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The question is to calculate, given $n$ a positive integer

$$S_n:=\sum_{k=1}^n\cos\frac{k\pi}{n+1},$$ and

$$P_n:=\prod_{k=1}^n(-2)\cos\frac{k\pi}{n+1}.$$

By the way, if we define the matrix $\mathbf{A}_n=[a_{ij}^n]_{\forall\,i,j}$ as follows:

$$ a_{ij}^n= \begin{cases} 1, \text{ if $i=j+1$ or $j = i+1$}\\ 0, \text{ otherwise} \end{cases}, $$ or simply

$$ \mathbf{A}_n=\begin{bmatrix} 0 & 1 & 0 & \cdots & \cdots & \cdots & 0 \\ 1 & 0 & 1 & \ddots & & & \vdots\\ 0 & 1 & 0 & \ddots & \ddots & & \vdots \\ \vdots & \ddots & \ddots & \ddots & \ddots & \ddots & \vdots \\ \vdots & & \ddots & \ddots & 0 & 1 & 0 \\ \vdots & & & \ddots & 1 & 0 & 1 \\ 0 & \cdots & \cdots & \cdots & 0 & 1 & 0 \\ \end{bmatrix},$$

then one can show that

$$S_n=\operatorname{tr}(\mathbf{A}_n),\tag{E1}$$

and

$$P_n=\det(\mathbf{A}_n).\tag{E2}$$

Is it easy to find the determinant of $\mathbf{A}_n$?

Without proving $(E1)$ and $(E2)$, I can say that

  • $S_n=0$ for all $n$; and
  • if $n$ is odd then $P_n=0$
  • if $n$ is even then $P_n=\ldots$.
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    Notice that $P_n$ is not an integer for even $n$ (it equals $0$ for odd $n$), so it cannot be equal to the determinant. – 2017-01-01
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    I was told to calculate $P_n(x):=\det(\mathbf{A}_n-xI_n)$ which I found equals to $\dfrac{\sin(n+1)\theta}{\sin\theta}$ for $\theta\in(0,\pi)$ and $x=-2\cos\theta$. After that, $P_n(0)=\det(\mathbf{A}_n)$ which is true for $\theta=\dfrac{\pi}{2}$. – 2017-01-01
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    What can I tell you, $P_n$ is visibly non-integral. $P_2=\cos(\pi/3) \cos(2\pi/3 = -\frac14.$ – 2017-01-01
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    I missed a $-2$ inside the product. – 2017-01-01
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    Show $P_n(x)$ is a $n$ degree polynomial that vanishes when $ \theta_k = \frac{k\pi}{n+1}$ for $k=1,2,3,\ldots,n$. This gives you the roots $x_k(\theta_k)$. From Vieta's formula you know that $P_n(x) = A[x^n - \sum x_k x^{n-1} + \ldots + (-1)^n x_1\cdots x_n]$. Taking $x=0$ you get $P_n(0) = \det A_n$ which gives you the relation between the product of the roots and $\det A_n$. – 2017-01-01
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    The formula for $P_n$ is false for $n=2$: it yields $\det A_2=\dfrac12$; where as a simple calculation says it is $-1$. – 2017-01-01
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    For $n=2$, $\det\mathbf{A}_2=-1$ and $P_2=(-2\cos \pi/3)(-2\cos2\pi/3)=-1$, no? – 2017-01-01

3 Answers 3

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For $S_n$, you can use the factorisation formula, analogous to formula for the sum of consecutive terms of an arithmetic sequence: $$\cos\theta+\cos2\theta+\dots+\cos n\theta=\frac{\sin\dfrac{n\theta}2}{\sin \dfrac{\theta\mathstrut}2}\,\cos\dfrac{(n+1)\theta}2$$ with $\;\theta=\dfrac{\pi}{n+1}$.

For the determinant, it is a tridiagonal determinant. These can be calculated by induction: if $A_k$ $\;(k\le n)$ is the leading principal minor, one has the relation: $$A_n=a_{n,\mkern1mu n}A_{n-1} -a_{n,\mkern1mu n-1}a_{n-1,\mkern1mu n}A_{n-2},$$ which gives i, the present case $$A_n=-A_{n-2},$$ with initial conditions $A_1=0$, $\; A_2=-1$. Hence $$A_{2n+1}=0,\quad A_{2n}=(-1)^n.$$

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    The OP knows it's a determinant, he wants a determinant-free way of computing it. – 2017-01-01
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    This was not very clear to me. Anyway, it gives a way to check other ways of computing. – 2017-01-01
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$S_n$ is the real part of $\sum_{k=1}^n e^{\frac{k i \pi}{n+1}}.$ The second identity is false.

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    @SimpleArt Thanks for proofreading. – 2017-01-01
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    It would be better if you mentioned roots of unity. – 2017-01-01
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    @SimpleArt Why? I am sure the OP can sum a geometric series whether or not it has anything to do with the roots of unity. – 2017-01-01
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    The sum of roots of unity are well known, and it'd make all of this more beautiful IMO. Have you checked what your sum evaluates to? – 2017-01-01
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    @SimpleArt I don't think that your idea of beauty is universal. And the second part of your comment makes no sense. If you have a different answer, by all means, post it. – 2017-01-01
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    For which $n$ the second identity is false? – 2017-01-01
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    I don't wish to post an answer that is near identical to yours, though since you asked for it, I may as well. – 2017-01-01
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    @Azzo $n=2.$ Notice that the RHS equals $-1,$ while the LHS is strictly less than $1$ in absolute value. – 2017-01-01
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    @SimpleArt The point of comments is to make informative remarks. Your remark is "I don't like your answer on aesthetic grounds". That is your right, but the information content is zero, since your taste is yours, and yours alone. – 2017-01-01
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    It provides an interesting piece of information: You could consider this geometrically. That is, if you can interpret what that means. – 2017-01-01
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    @IgorRivin You are right. I am sure I missed something. I will edit the question. – 2017-01-01
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Hint: for the first use that $$\cos(\pi-x)=-\cos(x),$$ which can be easily visualized by using the unit circle:

enter image description here

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    Well, you took my answer. +1 when I find votes to give you. – 2017-01-01
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    @SimpleArt in the sense that you were about to answer in the same way? – 2017-01-01
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    More or less yes. – 2017-01-01