Cfrgtkky

Consider the Taylor series of function $$frac{2e^x}{e^{2x}+1}=sum_{n=0}^infty frac{E_n}{n!}x^n$$
Prove that
$$E_0=1, E_{2n-1}=0, E_{2n}=-sum_{l=0}^{n-1}binom{2n}{2l}E_{2l}, ngeq 1$$.

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?

You want
$$
left(sum_{n=0}^inftyfrac{E_n}{n!}x^nright)cosh x=1
$$
i.e.
$$
left(sum_{n=0}^inftyfrac{E_n}{n!}x^nright)
left(sum_{m=0}^inftyfrac1{(2m)!}x^{2m}right)=1.
$$
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 $E_nneq 0$. 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
$$
sum_{m=0}^nfrac{E_{2(n-m)}}{[2(n-m)]!(2m)!}=0
$$
which rearrange to the recurrence equation (remember $frac{(2n)!}{[2(n-m)]!(2m)!}=binom{2n}{2(n-m)}$).

note that:
$$arctan(y)=sum_{n=0}^inftyfrac{(-1)^n}{2n+1}x^{2n+1}$$
and:
$$frac{2e^x}{e^{2x}+1}=frac{d}{dx}left[arctan(e^x)right]$$
so:
$$arctan(e^x)=sum_{n=0}^inftyfrac{(-1)^n}{2n+1}e^{(2n+1)x}$$
so:
$$frac{2e^x}{e^{2x}+1}=sum_{n=0}^infty(-1)^ne^{(2n+1)x}$$
and we know:
$$e^x=sum_{m=0}^inftyfrac{x^m}{m!}$$
now we can say:
$$frac{2e^x}{e^{2x}+1}=sum_{n=0}^inftysum_{m=0}^inftyfrac{(-1)^nleft[(2n+1)xright]^m}{m!}$$