Fourier Series Integration in terms of Pi

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Bob Gill
Bob Gill 2023-4-17
编辑： VBBV 2024-7-15
Hello,
The following code is just to check my integration of a fourier series transform, but the output doesn't seem to be right for bn. It displays a large number at the end of the bn output.
syms t
syms n 'integer'
an = (1/pi)*(int(-1*cos(n*pi*t/pi),-pi,-pi/2)+int(0*cos(n*pi*t/pi), -pi/2, pi/2)+int(1*cos(n*pi*t/pi), pi/2, pi))
bn = (1/pi)*(int(-1*sin(n*pi*t/pi),-pi,-pi/2)+int(0*sin(n*pi*t/pi), -pi/2, pi/2)+int(1*sin(n*pi*t/pi), pi/2, pi))
pretty(an)
pretty(bn)

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回答（1 个）

VBBV
VBBV 2023-4-17
编辑：VBBV 2024-7-15
Ran in:
Hi @Bob Gill, the value of bn can be computed as follows
syms t n 'integer'
an = (1/pi)*(int(-1*cos(n*pi*t/pi),-pi,-pi/2)+int(0*cos(n*pi*t/pi), -pi/2, pi/2)+int(-1*cos(n*pi*t/pi), pi/2, pi))
an = 
bn = (1/pi)*(int(-1*sin(n*pi*t/pi),-pi,-pi/2)+int(0*sin(n*pi*t/pi), -pi/2, pi/2)+int(-1*sin(n*pi*t/pi), pi/2, pi))
bn = 
0
vpa(an,2)
ans = 
vpa(bn,2)
ans = 
0.0
  1 个评论
VBBV
VBBV 2023-4-17
编辑：VBBV 2023-4-17
Use vpa to ccompute the bn value. You can also consider the vpaintegral function to compute the values for an and bn

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