Recursive formula of even indexed Euler numbers

Let ${\rm E}_{2n}$ denote the even indexed Euler numbers. Prove that:

$${\rm E}_0=1, \;\; {\rm E}_{2n}=-\sum_{k=0}^{n-1}\binom{2n}{2k}{\rm E}_{2k}, \; n\geq 1$$

Solution



We are invoking the Taylor series of $\sec $ , that is the formula:

$$\frac{1}{\cos{x}}=\sum_{n=0}^{\infty}{\frac{(-1)^{n}{\rm{E}}_{2n}x^{2n}}{(2n)!}}\,,\quad |x|<\frac{\pi}{2}$$

Now, since $\cos x \sec x =1$ we get that:

$$ \begin{aligned} 1 &=\cos x \sec x \\ &= \sum_{n=0}^{\infty}\frac{(-1)^n{\rm E}_{2n}x^{2n}}{(2n)!}\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n}}{\left ( 2n \right )!}\\ &= \sum_{n=0}^{\infty}\left ( \sum_{k=0}^{n}\frac{1}{\left ( 2n-2k \right )!}\frac{{\rm E}_{2k}}{(2k)!} \right )(-1)^n x^{2n}\\ &=\sum_{n=0}^{\infty}\left ( \sum_{k=0}^{n}\binom{2n}{2k}{\rm E}_{2k} \right )\frac{(-1)^n x^{2n}}{(2n)!} \\ \end{aligned}$$

Equating coefficients we get the wanted result.

P.S: Going one step further one can prove , from the recursive formula , that:

$${\rm{E}}_{2n}=(-1)^n(2n)!\left|{\begin{array}{ccccccc} \frac{1}{2!} & 1 & 0 & 0 & \cdots & 0 & 0\\ \\ \frac{1}{4!} & \frac{1}{2!} & 1 & 0 & \cdots & 0 & 0\\ \\ \frac{1}{6!} & \frac{1}{4!} & \frac{1}{2!} & 1 & \cdots & 0 & 0\\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ \\ \frac{1}{(2n-2)!} & \frac{1}{(2n-4)!} & \frac{1}{(2n-6)!} & \frac{1}{(2n-8)!} & \cdots & \frac{1}{2!} & 1\\\\ \frac{1}{(2n)!} &\frac{1}{(2n-2)!} & \frac{1}{(2n-4)!} & \frac{1}{(2n-6)!} & \cdots & \frac{1}{4!} & \frac{1}{2!}\end{array}}\right|$$