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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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You mean, how did anybody come up with them in the first place? –  Arturo Magidin Sep 23 '10 at 3:27
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Personally, I wonder who first thought of decomposing $\exp(x)$ as $\sinh(x)+\cosh(x)$ –  J. M. Sep 23 '10 at 3:40
    
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): $$A=(A+A^t)/2 + (A-A^t)/2$$. –  Hans Lundmark Sep 23 '10 at 6:43
    
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. –  Qiaochu Yuan Sep 23 '10 at 7:38
    
More generally see my post below for bisections and multisections of series math.stackexchange.com/questions/3510/… –  Bill Dubuque Sep 24 '10 at 3:08
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3 Answers 3

up vote 24 down vote accepted

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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Thanks so much! –  Mana Sep 23 '10 at 3:59
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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, thinking of this as an equation with two unknowns helped me out a lot. –  Mana Sep 23 '10 at 4:00
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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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