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Self adjoint operators, eigenfunctions & eigenvalues
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[QUOTE="Incand, post: 5265608, member: 525632"] Thanks for taking the time replying! You're absolutely right about the index in that sum and thanks for clarifying that I can't have those constants from the function as an eigenvalue. If I write out ##Df(\theta)## I have ##\sum_{-N}^N \left(c_k e^{ik\theta} + c_ke^{-ik\theta} \right) = \sum_{-N}^N (c_k + c_{-k})e^{ik\theta}## I'm not sure I get all eigenfunctions here but I was thinking if I take eigenfunctions of the form ##f\in P_n## with the added constraint that ##c_k+c_{-k}=0## I have the eigenvalue zero? I also possibly see other eigenvalues For example if I choose the constrant that ##c_{-k} = ac_{k}## with ##a## being a complex number I have the eigenvalue ##(1+a)##. But not sure if I'm allowed too do this? The eigenvalue doesn't vary like in the last case(which was obviously wrong) but I do use properties from the function I guess. [/QUOTE]
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Self adjoint operators, eigenfunctions & eigenvalues
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