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Even and odd Functions question

  1. Sep 13, 2006 #1
    My prof said "every function is the sum of an even and an odd function, explain."

    ive spent about 2 hours off and on thinking about this and i havent come up with anything really.

    is it because f(g(x)) is an even function if either f(x) or g(x) is even, and you can just split every function into simpler functions , one being even and the other odd? or is that just proving that every function is a product of an even and an odd?
    Last edited: Sep 13, 2006
  2. jcsd
  3. Sep 13, 2006 #2
    Suppose you write f(x) as
    [tex] f(x) = \frac{f(x)+f(-x)}{2} + \frac{f(x)-f(-x)}{2}[/tex]
    Can you say anything about the evenness or oddness of (f(x)+f(-x)) and (f(x)-f(-x))?
  4. Sep 13, 2006 #3
    is (f(x) + f(-x)) an even function because even function, f(x)= f(-x) and
    is (f(x) - f(-x)) an odd function because even functions, f(x) = -f(x)?

    im not really sure
    Last edited: Sep 13, 2006
  5. Sep 13, 2006 #4
    First, do you agree that I can write
    [tex] f(x) = \frac{f(x)+f(-x)}{2} + \frac{f(x)-f(-x)}{2}[/tex]
    for any f(x)?

    Second, do you know the definition of even and odd functions?
    For an even function g(x): g(x) = g(-x)
    For an odd function g(x): g(-x) = -g(x)

    Now let h(x) = f(x)+f(-x)
    and let k(x) = f(x) - f(-x)

    Check to see if h or k are even or odd.
  6. Sep 13, 2006 #5
    umm i agree with your second point but im still not understanding the first one.
  7. Sep 13, 2006 #6
    Ok, the point I'm trying to make is that for any f(x), I can always write

    [tex] f(x) = \frac{f(x)+f(-x)}{2} + \frac{f(x)-f(-x)}{2}[/tex]

    Now I'm going to define

    [tex] h(x) = \frac{f(x)+f(-x)}{2} \;,\;\; k(x) = \frac{f(x)-f(-x)}{2}[/tex]

    By our definitions of h and k, it is clearly true that
    f(x) = h(x) + k(x)
    So if you can show that h(x) is always even and k(x) is always odd, then you are done. To do this, apply the definition of even and odd functions.
    h(x) = 1/2(f(x) + f(-x))
    h(-x) = 1/2(f(-x) + f(--x)) = 1/2(f(-x)+f(x)) = ?

    Then do the same for k(x). If you find that h(x)=h(-x) or h(x)=-h(-x), then you can say h is even or odd.
  8. Sep 13, 2006 #7
    Very Nice, now every thing makes sense thank you soo much but the only thing im still wondering is where did the the 1/2 come from?

    h(x) is the even function and k(x) is the odd function right?

    but that doesnt really show that all functions are a sum of......
    Last edited: Sep 13, 2006
  9. Sep 13, 2006 #8
    A lot of times it helps to rewrite things by cleverly adding 0 or multiplying by 1.
    [tex] f(x) = \frac{2}{2}f(x) = \frac{1}{2}(f(x)+f(x)) = \frac{f(x)}{2}+\frac{f(x)}{2} = \frac{f(x)}{2}+\frac{f(x)}{2} + 0 = \frac{f(x)}{2}+\frac{f(x)}{2} + \left( f(-x) - f(-x) \right) =[/tex]

    [tex] \frac{f(x)}{2}+\frac{f(x)}{2} + \frac{1}{2}\left( f(-x) - f(-x) \right) =[/tex]

    [tex] \frac{f(x)}{2}+\frac{f(x)}{2} + \frac{f(-x)}{2}-\frac{f(-x)}{2} = \left(\frac{f(x)}{2}+\frac{f(-x)}{2} \right) + \left(\frac{f(x)}{2} -\frac{f(-x)}{2} \right)[/tex]
  10. Sep 13, 2006 #9
    HaHa(giddy laugh)

    LeBrad you are a King, thank you very much for you help.
  11. Sep 14, 2006 #10


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    Science Advisor

    By the way, sine and cosine are already odd and even functions (respectively) but ex is not. It's even and odd "parts"
    [tex]\frac{e^x+ e^{-x}}{2}= cosh(x)[/tex]
    [tex]\frac{e^x- e^{-x}}{2}= sinh(x)[/tex]
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