Expansion in eigenbasis of the discrete Laplacian.

Question

Let $
\mathbf{L}=\begin{bmatrix}
\delta&\tau&&&&O\\
\sigma&\delta&\tau&&&&\\
&\sigma&\cdot&\cdot&&\\
&&\cdot&\cdot&\cdot&\\
&&&\cdot&\cdot&\tau\\
O&&&&\sigma&\delta
\end{bmatrix}
$ be the $n\times n$ discrete Laplacian matrix. This is a tridiagonal Toeplitz matrix hence we know its eigenvalues and eigenvectors. The $k$-th element of the $j$-th eigenvector is thus given by
$$
v_{j,k}=\sqrt{\frac{2}{n+1}}\sin\bigg(\frac{jk\pi}{n+1}\bigg)
$$
Also, $\mathbf{L}$ has $n$ distinct non-zero eigenvalues hence it is invertible and has a basis of eigenvectors.
Now I have the vector $\mathbf{f}=\begin{bmatrix}1,1,…,1\end{bmatrix}^T$. I want to expand $\mathbf{f}$ in the eigenbasis of $\mathbf{L}$, so $\mathbf{f}=\sum_{j=1}^n\alpha_j\mathbf{v}_j$. Is there an easy trick to construct the coefficients $\alpha_j$? I was maybe thinking to exploit the orthogonality of the eigenvectors but I am not sure this is going to work.

Answer 1

Based on the comments, it seems that you're ultimately looking for closed form for the expression $$ \alpha_j=\sum_{k=1}^n \sin\left(\frac{jk\pi}{n+1}\right). $$ Letting $\omega = \exp(\frac{i\pi}{n+1})$, we have $$ \alpha_j = \sum_{k=1}^n \operatorname{Im}[\omega^{jk}] = \operatorname{Im}\sum_{k=1}^n \omega^{jk} = \operatorname{Im}\left( \omega \cdot \sum_{k=0}^{n-1} (\omega^{j})^k\right) = \operatorname{Im} \left(\omega \cdot \frac{1 - \omega^{jn}}{1 - \omega^j}\right)\\ = \operatorname{Im} \left(\frac{\omega^{j} - \omega^{j(n+1)}}{1 - \omega^j}\right) = \operatorname{Im} \left(\frac{\omega^{j} - [\omega^{(n+1)}]^j}{1 - \omega^j}\right) = \operatorname{Im} \frac{\omega^{j} - (-1)^j }{1 - \omega^j} $$ In the case that $j$ is even, this becomes $$ \alpha_j = \operatorname{Im} \frac{\omega^{j} - 1 }{1 - \omega^j} = -\operatorname{Im} \frac{1 - \omega^j}{1 - \omega^j} = -\operatorname{Im} 1 = 0. $$ When $j$ is odd, we have $$ \operatorname{Im} \frac{1 + \omega^{j}}{1 - \omega^j}\\ = \operatorname{Im} \frac{(1 + \omega^{j})(1 - \omega^{-j})}{(1 - \omega^j)(1 - \omega^{-j})} = \operatorname{Im} \frac{\omega^j - \omega^{-j}}{|1 - \omega^j|^2} =\frac{2 \operatorname{Im}(\omega^j)}{|1 - \omega^j|^2}. $$

We then have $\operatorname{Im}(\omega^j) = \sin(j\pi/(n+1))$ and $$ |1 - \omega^j|^2 = [1-\cos(j\pi/(n+1))]^2 + \sin^2(j\pi/(n+1))\\ = 2 - 2\cos(j\pi/(n+1)). $$ All together: for odd $j$, $$ \alpha_j = \frac{\sin(j\pi/(n+1))}{1 - \cos(j\pi/(n+1))}. $$ In fact, we have the identity $\frac{\sin x}{1 - \cos x} = \cot(x/2)$, so the above can be written as $$ \alpha_j = \cot\left(\frac{j\pi}{2(n+1)}\right) = \cot\left(\frac{j}{n+1}\cdot\frac{\pi}{2}\right). $$ Since $\cot(x)$ is decreasing on the interval $(0,\pi/2]$, we see that for odd $j$, the $\alpha_j$ decrease as $j$ increases.