Originally Posted by

**Fabio010** I am going to try to traduce the problem in English. :/

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

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

Why is that solution?

We just need to show that $\displaystyle f(x) = f(x)$?

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

Comment #2.

The **purpose** of the restriction regarding the interval, $\displaystyle -l<x<l\,,$ 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, $\displaystyle -l<x<l\,,$ then the functions, g(x) and h(x) mentioned by **Deveno**, should also include a coordinate translation.

Added in **Edit**:

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