1. ## Power Series Expansion

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

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

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

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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:

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

2. Regarding the power series which is defined as

$\displaystyle f(q) = f(a) + f'(a)(x-a)+....$ 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?

2. Originally Posted by Paperwings
Problem
===========================
Let $\displaystyle f(q) = \frac{1}{q} [1-(1-2q)^{-3/2}]$ (*)

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

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

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

2. Regarding the power series which is defined as

$\displaystyle f(q) = f(a) + f'(a)(x-a)+....$ 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?

The Matlab expansion is correct. Your problem statement has an error in the $\displaystyle q$ 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 $\displaystyle \frac{x^2}{x}$; it only "looks like" there is a problem at 0.