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In a power series

$$\sum c_n (x-a)^n$$

What qualifies to be in the $(x-a)$ part?

eg.

  • $2-x$
  • $5x-2$
  • $\frac{x-3}{2}$
  • A combination of the above? $\frac{3-2x}{6}$

In a question I am working on, finding the radius of convergence of

$$\sum_{n=0}^{\infty} \frac{(5-4x)^{2n}}{9^n n^{5/4}}$$

the model answer has

$$c_{2n} = \frac{4^{2n}}{9^n n^{5/4}}$$

But what I actually did:

$$\sum_{n=0}^{\infty} \frac{(5-4x)^{2n}}{9^n n^{5/4}} \\ =\frac{(-1)^{2n}(4x-5)^{2n}}{9^nn^{5/4}} \\ =\frac{(-1)^{2n}(x-\frac{5}{4})^{2n}}{4^{2n} 9^n n^{5/4}}$$

So my $$c_n=\frac{(-1)^{2n}}{4^{2n} 9^n n^{5/4}}$$

Why isit in the given answer ($c_N$), its not an alternating series, and why is $4^{2n}$ in the numerator?

UPDATE: OK about the $4$ in denominator question, that was my mistake.

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You factored $4$ out of $(5-4x)^{2n}$ incorrectly (why does it end up in the denominator in your computation?). The first question about the constant part seems to be a wholly different question. –  anon Apr 22 '12 at 8:44
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1 Answer

up vote 1 down vote accepted

The problem with the factor $4^{2n}$ has already been resolved. For the remaining discrepancy between your result and the model answer, note that $(-1)^{2n}=1$ for integer $n$. Apart from these two oversights and the presentational error that you have a sum in the first expression and then just drop the sum, thus equating the entire sum to one of its coefficients, your approach to calculating the coefficients of the power series is correct.

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