Given an exponential generating function, is it possible to isolate only the even terms? Suppose you have an exponential generating function
$$
F(x)=F_0+F_1x+F_2\frac{x^2}{2!}+\cdots+F_n\frac{x^n}{n!}+\cdots
$$
and you want to get only the even terms
$$
F_e(x)=F_0+F_2\frac{x^2}{2!}+F_4\frac{x^4}{4!}+\cdots
$$
Is it possible to write $F_e(x)$ in terms of $F(x)$? I noticed that taking the derivative of $F(x)$ seems to shift the coefficients down a term, but couldn't get much further than that.
I'm also interested in the odd case, but I guess that follows easily since $F(x)-F_e(x)=F_o(x)$. 
 A: The general technique of isolating the terms which are congruent to $a \bmod n$ is known in the US high school competition circuit as the "roots of unity filter" and known to most others as the discrete Fourier transform.
The abstract idea is that the cyclic group $\mathbb{Z}/n\mathbb{Z}$ acts on the space of, say, formal power series in one variable $z$ over $\mathbb{C}$ via $z \mapsto e^{\frac{2 \pi i}{n} } z$. This group representation has $n$ isotypic components (subspaces where the group acts the same way) corresponding to the different residues $\bmod n$, and what we are interested in computing is the projections to these components, which can be done quite explicitly by averaging appropriately over the group action. 
For example, to isolate the terms divisible by $3$ you use $\frac{F(z) + F(\omega z) + F(\omega^2 z)}{3}$ where $\omega = e^{ \frac{2 \pi i}{3} }$. The fact that all the terms cancel out nicely when you do this is a special case of the orthogonality relations in representation theory. 
A: $F_e(x)=\frac{1}{2}(F(x)+F(-x))$.
$F_o(x)=\frac{1}{2}(F(x)-F(-x))$.
See also How do I divide a function into even and odd sections?
A: They're called series bisections, e.g. bisecting into even and odd parts the power series for $\:\rm e^{\,{\it i}\,x} \:,\;$ 
$$\begin{align}
\rm f(x) \  &= \ \rm\frac{f(x)+f(-x)}{2} \;+\; \frac{f(x)-f(-x)}{2} \\[.2em]
\Rightarrow\quad \rm e^{\,{\it i}\,x} \  &= \ \rm\cos(x) \ +\ {\it i} \ \sin(x)  \end{align}\qquad$$
Similarly one can perform multisections into $\rm\:n\:$ parts using $\rm\:n\:$'th roots of unity - see my post here for some examples and see Riordan's classic 
textbook Combinatorial Identities for many applications. Briefly, 
with  $\rm\:\zeta\ $ a primitive $\rm\:n$'th root of unity,  the $\rm\:m$'th $\rm\:n$-section selects the linear progression of $\rm\: m+k\:n\:$ indexed terms from a series $\rm\ f(x)\ =\ a_0 + a_1\ x + a_2\ x^2 +\:\cdots\ $ as follows
$\rm\quad\quad\quad\quad  a_m\ x^m\ +\ a_{m+n}\ x^{m+n}\ +\ a_{m+2\:n}\ x^{m+2\:n}\ +\:\cdots $
$\rm\quad\quad =\ \frac{1}{n} \big(f(x)\ +\ f(x\zeta)\ \zeta^{-m}\ +\ f(x\zeta^{\:2})\ \zeta^{-2m}\ +\:\cdots\: +\ f(x\zeta^{\ n-1})\ \zeta^{\ (1-n)\:m}\big)$
Exercise $\;$ Use multisections to give elegant proofs of the following
$\quad\quad\rm\displaystyle sin(x)/e^{x} \quad\:$ has every $\rm 4\ k\;$'th term zero in its power series
$\quad\quad\rm\displaystyle cos(x)/e^{x} \quad$ has every $\rm 4k\!+\!2\;$'th term zero in its power series
See the posts in this thread for various solutions and more on multisections. When you later study representation theory of groups you will learn that this is a special case of much more general results, with relations to Fourier and other transforms. It's also closely related to various Galois-theoretic results on modules, e.g. see my remark about Hilbert's Theorem 90 in the linked thread.
