# Thread: Even and odd functions.

1. ## Even and odd functions.

I am going to try to traduce the problem in english. :/

Demonstrate that any function $f(x)$, in interval $-l , can be represented as the sum of even and odd functions.

Solution is: $f(x)=$ $\frac{1}{2}$ $[f(x)+f(-x)]$+ $\frac{1}{2}$ $[f(x)-f(-x)]$

Why is that solution?
We just need to show that $f(x) = f(x)$?

2. ## Re: Even and odd functions.

of course, the two sides are equal.

the point is, if we define:

$g(x) = \frac{f(x) + f(-x)}{2},\ h(x) = \frac{f(x) - f(-x)}{2}$

then $f(x) = g(x) + h(x)$, as you observed, but the "real interesting" part is:

$g(-x) = \frac{f(-x) + f(-(-x))}{2} = \frac{f(-x) + f(x)}{2} = g(x)$ so g is even, and:

$h(-x) = \frac{f(-x) - f(-(-x))}{2} = \frac{f(-x) - f(x)}{2} = -h(x)$ so h is odd.

3. ## Re: Even and odd functions.

So the functions can be expressed with the sum of even and odd functions. as problem say.

Thanks for the explanation.

4. ## Re: Even and odd functions.

Originally Posted by Fabio010
I am going to try to traduce the problem in English. :/

Demonstrate that any function $f(x)$, in interval $-l , can be represented as the sum of even and odd functions.

Solution is: $f(x)=\frac{1}{2}[f(x)+(f-x)]+\frac{1}{2}[f(x)-(f-x)]$

Why is that solution?
We just need to show that $f(x) = f(x)$?
Comment #1:

You have a mistake in your solution, not caused but translation into English; what is known as a "typo" -- a typographical error.

The solution should be:
$f(x)=\frac{1}{2}[f(x)+f(-x)] \frac{1}{2}[f(x)-f(-x)]$
Comment #2.

The purpose of the restriction regarding the interval, $-l is not at all clear to me. If by any chance the functions are supposed to be even or odd with respect to the center of the interval, $-l then the functions, g(x) and h(x) mentioned by Deveno, should also include a coordinate translation.

The above equation is missing a plus sign. It should be:
$f(x)=\frac{1}{2}[f(x)+f(-x)]+ \frac{1}{2}[f(x)-f(-x)]$

5. ## Re: Even and odd functions.

Originally Posted by SammyS
Comment #1:

You have a mistake in your solution, not caused but translation into English; what is known as a "typo" -- a typographical error.

The solution should be:
$f(x)=\frac{1}{2}[f(x)+f(-x)] \frac{1}{2}[f(x)-f(-x)]$

Comment #2.

The purpose of the restriction regarding the interval, $-l is not at all clear to me. If by any chance the functions are supposed to be even or odd with respect to the center of the interval, $-l then the functions, g(x) and h(x) mentioned by Deveno, should also include a coordinate translation.
1) the center of the interval $\ (-l,l)$ is 0.

2) the product you give is not equal to f. to see this, suppose we have f(x) = x (a perfectly ordinary function). then:

$\frac{1}{2}[f(x)+f(-x)]\frac{1}{2}[f(x) - f(-x)] = \frac{1}{4}(x + (-x))(x - (-x))$

$= \frac{1}{4}(0)(2x) = 0$ which is certainly NOT equal to f(x) = x.

6. ## Re: Even and odd functions.

Deveno explained me the solution correctly.
Topic is Solved. Thanks for the help.

7. ## Re: Even and odd functions.

Originally Posted by Deveno
1) the center of the interval $\ (-l,l)$ is 0.

2) the product you give is not equal to f. to see this, suppose we have f(x) = x (a perfectly ordinary function). then:

$\frac{1}{2}[f(x)+f(-x)]\frac{1}{2}[f(x) - f(-x)] = \frac{1}{4}(x + (-x))(x - (-x))$

$= \frac{1}{4}(0)(2x) = 0$ which is certainly NOT equal to f(x) = x.
I had a typo, correcting Fabio's typo is all.

Before Fabio edited his Original Post, he ha a couple of typos (See this quote in post #3). Then I inadvertently dropped the + sign between the two functions. I did a lot of cutting & pasting, because there are a lot of [​TEX​] [​/TEX​] tags in his equations.

8. ## Re: Even and odd functions.

Yes, i wrote $(f-x)$ instead of $f(-x)$
but i corrected before your post.