# Possible errata for Paul J. Nahin's An Imaginary Tale

On pp. 71—72 of Paul J. Nahin's An Imaginary Tale: the story of $\sqrt{-1}$, Princeton University Press (2016), the author derives the following formula:

$\displaystyle \prod_{k=1}^{n-1} \cos \left(\frac{k \pi}{2n} \right) = \frac{\sqrt{n}}{2^{n-1}}$

He then writes on p. 72:

[...] if you simply define a new index variable on the product as $j = n - k$ (so that as $k$ runs from $1$ to $n-1$ then $j$ runs $n-1$ to $1$), and if you recall that $\sin(\theta) = \cos(\frac{1}{2}\pi - \theta)$, then it is easy to see that we also have the result

$\displaystyle \prod_{k=1}^{n-1} \sin \left(\frac{k \pi}{2n} \right) = \frac{\sqrt{n}}{2^{n-1}}$

This last formula seems identical to me as the first one above, only having replaced the cosine for a sine. The new index variable j does not seem to be used, nor does that relation between $\sin(\theta)$ and the cosine function. Supposing this is an errata, what would be the correct form for the second formula?

• Having Mathematica compute for some random values of $n$ does not show a counter example to Nahin's results. Commented Aug 31, 2016 at 9:56

There is no error. Here is a more detailed derivation. We start with $$\prod_{k=1}^{n-1} \cos \left(\frac{k \pi}{2n} \right)=\prod_{k=1}^{n-1} \sin \left(\frac{\pi}{2}-\frac{k \pi}{2n} \right)=\prod_{k=1}^{n-1} \sin \left(\frac{(n-k) \pi}{2n} \right).$$ Now make a change of variables $j=n-k$ to obtain $$\prod_{j} \sin \left(\frac{j \pi}{2n} \right),$$ where in the product $j$ ranges over all integers such that $k=n-j$ is between $1$ and $n-1$. But such integers $j$ are just all the integers between $1$ and $n-1$. So in fact this product is the same as $$\prod_{j=1}^{n-1} \sin \left(\frac{j \pi}{2n} \right).$$ Now we just rename the dummy variable $j$ to $k$ to get $$\prod_{k=1}^{n-1} \sin \left(\frac{k \pi}{2n} \right).$$
I believe you are correct, the index on the product should be $j$ since forming the product form $1$ to $n-1$ (over $k$) is the same as forming the product from $n-1$ to $1$ (over $j$).
Now, for the $\sin \cos$ extravaganza, viz \begin{align} \frac{\pi}{2}-\frac{k \pi}{2n} &= \frac{\pi}{2}\left( 1-\frac{k}{n}\right) \\ &= \frac{\pi}{2}\left(\frac{n-k}{n} \right) \\ &= \frac{\pi}{2}\frac{j}{n} \end{align} And thus the line should read $$\prod_{j=1}^{n-1} \sin \left(\frac{j \pi}{2n}\right) = \frac{\sqrt{n}}{2^{n-1}}$$
• Thanks for the reply. It turns out that the last equation in your post using $j$ as index variable for the product is the same as using $k$ (see Eric's answer above). But I am just writing because there is a minor problem in your explanation of the "sin cos extravaganza", in the first line you should end with $\frac{k}{n}$ and not $\frac{n}{k}$, right? Commented Aug 31, 2016 at 15:50
• @gilberto.agostinho.f Yes, a slight typo - I agree with Eric too, my answer was meant to elicit this fact, that the dummy variable allows for either indexation through $j$ or $k$. Thanks.