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Math Help - Even/Odd proof involving Taylor Series

  1. #1
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    Even/Odd proof involving Taylor Series

    I do not understand how to go about the following proof, any suggestions?

    Suppose f(x) = a0 + a1x + a2x2 + a3x3 + a4x4 + ... for x in an interval
    (-R, R).

    If f is even, prove that:
    0 = a1 = a3 = a5 = a7 = ...

    If f is odd, prove that:
    0 = a0 = a2 = a4 = a6 = ...

    Hint. The coefficients of a power series converging on an interval have to
    be the Taylor coefficients of the function that the series converges to. Thus
    it is not possible for different power series to converge to the same function
    on an interval (-R, R).
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  2. #2
    Super Member Deadstar's Avatar
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    Perhaps this is the wrong method but I'll post it up...

    For the case when f is even, f(x) = f(-x).

    f(x) = a_0 + a_1 x + a_2x^2 + a_3 x^3 + \dots

    f(-x) = a_0 - a_1 x + a_2 x^2 - a_3 x^3 + \dots

    So equating the series you get...

    a_0 + a_1 x + a_2x^2 + a_3 x^3 + \dots = a_0 - a_1 x + a_2x^2 - a_3 x^3 + \dots

    Canceling gives you...
    a_1 x + a_3 x^3 + \dots = - a_1 x - a_3 x^3 - \dots

    Hence equating coefficients gives...

    a_1 = -a_1, a_1 = 0,

    a_3 = -a_3, a_3 = 0,

    etc...

    Similar for the f is odd case (in which f(x) = -f(-x))
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  3. #3
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    Ok, that makes sense. Thanks for the quick reply .
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  4. #4
    Super Member Deadstar's Avatar
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    Quote Originally Posted by crymorenoobs View Post
    Ok, that makes sense. Thanks for the quick reply .
    Like I said it might not be the right method but was the first thing I thought of when I looked at it. Seems right enough though...
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Deadstar View Post
    Perhaps this is the wrong method but I'll post it up...

    For the case when f is even, f(x) = f(-x).

    f(x) = a_0 + a_1 x + a_2x^2 + a_3 x^3 + \dots

    f(-x) = a_0 - a_1 x + a_2 x^2 - a_3 x^3 + \dots

    So equating the series you get...

    a_0 + a_1 x + a_2x^2 + a_3 x^3 + \dots = a_0 - a_1 x + a_2x^2 - a_3 x^3 + \dots

    Canceling gives you...
    a_1 x + a_3 x^3 + \dots = - a_1 x - a_3 x^3 - \dots

    Hence equating coefficients gives...

    a_1 = -a_1, a_1 = 0,

    a_3 = -a_3, a_3 = 0,

    etc...

    Similar for the f is odd case (in which f(x) = -f(-x))
    Correct method although maybe a cleaner way to say it:

    For the even case 2f(x)=f(x)+f(-x)=2\sum_{j=0}^{\infty}a_{2j}\implies a_{2j-1}=0
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