Fourier series of $f(x)=pi-x$
$begingroup$
$$f(x)=pi-x qquad x in [0,2 pi[$$
$$a_0=frac{1}{pi} int_0^{2 pi}f(x) dx=frac{1}{pi} int_0^{2 pi} (pi-x) dx=0$$
$$a_n=frac{1}{pi} int_0^{2 pi} cos(nx) dx=frac{1}{pi} int_0^{2 pi}(pi cos(nx)-x cos(nx)) dx=$$
$$frac{1}{pi} Big( Big[frac{pi}{n} sin(nx) Big]_0^{2 pi}-Big[frac{x}{n} sin(nx)+frac{1}{n^2} cos(nx) Big]_0^{2 pi} Big)=0 $$
$$b_n=frac{1}{pi} Big( Big[ -frac{pi}{n} cos(nx) Big]_0^{2 pi}-Big[-frac{x}{n} cos(nx)+frac{1}{n^2} sin(nx) Big]_0^{2 pi} Big)=frac{2}{n}$$
$$f(x)=2 sum_{n=1}^{+infty} frac{1}{n} sin(nx)$$
Is it correct?
fourier-series
asked Sep 30 '16 at 14:33
ElsaElsa
82111
$endgroup$
add a comment |
$begingroup$
$$f(x)=pi-x qquad x in [0,2 pi[$$
$$a_0=frac{1}{pi} int_0^{2 pi}f(x) dx=frac{1}{pi} int_0^{2 pi} (pi-x) dx=0$$
$$a_n=frac{1}{pi} int_0^{2 pi} cos(nx) dx=frac{1}{pi} int_0^{2 pi}(pi cos(nx)-x cos(nx)) dx=$$
$$frac{1}{pi} Big( Big[frac{pi}{n} sin(nx) Big]_0^{2 pi}-Big[frac{x}{n} sin(nx)+frac{1}{n^2} cos(nx) Big]_0^{2 pi} Big)=0 $$
$$b_n=frac{1}{pi} Big( Big[ -frac{pi}{n} cos(nx) Big]_0^{2 pi}-Big[-frac{x}{n} cos(nx)+frac{1}{n^2} sin(nx) Big]_0^{2 pi} Big)=frac{2}{n}$$
$$f(x)=2 sum_{n=1}^{+infty} frac{1}{n} sin(nx)$$
Is it correct?
fourier-series
asked Sep 30 '16 at 14:33
ElsaElsa
82111
$endgroup$
2
$begingroup$
That is the correct approach. A plot of your answer also shows that the approximation is tending to the desired function wolframalpha.com/input/…
$endgroup$
– Hugh
Sep 30 '16 at 15:14
add a comment |
$begingroup$
$$f(x)=pi-x qquad x in [0,2 pi[$$
$$a_0=frac{1}{pi} int_0^{2 pi}f(x) dx=frac{1}{pi} int_0^{2 pi} (pi-x) dx=0$$
$$a_n=frac{1}{pi} int_0^{2 pi} cos(nx) dx=frac{1}{pi} int_0^{2 pi}(pi cos(nx)-x cos(nx)) dx=$$
$$frac{1}{pi} Big( Big[frac{pi}{n} sin(nx) Big]_0^{2 pi}-Big[frac{x}{n} sin(nx)+frac{1}{n^2} cos(nx) Big]_0^{2 pi} Big)=0 $$
$$b_n=frac{1}{pi} Big( Big[ -frac{pi}{n} cos(nx) Big]_0^{2 pi}-Big[-frac{x}{n} cos(nx)+frac{1}{n^2} sin(nx) Big]_0^{2 pi} Big)=frac{2}{n}$$
$$f(x)=2 sum_{n=1}^{+infty} frac{1}{n} sin(nx)$$
Is it correct?
fourier-series
asked Sep 30 '16 at 14:33
ElsaElsa
82111
$endgroup$
$$f(x)=pi-x qquad x in [0,2 pi[$$
$$a_0=frac{1}{pi} int_0^{2 pi}f(x) dx=frac{1}{pi} int_0^{2 pi} (pi-x) dx=0$$
$$a_n=frac{1}{pi} int_0^{2 pi} cos(nx) dx=frac{1}{pi} int_0^{2 pi}(pi cos(nx)-x cos(nx)) dx=$$
$$frac{1}{pi} Big( Big[frac{pi}{n} sin(nx) Big]_0^{2 pi}-Big[frac{x}{n} sin(nx)+frac{1}{n^2} cos(nx) Big]_0^{2 pi} Big)=0 $$
$$b_n=frac{1}{pi} Big( Big[ -frac{pi}{n} cos(nx) Big]_0^{2 pi}-Big[-frac{x}{n} cos(nx)+frac{1}{n^2} sin(nx) Big]_0^{2 pi} Big)=frac{2}{n}$$
$$f(x)=2 sum_{n=1}^{+infty} frac{1}{n} sin(nx)$$
Is it correct?
fourier-series
fourier-series
asked Sep 30 '16 at 14:33
ElsaElsa
82111
asked Sep 30 '16 at 14:33
ElsaElsa
82111
asked Sep 30 '16 at 14:33
ElsaElsa
82111
asked Sep 30 '16 at 14:33
ElsaElsa
82111
asked Sep 30 '16 at 14:33
ElsaElsa
82111
82111
2
$begingroup$
That is the correct approach. A plot of your answer also shows that the approximation is tending to the desired function wolframalpha.com/input/…
$endgroup$
– Hugh
Sep 30 '16 at 15:14
add a comment |
2
$begingroup$
That is the correct approach. A plot of your answer also shows that the approximation is tending to the desired function wolframalpha.com/input/…
$endgroup$
– Hugh
Sep 30 '16 at 15:14
2
2
$begingroup$
That is the correct approach. A plot of your answer also shows that the approximation is tending to the desired function wolframalpha.com/input/…
$endgroup$
– Hugh
Sep 30 '16 at 15:14
$begingroup$
That is the correct approach. A plot of your answer also shows that the approximation is tending to the desired function wolframalpha.com/input/…
$endgroup$
– Hugh
Sep 30 '16 at 15:14
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Slight errors:
$$
begin{align}
%
a_{0} &= frac{1}{pi} int_{-pi }^{pi } f(x) , dx = 2pi \[5pt]
%
b_{k} &= frac{1}{pi } int_{-pi }^{pi } f(x) sin (k x) , dx = (-1)^k frac{2}{k pi}
%
end{align}
$$
The decay of the amplitudes is linear:
The series expansion looks like
$$
pi - x = pi - 2 sin (x) + sin (2x) - frac{2}{3} sin (3x) + frac{1}{2} sin left( 4x right) - dots
$$
Convergence sequence:
answered Apr 6 '17 at 15:30
dantopadantopa
6,69442245
$endgroup$
2
$begingroup$
Nice illustration of the Gibbs Phenomenon in this answer, too.
$endgroup$
– Dan Sheppard
Dec 29 '18 at 23:06
add a comment |
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1 Answer
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$begingroup$
Slight errors:
$$
begin{align}
%
a_{0} &= frac{1}{pi} int_{-pi }^{pi } f(x) , dx = 2pi \[5pt]
%
b_{k} &= frac{1}{pi } int_{-pi }^{pi } f(x) sin (k x) , dx = (-1)^k frac{2}{k pi}
%
end{align}
$$
The decay of the amplitudes is linear:
The series expansion looks like
$$
pi - x = pi - 2 sin (x) + sin (2x) - frac{2}{3} sin (3x) + frac{1}{2} sin left( 4x right) - dots
$$
Convergence sequence:
answered Apr 6 '17 at 15:30
dantopadantopa
6,69442245
$endgroup$
2
$begingroup$
Nice illustration of the Gibbs Phenomenon in this answer, too.
$endgroup$
– Dan Sheppard
Dec 29 '18 at 23:06
add a comment |
$begingroup$
Slight errors:
$$
begin{align}
%
a_{0} &= frac{1}{pi} int_{-pi }^{pi } f(x) , dx = 2pi \[5pt]
%
b_{k} &= frac{1}{pi } int_{-pi }^{pi } f(x) sin (k x) , dx = (-1)^k frac{2}{k pi}
%
end{align}
$$
The decay of the amplitudes is linear:
The series expansion looks like
$$
pi - x = pi - 2 sin (x) + sin (2x) - frac{2}{3} sin (3x) + frac{1}{2} sin left( 4x right) - dots
$$
Convergence sequence:
answered Apr 6 '17 at 15:30
dantopadantopa
6,69442245
$endgroup$
2
$begingroup$
Nice illustration of the Gibbs Phenomenon in this answer, too.
$endgroup$
– Dan Sheppard
Dec 29 '18 at 23:06
add a comment |
$begingroup$
Slight errors:
$$
begin{align}
%
a_{0} &= frac{1}{pi} int_{-pi }^{pi } f(x) , dx = 2pi \[5pt]
%
b_{k} &= frac{1}{pi } int_{-pi }^{pi } f(x) sin (k x) , dx = (-1)^k frac{2}{k pi}
%
end{align}
$$
The decay of the amplitudes is linear:
The series expansion looks like
$$
pi - x = pi - 2 sin (x) + sin (2x) - frac{2}{3} sin (3x) + frac{1}{2} sin left( 4x right) - dots
$$
Convergence sequence:
answered Apr 6 '17 at 15:30
dantopadantopa
6,69442245
$endgroup$
Slight errors:
$$
begin{align}
%
a_{0} &= frac{1}{pi} int_{-pi }^{pi } f(x) , dx = 2pi \[5pt]
%
b_{k} &= frac{1}{pi } int_{-pi }^{pi } f(x) sin (k x) , dx = (-1)^k frac{2}{k pi}
%
end{align}
$$
The decay of the amplitudes is linear:
The series expansion looks like
$$
pi - x = pi - 2 sin (x) + sin (2x) - frac{2}{3} sin (3x) + frac{1}{2} sin left( 4x right) - dots
$$
Convergence sequence:
answered Apr 6 '17 at 15:30
dantopadantopa
6,69442245
answered Apr 6 '17 at 15:30
dantopadantopa
6,69442245
answered Apr 6 '17 at 15:30
dantopadantopa
6,69442245
answered Apr 6 '17 at 15:30
dantopadantopa
6,69442245
6,69442245
2
$begingroup$
Nice illustration of the Gibbs Phenomenon in this answer, too.
$endgroup$
– Dan Sheppard
Dec 29 '18 at 23:06
add a comment |
2
$begingroup$
Nice illustration of the Gibbs Phenomenon in this answer, too.
$endgroup$
– Dan Sheppard
Dec 29 '18 at 23:06
2
2
$begingroup$
Nice illustration of the Gibbs Phenomenon in this answer, too.
$endgroup$
– Dan Sheppard
Dec 29 '18 at 23:06
$begingroup$
Nice illustration of the Gibbs Phenomenon in this answer, too.
$endgroup$
– Dan Sheppard
Dec 29 '18 at 23:06
add a comment |
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$begingroup$
That is the correct approach. A plot of your answer also shows that the approximation is tending to the desired function wolframalpha.com/input/…
$endgroup$
– Hugh
Sep 30 '16 at 15:14