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Derivative of a symetrical function

  1. May 5, 2005 #1

    quasar987

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    If f(x) is symetrical, then

    [tex]f^{(n)}(x) =0, \ \ \ n=1,3,5,...[/tex]

    What would be a proof of that?
     
  2. jcsd
  3. May 5, 2005 #2

    quasar987

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    ummm, no that doesn't make sense, does it. How about

    If f(x) is symetrical and [itex]C^{\infty}[/itex], then the terms of odd powers in the Taylor expansion vanish.

    What would be a proof of that?
     
    Last edited: May 5, 2005
  4. May 6, 2005 #3
    Suppose that the coefficients of the odd terms are not all zero. Look at [itex]f(-x)[/itex] and use the fact that (convergent) Taylor series are equal in an interval iff the coefficients are equal.
     
    Last edited: May 6, 2005
  5. May 6, 2005 #4

    quasar987

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    Great, thank you Data.
     
  6. May 6, 2005 #5

    dextercioby

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    So you meant "even" instead of "symmetric"...The function is "even",its graph is "symmetric" wrt the vertically chosen Oy axis...:wink:

    Daniel.
     
  7. May 6, 2005 #6

    HallsofIvy

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    One can also show that any odd derivative (1, 3, 5, etc.) of an even function is an odd function. Of course, for any odd function, f, f(0)= 0.
     
  8. May 6, 2005 #7

    matt grime

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    And odd functions are also symmetric, rotationally,
     
  9. May 6, 2005 #8

    cepheid

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    Yeah, what does rotational symmetry, or symmetry about the origin mean again? Thinking of graphs of odd functions I can remember (e.g. y = x3), the only thing I can think of is that it means rotating the graph of the function 180o about the origin, gives an image the same as the original.
     
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