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Fourier sine series simplification
$begingroup$
I'm having trouble simplifying a Fourier sine expansion for the following function:
$$f(x) = max{{frac{pi}{2}, x}}$$ on the interval of $[0,pi]$. Since we're doing a sine series then $a_n = 0$ and the function collapses down to $f(x) = sum_{n=1}^{infty}b_nsin(nx)$, where
$$b_n = frac{2}{pi}int_{0}^{pi}f(x)sin(nx)dx = frac{2}{pi}bigg(int_{0}^{pi/2}frac{pi}{2}sin(nx) dx quad+ quadint_{pi/2}^{pi}xsin(nx)dx)bigg)$$ which I've managed to calculate (hopefully correctly) as:
$$b_n = frac{2}{pi}bigg(frac{pi}{2n} - frac{pi}{n}cdot(-1)^n - frac{1}{n^2}sinbig(frac{npi}{2}big) bigg)$$
Now I'm kinda stuck. I mean I could just put this into the final formula but I think there is a way to simplify it, just can't quite find it. The problematic part is obviously $sinbig(frac{npi}{2}big)$ since for $n = 2k$ it is equal to zero. But if $n$ is an odd number it's either $-1$ or $1$. I've tried to get rid of the even elements in the sum (I had hoped they would be zero), but:
$$b_{2n}= frac{2}{pi}bigg(frac{pi}{4n} - frac{pi}{2n}cdot(-1)^{2n}-frac{1}{(2n)^2}sinbig(npibig)bigg) = frac{2}{pi}bigg(frac{pi}{4n} - frac{pi}{2n}bigg) = frac{2}{pi}cdotbigg(-frac{pi}{4n}bigg) = -frac{1}{2n}$$
Looks pretty neat but still doesn't help me all that much.
Any ideas on how to simplify this?
Thanks.
real-analysis trigonometry fourier-series
asked Jan 20 at 12:45
KoyKoy
35116
$endgroup$
add a comment |
$begingroup$
I'm having trouble simplifying a Fourier sine expansion for the following function:
$$f(x) = max{{frac{pi}{2}, x}}$$ on the interval of $[0,pi]$. Since we're doing a sine series then $a_n = 0$ and the function collapses down to $f(x) = sum_{n=1}^{infty}b_nsin(nx)$, where
$$b_n = frac{2}{pi}int_{0}^{pi}f(x)sin(nx)dx = frac{2}{pi}bigg(int_{0}^{pi/2}frac{pi}{2}sin(nx) dx quad+ quadint_{pi/2}^{pi}xsin(nx)dx)bigg)$$ which I've managed to calculate (hopefully correctly) as:
$$b_n = frac{2}{pi}bigg(frac{pi}{2n} - frac{pi}{n}cdot(-1)^n - frac{1}{n^2}sinbig(frac{npi}{2}big) bigg)$$
Now I'm kinda stuck. I mean I could just put this into the final formula but I think there is a way to simplify it, just can't quite find it. The problematic part is obviously $sinbig(frac{npi}{2}big)$ since for $n = 2k$ it is equal to zero. But if $n$ is an odd number it's either $-1$ or $1$. I've tried to get rid of the even elements in the sum (I had hoped they would be zero), but:
$$b_{2n}= frac{2}{pi}bigg(frac{pi}{4n} - frac{pi}{2n}cdot(-1)^{2n}-frac{1}{(2n)^2}sinbig(npibig)bigg) = frac{2}{pi}bigg(frac{pi}{4n} - frac{pi}{2n}bigg) = frac{2}{pi}cdotbigg(-frac{pi}{4n}bigg) = -frac{1}{2n}$$
Looks pretty neat but still doesn't help me all that much.
Any ideas on how to simplify this?
Thanks.
real-analysis trigonometry fourier-series
asked Jan 20 at 12:45
KoyKoy
35116
$endgroup$
add a comment |
$begingroup$
I'm having trouble simplifying a Fourier sine expansion for the following function:
$$f(x) = max{{frac{pi}{2}, x}}$$ on the interval of $[0,pi]$. Since we're doing a sine series then $a_n = 0$ and the function collapses down to $f(x) = sum_{n=1}^{infty}b_nsin(nx)$, where
$$b_n = frac{2}{pi}int_{0}^{pi}f(x)sin(nx)dx = frac{2}{pi}bigg(int_{0}^{pi/2}frac{pi}{2}sin(nx) dx quad+ quadint_{pi/2}^{pi}xsin(nx)dx)bigg)$$ which I've managed to calculate (hopefully correctly) as:
$$b_n = frac{2}{pi}bigg(frac{pi}{2n} - frac{pi}{n}cdot(-1)^n - frac{1}{n^2}sinbig(frac{npi}{2}big) bigg)$$
Now I'm kinda stuck. I mean I could just put this into the final formula but I think there is a way to simplify it, just can't quite find it. The problematic part is obviously $sinbig(frac{npi}{2}big)$ since for $n = 2k$ it is equal to zero. But if $n$ is an odd number it's either $-1$ or $1$. I've tried to get rid of the even elements in the sum (I had hoped they would be zero), but:
$$b_{2n}= frac{2}{pi}bigg(frac{pi}{4n} - frac{pi}{2n}cdot(-1)^{2n}-frac{1}{(2n)^2}sinbig(npibig)bigg) = frac{2}{pi}bigg(frac{pi}{4n} - frac{pi}{2n}bigg) = frac{2}{pi}cdotbigg(-frac{pi}{4n}bigg) = -frac{1}{2n}$$
Looks pretty neat but still doesn't help me all that much.
Any ideas on how to simplify this?
Thanks.
real-analysis trigonometry fourier-series
asked Jan 20 at 12:45
KoyKoy
35116
$endgroup$
I'm having trouble simplifying a Fourier sine expansion for the following function:
$$f(x) = max{{frac{pi}{2}, x}}$$ on the interval of $[0,pi]$. Since we're doing a sine series then $a_n = 0$ and the function collapses down to $f(x) = sum_{n=1}^{infty}b_nsin(nx)$, where
$$b_n = frac{2}{pi}int_{0}^{pi}f(x)sin(nx)dx = frac{2}{pi}bigg(int_{0}^{pi/2}frac{pi}{2}sin(nx) dx quad+ quadint_{pi/2}^{pi}xsin(nx)dx)bigg)$$ which I've managed to calculate (hopefully correctly) as:
$$b_n = frac{2}{pi}bigg(frac{pi}{2n} - frac{pi}{n}cdot(-1)^n - frac{1}{n^2}sinbig(frac{npi}{2}big) bigg)$$
Now I'm kinda stuck. I mean I could just put this into the final formula but I think there is a way to simplify it, just can't quite find it. The problematic part is obviously $sinbig(frac{npi}{2}big)$ since for $n = 2k$ it is equal to zero. But if $n$ is an odd number it's either $-1$ or $1$. I've tried to get rid of the even elements in the sum (I had hoped they would be zero), but:
$$b_{2n}= frac{2}{pi}bigg(frac{pi}{4n} - frac{pi}{2n}cdot(-1)^{2n}-frac{1}{(2n)^2}sinbig(npibig)bigg) = frac{2}{pi}bigg(frac{pi}{4n} - frac{pi}{2n}bigg) = frac{2}{pi}cdotbigg(-frac{pi}{4n}bigg) = -frac{1}{2n}$$
Looks pretty neat but still doesn't help me all that much.
Any ideas on how to simplify this?
Thanks.
real-analysis trigonometry fourier-series
real-analysis trigonometry fourier-series
asked Jan 20 at 12:45
KoyKoy
35116
asked Jan 20 at 12:45
KoyKoy
35116
asked Jan 20 at 12:45
KoyKoy
35116
asked Jan 20 at 12:45
KoyKoy
35116
asked Jan 20 at 12:45
KoyKoy
35116
35116
add a comment |
add a comment |
1 Answer
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$begingroup$
So you know that $sinleft(dfrac{npi}{2}right)$ is $0$ if $n$ is even, and $-1$ or $1$ if $n$ is odd. So why not separate cases even furthur: If$n=2k+1$, $k$ can be even or odd, so you have $n=4k+1$ or $n=4k+3$. So do the sum like this:
$$f(x)=sum_{n=1}^infty b_{2n}sin(2nx)+sum_{n=1}^infty b_{4n+1}sin((4n+1)x)+sum_{n=1}^infty b_{4n+3}sin((4n+3)x).$$
answered Jan 20 at 13:22
ScientificaScientifica
6,79641335
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$begingroup$
So you know that $sinleft(dfrac{npi}{2}right)$ is $0$ if $n$ is even, and $-1$ or $1$ if $n$ is odd. So why not separate cases even furthur: If$n=2k+1$, $k$ can be even or odd, so you have $n=4k+1$ or $n=4k+3$. So do the sum like this:
$$f(x)=sum_{n=1}^infty b_{2n}sin(2nx)+sum_{n=1}^infty b_{4n+1}sin((4n+1)x)+sum_{n=1}^infty b_{4n+3}sin((4n+3)x).$$
answered Jan 20 at 13:22
ScientificaScientifica
6,79641335
$endgroup$
add a comment |
$begingroup$
So you know that $sinleft(dfrac{npi}{2}right)$ is $0$ if $n$ is even, and $-1$ or $1$ if $n$ is odd. So why not separate cases even furthur: If$n=2k+1$, $k$ can be even or odd, so you have $n=4k+1$ or $n=4k+3$. So do the sum like this:
$$f(x)=sum_{n=1}^infty b_{2n}sin(2nx)+sum_{n=1}^infty b_{4n+1}sin((4n+1)x)+sum_{n=1}^infty b_{4n+3}sin((4n+3)x).$$
answered Jan 20 at 13:22
ScientificaScientifica
6,79641335
$endgroup$
add a comment |
$begingroup$
So you know that $sinleft(dfrac{npi}{2}right)$ is $0$ if $n$ is even, and $-1$ or $1$ if $n$ is odd. So why not separate cases even furthur: If$n=2k+1$, $k$ can be even or odd, so you have $n=4k+1$ or $n=4k+3$. So do the sum like this:
$$f(x)=sum_{n=1}^infty b_{2n}sin(2nx)+sum_{n=1}^infty b_{4n+1}sin((4n+1)x)+sum_{n=1}^infty b_{4n+3}sin((4n+3)x).$$
answered Jan 20 at 13:22
ScientificaScientifica
6,79641335
$endgroup$
So you know that $sinleft(dfrac{npi}{2}right)$ is $0$ if $n$ is even, and $-1$ or $1$ if $n$ is odd. So why not separate cases even furthur: If$n=2k+1$, $k$ can be even or odd, so you have $n=4k+1$ or $n=4k+3$. So do the sum like this:
$$f(x)=sum_{n=1}^infty b_{2n}sin(2nx)+sum_{n=1}^infty b_{4n+1}sin((4n+1)x)+sum_{n=1}^infty b_{4n+3}sin((4n+3)x).$$
answered Jan 20 at 13:22
ScientificaScientifica
6,79641335
answered Jan 20 at 13:22
ScientificaScientifica
6,79641335
answered Jan 20 at 13:22
ScientificaScientifica
6,79641335
answered Jan 20 at 13:22
ScientificaScientifica
6,79641335
6,79641335
add a comment |
add a comment |
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