Bernoulli Number analog using Cosine

Question

I know that Bernoulli Numbers can be found with the generating function $$\frac{x}{e^x-1}=\sum_{n=0}^{\infty}\frac{B_n}{n!}x^n$$

I was wondering if any work has been done using a similar equation $$\frac{x^2}{\cos{x}-1}=\sum_{n=0}^{\infty}\frac{C_n}{n!}x^{2n}$$

I'm particularly interested in the $C_n$. Can anyone help me with a reference to work with this particular generating function or to the coefficients $C_n$?

Answer 1

The general formula you're looking for is

$$\dfrac{x^2}{\cos x-1}~=~\sum_{n=0}^\infty(-1)^n\frac{B_{2n}}{(2n-2)!~n}~x^{2n}$$

where $B_k$ is the $k^{th}$ Bernoulli number. Of course, the $0^{th}$ term has to be computed by evaluating the limit $\displaystyle\lim_{n\to0}\Big[(2n-2)!~n\Big]=-\frac12.$

Answer 2

Your sequence (negated) starts $$ 2, \frac{1}{6}, \frac{1}{60}, \frac{1}{504}, \frac{1}{3600}, \frac{1}{22176}, \frac{691}{82555200}, \frac{1}{570240}, \frac{3617}{8821612800}, \frac{43867}{414096883200}, \frac{174611}{5822264448000}, \ldots $$

We know that $$ \frac{1-\cos x}{x^2} = \frac{1}{2!} - \frac{1}{4!}x^2 + \frac{1}{6!}x^4 - \frac{1}{8!}x^6 + \cdots. $$ Let $c_n = C_n/n!$. The formula for the inverse of a power series gives $$ c_0 = 2, \quad c_n = 2\frac{c_{n-1}}{4!} - 2\frac{c_{n-2}}{6!} + 2\frac{c_{n-3}}{8!} - 2\frac{c_{n-4}}{10!} + \cdots. $$

Answer 3

I don't have a reference for this formula, which was given in Lucian's answer (and also suggested numerically by Yuval Filmus's answer): $$ \frac{x^2}{1 - \cos x} = 2\sum_{n = 0}^\infty (-1)^{n - 1}\frac{(2n - 1)B_{2n}}{(2n)!}x^{2n}. $$ Nor do I know how it would naturally be derived. What follows, meanwhile, is a derivation that's rather unnatural, because it requires knowledge in advance of what is to be proved.

The familiar recurrence relation for the Bernoulli numbers, obtained (for example) by setting the Cauchy product of the series for $(e^x - 1)/x$ and $x/(e^x - 1)$ equal to $1$, is: $$ \sum_{r = 0}^{m - 1} \binom{m}{r}B_r = 0 \qquad (m > 1). $$ I'll also be taking for granted these familiar facts: $B_0 = 1$, $B_1 = -\frac{1}{2}$, and $B_r = 0$ for all odd $r \geqslant 3$. Taking in turn $m = 2n + 1$ and $m = 2n + 2$ in the recurrence relation above, and rearranging the terms involving $B_0$ and $B_1$, we get, for all $n \geqslant 1$: \begin{align*} \sum_{j = 1}^n \binom{2n + 1}{2j}B_{2j} & = n - \frac{1}{2}, \\ \sum_{j = 1}^n \binom{2n + 2}{2j}B_{2j} & = n. \end{align*} Either of these relations can be used to express $B_{2n}$ in terms of $B_2, B_4, \ldots, B_{2n-2}$ (and of course to get $B_2 = \frac{1}{6}$ to begin with).

If we multiply out the Cauchy product of the series for $(1 - \cos x)/x^2$ and $x^2/(1 - \cos x),$ we infer a different recurrence relation from either of those above, to wit: $$ \sum_{j = 1}^n (2j - 1)\binom{2n + 2}{2j}B_{2j} = 1, $$ so it would seem that one possible way to prove the formula given by Lucian would be to prove this recurrence relation.

To do so, take the relations already obtained, subtract the first from the second, and use the Pascal's Triangle relation between binomial coefficients, to get yet another recurrence relation: $$ \sum_{j = 1}^n \binom{2n + 1}{2j - 1}B_{2j} = \frac{1}{2}. $$ Hence (one could have taken a different linear combination of the two relations in the first place, but it seemed neater to prove the foregoing identity first): \begin{gather*} \sum_{j = 1}^n (2j - 1)\binom{2n + 2}{2j}B_{2j} \\ = \sum_{j = 1}^n 2j\binom{2n + 2}{2j}B_{2j} - \sum_{j = 1}^n \binom{2n + 2}{2j}B_{2j} \\ = (2n + 2)\sum_{j = 1}^n \binom{2n + 1}{2j - 1}B_{2j} - n \\ = (n + 1) - n = 1, \end{gather*} as required.

Answer 4

Here is a more natural derivation of the formula: $$ \frac{z^2}{1 - \cos z} = 2\sum_{n = 0}^\infty (-1)^{n - 1}\frac{(2n - 1)B_{2n}}{(2n)!}z^{2n} \ \ \ (\left\lvert{z}\right\rvert < 2\pi). $$

We start with the defining result that for all $z \in \mathbb{C}$ such that $\left\lvert{z}\right\rvert < 2\pi$, $$ \frac{z}{e^z - 1} = \sum_{m = 0}^\infty \frac{B_m}{m!}z^m. $$

Then: \begin{gather*} \coth\frac{z}{2} = \frac{\cosh\frac{z}{2}}{\sinh\frac{z}{2}} = \frac{e^z + 1}{e^z - 1} = \frac{2}{e^z - 1} + 1, \\ \therefore\ \frac{z}{2}\coth\frac{z}{2} = \frac{z}{e^z - 1} + \frac{z}{2}, \\ \therefore\ \frac{z}{2}\cot\frac{z}{2} = \frac{iz}{2}\coth\frac{iz}{2} = \frac{iz}{e^{iz} - 1} + \frac{iz}{2} \\ = \sum_{n = 0}^\infty (-1)^n\frac{B_{2n}}{(2n)!}z^{2n} + iz\sum_{n = 0}^\infty (-1)^n\frac{B_{2n + 1}}{(2n + 1)!}z^{2n} + \frac{iz}{2} \end{gather*} But this is a real-valued function of $z$ when $z$ is real, therefore: \begin{gather*} B_1 = -\frac{1}{2}; \ \ \ B_{2n + 1} = 0 \ \ \ (n \geqslant 1); \\ \frac{z}{2}\cot\frac{z}{2} = \sum_{n = 0}^\infty (-1)^n\frac{B_{2n}}{(2n)!}z^{2n} \ \ \ (\left\lvert{z}\right\rvert < 2\pi). \end{gather*} (We could also have argued, earlier, that $\frac{z}{2}\coth\frac{z}{2}$ is an even function of $z$.) Differentiating the series term-by-term, and multiplying by $z$: \begin{gather*} \sum_{n = 0}^\infty (-1)^n\frac{2nB_{2n}}{(2n)!}z^{2n} = z\frac{d}{dz}\left(\frac{z}{2}\cot\frac{z}{2}\right) \\ = \frac{z}{2}\cot\frac{z}{2} - \frac{z^2}{4}\mathrm{cosec}^2\frac{z}{2} = \frac{z}{2}\cot\frac{z}{2} - \frac{1}{2}\cdot\frac{z^2}{1 - \cos z}, \end{gather*} which gives the required result.