Taylor series of functions

Question

Consider the Taylor series of the function

$$\frac{2e^x}{e^{2x}+1} = \sum_{n=0}^{\infty} \frac{E_n}{n!} x^n$$

Prove that $E_0 = 1, E_{2n-1} = 0$ and, for $n \ge 1$, $$E_{2n} = - \sum_{l=0}^{n-1} C_{2l}^{2n} {E_{2l}}$$

The hint given was to consider $\cosh(x)$.

I understand that the function is actually the reciprocal of $\cosh(x)$, however, I have no idea how to proceed.

Answer 1

We use the Taylor series expansion of $\cosh x$ which you can find e.g. here. \begin{align*} \cosh x = \frac{e^{2x}+1}{2e^x} = \sum_{l=0}^\infty \frac{x^{2l}}{(2l)!} \end{align*}

We obtain \begin{align*} \color{blue}{1}&=\left(\sum_{k=0}^\infty\frac{E_k}{k!}x^k\right)\frac{e^{2x}+1}{2e^x}\tag{1}\\ &=\left(\sum_{k=0}^\infty\frac{E_k}{k!}x^k\right)\left(\sum_{l=0}^\infty\frac{x^{2l}}{(2l)!}\right)\\ &=\left(\sum_{k=0}^\infty\frac{E_k}{k!}x^k\right)\left(\sum_{l=0}^\infty\frac{1+(-1)^l}{2}\frac{x^l}{l!}\right)\\ &=\sum_{n=0}^\infty\left(\sum_{{k+l=n}\atop{k,l\geq 0}}\frac{E_k}{k!}\cdot\frac{1+(-1)^l}{2l!}\right)x^n\\ &\,\,\color{blue}{=\sum_{n=0}^\infty\left(\sum_{k=0}^n\binom{n}{k}\frac{1+(-1)^{n-k}}{2}E_k\right)\frac{x^n}{n!}}\tag{2} \end{align*}

Coefficient comparison of (1) and (2) gives \begin{align*} E_0=1\quad \text{and}\quad E_n=\sum_{k=0}^n\binom{n}{k}\frac{1+(-1)^{n-k}}{2}E_k\qquad\qquad n\geq 1\tag{3} \end{align*}

From (3) we get for even index \begin{align*} \color{blue}{E_{2n}}&=-\sum_{k=0}^{2n-1}\binom{2n}{k}\frac{1+(-1)^{k}}{2}E_k\\ &\,\,\color{blue}{=-\sum_{k=0}^{n-1}\binom{2n}{2k}E_{2k}\qquad\qquad\qquad\qquad\qquad n\geq 1} \end{align*}

Since \begin{align*} \frac{2e^{-x}}{e^{-2x}+1}=\frac{2e^x}{e^{2x}+1}\\ \end{align*} we see the function $\frac{2e^x}{e^{2x}+1}$ is even and we conclude $\color{blue}{E_{2n-1}=0, n\geq 1}$.

Answer 2

Because the OP asked for some clarification and because I haven't done this type of thing very often (ever?) and it's good to practice I have tried my best here. This isn't a complete answer just a supplement. This really is just the details of the $4$th equality in Markus Scheuer's answer box above. The one that says:

$$ \left(\sum_{k=0}^\infty\frac{E_k}{k!}x^k\right)\left(\sum_{l=0}^\infty\frac{1+(-1)^l}{2}\frac{x^l}{l!}\right) =\sum_{n=0}^\infty\left(\sum_{{k+l=n}\atop{k,l\geq 0}}\frac{E_k}{k!}\cdot\frac{1+(-1)^l}{2l!}\right)x^n $$

The details may be tricky for someone (like me) who hasn't done this often before so here they are in their details. We note that

$$\cosh(x)=\sum_{n=0}^\infty \frac{x^{2n}}{(2n)!}=\sum_{n=0}^\infty \frac{A_n}{n!}x^n$$

Where $A_n=0$ when $n$ is odd and otherwise $A_n=1$. Markus Scheuer cleverly writes this as $\frac{1+(-1)^n}{2}$ We are motivated to rewrite the series like this so it is in the format of a typical power series so we can easily apply the Cauchy product. As seen in the wiki:

$$ \left(\sum_{i=0}^\infty a_i x^i\right) \cdot \left(\sum_{j=0}^\infty b_j x^j\right) = \sum_{k=0}^\infty c_k x^k \text{ where } c_k=\sum_{l=0}^k a_l b_{k-l}$$

So in this case we have $a_i=\frac{A_i}{i!}$ and $b_j=\frac{E_j}{j!}$ and then $c_k =\sum_{l=0}^k \frac{A_lE_{k-l}}{l!(k-l)!}$

To make this a little cleaner we can multiply by $\frac{k!}{k!}$ so that we can make the ${k\choose l} =\frac{k!}{l!(k-l)!}$ appear. The other motivation for doing this trick would be that we want a binomial coefficient to appear because it's in your given prompt. Then we have $c_k=\sum_{l=0}^k {k \choose l} \frac{A_l E_{k-l}}{k!}$. And the rest is shown in the other answer.