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Calculus II - Taylor Series Question

  1. Aug 22, 2011 #1
    1. The problem statement, all variables and given/known data

    Find the power series for f(x) using the definition of taylor series expansion about a=9. f(x)=1/sqrt(x)

    2. Relevant equations



    3. The attempt at a solution

    Find the power series for f(x) using the definition of taylor series expansion about a=9. f(x)=1/sqrt(x)

    f(x) = 1/sqrt(x) => f(9) = 1/3
    f'(x) = - 1/(2 x^(3/2)) => f'(9) = -1/54
    f''(x) = (1*3)/(2^2 x^(5/2)) => f''(9) = 1/324
    f'''(x) = -(1*3*5)/(2^3 x^(7/2)) => f'''(9) = 5/5852
    ...

    The patter that seems to be developing is

    f^(k)(x) = ( (-1)^k ?)/(2^k x^(1/2 + k) )
    but I'm lost as to how to express the pattern that appears in the numerator, hence the question mark, of each order of the function
    for f(x) we have 1
    for f'(x) we have 1
    for f''(x) we have 1*3
    for f'''(x) we have 1*3*5
    ...
    The patter that seems to appear after the 1st derivative in a sort of backwards batter, 5*3*1 instead of 1*3*5, is (2k-1)(2k-3)(2k-5)... until the last term is equal to 1

    I'm sort of lost... thanks for any help!
     
  2. jcsd
  3. Aug 22, 2011 #2

    micromass

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    Why are you lost? You found the answer?

    The thing you might do first is to prove (by induction) that this is indeed the pattern that is emerging in the numerator!
     
  4. Aug 22, 2011 #3

    lanedance

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    i think you've identified the pattern
    n=0, 1
    n=1, 1
    n=2, 1.3
    n=3, 1.3.5
    ..
    so for general n,
    [tex] 1.3...(2n-1) = \Pi_{i=1}^{n}(2i-1)[/tex]

    I don't think there is any other simple way to write this, though it can be represented as a double factorial
    http://en.wikipedia.org/wiki/Factorial#Double_factorial
     
  5. Aug 22, 2011 #4
    The problem is that I don't know how to come up with the series representation of the function because I don't know how to describe the pattern in the numerator with mathematical functions

    inf
    sigma ( (-1)^k*1*3*(2k-1)*(x-9)^k)/(k!*2^k x^(1/2 + k) )
    k=0

    I don't know what to do about the question mark because I don't know how to describe the pattern with mathematical functions... maybe I'm over thinking it
     
    Last edited: Aug 22, 2011
  6. Aug 22, 2011 #5
    How would I write the series representation without a double factorial?
     
  7. Aug 22, 2011 #6

    lanedance

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    how about as I did?
     
  8. Aug 22, 2011 #7

    lanedance

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    or how about
    [tex] 1.3...(2n-1) = \Pi_{i=1}^{n}(2i-1) = \frac{(2n-1)!}{2^{n-1}(n-1)!} = (2n-1)!![/tex]
     
  9. Aug 22, 2011 #8
    inf
    sigma ( (-1)^k*1*3*(2k-1)*(x-9)^k)/(k!*2^k x^(1/2 + k) )
    k=0

    for k =0

    ( (-1)^0*1*3*...*(2*0-1)*(x-9)^0)/(0!*2^0 x^(1/2 + 0) )
    (1*3*-1)/sqrt(x)
    -3/sqrt(x)
    it doesn't seem to work because of the 2*0-1 is negative 1 and doesn't cancel out the 1*3 to leave me with one in the numerator to give me f^(0)(x) = 1/sqrt(x) but instead -3/sqrt(x)?
     
  10. Aug 22, 2011 #9
    I guess I should learn double factorials than but it's not a standard calculus II topic so that's why I was lead to believe I was doing something wrong but I guess I'm not because I had no clue how to represent the pattern in the numerator
     
  11. Aug 23, 2011 #10

    lanedance

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    You may have to write the sum from k=1 onwards for that reason though i haven't checked.

    If you read post #7 there is a couple of different ways to write it. The double factorial is just a nice shorthand for these, but doesn't get used very often in my experience. If I were marking, any of the forms in post #7 would be fine.

    The main thing is you recognised the pattern. As micromass points out if you want to be rigorous you should use induction to prove this is the pattern. However looking at the action of the derivative it should be easy to convince yourself it is correct.
     
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