Mixing
Fourier expansion, mistake?
$begingroup$
Give the Fourier sinus series of $x(pi -x ) quad (0<x<pi)$.
I started calculating the coefficients $b_n$:
$b_n = frac2pi int_0^pi x(pi-x)sin(2nx) dx = frac2pi int_0^pi (xpi-x^2)sin(2nx) dx = frac2pi left( frac{-1}{2n}[(xpi-x^2)cos(2nx)]_0^{pi}
+ frac{1}{2n}int_0^pi (-2x+pi)cos(2nx)dxright) $ $= frac1{pi n}int_0^pi(-2x+pi)cos(2nx)dx = frac{1}{pi n} left( frac{1}{2n}[(-2x+pi)sin(2nx)]_0^pi+ frac{1}{n}int_0^pi sin(2nx)dxright)$ $= frac{-1}{2pi n^3}[cos(2nx)]_0^pi = 0.$
How come I get $0$? Does anybody see my mistake?
I tried the same for $a_n$ and $a_0$. I found $a_n = 0$ and $a_0 = frac{pi^2}{3}$. So assuming these values to be true, would imply that the series is just the constant $frac{pi^2}{6}$. What did I do wrong?
real-analysis
asked Jan 13 at 12:17
ZacharyZachary
1559
$endgroup$
|
show 2 more comments
$begingroup$
Give the Fourier sinus series of $x(pi -x ) quad (0<x<pi)$.
I started calculating the coefficients $b_n$:
$b_n = frac2pi int_0^pi x(pi-x)sin(2nx) dx = frac2pi int_0^pi (xpi-x^2)sin(2nx) dx = frac2pi left( frac{-1}{2n}[(xpi-x^2)cos(2nx)]_0^{pi}
+ frac{1}{2n}int_0^pi (-2x+pi)cos(2nx)dxright) $ $= frac1{pi n}int_0^pi(-2x+pi)cos(2nx)dx = frac{1}{pi n} left( frac{1}{2n}[(-2x+pi)sin(2nx)]_0^pi+ frac{1}{n}int_0^pi sin(2nx)dxright)$ $= frac{-1}{2pi n^3}[cos(2nx)]_0^pi = 0.$
How come I get $0$? Does anybody see my mistake?
I tried the same for $a_n$ and $a_0$. I found $a_n = 0$ and $a_0 = frac{pi^2}{3}$. So assuming these values to be true, would imply that the series is just the constant $frac{pi^2}{6}$. What did I do wrong?
real-analysis
asked Jan 13 at 12:17
ZacharyZachary
1559
$endgroup$
$begingroup$
Why do you use $sin 2nx$ instead of $sin nx$? It does not form an orthonormal basis. Note that we can extend the given function uniquely to an odd function on $(-pi,pi)$.
$endgroup$
– Song
Jan 13 at 13:12
$begingroup$
Isn't the formula $b_n = frac{1}{L} int_{-L}^L f(x) sin(frac{npi x}{L})dx$? In the example $L$ equals $frac{pi}{2}$.
$endgroup$
– Zachary
Jan 13 at 13:15
$begingroup$
Umm ... you are missing that we need cosine terms when $L=frac{pi}{2}$. As I said, extend $f$ to $(-pi,pi)$ and regard as if $L=pi$.
$endgroup$
– Song
Jan 13 at 13:16
$begingroup$
$lim_{xto 0+} x(pi-x) = 0$. So when extending the function, can I pick $F(x) = 0$ for $-pi < x < 0$?
$endgroup$
– Zachary
Jan 13 at 13:19
1
$begingroup$
I'll give you more details: Our goal is to write $f=sum_n b_n sin nx$ for $0<x<pi$. Since we exclude cosine series, the RHS is an odd function. Hence if we can write $f$ as a sine seires, then it should also be true that $tilde{f}(x) = sum_n b_n sin nx$ for $|x|<pi$ where $tilde{f}$ is an odd extension of $f$. Now, $b_n =1/pi int_{-pi}^pi tilde{f}(x)sin nxdx = 2/pi int_{0}^pi f(x)sin nxdx.$
$endgroup$
– Song
Jan 13 at 13:28
|
show 2 more comments
$begingroup$
Give the Fourier sinus series of $x(pi -x ) quad (0<x<pi)$.
I started calculating the coefficients $b_n$:
$b_n = frac2pi int_0^pi x(pi-x)sin(2nx) dx = frac2pi int_0^pi (xpi-x^2)sin(2nx) dx = frac2pi left( frac{-1}{2n}[(xpi-x^2)cos(2nx)]_0^{pi}
+ frac{1}{2n}int_0^pi (-2x+pi)cos(2nx)dxright) $ $= frac1{pi n}int_0^pi(-2x+pi)cos(2nx)dx = frac{1}{pi n} left( frac{1}{2n}[(-2x+pi)sin(2nx)]_0^pi+ frac{1}{n}int_0^pi sin(2nx)dxright)$ $= frac{-1}{2pi n^3}[cos(2nx)]_0^pi = 0.$
How come I get $0$? Does anybody see my mistake?
I tried the same for $a_n$ and $a_0$. I found $a_n = 0$ and $a_0 = frac{pi^2}{3}$. So assuming these values to be true, would imply that the series is just the constant $frac{pi^2}{6}$. What did I do wrong?
real-analysis
asked Jan 13 at 12:17
ZacharyZachary
1559
$endgroup$
Give the Fourier sinus series of $x(pi -x ) quad (0<x<pi)$.
I started calculating the coefficients $b_n$:
$b_n = frac2pi int_0^pi x(pi-x)sin(2nx) dx = frac2pi int_0^pi (xpi-x^2)sin(2nx) dx = frac2pi left( frac{-1}{2n}[(xpi-x^2)cos(2nx)]_0^{pi}
+ frac{1}{2n}int_0^pi (-2x+pi)cos(2nx)dxright) $ $= frac1{pi n}int_0^pi(-2x+pi)cos(2nx)dx = frac{1}{pi n} left( frac{1}{2n}[(-2x+pi)sin(2nx)]_0^pi+ frac{1}{n}int_0^pi sin(2nx)dxright)$ $= frac{-1}{2pi n^3}[cos(2nx)]_0^pi = 0.$
How come I get $0$? Does anybody see my mistake?
I tried the same for $a_n$ and $a_0$. I found $a_n = 0$ and $a_0 = frac{pi^2}{3}$. So assuming these values to be true, would imply that the series is just the constant $frac{pi^2}{6}$. What did I do wrong?
real-analysis
real-analysis
asked Jan 13 at 12:17
ZacharyZachary
1559
asked Jan 13 at 12:17
ZacharyZachary
1559
edited Jan 13 at 12:35
Zachary
asked Jan 13 at 12:17
ZacharyZachary
1559
asked Jan 13 at 12:17
ZacharyZachary
1559
asked Jan 13 at 12:17
ZacharyZachary
1559
1559
$begingroup$
Why do you use $sin 2nx$ instead of $sin nx$? It does not form an orthonormal basis. Note that we can extend the given function uniquely to an odd function on $(-pi,pi)$.
$endgroup$
– Song
Jan 13 at 13:12
$begingroup$
Isn't the formula $b_n = frac{1}{L} int_{-L}^L f(x) sin(frac{npi x}{L})dx$? In the example $L$ equals $frac{pi}{2}$.
$endgroup$
– Zachary
Jan 13 at 13:15
$begingroup$
Umm ... you are missing that we need cosine terms when $L=frac{pi}{2}$. As I said, extend $f$ to $(-pi,pi)$ and regard as if $L=pi$.
$endgroup$
– Song
Jan 13 at 13:16
$begingroup$
$lim_{xto 0+} x(pi-x) = 0$. So when extending the function, can I pick $F(x) = 0$ for $-pi < x < 0$?
$endgroup$
– Zachary
Jan 13 at 13:19
1
$begingroup$
I'll give you more details: Our goal is to write $f=sum_n b_n sin nx$ for $0<x<pi$. Since we exclude cosine series, the RHS is an odd function. Hence if we can write $f$ as a sine seires, then it should also be true that $tilde{f}(x) = sum_n b_n sin nx$ for $|x|<pi$ where $tilde{f}$ is an odd extension of $f$. Now, $b_n =1/pi int_{-pi}^pi tilde{f}(x)sin nxdx = 2/pi int_{0}^pi f(x)sin nxdx.$
$endgroup$
– Song
Jan 13 at 13:28
|
show 2 more comments
$begingroup$
Why do you use $sin 2nx$ instead of $sin nx$? It does not form an orthonormal basis. Note that we can extend the given function uniquely to an odd function on $(-pi,pi)$.
$endgroup$
– Song
Jan 13 at 13:12
$begingroup$
Isn't the formula $b_n = frac{1}{L} int_{-L}^L f(x) sin(frac{npi x}{L})dx$? In the example $L$ equals $frac{pi}{2}$.
$endgroup$
– Zachary
Jan 13 at 13:15
$begingroup$
Umm ... you are missing that we need cosine terms when $L=frac{pi}{2}$. As I said, extend $f$ to $(-pi,pi)$ and regard as if $L=pi$.
$endgroup$
– Song
Jan 13 at 13:16
$begingroup$
$lim_{xto 0+} x(pi-x) = 0$. So when extending the function, can I pick $F(x) = 0$ for $-pi < x < 0$?
$endgroup$
– Zachary
Jan 13 at 13:19
1
$begingroup$
I'll give you more details: Our goal is to write $f=sum_n b_n sin nx$ for $0<x<pi$. Since we exclude cosine series, the RHS is an odd function. Hence if we can write $f$ as a sine seires, then it should also be true that $tilde{f}(x) = sum_n b_n sin nx$ for $|x|<pi$ where $tilde{f}$ is an odd extension of $f$. Now, $b_n =1/pi int_{-pi}^pi tilde{f}(x)sin nxdx = 2/pi int_{0}^pi f(x)sin nxdx.$
$endgroup$
– Song
Jan 13 at 13:28
$begingroup$
Why do you use $sin 2nx$ instead of $sin nx$? It does not form an orthonormal basis. Note that we can extend the given function uniquely to an odd function on $(-pi,pi)$.
$endgroup$
– Song
Jan 13 at 13:12
$begingroup$
Why do you use $sin 2nx$ instead of $sin nx$? It does not form an orthonormal basis. Note that we can extend the given function uniquely to an odd function on $(-pi,pi)$.
$endgroup$
– Song
Jan 13 at 13:12
$begingroup$
Isn't the formula $b_n = frac{1}{L} int_{-L}^L f(x) sin(frac{npi x}{L})dx$? In the example $L$ equals $frac{pi}{2}$.
$endgroup$
– Zachary
Jan 13 at 13:15
$begingroup$
Isn't the formula $b_n = frac{1}{L} int_{-L}^L f(x) sin(frac{npi x}{L})dx$? In the example $L$ equals $frac{pi}{2}$.
$endgroup$
– Zachary
Jan 13 at 13:15
$begingroup$
Umm ... you are missing that we need cosine terms when $L=frac{pi}{2}$. As I said, extend $f$ to $(-pi,pi)$ and regard as if $L=pi$.
$endgroup$
– Song
Jan 13 at 13:16
$begingroup$
Umm ... you are missing that we need cosine terms when $L=frac{pi}{2}$. As I said, extend $f$ to $(-pi,pi)$ and regard as if $L=pi$.
$endgroup$
– Song
Jan 13 at 13:16
$begingroup$
$lim_{xto 0+} x(pi-x) = 0$. So when extending the function, can I pick $F(x) = 0$ for $-pi < x < 0$?
$endgroup$
– Zachary
Jan 13 at 13:19
$begingroup$
$lim_{xto 0+} x(pi-x) = 0$. So when extending the function, can I pick $F(x) = 0$ for $-pi < x < 0$?
$endgroup$
– Zachary
Jan 13 at 13:19
1
1
$begingroup$
I'll give you more details: Our goal is to write $f=sum_n b_n sin nx$ for $0<x<pi$. Since we exclude cosine series, the RHS is an odd function. Hence if we can write $f$ as a sine seires, then it should also be true that $tilde{f}(x) = sum_n b_n sin nx$ for $|x|<pi$ where $tilde{f}$ is an odd extension of $f$. Now, $b_n =1/pi int_{-pi}^pi tilde{f}(x)sin nxdx = 2/pi int_{0}^pi f(x)sin nxdx.$
$endgroup$
– Song
Jan 13 at 13:28
$begingroup$
I'll give you more details: Our goal is to write $f=sum_n b_n sin nx$ for $0<x<pi$. Since we exclude cosine series, the RHS is an odd function. Hence if we can write $f$ as a sine seires, then it should also be true that $tilde{f}(x) = sum_n b_n sin nx$ for $|x|<pi$ where $tilde{f}$ is an odd extension of $f$. Now, $b_n =1/pi int_{-pi}^pi tilde{f}(x)sin nxdx = 2/pi int_{0}^pi f(x)sin nxdx.$
$endgroup$
– Song
Jan 13 at 13:28
|
show 2 more comments
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$begingroup$
Why do you use $sin 2nx$ instead of $sin nx$? It does not form an orthonormal basis. Note that we can extend the given function uniquely to an odd function on $(-pi,pi)$.
$endgroup$
– Song
Jan 13 at 13:12
$begingroup$
Isn't the formula $b_n = frac{1}{L} int_{-L}^L f(x) sin(frac{npi x}{L})dx$? In the example $L$ equals $frac{pi}{2}$.
$endgroup$
– Zachary
Jan 13 at 13:15
$begingroup$
Umm ... you are missing that we need cosine terms when $L=frac{pi}{2}$. As I said, extend $f$ to $(-pi,pi)$ and regard as if $L=pi$.
$endgroup$
– Song
Jan 13 at 13:16
$begingroup$
$lim_{xto 0+} x(pi-x) = 0$. So when extending the function, can I pick $F(x) = 0$ for $-pi < x < 0$?
$endgroup$
– Zachary
Jan 13 at 13:19
1
$begingroup$
I'll give you more details: Our goal is to write $f=sum_n b_n sin nx$ for $0<x<pi$. Since we exclude cosine series, the RHS is an odd function. Hence if we can write $f$ as a sine seires, then it should also be true that $tilde{f}(x) = sum_n b_n sin nx$ for $|x|<pi$ where $tilde{f}$ is an odd extension of $f$. Now, $b_n =1/pi int_{-pi}^pi tilde{f}(x)sin nxdx = 2/pi int_{0}^pi f(x)sin nxdx.$
$endgroup$
– Song
Jan 13 at 13:28