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For this series, find the radius of convergence and write it as a geometric series and give a formula if $x>3$

$$\sum_{n=0}^{\infty} \frac{1}{2^{n+1}}(x-3)^n$$

Now finding the radius of convergence wasn't too difficult and I'll save you guys the trouble of doing it because the interval of convergence is $ -1 = 3 - 4< x < 4 + 3 = 7$ and so the radius of convergence is 2

The second part confuses me because I don't understand (if it is even possible) to convert a power series to a geometric series.

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You are confusing the radius of curvature with the radius of convergence, and you got the wrong answer. –  Gerry Myerson Jun 18 '12 at 1:34
    
What if we set $\dfrac{x-3} 2 $ as $r$, and take $1/2$ out? –  Pedro Tamaroff Jun 18 '12 at 1:37
    
Oh shoot, wrong term. I have no idea why i did that Gerry. –  sidht Jun 18 '12 at 1:38
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You still have the wrong answer for the radius of convergence. –  Gerry Myerson Jun 18 '12 at 1:41
    
Yes, I made a careless mistake. Thank you for catching that Gerry again. –  sidht Jun 18 '12 at 1:43
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2 Answers

up vote 3 down vote accepted

For this particular function, shouldn't one note that $$ \sum_{n=0}^{\infty} \frac{1}{2^{n+1}}(x-3)^n = \frac{1}{2}\sum_{n=0}^{\infty} \frac{1}{2^{n}}(x-3)^n = \frac{1}{2}\sum^{\infty}_{n=0}\left(\frac{x-3}{2}\right)^{n}$$ which is a geometric series!

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You haven't left OP much to do, have you? –  Gerry Myerson Jun 18 '12 at 1:40
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As for the radius of convergence, we need $$ \biggl| \dfrac{x - 3}{2} \biggr| < 1. $$ –  Eugene Jun 18 '12 at 1:40
    
@Eugene, I basically also got what Alex has, but why are you assuming $\biggl| \dfrac{x - 3}{2} \biggr| < 1$ instead of $|x - 3| < 2$? I realize you can't use $1/(1 - x)$ if I don't assume that the ratio is less than 1. But when x > 3, doesn't the sum diverge? –  sidht Jun 18 '12 at 1:53
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@jak What you wrote is the same thing as what I wrote just so you know. –  Eugene Jun 18 '12 at 1:54
    
Oh It's because you substituted $r = \biggl| \dfrac{x - 3}{2} \biggr| < 1$. Okay gotcha –  sidht Jun 18 '12 at 1:56
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It is not always possible to convert a power series to a geometric series, but in this case it can be done. A geometric series is $a+ar+ar^2+\dots$, all you have to do is a little pattern matching to figure out what $a$ and $r$ have to be in your example.

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