Let’s dissecate this :
>>> from sympy import symbols, fourier_series>>> t=symbols("t", real=True)>>>
>>> T=symbols("T", positive=True)>>> foo=fourier_series(1/t, (t, -T/2, T/2)) ;
>>> foo
FourierSeries(1/t, (t, -T/2, T/2), (0, SeqFormula(0, (_k, 1, oo)),
SeqFormula(4*sin(2*_n*pi*t/T)*Si(_n*pi)/T, (_n, 1, oo))))
The constant term tof this series is taken to be 0. The even terms are all
0. The coefficient of the nth odd term is Si(n*pi). These coefficients do
*not* converge to 0 :
>>> k=symbols("k", integer=True)>>> limit(Si(k*pi), k, oo)
pi/2
therefore their sum does not necessarily converge.
Therefore, the expressions returned by fourier_transform are analitically
correct, but their existence *do not* imply their convergence.
In other word, fourier series return the elements allowing computation of
the Fourier series of the function *if such a series exist*, but *do not
assess its existence. *
HTH,
Le mercredi 14 décembre 2022 à 19:15:24 UTC+1, antonv...@gmail.com a écrit :
> Hi, I'm trying to find fourier series decomposition of 1/x function on
> [-3; 3] interval. IMHO, such a decomposition doesn't exist, because the a0
> и an doesn't exist (corresponding integrals doesn't converge). But sympy
> function fourier_series, returns the answer (omitting the a0, an
> coefficients and only taking into account the bn coefficients). Is there an
> error in fourier_series function?
> fourier_series(1/x,(x,-3,3)) =
> (2/3)*sin(pi*x/3)*Si(pi)+(2/3)*sin(2*pi*x/3)*Si(2*pi)+(2/3)*sin(pi*x)*Si(3*pi)+...
>
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