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I'm having problems with the following calculation.

Let $a >0$

$$ \begin{align} & \sum_{n=1}^\infty \arctan \left(\frac{2a^2}{n^2}\right) = \text{Im} \sum_{n=1}^\infty \log \left( 1 + \frac{2ia^2}{n^2} \right) \\ & \text{(where I probably should be using the principal branch of log)} \\ \\ & = \text{Im} \log \prod_{n=1}^\infty \left(1 + \frac{2ia^2}{n^2} \right) \\ \\ & = \text{Im} \log \prod_{n=1}^\infty \left(1 - \frac{(\sqrt{-2i}a)^2}{n^2} \right) \\ \\ & =\text{Im} \log \frac{\sin (\pi \sqrt{-2i}a)}{\pi \sqrt{-2i}a} \\ \\ & = \text{Im} \log \frac{\sin \left(\pi (1-i)a\right)}{\pi (1-i)a} \\ \\ & = \text{Im} \log \frac{\sin (\pi a) \cos (i\pi a) - \cos (\pi a) \sin (i \pi a)}{\pi (1-i)a} \\ \\ & = \text{Im} \log \frac{\sin (\pi a) \cosh (\pi a) - i\cos (\pi a) \sinh (\pi a)}{\pi (1-i)a} \\ \\ & = \text{Im} \log \frac{\sin (\pi a) \cosh (\pi a) + \cos(\pi a) \sinh (\pi a) + i \big( \sin (\pi a) \cosh (\pi a) - \cos (\pi a) \sinh ( \pi a) \big)}{2\pi a} \\ \\ & = \text{Arg} \ \frac{\sin (\pi a) \cosh (\pi a) + \cos(\pi a) \sinh (\pi a)+ i \big( \sin (\pi a) \cosh (\pi a) - \cos (\pi a) \sinh ( \pi a) \big)}{2\pi a} \end{align} $$

but then for many values of $a$, the answer is off by a multiple of $\pi$

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$\sin(A+B)=\sin(A)\cos(B)+\cos(A)\sin(B)$. It seems like you have a minus sign that doesn't belong there, if I read this right. I don't know if that fixes the issue though. –  alex.jordan Apr 5 '12 at 2:38
    
What allows you to replace $\operatorname{Im}\,\log(X)$ with $\operatorname{Arg}(X)$ at the last step? This is only true if $\log(X)$ is imaginary. –  alex.jordan Apr 5 '12 at 2:45
    
I used the identity for $\sin (A-B)$. And I don't quite understand your second point. $\log(z) = |z| + i \text{Arg}(z)$. –  Random Variable Apr 5 '12 at 2:50
    
Then you should not have $-i\pi a$ as the argument in $\cos$ and $\sin$ on the next line. Rather you should have $i\pi a$. –  alex.jordan Apr 5 '12 at 2:53
    
I think the first line of my "solution" might me restrict me to only certain values of $a$. –  Random Variable Apr 5 '12 at 2:56

1 Answer 1

up vote 4 down vote accepted

I think I understand the issue now. In this setting, we cannot say that $\log(A)+\log(B)$ is equal to $\log(AB)$. This (generally) invalidates your exchange of an infinite sum for an infinite product. But keep reading to the end to see a fix.

Let's write $\log$ for the principle branched logarithm, and $\operatorname{lug}$ for a potentially different branched logarithm. For fixed $A$ and $B$, there is some $\operatorname{lug}$ for which $$\log(A)+\log(B)=\operatorname{lug}(AB)$$ but it might have a different branch.

Consider $A=B=\exp(i3\pi/4)$. Then $$\begin{align}\log(A)+\log(B)&=i3\pi/4+i3\pi/4\\&=i3\pi/2\end{align}$$ But this addition has crossed over the negative real axis in an angular sense. So $\log(AB)$ is something different: $$\log(AB)=\log(\exp(i3\pi/2))=-i\pi/2$$

So it is not exactly that some values of $a$ cause a problem with your first line, but rather with your second line. If $a$ is small enough such that the infinite summation in the first line never surpasses $\pi i$, then the final result should be OK. For other $a$, you would want to find how many times that sum crossed the negative reals in an angular sense, and account for so many multiples of $2\pi i$.

$\operatorname{Arg}$ is discontinuous at $x+iy$ with negative $x$ and $y=0$. The discontinuities in $\operatorname{Arg}$ in your final line will correspond to values of $a$ where the original sum passes $\pi$, $3\pi$, etc. So let's solve for when the imaginary part of the input to $\operatorname{Arg}$ in your final line equals zero.

$$\begin{align} \sin(\pi a)\cosh(\pi a)−\cos(\pi a)\sinh(\pi a) & = 0\\ \sin(\pi a)\cosh(\pi a)&=\cos(\pi a)\sinh(\pi a)\\ \tan(\pi a)&=\tanh(\pi a) \end{align} $$

The positive solutions to this equation can be solved numerically: $a\approx1.249\ldots,2.249\ldots,3.250\ldots,4.25\ldots$. Only every other solution actually corresponds to a discontinuity in $\operatorname{Arg}$, since we also require the real part of the input to $\operatorname{Arg}$ in your final line to be negative.

So rounding just a little, your result is valid as is for $|a|<a_0=1.25$. After that, we can count how many times the original sum has passed $\pi$, $3\pi$, $5\pi$, $\ldots$ with $\left\lceil\frac{a-a_0}{2}\right\rceil$. So to your result, let's add $2\pi\left\lceil\frac{a-a_0}{2}\right\rceil$. Of course, there has been some rounding here, so this won't always be right for values of $a$ very near to the critical $a$ values $\approx 1.25, 3.25, 5.25\ldots$.

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That never crossed my mind, but is indeed true. But then what values of $a$ is my solution valid for? –  Random Variable Apr 5 '12 at 3:51
    
Only $a$ where $\sum\log\left(1+\frac{2ia^2}{n^2}\right)$ has an imaginary part less than $\pi$. Roughly, $\log\left(1+\frac{2ia^2}{n^2}\right)\approx\frac{2ia^2}{n^2}$ when $a$ is small. So for small $a$, the sum is approximated by $2ia^2\frac{\pi^2}{6}$. Setting that equal to $i\pi$ and solving for $a$ yields $a=\sqrt{3/\pi}$. So, roughly any positive $a$ below this will allow your formula to work. But then again that bound is close to $1$, and I'm not sure the approximation $\log\left(1+\frac{2ia^2}{n^2}\right)\approx\frac{2ia^2}{n^2}$ is still any good in that region. –  alex.jordan Apr 5 '12 at 4:01
    
So this is a bad approach to the problem since you don't really know when the value of $a$ is small enough. And when it's too big, it's never exactly clear how many $2 \pi$'s you should add to the final answer. –  Random Variable Apr 5 '12 at 4:09
    
Wolfram Alpha computes that if $a=1$, the original arctangent sum equals $3\pi/4$, shy of $\pi$. So if you can establish this independently of WA, then your result holds for $|a|\leq1$ (and also for larger $a$ but it's unclear where that boundary lies.) –  alex.jordan Apr 5 '12 at 4:26
    
No I think it's a great approach. I'm about to edit my answer to show how to determine the critical $a$ values. –  alex.jordan Apr 5 '12 at 5:08

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