
#1
Jan1504, 10:49 AM

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 



#2
Jan1504, 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.




#3
Jan1504, 01:05 PM

P: 29

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. 



#4
Jan1504, 01:08 PM

Sci Advisor
HW Helper
P: 2,538

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] 



#5
Jan1504, 02:41 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,879

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. 


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