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Cauchy summation in maple

  1. Sep 10, 2007 #1

    Simfish

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    So...

    I want to find the Cauchy sum of the Taylor polynomial of [tex]\exp x \sin x[/tex]. I know how to do this with maple, which only requires the command
    taylor(sin(x)*exp(x), x = 0, n). I can also try the good old [tex]f(a)+\frac{f'(a)}{1!}(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+\cdots[/tex] formula, but that isn't the learning objective. But I want to see if I did the Cauchy summation correctly with maple, and maple has different commands for summing up polynomial series. So I use the maple command.

    sum((-1)^k*x^(n+k+1)/(factorial(2*k+1)*factorial(n-k)), k = 0 .. n)

    Problem is, how can I set a value to n?

    Is [tex]\sum_{k=0}^n {\frac { \left( -1 \right) ^{k}{x}^{n+k+1}}{ \left( 2\,k+1 \right) !\,\left( n-k \right)!}}[/tex] the right Cauchy sum of this series anyhow?
     
    Last edited: Sep 10, 2007
  2. jcsd
  3. Sep 10, 2007 #2

    Simfish

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    Gold Member

    test...

    The problem here is that this is when [tex]a_n = \sin x , b_n = \exp x[/tex]. When I set [tex]b_n = \sin x, a_n = \exp x[/tex], I get [tex]\sum_{k=0}^n {\frac { \left( -1 \right) ^{n-k}{x}^{n-2k+1}}{ \left( 2\,n - 2k + 1 \right) !\,\left( n)!}}[/tex]

    The other problem is that in the Cauchy formula [tex]c_n=\sum_{k=0}^n a_k b_{n-k}[/tex] I expect [tex]a_n[/tex] and [tex]b_n[/tex] to be symmetrical. But yet when I set u = n - k, where [tex]k \in [0,n], u \in [n, 0] [/tex], which would be a backwards summation. Is there a better way to ensure symmetry of the two terms?
     
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