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If

$$s(y) = \begin{cases} 2\sin(y/2)/y & \text{if $y \neq 0$} \\ 1 & \text{otherwise} \end{cases}$$

why is the taylor expansion of $g(y) = \frac{1}{s(y)}$:

$$ \frac{1}{s(y)} = 1 + \sum_{k=1}^{+\infty} \frac{(2^{k-1} - 1)B_k}{2^{2k}(2k)!}y^{2k},$$ with $B_k$ Bernoulli numbers.

I just copied the formulas from a paper, where there's no proof. I tried to derive the expansion by myself, but i get confused with the several derivatives.

Thank you (Any reference or derivation would be highly appreciated).

Update...

Trying to develop the product $g(y)s(y) = 1$

It means that:

$$g(y)s(y) = \left( \sum_{j=0}^{+\infty} \frac{g^{(j)}(0)}{j!}\right) \left( \sum_{j=0}^{+\infty} (-1)^j\frac{y^{2j}}{2^{2j}(2j+1)!}\right) $$,

using the cauchy product formula I have:

$$ g(y)s(y) = \sum_{j=0}^{+\infty} \sum_{k=0}^{j} (-1)^{j-k}\frac{g^{(k)}(0)y^{2j-k}}{2^{2j-2k}k!(2j-2k+1)!} = 1, $$

So I guess I have to use the constraint on the product to derive some recurrence relationship, is it?

Update 2.

What I'd try is to apply the following transformation:

$$ T(j,k) = \left\{ \begin{array}{l} q = 2j - k \\ p = j \end{array} \right. $$

which I is a bijection, but I'm not sure how to derive the boundaries for $p$, since $0 \leq q \leq +\infty$

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  • $\begingroup$ Hint: develop the product of the formal series $g(y)s(y)$ where the coefficients $g_k$ are unknown. You should see a recurrence appear that leads to the Bernoulli numbers. $\endgroup$ – Yves Daoust Jan 26 '16 at 8:55
  • $\begingroup$ I still get confused... i'm sorry but I need something more elaborated. $\endgroup$ – user8469759 Jan 26 '16 at 9:10
  • $\begingroup$ Did you try to develop the product ? $\endgroup$ – Yves Daoust Jan 26 '16 at 9:19
  • $\begingroup$ See the question... $\endgroup$ – user8469759 Jan 26 '16 at 10:07
  • $\begingroup$ Now you can group the terms with $2j-k$ constant and express that they sum to $0$ (or $1$ for $2j-k=0)$. $\endgroup$ – Yves Daoust Jan 26 '16 at 10:47

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