# 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.

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The first thing to realise is that $E(x)$ and $O(x)$ aren't the only odd/even functions in existence; thye are related to $f$. Indeed the relation sought between these is that $f(x)=E(x)+O(x)$ for all $x$. With this requirement plus the fact that $E$ and $O$ are even resp. odd (and not $f$, as it seems in your question), you can derive their expressions. – Marc van Leeuwen Mar 18 '12 at 18:02
do you mean to ask why, if $E(x)$ is an even function, then there exists a function $f(x)$ such that $E(x) = (f(x)+f(-x))/2$ ? – mercio Mar 18 '12 at 18:04
what is $E(x)$? – quartz Mar 18 '12 at 18:08

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