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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?

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    More generally see my post below for **bisections** and **multisections** of series http://math.stackexchange.com/questions/3510/how-to-prove-eulers-formula-expi-t-costi-sint/3514#35142010-09-24

4 Answers 4

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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.

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    (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.2017-01-14
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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 $h$.

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    Thanks. Yeah, thinkin$g$ of $t$his as an equation with two unknowns helped me out a lot.2010-09-23
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From intuition:

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

Division by 2 is performed to normalize.

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

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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

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    All this was already explained clearly in the accepted answer. I don't think it is necessary to add one.2017-11-22