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Thread: Trying to grasp radius of convergence for a complex series - example

  1. #1
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    Trying to grasp radius of convergence for a complex series - example

    $\displaystyle \displaystyle\sum\limits_{n\geq 1}^\infty \frac{(4iz - 2)^{2n}}{5^n(n+2)}
    $

    The first thing that popped up in my mind was this, absolute value of $\displaystyle \frac{1}{r} = lim |(a_{n+1})/a_n|$. Furthermore, look at start and end points. So I started tinkering with that, but I don't get it...

    Convergence radius for n
    $\displaystyle \displaystyle \frac { \frac{(z+\frac{i}{2})^{2(n+1)}}{5^{n+1}(n+1+2)}} {\frac{(z+\frac{i}{2})^{2n}}{5^{n}(n+2)}}$ = $\displaystyle \displaystyle \frac { \frac{(z+\frac{i}{2})^{2n} (z+\frac{i}{2})^2}{5^n 5(n+3)}} {\frac{(z+\frac{i}{2})^{2n}}{5^{n}(n+2)}}$ = $\displaystyle \displaystyle \frac { \frac{ (z+\frac{i}{2})^2}{ 5(n+3)}} {\frac{1}{(n+2)}}$ = $\displaystyle \displaystyle (z+\frac{i}{2})^2 \frac{n+2}{5(n+3)}$

    Starting point n = 1
    $\displaystyle \frac{(z+\frac{i}{2})^{2}}{{5^{1}(1+2)}}$ = $\displaystyle \frac{(z+\frac{i}{2})^2}{15}$

    ...

    I'd really appreciate if someone pointed me in the right direction...
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  2. #2
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    Quote Originally Posted by liquidFuzz View Post
    $\displaystyle \displaystyle\sum\limits_{n\geq 1}^\infty \frac{(4iz - 2)^{2n}}{5^n(n+2)}
    $

    The first thing that popped up in my mind was this, absolute value of $\displaystyle \frac{1}{r} = lim |(a_{n+1})/a_n|$. Furthermore, look at start and end points. So I started tinkering with that, but I don't get it...

    Convergence radius for n
    $\displaystyle \displaystyle \frac { \frac{(z+\frac{i}{2})^{2(n+1)}}{5^{n+1}(n+1+2)}} {\frac{(z+\frac{i}{2})^{2n}}{5^{n}(n+2)}}$ = $\displaystyle \displaystyle \frac { \frac{(z+\frac{i}{2})^{2n} (z+\frac{i}{2})^2}{5^n 5(n+3)}} {\frac{(z+\frac{i}{2})^{2n}}{5^{n}(n+2)}}$ = $\displaystyle \displaystyle \frac { \frac{ (z+\frac{i}{2})^2}{ 5(n+3)}} {\frac{1}{(n+2)}}$ = $\displaystyle \displaystyle (z+\frac{i}{2})^2 \frac{n+2}{5(n+3)}$
    Looks good so far.

    Starting point n = 1
    $\displaystyle \frac{(z+\frac{i}{2})^{2}}{{5^{1}(1+2)}}$ = $\displaystyle \frac{(z+\frac{i}{2})^2}{15}$
    I don't see your point here. What is important is the limit as n goes to infinity. What happens when n= 1 is irrelevant.

    $\displaystyle \lim_{n\to\infty}\frac{n+2}{5(n+3}= \lim_{n\to\infty}\frac{1+\frac{2}{n}}{5(1+ \frac{3}{n})}$
    where I have divided both numerator and denominator by n. As n goes to infinity, both of those fractions, with n in the denominator, go to 0. The limit is $\displaystyle \frac{1}{5}$. Also, everything should be in absolute value signs. In the limit, you have $\displaystyle \frac{1}{5}|(z+ \frac{i}{2})^2|$ and, by the "ratio test" that must be less than 1.

    $\displaystyle \frac{1}{5}|(z+ \frac{i}{2})^2|< 1$
    gives $\displaystyle |z+ \frac{i}{2}|^2< 5$
    so
    $\displaystyle |z+ \frac{i}{2}|< \sqrt{5}$.

    That means that the radius of convergence is $\displaystyle \sqrt{5}$ and the series converges (uniformly) inside the circle centered at $\displaystyle \frac{i}{2}$ with radius $\displaystyle \sqrt{5}$.

    ...

    I'd really appreciate if someone pointed me in the right direction...
    Last edited by HallsofIvy; Sep 16th 2010 at 11:53 AM.
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  3. #3
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    Thank you for a concise explanation and most of all, spending time on dudes like me!

    I only just picked up math again after some 10 years. You're post really helped. However, I don't really get the difference between uniform and absolute convergence. I there any relation between them?
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  4. #4
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    Quote Originally Posted by liquidFuzz View Post
    Thank you for a concise explanation and most of all, spending time on dudes like me!

    I only just picked up math again after some 10 years. You're post really helped. However, I don't really get the difference between uniform and absolute convergence. I there any relation between them?
    "Absolute" convergence simply means of a series of numbers simply means that $\displaystyle \sum |a_n|$ converges. It is easy to prove that if a series "converges absolutely" then the series converges. The other way is not always true. For example, the series $\displaystyle \sum_{n=1}^\infty \frac{(-1)^n}{n}$ converges because it the signs alternate and the sequence of term $\displaystyle \frac{1}{n}$ decreases to 0. But taking the absolute value of each term gives $\displaystyle \sum_{n=1}^\infty \frac{1}{n}$, the "harmonic series"which is well known (perhaps by the integral test) not to converge. In a sense, the harmonic series is the "boundary" between "convergent" and "divergent" series: [tex]\sum_{n=1}^\infty \frac{1}{n^r}[tex] converges for r> 1, diverges for $\displaystyle r\le 1$.

    "Uniform convergence" is a deeper subject! In ordinary convergence of sequences (not series) of functions, we are looking at convergence at individual values of x and we assume a specific value of x when we are looking for the "$\displaystyle \delta$" to correspond to a given "$\displaystyle \epsilon$". Uniform convergence is always on a given set and, given [tex]\epsilon> 0, we can find a value of $\displaystyle \delta$ that will work for any value of x in the set. That allows us to work with properties that necessarily involve more than one value of x- for example, continuity.

    Consider the sequence of functions $\displaystyle \{f_n(x)\}$ defined by "f(x)= 0 if x< 0, $\displaystyle f(x)= nx$ for $\displaystyle 0\le x\le \frac{1}{n}$, [tex]f(x)= 1[tex] for x> 1/n. For example, [tex]f_1(x)= 0 for x< 0, $\displaystyle f_1(x)= x$ for x between 0 and 1, and $\displaystyle f_1(x)= 1$ for x> 1. We have $\displaystyle \lim_{x\to 0} f(x)= 0$ as we come from either above or below and $\displaystyle lim_{x\to 1} f_1(x)= 1$ from either above or below. $\displaystyle f_2(x)= 0$ for x< 0, $\displaystyle f_2(x)= 2x$ for between 0 and 1/2, and $\displaystyle f_2(x)= 1$ for x> 2. Again, we have $\displaystyle \lim_{x\to 0}f_2(x)= 0$ as we approach 0 from either above or below and $\displaystyle \lim_{x\to 1/2} f_2(x)= 1$ from either above or below. Similarly for all n: $\displaystyle f_n(x)$ is continuous for all x.

    Of course, if x< 0, the $\displaystyle f_n(x)= 0$ for all n so for x< 0, $\displaystyle \lim_{n\to\infty} f(x)= 0$. If x= 0, the $\displaystyle f_n(0)= n(0)= 0$ for all x so, in fact, for all $\displaystyle x\le 0$ the limit function is 0. But if x> 0, then for some positive integer, N, 1/N< x so for n> N, $\displaystyle f_n(x)= 1$ and the limit is 1.

    That is, $\displaystyle \lim_{n\to\infty} f_n(x)$ is equal to 0 for $\displaystyle x\le 0$, 1 for $\displaystyle x> 0$. The limit function is not continuous. That convergence is NOT 'uniform'.
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