How might I find $\sum\limits_{n=-\infty}^\infty {z^n\over n}-\sum\limits_{n=-\infty}^\infty {z^{2n}\over 2n}$ where $z=Re^{i\theta}$? Thanks.
How to sum this complex series?
-
1It'll diverge... unless you intended $n$ to be the index instead of $m$. – 2011-11-02
-
0Thanks, J.M. and mixedmath. You are quite right in spotting the typo. Thanks! Edited. Assuming that $|R|<1$, what would the series converge to? Thanks. – 2011-11-02
-
0It can't converge for $|R|<1$ since your exponent goes from $-\infty$. It might converge for $|R|=1$ when $z\neq 1,-1$, in which case the value will be $\log{|\frac{1+z}{1-z}|}$ – 2011-11-02
-
0@saxa: If $|R|<1$ the negative index terms will diverge. So your only hope is $|R|=1$. Even so, the $n=0$ terms are troubling. Generally for a series like this, you can use $\sum\frac{z^n}{n}=\sum \frac{1}{z}\int z^n\;dz$ if it converges – 2011-11-02
-
0@RossMillikan - I was assuming that the two $n=0$ terms cancelled, and that you could rewrite this as one series, $\sum_{n=-\infty}^{\infty} \frac{z^{2n+1}}{2n+1}$ – 2011-11-02
-
1Actually, this sort of sum shows up in the theory of vertex algebras all the time. Usually it would be taken to be $\log | \frac{1+z}{1-z} |$, with the understanding that one is not actually performing the sum, one is simply setting the commutator(?) of two current operators to be this function, which makes the rest of the mathematics consistent. – 2011-11-02
-
0@ThomasAndrews: You incorrectly summed the series, a small sign error. See my answer. We have that $$\sum_{n=-\infty}^\infty=\frac{1}{2}\log\left(\frac{1+z}{1-z}\right)-\frac{1}{2}\log\left(\frac{1+z^{-1}}{1-z^{-1}}\right).$$ What you wrote above is $$\frac{1}{2}\log\left(\frac{1+z}{1-z}\right)+\frac{1}{2}\log\left(\frac{1+z^{-1}}{1-z^{-1}}\right)=\log\sqrt{\frac{(1+z)(1+z^{-1})}{(1-z)(1-z^{-1}}}=\log\biggr|\frac{1+z}{1-z}\biggr|$$ since $z^{-1}=\overline{z}$. I think the reason for this sign error is forgetting that the negative $n$ contribute a negative sign to part of the series. – 2011-11-02
2 Answers
To get rid of the problem with the index $n=0$, we can rewrite the series, as you may have intended, $$ \sum_{k=-\infty}^\infty\frac{z^{2k+1}}{2k+1}\tag{1} $$ If $|z|\not=1$, the double-ended series diverges, so I will assume that $R=1$. Using the series $$ \sum_{k=0}^\infty\frac{z^{2k+1}}{2k+1}=\frac{1}{2}\log\left(\frac{1+z}{1-z}\right)\tag{2} $$ the series $(1)$ becomes $$ \begin{align} \sum_{k=-\infty}^\infty\frac{e^{i(2k+1)\theta}}{2k+1} &=\sum_{k=0}^\infty\frac{e^{i(2k+1)\theta}-e^{-i(2k+1)\theta}}{2k+1}\\ &=\frac{1}{2}\log\left(\frac{1+e^{i\theta}}{1-e^{i\theta}}\right)-\frac{1}{2}\log\left(\frac{1+e^{-i\theta}}{1-e^{-i\theta}}\right)\\ &=\frac{1}{2}\log\left(\frac{e^{-i\theta/2}+e^{i\theta/2}}{e^{-i\theta/2}-e^{i\theta/2}}\right)-\frac{1}{2}\log\left(\frac{e^{i\theta/2}+e^{-i\theta/2}}{e^{i\theta/2}-e^{-i\theta/2}}\right)\\ &=\frac{1}{2}\log(i\cot(\theta/2))-\frac{1}{2}\log(-i\cot(\theta/2))\\ &=\left\{\begin{array}{}+\frac{\pi}{2}i&\text{ when }0<\theta<\pi\\-\frac{\pi}{2}i&\text{ when }-\pi<\theta<0\end{array}\right.\tag{3} \end{align} $$ This agrees with Eric Naslund's answer.
First, a few things: The index of summation appears to be wrong, and I hope we are not summing when $n=0$. I assume we are looking at $$f(z)=\sum_{n\in\mathbb{Z}\backslash \{0\}}\frac{z^n}{n}-\sum_{n\in\mathbb{Z}\backslash \{0\}}\frac{z^{2n}}{2n}.$$
We have to be very careful. First, if $|z|\neq 1$, this will diverge by the divergence test. Both $z=1$ and $z=-1$ are singularities, and it is not clear what kind. From now on, lets assume that $|z|=1$ and $z\neq \pm 1$.
The Answer: The result is tricky and we will have $f(z)=\frac{\pi i}{2}$ when $|z|=1,\ \text{Im}(z)>0$ and $f(z)=-\frac{\pi i}{2}$ when $|z|=1,\ \text{Im}(z)>0$. This can be extended to the unit disk, but it is very discontinuous. The reason for this is subtleties with branch cuts, and the problems that they cause.
Proof: All the series conditionally converge, and we can justify rewriting $f(z)$ as
$$f(z)=\sum_{n=1}^\infty \frac{z^n}{n}-\sum_{n=1}^\infty\frac{z^{-n}}{n}-\sum_{n=1}^\infty \frac{z^{2n}}{2n}+\sum_{n=1}^\infty \frac{z^{-2n}}{2n}.$$
Using the fact that $$\sum_{n=1}^\infty \frac{z^n}{n}=-\log(1-z)$$ we see that the above is $$-\log(1-z)+\log\left(1-z^{-1}\right)+\frac{1}{2}\log\left(1-z^{2}\right)-\frac{1}{2}\log\left(1-z^{-2}\right).$$
Caution with branches: Notice that $f(z)=-f(z^{-1})$, and since we are working on the unit circle and $z^{-1}=\overline{z}$ we see that the function will take opposite values on the top and bottom halves.
Combining some terms this is
$$\frac{1}{2}\log\left(\frac{\left(1-z^{-1}\right)^2\left(1-z^{2}\right)}{\left(1-z\right)^2\left(1-z^{-2}\right)}\right)=\frac{1}{2}\log\left(\frac{\left(1-z^{-1}\right)\left(1+z\right)}{\left(1-z\right)\left(1+z^{-1}\right)}\right)=\frac{1}{2}\log(-1).$$ Hence the answer is in some sense $\frac{1}{2}\log(-1)$, but not quite. We have to be very careful with Branch Cuts. When we combined the above terms, the argument secretely changed by a whole factor of $2\pi$ because we were not keeping track. If we very carefully work things out, we arrive at:
$$f(z)=\frac{\pi i}{2}\ \text{for} \ |z|=1,\ \text{Im}(z)>0\ \text{and} \ f(z)=-\frac{\pi i}{2}\ \text{for} \ |z|=1,\ \text{Im}(z)<0.$$
-
0If $z=1$ everything does not cancel, because $\frac{1}{n}$ is not the same as $\frac{1}{2n}$ – 2011-11-02
-
0@ThomasAndrews: Well depending on how you look at it, yes it does. The series $\sum_{n=-M,\ n\neq =0}^{n=M}\frac{z^n}{n}$ literally equals zero for $z=1$.... It has a singularity, but sure it is not clear what value should be assigned there. – 2011-11-02
-
0So, you are defining the difference of two diverging series to be $0$? That seems odd. – 2011-11-02
-
0Also notice that if $|z|=1$, then $z^{-1}=\bar{z}$. So the only meaningful values for these series are real numbers when $|z|=1$, but your answer is imaginary. You've made a mistake somewhere. – 2011-11-02
-
0Also, if $z=-1$, the first series (of $z^n/n$) converges, but $z^{2n}/2n$ diverges. – 2011-11-02
-
0@ThomasAndrews: I don't mean to be rude, but I never defined the value to be imaginary for $z=\pm 1$. It quite clearly says if $\text{Im}(z)<0$ or $>0$, both of which are strict inequalities. As I mentioned, there are singularities at those points. Also, if you had looked at what I wrote, I mentioned that the left hand series diverged anyway for $z=-1$, I never said it canceled out there. – 2011-11-02
-
0But the value at *all* $|z|=1$ has to be real, not just $z=\pm 1$. In general, if $f(z)=\sum_{i=1}^\infty a_i z^i$ with all $a_i$ real, then $f(z)+f(z^{-1})$ must be real when $|z|=1$. This difference is the difference of two such series, so the value on all $|z|=1$ must be real. – 2011-11-02
-
0@ThomasAndrews: Ok, I understand your complaint. Please read this: Your series posted as a comment to the question is incorrect by a minor sign error. The series is $\tanh^{-1}(z)-\tanh^{-1}\left(\frac{1}{z}\right)$, NOT $\tanh^{-1}(z)+\tanh^{-1}\left(\frac{1}{z}\right)$. The thing is here, because it is of the form $f(z)-f(z^{-1})$ it must be imaginary. I think the problem is that when you computed the series, the negative sign coming from the negative $n$ was missed. – 2011-11-02
-
0So, are you saying that if $f$ has a power series with only real coefficients, then $f(z)+f(z^{-1})$ can be non-real for some value $|z|=1$? Because that is your result, and it is not possible. – 2011-11-02
-
0@ThomasAndrews: Please read my comments! It is of the form $f(z)-f(z^{-1})$ NOT $f(z)+f(z^{-1})$, it is very different!! The series is $$\sum_{n=-\infty}^\infty \frac{z^{2n+1}}{2n+1}$$ agreed? Rearranging this is $$\sum_{n=0}^\infty \frac{z^{2n+1}}{2n+1}+\sum_{n=-\infty}^{-1} \frac{z^{2n+1}}{2n+1}=\sum_{n=0}^\infty \frac{z^{2n+1}}{2n+1}-\sum_{n=0}^\infty \frac{z^{2n+1}}{2n+1}$$ because a negative sign comes out of the denominator $2n+1$ when we rearrange. Please, do you see why it is minus and not plus? That is basically all we disagree on, you summed the series incorrectly to start! – 2011-11-02
-
0Okay, you are right. I missed that it was difference. Got blinded by the fact that Craig came up with the same answer that I did in comments above. – 2011-11-02
-
0`this will diverge by the divergence test`. Did you perhaps mean the [Term Test](http://en.wikipedia.org/wiki/Term_test)? – 2011-11-02
-
0@robjohn: Ya, same thing. I just remember my prof in first year always called it the divergence test, so that name stuck with me. – 2011-11-02