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Mathematics Wavelets and Fourier Series

Cosine coefficient of the Fourier serie of $\sqrt{1 - k^2 \sin{t}^2}$

afreeka b

Posted 11 years ago

Hi Everyone, I'm curently working a physical model based on the Kernel Poisson. I'm struggling with a Fourier serie I need to find the Fourier series of the following function: $f(t)=\left[\sqrt{1-k^2\sin^2t}\,\right]$ (eq.1) The function is even and $\pi$-periodic. The Fourier serie should be in this form: $f(t)=\frac{a_0}2+\sum\limits_{i=0}^\infty a_n\cos[2nt]$ (eq.2) Then, at $t\to0$, the Taylor serie is: $f(t)=\left[\frac{2E[k^2]}\pi+\sum_{i=0}^\infty\frac1{2^{2i-1}}\pmatrix{1/2\\ i}(k)^{2i}\sum_{j=0}^{i-1}(-1)^j\pmatrix{2i\\j}\cos(2(i-j)t) \right]$ (eq.3) It's pretty close to the final Fourier serie but I cannot identify the coefficient $a_n$ from (eq.2) and (eq.3). Can someone give a help on this? Any support is appreciated. Thanks in advance. PS: $k<<1$ is real postive and E is the complete elliptic integral of the 2nd kind.

POSTED BY: afreeka b

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Udo Krause

Udo Krause, NOrganization

Posted 11 years ago

You could use directly In[25]:= FourierCosSeries[Sqrt[1 - k^2 Sin[t]^2], t, 8] Out[25]= (2 EllipticE[k^2])/\[Pi] + ( 2 Cos[2 t] (-2 (-2 + k^2) EllipticE[k^2] + 4 (-1 + k^2) EllipticK[k^2]))/(3 k^2 \[Pi]) + ( 2 Cos[4 t] (-2 (16 - 16 k^2 + k^4) EllipticE[k^2] + 16 (2 - 3 k^2 + k^4) EllipticK[k^2]))/(15 k^4 \[Pi]) + ( 2 Cos[6 t] ((512 - 768 k^2 + 268 k^4 - 6 k^6) EllipticE[k^2] + 4 (-128 + 256 k^2 - 155 k^4 + 27 k^6) EllipticK[k^2]))/( 105 k^6 \[Pi]) + ( 2 Cos[8 t] (-2 (2048 - 4096 k^2 + 2496 k^4 - 448 k^6 + 5 k^8) EllipticE[k^2] + 64 (64 - 160 k^2 + 138 k^4 - 47 k^6 + 5 k^8) EllipticK[k^2]))/( 315 k^8 \[Pi]) In[27]:= FourierCosCoefficient[Sqrt[1 - k^2 Sin[t]^2], t, 8] Out[27]= (2 (-2 (2048 - 4096 k^2 + 2496 k^4 - 448 k^6 + 5 k^8) EllipticE[k^2] + 64 (64 - 160 k^2 + 138 k^4 - 47 k^6 + 5 k^8) EllipticK[ k^2]))/(315 k^8 \[Pi]) Now you have to reconstruct the denominator and the polynomials in front of EllipticE as well as in front of EllipticK in dependence from n. Possibly you can do so by understanding how the elliptic integrals do appear in the formulae.

You could use directly

In[25]:= FourierCosSeries[Sqrt[1 - k^2 Sin[t]^2], t, 8]
Out[25]= (2 EllipticE[k^2])/\[Pi] + (
 2 Cos[2 t] (-2 (-2 + k^2) EllipticE[k^2] + 
    4 (-1 + k^2) EllipticK[k^2]))/(3 k^2 \[Pi]) + (
 2 Cos[4 t] (-2 (16 - 16 k^2 + k^4) EllipticE[k^2] + 
    16 (2 - 3 k^2 + k^4) EllipticK[k^2]))/(15 k^4 \[Pi]) + (
 2 Cos[6 t] ((512 - 768 k^2 + 268 k^4 - 6 k^6) EllipticE[k^2] + 
    4 (-128 + 256 k^2 - 155 k^4 + 27 k^6) EllipticK[k^2]))/(
 105 k^6 \[Pi]) + (
 2 Cos[8 t] (-2 (2048 - 4096 k^2 + 2496 k^4 - 448 k^6 + 
       5 k^8) EllipticE[k^2] + 
    64 (64 - 160 k^2 + 138 k^4 - 47 k^6 + 5 k^8) EllipticK[k^2]))/(
 315 k^8 \[Pi])

In[27]:= FourierCosCoefficient[Sqrt[1 - k^2 Sin[t]^2], t, 8]
Out[27]= (2 (-2 (2048 - 4096 k^2 + 2496 k^4 - 448 k^6 + 
      5 k^8) EllipticE[k^2] + 
   64 (64 - 160 k^2 + 138 k^4 - 47 k^6 + 5 k^8) EllipticK[
     k^2]))/(315 k^8 \[Pi])

Now you have to reconstruct the denominator and the polynomials in front of EllipticE as well as in front of EllipticK in dependence from n.

Possibly you can do so by understanding how the elliptic integrals do appear in the formulae.

POSTED BY: Udo Krause

afreeka b

Posted 11 years ago

Thanks a lot. it was very helpful. Your comment gave me some hint and I found the analytical form of the Fourier transform which is a sum of Gauss's hypergeometric fonctions.:)

POSTED BY: afreeka b

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