Deriving even odd function expressions What is the logic/thinking process behind deriving an expression for even and odd functions in terms of $f(x)$ and $f(-x)$?
I've been pondering about it for a few hours now, and I'm still not sure how one proceeds from the properties of even and odd functions to derive:
$$\begin{align*}
E(x) &= \frac{f(x) + f(-x)}{2}\\
O(x) &= \frac{f(x) - f(-x)}{2}
\end{align*}$$
What is the logic and thought process from using the respective even and odd properties,
$$\begin{align*}
f(-x) &= f(x)\\   
f(-x) &= -f(x)
\end{align*}$$ 
to derive $E(x)$ and $O(x)$?  
The best I get to is:
For even: $f(x)-f(-x)=0$ and for odd: $f(x)+f(-x)=0$
Given the definition of $E(x)$ and $O(x)$, it makes a lot of sense (hindsight usually is) but starting from just the properties. Wow, I feel I'm missing something crucial.

Although I'm saying nothing new with respect to previous answers, I'll try to cut things down to their bare essence, which is just equations solving; the function point of view is hardly used.
You want $E$ and $O$ to be even and odd functions respectively, so
$$ E(x)=E(-x) \qquad\text{and}\qquad O(x)=-O(-x) $$
and you want $f$ to be their sum, so that
$$ f(x)=E(x)+O(x) \qquad\text{and also}\qquad f(-x)=E(-x)+O(-x). $$
Now forget about $x$, all you need is $E(x)$ and $O(x)$ when $f(x)$ and $f(-x)$ are given. Calling these unknowns $a=E(x)$ and $b=O(x)$ (too bad the name $x$ was already taken), we can immediately express $E(-x)=a$ and $O(-x)=-b$, so it is not necessary to have separate unknowns for them.
Now the second set of equations gives
$$ a+b=f(x) \qquad\text{and}\qquad a-b=f(-x), $$
where the (now) right hand sides are known values. All that remains is to solve this for $a$ and $b$, which by any of the many known methods gives
$$
  a=\frac{f(x)+f(-x)}2\qquad\text{and}\qquad b=\frac{f(x)-f(-x)}2
$$
in other words
$$
  E(x)=\frac{f(x)+f(-x)}2\qquad\text{and}\qquad O(x)=\frac{f(x)-f(-x)}2.
$$
This should hold for any $x$, which defines $E,O$ as functions.
