# Proof help

1. Jan 15, 2004

### gimpy

Ok im taking an analysis course and im having trouble with one of these proofs.

Prove any function f(x) can be written uniquely as f(x) = E(x) + O(x) when E is and even function and O is an odd function.

So to try and prove it i did this:
f(-x) = E(-x) + O(-x)
E(-x) = E(x) since E is an even function
O(-x) = -O(x) since O is an odd function
therefore
E(-x) + O(-x) = E(x) + (-O(x)) = E(x) - O(x)

Im sure that is right i just don't see how this could be a proof that any function f(x) can be written uniquely as f(x) = E(x) + O(x) when E is and even function and O is an odd function.

So i did some more calculations:
Suppose E and O are are both even functions, then:
f(-x) = E(-x) + O(-x) = E(x) + O(x)
Suppose E and O are both odd functions, then:
f(-x) = E(-x) + O(-x) = -E(x) + (-O(x)) = -E(x) - O(x) = -(O(x) + E(x))
and finally suppose E is odd and O is even:
f(-x) = E(-x) + O(-x) = -E(x) + O(x)

Am on on the right track of did i completely miss something?

Thanks

Last edited: Jan 15, 2004
2. Jan 15, 2004

### Lonewolf

Try considering the function $$E(x) = 1/2[F(x) + F(-x)]$$ and consider what $$O(x)$$ could be.

3. Jan 15, 2004

### gimpy

Ok so
$$E(x) = 1/2[F(x) + F(-x)]$$
$$O(x) = 1/2[F(x) - F(-x)]$$

then
$$F(x) = E(x) + O(x)$$
$$= 1/2[F(x) + F(-x)] + 1/2[F(x) - F(-x)]$$
$$= 1/2[F(x) + F(-x) + F(x) - F(-x)]$$
$$= 1/2[2F(x)]$$
$$= F(x)$$

Ok this is what i have worked out so far. But i still don't see how it solves my problem. I must be missing something.

4. Jan 15, 2004

### NateTG

Well, there are functions that can't be written that way. For example $$\ln(x)$$ or $$x!$$ cannot be written as the sum of an even function and an odd function because neither of them is defined when $$x<0$$.

A good tactic for situations like this can be to attempt to find problems with the statement.

Let's assume - for now - that a function $$f$$ can be written as:
$$f(x)=E(x)+O(x)$$
where $$E$$ and $$O$$ are even and odd respectively. Since we're dealing with even and odd, the natural inclination is to look at $$f(-x)$$. Clearly
$$f(-x)=E(-x)+O(-x)$$
Now, using the properties of odd and even we get:
$$f(-x)=E(x)-O(x)$$
This gives us two equations, and two unknowns $$E$$ and $$O$$.
Solving for $$E$$ and $$O$$ will give the answer.
If you need an even stronger hint, try writing $$f(x)+f(-x)$$ in terms of $$E(x)$$ and $$O(x)$$

5. Jan 15, 2004

### HallsofIvy

First state the "theorem" clearly:

Any function, f, defined for all real numbers, can be written f(x)= E(x)+ O(x) where E is even and O is odd. If f is not defined for all real numbers, this may not be true.

The problem with your "proof" is that you start by assuming that f(x)= E(x)+ O(x) which is what you are asked to prove.

What LoneWolf suggested was that you define E(x)= (f(x)+ f(-x))/2 and O(x)= (f(x)- f(-x))/2. You showed in your response to that that
f(x)= E(x)+ O(x). The part you are "missing" is that

E(-x)= (f(-x)+ f(-(-x))/2= (f(-x)+ f(x))/2= E(x) and
O(-x)= (f(-x)- f(-(-x))/2= (f(-x)- f(x))/2= -(f(x)- f(-x))/2= -O(x) so that these are even and odd functions.

6. Jan 15, 2004

### gimpy

cool, thanks guys, i didn't understand what i was trying to prove.

Ok but does this proove that it is unique?

Last edited: Jan 15, 2004