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Math Help - Taylor Polynomials

  1. #1
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    Taylor Polynomials

    Now that CaptainBlack and Jhevon have helped me out with Latex notation I can ask my question.

    I was working on taylor polynomials and I ran into a little snag which has me stumped.

    For the function f(x) = \cos (x) we get the taylor series:
    \cos (x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - ... + \frac{(-1)^n f^{(n)} (0) x^{2n}}{(2n)!}
    So for \cos (x^6) we get
    \cos (x^6) = 1 - \frac{x^{12}}{2!} + \frac{x^{24}}{4!} - ... + \frac{(-1)^n f^{(n)} (0) x^{12n}}{(2n)!}

    Yet if we construct the taylor polynomial for  \cos (x^6)
    from scratch:
     f (x) = \cos (x^6)
     f^{'} (x) = -6x^5\sin (x^6)
     f^{''} (x) = -30x^4\sin (x^6) - 36x^{10}\cos (x^6)
     f^{'''} (x) = -120x^3\sin (x^6) - 180x^9\cos (x^6) - 360x^9\cos (x^6) + 216x^{15}\sin (x^6)

    for x= 0
     f (0) = \cos (0^6) = 1
     f^{'} (0) = -6(0^5)\sin (0^6) = 0
     f^{''} (0) = -30(0^4)\sin (0^6) - 36(0^{10})\cos (0^6) = 0
     f^{'''} (0) = -120(0^3)\sin (0^6) - 180(0^9)\cos (0^6) - 360(0^9)\cos (0^6) + 216(0^{15})\sin (0^6)= 0
    in fact we dont see a non zero constant term until the 12th order differential:
    Which looks quite crazy I've just realised. Some of the terms are -36.10! - 18.10! - 8.10! - 3.10! .
    So if we get this, then the numerator on the second term of the taylor polynomial should be massive. So why does letting x = x^6 work? I understand the logic behind it I suppose, as the taylor polynomial work for any value of x and x^6 is just a finite number (for any finite x and for this  cos function, I know it doesn't hold for all functions). But why does the method of constructing the taylor polynomial from the formula:
    \sum_{a=0}^{\infty} \frac{f^{(a)} (0) x^{(a)}}{a!} fail?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by behemoth100 View Post
    Now that CaptainBlack and Jhevon have helped me out with Latex notation I can ask my question.

    I was working on taylor polynomials and I ran into a little snag which has me stumped.

    For the function f(x) = \cos (x) we get the taylor series:
    \cos (x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - ... + \frac{(-1)^n f^{(n)} (0) x^{2n}}{(2n)!}
    So for \cos (x^6) we get
    \cos (x^6) = 1 - \frac{x^{12}}{2!} + \frac{x^{24}}{4!} - ... + \frac{(-1)^n f^{(n)} (0) x^{12n}}{(2n)!}

    Yet if we construct the taylor polynomial for  \cos (x^6)
    from scratch:
     f (x) = \cos (x^6)
     f^{'} (x) = -6x^5\sin (x^6)
     f^{''} (x) = -30x^4\sin (x^6) - 36x^{10}\cos (x^6)
     f^{'''} (x) = -120x^3\sin (x^6) - 180x^9\cos (x^6) - 360x^9\cos (x^6) + 216x^{15}\sin (x^6)

    for x= 0
     f (0) = \cos (0^6) = 1
     f^{'} (0) = -6(0^5)\sin (0^6) = 0
     f^{''} (0) = -30(0^4)\sin (0^6) - 36(0^{10})\cos (0^6) = 0
     f^{'''} (0) = -120(0^3)\sin (0^6) - 180(0^9)\cos (0^6) - 360(0^9)\cos (0^6) + 216(0^{15})\sin (0^6)= 0
    in fact we dont see a non zero constant term until the 12th order differential:
    Which looks quite crazy I've just realised. Some of the terms are -36.10! - 18.10! - 8.10! - 3.10! .
    So if we get this, then the numerator on the second term of the taylor polynomial should be massive. So why does letting x = x^6 work? I understand the logic behind it I suppose, as the taylor polynomial work for any value of x and x^6 is just a finite number (for any finite x and for this  cos function, I know it doesn't hold for all functions). But why does the method of constructing the taylor polynomial from the formula:
    \sum_{a=0}^{\infty} \frac{f^{(a)} (0) x^{(a)}}{a!} fail?
    It does not fail, it gives exactly the same results as substituting x^6 into the expansion of \cos(x).

    Now the direct computation is long winded and it is error prone, and depends
    on the cancellation of large factors, but it does give the same reslt.

    RonL
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  3. #3
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    Hmmm... but from my working you have... oh wait are you saying that the numerator will equal \frac{12!}{2!}?
    Last edited by behemoth100; May 28th 2007 at 03:36 AM.
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by behemoth100 View Post
    Hmmm... but from my working you have... oh wait are you saying that the numerator will equal \frac{12!}{2!}?
    When I run this on a Computer Algebra System, the coefficients of the term
    in x^{12} does indeed cancels down to -1/2

    RonL
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