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What is the Fourier series of the function $$ f(t) = \frac{1}{1+ a t^2}$$ over $[0,1]$, where $a >0$ is some constant? I mean, are the coefficients known?

passerby51
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  • Why do you ask? – hmakholm left over Monica Mar 08 '13 at 14:30
  • Henning Makholm, because I need to know! – passerby51 Mar 08 '13 at 14:35
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    What do you "need" to know it for? (In other words: Is this homework? If it is, then what are your thoughts?) – hmakholm left over Monica Mar 08 '13 at 14:37
  • This function is what physicists say spectra are made of. Knowing the Fourier series of it allows me to build a reproducing kernel Hilbert space (RKHS), via http://en.wikipedia.org/wiki/Mercer's_theorem . Knowing the coefficients of the Fourier series is the same as knowing the eigenvalues of the kernel operator, which can be used to relate the norm of the RKHS to the L^2 norm. – passerby51 Mar 08 '13 at 14:41
  • ... the kernel will be $K(x,y) = f(x-y)$. – passerby51 Mar 08 '13 at 14:42
  • This is not homework and my Fourier series knowledge (and integration) is rusty. My thoughts on the subject are that someone should know whether $\int_0^1 \frac{\cos(kt)}{1+t^2} dt$ has a closed form. – passerby51 Mar 08 '13 at 14:45
  • It has a closed form, but in terms of special functions. – Mhenni Benghorbal Mar 08 '13 at 15:00
  • Mhenni Benghorbal, thanks, it might still be useful to know. – passerby51 Mar 08 '13 at 15:48
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    It's kind of a strange function you are asking about. I mean, a Lorentzian between $[0,1]$, and then discontinuously repeating over the real line. I do know a thing or two about lasers, and I do know that the spectra are typically continuous. In that case, you are better off with a Fourier transform over the entire real line. And the spectra of a Lorentzian is very well known. – Ron Gordon Mar 08 '13 at 17:26
  • Thanks. You might be right, it might be easier to think of it as a function on $\mathbb{R}$. – passerby51 Mar 09 '13 at 02:19

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