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## Cauchy summation in maple

So...

I want to find the Cauchy sum of the Taylor polynomial of $$\exp x \sin x$$. 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 $$f(a)+\frac{f'(a)}{1!}(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+\cdots$$ 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 $$\sum_{k=0}^n {\frac { \left( -1 \right) ^{k}{x}^{n+k+1}}{ \left( 2\,k+1 \right) !\,\left( n-k \right)!}}$$ the right Cauchy sum of this series anyhow?
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 Blog Entries: 1 Recognitions: Gold Member test... The problem here is that this is when $$a_n = \sin x , b_n = \exp x$$. When I set $$b_n = \sin x, a_n = \exp x$$, I get $$\sum_{k=0}^n {\frac { \left( -1 \right) ^{n-k}{x}^{n-2k+1}}{ \left( 2\,n - 2k + 1 \right) !\,\left( n)!}}$$ The other problem is that in the Cauchy formula $$c_n=\sum_{k=0}^n a_k b_{n-k}$$ I expect $$a_n$$ and $$b_n$$ to be symmetrical. But yet when I set u = n - k, where $$k \in [0,n], u \in [n, 0]$$, which would be a backwards summation. Is there a better way to ensure symmetry of the two terms?