Fourier series of $f(x)=pi-x$












1












$begingroup$


$$f(x)=pi-x qquad x in [0,2 pi[$$


$$a_0=frac{1}{pi} int_0^{2 pi}f(x) dx=frac{1}{pi} int_0^{2 pi} (pi-x) dx=0$$






$$a_n=frac{1}{pi} int_0^{2 pi} cos(nx) dx=frac{1}{pi} int_0^{2 pi}(pi cos(nx)-x cos(nx)) dx=$$
$$frac{1}{pi} Big( Big[frac{pi}{n} sin(nx) Big]_0^{2 pi}-Big[frac{x}{n} sin(nx)+frac{1}{n^2} cos(nx) Big]_0^{2 pi} Big)=0 $$





$$b_n=frac{1}{pi} Big( Big[ -frac{pi}{n} cos(nx) Big]_0^{2 pi}-Big[-frac{x}{n} cos(nx)+frac{1}{n^2} sin(nx) Big]_0^{2 pi} Big)=frac{2}{n}$$



$$f(x)=2 sum_{n=1}^{+infty} frac{1}{n} sin(nx)$$



Is it correct?






fourier-series







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$endgroup$








  • 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


















1












$begingroup$


$$f(x)=pi-x qquad x in [0,2 pi[$$


$$a_0=frac{1}{pi} int_0^{2 pi}f(x) dx=frac{1}{pi} int_0^{2 pi} (pi-x) dx=0$$






$$a_n=frac{1}{pi} int_0^{2 pi} cos(nx) dx=frac{1}{pi} int_0^{2 pi}(pi cos(nx)-x cos(nx)) dx=$$
$$frac{1}{pi} Big( Big[frac{pi}{n} sin(nx) Big]_0^{2 pi}-Big[frac{x}{n} sin(nx)+frac{1}{n^2} cos(nx) Big]_0^{2 pi} Big)=0 $$





$$b_n=frac{1}{pi} Big( Big[ -frac{pi}{n} cos(nx) Big]_0^{2 pi}-Big[-frac{x}{n} cos(nx)+frac{1}{n^2} sin(nx) Big]_0^{2 pi} Big)=frac{2}{n}$$



$$f(x)=2 sum_{n=1}^{+infty} frac{1}{n} sin(nx)$$



Is it correct?






fourier-series







share|cite|improve this question









$endgroup$








  • 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
















1












1








1





$begingroup$


$$f(x)=pi-x qquad x in [0,2 pi[$$


$$a_0=frac{1}{pi} int_0^{2 pi}f(x) dx=frac{1}{pi} int_0^{2 pi} (pi-x) dx=0$$






$$a_n=frac{1}{pi} int_0^{2 pi} cos(nx) dx=frac{1}{pi} int_0^{2 pi}(pi cos(nx)-x cos(nx)) dx=$$
$$frac{1}{pi} Big( Big[frac{pi}{n} sin(nx) Big]_0^{2 pi}-Big[frac{x}{n} sin(nx)+frac{1}{n^2} cos(nx) Big]_0^{2 pi} Big)=0 $$





$$b_n=frac{1}{pi} Big( Big[ -frac{pi}{n} cos(nx) Big]_0^{2 pi}-Big[-frac{x}{n} cos(nx)+frac{1}{n^2} sin(nx) Big]_0^{2 pi} Big)=frac{2}{n}$$



$$f(x)=2 sum_{n=1}^{+infty} frac{1}{n} sin(nx)$$



Is it correct?






fourier-series







share|cite|improve this question









$endgroup$




$$f(x)=pi-x qquad x in [0,2 pi[$$


$$a_0=frac{1}{pi} int_0^{2 pi}f(x) dx=frac{1}{pi} int_0^{2 pi} (pi-x) dx=0$$






$$a_n=frac{1}{pi} int_0^{2 pi} cos(nx) dx=frac{1}{pi} int_0^{2 pi}(pi cos(nx)-x cos(nx)) dx=$$
$$frac{1}{pi} Big( Big[frac{pi}{n} sin(nx) Big]_0^{2 pi}-Big[frac{x}{n} sin(nx)+frac{1}{n^2} cos(nx) Big]_0^{2 pi} Big)=0 $$





$$b_n=frac{1}{pi} Big( Big[ -frac{pi}{n} cos(nx) Big]_0^{2 pi}-Big[-frac{x}{n} cos(nx)+frac{1}{n^2} sin(nx) Big]_0^{2 pi} Big)=frac{2}{n}$$



$$f(x)=2 sum_{n=1}^{+infty} frac{1}{n} sin(nx)$$



Is it correct?






fourier-series




fourier-series






share|cite|improve this question













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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
















  • 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












1 Answer
1






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oldest

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1












$begingroup$

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:



amps



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:



1210100






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    Nice illustration of the Gibbs Phenomenon in this answer, too.
    $endgroup$
    – Dan Sheppard
    Dec 29 '18 at 23:06












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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

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:



amps



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:



1210100






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    Nice illustration of the Gibbs Phenomenon in this answer, too.
    $endgroup$
    – Dan Sheppard
    Dec 29 '18 at 23:06
















1












$begingroup$

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:



amps



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:



1210100






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    Nice illustration of the Gibbs Phenomenon in this answer, too.
    $endgroup$
    – Dan Sheppard
    Dec 29 '18 at 23:06














1












1








1





$begingroup$

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:



amps



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:



1210100






share|cite|improve this answer









$endgroup$



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:



amps



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:



1210100







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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














  • 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


















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