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Free alternative to Mathematica?

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  1. Jun 5, 2014 #1
    In Mathematica, I computed the integral ##\lim\limits_{n\to\infty} \frac{1}{n^2} \int\limits_{0}^{\frac{\pi}{2}} \frac{\sin((2n+1)x)}{\sinh(x)} \ dx## and it incorrectly output the answer ##-\frac{3\pi}{2}##. I wanted to try a different CAS software that might provide a different answer, but so far wxMaxima has not worked and I'm not sure what else I can use. I'm also looking for a free alternative because I want to use it at home, since I can only use Mathematica in school and I can't afford a personal license for Mathematica.
     
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  3. Jun 5, 2014 #2

    phyzguy

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    Science Advisor

    There is an open source free software called Sage that does symbolic mathematics. It's not as good as Mathematica, but it's free.
     
  4. Jun 5, 2014 #3
    This software takes up 7 gigabytes :eek: but if it can compute this integral, then I'll keep it around.

    Edit: Nope, it doesn't work :(
     
    Last edited: Jun 6, 2014
  5. Jun 6, 2014 #4
    You might try Axiom. It can be demanding when it comes to details, but it might get you an answer if you carefully follow all the rules.

    I'm puzzled how you ended up with -3Pi/2 from Mathematica. When I try

    Code (Text):
    Limit[1/n^2*Integrate[Sin[(2*n+1)*x]/Sinh[x], {x, 0, Pi/2}], n->Infinity]
    it can do the integration but can't seem to find the last step to resolve the limit.
     
  6. Jun 6, 2014 #5

    phyzguy

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    When I ran this with my copy of Mathematica 7, it gave me the (I think correct) answer of 0. Are you sure you entered it into Mathematica correctly?

    Code (Text):

    In[21]:= Limit[
     n^-2*Integrate[Sin[(2 n + 1) x]/Sinh[x], {x, 0, \[Pi]/2}],
     n -> \[Infinity]]

    Out[21]= 0

     
     
  7. Jun 6, 2014 #6
    Code (Text):
    In[1]:= $Version

    Out[1]= "9.0 for Microsoft Windows (64-bit) (January 25, 2013)"

    In[2]:= Limit[1/n^2*Integrate[Sin[(2*n + 1)*x]/Sinh[x], {x, 0, Pi/2}], n -> Infinity]

    Out[2]= Limit[-((I (Beta[E^Pi, (1/2-I/2)-I n, 0] - Beta[E^Pi, (1/2+I/2)+I n, 0] -
         I Pi Tanh[(1/2+n) Pi]))/(2 n^2)), n->Infinity]
     
  8. Jun 6, 2014 #7

    phyzguy

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    Science Advisor

    Sorry, I meant to address my question to the OP. E7.5, can you show us the Mathematica code you used? I think the integral itself approaches pi/2 as n gets large, so when divided by n^2, the limit should be zero, which is what my version of Mathematica gave.
     
  9. Jun 11, 2014 #8
    I think Octave is what you want. It is an open source clone of MatLab and shares much of the same language as MatLab. I recently installed it on my virtual machine and was able to run most of the scripts I wrote in MatLab that I use for my research on Octave without too much debugging.

    http://www.gnu.org/software/octave/
     
  10. Jun 11, 2014 #9

    DrClaude

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    Staff: Mentor

    Matlab is not Mathematica. Octave can't do symbolic math.
     
  11. Jun 11, 2014 #10
    Oops, should have read more closely, my bad.
     
  12. Jan 4, 2015 #11
    try in Mathematica

    In[15]= FullSimplify[Limit[n^-2*Integrate[Sin[(2 n+1) x]/Sinh[x],{x,0,π/2}],n->∞]]

    Out[15]= 0

    whereas

    Limit[1/n^2*Integrate[Sin[(2*n+1)*x]/Sinh[x],{x,0,Pi/2}],n->Infinity]

    gives an incomplete answer.


     
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