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Homework Help: Problem with taylor series

  1. Dec 6, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the taylor series of f(x)=1/(x)^(1/2) ; a=9

    2. The attempt at a solution

    f(x) = (x)^(-1/2)
    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)
    f''''(x) = (1/2)*(3/2)*(5/2)*(7/2)*x^(-11/2)

    f(9) = (1/3)
    f'(9) = -(1/(3^3))*(1/2)
    f''(9) = (1/(3^5))*(1/(2^2))
    f'''(9) = -(1/(3^7))*(1/(2^3))
    f''''(9) = (1/(3^9))*(1/(2^4))

    pluging to the taylor series,

    =f(9) + f'(9)*(x-9) + (f"(9)*(x-9)^2)/2! + (f"'(9)*(x-9)^3)/3! + ...

    = (1/3) - (1/(3^3))*(1/2)*(x-9) + ((1/(3^5))*(1/(2^2))*(x-9)^2)/2! - ((1/(3^7))*(1/(2^3))*(x-9)^3)/3! +...+ ((-1)^1* (1)(3)(5)...(2n-1)*(x- 9)^n)/((3^(2n+1))*(2^n)*(n!))

    =1/3 + sum(n=1 to infinity) ((-1)^n* (1)(3)(5)...(2n-1)*(x-9)^n)/((3^(2n+1))*(2^n)*(n!))

    but the answer at the back of the book is sum(n=0 to infinity) ((-1)^n* (1)(3)(5)...(2n-1)*(x-9)^n)/((3^(2n+1))*(2^n)*(n!))

    i'm not convinced because when you plug n=0 to ((-1)^n* (1)(3)(5)...(2n-1)*(x 9)^n)/((3^(2n+1))*(2^n)*(n!)),

    the first terms turns out to be -1/3 which is negative that is, 2n-1 => 2(0)-1 = -1 which makes 1/3 to be negative. The expanded term above is 1/3 which is positive. Pls can someone explain this? please thanks
  2. jcsd
  3. Dec 6, 2009 #2
    It looks like you're right - the book should have been a little more careful to distinguish the first coefficient as a special case from the general expression. I would have said the following:

    f(x) = \alpha_0} + \sum_{n=1}^{\infty} \alpha_n \cdot (x-9)^n


    \alpha_0 = \frac{1}{3}


    \alpha_n = (-1)^n \cdot \left ( \frac{1}{n!} \right ) \cdot \left ( \frac{1}{2} \right )^n \cdot \left ( \frac{1}{3} \right )^{2n+1} (1)(3) \cdots (2n-1) \quad \text{for n = 1, 2, \dots}

    Note the restriction on [itex]n[/itex] for the general term.

    During the construction, this would have been obvious, but it is still sloppy to drop a general expression that doesn't intuitively work for all implied cases.

    And, of course, if they really wanted to confuse you, they could have correctly replaced the repeated multiplications with the "double-factorial" term
    (|2 \cdot n -1| )!!

    but I'll leave it at that.
  4. Dec 6, 2009 #3
    thank you very much :)
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