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Prove derivative of an odd function is even and vise versa.

  1. Mar 24, 2010 #1
    1. The problem statement, all variables and given/known data
    Prove if f(x) is defined on -L< x < L, and if f(x) is odd function on (-L,L), the f'(x) is even function. and vise versa.


    2. Relevant equations

    Using Fourier series expansion, f(x) is odd function on (-L,L) can be represented by Fourier sine series expansion. If f(x) is even function on (-L,L), the it can be represented by cosine series.



    3. The attempt at a solution

    By series expansion of odd and even function, if f(x) is odd and represented by sine series, then the derivative of the sine series is cosine series and it become an even function. And the reverse is true from even to odd by derivative.


    Anyone have a better way of proving this than what I have instead of refering to series expansion?
     
  2. jcsd
  3. Mar 24, 2010 #2

    gabbagabbahey

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    You could use the limit definition of f'(-x) to show that it is equal to f'(x)
     
  4. Mar 24, 2010 #3
    Let [itex]n(x) = -x[/itex] so basically just a "negation function". To say that f is even is to say f(x)=f(-x) or in other words [itex]f = f \circ n[/itex], and to say it's odd is to say -f(x)=f(-x) or in other words [itex]-f = f\circ n[/itex].

    Now use:
    [tex](f \circ n)(x)' = f'(n(x)) n'(x) = -f'(-x)[/tex]
    in both cases.
     
  5. Mar 24, 2010 #4
    Is [tex]f\circ n = f(n(x))=f(-x)?[/tex]
     
    Last edited: Mar 24, 2010
  6. Mar 24, 2010 #5

    Office_Shredder

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    Yes it is
     
  7. Mar 24, 2010 #6
    Let me clarify:

    A) Let f(x) be even:

    [tex]\Rightarrow f(-x)=f(x)[/tex]

    [tex]f(-x)= f(n(x)) =f(x)\Rightarrow [f(n(x))]'=f'(x)[/tex] (1)

    [tex] [df(n(x))]' =f'(nx)n'(x)=-f'(n(x))=-f'(-x)[/tex] (2)

    (1) and (2)[tex]\Rightarrow f'(x)=-f'(-x)[/tex]

    Therefore if f(x) is even, f'(x) is odd.




    B) Let f(x) be odd:

    [tex]\Rightarrow f(-x)=-f(x)[/tex]

    [tex]f(-x)= f(n(x)) =-f(x)\Rightarrow [f(n(x))]'=-f'(x)[/tex] (3)

    [tex] [df(n(x))]' =f'(nx)n'(x)=-f'(n(x))=-f'(-x)[/tex] (4)

    (3) and (4)[tex]\Rightarrow -f'(x)=-f'(-x)[/tex]

    Therefore if f(x) is odd, f'(x) is even



    Tell me whether I got this right? Does this apply to integration that integrate of an odd function is an even function plus a constant and vise versa?

    Thanks

    Alan
     
  8. Mar 24, 2010 #7
    Yes you did.

    Given a function f(x) that is integrable on the appropriate intervals define:
    [tex]F(x) = \int_0^x f(t) dt[/tex]
    Then you wish to show that f(x) is odd imply F(x) even, and f(x) even imply F(x) odd. In general we have the identity:
    [tex]\int_a^b f(t) dt = -\int_{-a}^{-b}f(-t)dt[/tex]
    You can use that to write:
    [tex]F(-x) = \int_0^{-x} f(t)dt = -\int_0^x f(-t) dt[/tex]
    Now you can use f(-t)=f(t) or f(-t)=-f(t) depending on whether you're in the odd or even case to investigate how F(-x) relates to F(x).
     
  9. Mar 25, 2010 #8
    Thanks for your help, I need to verify again.


    Let u=-t, therefore du=-dt. Also the integration of t from -a to -b imply integration of u from a to b.

    [tex]\Rightarrow -\int_{-a}^{-b}f(-t)dt = \int_{a}^{b}f(u)du = F(b)-F(a)[/tex]


    [tex]\int_{a}^{b}f(t)dt = F(b)-F(a)\Rightarrow \int_a^b f(t) dt = -\int_{-a}^{-b}f(-t)dt = F(b)-F(a)[/tex]

    [tex]\Rightarrow F(x)=\int_0^x f(t)dt = -\int _0^{-x}f(-t)dt[/tex]

    If f(x) is even, f(-t)=f(t):

    [tex]\Rightarrow F(x)=\int_0^x f(t)dt = -\int _0^{-x}f(-t)dt = -\int _0^{-x}f(t)dt = -[F(-x)-F(0)] = -F(-x)[/tex]

    This give F(x)=-F(-x) which imply if f(x) is even, F(x) is odd. Therefore integration of an even function gives an odd function.

    There other way would be the same that integrate of an odd function give an even function.

    Am I correct?
     
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