(x^i + 1), x), i, 0, 1) => log(x+1) + x/2 ex 7: f[i](x):=x^i$ g[i](x):=x^i$ h[i](x):=x^i$ /* reference f[i] and g[i]  see 740134 for the effect this has on previous defn of sum */ f[i]$ g[i](t)$ sum (f[i](x), i, 0, n) => 'sum (x^i, i, 0, n) sum (g[i](x), i, 0, n) => 'sum (x^i, i, 0, n) sum (h[i](x), i, 0, n) => 'sum (x^i, i, 0, n) ex 8: sum (integrate (x^i, x), i, 0, n) => 'sum (x^(i+1) / (i+1), i, 0, n)  Comment By: Barton Willis (willisbl) Date: 20050613 05:30 Message: Logged In: YES user_id=895922 If we decide to use Robert's $sum function, the translate property for sum will need to be changed. Consider: (1) Redefine $sum as Robert suggested. (2) Try this: (%i1) ak : k^2$ (%i2) g(a,n) := sum(a,k,1,n)$ (%i3) g(ak,5); (%o3) 55 < correct for Robert's $sum function (%i4) translate(g); (%o4) [g] (%i5) g(ak,5); (%o5) 5*k^2 < yeech There might be other things that need fixing: (%i8) properties(sum); (%o8) [Special Evaluation Form,OUTATIVE,NOUN,RULE] Barton  Comment By: Robert Dodier (robert_dodier) Date: 20050612 20:34 Message: Logged In: YES user_id=501686 OK, for the record, here is a definition of $sum which implements one of the ideas that has been proposed, namely this policy: evaluate the summand after binding the summation variable to itself. (defmspec $sum (l) (setq l (cdr l)) (if (= (length l) 4) (progv (list (cadr l)) (list (cadr l)) (dosum (meval (car l)) (cadr l) (meval (caddr l)) (meval (cadddr l)) t)) (wnaerr '$sum))) This version yields the results expected by the person who originated this bug report, and yields the results predicted by S Macrakis in the email referenced below (http://www.math.utexas.edu/pipermail/maxima/2003/004869.html), and run_testsuite() runs to completion with no errors. I'm in favor of making this change, but I'm just recording this code snippet here for future reference; no changes planned at the moment. For comparison here is the current version of $sum: (defmspec $sum (l) (setq l (cdr l)) (if (= (length l) 4) (dosum (car l) (cadr l) (meval (caddr l)) (meval (cadddr l)) t) (wnaerr '$sum)))  Comment By: Barton Willis (willisbl) Date: 20050501 06:09 Message: Logged In: YES user_id=895922 We've discussed this before; see http://www.math.utexas.edu/pipermail/maxima/2003/004869.html http://www.math.utexas.edu/pipermail/maxima/2003/004870.html Maybe a simple workaround such as mysum(f,v,lo,hi):=block([acc:0], if integerp(lo) and integerp(hi) then for i from lo thru hi do acc:acc+substitute(i,v,f) else acc:funmake('mysum,[f,v,lo,hi]), acc) or mysum(f,lo,hi):= block([acc:0], if integerp(lo) and integerp(hi) then for i : lo thru hi do acc : acc + apply(f,[i]) else acc : funmake('mysum,[f,lo,hi]), acc); might work for you, Barton  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1192935&group_id=4933 
From: SourceForge.net <noreply@so...>  20050615 14:37:16

Bugs item #1192935, was opened at 20050430 05:10 Message generated for change (Comment added) made by robert_dodier You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1192935&group_id=4933 Please note that this message will contain a full copy of the comment thread, including the initial issue submission, for this request, not just the latest update. Category: Lisp Core Group: None Status: Open Resolution: None Priority: 5 Submitted By: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: termsubstitution in function sum is too late Initial Comment: (%i1) "Lower RiemannSum, Intervall [0,1], 10 parts"$ (%i2) /*Function*/ f(x):=x^2$ (%i3) /*area width*/ b: 1/10$ (%i4) /*area depending on k=0,...,9*/ Ak: b*f(k*b)$ (%i5) Ak; 2 k (%o5)  1000 (%i6) /* wrong */ sum(Ak,k,0,9); 2 k (%o6)  100 obviously, we first have the summation and then the symbolsubstitution (%i7) sum(''Ak,k,0,9); 57 (%o7)  200 finally, this is ok, but the usage of the doublequote is hard to explain to beginners. it gets even worse, if you use a function A(k):=k*f(k*b)$ for a sum with n partitions: (%i8) "Lower RiemannSum, Intervall [0,1], n parts"$ (%i9) /*area width*/ b:1/n$ (%i10) /*area depending on k=0,...,n1*/ A(k):=k*f(k*b)$ (%i11) A(k); 3 k (%o11)  2 n (%i12) sum(A(k),k,0,n1); n  1 ==== (%o12) > A(k) / ==== k = 0 in that case you need ''(A(k)), a bracket and the doublequote. is it possible to patch the sumfunction in that way, that we first have the term or symbolsubstitution and second the summation?  >Comment By: Robert Dodier (robert_dodier) Date: 20050615 08:37 Message: Logged In: YES user_id=501686 I'm attaching a list of test cases for sum, which can be executed by batch ("rtestsum.mac", test) . Doubtless there will be some discussion as to what the expected output of each test should be. All tests pass after making these 2 changes: ;; ORIGINAL: MEVAL ONCE ;; (mbinding (lind l*i) (meval exp)) ;; TRIED THIS: MEVAL TWICE ;; (mbinding (lind l*i) (meval (meval exp))) ;; THIRD TIME IS A CHARM ?? ($substitute *i ind (mbinding (lind l*i) (meval exp))) ;; ORIGINAL: MEVALATOMS ;; (resimplify (mevalatoms (if (and (not (atom exp)) (resimplify (meval (if (and (not (atom exp))  Comment By: Robert Dodier (robert_dodier) Date: 20050614 08:42 Message: Logged In: YES user_id=501686 OK, here is another attempt. I think I now understand this comment by Stavros in bug report # 740134  "First evaluate foo with i bound to itself, and check if that result is free of i. If so, return the product. If not, *substitute* (don't evaluate) i=lowerlimit, i=lowerlimit+1, etc." ;; ORIGINAL: MEVAL ONCE ;; (mbinding (lind l*i) (meval exp)) ;; TRIED THIS: MEVAL TWICE ;; (mbinding (lind l*i) (meval (meval exp))) ;; THIRD TIME IS A CHARM ?? ($substitute *i ind (mbinding (lind l*i) (meval exp))) This version now yields this result  ex 9: f(x) := sum (x, i, 1, 3); f(x) => 3 x as expected, as well as the same results for ex 1 through ex 8.  Comment By: Robert Dodier (robert_dodier) Date: 20050613 23:00 Message: Logged In: YES user_id=501686 Actually, after looking at the code some more (specifically DOSUM in src/asum.lisp) it looks to me like the intent is indeed to evaluate the summand after binding the summation index to itself. In addition, there is code to assume the index is between its lower and upper limits (although assume/forget is handled incorrectly, see 851765). I believe that the observed behavior is a bug, in the narrow sense that the observed behavior is different from what the code author intended. See also my comments on SF bug # 740134. Here is a different patch. I like this one better than the previous one. In DOSUM: ;; ORIGINAL: MEVAL ONCE ;; (mbinding (lind l*i) (meval exp)) (mbinding (lind l*i) (meval (meval exp))) In MEVALSUMARG: ;; ORIGINAL: MEVALATOMS ;; (resimplify (mevalatoms (if (and (not (atom exp)) (resimplify (meval (if (and (not (atom exp)) With these changes, the results are as expected for the examples cited by the original poster, and also the example about the translated function produces the correct result, and some examples adapted from another bug report (740134) yield correct results. ex 1: f(x):=x^2$ b: 1/10$ Ak: b*f(k*b)$ sum(Ak,k,0,9); => 57/200 ex 2: A(k):=k*f(k*b)$ b:1/n$ A(k) => k^3/n^2 sum(A(k),k,0,n1) => ('sum(k^3,k,0,n1))/n^2 sum(A(k),k,0,n1), simpsum => ((n1)^4 + 2*(n1)^3 + (n1)^2) / (4*n^2) ex 3: ak : k^2$ g(a,n) := sum(a,k,1,n)$ g(ak,5) => 55 translate (g)$ g(ak,5) => 55 some other examples, adapted from bug # 740134: ex 4: sum (print (i), i, 1, 3) => prints 1, 2, 3 then returns 6 ex 5: sum (integrate (x^i ,x), i, 0, 2) => x^3/3 + x^2/2 + x ex 6: sum (integrate (1/(x^i + 1), x), i, 0, 1) => log(x+1) + x/2 ex 7: f[i](x):=x^i$ g[i](x):=x^i$ h[i](x):=x^i$ /* reference f[i] and g[i]  see 740134 for the effect this has on previous defn of sum */ f[i]$ g[i](t)$ sum (f[i](x), i, 0, n) => 'sum (x^i, i, 0, n) sum (g[i](x), i, 0, n) => 'sum (x^i, i, 0, n) sum (h[i](x), i, 0, n) => 'sum (x^i, i, 0, n) ex 8: sum (integrate (x^i, x), i, 0, n) => 'sum (x^(i+1) / (i+1), i, 0, n)  Comment By: Barton Willis (willisbl) Date: 20050613 05:30 Message: Logged In: YES user_id=895922 If we decide to use Robert's $sum function, the translate property for sum will need to be changed. Consider: (1) Redefine $sum as Robert suggested. (2) Try this: (%i1) ak : k^2$ (%i2) g(a,n) := sum(a,k,1,n)$ (%i3) g(ak,5); (%o3) 55 < correct for Robert's $sum function (%i4) translate(g); (%o4) [g] (%i5) g(ak,5); (%o5) 5*k^2 < yeech There might be other things that need fixing: (%i8) properties(sum); (%o8) [Special Evaluation Form,OUTATIVE,NOUN,RULE] Barton  Comment By: Robert Dodier (robert_dodier) Date: 20050612 20:34 Message: Logged In: YES user_id=501686 OK, for the record, here is a definition of $sum which implements one of the ideas that has been proposed, namely this policy: evaluate the summand after binding the summation variable to itself. (defmspec $sum (l) (setq l (cdr l)) (if (= (length l) 4) (progv (list (cadr l)) (list (cadr l)) (dosum (meval (car l)) (cadr l) (meval (caddr l)) (meval (cadddr l)) t)) (wnaerr '$sum))) This version yields the results expected by the person who originated this bug report, and yields the results predicted by S Macrakis in the email referenced below (http://www.math.utexas.edu/pipermail/maxima/2003/004869.html), and run_testsuite() runs to completion with no errors. I'm in favor of making this change, but I'm just recording this code snippet here for future reference; no changes planned at the moment. For comparison here is the current version of $sum: (defmspec $sum (l) (setq l (cdr l)) (if (= (length l) 4) (dosum (car l) (cadr l) (meval (caddr l)) (meval (cadddr l)) t) (wnaerr '$sum)))  Comment By: Barton Willis (willisbl) Date: 20050501 06:09 Message: Logged In: YES user_id=895922 We've discussed this before; see http://www.math.utexas.edu/pipermail/maxima/2003/004869.html http://www.math.utexas.edu/pipermail/maxima/2003/004870.html Maybe a simple workaround such as mysum(f,v,lo,hi):=block([acc:0], if integerp(lo) and integerp(hi) then for i from lo thru hi do acc:acc+substitute(i,v,f) else acc:funmake('mysum,[f,v,lo,hi]), acc) or mysum(f,lo,hi):= block([acc:0], if integerp(lo) and integerp(hi) then for i : lo thru hi do acc : acc + apply(f,[i]) else acc : funmake('mysum,[f,lo,hi]), acc); might work for you, Barton  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1192935&group_id=4933 
From: SourceForge.net <noreply@so...>  20050618 17:42:45

Bugs item #1192935, was opened at 20050430 13:10 Message generated for change (Comment added) made by van_nek You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1192935&group_id=4933 Please note that this message will contain a full copy of the comment thread, including the initial issue submission, for this request, not just the latest update. Category: Lisp Core Group: None Status: Open Resolution: None Priority: 5 Submitted By: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: termsubstitution in function sum is too late Initial Comment: (%i1) "Lower RiemannSum, Intervall [0,1], 10 parts"$ (%i2) /*Function*/ f(x):=x^2$ (%i3) /*area width*/ b: 1/10$ (%i4) /*area depending on k=0,...,9*/ Ak: b*f(k*b)$ (%i5) Ak; 2 k (%o5)  1000 (%i6) /* wrong */ sum(Ak,k,0,9); 2 k (%o6)  100 obviously, we first have the summation and then the symbolsubstitution (%i7) sum(''Ak,k,0,9); 57 (%o7)  200 finally, this is ok, but the usage of the doublequote is hard to explain to beginners. it gets even worse, if you use a function A(k):=k*f(k*b)$ for a sum with n partitions: (%i8) "Lower RiemannSum, Intervall [0,1], n parts"$ (%i9) /*area width*/ b:1/n$ (%i10) /*area depending on k=0,...,n1*/ A(k):=k*f(k*b)$ (%i11) A(k); 3 k (%o11)  2 n (%i12) sum(A(k),k,0,n1); n  1 ==== (%o12) > A(k) / ==== k = 0 in that case you need ''(A(k)), a bracket and the doublequote. is it possible to patch the sumfunction in that way, that we first have the term or symbolsubstitution and second the summation?  Comment By: van_Nek (van_nek) Date: 20050618 19:42 Message: Logged In: YES user_id=1269745 Dear Maximadevelopers, thank you for your support. I tried the last fix from Robert Dodier and it works. Volker van Nek original (anonymous) poster  Comment By: Robert Dodier (robert_dodier) Date: 20050615 16:37 Message: Logged In: YES user_id=501686 I'm attaching a list of test cases for sum, which can be executed by batch ("rtestsum.mac", test) . Doubtless there will be some discussion as to what the expected output of each test should be. All tests pass after making these 2 changes: ;; ORIGINAL: MEVAL ONCE ;; (mbinding (lind l*i) (meval exp)) ;; TRIED THIS: MEVAL TWICE ;; (mbinding (lind l*i) (meval (meval exp))) ;; THIRD TIME IS A CHARM ?? ($substitute *i ind (mbinding (lind l*i) (meval exp))) ;; ORIGINAL: MEVALATOMS ;; (resimplify (mevalatoms (if (and (not (atom exp)) (resimplify (meval (if (and (not (atom exp))  Comment By: Robert Dodier (robert_dodier) Date: 20050614 16:42 Message: Logged In: YES user_id=501686 OK, here is another attempt. I think I now understand this comment by Stavros in bug report # 740134  "First evaluate foo with i bound to itself, and check if that result is free of i. If so, return the product. If not, *substitute* (don't evaluate) i=lowerlimit, i=lowerlimit+1, etc." ;; ORIGINAL: MEVAL ONCE ;; (mbinding (lind l*i) (meval exp)) ;; TRIED THIS: MEVAL TWICE ;; (mbinding (lind l*i) (meval (meval exp))) ;; THIRD TIME IS A CHARM ?? ($substitute *i ind (mbinding (lind l*i) (meval exp))) This version now yields this result  ex 9: f(x) := sum (x, i, 1, 3); f(x) => 3 x as expected, as well as the same results for ex 1 through ex 8.  Comment By: Robert Dodier (robert_dodier) Date: 20050614 07:00 Message: Logged In: YES user_id=501686 Actually, after looking at the code some more (specifically DOSUM in src/asum.lisp) it looks to me like the intent is indeed to evaluate the summand after binding the summation index to itself. In addition, there is code to assume the index is between its lower and upper limits (although assume/forget is handled incorrectly, see 851765). I believe that the observed behavior is a bug, in the narrow sense that the observed behavior is different from what the code author intended. See also my comments on SF bug # 740134. Here is a different patch. I like this one better than the previous one. In DOSUM: ;; ORIGINAL: MEVAL ONCE ;; (mbinding (lind l*i) (meval exp)) (mbinding (lind l*i) (meval (meval exp))) In MEVALSUMARG: ;; ORIGINAL: MEVALATOMS ;; (resimplify (mevalatoms (if (and (not (atom exp)) (resimplify (meval (if (and (not (atom exp)) With these changes, the results are as expected for the examples cited by the original poster, and also the example about the translated function produces the correct result, and some examples adapted from another bug report (740134) yield correct results. ex 1: f(x):=x^2$ b: 1/10$ Ak: b*f(k*b)$ sum(Ak,k,0,9); => 57/200 ex 2: A(k):=k*f(k*b)$ b:1/n$ A(k) => k^3/n^2 sum(A(k),k,0,n1) => ('sum(k^3,k,0,n1))/n^2 sum(A(k),k,0,n1), simpsum => ((n1)^4 + 2*(n1)^3 + (n1)^2) / (4*n^2) ex 3: ak : k^2$ g(a,n) := sum(a,k,1,n)$ g(ak,5) => 55 translate (g)$ g(ak,5) => 55 some other examples, adapted from bug # 740134: ex 4: sum (print (i), i, 1, 3) => prints 1, 2, 3 then returns 6 ex 5: sum (integrate (x^i ,x), i, 0, 2) => x^3/3 + x^2/2 + x ex 6: sum (integrate (1/(x^i + 1), x), i, 0, 1) => log(x+1) + x/2 ex 7: f[i](x):=x^i$ g[i](x):=x^i$ h[i](x):=x^i$ /* reference f[i] and g[i]  see 740134 for the effect this has on previous defn of sum */ f[i]$ g[i](t)$ sum (f[i](x), i, 0, n) => 'sum (x^i, i, 0, n) sum (g[i](x), i, 0, n) => 'sum (x^i, i, 0, n) sum (h[i](x), i, 0, n) => 'sum (x^i, i, 0, n) ex 8: sum (integrate (x^i, x), i, 0, n) => 'sum (x^(i+1) / (i+1), i, 0, n)  Comment By: Barton Willis (willisbl) Date: 20050613 13:30 Message: Logged In: YES user_id=895922 If we decide to use Robert's $sum function, the translate property for sum will need to be changed. Consider: (1) Redefine $sum as Robert suggested. (2) Try this: (%i1) ak : k^2$ (%i2) g(a,n) := sum(a,k,1,n)$ (%i3) g(ak,5); (%o3) 55 < correct for Robert's $sum function (%i4) translate(g); (%o4) [g] (%i5) g(ak,5); (%o5) 5*k^2 < yeech There might be other things that need fixing: (%i8) properties(sum); (%o8) [Special Evaluation Form,OUTATIVE,NOUN,RULE] Barton  Comment By: Robert Dodier (robert_dodier) Date: 20050613 04:34 Message: Logged In: YES user_id=501686 OK, for the record, here is a definition of $sum which implements one of the ideas that has been proposed, namely this policy: evaluate the summand after binding the summation variable to itself. (defmspec $sum (l) (setq l (cdr l)) (if (= (length l) 4) (progv (list (cadr l)) (list (cadr l)) (dosum (meval (car l)) (cadr l) (meval (caddr l)) (meval (cadddr l)) t)) (wnaerr '$sum))) This version yields the results expected by the person who originated this bug report, and yields the results predicted by S Macrakis in the email referenced below (http://www.math.utexas.edu/pipermail/maxima/2003/004869.html), and run_testsuite() runs to completion with no errors. I'm in favor of making this change, but I'm just recording this code snippet here for future reference; no changes planned at the moment. For comparison here is the current version of $sum: (defmspec $sum (l) (setq l (cdr l)) (if (= (length l) 4) (dosum (car l) (cadr l) (meval (caddr l)) (meval (cadddr l)) t) (wnaerr '$sum)))  Comment By: Barton Willis (willisbl) Date: 20050501 14:09 Message: Logged In: YES user_id=895922 We've discussed this before; see http://www.math.utexas.edu/pipermail/maxima/2003/004869.html http://www.math.utexas.edu/pipermail/maxima/2003/004870.html Maybe a simple workaround such as mysum(f,v,lo,hi):=block([acc:0], if integerp(lo) and integerp(hi) then for i from lo thru hi do acc:acc+substitute(i,v,f) else acc:funmake('mysum,[f,v,lo,hi]), acc) or mysum(f,lo,hi):= block([acc:0], if integerp(lo) and integerp(hi) then for i : lo thru hi do acc : acc + apply(f,[i]) else acc : funmake('mysum,[f,lo,hi]), acc); might work for you, Barton  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1192935&group_id=4933 
From: SourceForge.net <noreply@so...>  20050619 16:29:05

Bugs item #1192935, was opened at 20050430 05:10 Message generated for change (Comment added) made by robert_dodier You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1192935&group_id=4933 Please note that this message will contain a full copy of the comment thread, including the initial issue submission, for this request, not just the latest update. Category: Lisp Core Group: None Status: Open Resolution: None Priority: 5 Submitted By: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: termsubstitution in function sum is too late Initial Comment: (%i1) "Lower RiemannSum, Intervall [0,1], 10 parts"$ (%i2) /*Function*/ f(x):=x^2$ (%i3) /*area width*/ b: 1/10$ (%i4) /*area depending on k=0,...,9*/ Ak: b*f(k*b)$ (%i5) Ak; 2 k (%o5)  1000 (%i6) /* wrong */ sum(Ak,k,0,9); 2 k (%o6)  100 obviously, we first have the summation and then the symbolsubstitution (%i7) sum(''Ak,k,0,9); 57 (%o7)  200 finally, this is ok, but the usage of the doublequote is hard to explain to beginners. it gets even worse, if you use a function A(k):=k*f(k*b)$ for a sum with n partitions: (%i8) "Lower RiemannSum, Intervall [0,1], n parts"$ (%i9) /*area width*/ b:1/n$ (%i10) /*area depending on k=0,...,n1*/ A(k):=k*f(k*b)$ (%i11) A(k); 3 k (%o11)  2 n (%i12) sum(A(k),k,0,n1); n  1 ==== (%o12) > A(k) / ==== k = 0 in that case you need ''(A(k)), a bracket and the doublequote. is it possible to patch the sumfunction in that way, that we first have the term or symbolsubstitution and second the summation?  >Comment By: Robert Dodier (robert_dodier) Date: 20050619 10:28 Message: Logged In: YES user_id=501686 I'm removing rtestsum.mac, in anticipation of attaching a new version in a minute.  Comment By: van_Nek (van_nek) Date: 20050618 11:42 Message: Logged In: YES user_id=1269745 Dear Maximadevelopers, thank you for your support. I tried the last fix from Robert Dodier and it works. Volker van Nek original (anonymous) poster  Comment By: Robert Dodier (robert_dodier) Date: 20050615 08:37 Message: Logged In: YES user_id=501686 I'm attaching a list of test cases for sum, which can be executed by batch ("rtestsum.mac", test) . Doubtless there will be some discussion as to what the expected output of each test should be. All tests pass after making these 2 changes: ;; ORIGINAL: MEVAL ONCE ;; (mbinding (lind l*i) (meval exp)) ;; TRIED THIS: MEVAL TWICE ;; (mbinding (lind l*i) (meval (meval exp))) ;; THIRD TIME IS A CHARM ?? ($substitute *i ind (mbinding (lind l*i) (meval exp))) ;; ORIGINAL: MEVALATOMS ;; (resimplify (mevalatoms (if (and (not (atom exp)) (resimplify (meval (if (and (not (atom exp))  Comment By: Robert Dodier (robert_dodier) Date: 20050614 08:42 Message: Logged In: YES user_id=501686 OK, here is another attempt. I think I now understand this comment by Stavros in bug report # 740134  "First evaluate foo with i bound to itself, and check if that result is free of i. If so, return the product. If not, *substitute* (don't evaluate) i=lowerlimit, i=lowerlimit+1, etc." ;; ORIGINAL: MEVAL ONCE ;; (mbinding (lind l*i) (meval exp)) ;; TRIED THIS: MEVAL TWICE ;; (mbinding (lind l*i) (meval (meval exp))) ;; THIRD TIME IS A CHARM ?? ($substitute *i ind (mbinding (lind l*i) (meval exp))) This version now yields this result  ex 9: f(x) := sum (x, i, 1, 3); f(x) => 3 x as expected, as well as the same results for ex 1 through ex 8.  Comment By: Robert Dodier (robert_dodier) Date: 20050613 23:00 Message: Logged In: YES user_id=501686 Actually, after looking at the code some more (specifically DOSUM in src/asum.lisp) it looks to me like the intent is indeed to evaluate the summand after binding the summation index to itself. In addition, there is code to assume the index is between its lower and upper limits (although assume/forget is handled incorrectly, see 851765). I believe that the observed behavior is a bug, in the narrow sense that the observed behavior is different from what the code author intended. See also my comments on SF bug # 740134. Here is a different patch. I like this one better than the previous one. In DOSUM: ;; ORIGINAL: MEVAL ONCE ;; (mbinding (lind l*i) (meval exp)) (mbinding (lind l*i) (meval (meval exp))) In MEVALSUMARG: ;; ORIGINAL: MEVALATOMS ;; (resimplify (mevalatoms (if (and (not (atom exp)) (resimplify (meval (if (and (not (atom exp)) With these changes, the results are as expected for the examples cited by the original poster, and also the example about the translated function produces the correct result, and some examples adapted from another bug report (740134) yield correct results. ex 1: f(x):=x^2$ b: 1/10$ Ak: b*f(k*b)$ sum(Ak,k,0,9); => 57/200 ex 2: A(k):=k*f(k*b)$ b:1/n$ A(k) => k^3/n^2 sum(A(k),k,0,n1) => ('sum(k^3,k,0,n1))/n^2 sum(A(k),k,0,n1), simpsum => ((n1)^4 + 2*(n1)^3 + (n1)^2) / (4*n^2) ex 3: ak : k^2$ g(a,n) := sum(a,k,1,n)$ g(ak,5) => 55 translate (g)$ g(ak,5) => 55 some other examples, adapted from bug # 740134: ex 4: sum (print (i), i, 1, 3) => prints 1, 2, 3 then returns 6 ex 5: sum (integrate (x^i ,x), i, 0, 2) => x^3/3 + x^2/2 + x ex 6: sum (integrate (1/(x^i + 1), x), i, 0, 1) => log(x+1) + x/2 ex 7: f[i](x):=x^i$ g[i](x):=x^i$ h[i](x):=x^i$ /* reference f[i] and g[i]  see 740134 for the effect this has on previous defn of sum */ f[i]$ g[i](t)$ sum (f[i](x), i, 0, n) => 'sum (x^i, i, 0, n) sum (g[i](x), i, 0, n) => 'sum (x^i, i, 0, n) sum (h[i](x), i, 0, n) => 'sum (x^i, i, 0, n) ex 8: sum (integrate (x^i, x), i, 0, n) => 'sum (x^(i+1) / (i+1), i, 0, n)  Comment By: Barton Willis (willisbl) Date: 20050613 05:30 Message: Logged In: YES user_id=895922 If we decide to use Robert's $sum function, the translate property for sum will need to be changed. Consider: (1) Redefine $sum as Robert suggested. (2) Try this: (%i1) ak : k^2$ (%i2) g(a,n) := sum(a,k,1,n)$ (%i3) g(ak,5); (%o3) 55 < correct for Robert's $sum function (%i4) translate(g); (%o4) [g] (%i5) g(ak,5); (%o5) 5*k^2 < yeech There might be other things that need fixing: (%i8) properties(sum); (%o8) [Special Evaluation Form,OUTATIVE,NOUN,RULE] Barton  Comment By: Robert Dodier (robert_dodier) Date: 20050612 20:34 Message: Logged In: YES user_id=501686 OK, for the record, here is a definition of $sum which implements one of the ideas that has been proposed, namely this policy: evaluate the summand after binding the summation variable to itself. (defmspec $sum (l) (setq l (cdr l)) (if (= (length l) 4) (progv (list (cadr l)) (list (cadr l)) (dosum (meval (car l)) (cadr l) (meval (caddr l)) (meval (cadddr l)) t)) (wnaerr '$sum))) This version yields the results expected by the person who originated this bug report, and yields the results predicted by S Macrakis in the email referenced below (http://www.math.utexas.edu/pipermail/maxima/2003/004869.html), and run_testsuite() runs to completion with no errors. I'm in favor of making this change, but I'm just recording this code snippet here for future reference; no changes planned at the moment. For comparison here is the current version of $sum: (defmspec $sum (l) (setq l (cdr l)) (if (= (length l) 4) (dosum (car l) (cadr l) (meval (caddr l)) (meval (cadddr l)) t) (wnaerr '$sum)))  Comment By: Barton Willis (willisbl) Date: 20050501 06:09 Message: Logged In: YES user_id=895922 We've discussed this before; see http://www.math.utexas.edu/pipermail/maxima/2003/004869.html http://www.math.utexas.edu/pipermail/maxima/2003/004870.html Maybe a simple workaround such as mysum(f,v,lo,hi):=block([acc:0], if integerp(lo) and integerp(hi) then for i from lo thru hi do acc:acc+substitute(i,v,f) else acc:funmake('mysum,[f,v,lo,hi]), acc) or mysum(f,lo,hi):= block([acc:0], if integerp(lo) and integerp(hi) then for i : lo thru hi do acc : acc + apply(f,[i]) else acc : funmake('mysum,[f,lo,hi]), acc); might work for you, Barton  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1192935&group_id=4933 
From: SourceForge.net <noreply@so...>  20050619 16:33:41

Bugs item #1192935, was opened at 20050430 05:10 Message generated for change (Comment added) made by robert_dodier You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1192935&group_id=4933 Please note that this message will contain a full copy of the comment thread, including the initial issue submission, for this request, not just the latest update. Category: Lisp Core Group: None Status: Open Resolution: None Priority: 5 Submitted By: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: termsubstitution in function sum is too late Initial Comment: (%i1) "Lower RiemannSum, Intervall [0,1], 10 parts"$ (%i2) /*Function*/ f(x):=x^2$ (%i3) /*area width*/ b: 1/10$ (%i4) /*area depending on k=0,...,9*/ Ak: b*f(k*b)$ (%i5) Ak; 2 k (%o5)  1000 (%i6) /* wrong */ sum(Ak,k,0,9); 2 k (%o6)  100 obviously, we first have the summation and then the symbolsubstitution (%i7) sum(''Ak,k,0,9); 57 (%o7)  200 finally, this is ok, but the usage of the doublequote is hard to explain to beginners. it gets even worse, if you use a function A(k):=k*f(k*b)$ for a sum with n partitions: (%i8) "Lower RiemannSum, Intervall [0,1], n parts"$ (%i9) /*area width*/ b:1/n$ (%i10) /*area depending on k=0,...,n1*/ A(k):=k*f(k*b)$ (%i11) A(k); 3 k (%o11)  2 n (%i12) sum(A(k),k,0,n1); n  1 ==== (%o12) > A(k) / ==== k = 0 in that case you need ''(A(k)), a bracket and the doublequote. is it possible to patch the sumfunction in that way, that we first have the term or symbolsubstitution and second the summation?  >Comment By: Robert Dodier (robert_dodier) Date: 20050619 10:33 Message: Logged In: YES user_id=501686 Attaching a longer list of test cases. batch ("rtestsum.mac", test); executes the test. As always the "correct" results (2nd line in each pair) are subject to review.  Comment By: Robert Dodier (robert_dodier) Date: 20050619 10:28 Message: Logged In: YES user_id=501686 I'm removing rtestsum.mac, in anticipation of attaching a new version in a minute.  Comment By: van_Nek (van_nek) Date: 20050618 11:42 Message: Logged In: YES user_id=1269745 Dear Maximadevelopers, thank you for your support. I tried the last fix from Robert Dodier and it works. Volker van Nek original (anonymous) poster  Comment By: Robert Dodier (robert_dodier) Date: 20050615 08:37 Message: Logged In: YES user_id=501686 I'm attaching a list of test cases for sum, which can be executed by batch ("rtestsum.mac", test) . Doubtless there will be some discussion as to what the expected output of each test should be. All tests pass after making these 2 changes: ;; ORIGINAL: MEVAL ONCE ;; (mbinding (lind l*i) (meval exp)) ;; TRIED THIS: MEVAL TWICE ;; (mbinding (lind l*i) (meval (meval exp))) ;; THIRD TIME IS A CHARM ?? ($substitute *i ind (mbinding (lind l*i) (meval exp))) ;; ORIGINAL: MEVALATOMS ;; (resimplify (mevalatoms (if (and (not (atom exp)) (resimplify (meval (if (and (not (atom exp))  Comment By: Robert Dodier (robert_dodier) Date: 20050614 08:42 Message: Logged In: YES user_id=501686 OK, here is another attempt. I think I now understand this comment by Stavros in bug report # 740134  "First evaluate foo with i bound to itself, and check if that result is free of i. If so, return the product. If not, *substitute* (don't evaluate) i=lowerlimit, i=lowerlimit+1, etc." ;; ORIGINAL: MEVAL ONCE ;; (mbinding (lind l*i) (meval exp)) ;; TRIED THIS: MEVAL TWICE ;; (mbinding (lind l*i) (meval (meval exp))) ;; THIRD TIME IS A CHARM ?? ($substitute *i ind (mbinding (lind l*i) (meval exp))) This version now yields this result  ex 9: f(x) := sum (x, i, 1, 3); f(x) => 3 x as expected, as well as the same results for ex 1 through ex 8.  Comment By: Robert Dodier (robert_dodier) Date: 20050613 23:00 Message: Logged In: YES user_id=501686 Actually, after looking at the code some more (specifically DOSUM in src/asum.lisp) it looks to me like the intent is indeed to evaluate the summand after binding the summation index to itself. In addition, there is code to assume the index is between its lower and upper limits (although assume/forget is handled incorrectly, see 851765). I believe that the observed behavior is a bug, in the narrow sense that the observed behavior is different from what the code author intended. See also my comments on SF bug # 740134. Here is a different patch. I like this one better than the previous one. In DOSUM: ;; ORIGINAL: MEVAL ONCE ;; (mbinding (lind l*i) (meval exp)) (mbinding (lind l*i) (meval (meval exp))) In MEVALSUMARG: ;; ORIGINAL: MEVALATOMS ;; (resimplify (mevalatoms (if (and (not (atom exp)) (resimplify (meval (if (and (not (atom exp)) With these changes, the results are as expected for the examples cited by the original poster, and also the example about the translated function produces the correct result, and some examples adapted from another bug report (740134) yield correct results. ex 1: f(x):=x^2$ b: 1/10$ Ak: b*f(k*b)$ sum(Ak,k,0,9); => 57/200 ex 2: A(k):=k*f(k*b)$ b:1/n$ A(k) => k^3/n^2 sum(A(k),k,0,n1) => ('sum(k^3,k,0,n1))/n^2 sum(A(k),k,0,n1), simpsum => ((n1)^4 + 2*(n1)^3 + (n1)^2) / (4*n^2) ex 3: ak : k^2$ g(a,n) := sum(a,k,1,n)$ g(ak,5) => 55 translate (g)$ g(ak,5) => 55 some other examples, adapted from bug # 740134: ex 4: sum (print (i), i, 1, 3) => prints 1, 2, 3 then returns 6 ex 5: sum (integrate (x^i ,x), i, 0, 2) => x^3/3 + x^2/2 + x ex 6: sum (integrate (1/(x^i + 1), x), i, 0, 1) => log(x+1) + x/2 ex 7: f[i](x):=x^i$ g[i](x):=x^i$ h[i](x):=x^i$ /* reference f[i] and g[i]  see 740134 for the effect this has on previous defn of sum */ f[i]$ g[i](t)$ sum (f[i](x), i, 0, n) => 'sum (x^i, i, 0, n) sum (g[i](x), i, 0, n) => 'sum (x^i, i, 0, n) sum (h[i](x), i, 0, n) => 'sum (x^i, i, 0, n) ex 8: sum (integrate (x^i, x), i, 0, n) => 'sum (x^(i+1) / (i+1), i, 0, n)  Comment By: Barton Willis (willisbl) Date: 20050613 05:30 Message: Logged In: YES user_id=895922 If we decide to use Robert's $sum function, the translate property for sum will need to be changed. Consider: (1) Redefine $sum as Robert suggested. (2) Try this: (%i1) ak : k^2$ (%i2) g(a,n) := sum(a,k,1,n)$ (%i3) g(ak,5); (%o3) 55 < correct for Robert's $sum function (%i4) translate(g); (%o4) [g] (%i5) g(ak,5); (%o5) 5*k^2 < yeech There might be other things that need fixing: (%i8) properties(sum); (%o8) [Special Evaluation Form,OUTATIVE,NOUN,RULE] Barton  Comment By: Robert Dodier (robert_dodier) Date: 20050612 20:34 Message: Logged In: YES user_id=501686 OK, for the record, here is a definition of $sum which implements one of the ideas that has been proposed, namely this policy: evaluate the summand after binding the summation variable to itself. (defmspec $sum (l) (setq l (cdr l)) (if (= (length l) 4) (progv (list (cadr l)) (list (cadr l)) (dosum (meval (car l)) (cadr l) (meval (caddr l)) (meval (cadddr l)) t)) (wnaerr '$sum))) This version yields the results expected by the person who originated this bug report, and yields the results predicted by S Macrakis in the email referenced below (http://www.math.utexas.edu/pipermail/maxima/2003/004869.html), and run_testsuite() runs to completion with no errors. I'm in favor of making this change, but I'm just recording this code snippet here for future reference; no changes planned at the moment. For comparison here is the current version of $sum: (defmspec $sum (l) (setq l (cdr l)) (if (= (length l) 4) (dosum (car l) (cadr l) (meval (caddr l)) (meval (cadddr l)) t) (wnaerr '$sum)))  Comment By: Barton Willis (willisbl) Date: 20050501 06:09 Message: Logged In: YES user_id=895922 We've discussed this before; see http://www.math.utexas.edu/pipermail/maxima/2003/004869.html http://www.math.utexas.edu/pipermail/maxima/2003/004870.html Maybe a simple workaround such as mysum(f,v,lo,hi):=block([acc:0], if integerp(lo) and integerp(hi) then for i from lo thru hi do acc:acc+substitute(i,v,f) else acc:funmake('mysum,[f,v,lo,hi]), acc) or mysum(f,lo,hi):= block([acc:0], if integerp(lo) and integerp(hi) then for i : lo thru hi do acc : acc + apply(f,[i]) else acc : funmake('mysum,[f,lo,hi]), acc); might work for you, Barton  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1192935&group_id=4933 
From: SourceForge.net <noreply@so...>  20050619 16:52:28

Bugs item #1192935, was opened at 20050430 05:10 Message generated for change (Comment added) made by robert_dodier You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1192935&group_id=4933 Please note that this message will contain a full copy of the comment thread, including the initial issue submission, for this request, not just the latest update. Category: Lisp Core Group: None Status: Open Resolution: None Priority: 5 Submitted By: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: termsubstitution in function sum is too late Initial Comment: (%i1) "Lower RiemannSum, Intervall [0,1], 10 parts"$ (%i2) /*Function*/ f(x):=x^2$ (%i3) /*area width*/ b: 1/10$ (%i4) /*area depending on k=0,...,9*/ Ak: b*f(k*b)$ (%i5) Ak; 2 k (%o5)  1000 (%i6) /* wrong */ sum(Ak,k,0,9); 2 k (%o6)  100 obviously, we first have the summation and then the symbolsubstitution (%i7) sum(''Ak,k,0,9); 57 (%o7)  200 finally, this is ok, but the usage of the doublequote is hard to explain to beginners. it gets even worse, if you use a function A(k):=k*f(k*b)$ for a sum with n partitions: (%i8) "Lower RiemannSum, Intervall [0,1], n parts"$ (%i9) /*area width*/ b:1/n$ (%i10) /*area depending on k=0,...,n1*/ A(k):=k*f(k*b)$ (%i11) A(k); 3 k (%o11)  2 n (%i12) sum(A(k),k,0,n1); n  1 ==== (%o12) > A(k) / ==== k = 0 in that case you need ''(A(k)), a bracket and the doublequote. is it possible to patch the sumfunction in that way, that we first have the term or symbolsubstitution and second the summation?  >Comment By: Robert Dodier (robert_dodier) Date: 20050619 10:52 Message: Logged In: YES user_id=501686 I'm attaching my latest and greatest revision (tmpsum5.lisp) of DOSUM, DO%SUM, and MEVALSUMARGS. I find that load ("tmpsum5.lisp"); batch ("rtestsum.mac", test); where rtestsum.mac is the currently attached version, yields all tests passed for Maxima 5.9.1cvs built with clisp 2.33.2 and with gcl 2.6.6. This revision uses a gensym index for cases handled by MEVALSUMARGS (i.e., cases for which (upper limit) minus (lower limit) is not an integer); the gensym index is the argument for assume/forget. When upper minus lower is an integer, the index is bound to lower, lower+1, lower+2, etc., the summand is evaluated, and then lower, lower+1, lower+2, etc. is substituted for any remaining instances of index.  Comment By: Robert Dodier (robert_dodier) Date: 20050619 10:33 Message: Logged In: YES user_id=501686 Attaching a longer list of test cases. batch ("rtestsum.mac", test); executes the test. As always the "correct" results (2nd line in each pair) are subject to review.  Comment By: Robert Dodier (robert_dodier) Date: 20050619 10:28 Message: Logged In: YES user_id=501686 I'm removing rtestsum.mac, in anticipation of attaching a new version in a minute.  Comment By: van_Nek (van_nek) Date: 20050618 11:42 Message: Logged In: YES user_id=1269745 Dear Maximadevelopers, thank you for your support. I tried the last fix from Robert Dodier and it works. Volker van Nek original (anonymous) poster  Comment By: Robert Dodier (robert_dodier) Date: 20050615 08:37 Message: Logged In: YES user_id=501686 I'm attaching a list of test cases for sum, which can be executed by batch ("rtestsum.mac", test) . Doubtless there will be some discussion as to what the expected output of each test should be. All tests pass after making these 2 changes: ;; ORIGINAL: MEVAL ONCE ;; (mbinding (lind l*i) (meval exp)) ;; TRIED THIS: MEVAL TWICE ;; (mbinding (lind l*i) (meval (meval exp))) ;; THIRD TIME IS A CHARM ?? ($substitute *i ind (mbinding (lind l*i) (meval exp))) ;; ORIGINAL: MEVALATOMS ;; (resimplify (mevalatoms (if (and (not (atom exp)) (resimplify (meval (if (and (not (atom exp))  Comment By: Robert Dodier (robert_dodier) Date: 20050614 08:42 Message: Logged In: YES user_id=501686 OK, here is another attempt. I think I now understand this comment by Stavros in bug report # 740134  "First evaluate foo with i bound to itself, and check if that result is free of i. If so, return the product. If not, *substitute* (don't evaluate) i=lowerlimit, i=lowerlimit+1, etc." ;; ORIGINAL: MEVAL ONCE ;; (mbinding (lind l*i) (meval exp)) ;; TRIED THIS: MEVAL TWICE ;; (mbinding (lind l*i) (meval (meval exp))) ;; THIRD TIME IS A CHARM ?? ($substitute *i ind (mbinding (lind l*i) (meval exp))) This version now yields this result  ex 9: f(x) := sum (x, i, 1, 3); f(x) => 3 x as expected, as well as the same results for ex 1 through ex 8.  Comment By: Robert Dodier (robert_dodier) Date: 20050613 23:00 Message: Logged In: YES user_id=501686 Actually, after looking at the code some more (specifically DOSUM in src/asum.lisp) it looks to me like the intent is indeed to evaluate the summand after binding the summation index to itself. In addition, there is code to assume the index is between its lower and upper limits (although assume/forget is handled incorrectly, see 851765). I believe that the observed behavior is a bug, in the narrow sense that the observed behavior is different from what the code author intended. See also my comments on SF bug # 740134. Here is a different patch. I like this one better than the previous one. In DOSUM: ;; ORIGINAL: MEVAL ONCE ;; (mbinding (lind l*i) (meval exp)) (mbinding (lind l*i) (meval (meval exp))) In MEVALSUMARG: ;; ORIGINAL: MEVALATOMS ;; (resimplify (mevalatoms (if (and (not (atom exp)) (resimplify (meval (if (and (not (atom exp)) With these changes, the results are as expected for the examples cited by the original poster, and also the example about the translated function produces the correct result, and some examples adapted from another bug report (740134) yield correct results. ex 1: f(x):=x^2$ b: 1/10$ Ak: b*f(k*b)$ sum(Ak,k,0,9); => 57/200 ex 2: A(k):=k*f(k*b)$ b:1/n$ A(k) => k^3/n^2 sum(A(k),k,0,n1) => ('sum(k^3,k,0,n1))/n^2 sum(A(k),k,0,n1), simpsum => ((n1)^4 + 2*(n1)^3 + (n1)^2) / (4*n^2) ex 3: ak : k^2$ g(a,n) := sum(a,k,1,n)$ g(ak,5) => 55 translate (g)$ g(ak,5) => 55 some other examples, adapted from bug # 740134: ex 4: sum (print (i), i, 1, 3) => prints 1, 2, 3 then returns 6 ex 5: sum (integrate (x^i ,x), i, 0, 2) => x^3/3 + x^2/2 + x ex 6: sum (integrate (1/(x^i + 1), x), i, 0, 1) => log(x+1) + x/2 ex 7: f[i](x):=x^i$ g[i](x):=x^i$ h[i](x):=x^i$ /* reference f[i] and g[i]  see 740134 for the effect this has on previous defn of sum */ f[i]$ g[i](t)$ sum (f[i](x), i, 0, n) => 'sum (x^i, i, 0, n) sum (g[i](x), i, 0, n) => 'sum (x^i, i, 0, n) sum (h[i](x), i, 0, n) => 'sum (x^i, i, 0, n) ex 8: sum (integrate (x^i, x), i, 0, n) => 'sum (x^(i+1) / (i+1), i, 0, n)  Comment By: Barton Willis (willisbl) Date: 20050613 05:30 Message: Logged In: YES user_id=895922 If we decide to use Robert's $sum function, the translate property for sum will need to be changed. Consider: (1) Redefine $sum as Robert suggested. (2) Try this: (%i1) ak : k^2$ (%i2) g(a,n) := sum(a,k,1,n)$ (%i3) g(ak,5); (%o3) 55 < correct for Robert's $sum function (%i4) translate(g); (%o4) [g] (%i5) g(ak,5); (%o5) 5*k^2 < yeech There might be other things that need fixing: (%i8) properties(sum); (%o8) [Special Evaluation Form,OUTATIVE,NOUN,RULE] Barton  Comment By: Robert Dodier (robert_dodier) Date: 20050612 20:34 Message: Logged In: YES user_id=501686 OK, for the record, here is a definition of $sum which implements one of the ideas that has been proposed, namely this policy: evaluate the summand after binding the summation variable to itself. (defmspec $sum (l) (setq l (cdr l)) (if (= (length l) 4) (progv (list (cadr l)) (list (cadr l)) (dosum (meval (car l)) (cadr l) (meval (caddr l)) (meval (cadddr l)) t)) (wnaerr '$sum))) This version yields the results expected by the person who originated this bug report, and yields the results predicted by S Macrakis in the email referenced below (http://www.math.utexas.edu/pipermail/maxima/2003/004869.html), and run_testsuite() runs to completion with no errors. I'm in favor of making this change, but I'm just recording this code snippet here for future reference; no changes planned at the moment. For comparison here is the current version of $sum: (defmspec $sum (l) (setq l (cdr l)) (if (= (length l) 4) (dosum (car l) (cadr l) (meval (caddr l)) (meval (cadddr l)) t) (wnaerr '$sum)))  Comment By: Barton Willis (willisbl) Date: 20050501 06:09 Message: Logged In: YES user_id=895922 We've discussed this before; see http://www.math.utexas.edu/pipermail/maxima/2003/004869.html http://www.math.utexas.edu/pipermail/maxima/2003/004870.html Maybe a simple workaround such as mysum(f,v,lo,hi):=block([acc:0], if integerp(lo) and integerp(hi) then for i from lo thru hi do acc:acc+substitute(i,v,f) else acc:funmake('mysum,[f,v,lo,hi]), acc) or mysum(f,lo,hi):= block([acc:0], if integerp(lo) and integerp(hi) then for i : lo thru hi do acc : acc + apply(f,[i]) else acc : funmake('mysum,[f,lo,hi]), acc); might work for you, Barton  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1192935&group_id=4933 
From: SourceForge.net <noreply@so...>  20050620 01:59:26

Bugs item #1192935, was opened at 20050430 05:10 Message generated for change (Comment added) made by robert_dodier You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1192935&group_id=4933 Please note that this message will contain a full copy of the comment thread, including the initial issue submission, for this request, not just the latest update. Category: Lisp Core Group: None Status: Open Resolution: None Priority: 5 Submitted By: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: termsubstitution in function sum is too late Initial Comment: (%i1) "Lower RiemannSum, Intervall [0,1], 10 parts"$ (%i2) /*Function*/ f(x):=x^2$ (%i3) /*area width*/ b: 1/10$ (%i4) /*area depending on k=0,...,9*/ Ak: b*f(k*b)$ (%i5) Ak; 2 k (%o5)  1000 (%i6) /* wrong */ sum(Ak,k,0,9); 2 k (%o6)  100 obviously, we first have the summation and then the symbolsubstitution (%i7) sum(''Ak,k,0,9); 57 (%o7)  200 finally, this is ok, but the usage of the doublequote is hard to explain to beginners. it gets even worse, if you use a function A(k):=k*f(k*b)$ for a sum with n partitions: (%i8) "Lower RiemannSum, Intervall [0,1], n parts"$ (%i9) /*area width*/ b:1/n$ (%i10) /*area depending on k=0,...,n1*/ A(k):=k*f(k*b)$ (%i11) A(k); 3 k (%o11)  2 n (%i12) sum(A(k),k,0,n1); n  1 ==== (%o12) > A(k) / ==== k = 0 in that case you need ''(A(k)), a bracket and the doublequote. is it possible to patch the sumfunction in that way, that we first have the term or symbolsubstitution and second the summation?  >Comment By: Robert Dodier (robert_dodier) Date: 20050619 19:59 Message: Logged In: YES user_id=501686 Removing rtestsum.mac, will upload a new version momentarily.  Comment By: Robert Dodier (robert_dodier) Date: 20050619 10:52 Message: Logged In: YES user_id=501686 I'm attaching my latest and greatest revision (tmpsum5.lisp) of DOSUM, DO%SUM, and MEVALSUMARGS. I find that load ("tmpsum5.lisp"); batch ("rtestsum.mac", test); where rtestsum.mac is the currently attached version, yields all tests passed for Maxima 5.9.1cvs built with clisp 2.33.2 and with gcl 2.6.6. This revision uses a gensym index for cases handled by MEVALSUMARGS (i.e., cases for which (upper limit) minus (lower limit) is not an integer); the gensym index is the argument for assume/forget. When upper minus lower is an integer, the index is bound to lower, lower+1, lower+2, etc., the summand is evaluated, and then lower, lower+1, lower+2, etc. is substituted for any remaining instances of index.  Comment By: Robert Dodier (robert_dodier) Date: 20050619 10:33 Message: Logged In: YES user_id=501686 Attaching a longer list of test cases. batch ("rtestsum.mac", test); executes the test. As always the "correct" results (2nd line in each pair) are subject to review.  Comment By: Robert Dodier (robert_dodier) Date: 20050619 10:28 Message: Logged In: YES user_id=501686 I'm removing rtestsum.mac, in anticipation of attaching a new version in a minute.  Comment By: van_Nek (van_nek) Date: 20050618 11:42 Message: Logged In: YES user_id=1269745 Dear Maximadevelopers, thank you for your support. I tried the last fix from Robert Dodier and it works. Volker van Nek original (anonymous) poster  Comment By: Robert Dodier (robert_dodier) Date: 20050615 08:37 Message: Logged In: YES user_id=501686 I'm attaching a list of test cases for sum, which can be executed by batch ("rtestsum.mac", test) . Doubtless there will be some discussion as to what the expected output of each test should be. All tests pass after making these 2 changes: ;; ORIGINAL: MEVAL ONCE ;; (mbinding (lind l*i) (meval exp)) ;; TRIED THIS: MEVAL TWICE ;; (mbinding (lind l*i) (meval (meval exp))) ;; THIRD TIME IS A CHARM ?? ($substitute *i ind (mbinding (lind l*i) (meval exp))) ;; ORIGINAL: MEVALATOMS ;; (resimplify (mevalatoms (if (and (not (atom exp)) (resimplify (meval (if (and (not (atom exp))  Comment By: Robert Dodier (robert_dodier) Date: 20050614 08:42 Message: Logged In: YES user_id=501686 OK, here is another attempt. I think I now understand this comment by Stavros in bug report # 740134  "First evaluate foo with i bound to itself, and check if that result is free of i. If so, return the product. If not, *substitute* (don't evaluate) i=lowerlimit, i=lowerlimit+1, etc." ;; ORIGINAL: MEVAL ONCE ;; (mbinding (lind l*i) (meval exp)) ;; TRIED THIS: MEVAL TWICE ;; (mbinding (lind l*i) (meval (meval exp))) ;; THIRD TIME IS A CHARM ?? ($substitute *i ind (mbinding (lind l*i) (meval exp))) This version now yields this result  ex 9: f(x) := sum (x, i, 1, 3); f(x) => 3 x as expected, as well as the same results for ex 1 through ex 8.  Comment By: Robert Dodier (robert_dodier) Date: 20050613 23:00 Message: Logged In: YES user_id=501686 Actually, after looking at the code some more (specifically DOSUM in src/asum.lisp) it looks to me like the intent is indeed to evaluate the summand after binding the summation index to itself. In addition, there is code to assume the index is between its lower and upper limits (although assume/forget is handled incorrectly, see 851765). I believe that the observed behavior is a bug, in the narrow sense that the observed behavior is different from what the code author intended. See also my comments on SF bug # 740134. Here is a different patch. I like this one better than the previous one. In DOSUM: ;; ORIGINAL: MEVAL ONCE ;; (mbinding (lind l*i) (meval exp)) (mbinding (lind l*i) (meval (meval exp))) In MEVALSUMARG: ;; ORIGINAL: MEVALATOMS ;; (resimplify (mevalatoms (if (and (not (atom exp)) (resimplify (meval (if (and (not (atom exp)) With these changes, the results are as expected for the examples cited by the original poster, and also the example about the translated function produces the correct result, and some examples adapted from another bug report (740134) yield correct results. ex 1: f(x):=x^2$ b: 1/10$ Ak: b*f(k*b)$ sum(Ak,k,0,9); => 57/200 ex 2: A(k):=k*f(k*b)$ b:1/n$ A(k) => k^3/n^2 sum(A(k),k,0,n1) => ('sum(k^3,k,0,n1))/n^2 sum(A(k),k,0,n1), simpsum => ((n1)^4 + 2*(n1)^3 + (n1)^2) / (4*n^2) ex 3: ak : k^2$ g(a,n) := sum(a,k,1,n)$ g(ak,5) => 55 translate (g)$ g(ak,5) => 55 some other examples, adapted from bug # 740134: ex 4: sum (print (i), i, 1, 3) => prints 1, 2, 3 then returns 6 ex 5: sum (integrate (x^i ,x), i, 0, 2) => x^3/3 + x^2/2 + x ex 6: sum (integrate (1/(x^i + 1), x), i, 0, 1) => log(x+1) + x/2 ex 7: f[i](x):=x^i$ g[i](x):=x^i$ h[i](x):=x^i$ /* reference f[i] and g[i]  see 740134 for the effect this has on previous defn of sum */ f[i]$ g[i](t)$ sum (f[i](x), i, 0, n) => 'sum (x^i, i, 0, n) sum (g[i](x), i, 0, n) => 'sum (x^i, i, 0, n) sum (h[i](x), i, 0, n) => 'sum (x^i, i, 0, n) ex 8: sum (integrate (x^i, x), i, 0, n) => 'sum (x^(i+1) / (i+1), i, 0, n)  Comment By: Barton Willis (willisbl) Date: 20050613 05:30 Message: Logged In: YES user_id=895922 If we decide to use Robert's $sum function, the translate property for sum will need to be changed. Consider: (1) Redefine $sum as Robert suggested. (2) Try this: (%i1) ak : k^2$ (%i2) g(a,n) := sum(a,k,1,n)$ (%i3) g(ak,5); (%o3) 55 < correct for Robert's $sum function (%i4) translate(g); (%o4) [g] (%i5) g(ak,5); (%o5) 5*k^2 < yeech There might be other things that need fixing: (%i8) properties(sum); (%o8) [Special Evaluation Form,OUTATIVE,NOUN,RULE] Barton  Comment By: Robert Dodier (robert_dodier) Date: 20050612 20:34 Message: Logged In: YES user_id=501686 OK, for the record, here is a definition of $sum which implements one of the ideas that has been proposed, namely this policy: evaluate the summand after binding the summation variable to itself. (defmspec $sum (l) (setq l (cdr l)) (if (= (length l) 4) (progv (list (cadr l)) (list (cadr l)) (dosum (meval (car l)) (cadr l) (meval (caddr l)) (meval (cadddr l)) t)) (wnaerr '$sum))) This version yields the results expected by the person who originated this bug report, and yields the results predicted by S Macrakis in the email referenced below (http://www.math.utexas.edu/pipermail/maxima/2003/004869.html), and run_testsuite() runs to completion with no errors. I'm in favor of making this change, but I'm just recording this code snippet here for future reference; no changes planned at the moment. For comparison here is the current version of $sum: (defmspec $sum (l) (setq l (cdr l)) (if (= (length l) 4) (dosum (car l) (cadr l) (meval (caddr l)) (meval (cadddr l)) t) (wnaerr '$sum)))  Comment By: Barton Willis (willisbl) Date: 20050501 06:09 Message: Logged In: YES user_id=895922 We've discussed this before; see http://www.math.utexas.edu/pipermail/maxima/2003/004869.html http://www.math.utexas.edu/pipermail/maxima/2003/004870.html Maybe a simple workaround such as mysum(f,v,lo,hi):=block([acc:0], if integerp(lo) and integerp(hi) then for i from lo thru hi do acc:acc+substitute(i,v,f) else acc:funmake('mysum,[f,v,lo,hi]), acc) or mysum(f,lo,hi):= block([acc:0], if integerp(lo) and integerp(hi) then for i : lo thru hi do acc : acc + apply(f,[i]) else acc : funmake('mysum,[f,lo,hi]), acc); might work for you, Barton  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1192935&group_id=4933 
From: SourceForge.net <noreply@so...>  20050620 02:04:50

Bugs item #1192935, was opened at 20050430 05:10 Message generated for change (Comment added) made by robert_dodier You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1192935&group_id=4933 Please note that this message will contain a full copy of the comment thread, including the initial issue submission, for this request, not just the latest update. Category: Lisp Core Group: None Status: Open Resolution: None Priority: 5 Submitted By: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: termsubstitution in function sum is too late Initial Comment: (%i1) "Lower RiemannSum, Intervall [0,1], 10 parts"$ (%i2) /*Function*/ f(x):=x^2$ (%i3) /*area width*/ b: 1/10$ (%i4) /*area depending on k=0,...,9*/ Ak: b*f(k*b)$ (%i5) Ak; 2 k (%o5)  1000 (%i6) /* wrong */ sum(Ak,k,0,9); 2 k (%o6)  100 obviously, we first have the summation and then the symbolsubstitution (%i7) sum(''Ak,k,0,9); 57 (%o7)  200 finally, this is ok, but the usage of the doublequote is hard to explain to beginners. it gets even worse, if you use a function A(k):=k*f(k*b)$ for a sum with n partitions: (%i8) "Lower RiemannSum, Intervall [0,1], n parts"$ (%i9) /*area width*/ b:1/n$ (%i10) /*area depending on k=0,...,n1*/ A(k):=k*f(k*b)$ (%i11) A(k); 3 k (%o11)  2 n (%i12) sum(A(k),k,0,n1); n  1 ==== (%o12) > A(k) / ==== k = 0 in that case you need ''(A(k)), a bracket and the doublequote. is it possible to patch the sumfunction in that way, that we first have the term or symbolsubstitution and second the summation?  >Comment By: Robert Dodier (robert_dodier) Date: 20050619 20:04 Message: Logged In: YES user_id=501686 New version of rtestsum.mac. Two new test cases, sum (lambda ([i], i^2, 1, 3) and sum (lambda ([i], i^2), i, 1, n).  Comment By: Robert Dodier (robert_dodier) Date: 20050619 19:59 Message: Logged In: YES user_id=501686 Removing rtestsum.mac, will upload a new version momentarily.  Comment By: Robert Dodier (robert_dodier) Date: 20050619 10:52 Message: Logged In: YES user_id=501686 I'm attaching my latest and greatest revision (tmpsum5.lisp) of DOSUM, DO%SUM, and MEVALSUMARGS. I find that load ("tmpsum5.lisp"); batch ("rtestsum.mac", test); where rtestsum.mac is the currently attached version, yields all tests passed for Maxima 5.9.1cvs built with clisp 2.33.2 and with gcl 2.6.6. This revision uses a gensym index for cases handled by MEVALSUMARGS (i.e., cases for which (upper limit) minus (lower limit) is not an integer); the gensym index is the argument for assume/forget. When upper minus lower is an integer, the index is bound to lower, lower+1, lower+2, etc., the summand is evaluated, and then lower, lower+1, lower+2, etc. is substituted for any remaining instances of index.  Comment By: Robert Dodier (robert_dodier) Date: 20050619 10:33 Message: Logged In: YES user_id=501686 Attaching a longer list of test cases. batch ("rtestsum.mac", test); executes the test. As always the "correct" results (2nd line in each pair) are subject to review.  Comment By: Robert Dodier (robert_dodier) Date: 20050619 10:28 Message: Logged In: YES user_id=501686 I'm removing rtestsum.mac, in anticipation of attaching a new version in a minute.  Comment By: van_Nek (van_nek) Date: 20050618 11:42 Message: Logged In: YES user_id=1269745 Dear Maximadevelopers, thank you for your support. I tried the last fix from Robert Dodier and it works. Volker van Nek original (anonymous) poster  Comment By: Robert Dodier (robert_dodier) Date: 20050615 08:37 Message: Logged In: YES user_id=501686 I'm attaching a list of test cases for sum, which can be executed by batch ("rtestsum.mac", test) . Doubtless there will be some discussion as to what the expected output of each test should be. All tests pass after making these 2 changes: ;; ORIGINAL: MEVAL ONCE ;; (mbinding (lind l*i) (meval exp)) ;; TRIED THIS: MEVAL TWICE ;; (mbinding (lind l*i) (meval (meval exp))) ;; THIRD TIME IS A CHARM ?? ($substitute *i ind (mbinding (lind l*i) (meval exp))) ;; ORIGINAL: MEVALATOMS ;; (resimplify (mevalatoms (if (and (not (atom exp)) (resimplify (meval (if (and (not (atom exp))  Comment By: Robert Dodier (robert_dodier) Date: 20050614 08:42 Message: Logged In: YES user_id=501686 OK, here is another attempt. I think I now understand this comment by Stavros in bug report # 740134  "First evaluate foo with i bound to itself, and check if that result is free of i. If so, return the product. If not, *substitute* (don't evaluate) i=lowerlimit, i=lowerlimit+1, etc." ;; ORIGINAL: MEVAL ONCE ;; (mbinding (lind l*i) (meval exp)) ;; TRIED THIS: MEVAL TWICE ;; (mbinding (lind l*i) (meval (meval exp))) ;; THIRD TIME IS A CHARM ?? ($substitute *i ind (mbinding (lind l*i) (meval exp))) This version now yields this result  ex 9: f(x) := sum (x, i, 1, 3); f(x) => 3 x as expected, as well as the same results for ex 1 through ex 8.  Comment By: Robert Dodier (robert_dodier) Date: 20050613 23:00 Message: Logged In: YES user_id=501686 Actually, after looking at the code some more (specifically DOSUM in src/asum.lisp) it looks to me like the intent is indeed to evaluate the summand after binding the summation index to itself. In addition, there is code to assume the index is between its lower and upper limits (although assume/forget is handled incorrectly, see 851765). I believe that the observed behavior is a bug, in the narrow sense that the observed behavior is different from what the code author intended. See also my comments on SF bug # 740134. Here is a different patch. I like this one better than the previous one. In DOSUM: ;; ORIGINAL: MEVAL ONCE ;; (mbinding (lind l*i) (meval exp)) (mbinding (lind l*i) (meval (meval exp))) In MEVALSUMARG: ;; ORIGINAL: MEVALATOMS ;; (resimplify (mevalatoms (if (and (not (atom exp)) (resimplify (meval (if (and (not (atom exp)) With these changes, the results are as expected for the examples cited by the original poster, and also the example about the translated function produces the correct result, and some examples adapted from another bug report (740134) yield correct results. ex 1: f(x):=x^2$ b: 1/10$ Ak: b*f(k*b)$ sum(Ak,k,0,9); => 57/200 ex 2: A(k):=k*f(k*b)$ b:1/n$ A(k) => k^3/n^2 sum(A(k),k,0,n1) => ('sum(k^3,k,0,n1))/n^2 sum(A(k),k,0,n1), simpsum => ((n1)^4 + 2*(n1)^3 + (n1)^2) / (4*n^2) ex 3: ak : k^2$ g(a,n) := sum(a,k,1,n)$ g(ak,5) => 55 translate (g)$ g(ak,5) => 55 some other examples, adapted from bug # 740134: ex 4: sum (print (i), i, 1, 3) => prints 1, 2, 3 then returns 6 ex 5: sum (integrate (x^i ,x), i, 0, 2) => x^3/3 + x^2/2 + x ex 6: sum (integrate (1/(x^i + 1), x), i, 0, 1) => log(x+1) + x/2 ex 7: f[i](x):=x^i$ g[i](x):=x^i$ h[i](x):=x^i$ /* reference f[i] and g[i]  see 740134 for the effect this has on previous defn of sum */ f[i]$ g[i](t)$ sum (f[i](x), i, 0, n) => 'sum (x^i, i, 0, n) sum (g[i](x), i, 0, n) => 'sum (x^i, i, 0, n) sum (h[i](x), i, 0, n) => 'sum (x^i, i, 0, n) ex 8: sum (integrate (x^i, x), i, 0, n) => 'sum (x^(i+1) / (i+1), i, 0, n)  Comment By: Barton Willis (willisbl) Date: 20050613 05:30 Message: Logged In: YES user_id=895922 If we decide to use Robert's $sum function, the translate property for sum will need to be changed. Consider: (1) Redefine $sum as Robert suggested. (2) Try this: (%i1) ak : k^2$ (%i2) g(a,n) := sum(a,k,1,n)$ (%i3) g(ak,5); (%o3) 55 < correct for Robert's $sum function (%i4) translate(g); (%o4) [g] (%i5) g(ak,5); (%o5) 5*k^2 < yeech There might be other things that need fixing: (%i8) properties(sum); (%o8) [Special Evaluation Form,OUTATIVE,NOUN,RULE] Barton  Comment By: Robert Dodier (robert_dodier) Date: 20050612 20:34 Message: Logged In: YES user_id=501686 OK, for the record, here is a definition of $sum which implements one of the ideas that has been proposed, namely this policy: evaluate the summand after binding the summation variable to itself. (defmspec $sum (l) (setq l (cdr l)) (if (= (length l) 4) (progv (list (cadr l)) (list (cadr l)) (dosum (meval (car l)) (cadr l) (meval (caddr l)) (meval (cadddr l)) t)) (wnaerr '$sum))) This version yields the results expected by the person who originated this bug report, and yields the results predicted by S Macrakis in the email referenced below (http://www.math.utexas.edu/pipermail/maxima/2003/004869.html), and run_testsuite() runs to completion with no errors. I'm in favor of making this change, but I'm just recording this code snippet here for future reference; no changes planned at the moment. For comparison here is the current version of $sum: (defmspec $sum (l) (setq l (cdr l)) (if (= (length l) 4) (dosum (car l) (cadr l) (meval (caddr l)) (meval (cadddr l)) t) (wnaerr '$sum)))  Comment By: Barton Willis (willisbl) Date: 20050501 06:09 Message: Logged In: YES user_id=895922 We've discussed this before; see http://www.math.utexas.edu/pipermail/maxima/2003/004869.html http://www.math.utexas.edu/pipermail/maxima/2003/004870.html Maybe a simple workaround such as mysum(f,v,lo,hi):=block([acc:0], if integerp(lo) and integerp(hi) then for i from lo thru hi do acc:acc+substitute(i,v,f) else acc:funmake('mysum,[f,v,lo,hi]), acc) or mysum(f,lo,hi):= block([acc:0], if integerp(lo) and integerp(hi) then for i : lo thru hi do acc : acc + apply(f,[i]) else acc : funmake('mysum,[f,lo,hi]), acc); might work for you, Barton  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1192935&group_id=4933 
From: SourceForge.net <noreply@so...>  20050620 02:28:38

Bugs item #1192935, was opened at 20050430 05:10 Message generated for change (Comment added) made by robert_dodier You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1192935&group_id=4933 Please note that this message will contain a full copy of the comment thread, including the initial issue submission, for this request, not just the latest update. Category: Lisp Core Group: None Status: Open Resolution: None Priority: 5 Submitted By: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: termsubstitution in function sum is too late Initial Comment: (%i1) "Lower RiemannSum, Intervall [0,1], 10 parts"$ (%i2) /*Function*/ f(x):=x^2$ (%i3) /*area width*/ b: 1/10$ (%i4) /*area depending on k=0,...,9*/ Ak: b*f(k*b)$ (%i5) Ak; 2 k (%o5)  1000 (%i6) /* wrong */ sum(Ak,k,0,9); 2 k (%o6)  100 obviously, we first have the summation and then the symbolsubstitution (%i7) sum(''Ak,k,0,9); 57 (%o7)  200 finally, this is ok, but the usage of the doublequote is hard to explain to beginners. it gets even worse, if you use a function A(k):=k*f(k*b)$ for a sum with n partitions: (%i8) "Lower RiemannSum, Intervall [0,1], n parts"$ (%i9) /*area width*/ b:1/n$ (%i10) /*area depending on k=0,...,n1*/ A(k):=k*f(k*b)$ (%i11) A(k); 3 k (%o11)  2 n (%i12) sum(A(k),k,0,n1); n  1 ==== (%o12) > A(k) / ==== k = 0 in that case you need ''(A(k)), a bracket and the doublequote. is it possible to patch the sumfunction in that way, that we first have the term or symbolsubstitution and second the summation?  >Comment By: Robert Dodier (robert_dodier) Date: 20050619 20:28 Message: Logged In: YES user_id=501686 Yet another revision (attached file tmpsum6.lisp) of DOSUM and friends. This version passes run_testsuite() and batch ("rtestsum.mac", test) using the currently attached version of rtestsum.mac. This version avoids substituting for the index in expressions which are free of the index. This version also has an alternate definition of the function FREE. The simplification code (and other code) calls FREE instead of FREEOF. FREEOF looks for dummy variables, but FREE does not, so with the original defn of FREE, sum (lambda([i], i^2), i, 1, n) does not simplify to n lambda ([i], i^2), i, 1, n). I'm not very happy about changing FREE since it is called from all over the place, but I'm also not very happy about the presence of FREE, either; why not call FREEOF ? I don't think changing FREE at this point is such a good idea; it seems to me the two choices are (1) inspect the sum/product simplification code and change FREE to FREEOF where possible. That would be a lot of work. (2) Leave FREE as it is, and just accept that expressions with dummy variables can't be factored out of summations. At this point I think I'd rather leave FREE alone, and file a separate bug report about the inability to factor out expressions with dummy variables, and just leave it for another day.  Comment By: Robert Dodier (robert_dodier) Date: 20050619 20:04 Message: Logged In: YES user_id=501686 New version of rtestsum.mac. Two new test cases, sum (lambda ([i], i^2, 1, 3) and sum (lambda ([i], i^2), i, 1, n).  Comment By: Robert Dodier (robert_dodier) Date: 20050619 19:59 Message: Logged In: YES user_id=501686 Removing rtestsum.mac, will upload a new version momentarily.  Comment By: Robert Dodier (robert_dodier) Date: 20050619 10:52 Message: Logged In: YES user_id=501686 I'm attaching my latest and greatest revision (tmpsum5.lisp) of DOSUM, DO%SUM, and MEVALSUMARGS. I find that load ("tmpsum5.lisp"); batch ("rtestsum.mac", test); where rtestsum.mac is the currently attached version, yields all tests passed for Maxima 5.9.1cvs built with clisp 2.33.2 and with gcl 2.6.6. This revision uses a gensym index for cases handled by MEVALSUMARGS (i.e., cases for which (upper limit) minus (lower limit) is not an integer); the gensym index is the argument for assume/forget. When upper minus lower is an integer, the index is bound to lower, lower+1, lower+2, etc., the summand is evaluated, and then lower, lower+1, lower+2, etc. is substituted for any remaining instances of index.  Comment By: Robert Dodier (robert_dodier) Date: 20050619 10:33 Message: Logged In: YES user_id=501686 Attaching a longer list of test cases. batch ("rtestsum.mac", test); executes the test. As always the "correct" results (2nd line in each pair) are subject to review.  Comment By: Robert Dodier (robert_dodier) Date: 20050619 10:28 Message: Logged In: YES user_id=501686 I'm removing rtestsum.mac, in anticipation of attaching a new version in a minute.  Comment By: van_Nek (van_nek) Date: 20050618 11:42 Message: Logged In: YES user_id=1269745 Dear Maximadevelopers, thank you for your support. I tried the last fix from Robert Dodier and it works. Volker van Nek original (anonymous) poster  Comment By: Robert Dodier (robert_dodier) Date: 20050615 08:37 Message: Logged In: YES user_id=501686 I'm attaching a list of test cases for sum, which can be executed by batch ("rtestsum.mac", test) . Doubtless there will be some discussion as to what the expected output of each test should be. All tests pass after making these 2 changes: ;; ORIGINAL: MEVAL ONCE ;; (mbinding (lind l*i) (meval exp)) ;; TRIED THIS: MEVAL TWICE ;; (mbinding (lind l*i) (meval (meval exp))) ;; THIRD TIME IS A CHARM ?? ($substitute *i ind (mbinding (lind l*i) (meval exp))) ;; ORIGINAL: MEVALATOMS ;; (resimplify (mevalatoms (if (and (not (atom exp)) (resimplify (meval (if (and (not (atom exp))  Comment By: Robert Dodier (robert_dodier) Date: 20050614 08:42 Message: Logged In: YES user_id=501686 OK, here is another attempt. I think I now understand this comment by Stavros in bug report # 740134  "First evaluate foo with i bound to itself, and check if that result is free of i. If so, return the product. If not, *substitute* (don't evaluate) i=lowerlimit, i=lowerlimit+1, etc." ;; ORIGINAL: MEVAL ONCE ;; (mbinding (lind l*i) (meval exp)) ;; TRIED THIS: MEVAL TWICE ;; (mbinding (lind l*i) (meval (meval exp))) ;; THIRD TIME IS A CHARM ?? ($substitute *i ind (mbinding (lind l*i) (meval exp))) This version now yields this result  ex 9: f(x) := sum (x, i, 1, 3); f(x) => 3 x as expected, as well as the same results for ex 1 through ex 8.  Comment By: Robert Dodier (robert_dodier) Date: 20050613 23:00 Message: Logged In: YES user_id=501686 Actually, after looking at the code some more (specifically DOSUM in src/asum.lisp) it looks to me like the intent is indeed to evaluate the summand after binding the summation index to itself. In addition, there is code to assume the index is between its lower and upper limits (although assume/forget is handled incorrectly, see 851765). I believe that the observed behavior is a bug, in the narrow sense that the observed behavior is different from what the code author intended. See also my comments on SF bug # 740134. Here is a different patch. I like this one better than the previous one. In DOSUM: ;; ORIGINAL: MEVAL ONCE ;; (mbinding (lind l*i) (meval exp)) (mbinding (lind l*i) (meval (meval exp))) In MEVALSUMARG: ;; ORIGINAL: MEVALATOMS ;; (resimplify (mevalatoms (if (and (not (atom exp)) (resimplify (meval (if (and (not (atom exp)) With these changes, the results are as expected for the examples cited by the original poster, and also the example about the translated function produces the correct result, and some examples adapted from another bug report (740134) yield correct results. ex 1: f(x):=x^2$ b: 1/10$ Ak: b*f(k*b)$ sum(Ak,k,0,9); => 57/200 ex 2: A(k):=k*f(k*b)$ b:1/n$ A(k) => k^3/n^2 sum(A(k),k,0,n1) => ('sum(k^3,k,0,n1))/n^2 sum(A(k),k,0,n1), simpsum => ((n1)^4 + 2*(n1)^3 + (n1)^2) / (4*n^2) ex 3: ak : k^2$ g(a,n) := sum(a,k,1,n)$ g(ak,5) => 55 translate (g)$ g(ak,5) => 55 some other examples, adapted from bug # 740134: ex 4: sum (print (i), i, 1, 3) => prints 1, 2, 3 then returns 6 ex 5: sum (integrate (x^i ,x), i, 0, 2) => x^3/3 + x^2/2 + x ex 6: sum (integrate (1/(x^i + 1), x), i, 0, 1) => log(x+1) + x/2 ex 7: f[i](x):=x^i$ g[i](x):=x^i$ h[i](x):=x^i$ /* reference f[i] and g[i]  see 740134 for the effect this has on previous defn of sum */ f[i]$ g[i](t)$ sum (f[i](x), i, 0, n) => 'sum (x^i, i, 0, n) sum (g[i](x), i, 0, n) => 'sum (x^i, i, 0, n) sum (h[i](x), i, 0, n) => 'sum (x^i, i, 0, n) ex 8: sum (integrate (x^i, x), i, 0, n) => 'sum (x^(i+1) / (i+1), i, 0, n)  Comment By: Barton Willis (willisbl) Date: 20050613 05:30 Message: Logged In: YES user_id=895922 If we decide to use Robert's $sum function, the translate property for sum will need to be changed. Consider: (1) Redefine $sum as Robert suggested. (2) Try this: (%i1) ak : k^2$ (%i2) g(a,n) := sum(a,k,1,n)$ (%i3) g(ak,5); (%o3) 55 < correct for Robert's $sum function (%i4) translate(g); (%o4) [g] (%i5) g(ak,5); (%o5) 5*k^2 < yeech There might be other things that need fixing: (%i8) properties(sum); (%o8) [Special Evaluation Form,OUTATIVE,NOUN,RULE] Barton  Comment By: Robert Dodier (robert_dodier) Date: 20050612 20:34 Message: Logged In: YES user_id=501686 OK, for the record, here is a definition of $sum which implements one of the ideas that has been proposed, namely this policy: evaluate the summand after binding the summation variable to itself. (defmspec $sum (l) (setq l (cdr l)) (if (= (length l) 4) (progv (list (cadr l)) (list (cadr l)) (dosum (meval (car l)) (cadr l) (meval (caddr l)) (meval (cadddr l)) t)) (wnaerr '$sum))) This version yields the results expected by the person who originated this bug report, and yields the results predicted by S Macrakis in the email referenced below (http://www.math.utexas.edu/pipermail/maxima/2003/004869.html), and run_testsuite() runs to completion with no errors. I'm in favor of making this change, but I'm just recording this code snippet here for future reference; no changes planned at the moment. For comparison here is the current version of $sum: (defmspec $sum (l) (setq l (cdr l)) (if (= (length l) 4) (dosum (car l) (cadr l) (meval (caddr l)) (meval (cadddr l)) t) (wnaerr '$sum)))  Comment By: Barton Willis (willisbl) Date: 20050501 06:09 Message: Logged In: YES user_id=895922 We've discussed this before; see http://www.math.utexas.edu/pipermail/maxima/2003/004869.html http://www.math.utexas.edu/pipermail/maxima/2003/004870.html Maybe a simple workaround such as mysum(f,v,lo,hi):=block([acc:0], if integerp(lo) and integerp(hi) then for i from lo thru hi do acc:acc+substitute(i,v,f) else acc:funmake('mysum,[f,v,lo,hi]), acc) or mysum(f,lo,hi):= block([acc:0], if integerp(lo) and integerp(hi) then for i : lo thru hi do acc : acc + apply(f,[i]) else acc : funmake('mysum,[f,lo,hi]), acc); might work for you, Barton  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1192935&group_id=4933 
From: SourceForge.net <noreply@so...>  20050620 03:36:23

Bugs item #1192935, was opened at 20050430 06:10 Message generated for change (Comment added) made by willisbl You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1192935&group_id=4933 Please note that this message will contain a full copy of the comment thread, including the initial issue submission, for this request, not just the latest update. Category: Lisp Core Group: None Status: Open Resolution: None Priority: 5 Submitted By: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: termsubstitution in function sum is too late Initial Comment: (%i1) "Lower RiemannSum, Intervall [0,1], 10 parts"$ (%i2) /*Function*/ f(x):=x^2$ (%i3) /*area width*/ b: 1/10$ (%i4) /*area depending on k=0,...,9*/ Ak: b*f(k*b)$ (%i5) Ak; 2 k (%o5)  1000 (%i6) /* wrong */ sum(Ak,k,0,9); 2 k (%o6)  100 obviously, we first have the summation and then the symbolsubstitution (%i7) sum(''Ak,k,0,9); 57 (%o7)  200 finally, this is ok, but the usage of the doublequote is hard to explain to beginners. it gets even worse, if you use a function A(k):=k*f(k*b)$ for a sum with n partitions: (%i8) "Lower RiemannSum, Intervall [0,1], n parts"$ (%i9) /*area width*/ b:1/n$ (%i10) /*area depending on k=0,...,n1*/ A(k):=k*f(k*b)$ (%i11) A(k); 3 k (%o11)  2 n (%i12) sum(A(k),k,0,n1); n  1 ==== (%o12) > A(k) / ==== k = 0 in that case you need ''(A(k)), a bracket and the doublequote. is it possible to patch the sumfunction in that way, that we first have the term or symbolsubstitution and second the summation?  >Comment By: Barton Willis (willisbl) Date: 20050619 22:36 Message: Logged In: YES user_id=895922 Does tmpsum6.lisp need a ratdisrep somewhere? (%i1) load("c:/maxima/tmpsum6.lisp")$ (%i2) sum(k,k,1,3); (%o2) 6 (%i3) sum(rat(k),k,1,3); Maxima encountered a Lisp error: Error in COND [or a callee]: Caught fatal error [memory may be damaged] Automatically continuing. To reenable the Lisp debugger set *debuggerhook* to nil. (%i4) Barton  Comment By: Robert Dodier (robert_dodier) Date: 20050619 21:28 Message: Logged In: YES user_id=501686 Yet another revision (attached file tmpsum6.lisp) of DOSUM and friends. This version passes run_testsuite() and batch ("rtestsum.mac", test) using the currently attached version of rtestsum.mac. This version avoids substituting for the index in expressions which are free of the index. This version also has an alternate definition of the function FREE. The simplification code (and other code) calls FREE instead of FREEOF. FREEOF looks for dummy variables, but FREE does not, so with the original defn of FREE, sum (lambda([i], i^2), i, 1, n) does not simplify to n lambda ([i], i^2), i, 1, n). I'm not very happy about changing FREE since it is called from all over the place, but I'm also not very happy about the presence of FREE, either; why not call FREEOF ? I don't think changing FREE at this point is such a good idea; it seems to me the two choices are (1) inspect the sum/product simplification code and change FREE to FREEOF where possible. That would be a lot of work. (2) Leave FREE as it is, and just accept that expressions with dummy variables can't be factored out of summations. At this point I think I'd rather leave FREE alone, and file a separate bug report about the inability to factor out expressions with dummy variables, and just leave it for another day.  Comment By: Robert Dodier (robert_dodier) Date: 20050619 21:04 Message: Logged In: YES user_id=501686 New version of rtestsum.mac. Two new test cases, sum (lambda ([i], i^2, 1, 3) and sum (lambda ([i], i^2), i, 1, n).  Comment By: Robert Dodier (robert_dodier) Date: 20050619 20:59 Message: Logged In: YES user_id=501686 Removing rtestsum.mac, will upload a new version momentarily.  Comment By: Robert Dodier (robert_dodier) Date: 20050619 11:52 Message: Logged In: YES user_id=501686 I'm attaching my latest and greatest revision (tmpsum5.lisp) of DOSUM, DO%SUM, and MEVALSUMARGS. I find that load ("tmpsum5.lisp"); batch ("rtestsum.mac", test); where rtestsum.mac is the currently attached version, yields all tests passed for Maxima 5.9.1cvs built with clisp 2.33.2 and with gcl 2.6.6. This revision uses a gensym index for cases handled by MEVALSUMARGS (i.e., cases for which (upper limit) minus (lower limit) is not an integer); the gensym index is the argument for assume/forget. When upper minus lower is an integer, the index is bound to lower, lower+1, lower+2, etc., the summand is evaluated, and then lower, lower+1, lower+2, etc. is substituted for any remaining instances of index.  Comment By: Robert Dodier (robert_dodier) Date: 20050619 11:33 Message: Logged In: YES user_id=501686 Attaching a longer list of test cases. batch ("rtestsum.mac", test); executes the test. As always the "correct" results (2nd line in each pair) are subject to review.  Comment By: Robert Dodier (robert_dodier) Date: 20050619 11:28 Message: Logged In: YES user_id=501686 I'm removing rtestsum.mac, in anticipation of attaching a new version in a minute.  Comment By: van_Nek (van_nek) Date: 20050618 12:42 Message: Logged In: YES user_id=1269745 Dear Maximadevelopers, thank you for your support. I tried the last fix from Robert Dodier and it works. Volker van Nek original (anonymous) poster  Comment By: Robert Dodier (robert_dodier) Date: 20050615 09:37 Message: Logged In: YES user_id=501686 I'm attaching a list of test cases for sum, which can be executed by batch ("rtestsum.mac", test) . Doubtless there will be some discussion as to what the expected output of each test should be. All tests pass after making these 2 changes: ;; ORIGINAL: MEVAL ONCE ;; (mbinding (lind l*i) (meval exp)) ;; TRIED THIS: MEVAL TWICE ;; (mbinding (lind l*i) (meval (meval exp))) ;; THIRD TIME IS A CHARM ?? ($substitute *i ind (mbinding (lind l*i) (meval exp))) ;; ORIGINAL: MEVALATOMS ;; (resimplify (mevalatoms (if (and (not (atom exp)) (resimplify (meval (if (and (not (atom exp))  Comment By: Robert Dodier (robert_dodier) Date: 20050614 09:42 Message: Logged In: YES user_id=501686 OK, here is another attempt. I think I now understand this comment by Stavros in bug report # 740134  "First evaluate foo with i bound to itself, and check if that result is free of i. If so, return the product. If not, *substitute* (don't evaluate) i=lowerlimit, i=lowerlimit+1, etc." ;; ORIGINAL: MEVAL ONCE ;; (mbinding (lind l*i) (meval exp)) ;; TRIED THIS: MEVAL TWICE ;; (mbinding (lind l*i) (meval (meval exp))) ;; THIRD TIME IS A CHARM ?? ($substitute *i ind (mbinding (lind l*i) (meval exp))) This version now yields this result  ex 9: f(x) := sum (x, i, 1, 3); f(x) => 3 x as expected, as well as the same results for ex 1 through ex 8.  Comment By: Robert Dodier (robert_dodier) Date: 20050614 00:00 Message: Logged In: YES user_id=501686 Actually, after looking at the code some more (specifically DOSUM in src/asum.lisp) it looks to me like the intent is indeed to evaluate the summand after binding the summation index to itself. In addition, there is code to assume the index is between its lower and upper limits (although assume/forget is handled incorrectly, see 851765). I believe that the observed behavior is a bug, in the narrow sense that the observed behavior is different from what the code author intended. See also my comments on SF bug # 740134. Here is a different patch. I like this one better than the previous one. In DOSUM: ;; ORIGINAL: MEVAL ONCE ;; (mbinding (lind l*i) (meval exp)) (mbinding (lind l*i) (meval (meval exp))) In MEVALSUMARG: ;; ORIGINAL: MEVALATOMS ;; (resimplify (mevalatoms (if (and (not (atom exp)) (resimplify (meval (if (and (not (atom exp)) With these changes, the results are as expected for the examples cited by the original poster, and also the example about the translated function produces the correct result, and some examples adapted from another bug report (740134) yield correct results. ex 1: f(x):=x^2$ b: 1/10$ Ak: b*f(k*b)$ sum(Ak,k,0,9); => 57/200 ex 2: A(k):=k*f(k*b)$ b:1/n$ A(k) => k^3/n^2 sum(A(k),k,0,n1) => ('sum(k^3,k,0,n1))/n^2 sum(A(k),k,0,n1), simpsum => ((n1)^4 + 2*(n1)^3 + (n1)^2) / (4*n^2) ex 3: ak : k^2$ g(a,n) := sum(a,k,1,n)$ g(ak,5) => 55 translate (g)$ g(ak,5) => 55 some other examples, adapted from bug # 740134: ex 4: sum (print (i), i, 1, 3) => prints 1, 2, 3 then returns 6 ex 5: sum (integrate (x^i ,x), i, 0, 2) => x^3/3 + x^2/2 + x ex 6: sum (integrate (1/(x^i + 1), x), i, 0, 1) => log(x+1) + x/2 ex 7: f[i](x):=x^i$ g[i](x):=x^i$ h[i](x):=x^i$ /* reference f[i] and g[i]  see 740134 for the effect this has on previous defn of sum */ f[i]$ g[i](t)$ sum (f[i](x), i, 0, n) => 'sum (x^i, i, 0, n) sum (g[i](x), i, 0, n) => 'sum (x^i, i, 0, n) sum (h[i](x), i, 0, n) => 'sum (x^i, i, 0, n) ex 8: sum (integrate (x^i, x), i, 0, n) => 'sum (x^(i+1) / (i+1), i, 0, n)  Comment By: Barton Willis (willisbl) Date: 20050613 06:30 Message: Logged In: YES user_id=895922 If we decide to use Robert's $sum function, the translate property for sum will need to be changed. Consider: (1) Redefine $sum as Robert suggested. (2) Try this: (%i1) ak : k^2$ (%i2) g(a,n) := sum(a,k,1,n)$ (%i3) g(ak,5); (%o3) 55 < correct for Robert's $sum function (%i4) translate(g); (%o4) [g] (%i5) g(ak,5); (%o5) 5*k^2 < yeech There might be other things that need fixing: (%i8) properties(sum); (%o8) [Special Evaluation Form,OUTATIVE,NOUN,RULE] Barton  Comment By: Robert Dodier (robert_dodier) Date: 20050612 21:34 Message: Logged In: YES user_id=501686 OK, for the record, here is a definition of $sum which implements one of the ideas that has been proposed, namely this policy: evaluate the summand after binding the summation variable to itself. (defmspec $sum (l) (setq l (cdr l)) (if (= (length l) 4) (progv (list (cadr l)) (list (cadr l)) (dosum (meval (car l)) (cadr l) (meval (caddr l)) (meval (cadddr l)) t)) (wnaerr '$sum))) This version yields the results expected by the person who originated this bug report, and yields the results predicted by S Macrakis in the email referenced below (http://www.math.utexas.edu/pipermail/maxima/2003/004869.html), and run_testsuite() runs to completion with no errors. I'm in favor of making this change, but I'm just recording this code snippet here for future reference; no changes planned at the moment. For comparison here is the current version of $sum: (defmspec $sum (l) (setq l (cdr l)) (if (= (length l) 4) (dosum (car l) (cadr l) (meval (caddr l)) (meval (cadddr l)) t) (wnaerr '$sum)))  Comment By: Barton Willis (willisbl) Date: 20050501 07:09 Message: Logged In: YES user_id=895922 We've discussed this before; see http://www.math.utexas.edu/pipermail/maxima/2003/004869.html http://www.math.utexas.edu/pipermail/maxima/2003/004870.html Maybe a simple workaround such as mysum(f,v,lo,hi):=block([acc:0], if integerp(lo) and integerp(hi) then for i from lo thru hi do acc:acc+substitute(i,v,f) else acc:funmake('mysum,[f,v,lo,hi]), acc) or mysum(f,lo,hi):= block([acc:0], if integerp(lo) and integerp(hi) then for i : lo thru hi do acc : acc + apply(f,[i]) else acc : funmake('mysum,[f,lo,hi]), acc); might work for you, Barton  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1192935&group_id=4933 
From: SourceForge.net <noreply@so...>  20050620 05:27:42

Bugs item #1192935, was opened at 20050430 05:10 Message generated for change (Comment added) made by robert_dodier You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1192935&group_id=4933 Please note that this message will contain a full copy of the comment thread, including the initial issue submission, for this request, not just the latest update. Category: Lisp Core Group: None Status: Open Resolution: None Priority: 5 Submitted By: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: termsubstitution in function sum is too late Initial Comment: (%i1) "Lower RiemannSum, Intervall [0,1], 10 parts"$ (%i2) /*Function*/ f(x):=x^2$ (%i3) /*area width*/ b: 1/10$ (%i4) /*area depending on k=0,...,9*/ Ak: b*f(k*b)$ (%i5) Ak; 2 k (%o5)  1000 (%i6) /* wrong */ sum(Ak,k,0,9); 2 k (%o6)  100 obviously, we first have the summation and then the symbolsubstitution (%i7) sum(''Ak,k,0,9); 57 (%o7)  200 finally, this is ok, but the usage of the doublequote is hard to explain to beginners. it gets even worse, if you use a function A(k):=k*f(k*b)$ for a sum with n partitions: (%i8) "Lower RiemannSum, Intervall [0,1], n parts"$ (%i9) /*area width*/ b:1/n$ (%i10) /*area depending on k=0,...,n1*/ A(k):=k*f(k*b)$ (%i11) A(k); 3 k (%o11)  2 n (%i12) sum(A(k),k,0,n1); n  1 ==== (%o12) > A(k) / ==== k = 0 in that case you need ''(A(k)), a bracket and the doublequote. is it possible to patch the sumfunction in that way, that we first have the term or symbolsubstitution and second the summation?  >Comment By: Robert Dodier (robert_dodier) Date: 20050619 23:27 Message: Logged In: YES user_id=501686 About sum(rat(k),k,1,3), calling $FREEOF instead of FREEOF makes it happy again. Oh well. I think I'll just hang onto the current rev for a day or two to see what other changes accumulate, before posting a new version of tmpsum*.lisp. A related bit  sum(rat(k), k, 1, n) returns 'sum (k, k, 1, n), which isn't exactly wrong, but it does make me wonder what happened to the rat... it turns out this is what happened: subst (foo, k, rat(k)) => foo Not rat(foo) as I would have expected. Dunno what this is about.  Comment By: Barton Willis (willisbl) Date: 20050619 21:36 Message: Logged In: YES user_id=895922 Does tmpsum6.lisp need a ratdisrep somewhere? (%i1) load("c:/maxima/tmpsum6.lisp")$ (%i2) sum(k,k,1,3); (%o2) 6 (%i3) sum(rat(k),k,1,3); Maxima encountered a Lisp error: Error in COND [or a callee]: Caught fatal error [memory may be damaged] Automatically continuing. To reenable the Lisp debugger set *debuggerhook* to nil. (%i4) Barton  Comment By: Robert Dodier (robert_dodier) Date: 20050619 20:28 Message: Logged In: YES user_id=501686 Yet another revision (attached file tmpsum6.lisp) of DOSUM and friends. This version passes run_testsuite() and batch ("rtestsum.mac", test) using the currently attached version of rtestsum.mac. This version avoids substituting for the index in expressions which are free of the index. This version also has an alternate definition of the function FREE. The simplification code (and other code) calls FREE instead of FREEOF. FREEOF looks for dummy variables, but FREE does not, so with the original defn of FREE, sum (lambda([i], i^2), i, 1, n) does not simplify to n lambda ([i], i^2), i, 1, n). I'm not very happy about changing FREE since it is called from all over the place, but I'm also not very happy about the presence of FREE, either; why not call FREEOF ? I don't think changing FREE at this point is such a good idea; it seems to me the two choices are (1) inspect the sum/product simplification code and change FREE to FREEOF where possible. That would be a lot of work. (2) Leave FREE as it is, and just accept that expressions with dummy variables can't be factored out of summations. At this point I think I'd rather leave FREE alone, and file a separate bug report about the inability to factor out expressions with dummy variables, and just leave it for another day.  Comment By: Robert Dodier (robert_dodier) Date: 20050619 20:04 Message: Logged In: YES user_id=501686 New version of rtestsum.mac. Two new test cases, sum (lambda ([i], i^2, 1, 3) and sum (lambda ([i], i^2), i, 1, n).  Comment By: Robert Dodier (robert_dodier) Date: 20050619 19:59 Message: Logged In: YES user_id=501686 Removing rtestsum.mac, will upload a new version momentarily.  Comment By: Robert Dodier (robert_dodier) Date: 20050619 10:52 Message: Logged In: YES user_id=501686 I'm attaching my latest and greatest revision (tmpsum5.lisp) of DOSUM, DO%SUM, and MEVALSUMARGS. I find that load ("tmpsum5.lisp"); batch ("rtestsum.mac", test); where rtestsum.mac is the currently attached version, yields all tests passed for Maxima 5.9.1cvs built with clisp 2.33.2 and with gcl 2.6.6. This revision uses a gensym index for cases handled by MEVALSUMARGS (i.e., cases for which (upper limit) minus (lower limit) is not an integer); the gensym index is the argument for assume/forget. When upper minus lower is an integer, the index is bound to lower, lower+1, lower+2, etc., the summand is evaluated, and then lower, lower+1, lower+2, etc. is substituted for any remaining instances of index.  Comment By: Robert Dodier (robert_dodier) Date: 20050619 10:33 Message: Logged In: YES user_id=501686 Attaching a longer list of test cases. batch ("rtestsum.mac", test); executes the test. As always the "correct" results (2nd line in each pair) are subject to review.  Comment By: Robert Dodier (robert_dodier) Date: 20050619 10:28 Message: Logged In: YES user_id=501686 I'm removing rtestsum.mac, in anticipation of attaching a new version in a minute.  Comment By: van_Nek (van_nek) Date: 20050618 11:42 Message: Logged In: YES user_id=1269745 Dear Maximadevelopers, thank you for your support. I tried the last fix from Robert Dodier and it works. Volker van Nek original (anonymous) poster  Comment By: Robert Dodier (robert_dodier) Date: 20050615 08:37 Message: Logged In: YES user_id=501686 I'm attaching a list of test cases for sum, which can be executed by batch ("rtestsum.mac", test) . Doubtless there will be some discussion as to what the expected output of each test should be. All tests pass after making these 2 changes: ;; ORIGINAL: MEVAL ONCE ;; (mbinding (lind l*i) (meval exp)) ;; TRIED THIS: MEVAL TWICE ;; (mbinding (lind l*i) (meval (meval exp))) ;; THIRD TIME IS A CHARM ?? ($substitute *i ind (mbinding (lind l*i) (meval exp))) ;; ORIGINAL: MEVALATOMS ;; (resimplify (mevalatoms (if (and (not (atom exp)) (resimplify (meval (if (and (not (atom exp))  Comment By: Robert Dodier (robert_dodier) Date: 20050614 08:42 Message: Logged In: YES user_id=501686 OK, here is another attempt. I think I now understand this comment by Stavros in bug report # 740134  "First evaluate foo with i bound to itself, and check if that result is free of i. If so, return the product. If not, *substitute* (don't evaluate) i=lowerlimit, i=lowerlimit+1, etc." ;; ORIGINAL: MEVAL ONCE ;; (mbinding (lind l*i) (meval exp)) ;; TRIED THIS: MEVAL TWICE ;; (mbinding (lind l*i) (meval (meval exp))) ;; THIRD TIME IS A CHARM ?? ($substitute *i ind (mbinding (lind l*i) (meval exp))) This version now yields this result  ex 9: f(x) := sum (x, i, 1, 3); f(x) => 3 x as expected, as well as the same results for ex 1 through ex 8.  Comment By: Robert Dodier (robert_dodier) Date: 20050613 23:00 Message: Logged In: YES user_id=501686 Actually, after looking at the code some more (specifically DOSUM in src/asum.lisp) it looks to me like the intent is indeed to evaluate the summand after binding the summation index to itself. In addition, there is code to assume the index is between its lower and upper limits (although assume/forget is handled incorrectly, see 851765). I believe that the observed behavior is a bug, in the narrow sense that the observed behavior is different from what the code author intended. See also my comments on SF bug # 740134. Here is a different patch. I like this one better than the previous one. In DOSUM: ;; ORIGINAL: MEVAL ONCE ;; (mbinding (lind l*i) (meval exp)) (mbinding (lind l*i) (meval (meval exp))) In MEVALSUMARG: ;; ORIGINAL: MEVALATOMS ;; (resimplify (mevalatoms (if (and (not (atom exp)) (resimplify (meval (if (and (not (atom exp)) With these changes, the results are as expected for the examples cited by the original poster, and also the example about the translated function produces the correct result, and some examples adapted from another bug report (740134) yield correct results. ex 1: f(x):=x^2$ b: 1/10$ Ak: b*f(k*b)$ sum(Ak,k,0,9); => 57/200 ex 2: A(k):=k*f(k*b)$ b:1/n$ A(k) => k^3/n^2 sum(A(k),k,0,n1) => ('sum(k^3,k,0,n1))/n^2 sum(A(k),k,0,n1), simpsum => ((n1)^4 + 2*(n1)^3 + (n1)^2) / (4*n^2) ex 3: ak : k^2$ g(a,n) := sum(a,k,1,n)$ g(ak,5) => 55 translate (g)$ g(ak,5) => 55 some other examples, adapted from bug # 740134: ex 4: sum (print (i), i, 1, 3) => prints 1, 2, 3 then returns 6 ex 5: sum (integrate (x^i ,x), i, 0, 2) => x^3/3 + x^2/2 + x ex 6: sum (integrate (1/(x^i + 1), x), i, 0, 1) => log(x+1) + x/2 ex 7: f[i](x):=x^i$ g[i](x):=x^i$ h[i](x):=x^i$ /* reference f[i] and g[i]  see 740134 for the effect this has on previous defn of sum */ f[i]$ g[i](t)$ sum (f[i](x), i, 0, n) => 'sum (x^i, i, 0, n) sum (g[i](x), i, 0, n) => 'sum (x^i, i, 0, n) sum (h[i](x), i, 0, n) => 'sum (x^i, i, 0, n) ex 8: sum (integrate (x^i, x), i, 0, n) => 'sum (x^(i+1) / (i+1), i, 0, n)  Comment By: Barton Willis (willisbl) Date: 20050613 05:30 Message: Logged In: YES user_id=895922 If we decide to use Robert's $sum function, the translate property for sum will need to be changed. Consider: (1) Redefine $sum as Robert suggested. (2) Try this: (%i1) ak : k^2$ (%i2) g(a,n) := sum(a,k,1,n)$ (%i3) g(ak,5); (%o3) 55 < correct for Robert's $sum function (%i4) translate(g); (%o4) [g] (%i5) g(ak,5); (%o5) 5*k^2 < yeech There might be other things that need fixing: (%i8) properties(sum); (%o8) [Special Evaluation Form,OUTATIVE,NOUN,RULE] Barton  Comment By: Robert Dodier (robert_dodier) Date: 20050612 20:34 Message: Logged In: YES user_id=501686 OK, for the record, here is a definition of $sum which implements one of the ideas that has been proposed, namely this policy: evaluate the summand after binding the summation variable to itself. (defmspec $sum (l) (setq l (cdr l)) (if (= (length l) 4) (progv (list (cadr l)) (list (cadr l)) (dosum (meval (car l)) (cadr l) (meval (caddr l)) (meval (cadddr l)) t)) (wnaerr '$sum))) This version yields the results expected by the person who originated this bug report, and yields the results predicted by S Macrakis in the email referenced below (http://www.math.utexas.edu/pipermail/maxima/2003/004869.html), and run_testsuite() runs to completion with no errors. I'm in favor of making this change, but I'm just recording this code snippet here for future reference; no changes planned at the moment. For comparison here is the current version of $sum: (defmspec $sum (l) (setq l (cdr l)) (if (= (length l) 4) (dosum (car l) (cadr l) (meval (caddr l)) (meval (cadddr l)) t) (wnaerr '$sum)))  Comment By: Barton Willis (willisbl) Date: 20050501 06:09 Message: Logged In: YES user_id=895922 We've discussed this before; see http://www.math.utexas.edu/pipermail/maxima/2003/004869.html http://www.math.utexas.edu/pipermail/maxima/2003/004870.html Maybe a simple workaround such as mysum(f,v,lo,hi):=block([acc:0], if integerp(lo) and integerp(hi) then for i from lo thru hi do acc:acc+substitute(i,v,f) else acc:funmake('mysum,[f,v,lo,hi]), acc) or mysum(f,lo,hi):= block([acc:0], if integerp(lo) and integerp(hi) then for i : lo thru hi do acc : acc + apply(f,[i]) else acc : funmake('mysum,[f,lo,hi]), acc); might work for you, Barton  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1192935&group_id=4933 
From: SourceForge.net <noreply@so...>  20050621 06:03:20

Bugs item #1192935, was opened at 20050430 05:10 Message generated for change (Comment added) made by robert_dodier You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1192935&group_id=4933 Please note that this message will contain a full copy of the comment thread, including the initial issue submission, for this request, not just the latest update. Category: Lisp Core Group: None Status: Open Resolution: None Priority: 5 Submitted By: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: termsubstitution in function sum is too late Initial Comment: (%i1) "Lower RiemannSum, Intervall [0,1], 10 parts"$ (%i2) /*Function*/ f(x):=x^2$ (%i3) /*area width*/ b: 1/10$ (%i4) /*area depending on k=0,...,9*/ Ak: b*f(k*b)$ (%i5) Ak; 2 k (%o5)  1000 (%i6) /* wrong */ sum(Ak,k,0,9); 2 k (%o6)  100 obviously, we first have the summation and then the symbolsubstitution (%i7) sum(''Ak,k,0,9); 57 (%o7)  200 finally, this is ok, but the usage of the doublequote is hard to explain to beginners. it gets even worse, if you use a function A(k):=k*f(k*b)$ for a sum with n partitions: (%i8) "Lower RiemannSum, Intervall [0,1], n parts"$ (%i9) /*area width*/ b:1/n$ (%i10) /*area depending on k=0,...,n1*/ A(k):=k*f(k*b)$ (%i11) A(k); 3 k (%o11)  2 n (%i12) sum(A(k),k,0,n1); n  1 ==== (%o12) > A(k) / ==== k = 0 in that case you need ''(A(k)), a bracket and the doublequote. is it possible to patch the sumfunction in that way, that we first have the term or symbolsubstitution and second the summation?  >Comment By: Robert Dodier (robert_dodier) Date: 20050621 00:03 Message: Logged In: YES user_id=501686 Delete rtestsum.mac, going to upload a new one momentarily.  Comment By: Robert Dodier (robert_dodier) Date: 20050619 23:27 Message: Logged In: YES user_id=501686 About sum(rat(k),k,1,3), calling $FREEOF instead of FREEOF makes it happy again. Oh well. I think I'll just hang onto the current rev for a day or two to see what other changes accumulate, before posting a new version of tmpsum*.lisp. A related bit  sum(rat(k), k, 1, n) returns 'sum (k, k, 1, n), which isn't exactly wrong, but it does make me wonder what happened to the rat... it turns out this is what happened: subst (foo, k, rat(k)) => foo Not rat(foo) as I would have expected. Dunno what this is about.  Comment By: Barton Willis (willisbl) Date: 20050619 21:36 Message: Logged In: YES user_id=895922 Does tmpsum6.lisp need a ratdisrep somewhere? (%i1) load("c:/maxima/tmpsum6.lisp")$ (%i2) sum(k,k,1,3); (%o2) 6 (%i3) sum(rat(k),k,1,3); Maxima encountered a Lisp error: Error in COND [or a callee]: Caught fatal error [memory may be damaged] Automatically continuing. To reenable the Lisp debugger set *debuggerhook* to nil. (%i4) Barton  Comment By: Robert Dodier (robert_dodier) Date: 20050619 20:28 Message: Logged In: YES user_id=501686 Yet another revision (attached file tmpsum6.lisp) of DOSUM and friends. This version passes run_testsuite() and batch ("rtestsum.mac", test) using the currently attached version of rtestsum.mac. This version avoids substituting for the index in expressions which are free of the index. This version also has an alternate definition of the function FREE. The simplification code (and other code) calls FREE instead of FREEOF. FREEOF looks for dummy variables, but FREE does not, so with the original defn of FREE, sum (lambda([i], i^2), i, 1, n) does not simplify to n lambda ([i], i^2), i, 1, n). I'm not very happy about changing FREE since it is called from all over the place, but I'm also not very happy about the presence of FREE, either; why not call FREEOF ? I don't think changing FREE at this point is such a good idea; it seems to me the two choices are (1) inspect the sum/product simplification code and change FREE to FREEOF where possible. That would be a lot of work. (2) Leave FREE as it is, and just accept that expressions with dummy variables can't be factored out of summations. At this point I think I'd rather leave FREE alone, and file a separate bug report about the inability to factor out expressions with dummy variables, and just leave it for another day.  Comment By: Robert Dodier (robert_dodier) Date: 20050619 20:04 Message: Logged In: YES user_id=501686 New version of rtestsum.mac. Two new test cases, sum (lambda ([i], i^2, 1, 3) and sum (lambda ([i], i^2), i, 1, n).  Comment By: Robert Dodier (robert_dodier) Date: 20050619 19:59 Message: Logged In: YES user_id=501686 Removing rtestsum.mac, will upload a new version momentarily.  Comment By: Robert Dodier (robert_dodier) Date: 20050619 10:52 Message: Logged In: YES user_id=501686 I'm attaching my latest and greatest revision (tmpsum5.lisp) of DOSUM, DO%SUM, and MEVALSUMARGS. I find that load ("tmpsum5.lisp"); batch ("rtestsum.mac", test); where rtestsum.mac is the currently attached version, yields all tests passed for Maxima 5.9.1cvs built with clisp 2.33.2 and with gcl 2.6.6. This revision uses a gensym index for cases handled by MEVALSUMARGS (i.e., cases for which (upper limit) minus (lower limit) is not an integer); the gensym index is the argument for assume/forget. When upper minus lower is an integer, the index is bound to lower, lower+1, lower+2, etc., the summand is evaluated, and then lower, lower+1, lower+2, etc. is substituted for any remaining instances of index.  Comment By: Robert Dodier (robert_dodier) Date: 20050619 10:33 Message: Logged In: YES user_id=501686 Attaching a longer list of test cases. batch ("rtestsum.mac", test); executes the test. As always the "correct" results (2nd line in each pair) are subject to review.  Comment By: Robert Dodier (robert_dodier) Date: 20050619 10:28 Message: Logged In: YES user_id=501686 I'm removing rtestsum.mac, in anticipation of attaching a new version in a minute.  Comment By: van_Nek (van_nek) Date: 20050618 11:42 Message: Logged In: YES user_id=1269745 Dear Maximadevelopers, thank you for your support. I tried the last fix from Robert Dodier and it works. Volker van Nek original (anonymous) poster  Comment By: Robert Dodier (robert_dodier) Date: 20050615 08:37 Message: Logged In: YES user_id=501686 I'm attaching a list of test cases for sum, which can be executed by batch ("rtestsum.mac", test) . Doubtless there will be some discussion as to what the expected output of each test should be. All tests pass after making these 2 changes: ;; ORIGINAL: MEVAL ONCE ;; (mbinding (lind l*i) (meval exp)) ;; TRIED THIS: MEVAL TWICE ;; (mbinding (lind l*i) (meval (meval exp))) ;; THIRD TIME IS A CHARM ?? ($substitute *i ind (mbinding (lind l*i) (meval exp))) ;; ORIGINAL: MEVALATOMS ;; (resimplify (mevalatoms (if (and (not (atom exp)) (resimplify (meval (if (and (not (atom exp))  Comment By: Robert Dodier (robert_dodier) Date: 20050614 08:42 Message: Logged In: YES user_id=501686 OK, here is another attempt. I think I now understand this comment by Stavros in bug report # 740134  "First evaluate foo with i bound to itself, and check if that result is free of i. If so, return the product. If not, *substitute* (don't evaluate) i=lowerlimit, i=lowerlimit+1, etc." ;; ORIGINAL: MEVAL ONCE ;; (mbinding (lind l*i) (meval exp)) ;; TRIED THIS: MEVAL TWICE ;; (mbinding (lind l*i) (meval (meval exp))) ;; THIRD TIME IS A CHARM ?? ($substitute *i ind (mbinding (lind l*i) (meval exp))) This version now yields this result  ex 9: f(x) := sum (x, i, 1, 3); f(x) => 3 x as expected, as well as the same results for ex 1 through ex 8.  Comment By: Robert Dodier (robert_dodier) Date: 20050613 23:00 Message: Logged In: YES user_id=501686 Actually, after looking at the code some more (specifically DOSUM in src/asum.lisp) it looks to me like the intent is indeed to evaluate the summand after binding the summation index to itself. In addition, there is code to assume the index is between its lower and upper limits (although assume/forget is handled incorrectly, see 851765). I believe that the observed behavior is a bug, in the narrow sense that the observed behavior is different from what the code author intended. See also my comments on SF bug # 740134. Here is a different patch. I like this one better than the previous one. In DOSUM: ;; ORIGINAL: MEVAL ONCE ;; (mbinding (lind l*i) (meval exp)) (mbinding (lind l*i) (meval (meval exp))) In MEVALSUMARG: ;; ORIGINAL: MEVALATOMS ;; (resimplify (mevalatoms (if (and (not (atom exp)) (resimplify (meval (if (and (not (atom exp)) With these changes, the results are as expected for the examples cited by the original poster, and also the example about the translated function produces the correct result, and some examples adapted from another bug report (740134) yield correct results. ex 1: f(x):=x^2$ b: 1/10$ Ak: b*f(k*b)$ sum(Ak,k,0,9); => 57/200 ex 2: A(k):=k*f(k*b)$ b:1/n$ A(k) => k^3/n^2 sum(A(k),k,0,n1) => ('sum(k^3,k,0,n1))/n^2 sum(A(k),k,0,n1), simpsum => ((n1)^4 + 2*(n1)^3 + (n1)^2) / (4*n^2) ex 3: ak : k^2$ g(a,n) := sum(a,k,1,n)$ g(ak,5) => 55 translate (g)$ g(ak,5) => 55 some other examples, adapted from bug # 740134: ex 4: sum (print (i), i, 1, 3) => prints 1, 2, 3 then returns 6 ex 5: sum (integrate (x^i ,x), i, 0, 2) => x^3/3 + x^2/2 + x ex 6: sum (integrate (1/(x^i + 1), x), i, 0, 1) => log(x+1) + x/2 ex 7: f[i](x):=x^i$ g[i](x):=x^i$ h[i](x):=x^i$ /* reference f[i] and g[i]  see 740134 for the effect this has on previous defn of sum */ f[i]$ g[i](t)$ sum (f[i](x), i, 0, n) => 'sum (x^i, i, 0, n) sum (g[i](x), i, 0, n) => 'sum (x^i, i, 0, n) sum (h[i](x), i, 0, n) => 'sum (x^i, i, 0, n) ex 8: sum (integrate (x^i, x), i, 0, n) => 'sum (x^(i+1) / (i+1), i, 0, n)  Comment By: Barton Willis (willisbl) Date: 20050613 05:30 Message: Logged In: YES user_id=895922 If we decide to use Robert's $sum function, the translate property for sum will need to be changed. Consider: (1) Redefine $sum as Robert suggested. (2) Try this: (%i1) ak : k^2$ (%i2) g(a,n) := sum(a,k,1,n)$ (%i3) g(ak,5); (%o3) 55 < correct for Robert's $sum function (%i4) translate(g); (%o4) [g] (%i5) g(ak,5); (%o5) 5*k^2 < yeech There might be other things that need fixing: (%i8) properties(sum); (%o8) [Special Evaluation Form,OUTATIVE,NOUN,RULE] Barton  Comment By: Robert Dodier (robert_dodier) Date: 20050612 20:34 Message: Logged In: YES user_id=501686 OK, for the record, here is a definition of $sum which implements one of the ideas that has been proposed, namely this policy: evaluate the summand after binding the summation variable to itself. (defmspec $sum (l) (setq l (cdr l)) (if (= (length l) 4) (progv (list (cadr l)) (list (cadr l)) (dosum (meval (car l)) (cadr l) (meval (caddr l)) (meval (cadddr l)) t)) (wnaerr '$sum))) This version yields the results expected by the person who originated this bug report, and yields the results predicted by S Macrakis in the email referenced below (http://www.math.utexas.edu/pipermail/maxima/2003/004869.html), and run_testsuite() runs to completion with no errors. I'm in favor of making this change, but I'm just recording this code snippet here for future reference; no changes planned at the moment. For comparison here is the current version of $sum: (defmspec $sum (l) (setq l (cdr l)) (if (= (length l) 4) (dosum (car l) (cadr l) (meval (caddr l)) (meval (cadddr l)) t) (wnaerr '$sum)))  Comment By: Barton Willis (willisbl) Date: 20050501 06:09 Message: Logged In: YES user_id=895922 We've discussed this before; see http://www.math.utexas.edu/pipermail/maxima/2003/004869.html http://www.math.utexas.edu/pipermail/maxima/2003/004870.html Maybe a simple workaround such as mysum(f,v,lo,hi):=block([acc:0], if integerp(lo) and integerp(hi) then for i from lo thru hi do acc:acc+substitute(i,v,f) else acc:funmake('mysum,[f,v,lo,hi]), acc) or mysum(f,lo,hi):= block([acc:0], if integerp(lo) and integerp(hi) then for i : lo thru hi do acc : acc + apply(f,[i]) else acc : funmake('mysum,[f,lo,hi]), acc); might work for you, Barton  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1192935&group_id=4933 
From: SourceForge.net <noreply@so...>  20050621 06:06:50

Bugs item #1192935, was opened at 20050430 05:10 Message generated for change (Comment added) made by robert_dodier You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1192935&group_id=4933 Please note that this message will contain a full copy of the comment thread, including the initial issue submission, for this request, not just the latest update. Category: Lisp Core Group: None Status: Open Resolution: None Priority: 5 Submitted By: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: termsubstitution in function sum is too late Initial Comment: (%i1) "Lower RiemannSum, Intervall [0,1], 10 parts"$ (%i2) /*Function*/ f(x):=x^2$ (%i3) /*area width*/ b: 1/10$ (%i4) /*area depending on k=0,...,9*/ Ak: b*f(k*b)$ (%i5) Ak; 2 k (%o5)  1000 (%i6) /* wrong */ sum(Ak,k,0,9); 2 k (%o6)  100 obviously, we first have the summation and then the symbolsubstitution (%i7) sum(''Ak,k,0,9); 57 (%o7)  200 finally, this is ok, but the usage of the doublequote is hard to explain to beginners. it gets even worse, if you use a function A(k):=k*f(k*b)$ for a sum with n partitions: (%i8) "Lower RiemannSum, Intervall [0,1], n parts"$ (%i9) /*area width*/ b:1/n$ (%i10) /*area depending on k=0,...,n1*/ A(k):=k*f(k*b)$ (%i11) A(k); 3 k (%o11)  2 n (%i12) sum(A(k),k,0,n1); n  1 ==== (%o12) > A(k) / ==== k = 0 in that case you need ''(A(k)), a bracket and the doublequote. is it possible to patch the sumfunction in that way, that we first have the term or symbolsubstitution and second the summation?  >Comment By: Robert Dodier (robert_dodier) Date: 20050621 00:06 Message: Logged In: YES user_id=501686 New rtestsum.mac. A couple of new tests for sum, and also copied all sum tests and substituted product for sum (and adjusted expected outputs to suit).  Comment By: Robert Dodier (robert_dodier) Date: 20050621 00:03 Message: Logged In: YES user_id=501686 Delete rtestsum.mac, going to upload a new one momentarily.  Comment By: Robert Dodier (robert_dodier) Date: 20050619 23:27 Message: Logged In: YES user_id=501686 About sum(rat(k),k,1,3), calling $FREEOF instead of FREEOF makes it happy again. Oh well. I think I'll just hang onto the current rev for a day or two to see what other changes accumulate, before posting a new version of tmpsum*.lisp. A related bit  sum(rat(k), k, 1, n) returns 'sum (k, k, 1, n), which isn't exactly wrong, but it does make me wonder what happened to the rat... it turns out this is what happened: subst (foo, k, rat(k)) => foo Not rat(foo) as I would have expected. Dunno what this is about.  Comment By: Barton Willis (willisbl) Date: 20050619 21:36 Message: Logged In: YES user_id=895922 Does tmpsum6.lisp need a ratdisrep somewhere? (%i1) load("c:/maxima/tmpsum6.lisp")$ (%i2) sum(k,k,1,3); (%o2) 6 (%i3) sum(rat(k),k,1,3); Maxima encountered a Lisp error: Error in COND [or a callee]: Caught fatal error [memory may be damaged] Automatically continuing. To reenable the Lisp debugger set *debuggerhook* to nil. (%i4) Barton  Comment By: Robert Dodier (robert_dodier) Date: 20050619 20:28 Message: Logged In: YES user_id=501686 Yet another revision (attached file tmpsum6.lisp) of DOSUM and friends. This version passes run_testsuite() and batch ("rtestsum.mac", test) using the currently attached version of rtestsum.mac. This version avoids substituting for the index in expressions which are free of the index. This version also has an alternate definition of the function FREE. The simplification code (and other code) calls FREE instead of FREEOF. FREEOF looks for dummy variables, but FREE does not, so with the original defn of FREE, sum (lambda([i], i^2), i, 1, n) does not simplify to n lambda ([i], i^2), i, 1, n). I'm not very happy about changing FREE since it is called from all over the place, but I'm also not very happy about the presence of FREE, either; why not call FREEOF ? I don't think changing FREE at this point is such a good idea; it seems to me the two choices are (1) inspect the sum/product simplification code and change FREE to FREEOF where possible. That would be a lot of work. (2) Leave FREE as it is, and just accept that expressions with dummy variables can't be factored out of summations. At this point I think I'd rather leave FREE alone, and file a separate bug report about the inability to factor out expressions with dummy variables, and just leave it for another day.  Comment By: Robert Dodier (robert_dodier) Date: 20050619 20:04 Message: Logged In: YES user_id=501686 New version of rtestsum.mac. Two new test cases, sum (lambda ([i], i^2, 1, 3) and sum (lambda ([i], i^2), i, 1, n).  Comment By: Robert Dodier (robert_dodier) Date: 20050619 19:59 Message: Logged In: YES user_id=501686 Removing rtestsum.mac, will upload a new version momentarily.  Comment By: Robert Dodier (robert_dodier) Date: 20050619 10:52 Message: Logged In: YES user_id=501686 I'm attaching my latest and greatest revision (tmpsum5.lisp) of DOSUM, DO%SUM, and MEVALSUMARGS. I find that load ("tmpsum5.lisp"); batch ("rtestsum.mac", test); where rtestsum.mac is the currently attached version, yields all tests passed for Maxima 5.9.1cvs built with clisp 2.33.2 and with gcl 2.6.6. This revision uses a gensym index for cases handled by MEVALSUMARGS (i.e., cases for which (upper limit) minus (lower limit) is not an integer); the gensym index is the argument for assume/forget. When upper minus lower is an integer, the index is bound to lower, lower+1, lower+2, etc., the summand is evaluated, and then lower, lower+1, lower+2, etc. is substituted for any remaining instances of index.  Comment By: Robert Dodier (robert_dodier) Date: 20050619 10:33 Message: Logged In: YES user_id=501686 Attaching a longer list of test cases. batch ("rtestsum.mac", test); executes the test. As always the "correct" results (2nd line in each pair) are subject to review.  Comment By: Robert Dodier (robert_dodier) Date: 20050619 10:28 Message: Logged In: YES user_id=501686 I'm removing rtestsum.mac, in anticipation of attaching a new version in a minute.  Comment By: van_Nek (van_nek) Date: 20050618 11:42 Message: Logged In: YES user_id=1269745 Dear Maximadevelopers, thank you for your support. I tried the last fix from Robert Dodier and it works. Volker van Nek original (anonymous) poster  Comment By: Robert Dodier (robert_dodier) Date: 20050615 08:37 Message: Logged In: YES user_id=501686 I'm attaching a list of test cases for sum, which can be executed by batch ("rtestsum.mac", test) . Doubtless there will be some discussion as to what the expected output of each test should be. All tests pass after making these 2 changes: ;; ORIGINAL: MEVAL ONCE ;; (mbinding (lind l*i) (meval exp)) ;; TRIED THIS: MEVAL TWICE ;; (mbinding (lind l*i) (meval (meval exp))) ;; THIRD TIME IS A CHARM ?? ($substitute *i ind (mbinding (lind l*i) (meval exp))) ;; ORIGINAL: MEVALATOMS ;; (resimplify (mevalatoms (if (and (not (atom exp)) (resimplify (meval (if (and (not (atom exp))  Comment By: Robert Dodier (robert_dodier) Date: 20050614 08:42 Message: Logged In: YES user_id=501686 OK, here is another attempt. I think I now understand this comment by Stavros in bug report # 740134  "First evaluate foo with i bound to itself, and check if that result is free of i. If so, return the product. If not, *substitute* (don't evaluate) i=lowerlimit, i=lowerlimit+1, etc." ;; ORIGINAL: MEVAL ONCE ;; (mbinding (lind l*i) (meval exp)) ;; TRIED THIS: MEVAL TWICE ;; (mbinding (lind l*i) (meval (meval exp))) ;; THIRD TIME IS A CHARM ?? ($substitute *i ind (mbinding (lind l*i) (meval exp))) This version now yields this result  ex 9: f(x) := sum (x, i, 1, 3); f(x) => 3 x as expected, as well as the same results for ex 1 through ex 8.  Comment By: Robert Dodier (robert_dodier) Date: 20050613 23:00 Message: Logged In: YES user_id=501686 Actually, after looking at the code some more (specifically DOSUM in src/asum.lisp) it looks to me like the intent is indeed to evaluate the summand after binding the summation index to itself. In addition, there is code to assume the index is between its lower and upper limits (although assume/forget is handled incorrectly, see 851765). I believe that the observed behavior is a bug, in the narrow sense that the observed behavior is different from what the code author intended. See also my comments on SF bug # 740134. Here is a different patch. I like this one better than the previous one. In DOSUM: ;; ORIGINAL: MEVAL ONCE ;; (mbinding (lind l*i) (meval exp)) (mbinding (lind l*i) (meval (meval exp))) In MEVALSUMARG: ;; ORIGINAL: MEVALATOMS ;; (resimplify (mevalatoms (if (and (not (atom exp)) (resimplify (meval (if (and (not (atom exp)) With these changes, the results are as expected for the examples cited by the original poster, and also the example about the translated function produces the correct result, and some examples adapted from another bug report (740134) yield correct results. ex 1: f(x):=x^2$ b: 1/10$ Ak: b*f(k*b)$ sum(Ak,k,0,9); => 57/200 ex 2: A(k):=k*f(k*b)$ b:1/n$ A(k) => k^3/n^2 sum(A(k),k,0,n1) => ('sum(k^3,k,0,n1))/n^2 sum(A(k),k,0,n1), simpsum => ((n1)^4 + 2*(n1)^3 + (n1)^2) / (4*n^2) ex 3: ak : k^2$ g(a,n) := sum(a,k,1,n)$ g(ak,5) => 55 translate (g)$ g(ak,5) => 55 some other examples, adapted from bug # 740134: ex 4: sum (print (i), i, 1, 3) => prints 1, 2, 3 then returns 6 ex 5: sum (integrate (x^i ,x), i, 0, 2) => x^3/3 + x^2/2 + x ex 6: sum (integrate (1/(x^i + 1), x), i, 0, 1) => log(x+1) + x/2 ex 7: f[i](x):=x^i$ g[i](x):=x^i$ h[i](x):=x^i$ /* reference f[i] and g[i]  see 740134 for the effect this has on previous defn of sum */ f[i]$ g[i](t)$ sum (f[i](x), i, 0, n) => 'sum (x^i, i, 0, n) sum (g[i](x), i, 0, n) => 'sum (x^i, i, 0, n) sum (h[i](x), i, 0, n) => 'sum (x^i, i, 0, n) ex 8: sum (integrate (x^i, x), i, 0, n) => 'sum (x^(i+1) / (i+1), i, 0, n)  Comment By: Barton Willis (willisbl) Date: 20050613 05:30 Message: Logged In: YES user_id=895922 If we decide to use Robert's $sum function, the translate property for sum will need to be changed. Consider: (1) Redefine $sum as Robert suggested. (2) Try this: (%i1) ak : k^2$ (%i2) g(a,n) := sum(a,k,1,n)$ (%i3) g(ak,5); (%o3) 55 < correct for Robert's $sum function (%i4) translate(g); (%o4) [g] (%i5) g(ak,5); (%o5) 5*k^2 < yeech There might be other things that need fixing: (%i8) properties(sum); (%o8) [Special Evaluation Form,OUTATIVE,NOUN,RULE] Barton  Comment By: Robert Dodier (robert_dodier) Date: 20050612 20:34 Message: Logged In: YES user_id=501686 OK, for the record, here is a definition of $sum which implements one of the ideas that has been proposed, namely this policy: evaluate the summand after binding the summation variable to itself. (defmspec $sum (l) (setq l (cdr l)) (if (= (length l) 4) (progv (list (cadr l)) (list (cadr l)) (dosum (meval (car l)) (cadr l) (meval (caddr l)) (meval (cadddr l)) t)) (wnaerr '$sum))) This version yields the results expected by the person who originated this bug report, and yields the results predicted by S Macrakis in the email referenced below (http://www.math.utexas.edu/pipermail/maxima/2003/004869.html), and run_testsuite() runs to completion with no errors. I'm in favor of making this change, but I'm just recording this code snippet here for future reference; no changes planned at the moment. For comparison here is the current version of $sum: (defmspec $sum (l) (setq l (cdr l)) (if (= (length l) 4) (dosum (car l) (cadr l) (meval (caddr l)) (meval (cadddr l)) t) (wnaerr '$sum)))  Comment By: Barton Willis (willisbl) Date: 20050501 06:09 Message: Logged In: YES user_id=895922 We've discussed this before; see http://www.math.utexas.edu/pipermail/maxima/2003/004869.html http://www.math.utexas.edu/pipermail/maxima/2003/004870.html Maybe a simple workaround such as mysum(f,v,lo,hi):=block([acc:0], if integerp(lo) and integerp(hi) then for i from lo thru hi do acc:acc+substitute(i,v,f) else acc:funmake('mysum,[f,v,lo,hi]), acc) or mysum(f,lo,hi):= block([acc:0], if integerp(lo) and integerp(hi) then for i : lo thru hi do acc : acc + apply(f,[i]) else acc : funmake('mysum,[f,lo,hi]), acc); might work for you, Barton  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1192935&group_id=4933 
From: SourceForge.net <noreply@so...>  20050621 06:18:50

Bugs item #1192935, was opened at 20050430 05:10 Message generated for change (Comment added) made by robert_dodier You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1192935&group_id=4933 Please note that this message will contain a full copy of the comment thread, including the initial issue submission, for this request, not just the latest update. Category: Lisp Core Group: None Status: Open Resolution: None Priority: 5 Submitted By: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: termsubstitution in function sum is too late Initial Comment: (%i1) "Lower RiemannSum, Intervall [0,1], 10 parts"$ (%i2) /*Function*/ f(x):=x^2$ (%i3) /*area width*/ b: 1/10$ (%i4) /*area depending on k=0,...,9*/ Ak: b*f(k*b)$ (%i5) Ak; 2 k (%o5)  1000 (%i6) /* wrong */ sum(Ak,k,0,9); 2 k (%o6)  100 obviously, we first have the summation and then the symbolsubstitution (%i7) sum(''Ak,k,0,9); 57 (%o7)  200 finally, this is ok, but the usage of the doublequote is hard to explain to beginners. it gets even worse, if you use a function A(k):=k*f(k*b)$ for a sum with n partitions: (%i8) "Lower RiemannSum, Intervall [0,1], n parts"$ (%i9) /*area width*/ b:1/n$ (%i10) /*area depending on k=0,...,n1*/ A(k):=k*f(k*b)$ (%i11) A(k); 3 k (%o11)  2 n (%i12) sum(A(k),k,0,n1); n  1 ==== (%o12) > A(k) / ==== k = 0 in that case you need ''(A(k)), a bracket and the doublequote. is it possible to patch the sumfunction in that way, that we first have the term or symbolsubstitution and second the summation?  >Comment By: Robert Dodier (robert_dodier) Date: 20050621 00:18 Message: Logged In: YES user_id=501686 New DOSUM, etc, in tmpsum8.lisp. A few new sum tests have been added to rtestsum.mac, and all sum tests copied and turned into product tests. This revision passes all tests in current rtestsum.mac except for test 38: (assume(i < 0, j < 0, n > 0), product(product(abs(j) + abs(i), i, 1, j), j, 1, n)); => Lower bound to product is > upper bound. This emanates from SIMPROD1 (src/combin.lisp) which is called after the assumptions about i and j (actually, assumptions about the gensym standins for i and j) have been forgotten. This looks like a bug in combin.lisp  there is no assume/forget about the indices.  Comment By: Robert Dodier (robert_dodier) Date: 20050621 00:06 Message: Logged In: YES user_id=501686 New rtestsum.mac. A couple of new tests for sum, and also copied all sum tests and substituted product for sum (and adjusted expected outputs to suit).  Comment By: Robert Dodier (robert_dodier) Date: 20050621 00:03 Message: Logged In: YES user_id=501686 Delete rtestsum.mac, going to upload a new one momentarily.  Comment By: Robert Dodier (robert_dodier) Date: 20050619 23:27 Message: Logged In: YES user_id=501686 About sum(rat(k),k,1,3), calling $FREEOF instead of FREEOF makes it happy again. Oh well. I think I'll just hang onto the current rev for a day or two to see what other changes accumulate, before posting a new version of tmpsum*.lisp. A related bit  sum(rat(k), k, 1, n) returns 'sum (k, k, 1, n), which isn't exactly wrong, but it does make me wonder what happened to the rat... it turns out this is what happened: subst (foo, k, rat(k)) => foo Not rat(foo) as I would have expected. Dunno what this is about.  Comment By: Barton Willis (willisbl) Date: 20050619 21:36 Message: Logged In: YES user_id=895922 Does tmpsum6.lisp need a ratdisrep somewhere? (%i1) load("c:/maxima/tmpsum6.lisp")$ (%i2) sum(k,k,1,3); (%o2) 6 (%i3) sum(rat(k),k,1,3); Maxima encountered a Lisp error: Error in COND [or a callee]: Caught fatal error [memory may be damaged] Automatically continuing. To reenable the Lisp debugger set *debuggerhook* to nil. (%i4) Barton  Comment By: Robert Dodier (robert_dodier) Date: 20050619 20:28 Message: Logged In: YES user_id=501686 Yet another revision (attached file tmpsum6.lisp) of DOSUM and friends. This version passes run_testsuite() and batch ("rtestsum.mac", test) using the currently attached version of rtestsum.mac. This version avoids substituting for the index in expressions which are free of the index. This version also has an alternate definition of the function FREE. The simplification code (and other code) calls FREE instead of FREEOF. FREEOF looks for dummy variables, but FREE does not, so with the original defn of FREE, sum (lambda([i], i^2), i, 1, n) does not simplify to n lambda ([i], i^2), i, 1, n). I'm not very happy about changing FREE since it is called from all over the place, but I'm also not very happy about the presence of FREE, either; why not call FREEOF ? I don't think changing FREE at this point is such a good idea; it seems to me the two choices are (1) inspect the sum/product simplification code and change FREE to FREEOF where possible. That would be a lot of work. (2) Leave FREE as it is, and just accept that expressions with dummy variables can't be factored out of summations. At this point I think I'd rather leave FREE alone, and file a separate bug report about the inability to factor out expressions with dummy variables, and just leave it for another day.  Comment By: Robert Dodier (robert_dodier) Date: 20050619 20:04 Message: Logged In: YES user_id=501686 New version of rtestsum.mac. Two new test cases, sum (lambda ([i], i^2, 1, 3) and sum (lambda ([i], i^2), i, 1, n).  Comment By: Robert Dodier (robert_dodier) Date: 20050619 19:59 Message: Logged In: YES user_id=501686 Removing rtestsum.mac, will upload a new version momentarily.  Comment By: Robert Dodier (robert_dodier) Date: 20050619 10:52 Message: Logged In: YES user_id=501686 I'm attaching my latest and greatest revision (tmpsum5.lisp) of DOSUM, DO%SUM, and MEVALSUMARGS. I find that load ("tmpsum5.lisp"); batch ("rtestsum.mac", test); where rtestsum.mac is the currently attached version, yields all tests passed for Maxima 5.9.1cvs built with clisp 2.33.2 and with gcl 2.6.6. This revision uses a gensym index for cases handled by MEVALSUMARGS (i.e., cases for which (upper limit) minus (lower limit) is not an integer); the gensym index is the argument for assume/forget. When upper minus lower is an integer, the index is bound to lower, lower+1, lower+2, etc., the summand is evaluated, and then lower, lower+1, lower+2, etc. is substituted for any remaining instances of index.  Comment By: Robert Dodier (robert_dodier) Date: 20050619 10:33 Message: Logged In: YES user_id=501686 Attaching a longer list of test cases. batch ("rtestsum.mac", test); executes the test. As always the "correct" results (2nd line in each pair) are subject to review.  Comment By: Robert Dodier (robert_dodier) Date: 20050619 10:28 Message: Logged In: YES user_id=501686 I'm removing rtestsum.mac, in anticipation of attaching a new version in a minute.  Comment By: van_Nek (van_nek) Date: 20050618 11:42 Message: Logged In: YES user_id=1269745 Dear Maximadevelopers, thank you for your support. I tried the last fix from Robert Dodier and it works. Volker van Nek original (anonymous) poster  Comment By: Robert Dodier (robert_dodier) Date: 20050615 08:37 Message: Logged In: YES user_id=501686 I'm attaching a list of test cases for sum, which can be executed by batch ("rtestsum.mac", test) . Doubtless there will be some discussion as to what the expected output of each test should be. All tests pass after making these 2 changes: ;; ORIGINAL: MEVAL ONCE ;; (mbinding (lind l*i) (meval exp)) ;; TRIED THIS: MEVAL TWICE ;; (mbinding (lind l*i) (meval (meval exp))) ;; THIRD TIME IS A CHARM ?? ($substitute *i ind (mbinding (lind l*i) (meval exp))) ;; ORIGINAL: MEVALATOMS ;; (resimplify (mevalatoms (if (and (not (atom exp)) (resimplify (meval (if (and (not (atom exp))  Comment By: Robert Dodier (robert_dodier) Date: 20050614 08:42 Message: Logged In: YES user_id=501686 OK, here is another attempt. I think I now understand this comment by Stavros in bug report # 740134  "First evaluate foo with i bound to itself, and check if that result is free of i. If so, return the product. If not, *substitute* (don't evaluate) i=lowerlimit, i=lowerlimit+1, etc." ;; ORIGINAL: MEVAL ONCE ;; (mbinding (lind l*i) (meval exp)) ;; TRIED THIS: MEVAL TWICE ;; (mbinding (lind l*i) (meval (meval exp))) ;; THIRD TIME IS A CHARM ?? ($substitute *i ind (mbinding (lind l*i) (meval exp))) This version now yields this result  ex 9: f(x) := sum (x, i, 1, 3); f(x) => 3 x as expected, as well as the same results for ex 1 through ex 8.  Comment By: Robert Dodier (robert_dodier) Date: 20050613 23:00 Message: Logged In: YES user_id=501686 Actually, after looking at the code some more (specifically DOSUM in src/asum.lisp) it looks to me like the intent is indeed to evaluate the summand after binding the summation index to itself. In addition, there is code to assume the index is between its lower and upper limits (although assume/forget is handled incorrectly, see 851765). I believe that the observed behavior is a bug, in the narrow sense that the observed behavior is different from what the code author intended. See also my comments on SF bug # 740134. Here is a different patch. I like this one better than the previous one. In DOSUM: ;; ORIGINAL: MEVAL ONCE ;; (mbinding (lind l*i) (meval exp)) (mbinding (lind l*i) (meval (meval exp))) In MEVALSUMARG: ;; ORIGINAL: MEVALATOMS ;; (resimplify (mevalatoms (if (and (not (atom exp)) (resimplify (meval (if (and (not (atom exp)) With these changes, the results are as expected for the examples cited by the original poster, and also the example about the translated function produces the correct result, and some examples adapted from another bug report (740134) yield correct results. ex 1: f(x):=x^2$ b: 1/10$ Ak: b*f(k*b)$ sum(Ak,k,0,9); => 57/200 ex 2: A(k):=k*f(k*b)$ b:1/n$ A(k) => k^3/n^2 sum(A(k),k,0,n1) => ('sum(k^3,k,0,n1))/n^2 sum(A(k),k,0,n1), simpsum => ((n1)^4 + 2*(n1)^3 + (n1)^2) / (4*n^2) ex 3: ak : k^2$ g(a,n) := sum(a,k,1,n)$ g(ak,5) => 55 translate (g)$ g(ak,5) => 55 some other examples, adapted from bug # 740134: ex 4: sum (print (i), i, 1, 3) => prints 1, 2, 3 then returns 6 ex 5: sum (integrate (x^i ,x), i, 0, 2) => x^3/3 + x^2/2 + x ex 6: sum (integrate (1/(x^i + 1), x), i, 0, 1) => log(x+1) + x/2 ex 7: f[i](x):=x^i$ g[i](x):=x^i$ h[i](x):=x^i$ /* reference f[i] and g[i]  see 740134 for the effect this has on previous defn of sum */ f[i]$ g[i](t)$ sum (f[i](x), i, 0, n) => 'sum (x^i, i, 0, n) sum (g[i](x), i, 0, n) => 'sum (x^i, i, 0, n) sum (h[i](x), i, 0, n) => 'sum (x^i, i, 0, n) ex 8: sum (integrate (x^i, x), i, 0, n) => 'sum (x^(i+1) / (i+1), i, 0, n)  Comment By: Barton Willis (willisbl) Date: 20050613 05:30 Message: Logged In: YES user_id=895922 If we decide to use Robert's $sum function, the translate property for sum will need to be changed. Consider: (1) Redefine $sum as Robert suggested. (2) Try this: (%i1) ak : k^2$ (%i2) g(a,n) := sum(a,k,1,n)$ (%i3) g(ak,5); (%o3) 55 < correct for Robert's $sum function (%i4) translate(g); (%o4) [g] (%i5) g(ak,5); (%o5) 5*k^2 < yeech There might be other things that need fixing: (%i8) properties(sum); (%o8) [Special Evaluation Form,OUTATIVE,NOUN,RULE] Barton  Comment By: Robert Dodier (robert_dodier) Date: 20050612 20:34 Message: Logged In: YES user_id=501686 OK, for the record, here is a definition of $sum which implements one of the ideas that has been proposed, namely this policy: evaluate the summand after binding the summation variable to itself. (defmspec $sum (l) (setq l (cdr l)) (if (= (length l) 4) (progv (list (cadr l)) (list (cadr l)) (dosum (meval (car l)) (cadr l) (meval (caddr l)) (meval (cadddr l)) t)) (wnaerr '$sum))) This version yields the results expected by the person who originated this bug report, and yields the results predicted by S Macrakis in the email referenced below (http://www.math.utexas.edu/pipermail/maxima/2003/004869.html), and run_testsuite() runs to completion with no errors. I'm in favor of making this change, but I'm just recording this code snippet here for future reference; no changes planned at the moment. For comparison here is the current version of $sum: (defmspec $sum (l) (setq l (cdr l)) (if (= (length l) 4) (dosum (car l) (cadr l) (meval (caddr l)) (meval (cadddr l)) t) (wnaerr '$sum)))  Comment By: Barton Willis (willisbl) Date: 20050501 06:09 Message: Logged In: YES user_id=895922 We've discussed this before; see http://www.math.utexas.edu/pipermail/maxima/2003/004869.html http://www.math.utexas.edu/pipermail/maxima/2003/004870.html Maybe a simple workaround such as mysum(f,v,lo,hi):=block([acc:0], if integerp(lo) and integerp(hi) then for i from lo thru hi do acc:acc+substitute(i,v,f) else acc:funmake('mysum,[f,v,lo,hi]), acc) or mysum(f,lo,hi):= block([acc:0], if integerp(lo) and integerp(hi) then for i : lo thru hi do acc : acc + apply(f,[i]) else acc : funmake('mysum,[f,lo,hi]), acc); might work for you, Barton  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1192935&group_id=4933 
From: SourceForge.net <noreply@so...>  20050627 00:30:38

Bugs item #1192935, was opened at 20050430 05:10 Message generated for change (Comment added) made by robert_dodier You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1192935&group_id=4933 Please note that this message will contain a full copy of the comment thread, including the initial issue submission, for this request, not just the latest update. Category: Lisp Core Group: None Status: Open Resolution: None Priority: 5 Submitted By: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: termsubstitution in function sum is too late Initial Comment: (%i1) "Lower RiemannSum, Intervall [0,1], 10 parts"$ (%i2) /*Function*/ f(x):=x^2$ (%i3) /*area width*/ b: 1/10$ (%i4) /*area depending on k=0,...,9*/ Ak: b*f(k*b)$ (%i5) Ak; 2 k (%o5)  1000 (%i6) /* wrong */ sum(Ak,k,0,9); 2 k (%o6)  100 obviously, we first have the summation and then the symbolsubstitution (%i7) sum(''Ak,k,0,9); 57 (%o7)  200 finally, this is ok, but the usage of the doublequote is hard to explain to beginners. it gets even worse, if you use a function A(k):=k*f(k*b)$ for a sum with n partitions: (%i8) "Lower RiemannSum, Intervall [0,1], n parts"$ (%i9) /*area width*/ b:1/n$ (%i10) /*area depending on k=0,...,n1*/ A(k):=k*f(k*b)$ (%i11) A(k); 3 k (%o11)  2 n (%i12) sum(A(k),k,0,n1); n  1 ==== (%o12) > A(k) / ==== k = 0 in that case you need ''(A(k)), a bracket and the doublequote. is it possible to patch the sumfunction in that way, that we first have the term or symbolsubstitution and second the summation?  >Comment By: Robert Dodier (robert_dodier) Date: 20050626 18:30 Message: Logged In: YES user_id=501686 attempting to remove old version of rtestsum.mac and upload new version in the same update... let's see if it works.  Comment By: Robert Dodier (robert_dodier) Date: 20050621 00:18 Message: Logged In: YES user_id=501686 New DOSUM, etc, in tmpsum8.lisp. A few new sum tests have been added to rtestsum.mac, and all sum tests copied and turned into product tests. This revision passes all tests in current rtestsum.mac except for test 38: (assume(i < 0, j < 0, n > 0), product(product(abs(j) + abs(i), i, 1, j), j, 1, n)); => Lower bound to product is > upper bound. This emanates from SIMPROD1 (src/combin.lisp) which is called after the assumptions about i and j (actually, assumptions about the gensym standins for i and j) have been forgotten. This looks like a bug in combin.lisp  there is no assume/forget about the indices.  Comment By: Robert Dodier (robert_dodier) Date: 20050621 00:06 Message: Logged In: YES user_id=501686 New rtestsum.mac. A couple of new tests for sum, and also copied all sum tests and substituted product for sum (and adjusted expected outputs to suit).  Comment By: Robert Dodier (robert_dodier) Date: 20050621 00:03 Message: Logged In: YES user_id=501686 Delete rtestsum.mac, going to upload a new one momentarily.  Comment By: Robert Dodier (robert_dodier) Date: 20050619 23:27 Message: Logged In: YES user_id=501686 About sum(rat(k),k,1,3), calling $FREEOF instead of FREEOF makes it happy again. Oh well. I think I'll just hang onto the current rev for a day or two to see what other changes accumulate, before posting a new version of tmpsum*.lisp. A related bit  sum(rat(k), k, 1, n) returns 'sum (k, k, 1, n), which isn't exactly wrong, but it does make me wonder what happened to the rat... it turns out this is what happened: subst (foo, k, rat(k)) => foo Not rat(foo) as I would have expected. Dunno what this is about.  Comment By: Barton Willis (willisbl) Date: 20050619 21:36 Message: Logged In: YES user_id=895922 Does tmpsum6.lisp need a ratdisrep somewhere? (%i1) load("c:/maxima/tmpsum6.lisp")$ (%i2) sum(k,k,1,3); (%o2) 6 (%i3) sum(rat(k),k,1,3); Maxima encountered a Lisp error: Error in COND [or a callee]: Caught fatal error [memory may be damaged] Automatically continuing. To reenable the Lisp debugger set *debuggerhook* to nil. (%i4) Barton  Comment By: Robert Dodier (robert_dodier) Date: 20050619 20:28 Message: Logged In: YES user_id=501686 Yet another revision (attached file tmpsum6.lisp) of DOSUM and friends. This version passes run_testsuite() and batch ("rtestsum.mac", test) using the currently attached version of rtestsum.mac. This version avoids substituting for the index in expressions which are free of the index. This version also has an alternate definition of the function FREE. The simplification code (and other code) calls FREE instead of FREEOF. FREEOF looks for dummy variables, but FREE does not, so with the original defn of FREE, sum (lambda([i], i^2), i, 1, n) does not simplify to n lambda ([i], i^2), i, 1, n). I'm not very happy about changing FREE since it is called from all over the place, but I'm also not very happy about the presence of FREE, either; why not call FREEOF ? I don't think changing FREE at this point is such a good idea; it seems to me the two choices are (1) inspect the sum/product simplification code and change FREE to FREEOF where possible. That would be a lot of work. (2) Leave FREE as it is, and just accept that expressions with dummy variables can't be factored out of summations. At this point I think I'd rather leave FREE alone, and file a separate bug report about the inability to factor out expressions with dummy variables, and just leave it for another day.  Comment By: Robert Dodier (robert_dodier) Date: 20050619 20:04 Message: Logged In: YES user_id=501686 New version of rtestsum.mac. Two new test cases, sum (lambda ([i], i^2, 1, 3) and sum (lambda ([i], i^2), i, 1, n).  Comment By: Robert Dodier (robert_dodier) Date: 20050619 19:59 Message: Logged In: YES user_id=501686 Removing rtestsum.mac, will upload a new version momentarily.  Comment By: Robert Dodier (robert_dodier) Date: 20050619 10:52 Message: Logged In: YES user_id=501686 I'm attaching my latest and greatest revision (tmpsum5.lisp) of DOSUM, DO%SUM, and MEVALSUMARGS. I find that load ("tmpsum5.lisp"); batch ("rtestsum.mac", test); where rtestsum.mac is the currently attached version, yields all tests passed for Maxima 5.9.1cvs built with clisp 2.33.2 and with gcl 2.6.6. This revision uses a gensym index for cases handled by MEVALSUMARGS (i.e., cases for which (upper limit) minus (lower limit) is not an integer); the gensym index is the argument for assume/forget. When upper minus lower is an integer, the index is bound to lower, lower+1, lower+2, etc., the summand is evaluated, and then lower, lower+1, lower+2, etc. is substituted for any remaining instances of index.  Comment By: Robert Dodier (robert_dodier) Date: 20050619 10:33 Message: Logged In: YES user_id=501686 Attaching a longer list of test cases. batch ("rtestsum.mac", test); executes the test. As always the "correct" results (2nd line in each pair) are subject to review.  Comment By: Robert Dodier (robert_dodier) Date: 20050619 10:28 Message: Logged In: YES user_id=501686 I'm removing rtestsum.mac, in anticipation of attaching a new version in a minute.  Comment By: van_Nek (van_nek) Date: 20050618 11:42 Message: Logged In: YES user_id=1269745 Dear Maximadevelopers, thank you for your support. I tried the last fix from Robert Dodier and it works. Volker van Nek original (anonymous) poster  Comment By: Robert Dodier (robert_dodier) Date: 20050615 08:37 Message: Logged In: YES user_id=501686 I'm attaching a list of test cases for sum, which can be executed by batch ("rtestsum.mac", test) . Doubtless there will be some discussion as to what the expected output of each test should be. All tests pass after making these 2 changes: ;; ORIGINAL: MEVAL ONCE ;; (mbinding (lind l*i) (meval exp)) ;; TRIED THIS: MEVAL TWICE ;; (mbinding (lind l*i) (meval (meval exp))) ;; THIRD TIME IS A CHARM ?? ($substitute *i ind (mbinding (lind l*i) (meval exp))) ;; ORIGINAL: MEVALATOMS ;; (resimplify (mevalatoms (if (and (not (atom exp)) (resimplify (meval (if (and (not (atom exp))  Comment By: Robert Dodier (robert_dodier) Date: 20050614 08:42 Message: Logged In: YES user_id=501686 OK, here is another attempt. I think I now understand this comment by Stavros in bug report # 740134  "First evaluate foo with i bound to itself, and check if that result is free of i. If so, return the product. If not, *substitute* (don't evaluate) i=lowerlimit, i=lowerlimit+1, etc." ;; ORIGINAL: MEVAL ONCE ;; (mbinding (lind l*i) (meval exp)) ;; TRIED THIS: MEVAL TWICE ;; (mbinding (lind l*i) (meval (meval exp))) ;; THIRD TIME IS A CHARM ?? ($substitute *i ind (mbinding (lind l*i) (meval exp))) This version now yields this result  ex 9: f(x) := sum (x, i, 1, 3); f(x) => 3 x as expected, as well as the same results for ex 1 through ex 8.  Comment By: Robert Dodier (robert_dodier) Date: 20050613 23:00 Message: Logged In: YES user_id=501686 Actually, after looking at the code some more (specifically DOSUM in src/asum.lisp) it looks to me like the intent is indeed to evaluate the summand after binding the summation index to itself. In addition, there is code to assume the index is between its lower and upper limits (although assume/forget is handled incorrectly, see 851765). I believe that the observed behavior is a bug, in the narrow sense that the observed behavior is different from what the code author intended. See also my comments on SF bug # 740134. Here is a different patch. I like this one better than the previous one. In DOSUM: ;; ORIGINAL: MEVAL ONCE ;; (mbinding (lind l*i) (meval exp)) (mbinding (lind l*i) (meval (meval exp))) In MEVALSUMARG: ;; ORIGINAL: MEVALATOMS ;; (resimplify (mevalatoms (if (and (not (atom exp)) (resimplify (meval (if (and (not (atom exp)) With these changes, the results are as expected for the examples cited by the original poster, and also the example about the translated function produces the correct result, and some examples adapted from another bug report (740134) yield correct results. ex 1: f(x):=x^2$ b: 1/10$ Ak: b*f(k*b)$ sum(Ak,k,0,9); => 57/200 ex 2: A(k):=k*f(k*b)$ b:1/n$ A(k) => k^3/n^2 sum(A(k),k,0,n1) => ('sum(k^3,k,0,n1))/n^2 sum(A(k),k,0,n1), simpsum => ((n1)^4 + 2*(n1)^3 + (n1)^2) / (4*n^2) ex 3: ak : k^2$ g(a,n) := sum(a,k,1,n)$ g(ak,5) => 55 translate (g)$ g(ak,5) => 55 some other examples, adapted from bug # 740134: ex 4: sum (print (i), i, 1, 3) => prints 1, 2, 3 then returns 6 ex 5: sum (integrate (x^i ,x), i, 0, 2) => x^3/3 + x^2/2 + x ex 6: sum (integrate (1/(x^i + 1), x), i, 0, 1) => log(x+1) + x/2 ex 7: f[i](x):=x^i$ g[i](x):=x^i$ h[i](x):=x^i$ /* reference f[i] and g[i]  see 740134 for the effect this has on previous defn of sum */ f[i]$ g[i](t)$ sum (f[i](x), i, 0, n) => 'sum (x^i, i, 0, n) sum (g[i](x), i, 0, n) => 'sum (x^i, i, 0, n) sum (h[i](x), i, 0, n) => 'sum (x^i, i, 0, n) ex 8: sum (integrate (x^i, x), i, 0, n) => 'sum (x^(i+1) / (i+1), i, 0, n)  Comment By: Barton Willis (willisbl) Date: 20050613 05:30 Message: Logged In: YES user_id=895922 If we decide to use Robert's $sum function, the translate property for sum will need to be changed. Consider: (1) Redefine $sum as Robert suggested. (2) Try this: (%i1) ak : k^2$ (%i2) g(a,n) := sum(a,k,1,n)$ (%i3) g(ak,5); (%o3) 55 < correct for Robert's $sum function (%i4) translate(g); (%o4) [g] (%i5) g(ak,5); (%o5) 5*k^2 < yeech There might be other things that need fixing: (%i8) properties(sum); (%o8) [Special Evaluation Form,OUTATIVE,NOUN,RULE] Barton  Comment By: Robert Dodier (robert_dodier) Date: 20050612 20:34 Message: Logged In: YES user_id=501686 OK, for the record, here is a definition of $sum which implements one of the ideas that has been proposed, namely this policy: evaluate the summand after binding the summation variable to itself. (defmspec $sum (l) (setq l (cdr l)) (if (= (length l) 4) (progv (list (cadr l)) (list (cadr l)) (dosum (meval (car l)) (cadr l) (meval (caddr l)) (meval (cadddr l)) t)) (wnaerr '$sum))) This version yields the results expected by the person who originated this bug report, and yields the results predicted by S Macrakis in the email referenced below (http://www.math.utexas.edu/pipermail/maxima/2003/004869.html), and run_testsuite() runs to completion with no errors. I'm in favor of making this change, but I'm just recording this code snippet here for future reference; no changes planned at the moment. For comparison here is the current version of $sum: (defmspec $sum (l) (setq l (cdr l)) (if (= (length l) 4) (dosum (car l) (cadr l) (meval (caddr l)) (meval (cadddr l)) t) (wnaerr '$sum)))  Comment By: Barton Willis (willisbl) Date: 20050501 06:09 Message: Logged In: YES user_id=895922 We've discussed this before; see http://www.math.utexas.edu/pipermail/maxima/2003/004869.html http://www.math.utexas.edu/pipermail/maxima/2003/004870.html Maybe a simple workaround such as mysum(f,v,lo,hi):=block([acc:0], if integerp(lo) and integerp(hi) then for i from lo thru hi do acc:acc+substitute(i,v,f) else acc:funmake('mysum,[f,v,lo,hi]), acc) or mysum(f,lo,hi):= block([acc:0], if integerp(lo) and integerp(hi) then for i : lo thru hi do acc : acc + apply(f,[i]) else acc : funmake('mysum,[f,lo,hi]), acc); might work for you, Barton  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1192935&group_id=4933 
From: SourceForge.net <noreply@so...>  20050820 05:03:51

Bugs item #1192935, was opened at 20050430 05:10 Message generated for change (Comment added) made by robert_dodier You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1192935&group_id=4933 Please note that this message will contain a full copy of the comment thread, including the initial issue submission, for this request, not just the latest update. Category: Lisp Core Group: None Status: Open Resolution: None Priority: 5 Submitted By: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: termsubstitution in function sum is too late Initial Comment: (%i1) "Lower RiemannSum, Intervall [0,1], 10 parts"$ (%i2) /*Function*/ f(x):=x^2$ (%i3) /*area width*/ b: 1/10$ (%i4) /*area depending on k=0,...,9*/ Ak: b*f(k*b)$ (%i5) Ak; 2 k (%o5)  1000 (%i6) /* wrong */ sum(Ak,k,0,9); 2 k (%o6)  100 obviously, we first have the summation and then the symbolsubstitution (%i7) sum(''Ak,k,0,9); 57 (%o7)  200 finally, this is ok, but the usage of the doublequote is hard to explain to beginners. it gets even worse, if you use a function A(k):=k*f(k*b)$ for a sum with n partitions: (%i8) "Lower RiemannSum, Intervall [0,1], n parts"$ (%i9) /*area width*/ b:1/n$ (%i10) /*area depending on k=0,...,n1*/ A(k):=k*f(k*b)$ (%i11) A(k); 3 k (%o11)  2 n (%i12) sum(A(k),k,0,n1); n  1 ==== (%o12) > A(k) / ==== k = 0 in that case you need ''(A(k)), a bracket and the doublequote. is it possible to patch the sumfunction in that way, that we first have the term or symbolsubstitution and second the summation?  >Comment By: Robert Dodier (robert_dodier) Date: 20050819 23:03 Message: Logged In: YES user_id=501686 attaching a zip file containing unksum6.lisp, tmpsum13.lisp (two different sum reimplementations), rtestsum.mac, and mand.lisp (a reimplementation of mcondrelated stuff  needed for the one test which makes use of an unevaluated conditional).  Comment By: Robert Dodier (robert_dodier) Date: 20050626 18:30 Message: Logged In: YES user_id=501686 attempting to remove old version of rtestsum.mac and upload new version in the same update... let's see if it works.  Comment By: Robert Dodier (robert_dodier) Date: 20050621 00:18 Message: Logged In: YES user_id=501686 New DOSUM, etc, in tmpsum8.lisp. A few new sum tests have been added to rtestsum.mac, and all sum tests copied and turned into product tests. This revision passes all tests in current rtestsum.mac except for test 38: (assume(i < 0, j < 0, n > 0), product(product(abs(j) + abs(i), i, 1, j), j, 1, n)); => Lower bound to product is > upper bound. This emanates from SIMPROD1 (src/combin.lisp) which is called after the assumptions about i and j (actually, assumptions about the gensym standins for i and j) have been forgotten. This looks like a bug in combin.lisp  there is no assume/forget about the indices.  Comment By: Robert Dodier (robert_dodier) Date: 20050621 00:06 Message: Logged In: YES user_id=501686 New rtestsum.mac. A couple of new tests for sum, and also copied all sum tests and substituted product for sum (and adjusted expected outputs to suit).  Comment By: Robert Dodier (robert_dodier) Date: 20050621 00:03 Message: Logged In: YES user_id=501686 Delete rtestsum.mac, going to upload a new one momentarily.  Comment By: Robert Dodier (robert_dodier) Date: 20050619 23:27 Message: Logged In: YES user_id=501686 About sum(rat(k),k,1,3), calling $FREEOF instead of FREEOF makes it happy again. Oh well. I think I'll just hang onto the current rev for a day or two to see what other changes accumulate, before posting a new version of tmpsum*.lisp. A related bit  sum(rat(k), k, 1, n) returns 'sum (k, k, 1, n), which isn't exactly wrong, but it does make me wonder what happened to the rat... it turns out this is what happened: subst (foo, k, rat(k)) => foo Not rat(foo) as I would have expected. Dunno what this is about.  Comment By: Barton Willis (willisbl) Date: 20050619 21:36 Message: Logged In: YES user_id=895922 Does tmpsum6.lisp need a ratdisrep somewhere? (%i1) load("c:/maxima/tmpsum6.lisp")$ (%i2) sum(k,k,1,3); (%o2) 6 (%i3) sum(rat(k),k,1,3); Maxima encountered a Lisp error: Error in COND [or a callee]: Caught fatal error [memory may be damaged] Automatically continuing. To reenable the Lisp debugger set *debuggerhook* to nil. (%i4) Barton  Comment By: Robert Dodier (robert_dodier) Date: 20050619 20:28 Message: Logged In: YES user_id=501686 Yet another revision (attached file tmpsum6.lisp) of DOSUM and friends. This version passes run_testsuite() and batch ("rtestsum.mac", test) using the currently attached version of rtestsum.mac. This version avoids substituting for the index in expressions which are free of the index. This version also has an alternate definition of the function FREE. The simplification code (and other code) calls FREE instead of FREEOF. FREEOF looks for dummy variables, but FREE does not, so with the original defn of FREE, sum (lambda([i], i^2), i, 1, n) does not simplify to n lambda ([i], i^2), i, 1, n). I'm not very happy about changing FREE since it is called from all over the place, but I'm also not very happy about the presence of FREE, either; why not call FREEOF ? I don't think changing FREE at this point is such a good idea; it seems to me the two choices are (1) inspect the sum/product simplification code and change FREE to FREEOF where possible. That would be a lot of work. (2) Leave FREE as it is, and just accept that expressions with dummy variables can't be factored out of summations. At this point I think I'd rather leave FREE alone, and file a separate bug report about the inability to factor out expressions with dummy variables, and just leave it for another day.  Comment By: Robert Dodier (robert_dodier) Date: 20050619 20:04 Message: Logged In: YES user_id=501686 New version of rtestsum.mac. Two new test cases, sum (lambda ([i], i^2, 1, 3) and sum (lambda ([i], i^2), i, 1, n).  Comment By: Robert Dodier (robert_dodier) Date: 20050619 19:59 Message: Logged In: YES user_id=501686 Removing rtestsum.mac, will upload a new version momentarily.  Comment By: Robert Dodier (robert_dodier) Date: 20050619 10:52 Message: Logged In: YES user_id=501686 I'm attaching my latest and greatest revision (tmpsum5.lisp) of DOSUM, DO%SUM, and MEVALSUMARGS. I find that load ("tmpsum5.lisp"); batch ("rtestsum.mac", test); where rtestsum.mac is the currently attached version, yields all tests passed for Maxima 5.9.1cvs built with clisp 2.33.2 and with gcl 2.6.6. This revision uses a gensym index for cases handled by MEVALSUMARGS (i.e., cases for which (upper limit) minus (lower limit) is not an integer); the gensym index is the argument for assume/forget. When upper minus lower is an integer, the index is bound to lower, lower+1, lower+2, etc., the summand is evaluated, and then lower, lower+1, lower+2, etc. is substituted for any remaining instances of index.  Comment By: Robert Dodier (robert_dodier) Date: 20050619 10:33 Message: Logged In: YES user_id=501686 Attaching a longer list of test cases. batch ("rtestsum.mac", test); executes the test. As always the "correct" results (2nd line in each pair) are subject to review.  Comment By: Robert Dodier (robert_dodier) Date: 20050619 10:28 Message: Logged In: YES user_id=501686 I'm removing rtestsum.mac, in anticipation of attaching a new version in a minute.  Comment By: van_Nek (van_nek) Date: 20050618 11:42 Message: Logged In: YES user_id=1269745 Dear Maximadevelopers, thank you for your support. I tried the last fix from Robert Dodier and it works. Volker van Nek original (anonymous) poster  Comment By: Robert Dodier (robert_dodier) Date: 20050615 08:37 Message: Logged In: YES user_id=501686 I'm attaching a list of test cases for sum, which can be executed by batch ("rtestsum.mac", test) . Doubtless there will be some discussion as to what the expected output of each test should be. All tests pass after making these 2 changes: ;; ORIGINAL: MEVAL ONCE ;; (mbinding (lind l*i) (meval exp)) ;; TRIED THIS: MEVAL TWICE ;; (mbinding (lind l*i) (meval (meval exp))) ;; THIRD TIME IS A CHARM ?? ($substitute *i ind (mbinding (lind l*i) (meval exp))) ;; ORIGINAL: MEVALATOMS ;; (resimplify (mevalatoms (if (and (not (atom exp)) (resimplify (meval (if (and (not (atom exp))  Comment By: Robert Dodier (robert_dodier) Date: 20050614 08:42 Message: Logged In: YES user_id=501686 OK, here is another attempt. I think I now understand this comment by Stavros in bug report # 740134  "First evaluate foo with i bound to itself, and check if that result is free of i. If so, return the product. If not, *substitute* (don't evaluate) i=lowerlimit, i=lowerlimit+1, etc." ;; ORIGINAL: MEVAL ONCE ;; (mbinding (lind l*i) (meval exp)) ;; TRIED THIS: MEVAL TWICE ;; (mbinding (lind l*i) (meval (meval exp))) ;; THIRD TIME IS A CHARM ?? ($substitute *i ind (mbinding (lind l*i) (meval exp))) This version now yields this result  ex 9: f(x) := sum (x, i, 1, 3); f(x) => 3 x as expected, as well as the same results for ex 1 through ex 8.  Comment By: Robert Dodier (robert_dodier) Date: 20050613 23:00 Message: Logged In: YES user_id=501686 Actually, after looking at the code some more (specifically DOSUM in src/asum.lisp) it looks to me like the intent is indeed to evaluate the summand after binding the summation index to itself. In addition, there is code to assume the index is between its lower and upper limits (although assume/forget is handled incorrectly, see 851765). I believe that the observed behavior is a bug, in the narrow sense that the observed behavior is different from what the code author intended. See also my comments on SF bug # 740134. Here is a different patch. I like this one better than the previous one. In DOSUM: ;; ORIGINAL: MEVAL ONCE ;; (mbinding (lind l*i) (meval exp)) (mbinding (lind l*i) (meval (meval exp))) In MEVALSUMARG: ;; ORIGINAL: MEVALATOMS ;; (resimplify (mevalatoms (if (and (not (atom exp)) (resimplify (meval (if (and (not (atom exp)) With these changes, the results are as expected for the examples cited by the original poster, and also the example about the translated function produces the correct result, and some examples adapted from another bug report (740134) yield correct results. ex 1: f(x):=x^2$ b: 1/10$ Ak: b*f(k*b)$ sum(Ak,k,0,9); => 57/200 ex 2: A(k):=k*f(k*b)$ b:1/n$ A(k) => k^3/n^2 sum(A(k),k,0,n1) => ('sum(k^3,k,0,n1))/n^2 sum(A(k),k,0,n1), simpsum => ((n1)^4 + 2*(n1)^3 + (n1)^2) / (4*n^2) ex 3: ak : k^2$ g(a,n) := sum(a,k,1,n)$ g(ak,5) => 55 translate (g)$ g(ak,5) => 55 some other examples, adapted from bug # 740134: ex 4: sum (print (i), i, 1, 3) => prints 1, 2, 3 then returns 6 ex 5: sum (integrate (x^i ,x), i, 0, 2) => x^3/3 + x^2/2 + x ex 6: sum (integrate (1/(x^i + 1), x), i, 0, 1) => log(x+1) + x/2 ex 7: f[i](x):=x^i$ g[i](x):=x^i$ h[i](x):=x^i$ /* reference f[i] and g[i]  see 740134 for the effect this has on previous defn of sum */ f[i]$ g[i](t)$ sum (f[i](x), i, 0, n) => 'sum (x^i, i, 0, n) sum (g[i](x), i, 0, n) => 'sum (x^i, i, 0, n) sum (h[i](x), i, 0, n) => 'sum (x^i, i, 0, n) ex 8: sum (integrate (x^i, x), i, 0, n) => 'sum (x^(i+1) / (i+1), i, 0, n)  Comment By: Barton Willis (willisbl) Date: 20050613 05:30 Message: Logged In: YES user_id=895922 If we decide to use Robert's $sum function, the translate property for sum will need to be changed. Consider: (1) Redefine $sum as Robert suggested. (2) Try this: (%i1) ak : k^2$ (%i2) g(a,n) := sum(a,k,1,n)$ (%i3) g(ak,5); (%o3) 55 < correct for Robert's $sum function (%i4) translate(g); (%o4) [g] (%i5) g(ak,5); (%o5) 5*k^2 < yeech There might be other things that need fixing: (%i8) properties(sum); (%o8) [Special Evaluation Form,OUTATIVE,NOUN,RULE] Barton  Comment By: Robert Dodier (robert_dodier) Date: 20050612 20:34 Message: Logged In: YES user_id=501686 OK, for the record, here is a definition of $sum which implements one of the ideas that has been proposed, namely this policy: evaluate the summand after binding the summation variable to itself. (defmspec $sum (l) (setq l (cdr l)) (if (= (length l) 4) (progv (list (cadr l)) (list (cadr l)) (dosum (meval (car l)) (cadr l) (meval (caddr l)) (meval (cadddr l)) t)) (wnaerr '$sum))) This version yields the results expected by the person who originated this bug report, and yields the results predicted by S Macrakis in the email referenced below (http://www.math.utexas.edu/pipermail/maxima/2003/004869.html), and run_testsuite() runs to completion with no errors. I'm in favor of making this change, but I'm just recording this code snippet here for future reference; no changes planned at the moment. For comparison here is the current version of $sum: (defmspec $sum (l) (setq l (cdr l)) (if (= (length l) 4) (dosum (car l) (cadr l) (meval (caddr l)) (meval (cadddr l)) t) (wnaerr '$sum)))  Comment By: Barton Willis (willisbl) Date: 20050501 06:09 Message: Logged In: YES user_id=895922 We've discussed this before; see http://www.math.utexas.edu/pipermail/maxima/2003/004869.html http://www.math.utexas.edu/pipermail/maxima/2003/004870.html Maybe a simple workaround such as mysum(f,v,lo,hi):=block([acc:0], if integerp(lo) and integerp(hi) then for i from lo thru hi do acc:acc+substitute(i,v,f) else acc:funmake('mysum,[f,v,lo,hi]), acc) or mysum(f,lo,hi):= block([acc:0], if integerp(lo) and integerp(hi) then for i : lo thru hi do acc : acc + apply(f,[i]) else acc : funmake('mysum,[f,lo,hi]), acc); might work for you, Barton  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1192935&group_id=4933 
From: SourceForge.net <noreply@so...>  20050820 05:08:39

Bugs item #1192935, was opened at 20050430 05:10 Message generated for change (Comment added) made by robert_dodier You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1192935&group_id=4933 Please note that this message will contain a full copy of the comment thread, including the initial issue submission, for this request, not just the latest update. Category: Lisp Core Group: None Status: Open Resolution: None Priority: 5 Submitted By: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: termsubstitution in function sum is too late Initial Comment: (%i1) "Lower RiemannSum, Intervall [0,1], 10 parts"$ (%i2) /*Function*/ f(x):=x^2$ (%i3) /*area width*/ b: 1/10$ (%i4) /*area depending on k=0,...,9*/ Ak: b*f(k*b)$ (%i5) Ak; 2 k (%o5)  1000 (%i6) /* wrong */ sum(Ak,k,0,9); 2 k (%o6)  100 obviously, we first have the summation and then the symbolsubstitution (%i7) sum(''Ak,k,0,9); 57 (%o7)  200 finally, this is ok, but the usage of the doublequote is hard to explain to beginners. it gets even worse, if you use a function A(k):=k*f(k*b)$ for a sum with n partitions: (%i8) "Lower RiemannSum, Intervall [0,1], n parts"$ (%i9) /*area width*/ b:1/n$ (%i10) /*area depending on k=0,...,n1*/ A(k):=k*f(k*b)$ (%i11) A(k); 3 k (%o11)  2 n (%i12) sum(A(k),k,0,n1); n  1 ==== (%o12) > A(k) / ==== k = 0 in that case you need ''(A(k)), a bracket and the doublequote. is it possible to patch the sumfunction in that way, that we first have the term or symbolsubstitution and second the summation?  >Comment By: Robert Dodier (robert_dodier) Date: 20050819 23:08 Message: Logged In: YES user_id=501686 removing old versions of files  Comment By: Robert Dodier (robert_dodier) Date: 20050819 23:03 Message: Logged In: YES user_id=501686 attaching a zip file containing unksum6.lisp, tmpsum13.lisp (two different sum reimplementations), rtestsum.mac, and mand.lisp (a reimplementation of mcondrelated stuff  needed for the one test which makes use of an unevaluated conditional).  Comment By: Robert Dodier (robert_dodier) Date: 20050626 18:30 Message: Logged In: YES user_id=501686 attempting to remove old version of rtestsum.mac and upload new version in the same update... let's see if it works.  Comment By: Robert Dodier (robert_dodier) Date: 20050621 00:18 Message: Logged In: YES user_id=501686 New DOSUM, etc, in tmpsum8.lisp. A few new sum tests have been added to rtestsum.mac, and all sum tests copied and turned into product tests. This revision passes all tests in current rtestsum.mac except for test 38: (assume(i < 0, j < 0, n > 0), product(product(abs(j) + abs(i), i, 1, j), j, 1, n)); => Lower bound to product is > upper bound. This emanates from SIMPROD1 (src/combin.lisp) which is called after the assumptions about i and j (actually, assumptions about the gensym standins for i and j) have been forgotten. This looks like a bug in combin.lisp  there is no assume/forget about the indices.  Comment By: Robert Dodier (robert_dodier) Date: 20050621 00:06 Message: Logged In: YES user_id=501686 New rtestsum.mac. A couple of new tests for sum, and also copied all sum tests and substituted product for sum (and adjusted expected outputs to suit).  Comment By: Robert Dodier (robert_dodier) Date: 20050621 00:03 Message: Logged In: YES user_id=501686 Delete rtestsum.mac, going to upload a new one momentarily.  Comment By: Robert Dodier (robert_dodier) Date: 20050619 23:27 Message: Logged In: YES user_id=501686 About sum(rat(k),k,1,3), calling $FREEOF instead of FREEOF makes it happy again. Oh well. I think I'll just hang onto the current rev for a day or two to see what other changes accumulate, before posting a new version of tmpsum*.lisp. A related bit  sum(rat(k), k, 1, n) returns 'sum (k, k, 1, n), which isn't exactly wrong, but it does make me wonder what happened to the rat... it turns out this is what happened: subst (foo, k, rat(k)) => foo Not rat(foo) as I would have expected. Dunno what this is about.  Comment By: Barton Willis (willisbl) Date: 20050619 21:36 Message: Logged In: YES user_id=895922 Does tmpsum6.lisp need a ratdisrep somewhere? (%i1) load("c:/maxima/tmpsum6.lisp")$ (%i2) sum(k,k,1,3); (%o2) 6 (%i3) sum(rat(k),k,1,3); Maxima encountered a Lisp error: Error in COND [or a callee]: Caught fatal error [memory may be damaged] Automatically continuing. To reenable the Lisp debugger set *debuggerhook* to nil. (%i4) Barton  Comment By: Robert Dodier (robert_dodier) Date: 20050619 20:28 Message: Logged In: YES user_id=501686 Yet another revision (attached file tmpsum6.lisp) of DOSUM and friends. This version passes run_testsuite() and batch ("rtestsum.mac", test) using the currently attached version of rtestsum.mac. This version avoids substituting for the index in expressions which are free of the index. This version also has an alternate definition of the function FREE. The simplification code (and other code) calls FREE instead of FREEOF. FREEOF looks for dummy variables, but FREE does not, so with the original defn of FREE, sum (lambda([i], i^2), i, 1, n) does not simplify to n lambda ([i], i^2), i, 1, n). I'm not very happy about changing FREE since it is called from all over the place, but I'm also not very happy about the presence of FREE, either; why not call FREEOF ? I don't think changing FREE at this point is such a good idea; it seems to me the two choices are (1) inspect the sum/product simplification code and change FREE to FREEOF where possible. That would be a lot of work. (2) Leave FREE as it is, and just accept that expressions with dummy variables can't be factored out of summations. At this point I think I'd rather leave FREE alone, and file a separate bug report about the inability to factor out expressions with dummy variables, and just leave it for another day.  Comment By: Robert Dodier (robert_dodier) Date: 20050619 20:04 Message: Logged In: YES user_id=501686 New version of rtestsum.mac. Two new test cases, sum (lambda ([i], i^2, 1, 3) and sum (lambda ([i], i^2), i, 1, n).  Comment By: Robert Dodier (robert_dodier) Date: 20050619 19:59 Message: Logged In: YES user_id=501686 Removing rtestsum.mac, will upload a new version momentarily.  Comment By: Robert Dodier (robert_dodier) Date: 20050619 10:52 Message: Logged In: YES user_id=501686 I'm attaching my latest and greatest revision (tmpsum5.lisp) of DOSUM, DO%SUM, and MEVALSUMARGS. I find that load ("tmpsum5.lisp"); batch ("rtestsum.mac", test); where rtestsum.mac is the currently attached version, yields all tests passed for Maxima 5.9.1cvs built with clisp 2.33.2 and with gcl 2.6.6. This revision uses a gensym index for cases handled by MEVALSUMARGS (i.e., cases for which (upper limit) minus (lower limit) is not an integer); the gensym index is the argument for assume/forget. When upper minus lower is an integer, the index is bound to lower, lower+1, lower+2, etc., the summand is evaluated, and then lower, lower+1, lower+2, etc. is substituted for any remaining instances of index.  Comment By: Robert Dodier (robert_dodier) Date: 20050619 10:33 Message: Logged In: YES user_id=501686 Attaching a longer list of test cases. batch ("rtestsum.mac", test); executes the test. As always the "correct" results (2nd line in each pair) are subject to review.  Comment By: Robert Dodier (robert_dodier) Date: 20050619 10:28 Message: Logged In: YES user_id=501686 I'm removing rtestsum.mac, in anticipation of attaching a new version in a minute.  Comment By: van_Nek (van_nek) Date: 20050618 11:42 Message: Logged In: YES user_id=1269745 Dear Maximadevelopers, thank you for your support. I tried the last fix from Robert Dodier and it works. Volker van Nek original (anonymous) poster  Comment By: Robert Dodier (robert_dodier) Date: 20050615 08:37 Message: Logged In: YES user_id=501686 I'm attaching a list of test cases for sum, which can be executed by batch ("rtestsum.mac", test) . Doubtless there will be some discussion as to what the expected output of each test should be. All tests pass after making these 2 changes: ;; ORIGINAL: MEVAL ONCE ;; (mbinding (lind l*i) (meval exp)) ;; TRIED THIS: MEVAL TWICE ;; (mbinding (lind l*i) (meval (meval exp))) ;; THIRD TIME IS A CHARM ?? ($substitute *i ind (mbinding (lind l*i) (meval exp))) ;; ORIGINAL: MEVALATOMS ;; (resimplify (mevalatoms (if (and (not (atom exp)) (resimplify (meval (if (and (not (atom exp))  Comment By: Robert Dodier (robert_dodier) Date: 20050614 08:42 Message: Logged In: YES user_id=501686 OK, here is another attempt. I think I now understand this comment by Stavros in bug report # 740134  "First evaluate foo with i bound to itself, and check if that result is free of i. If so, return the product. If not, *substitute* (don't evaluate) i=lowerlimit, i=lowerlimit+1, etc." ;; ORIGINAL: MEVAL ONCE ;; (mbinding (lind l*i) (meval exp)) ;; TRIED THIS: MEVAL TWICE ;; (mbinding (lind l*i) (meval (meval exp))) ;; THIRD TIME IS A CHARM ?? ($substitute *i ind (mbinding (lind l*i) (meval exp))) This version now yields this result  ex 9: f(x) := sum (x, i, 1, 3); f(x) => 3 x as expected, as well as the same results for ex 1 through ex 8.  Comment By: Robert Dodier (robert_dodier) Date: 20050613 23:00 Message: Logged In: YES user_id=501686 Actually, after looking at the code some more (specifically DOSUM in src/asum.lisp) it looks to me like the intent is indeed to evaluate the summand after binding the summation index to itself. In addition, there is code to assume the index is between its lower and upper limits (although assume/forget is handled incorrectly, see 851765). I believe that the observed behavior is a bug, in the narrow sense that the observed behavior is different from what the code author intended. See also my comments on SF bug # 740134. Here is a different patch. I like this one better than the previous one. In DOSUM: ;; ORIGINAL: MEVAL ONCE ;; (mbinding (lind l*i) (meval exp)) (mbinding (lind l*i) (meval (meval exp))) In MEVALSUMARG: ;; ORIGINAL: MEVALATOMS ;; (resimplify (mevalatoms (if (and (not (atom exp)) (resimplify (meval (if (and (not (atom exp)) With these changes, the results are as expected for the examples cited by the original poster, and also the example about the translated function produces the correct result, and some examples adapted from another bug report (740134) yield correct results. ex 1: f(x):=x^2$ b: 1/10$ Ak: b*f(k*b)$ sum(Ak,k,0,9); => 57/200 ex 2: A(k):=k*f(k*b)$ b:1/n$ A(k) => k^3/n^2 sum(A(k),k,0,n1) => ('sum(k^3,k,0,n1))/n^2 sum(A(k),k,0,n1), simpsum => ((n1)^4 + 2*(n1)^3 + (n1)^2) / (4*n^2) ex 3: ak : k^2$ g(a,n) := sum(a,k,1,n)$ g(ak,5) => 55 translate (g)$ g(ak,5) => 55 some other examples, adapted from bug # 740134: ex 4: sum (print (i), i, 1, 3) => prints 1, 2, 3 then returns 6 ex 5: sum (integrate (x^i ,x), i, 0, 2) => x^3/3 + x^2/2 + x ex 6: sum (integrate (1/(x^i + 1), x), i, 0, 1) => log(x+1) + x/2 ex 7: f[i](x):=x^i$ g[i](x):=x^i$ h[i](x):=x^i$ /* reference f[i] and g[i]  see 740134 for the effect this has on previous defn of sum */ f[i]$ g[i](t)$ sum (f[i](x), i, 0, n) => 'sum (x^i, i, 0, n) sum (g[i](x), i, 0, n) => 'sum (x^i, i, 0, n) sum (h[i](x), i, 0, n) => 'sum (x^i, i, 0, n) ex 8: sum (integrate (x^i, x), i, 0, n) => 'sum (x^(i+1) / (i+1), i, 0, n)  Comment By: Barton Willis (willisbl) Date: 20050613 05:30 Message: Logged In: YES user_id=895922 If we decide to use Robert's $sum function, the translate property for sum will need to be changed. Consider: (1) Redefine $sum as Robert suggested. (2) Try this: (%i1) ak : k^2$ (%i2) g(a,n) := sum(a,k,1,n)$ (%i3) g(ak,5); (%o3) 55 < correct for Robert's $sum function (%i4) translate(g); (%o4) [g] (%i5) g(ak,5); (%o5) 5*k^2 < yeech There might be other things that need fixing: (%i8) properties(sum); (%o8) [Special Evaluation Form,OUTATIVE,NOUN,RULE] Barton  Comment By: Robert Dodier (robert_dodier) Date: 20050612 20:34 Message: Logged In: YES user_id=501686 OK, for the record, here is a definition of $sum which implements one of the ideas that has been proposed, namely this policy: evaluate the summand after binding the summation variable to itself. (defmspec $sum (l) (setq l (cdr l)) (if (= (length l) 4) (progv (list (cadr l)) (list (cadr l)) (dosum (meval (car l)) (cadr l) (meval (caddr l)) (meval (cadddr l)) t)) (wnaerr '$sum))) This version yields the results expected by the person who originated this bug report, and yields the results predicted by S Macrakis in the email referenced below (http://www.math.utexas.edu/pipermail/maxima/2003/004869.html), and run_testsuite() runs to completion with no errors. I'm in favor of making this change, but I'm just recording this code snippet here for future reference; no changes planned at the moment. For comparison here is the current version of $sum: (defmspec $sum (l) (setq l (cdr l)) (if (= (length l) 4) (dosum (car l) (cadr l) (meval (caddr l)) (meval (cadddr l)) t) (wnaerr '$sum)))  Comment By: Barton Willis (willisbl) Date: 20050501 06:09 Message: Logged In: YES user_id=895922 We've discussed this before; see http://www.math.utexas.edu/pipermail/maxima/2003/004869.html http://www.math.utexas.edu/pipermail/maxima/2003/004870.html Maybe a simple workaround such as mysum(f,v,lo,hi):=block([acc:0], if integerp(lo) and integerp(hi) then for i from lo thru hi do acc:acc+substitute(i,v,f) else acc:funmake('mysum,[f,v,lo,hi]), acc) or mysum(f,lo,hi):= block([acc:0], if integerp(lo) and integerp(hi) then for i : lo thru hi do acc : acc + apply(f,[i]) else acc : funmake('mysum,[f,lo,hi]), acc); might work for you, Barton  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1192935&group_id=4933 
From: SourceForge.net <noreply@so...>  20051124 08:39:46

Bugs item #1192935, was opened at 20050430 05:10 Message generated for change (Settings changed) made by robert_dodier You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1192935&group_id=4933 Please note that this message will contain a full copy of the comment thread, including the initial issue submission, for this request, not just the latest update. Category: Lisp Core Group: None >Status: Closed Resolution: None Priority: 5 Submitted By: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: termsubstitution in function sum is too late Initial Comment: (%i1) "Lower RiemannSum, Intervall [0,1], 10 parts"$ (%i2) /*Function*/ f(x):=x^2$ (%i3) /*area width*/ b: 1/10$ (%i4) /*area depending on k=0,...,9*/ Ak: b*f(k*b)$ (%i5) Ak; 2 k (%o5)  1000 (%i6) /* wrong */ sum(Ak,k,0,9); 2 k (%o6)  100 obviously, we first have the summation and then the symbolsubstitution (%i7) sum(''Ak,k,0,9); 57 (%o7)  200 finally, this is ok, but the usage of the doublequote is hard to explain to beginners. it gets even worse, if you use a function A(k):=k*f(k*b)$ for a sum with n partitions: (%i8) "Lower RiemannSum, Intervall [0,1], n parts"$ (%i9) /*area width*/ b:1/n$ (%i10) /*area depending on k=0,...,n1*/ A(k):=k*f(k*b)$ (%i11) A(k); 3 k (%o11)  2 n (%i12) sum(A(k),k,0,n1); n  1 ==== (%o12) > A(k) / ==== k = 0 in that case you need ''(A(k)), a bracket and the doublequote. is it possible to patch the sumfunction in that way, that we first have the term or symbolsubstitution and second the summation?  Comment By: Robert Dodier (robert_dodier) Date: 20050819 23:08 Message: Logged In: YES user_id=501686 removing old versions of files  Comment By: Robert Dodier (robert_dodier) Date: 20050819 23:03 Message: Logged In: YES user_id=501686 attaching a zip file containing unksum6.lisp, tmpsum13.lisp (two different sum reimplementations), rtestsum.mac, and mand.lisp (a reimplementation of mcondrelated stuff  needed for the one test which makes use of an unevaluated conditional).  Comment By: Robert Dodier (robert_dodier) Date: 20050626 18:30 Message: Logged In: YES user_id=501686 attempting to remove old version of rtestsum.mac and upload new version in the same update... let's see if it works.  Comment By: Robert Dodier (robert_dodier) Date: 20050621 00:18 Message: Logged In: YES user_id=501686 New DOSUM, etc, in tmpsum8.lisp. A few new sum tests have been added to rtestsum.mac, and all sum tests copied and turned into product tests. This revision passes all tests in current rtestsum.mac except for test 38: (assume(i < 0, j < 0, n > 0), product(product(abs(j) + abs(i), i, 1, j), j, 1, n)); => Lower bound to product is > upper bound. This emanates from SIMPROD1 (src/combin.lisp) which is called after the assumptions about i and j (actually, assumptions about the gensym standins for i and j) have been forgotten. This looks like a bug in combin.lisp  there is no assume/forget about the indices.  Comment By: Robert Dodier (robert_dodier) Date: 20050621 00:06 Message: Logged In: YES user_id=501686 New rtestsum.mac. A couple of new tests for sum, and also copied all sum tests and substituted product for sum (and adjusted expected outputs to suit).  Comment By: Robert Dodier (robert_dodier) Date: 20050621 00:03 Message: Logged In: YES user_id=501686 Delete rtestsum.mac, going to upload a new one momentarily.  Comment By: Robert Dodier (robert_dodier) Date: 20050619 23:27 Message: Logged In: YES user_id=501686 About sum(rat(k),k,1,3), calling $FREEOF instead of FREEOF makes it happy again. Oh well. I think I'll just hang onto the current rev for a day or two to see what other changes accumulate, before posting a new version of tmpsum*.lisp. A related bit  sum(rat(k), k, 1, n) returns 'sum (k, k, 1, n), which isn't exactly wrong, but it does make me wonder what happened to the rat... it turns out this is what happened: subst (foo, k, rat(k)) => foo Not rat(foo) as I would have expected. Dunno what this is about.  Comment By: Barton Willis (willisbl) Date: 20050619 21:36 Message: Logged In: YES user_id=895922 Does tmpsum6.lisp need a ratdisrep somewhere? (%i1) load("c:/maxima/tmpsum6.lisp")$ (%i2) sum(k,k,1,3); (%o2) 6 (%i3) sum(rat(k),k,1,3); Maxima encountered a Lisp error: Error in COND [or a callee]: Caught fatal error [memory may be damaged] Automatically continuing. To reenable the Lisp debugger set *debuggerhook* to nil. (%i4) Barton  Comment By: Robert Dodier (robert_dodier) Date: 20050619 20:28 Message: Logged In: YES user_id=501686 Yet another revision (attached file tmpsum6.lisp) of DOSUM and friends. This version passes run_testsuite() and batch ("rtestsum.mac", test) using the currently attached version of rtestsum.mac. This version avoids substituting for the index in expressions which are free of the index. This version also has an alternate definition of the function FREE. The simplification code (and other code) calls FREE instead of FREEOF. FREEOF looks for dummy variables, but FREE does not, so with the original defn of FREE, sum (lambda([i], i^2), i, 1, n) does not simplify to n lambda ([i], i^2), i, 1, n). I'm not very happy about changing FREE since it is called from all over the place, but I'm also not very happy about the presence of FREE, either; why not call FREEOF ? I don't think changing FREE at this point is such a good idea; it seems to me the two choices are (1) inspect the sum/product simplification code and change FREE to FREEOF where possible. That would be a lot of work. (2) Leave FREE as it is, and just accept that expressions with dummy variables can't be factored out of summations. At this point I think I'd rather leave FREE alone, and file a separate bug report about the inability to factor out expressions with dummy variables, and just leave it for another day.  Comment By: Robert Dodier (robert_dodier) Date: 20050619 20:04 Message: Logged In: YES user_id=501686 New version of rtestsum.mac. Two new test cases, sum (lambda ([i], i^2, 1, 3) and sum (lambda ([i], i^2), i, 1, n).  Comment By: Robert Dodier (robert_dodier) Date: 20050619 19:59 Message: Logged In: YES user_id=501686 Removing rtestsum.mac, will upload a new version momentarily.  Comment By: Robert Dodier (robert_dodier) Date: 20050619 10:52 Message: Logged In: YES user_id=501686 I'm attaching my latest and greatest revision (tmpsum5.lisp) of DOSUM, DO%SUM, and MEVALSUMARGS. I find that load ("tmpsum5.lisp"); batch ("rtestsum.mac", test); where rtestsum.mac is the currently attached version, yields all tests passed for Maxima 5.9.1cvs built with clisp 2.33.2 and with gcl 2.6.6. This revision uses a gensym index for cases handled by MEVALSUMARGS (i.e., cases for which (upper limit) minus (lower limit) is not an integer); the gensym index is the argument for assume/forget. When upper minus lower is an integer, the index is bound to lower, lower+1, lower+2, etc., the summand is evaluated, and then lower, lower+1, lower+2, etc. is substituted for any remaining instances of index.  Comment By: Robert Dodier (robert_dodier) Date: 20050619 10:33 Message: Logged In: YES user_id=501686 Attaching a longer list of test cases. batch ("rtestsum.mac", test); executes the test. As always the "correct" results (2nd line in each pair) are subject to review.  Comment By: Robert Dodier (robert_dodier) Date: 20050619 10:28 Message: Logged In: YES user_id=501686 I'm removing rtestsum.mac, in anticipation of attaching a new version in a minute.  Comment By: van_Nek (van_nek) Date: 20050618 11:42 Message: Logged In: YES user_id=1269745 Dear Maximadevelopers, thank you for your support. I tried the last fix from Robert Dodier and it works. Volker van Nek original (anonymous) poster  Comment By: Robert Dodier (robert_dodier) Date: 20050615 08:37 Message: Logged In: YES user_id=501686 I'm attaching a list of test cases for sum, which can be executed by batch ("rtestsum.mac", test) . Doubtless there will be some discussion as to what the expected output of each test should be. All tests pass after making these 2 changes: ;; ORIGINAL: MEVAL ONCE ;; (mbinding (lind l*i) (meval exp)) ;; TRIED THIS: MEVAL TWICE ;; (mbinding (lind l*i) (meval (meval exp))) ;; THIRD TIME IS A CHARM ?? ($substitute *i ind (mbinding (lind l*i) (meval exp))) ;; ORIGINAL: MEVALATOMS ;; (resimplify (mevalatoms (if (and (not (atom exp)) (resimplify (meval (if (and (not (atom exp))  Comment By: Robert Dodier (robert_dodier) Date: 20050614 08:42 Message: Logged In: YES user_id=501686 OK, here is another attempt. I think I now understand this comment by Stavros in bug report # 740134  "First evaluate foo with i bound to itself, and check if that result is free of i. If so, return the product. If not, *substitute* (don't evaluate) i=lowerlimit, i=lowerlimit+1, etc." ;; ORIGINAL: MEVAL ONCE ;; (mbinding (lind l*i) (meval exp)) ;; TRIED THIS: MEVAL TWICE ;; (mbinding (lind l*i) (meval (meval exp))) ;; THIRD TIME IS A CHARM ?? ($substitute *i ind (mbinding (lind l*i) (meval exp))) This version now yields this result  ex 9: f(x) := sum (x, i, 1, 3); f(x) => 3 x as expected, as well as the same results for ex 1 through ex 8.  Comment By: Robert Dodier (robert_dodier) Date: 20050613 23:00 Message: Logged In: YES user_id=501686 Actually, after looking at the code some more (specifically DOSUM in src/asum.lisp) it looks to me like the intent is indeed to evaluate the summand after binding the summation index to itself. In addition, there is code to assume the index is between its lower and upper limits (although assume/forget is handled incorrectly, see 851765). I believe that the observed behavior is a bug, in the narrow sense that the observed behavior is different from what the code author intended. See also my comments on SF bug # 740134. Here is a different patch. I like this one better than the previous one. In DOSUM: ;; ORIGINAL: MEVAL ONCE ;; (mbinding (lind l*i) (meval exp)) (mbinding (lind l*i) (meval (meval exp))) In MEVALSUMARG: ;; ORIGINAL: MEVALATOMS ;; (resimplify (mevalatoms (if (and (not (atom exp)) (resimplify (meval (if (and (not (atom exp)) With these changes, the results are as expected for the examples cited by the original poster, and also the example about the translated function produces the correct result, and some examples adapted from another bug report (740134) yield correct results. ex 1: f(x):=x^2$ b: 1/10$ Ak: b*f(k*b)$ sum(Ak,k,0,9); => 57/200 ex 2: A(k):=k*f(k*b)$ b:1/n$ A(k) => k^3/n^2 sum(A(k),k,0,n1) => ('sum(k^3,k,0,n1))/n^2 sum(A(k),k,0,n1), simpsum => ((n1)^4 + 2*(n1)^3 + (n1)^2) / (4*n^2) ex 3: ak : k^2$ g(a,n) := sum(a,k,1,n)$ g(ak,5) => 55 translate (g)$ g(ak,5) => 55 some other examples, adapted from bug # 740134: ex 4: sum (print (i), i, 1, 3) => prints 1, 2, 3 then returns 6 ex 5: sum (integrate (x^i ,x), i, 0, 2) => x^3/3 + x^2/2 + x ex 6: sum (integrate (1/(x^i + 1), x), i, 0, 1) => log(x+1) + x/2 ex 7: f[i](x):=x^i$ g[i](x):=x^i$ h[i](x):=x^i$ /* reference f[i] and g[i]  see 740134 for the effect this has on previous defn of sum */ f[i]$ g[i](t)$ sum (f[i](x), i, 0, n) => 'sum (x^i, i, 0, n) sum (g[i](x), i, 0, n) => 'sum (x^i, i, 0, n) sum (h[i](x), i, 0, n) => 'sum (x^i, i, 0, n) ex 8: sum (integrate (x^i, x), i, 0, n) => 'sum (x^(i+1) / (i+1), i, 0, n)  Comment By: Barton Willis (willisbl) Date: 20050613 05:30 Message: Logged In: YES user_id=895922 If we decide to use Robert's $sum function, the translate property for sum will need to be changed. Consider: (1) Redefine $sum as Robert suggested. (2) Try this: (%i1) ak : k^2$ (%i2) g(a,n) := sum(a,k,1,n)$ (%i3) g(ak,5); (%o3) 55 < correct for Robert's $sum function (%i4) translate(g); (%o4) [g] (%i5) g(ak,5); (%o5) 5*k^2 < yeech There might be other things that need fixing: (%i8) properties(sum); (%o8) [Special Evaluation Form,OUTATIVE,NOUN,RULE] Barton  Comment By: Robert Dodier (robert_dodier) Date: 20050612 20:34 Message: Logged In: YES user_id=501686 OK, for the record, here is a definition of $sum which implements one of the ideas that has been proposed, namely this policy: evaluate the summand after binding the summation variable to itself. (defmspec $sum (l) (setq l (cdr l)) (if (= (length l) 4) (progv (list (cadr l)) (list (cadr l)) (dosum (meval (car l)) (cadr l) (meval (caddr l)) (meval (cadddr l)) t)) (wnaerr '$sum))) This version yields the results expected by the person who originated this bug report, and yields the results predicted by S Macrakis in the email referenced below (http://www.math.utexas.edu/pipermail/maxima/2003/004869.html), and run_testsuite() runs to completion with no errors. I'm in favor of making this change, but I'm just recording this code snippet here for future reference; no changes planned at the moment. For comparison here is the current version of $sum: (defmspec $sum (l) (setq l (cdr l)) (if (= (length l) 4) (dosum (car l) (cadr l) (meval (caddr l)) (meval (cadddr l)) t) (wnaerr '$sum)))  Comment By: Barton Willis (willisbl) Date: 20050501 06:09 Message: Logged In: YES user_id=895922 We've discussed this before; see http://www.math.utexas.edu/pipermail/maxima/2003/004869.html http://www.math.utexas.edu/pipermail/maxima/2003/004870.html Maybe a simple workaround such as mysum(f,v,lo,hi):=block([acc:0], if integerp(lo) and integerp(hi) then for i from lo thru hi do acc:acc+substitute(i,v,f) else acc:funmake('mysum,[f,v,lo,hi]), acc) or mysum(f,lo,hi):= block([acc:0], if integerp(lo) and integerp(hi) then for i : lo thru hi do acc : acc + apply(f,[i]) else acc : funmake('mysum,[f,lo,hi]), acc); might work for you, Barton  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1192935&group_id=4933 
From: SourceForge.net <noreply@so...>  20051124 08:43:06

Bugs item #1192935, was opened at 20050430 05:10 Message generated for change (Comment added) made by robert_dodier You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1192935&group_id=4933 Please note that this message will contain a full copy of the comment thread, including the initial issue submission, for this request, not just the latest update. Category: Lisp Core Group: None Status: Closed Resolution: None Priority: 5 Submitted By: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: termsubstitution in function sum is too late Initial Comment: (%i1) "Lower RiemannSum, Intervall [0,1], 10 parts"$ (%i2) /*Function*/ f(x):=x^2$ (%i3) /*area width*/ b: 1/10$ (%i4) /*area depending on k=0,...,9*/ Ak: b*f(k*b)$ (%i5) Ak; 2 k (%o5)  1000 (%i6) /* wrong */ sum(Ak,k,0,9); 2 k (%o6)  100 obviously, we first have the summation and then the symbolsubstitution (%i7) sum(''Ak,k,0,9); 57 (%o7)  200 finally, this is ok, but the usage of the doublequote is hard to explain to beginners. it gets even worse, if you use a function A(k):=k*f(k*b)$ for a sum with n partitions: (%i8) "Lower RiemannSum, Intervall [0,1], n parts"$ (%i9) /*area width*/ b:1/n$ (%i10) /*area depending on k=0,...,n1*/ A(k):=k*f(k*b)$ (%i11) A(k); 3 k (%o11)  2 n (%i12) sum(A(k),k,0,n1); n  1 ==== (%o12) > A(k) / ==== k = 0 in that case you need ''(A(k)), a bracket and the doublequote. is it possible to patch the sumfunction in that way, that we first have the term or symbolsubstitution and second the summation?  Comment By: Robert Dodier (robert_dodier) Date: 20051124 01:43 Message: Logged In: YES user_id=501686 The reported bug is not present in the current cvs version of Maxima. Thank you for your report. If you see this bug in a later version of Maxima, please submit a new bug report.  Comment By: Robert Dodier (robert_dodier) Date: 20050819 23:08 Message: Logged In: YES user_id=501686 removing old versions of files  Comment By: Robert Dodier (robert_dodier) Date: 20050819 23:03 Message: Logged In: YES user_id=501686 attaching a zip file containing unksum6.lisp, tmpsum13.lisp (two different sum reimplementations), rtestsum.mac, and mand.lisp (a reimplementation of mcondrelated stuff  needed for the one test which makes use of an unevaluated conditional).  Comment By: Robert Dodier (robert_dodier) Date: 20050626 18:30 Message: Logged In: YES user_id=501686 attempting to remove old version of rtestsum.mac and upload new version in the same update... let's see if it works.  Comment By: Robert Dodier (robert_dodier) Date: 20050621 00:18 Message: Logged In: YES user_id=501686 New DOSUM, etc, in tmpsum8.lisp. A few new sum tests have been added to rtestsum.mac, and all sum tests copied and turned into product tests. This revision passes all tests in current rtestsum.mac except for test 38: (assume(i < 0, j < 0, n > 0), product(product(abs(j) + abs(i), i, 1, j), j, 1, n)); => Lower bound to product is > upper bound. This emanates from SIMPROD1 (src/combin.lisp) which is called after the assumptions about i and j (actually, assumptions about the gensym standins for i and j) have been forgotten. This looks like a bug in combin.lisp  there is no assume/forget about the indices.  Comment By: Robert Dodier (robert_dodier) Date: 20050621 00:06 Message: Logged In: YES user_id=501686 New rtestsum.mac. A couple of new tests for sum, and also copied all sum tests and substituted product for sum (and adjusted expected outputs to suit).  Comment By: Robert Dodier (robert_dodier) Date: 20050621 00:03 Message: Logged In: YES user_id=501686 Delete rtestsum.mac, going to upload a new one momentarily.  Comment By: Robert Dodier (robert_dodier) Date: 20050619 23:27 Message: Logged In: YES user_id=501686 About sum(rat(k),k,1,3), calling $FREEOF instead of FREEOF makes it happy again. Oh well. I think I'll just hang onto the current rev for a day or two to see what other changes accumulate, before posting a new version of tmpsum*.lisp. A related bit  sum(rat(k), k, 1, n) returns 'sum (k, k, 1, n), which isn't exactly wrong, but it does make me wonder what happened to the rat... it turns out this is what happened: subst (foo, k, rat(k)) => foo Not rat(foo) as I would have expected. Dunno what this is about.  Comment By: Barton Willis (willisbl) Date: 20050619 21:36 Message: Logged In: YES user_id=895922 Does tmpsum6.lisp need a ratdisrep somewhere? (%i1) load("c:/maxima/tmpsum6.lisp")$ (%i2) sum(k,k,1,3); (%o2) 6 (%i3) sum(rat(k),k,1,3); Maxima encountered a Lisp error: Error in COND [or a callee]: Caught fatal error [memory may be damaged] Automatically continuing. To reenable the Lisp debugger set *debuggerhook* to nil. (%i4) Barton  Comment By: Robert Dodier (robert_dodier) Date: 20050619 20:28 Message: Logged In: YES user_id=501686 Yet another revision (attached file tmpsum6.lisp) of DOSUM and friends. This version passes run_testsuite() and batch ("rtestsum.mac", test) using the currently attached version of rtestsum.mac. This version avoids substituting for the index in expressions which are free of the index. This version also has an alternate definition of the function FREE. The simplification code (and other code) calls FREE instead of FREEOF. FREEOF looks for dummy variables, but FREE does not, so with the original defn of FREE, sum (lambda([i], i^2), i, 1, n) does not simplify to n lambda ([i], i^2), i, 1, n). I'm not very happy about changing FREE since it is called from all over the place, but I'm also not very happy about the presence of FREE, either; why not call FREEOF ? I don't think changing FREE at this point is such a good idea; it seems to me the two choices are (1) inspect the sum/product simplification code and change FREE to FREEOF where possible. That would be a lot of work. (2) Leave FREE as it is, and just accept that expressions with dummy variables can't be factored out of summations. At this point I think I'd rather leave FREE alone, and file a separate bug report about the inability to factor out expressions with dummy variables, and just leave it for another day.  Comment By: Robert Dodier (robert_dodier) Date: 20050619 20:04 Message: Logged In: YES user_id=501686 New version of rtestsum.mac. Two new test cases, sum (lambda ([i], i^2, 1, 3) and sum (lambda ([i], i^2), i, 1, n).  Comment By: Robert Dodier (robert_dodier) Date: 20050619 19:59 Message: Logged In: YES user_id=501686 Removing rtestsum.mac, will upload a new version momentarily.  Comment By: Robert Dodier (robert_dodier) Date: 20050619 10:52 Message: Logged In: YES user_id=501686 I'm attaching my latest and greatest revision (tmpsum5.lisp) of DOSUM, DO%SUM, and MEVALSUMARGS. I find that load ("tmpsum5.lisp"); batch ("rtestsum.mac", test); where rtestsum.mac is the currently attached version, yields all tests passed for Maxima 5.9.1cvs built with clisp 2.33.2 and with gcl 2.6.6. This revision uses a gensym index for cases handled by MEVALSUMARGS (i.e., cases for which (upper limit) minus (lower limit) is not an integer); the gensym index is the argument for assume/forget. When upper minus lower is an integer, the index is bound to lower, lower+1, lower+2, etc., the summand is evaluated, and then lower, lower+1, lower+2, etc. is substituted for any remaining instances of index.  Comment By: Robert Dodier (robert_dodier) Date: 20050619 10:33 Message: Logged In: YES user_id=501686 Attaching a longer list of test cases. batch ("rtestsum.mac", test); executes the test. As always the "correct" results (2nd line in each pair) are subject to review.  Comment By: Robert Dodier (robert_dodier) Date: 20050619 10:28 Message: Logged In: YES user_id=501686 I'm removing rtestsum.mac, in anticipation of attaching a new version in a minute.  Comment By: van_Nek (van_nek) Date: 20050618 11:42 Message: Logged In: YES user_id=1269745 Dear Maximadevelopers, thank you for your support. I tried the last fix from Robert Dodier and it works. Volker van Nek original (anonymous) poster  Comment By: Robert Dodier (robert_dodier) Date: 20050615 08:37 Message: Logged In: YES user_id=501686 I'm attaching a list of test cases for sum, which can be executed by batch ("rtestsum.mac", test) . Doubtless there will be some discussion as to what the expected output of each test should be. All tests pass after making these 2 changes: ;; ORIGINAL: MEVAL ONCE ;; (mbinding (lind l*i) (meval exp)) ;; TRIED THIS: MEVAL TWICE ;; (mbinding (lind l*i) (meval (meval exp))) ;; THIRD TIME IS A CHARM ?? ($substitute *i ind (mbinding (lind l*i) (meval exp))) ;; ORIGINAL: MEVALATOMS ;; (resimplify (mevalatoms (if (and (not (atom exp)) (resimplify (meval (if (and (not (atom exp))  Comment By: Robert Dodier (robert_dodier) Date: 20050614 08:42 Message: Logged In: YES user_id=501686 OK, here is another attempt. I think I now understand this comment by Stavros in bug report # 740134  "First evaluate foo with i bound to itself, and check if that result is free of i. If so, return the product. If not, *substitute* (don't evaluate) i=lowerlimit, i=lowerlimit+1, etc." ;; ORIGINAL: MEVAL ONCE ;; (mbinding (lind l*i) (meval exp)) ;; TRIED THIS: MEVAL TWICE ;; (mbinding (lind l*i) (meval (meval exp))) ;; THIRD TIME IS A CHARM ?? ($substitute *i ind (mbinding (lind l*i) (meval exp))) This version now yields this result  ex 9: f(x) := sum (x, i, 1, 3); f(x) => 3 x as expected, as well as the same results for ex 1 through ex 8.  Comment By: Robert Dodier (robert_dodier) Date: 20050613 23:00 Message: Logged In: YES user_id=501686 Actually, after looking at the code some more (specifically DOSUM in src/asum.lisp) it looks to me like the intent is indeed to evaluate the summand after binding the summation index to itself. In addition, there is code to assume the index is between its lower and upper limits (although assume/forget is handled incorrectly, see 851765). I believe that the observed behavior is a bug, in the narrow sense that the observed behavior is different from what the code author intended. See also my comments on SF bug # 740134. Here is a different patch. I like this one better than the previous one. In DOSUM: ;; ORIGINAL: MEVAL ONCE ;; (mbinding (lind l*i) (meval exp)) (mbinding (lind l*i) (meval (meval exp))) In MEVALSUMARG: ;; ORIGINAL: MEVALATOMS ;; (resimplify (mevalatoms (if (and (not (atom exp)) (resimplify (meval (if (and (not (atom exp)) With these changes, the results are as expected for the examples cited by the original poster, and also the example about the translated function produces the correct result, and some examples adapted from another bug report (740134) yield correct results. ex 1: f(x):=x^2$ b: 1/10$ Ak: b*f(k*b)$ sum(Ak,k,0,9); => 57/200 ex 2: A(k):=k*f(k*b)$ b:1/n$ A(k) => k^3/n^2 sum(A(k),k,0,n1) => ('sum(k^3,k,0,n1))/n^2 sum(A(k),k,0,n1), simpsum => ((n1)^4 + 2*(n1)^3 + (n1)^2) / (4*n^2) ex 3: ak : k^2$ g(a,n) := sum(a,k,1,n)$ g(ak,5) => 55 translate (g)$ g(ak,5) => 55 some other examples, adapted from bug # 740134: ex 4: sum (print (i), i, 1, 3) => prints 1, 2, 3 then returns 6 ex 5: sum (integrate (x^i ,x), i, 0, 2) => x^3/3 + x^2/2 + x ex 6: sum (integrate (1/(x^i + 1), x), i, 0, 1) => log(x+1) + x/2 ex 7: f[i](x):=x^i$ g[i](x):=x^i$ h[i](x):=x^i$ /* reference f[i] and g[i]  see 740134 for the effect this has on previous defn of sum */ f[i]$ g[i](t)$ sum (f[i](x), i, 0, n) => 'sum (x^i, i, 0, n) sum (g[i](x), i, 0, n) => 'sum (x^i, i, 0, n) sum (h[i](x), i, 0, n) => 'sum (x^i, i, 0, n) ex 8: sum (integrate (x^i, x), i, 0, n) => 'sum (x^(i+1) / (i+1), i, 0, n)  Comment By: Barton Willis (willisbl) Date: 20050613 05:30 Message: Logged In: YES user_id=895922 If we decide to use Robert's $sum function, the translate property for sum will need to be changed. Consider: (1) Redefine $sum as Robert suggested. (2) Try this: (%i1) ak : k^2$ (%i2) g(a,n) := sum(a,k,1,n)$ (%i3) g(ak,5); (%o3) 55 < correct for Robert's $sum function (%i4) translate(g); (%o4) [g] (%i5) g(ak,5); (%o5) 5*k^2 < yeech There might be other things that need fixing: (%i8) properties(sum); (%o8) [Special Evaluation Form,OUTATIVE,NOUN,RULE] Barton  Comment By: Robert Dodier (robert_dodier) Date: 20050612 20:34 Message: Logged In: YES user_id=501686 OK, for the record, here is a definition of $sum which implements one of the ideas that has been proposed, namely this policy: evaluate the summand after binding the summation variable to itself. (defmspec $sum (l) (setq l (cdr l)) (if (= (length l) 4) (progv (list (cadr l)) (list (cadr l)) (dosum (meval (car l)) (cadr l) (meval (caddr l)) (meval (cadddr l)) t)) (wnaerr '$sum))) This version yields the results expected by the person who originated this bug report, and yields the results predicted by S Macrakis in the email referenced below (http://www.math.utexas.edu/pipermail/maxima/2003/004869.html), and run_testsuite() runs to completion with no errors. I'm in favor of making this change, but I'm just recording this code snippet here for future reference; no changes planned at the moment. For comparison here is the current version of $sum: (defmspec $sum (l) (setq l (cdr l)) (if (= (length l) 4) (dosum (car l) (cadr l) (meval (caddr l)) (meval (cadddr l)) t) (wnaerr '$sum)))  Comment By: Barton Willis (willisbl) Date: 20050501 06:09 Message: Logged In: YES user_id=895922 We've discussed this before; see http://www.math.utexas.edu/pipermail/maxima/2003/004869.html http://www.math.utexas.edu/pipermail/maxima/2003/004870.html Maybe a simple workaround such as mysum(f,v,lo,hi):=block([acc:0], if integerp(lo) and integerp(hi) then for i from lo thru hi do acc:acc+substitute(i,v,f) else acc:funmake('mysum,[f,v,lo,hi]), acc) or mysum(f,lo,hi):= block([acc:0], if integerp(lo) and integerp(hi) then for i : lo thru hi do acc : acc + apply(f,[i]) else acc : funmake('mysum,[f,lo,hi]), acc); might work for you, Barton  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1192935&group_id=4933 
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