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Math Help - Power Series Expansion

  1. #1
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    Power Series Expansion

    Problem
    ===========================
    Let  f(q) = \frac{1}{q} [1-(1-2q)^{-3/2}] (*)

    Evaluate this formula for  q = 10^{-8} using the power series expansion for f at q = 0:

     f(q) = -3 - \frac{3*5}{3!}q  - \frac{3*5*7}{3!}q^2 - .... (1)

    ===============================
    I have two questions.

    1.) Is this power series expansion correct? I tried inputting this function in MATLAB and it gives me a different power series expansion:

     f(q) = -3-\frac{15}{2}q-\frac{35}{2}q^2- \frac{315}{8}q^3 - \frac{693}{8}q^4 ... (2)

    2. Regarding the power series which is defined as

    f(q) = f(a) + f'(a)(x-a)+.... at q = a.

    Since this power series (1) I do not understand how a power series can be derived this way since at q = 0, then f(0) from (*) is undefined because we can not divide by 0, correct?

    Thank you for reading.
    Last edited by Paperwings; February 2nd 2009 at 01:38 PM.
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  2. #2
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    Quote Originally Posted by Paperwings View Post
    Problem
    ===========================
    Let  f(q) = \frac{1}{q} [1-(1-2q)^{-3/2}] (*)

    Evaluate this formula for  q = 10^{-8} using the power series expansion for f at q = 0:

     f(q) = -3 - \frac{3*5}{3!}q  - \frac{3*5*7}{3!}q^2 - .... (1)

    ===============================
    I have two questions.

    1.) Is this power series expansion correct? I tried inputting this function in MATLAB and it gives me a different power series expansion:

     f(q) = -3-\frac{15}{2}q-\frac{35}{2}q^2- \frac{315}{8}q^3 - \frac{693}{8}q^4 ... (2)

    2. Regarding the power series which is defined as

    f(q) = f(a) + f'(a)(x-a)+.... at q = a.

    Since this power series (1) I do not understand how a power series can be derived this way since at q = 0, then f(0) from (*) is undefined because we can not divide by 0, correct?

    Thank you for reading.
    The Matlab expansion is correct. Your problem statement has an error in the q term. Notice that the other terms agree.

    The singularity of f at 0 is "removable", so there is no problem with continuity at that point. (You might try evaluting the limit as q -> 0 by l'Hopitals rule, although it's obvious from the power series.) It's a little like the function \frac{x^2}{x}; it only "looks like" there is a problem at 0.
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