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From: SourceForge.net <noreply@so...>  20091221 12:27:22

Bugs item #719832, was opened at 20030411 19:53 Message generated for change (Comment added) made by crategus You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=719832&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  Limit Group: None Status: Open >Resolution: None Priority: 5 Private: No Submitted By: Stavros Macrakis (macrakis) Assigned to: Nobody/Anonymous (nobody) >Summary: limit(exp(x*%i)*x,x,inf) should give infinity Initial Comment: limit(exp(x*%i)*x,x,inf) => UND NO! Should be INFINITY  >Comment By: Dieter Kaiser (crategus) Date: 20091221 13:27 Message: Changing the title to reflect the issue better. Setting the resolution back to "None". Dieter Kaiser  Comment By: Stavros Macrakis (macrakis) Date: 20091221 12:54 Message: A noun form is certainly better than und, but the correct result is Infinity. I would have thought that in the Wolfram world, the correct result would be ComplexInfinity (which corresponds to Maxima's Infinity).  Comment By: Dieter Kaiser (crategus) Date: 20091221 02:44 Message: The example of this bug report gives no longer 'und but a noun form (Maxima 5.20post); (%i11) limit(exp(x*%i)*x,x,inf); (%o11) 'limit(x*%e^(%i*x),x,inf) I am not sure what is the right answer. Wolfram alpha gives a result in terms of an interval: E^((2 I) Interval[{0, Pi}]) Infinity A noun form is not a wrong result. Perhaps, we can close this bug report at this point. Further improvements of the limit routines might give a more complete answer. Setting the status to pending and the resolution to "works for me". Dieter Kaiser  Comment By: Stavros Macrakis (macrakis) Date: 20030412 03:55 Message: Logged In: YES user_id=588346 I believe that the definition of limit(f(x))=infinity is that for all N, there exists an X such that x>X implies abs(f(x))>N. That is satisfied in this case. In fact, you can choose X=N. The separate magnitudes of the real and imaginary parts are irrelevant. After all, limit(2+x*%i,x,inf) = infinity  Comment By: Barton Willis (willisb) Date: 20030412 03:51 Message: Logged In: YES user_id=570592 Let F : R > C and F(x) = x exp(i x) = x cos(x) + i x sin(x). Both the real and imaginary parts of F are oscillatory with linearly growing amplitudes; neither the real nor the imaginary parts have a limit towards infinity. I say the limit is UND. Barton  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=719832&group_id=4933 
From: SourceForge.net <noreply@so...>  20091221 11:54:31

Bugs item #719832, was opened at 20030411 13:53 Message generated for change (Comment added) made by macrakis You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=719832&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  Limit Group: None >Status: Open Resolution: Works For Me Priority: 5 Private: No Submitted By: Stavros Macrakis (macrakis) Assigned to: Nobody/Anonymous (nobody) Summary: limit(exp(x*%i)*x,x,inf) => UND Initial Comment: limit(exp(x*%i)*x,x,inf) => UND NO! Should be INFINITY  >Comment By: Stavros Macrakis (macrakis) Date: 20091221 06:54 Message: A noun form is certainly better than und, but the correct result is Infinity. I would have thought that in the Wolfram world, the correct result would be ComplexInfinity (which corresponds to Maxima's Infinity).  Comment By: Dieter Kaiser (crategus) Date: 20091220 20:44 Message: The example of this bug report gives no longer 'und but a noun form (Maxima 5.20post); (%i11) limit(exp(x*%i)*x,x,inf); (%o11) 'limit(x*%e^(%i*x),x,inf) I am not sure what is the right answer. Wolfram alpha gives a result in terms of an interval: E^((2 I) Interval[{0, Pi}]) Infinity A noun form is not a wrong result. Perhaps, we can close this bug report at this point. Further improvements of the limit routines might give a more complete answer. Setting the status to pending and the resolution to "works for me". Dieter Kaiser  Comment By: Stavros Macrakis (macrakis) Date: 20030411 21:55 Message: Logged In: YES user_id=588346 I believe that the definition of limit(f(x))=infinity is that for all N, there exists an X such that x>X implies abs(f(x))>N. That is satisfied in this case. In fact, you can choose X=N. The separate magnitudes of the real and imaginary parts are irrelevant. After all, limit(2+x*%i,x,inf) = infinity  Comment By: Barton Willis (willisb) Date: 20030411 21:51 Message: Logged In: YES user_id=570592 Let F : R > C and F(x) = x exp(i x) = x cos(x) + i x sin(x). Both the real and imaginary parts of F are oscillatory with linearly growing amplitudes; neither the real nor the imaginary parts have a limit towards infinity. I say the limit is UND. Barton  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=719832&group_id=4933 
From: SourceForge.net <noreply@so...>  20091221 01:44:26

Bugs item #719832, was opened at 20030411 19:53 Message generated for change (Comment added) made by crategus You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=719832&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  Limit Group: None >Status: Pending >Resolution: Works For Me Priority: 5 Private: No Submitted By: Stavros Macrakis (macrakis) Assigned to: Nobody/Anonymous (nobody) Summary: limit(exp(x*%i)*x,x,inf) => UND Initial Comment: limit(exp(x*%i)*x,x,inf) => UND NO! Should be INFINITY  >Comment By: Dieter Kaiser (crategus) Date: 20091221 02:44 Message: The example of this bug report gives no longer 'und but a noun form (Maxima 5.20post); (%i11) limit(exp(x*%i)*x,x,inf); (%o11) 'limit(x*%e^(%i*x),x,inf) I am not sure what is the right answer. Wolfram alpha gives a result in terms of an interval: E^((2 I) Interval[{0, Pi}]) Infinity A noun form is not a wrong result. Perhaps, we can close this bug report at this point. Further improvements of the limit routines might give a more complete answer. Setting the status to pending and the resolution to "works for me". Dieter Kaiser  Comment By: Stavros Macrakis (macrakis) Date: 20030412 03:55 Message: Logged In: YES user_id=588346 I believe that the definition of limit(f(x))=infinity is that for all N, there exists an X such that x>X implies abs(f(x))>N. That is satisfied in this case. In fact, you can choose X=N. The separate magnitudes of the real and imaginary parts are irrelevant. After all, limit(2+x*%i,x,inf) = infinity  Comment By: Barton Willis (willisb) Date: 20030412 03:51 Message: Logged In: YES user_id=570592 Let F : R > C and F(x) = x exp(i x) = x cos(x) + i x sin(x). Both the real and imaginary parts of F are oscillatory with linearly growing amplitudes; neither the real nor the imaginary parts have a limit towards infinity. I say the limit is UND. Barton  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=719832&group_id=4933 
From: SourceForge.net <noreply@so...>  20091221 00:19:33

Bugs item #655270, was opened at 20021217 17:41 Message generated for change (Comment added) made by crategus You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=655270&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  Integration Group: None >Status: Closed >Resolution: Fixed Priority: 5 Private: No Submitted By: Stavros Macrakis (macrakis) Assigned to: Nobody/Anonymous (nobody) Summary: Incomplete integration Initial Comment: integrate(sin(3*asin(x))) gives ((48*x^2*%E^(LOG(x+1)+LOG(1x))+48*x^448*x^2+12) *'INTEGRATE(xx^3,x)4*x^6*%E^(LOG(x+1)+LOG(1x)) 4*x^8+4*x^65*x^4+6*x^22)/(32*x^2*%E^(LOG(x+1) +LOG(1x))+32*x^432*x^2+8) which contains 'INTEGRATE(xx^3,x), which is obviously integrable. ev(...,integrate) successfully completes the integration, and ratsimp gets it down to something nice. Similar things happen with sin(a*asin(x)) for other a's. Note that trigexpand(sin(3*asin(x))) = 3*x*(1x^2)x^3. Low priority, because the answer is correct (if not ideal) and can easily be fixed.  >Comment By: Dieter Kaiser (crategus) Date: 20091221 01:19 Message: Because of revision 1.18 of risch.lisp and revision 1.53 of sin.lisp we get the following results: This is the example of this bug report: (%i2) integrate(sin(3*asin(x)),x); (%o2) (4*x^46*x^2+1)/4 One more example which is related and works too: (%i3) integrate(sin(4*asin(x)),x); (%o3) (15*((1x^2)^(3/2)/5x^2*(1x^2)^(3/2)/5) +sqrt(1x)*sqrt(x+1)*(93*x^4106*x^2+13)) /60 This integrals work in addition: (%i4) integrate(exp(asinh(x)),x); (%o4) log(sqrt(x^2+1)+x)/2+x*sqrt(x^2+1)/2+x^2/2 (%i5) integrate(exp(acosh(x)),x); (%o5) log(2*sqrt(x^21)+2*x)/2+x*sqrt(x^21)/2+(x^2/2+x)/2+(x^2/2x)/2 (%i6) integrate(exp(asech(x)),x); (%o6) log(2*sqrt(1x^2)/abs(x)+2/abs(x))+(log(x)+x)/2+(log(x)x)/2 +sqrt(1x^2) Closing this bug report as fixed. Dieter Kaiser  Comment By: Dieter Kaiser (crategus) Date: 20091108 16:20 Message: This is again the reported bug: (%i2) integrate(sin(3*asin(x)),x); (%o2) (12*'integrate(xx^3,x)5*x^4+6*x^22)/8 We get the complete result after an extra evaluation: (%i3) %,nouns; (%o3) (5*x^4+12*(x^2/2x^4/4)+6*x^22)/8 The risch algorithm adds up integrals which are not handled at some point of the algorithm. To do this the risch algorithm calls the routine rischnoun. The problem is that rischnoun generates solvable integrals. These integrals are part of the result of risch. In general it is not possible to call the integrator or risch again to avoid the construction of solvable integrals, because we can run into infinite loops. One small extension is possible immediately. We can look for a rational expression in rischnoun and call ratint to get the integral. The following code shows this: (defun rischnoun (exp1 &optional (exp2 exp1 exp2p)) (let (($logsimp t) ($%e_to_numlog t)) (unless exp2p (setq exp1 (rzero))) (if (ratp (setq exp2 (resimplify (disrep exp2))) intvar) ;; A rational expression which can be integrated by ratint. (setq exp2 (ratint exp2 intvar)) ;; A more general integrand. Return a noun form. (setq exp2 (list '(%integrate) exp2 intvar))) `(,exp1 ,exp2))) That is the example of the bug report: (%i5) integrate(sin(3*asin(x)),x); (%o5) (4*x^46*x^2+1)/4 More results for an odd factor in the integrand: (%i10) integrate(sin(5*asin(x)),x); (%o10) (32*x^660*x^4+30*x^21)/12 (%i11) integrate(sin(7*asin(x)),x); (%o11) (96*x^8224*x^6+168*x^442*x^2+1)/12 But it don't work completely for an even factor in the integrand. The reason is, that one of the integrands is not a rational expression: (%i12) integrate(sin(2*asin(x)),x); (%o12) (sqrt(1x)*sqrt(x+1)*(5*x^25)3*'integrate(sqrt(1x)*x*sqrt(x+1),x))/6 (%i13) integrate(sin(4*asin(x)),x); (%o13) (15*'integrate(sqrt(1x)*sqrt(x+1)*(x^3x),x) +sqrt(1x)*sqrt(x+1)*(93*x^4106*x^2+13)) /60 On the other hand we get more results, e.g. the following integrals now work: (%i7) integrate(exp(asinh(x)),x); (%o7) log(sqrt(x^2+1)+x)/2+x*sqrt(x^2+1)/2+x^2/2 (%i8) integrate(exp(acosh(x)),x); (%o8) log(2*sqrt(x^21)+2*x)/2+x*sqrt(x^21)/2+(x^2/2+x)/2+(x^2/2x)/2 (%i9) integrate(exp(atanh(x)),x); (%o9) %i*log(sqrt(1x^2)+%i*x)sqrt(1x^2) To get a more general solution we had to call the integrator in rischnoun again. To do this it is necessary to extend the integrator and risch with a mechanism to avoid infinite loops. Dieter Kaiser  Comment By: Raymond Toy (rtoy) Date: 20060410 19:39 Message: Logged In: YES user_id=28849 Tracing rischint indicates that this is a problem in the Risch integrator.  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=655270&group_id=4933 