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Proof help

by gimpy
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gimpy
#1
Jan15-04, 10:49 AM
gimpy's Avatar
P: 29
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
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Lonewolf
#2
Jan15-04, 11:13 AM
P: 333
Try considering the function [tex]E(x) = 1/2[F(x) + F(-x)][/tex] and consider what [tex]O(x)[/tex] could be.
gimpy
#3
Jan15-04, 01:05 PM
gimpy's Avatar
P: 29
Originally posted by Lonewolf
Try considering the function [tex]E(x) = 1/2[F(x) + F(-x)][/tex] and consider what [tex]O(x)[/tex] could be.

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

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

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.

NateTG
#4
Jan15-04, 01:08 PM
Sci Advisor
HW Helper
P: 2,537
Proof help

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

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 [tex]f[/tex] can be written as:
[tex]f(x)=E(x)+O(x)[/tex]
where [tex]E[/tex] and [tex]O[/tex] are even and odd respectively. Since we're dealing with even and odd, the natural inclination is to look at [tex]f(-x)[/tex]. Clearly
[tex]f(-x)=E(-x)+O(-x)[/tex]
Now, using the properties of odd and even we get:
[tex]f(-x)=E(x)-O(x)[/tex]
This gives us two equations, and two unknowns [tex]E[/tex] and [tex]O[/tex].
Solving for [tex]E[/tex] and [tex]O[/tex] will give the answer.
If you need an even stronger hint, try writing [tex]f(x)+f(-x)[/tex] in terms of [tex]E(x)[/tex] and [tex]O(x)[/tex]
HallsofIvy
#5
Jan15-04, 02:41 PM
Math
Emeritus
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Thanks
PF Gold
P: 39,683
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.
gimpy
#6
Jan15-04, 03:22 PM
gimpy's Avatar
P: 29
cool, thanks guys, i didn't understand what i was trying to prove.

Ok but does this proove that it is unique?


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