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Math Help - [SOLVED] Quick Analysis Proof: Sum of odd and even functions.

  1. #1
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    [SOLVED] Quick Analysis Proof: Sum of odd and even functions.

    Show f(x) can be expressed as the sum of E(x) and an odd function O(x).

    f(x) is defined for all x (assume domain D symmetric about 0)

    E(x) = \frac{f(x) + f(-x)}{2}

    f(x) is even if f(x) = f(-x)
    f(x) is odd if f(x) = -f(x)

    I'm not quite sure how to begin doing that.
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by Iceflash234 View Post
    Show f(x) can be expressed as the sum of E(x) and an odd function O(x).

    f(x) is defined for all x (assume domain D symmetric about 0)

    E(x) = \frac{f(x) + f(-x)}{2}

    f(x) is even if f(x) = f(-x)
    f(x) is odd if f(x) = -f(x)

    I'm not quite sure how to begin doing that.
    I'm not exactly sure what you are asked to do. But I can tell you this: every function whose domain is symmetric with respect to 0 (such that x is in the domain of f if and only if -x is in the domain of f) can be uniquely decomposed into an even function E(x) and an odd function O(x) such that f(x)=E(x)+O(x) holds for all x in the domain of f.

    Just define the even part of f(x) to be E(x) := \frac{f(x)+f(-x)}{2}, and the odd part of f(x) to be O(x) := \frac{f(x)-f(-x)}{2}.

    First, it is easy to show that E(x) is an even function. For this amounts to showing that for all x in the domain of E (and thus in the domain of f): E(-x)=E(x) holds. Proof:

    E(-x)=\frac{f(-x)+f(-(-x))}{2}=\frac{f(-x)+f(x)}{2}=\frac{f(x)+f(-x)}{2}=E(x).
    And note: we have done nothing fancier here than first substituting -x for x in the definition of E(x), and then applying very elementary algebra, until we ended up with E(x) again.

    Similarly, to show that O(x) is odd, we need to show that for all x in its domain: O(-x)=-O(x). Proof:

    O(-x)=\frac{f(-x)-f(-(-x))}{2}=\frac{f(-x)-f(x)}{2}=-\frac{f(x)-f(-x)}{2}=-O(x)

    If you now take two decompositions of f into even and odd functions, such as f(x)=E_1(x)+O_1(x) and f(x)=E_2(x)+O_2(x), then it follows that for all x in the domain of f
    E_1(x)+O_1(x)=E_2(x)+O_2(x) and thus E_1(x)-E_2(x)=O_2(x)-O_1(x).
    But the difference of two even functions is even, and the difference of two odd functions is odd. Thus E_1(x)-E_2(x) and O_2(x)-O_1(x) must be both even and odd functions at the same time, from which it follows that E_1(x)-E_2(x)=0 and O_2(x)-O_1(x)=0 for all x. Hence E_1=E_2 and O_1=O_2; which is to say that the decomposition of f(x) into an even and an odd part like this is unique.

    Maybe this helps - or maybe not...
    Last edited by Failure; April 8th 2010 at 11:30 AM.
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  3. #3
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    Thanks. That really helps a lot.

    I do have one question though. How does it look if f(x) = e^x?
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  4. #4
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    Quote Originally Posted by Iceflash234 View Post
    Thanks. That really helps a lot.

    I do have one question though. How does it look if f(x) = e^x?
    you get
    f(x)=e^x=\frac{e^x+e^{-x}}{2}+\frac{e^x-e^{-x}}{2}=\cosh(x)+\sinh(x)
    The even and odd part of e^x have special names: they are the hyperbolic cosine, and the hyperbolic sine, respectively.
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