[Maxima-bugs] [ maxima-Bugs-1192935 ] term-substitution in function sum is too late From: SourceForge.net - 2005-06-15 14:37:16 ```Bugs item #1192935, was opened at 2005-04-30 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: term-substitution in function sum is too late Initial Comment: (%i1) "Lower Riemann-Sum, 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 symbol-substitution (%i7) sum(''Ak,k,0,9); 57 (%o7) --- 200 finally, this is ok, but the usage of the double-quote 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 Riemann-Sum, Intervall [0,1], n parts"\$ (%i9) /*area width*/ b:1/n\$ (%i10) /*area depending on k=0,...,n-1*/ A(k):=k*f(k*b)\$ (%i11) A(k); 3 k (%o11) -- 2 n (%i12) sum(A(k),k,0,n-1); n - 1 ==== (%o12) > A(k) / ==== k = 0 in that case you need ''(A(k)), a bracket and the double-quote. is it possible to patch the sum-function in that way, that we first have the term- or symbol-substitution and second the summation? ---------------------------------------------------------------------- >Comment By: Robert Dodier (robert_dodier) Date: 2005-06-15 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: 2005-06-14 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: 2005-06-13 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,n-1) => ('sum(k^3,k,0,n-1))/n^2 sum(A(k),k,0,n-1), simpsum => ((n-1)^4 + 2*(n-1)^3 + (n-1)^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: 2005-06-13 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: 2005-06-12 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)) (wna-err '\$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) (wna-err '\$sum))) ---------------------------------------------------------------------- Comment By: Barton Willis (willisbl) Date: 2005-05-01 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 ```