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.

  • $\begingroup$ 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. $\endgroup$ Mar 18, 2012 at 18:02
  • $\begingroup$ do you mean to ask why, if $E(x)$ is an even function, then there exists a function $f(x)$ such that $E(x) = (f(x)+f(-x))/2$ ? $\endgroup$
    – mercio
    Mar 18, 2012 at 18:04
  • $\begingroup$ what is $E(x)$? $\endgroup$
    – quartz
    Mar 18, 2012 at 18:08
  • $\begingroup$ @mercio, kinda, the more I think about it, the more I'm trying to understand the link between $E(x)$, $O(x)$ and $f(x)$. $f(x)$ might be completely odd or even (or neither?), how then can we express $E(x)$ or $O(x) with $f(x)$? $\endgroup$
    – xlm
    Mar 18, 2012 at 18:23
  • $\begingroup$ @quartz, $E(x)$ denotes is an even function. Should have expressed as $f_{even}(x)$ and $O(x)$ as $f_{odd}(x)$. Apologies $\endgroup$
    – xlm
    Mar 18, 2012 at 18:24

4 Answers 4


A function $g(x)$ is said to be an even function if $\forall x \in \mathbb{R}$, we have $g(-x) = g(x)$. The naming even function arises from the fact that the functions $g(x) = x^{2n}$ where $n \in \mathbb{Z}$ i.e. the function which takes the even powers satisfy this condition since $(-x)^{2n} = x^{2n}$.

Similarly, a function $g(x)$ is said to be an odd function if $\forall x \in \mathbb{R}$, we have $g(-x) = -g(x)$. The naming odd function arises from the fact that the functions $g(x) = x^{2n+1}$ where $n \in \mathbb{Z}$ i.e. the function which takes the odd powers satisfy this condition since $(-x)^{2n+1} = -x^{2n+1}$.

The claim is that any function can be written as a sum of an even function and an odd function.

Note that $$f(x) = \left(\frac{f(x) + f(-x)}{2} \right) + \left(\frac{f(x) - f(-x)}{2} \right)$$

Now note that if we let $$E(x) = \left(\frac{f(x) + f(-x)}{2} \right)$$ then $E(x)$ is an even function since $$E(-x) = \left(\frac{f(-x) + f(-(-x))}{2} \right) = \left( \frac{f(-x) + f(x)}{2} \right) = \left(\frac{f(x) + f(-x)}{2} \right) = E(x)$$

Similarly, if we let $$O(x) = \left(\frac{f(x) - f(-x)}{2} \right)$$ then $O(x)$ is an odd function since $$O(-x) = \left(\frac{f(-x) - f(-(-x))}{2} \right) = \left( \frac{f(-x) - f(x)}{2} \right) = -\left(\frac{f(x) - f(-x)}{2} \right) = -O(x)$$

Hence, we have that $E(x)$ is an even function and $O(x)$ is an odd function such that $f(x) = E(x) + O(x)$. Hence, any function can be written as a sum of an even function and an odd function.

The thought process and the motivation is as follows. We want to write $f(x)$ as $E(x) + O(x)$, where $E(x)$ is an even function and $O(x)$ is an odd function. Hence, we have $E(x) + O(x) = f(x)$. Replacing $x$ by $-x$, we get that $E(-x) + O(-x) = f(-x)$. Since we enforce that $E(x)$ is even and $O(x)$ is odd, we get that $f(-x) = E(x) - O(x)$. Hence, we have that $$\begin{align} E(x) + O(x) & = f(x)\\ E(x) - O(x) & = f(-x) \end{align}$$ Solving the above gives us $$\begin{align}E(x) & = \frac{f(x) + f(-x)}{2}\\ O(x) & = \frac{f(x) - f(-x)}{2}\end{align}$$

  • $\begingroup$ Wow, thank you for laying out such a clear thought process - next time I get stuck I'll persevere longer and look back at this answer for motivation. Thanks again! $\endgroup$
    – xlm
    Mar 18, 2012 at 19:16

This is more intuitive if one views it in the special case of polynomials or power series expansions, where the even and odd parts correspond to the terms with even and odd exponents, e.g. bisecting into even and odd parts the power series for $\:\rm e^{{\it i}\:x} \:,\;$

$$\begin{align} \rm f(x) \ &= \ \rm\frac{f(x)+f(-x)}{2} \;+\; \frac{f(x)-f(-x)}{2} \\[.4em] \Rightarrow\quad \rm e^{\,{\large \it i}\,x} \ &= \ \rm\cos(x) \ +\ {\it i} \ \sin(x) \end{align}\qquad$$

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+2n}\ +\:\cdots $

$\rm\quad\quad\, =\,\ \frac{1}{n} \big(f(x)\ +\ f(x\zeta)\ \zeta^{-m}\ +\ f(x\zeta^{\:2})\ \zeta^{-2m}\ +\:\cdots\: +\ f(x\zeta^{\ n-1})\ \zeta^{\ (1-n)m}\big)$

Exercisse $\;$ Use multisections to give elegant proofs of the following

$\quad\quad\rm\displaystyle sin(x)/e^{x} \ \ $ has every $\rm 4k\,$'th term zero in its power series

$\quad\quad\rm\displaystyle cos(x)/e^{x} \ \, $ 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.

  • $\begingroup$ Nice generalization of even and odd functions. I've used this idea in different problems, but never extracted it into something like multisections. $\endgroup$
    – robjohn
    Mar 18, 2012 at 19:13
  • $\begingroup$ Whoa... everything past Euler's formula just blew out my brains. That being said, I really want to appreciate the intuition you describe - thanks for boosting my motivation to continue to study mathematics! Will be working towards solving those exercises! $\endgroup$
    – xlm
    Mar 18, 2012 at 19:28

This might be repeating parts of Sivaram's answer, but I think a reorganization might be enlightening.

Suppose we want to break $f$ into even and odd functions: $f(x)=E(x)+O(x)$ where $E(x)$ is even, that is $E(-x)=E(x)$, and $O(x)$ is odd, that is $O(-x)=-O(x)$. Simply from these considerations, we get $$ \begin{align} f(x)+f(-x) &=(E(x)+O(x))+(E(-x)+O(-x))\\ &=(E(x)+O(x))+(E(x)-O(x))\\ &=2E(x) \end{align} $$ and $$ \begin{align} f(x)-f(-x) &=(E(x)+O(x))-(E(-x)+O(-x))\\ &=(E(x)+O(x))-(E(x)-O(x))\\ &=2O(x) \end{align} $$ Therefore, $$ \begin{array}{} E(x)=\frac{f(x)+f(-x)}{2}&\text{and}&O(x)=\frac{f(x)-f(-x)}{2} \end{array} $$

  • $\begingroup$ Thanks, I really appreciate the different perspective. Will be studying over your thought process. $\endgroup$
    – xlm
    Mar 18, 2012 at 19:19

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.


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