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Proof help

  1. Jan 15, 2004 #1
    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
    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?

    Last edited: Jan 15, 2004
  2. jcsd
  3. Jan 15, 2004 #2
    Try considering the function [tex]E(x) = 1/2[F(x) + F(-x)][/tex] and consider what [tex]O(x)[/tex] could be.
  4. Jan 15, 2004 #3

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

    [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.
  5. Jan 15, 2004 #4


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    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:
    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
    Now, using the properties of odd and even we get:
    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]
  6. Jan 15, 2004 #5


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    Staff Emeritus
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    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.
  7. Jan 15, 2004 #6
    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
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