Axel Vogt

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20 years, 249 days
Munich, Bavaria, Germany

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These are answers submitted by Axel Vogt

Thx! Ok, I can follow. The other is something which worked in older versions (0
J:=int(whatever);
A:=unapply( J, x,y,z);
Then A is a function of x,y,z for the expression given by the integral
I mean not the prove, I mean the approach or ansatz
the identity is true (of course), but how you came up with this?
for me it works ... try to use 'restart', may be you have something else in your sheet?
  V:=convert(VAR,rational);
  simplify(V,size);
  length(V); # gives me 1298 characters using Maple 11
i would search for Fortran libs instead of trying to convert from Maple if wanting a "stand alone" version. have you looked at netlib, especially http://www.netlib.org/slatec/
it really would be helpful to support *easy* input of ASCII text with fixed fonts ...
do not try that: Maple is not for type setting ...
I always wanted to have two possible y-axes, one to the left, the other to the right. For example to plot a function approximation f(x) and its errors err(x) as one graphic. In Excel that's easy: mark a plotted function and choose the axis for displaying values.
I always hated this notation (and that for RootOf) ... 1) Is it possible to show all the strange symbols? Yes, try theIndets:=indets(indets(sol,atomic)), if sol is your solution 2) Does Maple use the assumptions given through the notations? Yes, try map( getassumptions, theIndets): op(%); 3) Can one extract and replace the symbols by more readable once (and 'norm' them, as repeated use changes through 'indexing')? Hm ... yes, but I have no friendly way ... and one has to add the assumptions for the new symbols
thanks for the answers and funny wordings :-) the reason is an "old" discussion on simplification by Carl Devore and Robert Israel (which is one of the rare cases where Google does not map the postings, here is the last post it needs inert forms in some cases Also check a possible bug in the routine, cf a posting by Walter Roberson, here Edited (10 Jun 2007) to beautify and point to an additional thread.
In my above posting one should use RootOf(f(x)-t=0,x, label=1): g:=unapply(%,t); Then it works for all reals, try it with plot(g(tau),tau=-10..10, color=blue). Adding to the other answers: you might google for Geddes "What is a closed form?" which gives additional insight for the 'philosophy' in Maple using RootsOf. Also http://front.math.ucdavis.edu/math.NT/9805045 might interest you: "What is a closed-form number ?" And a posting of Robert Israel in the Maple NG now comes to mind (last year? can not find it again ...): root-finding for elementary function is undecidable, so no hope for a general algorithm (he gave a proper citation) ...
It is not meaningless (even if the notation for RootOf is really ugly for my taste, since one can not see the according variable). As acer said: operational Maple handles it as a solution in most cases, just as if it would call the function "i_know_it_but_can_not_show_in_usual_terms". Let's try it, in your case it is the inverse w.r.t. composition: restart; Digits:=14; 5.*x*sqrt(1.+4.*x^2)+2.500000000*arcsinh(2.*x): convert(%,rational): f:=unapply(%,x); plot(f(x),x=-6..60); t = f(x); solve(%,x); g:=unapply(%,t); plot(g(t),t=0..35000, color=blue); fsolve( g(t) = 30, t=0..infinity); evalf(f(30)); or fsolve( f(z) = 1000, z); evalf(g(1000)); The solution is known to Maple, even for plotting, showing the inverse as reflection at the diagonal (does not work for t < -4, may be a bug or may be a problem with branch cuts?). Also t-f(g(t)); plot(%, t=0..1); and z-g(f(z)); plot(%,z=0..1); shows what I would expect: graphics gittering around 0 (well, it does not evaluate to 0 symbolically). It even would know the taylor series (but it would be better to use a different function which is not so time consuming, take 'erf' or so). And Maple behaves so for most cases: one can not expect that (locally) inverting a function can be given in known terms - that is already not the case for polynomials.
Thank you for both the answers! The reason for the question was how to terminate summing a series with reasonable fast decaying summands, I felt unsure for that.
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