While working on a proof showing that all functions limited to the domain of real numbers can be expressed as a sum of their odd and even components, I stumbled into a troublesome roadblock; namely, I had no clue how one divides the function into these even and odd parts.

Looking up a solution for the proof, I found these general formulas for the even and odd parts of a function $f(n)$:

$$\begin{align*} f_e(n)&\overset{\Delta}{=}\frac{f(n)+f(-n)}{2}\\ f_o(n)&\overset{\Delta}{=}\frac{f(n)-f(-n)}{2} \end{align*}$$

While I understand that in an even function $f(n) = f(-n)$ and that in an odd function $f(-n) = -f(n)$, I still don't get how these general formulas for the even and odd parts were obtained. Can someone guide me through the logic?

  • $\begingroup$ You mean, how did anybody come up with them in the first place? $\endgroup$ – Arturo Magidin Sep 23 '10 at 3:27
  • 5
    $\begingroup$ Personally, I wonder who first thought of decomposing $\exp(x)$ as $\sinh(x)+\cosh(x)$ $\endgroup$ – J. M. is a poor mathematician Sep 23 '10 at 3:40
  • 1
    $\begingroup$ One can do a similar thing for any involution (a linear transformation such that doing it twice brings you back where you started). For example, transposing a matrix is such an operation. This means that any matrix can be written as a symmetric part (which is unchanged by the operation) plus an antisymmetric part (which flips its sign): $$A=(A+A^t)/2 + (A-A^t)/2$$. $\endgroup$ – Hans Lundmark Sep 23 '10 at 6:43
  • 1
    $\begingroup$ I'd like to mention that this is a very special case of the discrete Fourier transform (en.wikipedia.org/wiki/Discrete_Fourier_transform); you should think of the even and odd parts as different "frequencies" into which you are decomposing f. $\endgroup$ – Qiaochu Yuan Sep 23 '10 at 7:38
  • $\begingroup$ More generally see my post below for bisections and multisections of series math.stackexchange.com/questions/3510/… $\endgroup$ – Bill Dubuque Sep 24 '10 at 3:08

Suppose you could write a function $f(x)$ as the sum of an even and an odd function; call them $E(x)$ and $O(x)$.

In particular, you would have \[f(x) = E(x)+O(x)\] and you would also have \[f(-x) = E(-x) + O(-x) = E(x) - O(x)\] with the latter equation because we are assuming $E$ is even and $O$ is odd, so $E(x)=E(-x)$ and $O(-x) = -O(x)$.

Adding both equations you get $f(x)+f(-x) = 2E(x)$. Subtracting the second equation from the first gives you $f(x)-f(-x)=2O(x)$. Now solve for $E(x)$ and $O(x)$, and you get the formulas you see in the solution. Then you check that the answer does indeed work (that is, you check that the formulas you found do give you an even and an odd function in all cases).

In other words: pretend you already know the answer, and try to deduce conditions that the answer must satisfy (these will be necessary conditions); if things go well, you'll get enough information about what they must be like to figure out what they are.

  • $\begingroup$ This doesn't make much sense to me. You are basically assuming that f(x) can be divided up into an even function and odd function, then you are finding the even/odd functions in terms of f(x). What does this accomplish, when you are trying to prove that f(x) can be written as an even + odd function? $\endgroup$ – Goldname Jan 14 '17 at 3:39
  • 2
    $\begingroup$ @Goldname: As I explained six and a half years ago, this gives you necessary conditions that the even and odd functions must satisfy. Having obtained these condition, you may check whether they are also sufficient; that is, you may verify that the given formulas do, in fact, produce an even and an odd function whose sum is $f$. By explicitly producing these two functions, you prove that $f$ can be so expressed. How you found those functions is irrelevant. $\endgroup$ – Arturo Magidin Jan 14 '17 at 4:09
  • $\begingroup$ So you're saying these formulas just give the conditions that the even and odd functions must satisfy, if they exist? You aren't actually proofing that they exist? $\endgroup$ – Goldname Jan 14 '17 at 4:35
  • 1
    $\begingroup$ I am proving they exist, because I am exhibiting them. If you are asked to prove that there exists an integer $x$ such that $x+4=5$, then you can establish its existence by exhibiting the number $1$ and showing it satisfies the given condition (when you add it to four, you get five). How do you find that number is irrelevant for the purposes of demonstrating its existence. In the matter at hand, we are assuming the existence of $E$ and $O$ in order to obtain necessary conditions they must satisfy; if such functions exist, then they must satisfy that $E(x)=(f(x)+f(-x))/2$ (cont) $\endgroup$ – Arturo Magidin Jan 14 '17 at 4:38
  • 1
    $\begingroup$ (cont) and $O(x) = (f(x)-f(-x))/2$. This, by itself, does not prove that such functions exist. But now, knowing that they must satisfy these conditions, I can say: Let the function $O$ be defined as $O(x) = (f(x)-f(-x))/2$, and let the function $E$ be defined as $E(x) = (f(x)+f(-x))/2$. Then we prove that $E(-x)=E(x)$ (so $E$ is even); that $O(-x)=-O(x)$ (so $O$ is odd), and that $f(x)=E(x)+O(x)$, thus proving, by explicitly producing the functions, that $f$ can be written as the sum of an even function and an odd one. $\endgroup$ – Arturo Magidin Jan 14 '17 at 4:40

Suppose $f(x)=g(x)+h(x)$ with $g$ even and $h$ odd. Then $f(-x)=g(-x)+h(-x)=g(x)-h(x)$. Think of $f(x)=g(x)+h(x)$ and $f(-x)=g(x)-h(x)$ as a system of two equations in two unknowns, $g(x)$ and $h(x)$, to solve for $g$ and $h$.

  • 1
    $\begingroup$ Thanks. Yeah, thinking of this as an equation with two unknowns helped me out a lot. $\endgroup$ – Mana Sep 23 '10 at 4:00

From intuition:

$$f_e(t) = \frac{f(t)+f(-t)}{2} $$

Division by 2 is performed to normalize.

$$f(t)=f_e(t)+f_o(t) \implies f_o(t)= f(t)-\frac{(f(t) +f(-t))}{2}=\frac{f(t)-f(-t)}{2}.$$


f(x)= fe(x)+fo(x).......(1)

The properties of fe(x) and fo(x) are:

fe(-x)= fe(x) And fo(-x)=-fo(x)

Now we have f(-x) = fe(-x) +fo(-x)...(2)

Adding eq (1) and (2) we have fe(x) =( f(x)+f(-x) )/2

Subtracting (1) and (2) we have fo(x) =( f(x)-f(-x) )/2

  • 2
    $\begingroup$ All this was already explained clearly in the accepted answer. I don't think it is necessary to add one. $\endgroup$ – Arnaud D. Nov 22 '17 at 10:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.