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Let $j \in Z_+$. Set $$ a_j^{(1)}=a_j:=\sum_{i=0}^j\frac{(-1)^{j-i}}{i!6^i(2(j-i)+1)!} $$ and $a_j^{(l+1)}=\sum_{i=0}^ja_ia_{j-i}^{(l)}$.

Find generating function $\sum_{j}a_jx^j$ so that allows to find all of $a_j^{\ell}$.

Thank you.

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1 Answer 1

up vote 2 down vote accepted

The coefficients $a_j$ are themselves Cauchy products. Indeed, define the sequences $(b_j)$ and $(c_j)$ by

$$ e^{x/6} = \sum_{j=0}^{\infty} \frac{1}{j! 6^j}x^j = \sum_{j=0}^{\infty} b_j x^j $$

and

$$ \frac{\sin \sqrt{x}}{\sqrt{x}} = \sum_{j=0}^{\infty} \frac{(-1)^j}{(2j+1)!}x^j = \sum_{j=0}^{\infty} c_j x^j. $$

Then

$$ \frac{e^{x/6} \sin \sqrt{x}}{\sqrt{x}} = \sum_{j=0}^{\infty} a_j x^j, $$

where

$$ a_j = \sum_{i=0}^{j} b_i c_{j-i} = \sum_{i=0}^{j} \frac{(-1)^{j-i}}{i! 6^i (2(j-i)+1)!}. $$

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@ Antonio Vargas: I am little bit confused. In order to find $a_j^l$, I have to differentiate the generating function at x=0, but at this point derivative does not exists. How to deal with this situation? Thank you. –  Michael Feb 8 '13 at 0:03
    
@Michael, after calculating the derivative, take the limit as $x \to 0$. This will give you the value of the derivative at $x=0$ even though the formula isn't defined there. –  Antonio Vargas Feb 8 '13 at 2:17
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