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So I have been asked a question for one of my problem sets but I haven't found a "good" way to approach it. Personally I was thinking of approaching the series it backwards but then again if I want to use a certain test(i.e. ratio test, comparison test, square root test, etc etc) that may be a much more difficult. Here it is,

For each interval given below, build a power series which has that interval as its interval of convergence. Justify your claims by showing that the interval of convergence of your series is as desired.

(a) [0,4)

(b) [-13,-6]

(c) (11,12)

If anybody can give me a good suggestion on how to approach the problem that would be great!

EhBabay
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2 Answers2

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Example (a): The midpoint of the interval is $2$, so we will be looking at something of shape $\sum a_n(x-2)^n$. We want radius of convergence $2$, so let us look for something of shape $\sum b_n \frac{(x-2)^n}{2^n}$.

We want convergence at $0$, and divergence at $4$. Look for something like $b_n=\frac{1}{n}$, possibly decorated with $(-1)^n$.

Let's check out $\sum \frac{1}{n2^n}(x-2)^n$. Now we can work in the standard direction, as in the usual find the interval of convergence question. Apply the Ratio Test. Then see what happens at $x=0$. Alternating series, convergence. At $x=4$, things go bad, we get the harmonic series.

Example (b): Use the same basic strategy, but use $b_n=\frac{1}{n^2}$ to ensure convergence at both ends.

The last one is left to you.

André Nicolas
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  • Ok I'm just wondering how you got the bn*(x-2)^2/2^n part in your explanation. – EhBabay Mar 30 '13 at 17:21
  • Typically for $\sum a_n (x-a)^n$, we get a certain radius of convergence, and we have convergence in $(a-r,a+r)$, and possibly at one endpoint or the other or both. So $a$ midway is clear, and we want $r=2$. What sort of choice of $a_n$ makes $\sum a_n y^n$ have radius of convergence $2$. The ratio test would give $a_{n+1}/a_n$ having limit $1/2$. So $a_n$ should sort of behave like $1/2^n$. Another way of thinking about it is that $\sum(y/2)^n$ converges for $-2\lt y\lt 2$, it is just a geometric series. – André Nicolas Mar 30 '13 at 17:30
  • Thank Andre thats what I thought. – EhBabay Mar 30 '13 at 17:32
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    @JohnCarpenter: The answers are very similar, but I think Alex Becker's answer is pedagogically better. – André Nicolas Mar 30 '13 at 17:39
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It is easy to translate and scale the interval of convergence of a power series, so all you really need to worry about is finding power series with the right behavior at the endpoints. For example, $\sum\limits_{n=0}^\infty x^n$ has interval of convergence $(-1,1)$, $\sum\limits_{n=0}^\infty \frac{x^n}{n}$ has interval of convergence $[-1,1)$ and $\sum\limits_{n=0}\frac{x^n}{n^2}$ has interval of convergence $[-1,1]$. All you need to do now is translate and scale. For a), we translate the interval of convergence of $\sum\limits_{n=0}^\infty \frac{x^n}{n}$ to the right by $1$ by substituting $x-1$ for $x$, and then scale this by $2$ by substituting $\frac{x}{2}$ for $x$, giving us $\sum\limits_{n=0}^\infty \frac{(x/2-1)^n}{n}$. The others can be solved similarly.

Alex Becker
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