For the function $f(x)$ we can write it as sum of even and odd functions:


My question is why it is important for us to write a function as sum of these two even and odd functions? Is there any application of that?

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    $\begingroup$ I think it may be useful in calculating definite integral. $\endgroup$ Commented Dec 12, 2020 at 18:03
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    $\begingroup$ I recently came across this video: youtube.com/watch?v=h7apO7q16V0&ab_channel=Reducible (the FFT algorithm), where even and oddness played a fundamental role. Perhaps it's too advanced for you for now, but the key reason why they were useful was because even and oddness both are a very special kind of symmetry, and the algorithm uses this symmetry to reduce the amount of work by HALF! $\endgroup$
    – D.R.
    Commented Dec 13, 2020 at 4:22
  • $\begingroup$ What book are you referring too @Soheil? These are some nice tricks that one doesn't see often $\endgroup$ Commented Dec 13, 2020 at 10:21
  • $\begingroup$ @Buraian: This is a well know equation in calculus I saw it in my notes and also in some of my translated books. $\endgroup$
    – Soheil
    Commented Dec 13, 2020 at 13:43
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    $\begingroup$ @Buraian I am not familiar with such books that contain calculus tricks. I myself try learning tricks by reading some elegant answers on this site and get idea from them (for example I learned how even functions would help to decompose a fraction by Bernard's answer on the question: math.stackexchange.com/questions/3615427/…) maybe other user have more experience about it. $\endgroup$
    – Soheil
    Commented Dec 13, 2020 at 16:04

4 Answers 4


When I was a high school student I thought that the even/odd decomposition you write about seemed kind of peculiar and not so fundamental. After learning more mathematics I realized the method behind it (extracting "symmetric pieces" by averaging and what you might call anti-averaging) is actually a very simple example of two important processes in mathematics: eigenspace decompositions and averaging over a group to extract symmetric pieces of a function (or vector, etc.). What I write below is not meant to give you new situations where your even/odd decomposition helps solve a calculus problem, but to show you many further examples of the same idea so you see it is quite broadly occurring in mathematics.

In nearly every situation where there is an operation that iterates twice to be the identity operation you get an analogue of the even/odd decomposition. Here are three examples.

  1. The matrix transpose (where $M^{\top\top} = M$) leads to the expression of a square matrix as a sum of matrices that are symmetric ($M^\top = M$) and skew-symmetric ($M^\top = -M$) $$ A = \frac{A + A^\top}{2} + \frac{A - A^\top}{2} $$

  2. Complex conjugation (where $\overline{\overline{z}} = z$) gives an "even/odd" type viewpoint on writing a complex number in standard form is $a+bi$, since this is the sum of a real number (fitting $\overline{w} = w$) and a purely imaginary number (fitting $\overline{w} = -w$): $$ z = \frac{z + \overline{z}}{2} + \frac{z - \overline{z}}{2} = a + bi $$ where $z = a + bi$ and $\overline{z} = a - bi$.

  3. The swap operator on functions ($f(x,y) \mapsto f(y,x)$) or tensors ($v \otimes w \mapsto w \otimes v$) leads to the expression of a function or tensor as a sum of symmetric and antisymmetric functions or tensors: $$ f(x,y) = \frac{f(x,y) + f(y,x)}{2} + \frac{f(x,y) - f(y,x)}{2} $$ and $$ v \otimes w = \frac{v \otimes w + w \otimes v}{2} + \frac{v \otimes w - w \otimes v}{2}. $$ This has a role in quantum mechanics, where it underlies the distinction between bosons (having symmetric wavefunctions) and fermions (having antisymmetric wavefunctions).

I said that in nearly every situation you get something like an even/odd decomposition because sometimes one of those parts is zero and thus uninteresting. For instance, a 180-degree rotation $R$ of the plane has $R(v) = -v$ for all $v$ in $\mathbf R^2$, so here the whole space "looks odd" under the effect of $R$. No vector in $\mathbf R^2$ is fixed by a 180-degree rotation except for the origin.

The use of "order $2$" here keeps the algebra very simple, but we can also consider higher-order symmetries rather than symmetries of order 2. Consider for each $n \geq 1$ trying to decompose a function $f:\mathbf C \to \mathbf C$ as a sum of functions $f_k(z)$ that get scaled by the $k$th power of each $n$th root of unity when it undergoes interior scaling by each $n$th root of unity: $f_k(\zeta z) = \zeta^k f_k(z)$ for all $n$th roots of unity $\zeta$ (or equivalently just $\zeta = e^{2\pi i/n}$) and all complex numbers $z$, where $0 \leq k \leq n-1$. The case $n=2$ is even/odd functions on $\mathbf C$ ($f_0(-z) = f_0(z)$ means $f_0$ is an even function and $f_1(-z) = -f_1(z)$ means $f_1$ is an odd function). Taking $n = 4$, we can try to decompose each function $f:\mathbf C \to \mathbf C$ as a sum of four functions $$ f(z) = f_0(z) + f_1(z) + f_2(z) + f_3(z) $$ where $f_0(iz) = f_0(z)$, $f_1(iz) = if_1(z)$, $f_2(iz) = -f_2(z)$, and $f_3(iz) = -if_3(z)$ for all $z \in \mathbf C$.Here are formulas for each of the functions: $$ f_0(z) = \frac{f(z) + f(iz) + f(-z) + f(-iz)}{4}, $$ $$ f_1(z) = \frac{f(z) - if(iz) - f(-z) + if(-iz)}{4}, $$ $$ f_2(z) = \frac{f(z) - f(iz) + f(-z) - f(-iz)}{4}, $$ $$ f_3(z) = \frac{f(z) + if(iz) - f(-z) - if(-iz)}{4}. $$ These averaging formulas are generalizations of the formulas you wrote for determining the even/odd parts of a function $\mathbf R \to \mathbf R$. And this is useful in Fourier analysis, since the Fourier transform on functions has order $4$.

The ideas presented here extend even further to the decomposition of a representation of a finite group as a sum of irreducible representations. For the cyclic group of order $2$ there are two irreducible representations, and that is reflected in the appearance of even functions and odd functions in your formula. So the even/odd decomposition for functions in your question is a special case of a really important idea in math. It is not just some "trick" to solve artificial calculus problems.

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    $\begingroup$ Thanks for the answer. but these are all about advanced math and I don't know about them unfortunately. $\endgroup$
    – Soheil
    Commented Dec 12, 2020 at 18:09
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    $\begingroup$ Please indicate in your question what your math background is if you want to get answers targeted to what you know now as opposed to what you may learn in the future if you go further in math. $\endgroup$
    – KCd
    Commented Dec 12, 2020 at 18:27
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    $\begingroup$ @KCd could you explain what a higher-order symmetry is? I found your answer fascinating but I am not familiar with all the terminology used. $\endgroup$ Commented Dec 13, 2020 at 10:19
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    $\begingroup$ @Buraian a higher-order symmetry is one that has to be iterated more than twice to return to the identity (fixing everything). While sending $x$ to $-x$ has order 2, sending $x$ to $ix$ has order 4, because repeated multiplication has the effect $x \leadsto ix \leadsto -x \leadsto -ix \leadsto x$, so (for nonzero $x$) we return to $x$ only after 4 iterations of this operation. $\endgroup$
    – KCd
    Commented Dec 13, 2020 at 10:29
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    $\begingroup$ @KCd This question of mine, and Moishe Kohan's answer, are very much in the spirit of your answer here. I thought you might enjoy it: math.stackexchange.com/q/755402/34287 $\endgroup$ Commented Dec 13, 2020 at 14:48

One really neat application for this decomposition (which I saw on the YouTube channel "Flammable Maths") is evaluating integrals of the form $$\int_{-a}^a\Bigg(\frac{E(x)}{1+t^{O(x)}}\Bigg)dx$$ where $t,a>0$ are constants, $E(x)$ is a (continuous) even function, and $O(x)$ is a (continuous) odd function. If you set $f(x)=\frac{E(x)}{1+t^{O(x)}}$ and write $$f(x)=\frac{f(x)+f(-x)}{2}+\frac{f(x)-f(-x)}{2}$$ you can say that $$\int_{-a}^a\Bigg(\frac{E(x)}{1+t^{O(x)}}\Bigg)dx=\int_{-a}^a\Bigg(\frac{f(x)+f(-x)}{2}\Bigg)dx+\int_{-a}^a\Bigg(\frac{f(x)-f(-x)}{2}\Bigg)dx$$ The last integral on the RHS vanishes since we're integrating an odd function on a symmetric domain. With a little algebra $\frac{f(x)+f(-x)}{2}=\frac{1}{2}E(x)$ giving us the awesome result $$\int_{-a}^a\frac{E(x)}{1+t^{O(x)}}dx=\int_{0}^aE(x)dx$$ which is really cool! This means we can say something like $$\int_{-1}^1\Bigg(\frac{x^4-x^2+1}{1+3^{\sin^2(x)\tan(x)+x^5+x}}\Bigg)dx=\int_0^1\big(x^4-x^2+1\big)dx=\frac{13}{15}$$ This can also be used to calculate some pretty nasty double integrals! $$\int_0^1 \int_{-x^2}^{x^2}\Bigg(\frac{xy^2+x^3}{1+3^{x\tan^{11}(y)+e^x\sin^7(y)}}\Bigg)dydx=\int_0^1 \int_0^{x^2}(xy^2+y^3)dydx=\frac{5}{24}$$ Love it.

Edit: This integration technique actually generalizes to integrals of the form $$\int_{-a}^a\Bigg(\frac{E_1(x)}{1+\big(E_2(x)\big)^{O(x)}}\Bigg)dx$$ where $E_1(x),E_2(x)$ are arbitrary (continuous) even functions while $O(x)$ is an arbitrary (continuous) odd function. Using the exact same procedure delineated above we can say $$\int_{-a}^a\Bigg(\frac{E_1(x)}{1+\big(E_2(x)\big)^{O(x)}}\Bigg)dx=\int_{0}^aE_1(x)dx$$ which means $$\int_{-1}^1\Bigg(\frac{x^4+x^2+1}{1+\big(x^2e^{-x^4}+\cos(x)\sin(x^2)\big)^{x^2\tan(x^3)+x}}\Bigg)dx=\int_0^1(x^4+x^2+1)dx=\frac{23}{15}$$

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    $\begingroup$ I was really confused by your notation, because I was reading $e(x)$ as $\exp(x)$ and couldn't understand why $e^x$ was even. Also, didn't grasp the $o$ in your denominator as a function, despite the text. Even though the definitions in your description are unambiguous, I think it costs some brain power to parse. (Disclaimer: I'm no mathematician.) $\endgroup$
    – EPrivat
    Commented Dec 13, 2020 at 4:40
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    $\begingroup$ Thank you for your comment. I just borrowing the notation from the video in which I saw this application. I will change my presentation to avoid further complications of this sort. $\endgroup$
    – user801306
    Commented Dec 13, 2020 at 4:43
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    $\begingroup$ I confirm that it is less confusing with capital letters. (Missed one in the integral on the third math display block.) $\endgroup$
    – EPrivat
    Commented Dec 13, 2020 at 4:49
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    $\begingroup$ Please use something like $\frak E$ or $\frak O$, or revert back to $E$ and $O$ as $\Bbb E$ denotes expectation, for example. $\endgroup$
    – TheSimpliFire
    Commented Dec 14, 2020 at 10:45

A famous example of a decomposition in odd and even functions is given by Euler's formula \begin{align*} \color{blue}{e^{iz}}&\color{blue}{=}\color{blue}{\cos z+i\sin z}\\ &=\frac{e^{iz}+e^{-iz}}{2}+\frac{e^{iz}-e^{-iz}}{2}\qquad\qquad z\in\mathbb{C}\\ \end{align*} which is used in many applications.


The answer by KCd mentions in passing what I'll talk about, but I'll elaborate on it: the short answer is Fourier analysis.

Splitting a function into odd and even components is an extremely useful problem-solving technique when working with the Fourier transform, and the associated Fourier series. A function that is purely even or purely odd is easier to find the Fourier transform/series of.

That may seem like a niche topic, but Fourier analysis is one of the most powerful and widely used mathematical techniques. You cannot go far into any STEM field without encountering it, and so making Fourier analysis easier is more significant than you might think.

There is a wealth of knowledge on the internet about what Fourier analysis is and how it works, so I won't reiterate it here. I've found this YouTube video as a good introduction to the topic.


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