# Understanding a taylor expansion

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 $0 \leq q \leq +\infty$

• 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. – Yves Daoust Jan 26 '16 at 8:55
• I still get confused... i'm sorry but I need something more elaborated. – user8469759 Jan 26 '16 at 9:10
• Did you try to develop the product ? – Yves Daoust Jan 26 '16 at 9:19
• See the question... – user8469759 Jan 26 '16 at 10:07
• Now you can group the terms with $2j-k$ constant and express that they sum to $0$ (or $1$ for $2j-k=0)$. – Yves Daoust Jan 26 '16 at 10:47