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I have a vector that looks like

$$ \begin{bmatrix} 2\sin(n\pi/N) - \sin(2n\pi/N) \\ - \sin(n\pi/N) + 2\sin(2n\pi/N) - \sin(3n\pi/N) \\ - \sin(2n\pi/N) + 2\sin(3n\pi/N) - \sin(4n\pi/N) \\ - \sin(3n\pi/N) + 2\sin(4n\pi/N) - \sin(5n\pi/N) \\ \vdots \\ - \sin((N-2)n\pi/N) + 2\sin((N-1)n\pi/N) \end{bmatrix} $$

Somehow I'm supposed to pull out $4 \sin^2(n \pi / (2N)$ from the vector and get

$$ \begin{bmatrix} \sin(n\pi/N) \\ \sin(2n\pi/N) \\ \sin(3n\pi/N) \\ \vdots \\ \sin((N-1)n\pi/N) \\ \end{bmatrix} $$

But it's not obvious to me how to do this...

financial_physician
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1 Answers1

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Here is an answer that attempts to show the "great picture" behind your (in fact classical) question ; the corresponding computations have already been done and are not necessarily to be re-invented. Nevertheless, see the Edit at the bottom of this answer giving a direct approach.

Let $\theta:=n\pi/N$, $$ V:=\begin{bmatrix} \sin(\theta) \\ \sin(2\theta) \\ \sin(3\theta) \\ \vdots \\ \sin((N-1)\theta) \end{bmatrix}$$

and $$ M:=\begin{bmatrix} 2&-1&0&0&\cdots &0&0&0\\ -1&2&-1&0&\cdots &0&0&0\\ 0&-1&2&-1&\cdots &0&0&0\\ &&\ddots&\ddots&\ddots&&& \\ 0&0&0&0&\cdots &-1&2&-1\\ 0&0&0&0&\cdots &0&-1&2\\ \end{bmatrix}$$

Your issue means that you have to prove that

$$MV=\lambda V \ \ \text{with} \ \ \lambda:=(2\sin \theta/2)^2$$

It is an eigenvalue-eigenvector classical problem (see references below).

References :

  1. It can be viewed as a consequence of the result given (and established) in this question with $a=2, b=c=-1$ giving eigenvalues (taking into account that their matrix is an $n \times n$ matrix whereas our matrix $M$ is $(N-1) \times (N-1)$:

$$\lambda_k=a+2\sqrt{bc}\cos k\frac{\pi}{N}=2(1+\cos k\frac{\pi}{N})=2 \times 2 \cos^2(k\frac{\pi}{2N})\tag{1}$$

Take now $k=N-1$ and you will find the result you are looking for:

$$\lambda_{n-1}=4 \sin^2 \frac{\pi}{2N}$$

(the fact one finds a $\cos^2$ in (1) instead of a $\sin^2$ isn't important : $\sin$ and $\cos$ exchange their values when $k$ varies from $1$ to $n-1$).

  1. Alternatively, see "Neumann case" with $k=1$ (smallest eigenvalue) at the bottom of this Wikipedia reference. This is because matrix $M$ can be considered as a discrete form of the opposite of the second derivative, which, not surprizingly, has a discretized version of the sine function (vector $V$) as an eigenvector.

Edit: Hint for a direct solution:

For the general case:

$-\sin (k-1)a+2 \sin ka-\sin (k+1)a$

$=2\ sin ka -(\sin (k-1)a)+2\sin (k+1)a)$

$=2\sin ka - 2 \sin \frac{(k-1)a+(k+1)a}{2}\cos \frac{(k-1)a-(k+1)a}{2}$

$=2 \sin ka-2 \sin ka \cos a = 2 \sin ka (1-\cos a)$

$=2 \sin ka \ \ 2 \sin^2 \frac{a}{2}$

knowing that you have to treat in a separate way the first and last term.

Jean Marie
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