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I think this question isn't that hard, but I am a bit confused:

Define $(Af)(x):=\int_{0}^{1}\cos(2\pi(x-y))f(y)dy.$ Then $A$ is an operator on functions. Find the eigenvalues and the eigenfunctions.

I can think of a lot of functions that give $0$, things like $f(x)=\cos(n2\pi x)$. Also one eigenfunction that gives eigenvalue $\frac {1}{2}$ (I think). My problem is I have no idea how to show that I have found all of them, and I don't know if I have found all of them.

Thanks for showing me!!

Edit: I havent really shown my work because there is a lot of it and it is all over the place. J.M. suggest that it is useful to write $Af(x)=\cos(2\pi x)\int_{0}^{1}\cos\left(2\pi y\right)f(y)dy+\sin\left(2\pi x\right)\int_{0}^{1}\sin\left(2\pi y\right)f(y)dy.$ I used this to find that $\cos(2\pi x)+\sin (2\pi x)$ is an eigenvector. I wasn't sure if this is the right track, and I am still confused about what to do from here. (Specifically, once we find a bunch of $\lambda$ how do we prove that is all of them?)

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    @J.M. How does one do this? And what is a secular equation? Thank you for your help.2011-04-20

1 Answers 1

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Let's start with where you're stuck ($\lambda$ is an eigenvalue):

$\lambda f(x)=\cos(2\pi x)\int_0^1\cos\left(2\pi y\right)f(y)\mathrm dy+\sin\left(2\pi x\right)\int_0^1\sin\left(2\pi y\right)f(y)\mathrm dy$

Let $c_1=\int_0^1\cos\left(2\pi y\right)f(y)\mathrm dy$ and $c_2=\int_0^1\sin\left(2\pi y\right)f(y)\mathrm dy$; $f(x)$ thus has the form

$f(x)=\frac1{\lambda}\left(c_1\cos(2\pi x)+c_2\sin\left(2\pi x\right)\right)$

We then assemble two equations: one where both sides of the equation are multiplied by $\cos(2\pi x)$, and one multiplied by $\sin(2\pi x)$, and then integrate both sides of the equation. I'll do it for $\cos(2\pi x)$:

$\cos(2\pi x)f(x)=\frac1{\lambda}\left(c_1\cos(2\pi x)\cos(2\pi x)+c_2\sin\left(2\pi x\right)\cos(2\pi x)\right)$

$c_1=\frac1{\lambda}\left(c_1\int_0^1\cos(2\pi x)\cos(2\pi x)\mathrm dx+c_2\int_0^1\sin\left(2\pi x\right)\cos(2\pi x)\mathrm dx\right)$

You should be able to recognize an (algebraic) eigenvalue equation at this point; the "secular equation" I was referring to in the comments is sometimes also referred to as the "characteristic polynomial".

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    @Fabian, thanks for covering for me. :) You nailed it.2011-04-24