The spectrum$(\mathbb{S}^1)=\{\lambda_k=k^2\ : k \in \mathbb{N}\}$, and the eigenfunctions $\mu_k(t)$ associated to the eigenvalues $\lambda_k$ are $a_k \cos kt + b_k \sin kt$ under the Laplacian operator. A things I know is $\mu_k$ is a periodic function on the interval $[0, 2 \pi)$, and I would like to find the set of normalized eigenfunction. How do I find out if I don't know the explicit values of $a_k$ and $b_k$


closed as unclear what you're asking by user7530, Claude Leibovici, user91500, John B, Davide Giraudo May 24 '16 at 9:53

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    $\begingroup$ I am a bit confused. You talk about the spectrum of a set but then talk about eigenvalues (presumably of some differential operator). Can you please be more clear? $\endgroup$ – Cameron Williams May 24 '16 at 0:41
  • $\begingroup$ @CameronWilliams Is it more clearer? $\endgroup$ – user332990 May 24 '16 at 0:44
  • $\begingroup$ So basically you just need to find values for $a_k$ and $b_k$? Try computing the integrals $\int_0^{2\pi} cos^2(kt)\,dt$ and likewise for $\sin$. (If I'm understanding you correctly.) You should have that $a_k^2\int_0^{2\pi}\cos^2(kt)\,dt = 1$. $\endgroup$ – Cameron Williams May 24 '16 at 0:46
  • $\begingroup$ @CameronWilliams Is it the property of normalization? Could you explain a bit more? $\endgroup$ – user332990 May 24 '16 at 0:51
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    $\begingroup$ Yes it is. You want to have that $\int_0^{2\pi} (a_k\cos^2(kt))^2\,dt = 1$ and likewise for $\sin$. This is what it means for the functions to be $L^2$-normalized. $\endgroup$ – Cameron Williams May 24 '16 at 0:52