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Math Forum / Math Software / Maple / October 2008Maple not able to do simple things
How can one make Maple do simple things without going to devices which should not be needed? The precise situation was that Maple was asked to differentiate .5*(coth(..5*x/(k+1) - coth(.5*x/k))/x with respect to k. It did this correctly, BUT left the derivative of coth as 1-coth^2, and did not cancel the x's in numerator and denominator. When I entered manually the simpler form, with these and multiplying the .5's together, it would not recognize them as equal. When asked to simplify, it came up with an awful mess, doing everything in terms of cosh. The simplified form of the two expressions it did recognize as equal. Maple does know about csch. Can a reasonable CAS not be able to handle anything this easy?  SignatureThis address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 > The precise situation was that Maple was asked to > differentiate [quoted text clipped - 4 lines] > the derivative of coth as 1-coth^2, and did not cancel > the x's in numerator and denominator. > When asked to simplify, it came up with an awful mess, > doing everything in terms of cosh. Is this more to your liking? diff(.5*(coth(.5*x/(k+1)) - coth(.5*x/k))/x, k): simplify(%, size); / 2 x 2 2 2 |.25 k coth(.5 -----) + (-.25 - .25 k - .50 k) coth(.5 x/k) \ k + 1 \ / 2 2 + .25 + .50 k| / ((k + 1) k ) / / Martin SignatureQuidquid latine scriptum est, altum videtur. >> The precise situation was that Maple was asked to >> differentiate >> .5*(coth(..5*x/(k+1) - coth(.5*x/k))/x >> with respect to k. It did this correctly, BUT left >> the derivative of coth as 1-coth^2, and did not cancel >> the x's in numerator and denominator. >> When asked to simplify, it came up with an awful mess, >> doing everything in terms of cosh. >Is this more to your liking? >diff(.5*(coth(.5*x/(k+1)) - coth(.5*x/k))/x, k): >simplify(%, size); > / 2 x 2 2 2 > |.25 k coth(.5 -----) + (-.25 - .25 k - .50 k) coth(.5 x/k) > \ k + 1 > \ / 2 2 > + .25 + .50 k| / ((k + 1) k ) > / / Not really. What I wanted is 0.5 x 2 0.5 x 2 0.25 csch(-----) 0.25 csch(-----) k + 1 k ----------------- - ----------------- 2 2 (k + 1) k which is the simple form to continue to be used, or even with the factor .25 multiplying the rest. Certainly replacing 1-coth^2 by -csch^2 is simplifying, and I do not see putting it over a common denominator as such. >Martin SignatureThis address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 >>> The precise situation was that Maple was asked to >>> differentiate >>> .5*(coth(..5*x/(k+1) - coth(.5*x/k))/x >>> with respect to k. It did this correctly, BUT left >>> the derivative of coth as 1-coth^2, and did not cancel >>> the x's in numerator and denominator. >>> When asked to simplify, it came up with an awful mess, >>> doing everything in terms of cosh. >>Is this more to your liking? >>diff(.5*(coth(.5*x/(k+1)) - coth(.5*x/k))/x, k): >>simplify(%, size); >> / 2 x 2 2 2 >> |.25 k coth(.5 -----) + (-.25 - .25 k - .50 k) coth(.5 x/k) >> \ k + 1 >> \ / 2 2 >> + .25 + .50 k| / ((k + 1) k ) >> / / >Not really. What I wanted is > 0.5 x 2 0.5 x 2 > 0.25 csch(-----) 0.25 csch(-----) > k + 1 k > ----------------- - ----------------- > 2 2 > (k + 1) k >which is the simple form to continue to be used, or even >with the factor .25 multiplying the rest. >Certainly replacing 1-coth^2 by -csch^2 is simplifying, >and I do not see putting it over a common denominator >as such. >>Martin I also tried again to see if Maple could see the equality of the expressions. The answer is no. SignatureThis address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 > I also tried again to see if Maple could see the > equality of the expressions. The answer is no. Here's how I would do that... > f := .5*(coth(.5*x/(k+1)) - coth(.5*x/k))/x; f := .5*(coth(.5*x/(k+1))-coth(.5*x/k))/x > maple_answer := diff(f,k); maple_answer := .5*(-.5*(1-coth(.5*x/(k+1))^2)*x/(k+1)^2+.5*(1-coth(.5*x/k)^2)*x/k^2)/x > rubin_answer := .25*csch(.5*x/(k+1))^2/(k+1)^2 - .25*csch(.5*x/k)^2/k^2; rubin_answer := .25*csch(.5*x/(k+1))^2/(k+1)^2-.25*csch(.5*x/k)^2/k^2 > simplify(maple_answer - rubin_answer); 0. [Actually, I would also use 1/2 and 1/4 instead of .5 and .25 ] SignatureG. A. Edgar http://www.math.ohio-state.edu/~edgar/ >>>> The precise situation was that Maple was asked to >>>> differentiate [quoted text clipped - 41 lines] > I also tried again to see if Maple could see the > equality of the expressions. The answer is no. 1. I do not actually know, what Maple does on trigonometric and hyperbolic functions, but it often seems difficult to force it into a desired result (especially if using 'simplify'). So I usually try to pass to exp and log. 2. To see where problems it in your case step in (from Maple's view) try diff( coth(t), t) as example: the answer is 1-coth(t)^2, and I guess you would expect -csch(t)^2. To enforce that one needs simplify(%); convert(%,csch); or some similar way. 3. For a complex expression one would try 'simplify', but it is almost impossible to guess the outcome (also because automatic simplification steps in and cases inbetween, where assumptions may be needed). 4. For the last - writing as a sum of terms - I do not know any general convenient way. However especially in your case one can by collect(h,x) [ where is your h = diff( ...) ], but working further destroys it again. 5. Equality ... that may depend on k<>0, k<> -1 and x<>0, but have not tries it seriously. Sometimes it is easier for Maple to recognize lhs/rhs= 1 or use conversions inbetween: hr=simplify(h, size); convert(%,expln); simplify(%); is(%); should do if your last expression is named hr. All that is not what actually covers your desire, yes. >2. >To see where problems it in your case step in (from Maple's view) [quoted text clipped - 3 lines] >To enforce that one needs simplify(%); convert(%,csch); or some >similar way. I thought this might be a similar way but Maple 11 didn't: > diff( coth(t), t); 2 1 - coth(t) > convert(%,csch); 4 csch(t) 1 - 1/4 ---------- 2 csch(2 t) > simplify(%); 1 - -------- 2 sinh(t) -- John Harper, School of Mathematics, Statistics and Computer Science, Victoria University, PO Box 600, Wellington 6140, New Zealand e-mail john.harper@vuw.ac.nz phone (+64)(4)463 6780 fax (+64)(4)463 5045 >> 2. >> To see where problems it in your case step in (from Maple's view) [quoted text clipped - 7 lines] > >> diff( coth(t), t); ... >> convert(%,csch); ... >> simplify(%); > 1 > - -------- > 2 > sinh(t) Yes, sorry. Finally one has to use convert again: 'diff(coth(t),t)': simplify(%): '%%'=convert(%,csch); d 2 -- coth(t) = -csch(t) dt |
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