How to show sum of an even and odd function is equal to value of a symmetric function

Question

So imagine i have a symmetric function with the domain -x, how can i prove that we can show this function a the sum of an even and odd function

Answer 1

I'm not sure what you mean by "a symmetric function with domain -x", but every function $f:\mathbb{R}\to \mathbb{R}$ is the sum of an even function and an odd function.

Simply let $g,h$ be defined by \begin{align*} g(x) &= \frac{f(x) + f(-x)}{2}\\[4pt] h(x) &= \frac{f(x) - f(-x)}{2}\\[4pt] \end{align*} Then $g$ is even, $h$ is odd, and $f=g+h$.

The same trick can be applied if instead of $f:\mathbb{R}\to \mathbb{R}$, we have $f:V\to W$, where $V,W$ are vector spaces over a field $K$ of characteristic not equal to $2$.

Update:

In response to the OP's clarification . . .

Note that the domain of $f$ need not be all of $V$.

If $f:D\to W$, where $D$ is a nonempty subset of $V$ such that $x\in D$ implies $-x\in D$, the construction given above still works.