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Math Help - Free computer algebra system

  1. #1
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    Free computer algebra system

    I am trying to integrate a very messy expression using Wolfram online integrator but I get the message:
    "You need to use Mathematica to evaluate this expression".
    The online integrator works only for relatively simple expressions.
    Unfortunately I do not have a license for Mathematica.
    Any proposal to overcome this?
    Is there a free computer algebra software that can be used as a replacement to Mathematica.
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  2. #2
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    Re: Free computer algebra system

    Quote Originally Posted by JulieK View Post
    I am trying to integrate a very messy expression using Wolfram online integrator but I get the message:
    "You need to use Mathematica to evaluate this expression".
    The online integrator works only for relatively simple expressions.
    Unfortunately I do not have a license for Mathematica.
    Any proposal to overcome this?
    Is there a free computer algebra software that can be used as a replacement to Mathematica.
    What is the expression you are trying to work out?

    -Dan
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  3. #3
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    Re: Free computer algebra system

    _{2}F_{1}(B;C;D;Ex^{2})\,Ax

    _{2}F_{1}(...) is the hypergeometric function, x is the independent variable and A, B, C, D, E and F are constants.
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  4. #4
    Forum Admin topsquark's Avatar
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    Re: Free computer algebra system

    Try the following input on Wolfram|Alpha: Hypergeometric2F1[1, 2, 1, x^2],for example.

    -Dan
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  5. #5
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    Re: Free computer algebra system

    This is not my expression.
    When I type an expression similar to my expression (A*x*Hypergeometric2F1[1,2,1,x^2]) in Wolfram Alpha I get a very simple answer which cannot be correct.
    I checked this by differentiating the answer.
    Last edited by JulieK; June 29th 2014 at 02:47 AM.
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  6. #6
    Forum Admin topsquark's Avatar
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    Re: Free computer algebra system

    Okay, first of all see here for the definition of the hypergeometric function as a series representation.

    Your answer is simple because the case I picked is simple. Here's the derivation:
    _2 F _1(1, 2, 1; x^2) = 1 + \sum _{n = 1}^{\infty} \frac{(1)_n \cdot (2)_n}{(1)_n} \frac{\left ( x^2 \right ) ^n}{n!}

    where (a)_n is the "Pochhammer symbol."

    = 1 + \sum_{n = 1}^{\infty} (2)_n \frac{x^{2n}}{n!}

    = 1 + \sum_{n = 1}^{\infty} \Gamma(2 + n)/ \Gamma(2) \frac{x^{2n}}{n!}

    = 1 + \sum_{n = 1}^{\infty} (n + 1) x^{2n}


    The summations are simple (I'll show those if you want) giving:

    = 1 + \frac{2x^2 - x^4}{(x^2 - 1)^2}

    _2 F _1(1, 2, 1; x^2) = \frac{1}{(x^2 - 1)^2}

    -Dan
    Thanks from JulieK
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  7. #7
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    Re: Free computer algebra system

    1. Many thanks!

    2. The problem with the power series is that it is defined only for |x|<1.

    3. What about my expression, i.e
    _{2}F_{1}(B;C;D;Ex^{2})\,Ax
    Can you do the same for this?
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  8. #8
    Forum Admin topsquark's Avatar
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    Re: Free computer algebra system

    Quote Originally Posted by JulieK View Post
    1. Many thanks!

    2. The problem with the power series is that it is defined only for |x|<1.

    3. What about my expression, i.e
    _{2}F_{1}(B;C;D;Ex^{2})\,Ax
    Can you do the same for this?
    I can't help you with the |x| < 1. You might be able to analytically continue it into regions where x is larger, but that is out of my league. As to Ax \times _2 F _1(B, C, D, Ex^2), that's just going to be
    \frac {Ax}{(x^2 - 1)^2}

    I'm not sure what you are trying to do here.

    -Dan
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