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