Mixing
Find the finite Fourier sine transform of: $F(x) = x^2$ where $0<x<l$.
$begingroup$
The question asks:
Find the finite Fourier sine transform of: $F(x) = x^2$ where $0<x<l$.
My tutors solution is simply:
$f_s(n) = int_{0}^{l}x^2 sin(frac{pi n x}{l})dx = frac{2pi l^3}{n^3 pi^3}(cos(npi-1)-frac{l^3}{npi}cos(npi)$.
Is there some trick that I'm missing? How the hell did he go from that integral to that so quickly - there's so many rules with Fourier transforms, so thats why I ask. So far, I think he just skipped a bunch of steps, which is annoying given I get stuck:
My workings:
We have:
$f_s(n) = int_{0}^{l}x^2 sin(frac{pi n x}{l})dx$
let $u = x^2, v = sin(frac{pi n x}{l})$ and use integration by parts:
$x^2sin(frac{pi n x}{l}) - int v dv$, where $frac{dv}{du} = cos(frac{pi n - sqrt{u}}{l}) + frac{sqrt{u}}{2}$, and then I get stuck. Any help would be appreciated.
fourier-analysis fourier-transform
asked Jan 19 at 2:57
Dr.DoofusDr.Doofus
12412
$endgroup$
add a comment |
$begingroup$
The question asks:
Find the finite Fourier sine transform of: $F(x) = x^2$ where $0<x<l$.
My tutors solution is simply:
$f_s(n) = int_{0}^{l}x^2 sin(frac{pi n x}{l})dx = frac{2pi l^3}{n^3 pi^3}(cos(npi-1)-frac{l^3}{npi}cos(npi)$.
Is there some trick that I'm missing? How the hell did he go from that integral to that so quickly - there's so many rules with Fourier transforms, so thats why I ask. So far, I think he just skipped a bunch of steps, which is annoying given I get stuck:
My workings:
We have:
$f_s(n) = int_{0}^{l}x^2 sin(frac{pi n x}{l})dx$
let $u = x^2, v = sin(frac{pi n x}{l})$ and use integration by parts:
$x^2sin(frac{pi n x}{l}) - int v dv$, where $frac{dv}{du} = cos(frac{pi n - sqrt{u}}{l}) + frac{sqrt{u}}{2}$, and then I get stuck. Any help would be appreciated.
fourier-analysis fourier-transform
asked Jan 19 at 2:57
Dr.DoofusDr.Doofus
12412
$endgroup$
add a comment |
$begingroup$
The question asks:
Find the finite Fourier sine transform of: $F(x) = x^2$ where $0<x<l$.
My tutors solution is simply:
$f_s(n) = int_{0}^{l}x^2 sin(frac{pi n x}{l})dx = frac{2pi l^3}{n^3 pi^3}(cos(npi-1)-frac{l^3}{npi}cos(npi)$.
Is there some trick that I'm missing? How the hell did he go from that integral to that so quickly - there's so many rules with Fourier transforms, so thats why I ask. So far, I think he just skipped a bunch of steps, which is annoying given I get stuck:
My workings:
We have:
$f_s(n) = int_{0}^{l}x^2 sin(frac{pi n x}{l})dx$
let $u = x^2, v = sin(frac{pi n x}{l})$ and use integration by parts:
$x^2sin(frac{pi n x}{l}) - int v dv$, where $frac{dv}{du} = cos(frac{pi n - sqrt{u}}{l}) + frac{sqrt{u}}{2}$, and then I get stuck. Any help would be appreciated.
fourier-analysis fourier-transform
asked Jan 19 at 2:57
Dr.DoofusDr.Doofus
12412
$endgroup$
The question asks:
Find the finite Fourier sine transform of: $F(x) = x^2$ where $0<x<l$.
My tutors solution is simply:
$f_s(n) = int_{0}^{l}x^2 sin(frac{pi n x}{l})dx = frac{2pi l^3}{n^3 pi^3}(cos(npi-1)-frac{l^3}{npi}cos(npi)$.
Is there some trick that I'm missing? How the hell did he go from that integral to that so quickly - there's so many rules with Fourier transforms, so thats why I ask. So far, I think he just skipped a bunch of steps, which is annoying given I get stuck:
My workings:
We have:
$f_s(n) = int_{0}^{l}x^2 sin(frac{pi n x}{l})dx$
let $u = x^2, v = sin(frac{pi n x}{l})$ and use integration by parts:
$x^2sin(frac{pi n x}{l}) - int v dv$, where $frac{dv}{du} = cos(frac{pi n - sqrt{u}}{l}) + frac{sqrt{u}}{2}$, and then I get stuck. Any help would be appreciated.
fourier-analysis fourier-transform
fourier-analysis fourier-transform
asked Jan 19 at 2:57
Dr.DoofusDr.Doofus
12412
asked Jan 19 at 2:57
Dr.DoofusDr.Doofus
12412
asked Jan 19 at 2:57
Dr.DoofusDr.Doofus
12412
asked Jan 19 at 2:57
Dr.DoofusDr.Doofus
12412
asked Jan 19 at 2:57
Dr.DoofusDr.Doofus
12412
12412
add a comment |
add a comment |
1 Answer
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$begingroup$
For this question, we use Integration by parts twice, begin{matrix}x^2&sindfrac{pi nx}{l}\2x&-dfrac l{pi n}cosdfrac{pi nx}{l}\2&-dfrac{l^2}{pi^2n^2}sindfrac{pi nx}{l}end{matrix}
Therefore the integral is then $$-dfrac{x^2l}{pi n}cosdfrac{pi nx}{l}+dfrac{2xl^2}{pi^2n^2}sindfrac{pi nx}{l}bigg|_0^l-dfrac{2l^2}{pi^2n^2}int_0^l sindfrac{pi nx}{l}dx$$
The rest should be easy for you.
answered Jan 19 at 3:15
kelvin hong 方kelvin hong 方
66718
$endgroup$
$begingroup$
Sorry, I don't understand the first 3 lines.
$endgroup$
– Dr.Doofus
Jan 19 at 3:38
$begingroup$
The left column stands for derivatives, and the right column stands for integration, this method applies to the problems which commonly needs several times of IBP. Check this link for more information: mechanicalpeacademy.com/… At the 1/3 of the webpage.
$endgroup$
– kelvin hong 方
Jan 19 at 3:41
1
$begingroup$
Thank you. Very smart, Kelvin!
$endgroup$
– Dr.Doofus
Jan 19 at 3:44
add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
For this question, we use Integration by parts twice, begin{matrix}x^2&sindfrac{pi nx}{l}\2x&-dfrac l{pi n}cosdfrac{pi nx}{l}\2&-dfrac{l^2}{pi^2n^2}sindfrac{pi nx}{l}end{matrix}
Therefore the integral is then $$-dfrac{x^2l}{pi n}cosdfrac{pi nx}{l}+dfrac{2xl^2}{pi^2n^2}sindfrac{pi nx}{l}bigg|_0^l-dfrac{2l^2}{pi^2n^2}int_0^l sindfrac{pi nx}{l}dx$$
The rest should be easy for you.
answered Jan 19 at 3:15
kelvin hong 方kelvin hong 方
66718
$endgroup$
$begingroup$
Sorry, I don't understand the first 3 lines.
$endgroup$
– Dr.Doofus
Jan 19 at 3:38
$begingroup$
The left column stands for derivatives, and the right column stands for integration, this method applies to the problems which commonly needs several times of IBP. Check this link for more information: mechanicalpeacademy.com/… At the 1/3 of the webpage.
$endgroup$
– kelvin hong 方
Jan 19 at 3:41
1
$begingroup$
Thank you. Very smart, Kelvin!
$endgroup$
– Dr.Doofus
Jan 19 at 3:44
add a comment |
$begingroup$
For this question, we use Integration by parts twice, begin{matrix}x^2&sindfrac{pi nx}{l}\2x&-dfrac l{pi n}cosdfrac{pi nx}{l}\2&-dfrac{l^2}{pi^2n^2}sindfrac{pi nx}{l}end{matrix}
Therefore the integral is then $$-dfrac{x^2l}{pi n}cosdfrac{pi nx}{l}+dfrac{2xl^2}{pi^2n^2}sindfrac{pi nx}{l}bigg|_0^l-dfrac{2l^2}{pi^2n^2}int_0^l sindfrac{pi nx}{l}dx$$
The rest should be easy for you.
answered Jan 19 at 3:15
kelvin hong 方kelvin hong 方
66718
$endgroup$
$begingroup$
Sorry, I don't understand the first 3 lines.
$endgroup$
– Dr.Doofus
Jan 19 at 3:38
$begingroup$
The left column stands for derivatives, and the right column stands for integration, this method applies to the problems which commonly needs several times of IBP. Check this link for more information: mechanicalpeacademy.com/… At the 1/3 of the webpage.
$endgroup$
– kelvin hong 方
Jan 19 at 3:41
1
$begingroup$
Thank you. Very smart, Kelvin!
$endgroup$
– Dr.Doofus
Jan 19 at 3:44
add a comment |
$begingroup$
For this question, we use Integration by parts twice, begin{matrix}x^2&sindfrac{pi nx}{l}\2x&-dfrac l{pi n}cosdfrac{pi nx}{l}\2&-dfrac{l^2}{pi^2n^2}sindfrac{pi nx}{l}end{matrix}
Therefore the integral is then $$-dfrac{x^2l}{pi n}cosdfrac{pi nx}{l}+dfrac{2xl^2}{pi^2n^2}sindfrac{pi nx}{l}bigg|_0^l-dfrac{2l^2}{pi^2n^2}int_0^l sindfrac{pi nx}{l}dx$$
The rest should be easy for you.
answered Jan 19 at 3:15
kelvin hong 方kelvin hong 方
66718
$endgroup$
For this question, we use Integration by parts twice, begin{matrix}x^2&sindfrac{pi nx}{l}\2x&-dfrac l{pi n}cosdfrac{pi nx}{l}\2&-dfrac{l^2}{pi^2n^2}sindfrac{pi nx}{l}end{matrix}
Therefore the integral is then $$-dfrac{x^2l}{pi n}cosdfrac{pi nx}{l}+dfrac{2xl^2}{pi^2n^2}sindfrac{pi nx}{l}bigg|_0^l-dfrac{2l^2}{pi^2n^2}int_0^l sindfrac{pi nx}{l}dx$$
The rest should be easy for you.
answered Jan 19 at 3:15
kelvin hong 方kelvin hong 方
66718
answered Jan 19 at 3:15
kelvin hong 方kelvin hong 方
66718
answered Jan 19 at 3:15
kelvin hong 方kelvin hong 方
66718
answered Jan 19 at 3:15
kelvin hong 方kelvin hong 方
66718
66718
$begingroup$
Sorry, I don't understand the first 3 lines.
$endgroup$
– Dr.Doofus
Jan 19 at 3:38
$begingroup$
The left column stands for derivatives, and the right column stands for integration, this method applies to the problems which commonly needs several times of IBP. Check this link for more information: mechanicalpeacademy.com/… At the 1/3 of the webpage.
$endgroup$
– kelvin hong 方
Jan 19 at 3:41
1
$begingroup$
Thank you. Very smart, Kelvin!
$endgroup$
– Dr.Doofus
Jan 19 at 3:44
add a comment |
$begingroup$
Sorry, I don't understand the first 3 lines.
$endgroup$
– Dr.Doofus
Jan 19 at 3:38
$begingroup$
The left column stands for derivatives, and the right column stands for integration, this method applies to the problems which commonly needs several times of IBP. Check this link for more information: mechanicalpeacademy.com/… At the 1/3 of the webpage.
$endgroup$
– kelvin hong 方
Jan 19 at 3:41
1
$begingroup$
Thank you. Very smart, Kelvin!
$endgroup$
– Dr.Doofus
Jan 19 at 3:44
$begingroup$
Sorry, I don't understand the first 3 lines.
$endgroup$
– Dr.Doofus
Jan 19 at 3:38
$begingroup$
Sorry, I don't understand the first 3 lines.
$endgroup$
– Dr.Doofus
Jan 19 at 3:38
$begingroup$
The left column stands for derivatives, and the right column stands for integration, this method applies to the problems which commonly needs several times of IBP. Check this link for more information: mechanicalpeacademy.com/… At the 1/3 of the webpage.
$endgroup$
– kelvin hong 方
Jan 19 at 3:41
$begingroup$
The left column stands for derivatives, and the right column stands for integration, this method applies to the problems which commonly needs several times of IBP. Check this link for more information: mechanicalpeacademy.com/… At the 1/3 of the webpage.
$endgroup$
– kelvin hong 方
Jan 19 at 3:41
1
1
$begingroup$
Thank you. Very smart, Kelvin!
$endgroup$
– Dr.Doofus
Jan 19 at 3:44
$begingroup$
Thank you. Very smart, Kelvin!
$endgroup$
– Dr.Doofus
Jan 19 at 3:44
add a comment |
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