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Find Taylor Series

  1. Sep 27, 2013 #1
    This is rather embarrassing, because I should have known how to do this for years.

    Question:

    Compute the Taylor Series of ##\frac{q}{\sqrt{1+x}}## about x = 0.

    Attempt at Solution:

    Term-wise, I have gotten...

    ##f(0)+f'(0)+f''(0)+... = 1+1\left(-\frac{1}{2}\right)x+1\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)x^2+1\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)x^3+...##

    I have gotten this to reduce to...

    ##\sum\limits_{k=0}^{\infty}x^k\left(-\frac{1}{2}\right)^k\frac{(2k-1)!!}{2^k}##

    There is definitely a better way to do this. I am not thinking clearly. Additionally, I am not that confident in my answer, given the time of night.
     
  2. jcsd
  3. Sep 27, 2013 #2

    Simon Bridge

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    At first glance - shouldn't there be a q in there?

    If your termwise relation is supposed to be the series, you seem to have missed some bits.

    Apart from that - it all seems pretty standard.
     
  4. Sep 27, 2013 #3
  5. Sep 27, 2013 #4

    HallsofIvy

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    Writing [itex]q/\sqrt{1+ x}[/itex] as [itex]q(1+ x)^{-1/2}[/itex] we have:
    The value at 0 is q. The first derivative is [itex]-(1/2)q(1+ x)^{-3/2}[/itex] so
    its value at 0 is -(1/2)q. The second derivtive is [itex](3/4)q(1+ x)^{-5/2}[/itex] so
    its value at 0 is (3/4)q. The third derivative is [itex]-(15/8)q(1+ x)^{-7/2}[/itex] so
    its value at 0 is -(15/8)q. The fourth derivative is [itex](105/16)q(1+ x)^{-9/2}[/itex] so
    its value at 0 is (105/16)q...

    That should be enough to see the pattern: it alternates sign with positive sign when n, the order of the derivative, is even so that is [itex](-1)^n[/itex]. The denominator of the fraction is a power of 2: [tex]2^n[/tex]. There is, obviously, a factor of "q". The numerator of the fraction is the only "tricky" part. It is 1(3)(5)(7)..., a sort of "factorial" except that the even integers are missing. We can write that as 1(3)(5)(7)= 1(2)(3)(4)(5)(6)(7)/[2(4)(6)]= 7!/([2(1)][(2)(2)][(2)(3)])= 7!/[2^3(3!)] (when n= 4). So that is (2n- 1)!/[2^{n-1}(n-1)!].

    Putting those together, the nth derivative, at x= 0, is
    [tex](-1)^n\frac{(2n-1)!}{2^{n-1}(2^n)(n- 1)!}q= (-1)^n\frac{(2n-1)!}{2^{2n-1}(n-1)!}q[/tex]
    and the coefficient of [itex]x^n[/itex] in the Taylor series expansion about x= 0 is that divided by n!.
     
    Last edited by a moderator: Sep 27, 2013
  6. Sep 27, 2013 #5

    mathman

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    You can get the same result, without calculating the derivatives, by getting the binomial expansion for (1+x)-1/2.
     
  7. Sep 30, 2013 #6
    Sorry guys. I meant to type 1 instead of q. My keyboard broke and the function key messed up. I did not catch that error.
     
  8. Sep 30, 2013 #7

    Simon Bridge

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    You have to know already that the binomial series is the Taylor series at x = 0 of the function f given by f(x) = (1 + x) α, where α ∈ ℂ is an arbitrary complex number.

    I suppose that answers the original question.
     
  9. Oct 1, 2013 #8

    mathman

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    The binomial theorem series and the Taylor series (for this case) are both power series around 0. Therefore they must be identical for the same function.
     
  10. Oct 1, 2013 #9

    Office_Shredder

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    If there was a simple explanation for why the binomial series worked for an arbitrary power then it would be a good answer, but as is it's basically just a way of telling someone what the answer is without explanation which isn't great.
     
  11. Oct 2, 2013 #10

    mathman

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