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Taylor series approximation of 1/sqrt x

  • 30-11-2011 8:53pm
    #1
    irishdude11
    Registered Users Posts: 377 ✭✭


    Im trying to find the taylor series approx of 1/sqrt x, centered at a = 9

    So here are my derivatives

    f'(x) = -1/2 . x ^-3/2
    f''(x) = 3/4 . x ^-5/2
    f'''(x) = -15/8 . x ^-7/2
    f''''(x) = 105/16 . x ^-9/2

    f'(9) = -1/2 . 1/(3^3)
    f''(9) = 3/4 . 1/(3^5)
    f'''(9) = -15/8 . 1/(3^7)
    f''''(9) = 105/16 . 1/(3^9)

    So in trying to find a sigma expression for the taylor series I have the following -

    Σ (-1)^n . ###/2^n . 1/3^(2n+1) . (x-9)^n

    The ### is the part I cant get - As you can see it is coming out like this -
    n = 1 ===> 1
    n = 2 ===> 3
    n = 3 ===> 15
    n = 4 ===> 105
    n = 5 ===> 945

    How can I put this in terms of n so I can complete my taylor series representation?


Comments

  • #2
    ray giraffe
    Registered Users Posts: 338 ✭✭ray giraffe


    irishdude11 wrote: »
    n = 1 ===> 1
    n = 2 ===> 3
    n = 3 ===> 15
    n = 4 ===> 105
    n = 5 ===> 945

    How can I put this in terms of n so I can complete my taylor series representation?

    1 -> 1
    2 -> 1x3
    3 -> 1x3x5
    4 -> 1x3x5x7
    5 -> 1x3x5x7x9
    6 -> 1x3x5x7x9x11
    ...
    ...
    n -> 1x3x5x ... x(?)

    Hope that helps!


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