Recursion relation for Euler numbers I am trying to solve the following:

The Euler numbers $E_n$ are defined by the power series expansion
$$\frac{1}{\cos z}=\sum_{n=0}^\infty \frac{E_n}{n!}z^n\text{ for }|z|<\pi/2$$
(a)  Show that $E_n=0$ when $n$  is an odd integer.
(b) Establish a recursion relation for the sequence.

Through research it seems that part a seems well established but I couldn't find a proof.
Part (b) I found a formula, but couldn't find how to establish it. Any help appreciated.
 A: The exponential generating function of the Euler numbers is $\mathrm{sech}(x)$. Since $\mathrm{sech}(x)$ is even, the odd Euler numbers are $0$. In this answer, it is shown that
$$
\begin{align}
1
&=\cosh(x)\,\mathrm{sech}(x)\\[9pt]
&=\sum_{k=0}^\infty\frac1{(2k)!}x^{2k}\sum_{n=0}^\infty\frac{\mathrm{E}_{2n}}{(2n)!}x^{2n}\\
&=\sum_{n=0}^\infty\left(\sum_{k=0}^n\frac1{(2n-2k)!}\frac{\mathrm{E}_{2k}}{(2k)!}\right)x^{2n}\\
&=\sum_{n=0}^\infty\left(\sum_{k=0}^n\binom{2n}{2k}\mathrm{E}_{2k}\right)\frac{x^{2n}}{(2n)!}\tag{1}
\end{align}
$$
Equating coefficients of $x^{2n}$, we get $\color{#C00000}{\mathrm{E}_0=1}$ and
$$
\color{#C00000}{\mathrm{E}_{2n}=-\sum_{k=0}^{n-1}\binom{2n}{2k}\mathrm{E}_{2k}}\tag{2}
$$

Since $\sec(x)=\mathrm{sech}(ix)$,
$$
\begin{align}
\sec(x)
&=\sum_{n=0}^\infty(-1)^n\frac{\mathrm{E}_{2n}}{(2n)!}x^{2n}\\
&=\sum_{n=0}^\infty\frac{\tilde{\mathrm{E}}_{2n}}{(2n)!}x^{2n}\tag{3}
\end{align}
$$
where $\tilde{\mathrm{E}}_{2n}=(-1)^n\mathrm{E}_{2n}$. Therefore, if $\sec(x)$ is the exponential generating function for $\tilde{\mathrm{E}}_n$, $(2)$ would become $\color{#00A000}{\tilde{\mathrm{E}}_0=1}$ and
$$
\color{#00A000}{\tilde{\mathrm{E}}_{2n}=\sum_{k=0}^{n-1}(-1)^{n-k+1}\binom{2n}{2k}\tilde{\mathrm{E}}_{2k}}\tag{4}
$$
A: For part (a): since $\cos(z)$ is an even function (that is, $\cos(z) = \cos(-z)$), then it follows that $\dfrac{1}{\cos(z)}$ is also even (why?).  You should be able to convince yourself that the Taylor series of an even function $f_e(z)$ can't have any odd terms (try differentiating both sides of the identity $f_e(-z) = f_e(z)$ at $z=0$ repeatedly using the chain rule).
For part (b): You know that $\cos(z)\times\dfrac{1}{\cos(z)} = 1$ identically; expand both terms on the LHS in terms of their Taylor series, multiply them out using the usual method for multiplying series, and then set the left and right sides of the equation to be equal as Taylor series — that is to say, look at the terms in $z^n$ on both sides (note that on the RHS, that term is $1$ if $n=0$ and $0$ otherwise).  You should find that this gives you a recursive formula for the Euler numbers.
A: Related problems: (I), (II). It should be 
$$\frac{1}{\cosh(z)} =\sum_{n=0}^\infty \frac{E_n}{n!}z^n,\quad|z|<\pi/2. $$ Here is few terms of Taylor series of $\frac{1}{\cosh(z)}$
$$ \frac{1}{\cosh(z)}=  1-{\frac {1}{2}}{z}^{2}+{\frac {5}{24}}{z}^{4}-{\frac {61}{720}}{z}^{
6}+O \left( {z}^{8} \right),$$
which they match your power series.
Added: Here is another representation. It may be of interest to some people 

$$\frac{1}{\cosh(z)} =\sum _{n=0}^{\infty }{\frac {\left( 
-1 \right)^{n}\left(\psi \left( 2\,n,\frac{3}{4}\right)-\psi \left( 2\,n,\frac{1}{4} \right) \right) {z}^{2\,n}}{ {4}^{n}{\pi }^{2\,n+1}\left( 2\,n \right) !}},$$

where $\psi(z)$ is the digamma function. Writing the OP's identity as
$$ \frac{1}{\cosh (z)}=\sum_{n=0}^\infty \frac{E_{2n}}{(2n)!}z^{2n} ,$$
one can readily see the following relation

$$ E_{2n} = {\frac {\left( 
-1 \right)^{n}\left(\psi \left( 2\,n,\frac{3}{4}\right)-\psi \left( 2\,n,\frac{1}{4} \right) \right) }{ {4}^{n}{\pi }^{2\,n+1}}}, \quad n\in \mathbb{N}\cup \{0\}.  $$    

