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Usually I see power series such as $\sum_{n=0}^\infty \frac{n!}{(n)^n}z^n$. Now, Im asked to find the radius of convergence for $\sum_{n=0}^\infty 2^{-n}z^{n^2}$ and also $\sum_{n=1}^\infty (3+4i)^n(z-4i)^n$.

How would I find the radius of convergence for those 2 power series?

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What do you expect them to be? For example, you know that if $z = 1$ in the first one, it (---). If $z >1$, then it (---). The second is funny-looking, but I suspect that you suspect that if z is near $4i$ then it converges. Note also that $|3 + 4i| = 25$. –  mixedmath Nov 10 '11 at 0:10
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@mixedmath Actually $|3+4i|=5$. –  Michael Hardy Nov 10 '11 at 0:23
    
One general way: make an educated guess, then try to prove said guess. Also note you should center the second series around $z=4i$, so maybe write $w=z-4i$. @mixedmath: 5 –  anon Nov 10 '11 at 0:23
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@Hardy: oh ... you're right... that's embarassing. And it's the most common right triangle. My pride hurts, so I'm going to go hide for a while. –  mixedmath Nov 10 '11 at 0:27

2 Answers 2

The second series is a geometric series with common ratio $(3+4i)(z-4i)$. It converges if the absolute value of the common ratio is less than $1$ and diverges if the common ratio is more than $1$. $$ |3+4i||z-4i| = \sqrt{3^2+4^2} |z-4i| = 5|z-4i|. $$ That is $<1$ iff $|z-4i|<1/5$, i.e. iff the distance between $z$ and $4i$ is less than $1/5$. The circle of convergence is therefore centered at $4i$ and has radius $1/5$.

If you apply the ratio test to the first series (recalling that $(n+1)^2-n^2=2n+1$), you get $$ \frac{|z|^{(n+1)^2}/2^{n+1}}{|z|^{n^2}/2^n} = \frac{|z|^{2n+1}}{2} \to 0\text{ as }n\to\infty\text{ if } |z|<1 \text{ and }\to\infty\text{ if } |z|>1. $$ So the center is $0$ and the radius is $1$.

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I didn't expect that the argument in favour of keeping $z$ would come up so soon! –  André Nicolas Nov 10 '11 at 0:27
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If $a$ and $b$ are real then $|a+bi|$, being the distance from $0$ to $a+bi$, is $\sqrt{a^2+b^2}$. Really just the Pythagorean theorem. You sometimes see it written as $\sqrt{(a+bi)(a-bi)}$. –  Michael Hardy Nov 10 '11 at 0:54

HINTS:

Look at $\displaystyle \sum \frac{1}{2^n} z^{n^2}$. Firstly, note that $z^{n^2}$ is a lot like $z^n$, but more-so. When $|z|=1$, though, it's a constant. And the resulting series is something we understand. I assume for ease of writing that $z$ is real. Then if $z > 1$, what do we know about $\lim_{n \to \infty} z^n$? Using that, consider what happens to $(z^n)^n$. I love geometric series.

Look at the second one. Argue by absolute values. $|zz'| = |z||z'|$.

Isn't it funny that after learning all those cute series techniques in calculus, everything reduces to geometric series?

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Thanks! Yeah it always seems the case when many areas of mathematics tend to be linked somehow to one another :D –  John Southall Nov 10 '11 at 0:51

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