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Sum of odd & even functions

  1. Jul 18, 2009 #1
    1. The problem statement, all variables and given/known data
    Suppose that the function f has domain all real numbers. Show that f can be written as the sum of an even function and an odd function.


    2. Relevant equations
    f(-x) = f(x) is even and f(-x)=-f(x) is odd


    3. The attempt at a solution
    If g(x) is an even function it can be written as [tex]g(x) = \frac{f(x)+f(-x)}{2} where f(x) is even and if h(x) is odd it can be written as h(x)=\frac{f(x)-f(-x)}{2} [/tex]. but how do you write f = g + h?
     
  2. jcsd
  3. Jul 18, 2009 #2
    [tex]
    f(x) = \frac{f(x)+f(-x)}{2} + \frac{f(x)-f(-x)}{2}
    [/tex]

    Now what can you say about [tex] \frac{f(x)+f(-x)}{2}[/tex] and [tex]
    \frac{f(x)-f(-x)}{2}[/tex] ?
     
  4. Jul 18, 2009 #3

    HallsofIvy

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    Your wording is awkward. It is not "if g is an even function then..." and "if h is odd...". You want to define g(x) to be (f(x)+ f(-x))/2 and show that it is even. Define h(x) to be (f(x)- f(-x))/2 and show that it is odd.

    [tex]g+h= \frac{f(x)+ f(-x)}{2}+ \frac{f(x)- f(-x)}{2}= \frac{f(x)+ f(-x)+ f(-x)+ f(x)- f(-x)}{2}[/tex]
    What is that equal to?
     
  5. Jul 19, 2009 #4
    How did Hallsofivy get five terms in the last equation divided by 2. I would only have got four divided by 2.
     
  6. Jul 19, 2009 #5
    It looks like a typo. The second [itex]+f(-x)[/itex] shouldn't be there.
     
  7. Jul 19, 2009 #6
    And the answer that those four terms divided by 2 is f(x). Now I am embarrassed asking the question, it is so simple.
     
  8. Jul 19, 2009 #7
    Haha. No worries. That was what VeeEight was getting at as well. It is a neat problem that is just a little trick of dividing by 2. No need to feel embarrassed.
     
  9. Jul 19, 2009 #8

    HallsofIvy

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    Yes, my eyes went squoggly for a moment. Thanks, n!kofeyn.
     
  10. Jul 19, 2009 #9
    Thanks all of you for your replies.
     
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