Consider the Taylor series of function $$\frac{2e^x}{e^{2x}+1}=\sum_{n=0}^\infty \frac{E_n}{n!}x^n$$ Prove that $$E_0=1, E_{2n-1}=0, E_{2n}=-\sum_{l=0}^{n-1}\binom{2n}{2l}E_{2l}, n\geq 1$$.
I am thinking to write the LHS as $\frac{1}{\frac{e^x+e^{-x}}{2}}=\frac{1}{\cosh x}$, but I am stuck here. How can I derive the above recurrence relation? Any enlightenment please?
On
note that: $$\arctan(y)=\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}x^{2n+1}$$ and: $$\frac{2e^x}{e^{2x}+1}=\frac{d}{dx}\left[\arctan(e^x)\right]$$ so: $$\arctan(e^x)=\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}e^{(2n+1)x}$$ so: $$\frac{2e^x}{e^{2x}+1}=\sum_{n=0}^\infty(-1)^ne^{(2n+1)x}$$ and we know: $$e^x=\sum_{m=0}^\infty\frac{x^m}{m!}$$ now we can say: $$\frac{2e^x}{e^{2x}+1}=\sum_{n=0}^\infty\sum_{m=0}^\infty\frac{(-1)^n\left[(2n+1)x\right]^m}{m!}$$
You want $$ \left(\sum_{n=0}^\infty\frac{E_n}{n!}x^n\right)\cosh x=1 $$ i.e. $$ \left(\sum_{n=0}^\infty\frac{E_n}{n!}x^n\right) \left(\sum_{m=0}^\infty\frac1{(2m)!}x^{2m}\right)=1. $$ Now expand the series in the LHS. Clearly all $E_n$ for $n$ odd must vanish, or there will be a least $n$, $n$ odd, such that $E_n\neq 0$. Then the coefficient of $x^n$ would not vanish.
So $E_0=1$, $E_{2n+1}=0$ for all $n$, and the coefficient of $x^{2n}$ gives $$ \sum_{m=0}^n\frac{E_{2(n-m)}}{[2(n-m)]!(2m)!}=0 $$ which rearrange to the recurrence equation (remember $\frac{(2n)!}{[2(n-m)]!(2m)!}=\binom{2n}{2(n-m)}$).