Mixing
How to find this series coefficients?
$begingroup$
I want to find the following series coefficients:
begin{equation}sum_2^infty B_{n}[sin(nx)-ncos(nx)]=sin(x)+frac{x^2}{pi}-xend{equation}
Interval is $[0,pi]$.
I tried:
begin{equation}B_{n}=frac{2}{pi}int_{0}^{pi} left(sin(nx)-ncos(nx)right)left(sin(x)+frac{x^2}{pi}-xright)dxend{equation}
So I found:
begin{equation}B_{n}=frac{2}{pi^2}left[frac{2+n^2(-2+pi)+(n^2(2+pi)-2)(-1)^n}{n^5-n^3}right]end{equation}
The problem is the denominator of $B_{n}$ which must be $n^7-n^3$ instead of
$n^5-n^3$.
The solution in my tutor's PDF is:
begin{equation}B_{n}=frac{2}{pi^2}left[frac{2+n^2(-2+pi)+(n^2(2+pi)-2)(-1)^n}{n^7-n^3}right]end{equation}
Where is the problem?
sequences-and-series
asked Jan 24 at 19:39
MyBBExpertMyBBExpert
32
$endgroup$
add a comment |
$begingroup$
I want to find the following series coefficients:
begin{equation}sum_2^infty B_{n}[sin(nx)-ncos(nx)]=sin(x)+frac{x^2}{pi}-xend{equation}
Interval is $[0,pi]$.
I tried:
begin{equation}B_{n}=frac{2}{pi}int_{0}^{pi} left(sin(nx)-ncos(nx)right)left(sin(x)+frac{x^2}{pi}-xright)dxend{equation}
So I found:
begin{equation}B_{n}=frac{2}{pi^2}left[frac{2+n^2(-2+pi)+(n^2(2+pi)-2)(-1)^n}{n^5-n^3}right]end{equation}
The problem is the denominator of $B_{n}$ which must be $n^7-n^3$ instead of
$n^5-n^3$.
The solution in my tutor's PDF is:
begin{equation}B_{n}=frac{2}{pi^2}left[frac{2+n^2(-2+pi)+(n^2(2+pi)-2)(-1)^n}{n^7-n^3}right]end{equation}
Where is the problem?
sequences-and-series
asked Jan 24 at 19:39
MyBBExpertMyBBExpert
32
$endgroup$
$begingroup$
$B_n$ is defined by $B_n=frac{1}{2pi}int_{-pi}^pi (sin(x)+frac{x^2}{pi}-x)sin nxrm dx = frac{1}{2npi}int_{-pi}^pi (sin(x)+frac{x^2}{pi}-x)cos nxrm dx$.
$endgroup$
– Nicolas FRANCOIS
Jan 24 at 19:45
$begingroup$
That is, if the interval on which the equality holds is $[-pi,pi]$. I didn't verified. But your way of computing $B_n$ is strange.
$endgroup$
– Nicolas FRANCOIS
Jan 24 at 19:46
$begingroup$
@NicolasFRANCOIS The interval is $[0,pi]$
$endgroup$
– MyBBExpert
Jan 24 at 19:57
add a comment |
$begingroup$
I want to find the following series coefficients:
begin{equation}sum_2^infty B_{n}[sin(nx)-ncos(nx)]=sin(x)+frac{x^2}{pi}-xend{equation}
Interval is $[0,pi]$.
I tried:
begin{equation}B_{n}=frac{2}{pi}int_{0}^{pi} left(sin(nx)-ncos(nx)right)left(sin(x)+frac{x^2}{pi}-xright)dxend{equation}
So I found:
begin{equation}B_{n}=frac{2}{pi^2}left[frac{2+n^2(-2+pi)+(n^2(2+pi)-2)(-1)^n}{n^5-n^3}right]end{equation}
The problem is the denominator of $B_{n}$ which must be $n^7-n^3$ instead of
$n^5-n^3$.
The solution in my tutor's PDF is:
begin{equation}B_{n}=frac{2}{pi^2}left[frac{2+n^2(-2+pi)+(n^2(2+pi)-2)(-1)^n}{n^7-n^3}right]end{equation}
Where is the problem?
sequences-and-series
asked Jan 24 at 19:39
MyBBExpertMyBBExpert
32
$endgroup$
I want to find the following series coefficients:
begin{equation}sum_2^infty B_{n}[sin(nx)-ncos(nx)]=sin(x)+frac{x^2}{pi}-xend{equation}
Interval is $[0,pi]$.
I tried:
begin{equation}B_{n}=frac{2}{pi}int_{0}^{pi} left(sin(nx)-ncos(nx)right)left(sin(x)+frac{x^2}{pi}-xright)dxend{equation}
So I found:
begin{equation}B_{n}=frac{2}{pi^2}left[frac{2+n^2(-2+pi)+(n^2(2+pi)-2)(-1)^n}{n^5-n^3}right]end{equation}
The problem is the denominator of $B_{n}$ which must be $n^7-n^3$ instead of
$n^5-n^3$.
The solution in my tutor's PDF is:
begin{equation}B_{n}=frac{2}{pi^2}left[frac{2+n^2(-2+pi)+(n^2(2+pi)-2)(-1)^n}{n^7-n^3}right]end{equation}
Where is the problem?
sequences-and-series
sequences-and-series
asked Jan 24 at 19:39
MyBBExpertMyBBExpert
32
asked Jan 24 at 19:39
MyBBExpertMyBBExpert
32
edited Jan 24 at 19:58
MyBBExpert
asked Jan 24 at 19:39
MyBBExpertMyBBExpert
32
asked Jan 24 at 19:39
MyBBExpertMyBBExpert
32
asked Jan 24 at 19:39
MyBBExpertMyBBExpert
32
32
$begingroup$
$B_n$ is defined by $B_n=frac{1}{2pi}int_{-pi}^pi (sin(x)+frac{x^2}{pi}-x)sin nxrm dx = frac{1}{2npi}int_{-pi}^pi (sin(x)+frac{x^2}{pi}-x)cos nxrm dx$.
$endgroup$
– Nicolas FRANCOIS
Jan 24 at 19:45
$begingroup$
That is, if the interval on which the equality holds is $[-pi,pi]$. I didn't verified. But your way of computing $B_n$ is strange.
$endgroup$
– Nicolas FRANCOIS
Jan 24 at 19:46
$begingroup$
@NicolasFRANCOIS The interval is $[0,pi]$
$endgroup$
– MyBBExpert
Jan 24 at 19:57
add a comment |
$begingroup$
$B_n$ is defined by $B_n=frac{1}{2pi}int_{-pi}^pi (sin(x)+frac{x^2}{pi}-x)sin nxrm dx = frac{1}{2npi}int_{-pi}^pi (sin(x)+frac{x^2}{pi}-x)cos nxrm dx$.
$endgroup$
– Nicolas FRANCOIS
Jan 24 at 19:45
$begingroup$
That is, if the interval on which the equality holds is $[-pi,pi]$. I didn't verified. But your way of computing $B_n$ is strange.
$endgroup$
– Nicolas FRANCOIS
Jan 24 at 19:46
$begingroup$
@NicolasFRANCOIS The interval is $[0,pi]$
$endgroup$
– MyBBExpert
Jan 24 at 19:57
$begingroup$
$B_n$ is defined by $B_n=frac{1}{2pi}int_{-pi}^pi (sin(x)+frac{x^2}{pi}-x)sin nxrm dx = frac{1}{2npi}int_{-pi}^pi (sin(x)+frac{x^2}{pi}-x)cos nxrm dx$.
$endgroup$
– Nicolas FRANCOIS
Jan 24 at 19:45
$begingroup$
$B_n$ is defined by $B_n=frac{1}{2pi}int_{-pi}^pi (sin(x)+frac{x^2}{pi}-x)sin nxrm dx = frac{1}{2npi}int_{-pi}^pi (sin(x)+frac{x^2}{pi}-x)cos nxrm dx$.
$endgroup$
– Nicolas FRANCOIS
Jan 24 at 19:45
$begingroup$
That is, if the interval on which the equality holds is $[-pi,pi]$. I didn't verified. But your way of computing $B_n$ is strange.
$endgroup$
– Nicolas FRANCOIS
Jan 24 at 19:46
$begingroup$
That is, if the interval on which the equality holds is $[-pi,pi]$. I didn't verified. But your way of computing $B_n$ is strange.
$endgroup$
– Nicolas FRANCOIS
Jan 24 at 19:46
$begingroup$
@NicolasFRANCOIS The interval is $[0,pi]$
$endgroup$
– MyBBExpert
Jan 24 at 19:57
$begingroup$
@NicolasFRANCOIS The interval is $[0,pi]$
$endgroup$
– MyBBExpert
Jan 24 at 19:57
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Note that the basis functions $u_n = sin(nx) - n cos(nx)$ are orthogonal on $[0,pi]$, with $int_0^pi u_n^2(x); dx = pi (n^2+1)/2$. Thus the coefficients for a function $f(x)$ that has such a series should be
$$ b_n = frac{2}{pi(n^2+1)} int_0^pi u_n(x) f(x); dx $$
In this case, if $n ne 1$,
$$ eqalign{B_n &= frac{2}{n^2+1} int_0^pi (sin(nx)-ncos(nx))left(sin(x)+frac{x^2}{pi}-xright); dx cr
&= {frac {2,pi, left( -1 right) ^{n}{n}^{2}+2,pi,{n}^{2}+4,{n}^
{2} left( -1 right) ^{n}-4,{n}^{2}-4, left( -1 right) ^{n}+4}{{n
}^{3}pi^2, left( {n}^{4}-1 right) }}
} $$
while
$$B_1 = frac{pi^2-8}{2pi^2}$$
answered Jan 24 at 20:24
Robert IsraelRobert Israel
327k23216469
$endgroup$
$begingroup$
Thank you! What about $B_{0}$ is that $frac{1}{pi}int_{0}^{pi} (sin(x)+frac{x^2}{pi}-x)dx$ or $frac{2}{pi}int_{0}^{pi} (sin(x)+frac{x^2}{pi}-x)dx$?
$endgroup$
– MyBBExpert
Jan 24 at 21:35
$begingroup$
Is there a $B_0$? For $n=0$, $sin(nx) - n cos(nx) = 0$.
$endgroup$
– Robert Israel
Jan 25 at 1:52
$begingroup$
Hm: these $f_n$ are eigenfunctions for $f''= lambda f$ with boundary conditions $f(0) = -f'(0)$, $f(pi) = -f'(pi)$. Besides $f_n$ for positive integers $n$, the other eigenfunction is $exp(-x)$. So that's what will give you another term in the expansion..
$endgroup$
– Robert Israel
Jan 25 at 2:22
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Note that the basis functions $u_n = sin(nx) - n cos(nx)$ are orthogonal on $[0,pi]$, with $int_0^pi u_n^2(x); dx = pi (n^2+1)/2$. Thus the coefficients for a function $f(x)$ that has such a series should be
$$ b_n = frac{2}{pi(n^2+1)} int_0^pi u_n(x) f(x); dx $$
In this case, if $n ne 1$,
$$ eqalign{B_n &= frac{2}{n^2+1} int_0^pi (sin(nx)-ncos(nx))left(sin(x)+frac{x^2}{pi}-xright); dx cr
&= {frac {2,pi, left( -1 right) ^{n}{n}^{2}+2,pi,{n}^{2}+4,{n}^
{2} left( -1 right) ^{n}-4,{n}^{2}-4, left( -1 right) ^{n}+4}{{n
}^{3}pi^2, left( {n}^{4}-1 right) }}
} $$
while
$$B_1 = frac{pi^2-8}{2pi^2}$$
answered Jan 24 at 20:24
Robert IsraelRobert Israel
327k23216469
$endgroup$
$begingroup$
Thank you! What about $B_{0}$ is that $frac{1}{pi}int_{0}^{pi} (sin(x)+frac{x^2}{pi}-x)dx$ or $frac{2}{pi}int_{0}^{pi} (sin(x)+frac{x^2}{pi}-x)dx$?
$endgroup$
– MyBBExpert
Jan 24 at 21:35
$begingroup$
Is there a $B_0$? For $n=0$, $sin(nx) - n cos(nx) = 0$.
$endgroup$
– Robert Israel
Jan 25 at 1:52
$begingroup$
Hm: these $f_n$ are eigenfunctions for $f''= lambda f$ with boundary conditions $f(0) = -f'(0)$, $f(pi) = -f'(pi)$. Besides $f_n$ for positive integers $n$, the other eigenfunction is $exp(-x)$. So that's what will give you another term in the expansion..
$endgroup$
– Robert Israel
Jan 25 at 2:22
add a comment |
$begingroup$
Note that the basis functions $u_n = sin(nx) - n cos(nx)$ are orthogonal on $[0,pi]$, with $int_0^pi u_n^2(x); dx = pi (n^2+1)/2$. Thus the coefficients for a function $f(x)$ that has such a series should be
$$ b_n = frac{2}{pi(n^2+1)} int_0^pi u_n(x) f(x); dx $$
In this case, if $n ne 1$,
$$ eqalign{B_n &= frac{2}{n^2+1} int_0^pi (sin(nx)-ncos(nx))left(sin(x)+frac{x^2}{pi}-xright); dx cr
&= {frac {2,pi, left( -1 right) ^{n}{n}^{2}+2,pi,{n}^{2}+4,{n}^
{2} left( -1 right) ^{n}-4,{n}^{2}-4, left( -1 right) ^{n}+4}{{n
}^{3}pi^2, left( {n}^{4}-1 right) }}
} $$
while
$$B_1 = frac{pi^2-8}{2pi^2}$$
answered Jan 24 at 20:24
Robert IsraelRobert Israel
327k23216469
$endgroup$
$begingroup$
Thank you! What about $B_{0}$ is that $frac{1}{pi}int_{0}^{pi} (sin(x)+frac{x^2}{pi}-x)dx$ or $frac{2}{pi}int_{0}^{pi} (sin(x)+frac{x^2}{pi}-x)dx$?
$endgroup$
– MyBBExpert
Jan 24 at 21:35
$begingroup$
Is there a $B_0$? For $n=0$, $sin(nx) - n cos(nx) = 0$.
$endgroup$
– Robert Israel
Jan 25 at 1:52
$begingroup$
Hm: these $f_n$ are eigenfunctions for $f''= lambda f$ with boundary conditions $f(0) = -f'(0)$, $f(pi) = -f'(pi)$. Besides $f_n$ for positive integers $n$, the other eigenfunction is $exp(-x)$. So that's what will give you another term in the expansion..
$endgroup$
– Robert Israel
Jan 25 at 2:22
add a comment |
$begingroup$
Note that the basis functions $u_n = sin(nx) - n cos(nx)$ are orthogonal on $[0,pi]$, with $int_0^pi u_n^2(x); dx = pi (n^2+1)/2$. Thus the coefficients for a function $f(x)$ that has such a series should be
$$ b_n = frac{2}{pi(n^2+1)} int_0^pi u_n(x) f(x); dx $$
In this case, if $n ne 1$,
$$ eqalign{B_n &= frac{2}{n^2+1} int_0^pi (sin(nx)-ncos(nx))left(sin(x)+frac{x^2}{pi}-xright); dx cr
&= {frac {2,pi, left( -1 right) ^{n}{n}^{2}+2,pi,{n}^{2}+4,{n}^
{2} left( -1 right) ^{n}-4,{n}^{2}-4, left( -1 right) ^{n}+4}{{n
}^{3}pi^2, left( {n}^{4}-1 right) }}
} $$
while
$$B_1 = frac{pi^2-8}{2pi^2}$$
answered Jan 24 at 20:24
Robert IsraelRobert Israel
327k23216469
$endgroup$
Note that the basis functions $u_n = sin(nx) - n cos(nx)$ are orthogonal on $[0,pi]$, with $int_0^pi u_n^2(x); dx = pi (n^2+1)/2$. Thus the coefficients for a function $f(x)$ that has such a series should be
$$ b_n = frac{2}{pi(n^2+1)} int_0^pi u_n(x) f(x); dx $$
In this case, if $n ne 1$,
$$ eqalign{B_n &= frac{2}{n^2+1} int_0^pi (sin(nx)-ncos(nx))left(sin(x)+frac{x^2}{pi}-xright); dx cr
&= {frac {2,pi, left( -1 right) ^{n}{n}^{2}+2,pi,{n}^{2}+4,{n}^
{2} left( -1 right) ^{n}-4,{n}^{2}-4, left( -1 right) ^{n}+4}{{n
}^{3}pi^2, left( {n}^{4}-1 right) }}
} $$
while
$$B_1 = frac{pi^2-8}{2pi^2}$$
answered Jan 24 at 20:24
Robert IsraelRobert Israel
327k23216469
answered Jan 24 at 20:24
Robert IsraelRobert Israel
327k23216469
answered Jan 24 at 20:24
Robert IsraelRobert Israel
327k23216469
answered Jan 24 at 20:24
Robert IsraelRobert Israel
327k23216469
327k23216469
$begingroup$
Thank you! What about $B_{0}$ is that $frac{1}{pi}int_{0}^{pi} (sin(x)+frac{x^2}{pi}-x)dx$ or $frac{2}{pi}int_{0}^{pi} (sin(x)+frac{x^2}{pi}-x)dx$?
$endgroup$
– MyBBExpert
Jan 24 at 21:35
$begingroup$
Is there a $B_0$? For $n=0$, $sin(nx) - n cos(nx) = 0$.
$endgroup$
– Robert Israel
Jan 25 at 1:52
$begingroup$
Hm: these $f_n$ are eigenfunctions for $f''= lambda f$ with boundary conditions $f(0) = -f'(0)$, $f(pi) = -f'(pi)$. Besides $f_n$ for positive integers $n$, the other eigenfunction is $exp(-x)$. So that's what will give you another term in the expansion..
$endgroup$
– Robert Israel
Jan 25 at 2:22
add a comment |
$begingroup$
Thank you! What about $B_{0}$ is that $frac{1}{pi}int_{0}^{pi} (sin(x)+frac{x^2}{pi}-x)dx$ or $frac{2}{pi}int_{0}^{pi} (sin(x)+frac{x^2}{pi}-x)dx$?
$endgroup$
– MyBBExpert
Jan 24 at 21:35
$begingroup$
Is there a $B_0$? For $n=0$, $sin(nx) - n cos(nx) = 0$.
$endgroup$
– Robert Israel
Jan 25 at 1:52
$begingroup$
Hm: these $f_n$ are eigenfunctions for $f''= lambda f$ with boundary conditions $f(0) = -f'(0)$, $f(pi) = -f'(pi)$. Besides $f_n$ for positive integers $n$, the other eigenfunction is $exp(-x)$. So that's what will give you another term in the expansion..
$endgroup$
– Robert Israel
Jan 25 at 2:22
$begingroup$
Thank you! What about $B_{0}$ is that $frac{1}{pi}int_{0}^{pi} (sin(x)+frac{x^2}{pi}-x)dx$ or $frac{2}{pi}int_{0}^{pi} (sin(x)+frac{x^2}{pi}-x)dx$?
$endgroup$
– MyBBExpert
Jan 24 at 21:35
$begingroup$
Thank you! What about $B_{0}$ is that $frac{1}{pi}int_{0}^{pi} (sin(x)+frac{x^2}{pi}-x)dx$ or $frac{2}{pi}int_{0}^{pi} (sin(x)+frac{x^2}{pi}-x)dx$?
$endgroup$
– MyBBExpert
Jan 24 at 21:35
$begingroup$
Is there a $B_0$? For $n=0$, $sin(nx) - n cos(nx) = 0$.
$endgroup$
– Robert Israel
Jan 25 at 1:52
$begingroup$
Is there a $B_0$? For $n=0$, $sin(nx) - n cos(nx) = 0$.
$endgroup$
– Robert Israel
Jan 25 at 1:52
$begingroup$
Hm: these $f_n$ are eigenfunctions for $f''= lambda f$ with boundary conditions $f(0) = -f'(0)$, $f(pi) = -f'(pi)$. Besides $f_n$ for positive integers $n$, the other eigenfunction is $exp(-x)$. So that's what will give you another term in the expansion..
$endgroup$
– Robert Israel
Jan 25 at 2:22
$begingroup$
Hm: these $f_n$ are eigenfunctions for $f''= lambda f$ with boundary conditions $f(0) = -f'(0)$, $f(pi) = -f'(pi)$. Besides $f_n$ for positive integers $n$, the other eigenfunction is $exp(-x)$. So that's what will give you another term in the expansion..
$endgroup$
– Robert Israel
Jan 25 at 2:22
add a comment |
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$begingroup$
$B_n$ is defined by $B_n=frac{1}{2pi}int_{-pi}^pi (sin(x)+frac{x^2}{pi}-x)sin nxrm dx = frac{1}{2npi}int_{-pi}^pi (sin(x)+frac{x^2}{pi}-x)cos nxrm dx$.
$endgroup$
– Nicolas FRANCOIS
Jan 24 at 19:45
$begingroup$
That is, if the interval on which the equality holds is $[-pi,pi]$. I didn't verified. But your way of computing $B_n$ is strange.
$endgroup$
– Nicolas FRANCOIS
Jan 24 at 19:46
$begingroup$
@NicolasFRANCOIS The interval is $[0,pi]$
$endgroup$
– MyBBExpert
Jan 24 at 19:57