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Math Help - Taylor Series Question

  1. #1
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    Taylor Series Question

    Got a taylor series question here, just want to see if I'm on the right track

    f(x) = x^(-1/2) at a = 9

    Took the derivative a bunch..

    f'(x) = (-1/2)x^-(3/2)
    f''(x) = (-1/2)(-3/2)x^-(5/2)
    f'''(x) = (-1/2)(-3/2)(-5/2)x^-(7/2)
    f4(f) = (-1/2)(-3/2)(-5/2)(-7/2)x^-(9/2)

    so the taylor series at a = 9 would be..

    [(-1)^n][(x-9)^(-1/2 - n)][2n-1]/[(n!)(2^n)]

    or am I way off?
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  2. #2
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    Quote Originally Posted by coolguy99 View Post
    Got a taylor series question here, just want to see if I'm on the right track

    f(x) = x^(-1/2) at a = 9

    Took the derivative a bunch..

    f'(x) = (-1/2)x^-(3/2)
    f''(x) = (-1/2)(-3/2)x^-(5/2)
    f'''(x) = (-1/2)(-3/2)(-5/2)x^-(7/2)
    f4(f) = (-1/2)(-3/2)(-5/2)(-7/2)x^-(9/2)

    so the taylor series at a = 9 would be..

    [(-1)^n][(x-9)^(-1/2 - n)][2n-1]/[(n!)(2^n)]

    or am I way off?
    Do you mean that is a coefficient? What you give is not a series and, since there is no "x" certainly not a Taylor's series.
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  3. #3
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    The question asked to find a taylor series for the equation f(x) = x^(-1/2) at a = 9... that's all the info it said, and I quoted it directly hahah
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