Solution to the Basel Problem in complex analysis pole issue.
up vote
3
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Solve:
$$sum_{n=1}^{infty} frac{1}{z^2}$$
Before you mark as duplicate, I have a problem with only the consideration of the pole, please read carefully!
Consider:
$$f(z) = frac{pi cot(pi z)}{z^2}$$
The obvious three order pole is at $z=0$
But there should be a pole at $z=n?$ because $frac{1}{tan(pi z)}$ doesnt exist for $z in mathbb{N}$
And there isnt any poles at $z=ni$
Wolframlpha: returns it as saying there is only one pole at $z=0$ I am confused now?
Consider a square:
How would we do this?
calculus real-analysis integration sequences-and-series complex-analysis
asked Jan 12 '15 at 14:38
Amad27
5,32321652
add a comment |
up vote
3
down vote
favorite
Solve:
$$sum_{n=1}^{infty} frac{1}{z^2}$$
Before you mark as duplicate, I have a problem with only the consideration of the pole, please read carefully!
Consider:
$$f(z) = frac{pi cot(pi z)}{z^2}$$
The obvious three order pole is at $z=0$
But there should be a pole at $z=n?$ because $frac{1}{tan(pi z)}$ doesnt exist for $z in mathbb{N}$
And there isnt any poles at $z=ni$
Wolframlpha: returns it as saying there is only one pole at $z=0$ I am confused now?
Consider a square:
How would we do this?
calculus real-analysis integration sequences-and-series complex-analysis
asked Jan 12 '15 at 14:38
Amad27
5,32321652
5
I don't know what you entered into WolframAlpha, but yes, $(picotpi z)/z^2$ has simple poles at all $ninmathbb{Z}setminus{0}$. The residue at $z=n$ is $1/n^2$ for $nneq 0$. (This is why one uses that trick with the cotangent.)
– mickep
Jan 12 '15 at 14:40
Is this account yours? If so, you should probably try to merge that account with this one by following the instructions on this page.
– robjohn♦
Jan 15 '15 at 0:26
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Solve:
$$sum_{n=1}^{infty} frac{1}{z^2}$$
Before you mark as duplicate, I have a problem with only the consideration of the pole, please read carefully!
Consider:
$$f(z) = frac{pi cot(pi z)}{z^2}$$
The obvious three order pole is at $z=0$
But there should be a pole at $z=n?$ because $frac{1}{tan(pi z)}$ doesnt exist for $z in mathbb{N}$
And there isnt any poles at $z=ni$
Wolframlpha: returns it as saying there is only one pole at $z=0$ I am confused now?
Consider a square:
How would we do this?
calculus real-analysis integration sequences-and-series complex-analysis
asked Jan 12 '15 at 14:38
Amad27
5,32321652
Solve:
$$sum_{n=1}^{infty} frac{1}{z^2}$$
Before you mark as duplicate, I have a problem with only the consideration of the pole, please read carefully!
Consider:
$$f(z) = frac{pi cot(pi z)}{z^2}$$
The obvious three order pole is at $z=0$
But there should be a pole at $z=n?$ because $frac{1}{tan(pi z)}$ doesnt exist for $z in mathbb{N}$
And there isnt any poles at $z=ni$
Wolframlpha: returns it as saying there is only one pole at $z=0$ I am confused now?
Consider a square:
How would we do this?
calculus real-analysis integration sequences-and-series complex-analysis
calculus real-analysis integration sequences-and-series complex-analysis
asked Jan 12 '15 at 14:38
Amad27
5,32321652
asked Jan 12 '15 at 14:38
Amad27
5,32321652
asked Jan 12 '15 at 14:38
Amad27
5,32321652
asked Jan 12 '15 at 14:38
Amad27
5,32321652
asked Jan 12 '15 at 14:38
Amad27
5,32321652
5,32321652
5
I don't know what you entered into WolframAlpha, but yes, $(picotpi z)/z^2$ has simple poles at all $ninmathbb{Z}setminus{0}$. The residue at $z=n$ is $1/n^2$ for $nneq 0$. (This is why one uses that trick with the cotangent.)
– mickep
Jan 12 '15 at 14:40
Is this account yours? If so, you should probably try to merge that account with this one by following the instructions on this page.
– robjohn♦
Jan 15 '15 at 0:26
add a comment |
5
I don't know what you entered into WolframAlpha, but yes, $(picotpi z)/z^2$ has simple poles at all $ninmathbb{Z}setminus{0}$. The residue at $z=n$ is $1/n^2$ for $nneq 0$. (This is why one uses that trick with the cotangent.)
– mickep
Jan 12 '15 at 14:40
Is this account yours? If so, you should probably try to merge that account with this one by following the instructions on this page.
– robjohn♦
Jan 15 '15 at 0:26
5
5
I don't know what you entered into WolframAlpha, but yes, $(picotpi z)/z^2$ has simple poles at all $ninmathbb{Z}setminus{0}$. The residue at $z=n$ is $1/n^2$ for $nneq 0$. (This is why one uses that trick with the cotangent.)
– mickep
Jan 12 '15 at 14:40
I don't know what you entered into WolframAlpha, but yes, $(picotpi z)/z^2$ has simple poles at all $ninmathbb{Z}setminus{0}$. The residue at $z=n$ is $1/n^2$ for $nneq 0$. (This is why one uses that trick with the cotangent.)
– mickep
Jan 12 '15 at 14:40
Is this account yours? If so, you should probably try to merge that account with this one by following the instructions on this page.
– robjohn♦
Jan 15 '15 at 0:26
Is this account yours? If so, you should probably try to merge that account with this one by following the instructions on this page.
– robjohn♦
Jan 15 '15 at 0:26
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
We have the Laurent expansion at $z=0$:
$$
begin{align}
frac{picot(pi z)}{z^2}
&=fracpi{z^2}frac{cos(pi z)}{sin(pi z)}\
&=fracpi{z^2}frac{1-frac{(pi z)^2}2+frac{(pi z)^4}{24}-dots}{pi zleft(1-frac{(pi z)^2}6+frac{(pi z)^4}{120}-dotsright)}\
&=frac1{z^3}left(1-frac{(pi z)^2}3-frac{(pi z)^4}{45}-dotsright)\[6pt]
&=frac1{z^3}bbox[5px,border:2px solid #C00000]{-frac{pi^2}{3z}}-frac{pi^4z}{45}-dots
end{align}
$$
Thus, the residue of $frac{picot(pi z)}{z^2}$ at $z=0$ is $-frac{pi^2}3$.
Since $picot(pi z)$ has residue $1$ at each integer, at $z=n$, where $n$ is a non-zero integer, the residue of $frac{picot(pi z)}{z^2}$ is $frac1{n^2}$.
Consider
$$
frac1{2pi i}int_gamma frac{picot(pi z)}{z^2},mathrm{d}z
$$
along the contour
$$
gamma=Re^{i[0,2pi]}cup re^{-i[0,2pi]}
$$
where $R=left(n+frac12right)pitoinfty$ where $ninmathbb{Z}$ and $rto0$.
The integral along $gamma$ is the sum of the residues of the singularities inside. The singularities inside the contour are those at the non-zero integers, zero being excluded by the small clockwise circle. That is,
$$
frac1{2pi i}int_gamma frac{picot(pi z)}{z^2},mathrm{d}z=2sum_{n=1}^inftyfrac1{n^2}
$$
The integral along the large circle tends to $0$ because $picot(pi z)$ is bounded on the whole of the large circle and $left|frac1{z^2}right|=frac1{R^2}$ on a circle of length $2pi R$.
The integral along the small clockwise circle is the negative of the residue of the singularity inside; that is, $frac{pi^2}3$. Thus,
$$
frac1{2pi i}int_gamma frac{picot(pi z)}{z^2},mathrm{d}z=frac{pi^2}3
$$
Putting these two together, we get
$$
sum_{n=1}^inftyfrac1{n^2}=frac{pi^2}6
$$
answered Jan 15 '15 at 1:16
robjohn♦
263k27301622
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
We have the Laurent expansion at $z=0$:
$$
begin{align}
frac{picot(pi z)}{z^2}
&=fracpi{z^2}frac{cos(pi z)}{sin(pi z)}\
&=fracpi{z^2}frac{1-frac{(pi z)^2}2+frac{(pi z)^4}{24}-dots}{pi zleft(1-frac{(pi z)^2}6+frac{(pi z)^4}{120}-dotsright)}\
&=frac1{z^3}left(1-frac{(pi z)^2}3-frac{(pi z)^4}{45}-dotsright)\[6pt]
&=frac1{z^3}bbox[5px,border:2px solid #C00000]{-frac{pi^2}{3z}}-frac{pi^4z}{45}-dots
end{align}
$$
Thus, the residue of $frac{picot(pi z)}{z^2}$ at $z=0$ is $-frac{pi^2}3$.
Since $picot(pi z)$ has residue $1$ at each integer, at $z=n$, where $n$ is a non-zero integer, the residue of $frac{picot(pi z)}{z^2}$ is $frac1{n^2}$.
Consider
$$
frac1{2pi i}int_gamma frac{picot(pi z)}{z^2},mathrm{d}z
$$
along the contour
$$
gamma=Re^{i[0,2pi]}cup re^{-i[0,2pi]}
$$
where $R=left(n+frac12right)pitoinfty$ where $ninmathbb{Z}$ and $rto0$.
The integral along $gamma$ is the sum of the residues of the singularities inside. The singularities inside the contour are those at the non-zero integers, zero being excluded by the small clockwise circle. That is,
$$
frac1{2pi i}int_gamma frac{picot(pi z)}{z^2},mathrm{d}z=2sum_{n=1}^inftyfrac1{n^2}
$$
The integral along the large circle tends to $0$ because $picot(pi z)$ is bounded on the whole of the large circle and $left|frac1{z^2}right|=frac1{R^2}$ on a circle of length $2pi R$.
The integral along the small clockwise circle is the negative of the residue of the singularity inside; that is, $frac{pi^2}3$. Thus,
$$
frac1{2pi i}int_gamma frac{picot(pi z)}{z^2},mathrm{d}z=frac{pi^2}3
$$
Putting these two together, we get
$$
sum_{n=1}^inftyfrac1{n^2}=frac{pi^2}6
$$
answered Jan 15 '15 at 1:16
robjohn♦
263k27301622
add a comment |
up vote
2
down vote
accepted
We have the Laurent expansion at $z=0$:
$$
begin{align}
frac{picot(pi z)}{z^2}
&=fracpi{z^2}frac{cos(pi z)}{sin(pi z)}\
&=fracpi{z^2}frac{1-frac{(pi z)^2}2+frac{(pi z)^4}{24}-dots}{pi zleft(1-frac{(pi z)^2}6+frac{(pi z)^4}{120}-dotsright)}\
&=frac1{z^3}left(1-frac{(pi z)^2}3-frac{(pi z)^4}{45}-dotsright)\[6pt]
&=frac1{z^3}bbox[5px,border:2px solid #C00000]{-frac{pi^2}{3z}}-frac{pi^4z}{45}-dots
end{align}
$$
Thus, the residue of $frac{picot(pi z)}{z^2}$ at $z=0$ is $-frac{pi^2}3$.
Since $picot(pi z)$ has residue $1$ at each integer, at $z=n$, where $n$ is a non-zero integer, the residue of $frac{picot(pi z)}{z^2}$ is $frac1{n^2}$.
Consider
$$
frac1{2pi i}int_gamma frac{picot(pi z)}{z^2},mathrm{d}z
$$
along the contour
$$
gamma=Re^{i[0,2pi]}cup re^{-i[0,2pi]}
$$
where $R=left(n+frac12right)pitoinfty$ where $ninmathbb{Z}$ and $rto0$.
The integral along $gamma$ is the sum of the residues of the singularities inside. The singularities inside the contour are those at the non-zero integers, zero being excluded by the small clockwise circle. That is,
$$
frac1{2pi i}int_gamma frac{picot(pi z)}{z^2},mathrm{d}z=2sum_{n=1}^inftyfrac1{n^2}
$$
The integral along the large circle tends to $0$ because $picot(pi z)$ is bounded on the whole of the large circle and $left|frac1{z^2}right|=frac1{R^2}$ on a circle of length $2pi R$.
The integral along the small clockwise circle is the negative of the residue of the singularity inside; that is, $frac{pi^2}3$. Thus,
$$
frac1{2pi i}int_gamma frac{picot(pi z)}{z^2},mathrm{d}z=frac{pi^2}3
$$
Putting these two together, we get
$$
sum_{n=1}^inftyfrac1{n^2}=frac{pi^2}6
$$
answered Jan 15 '15 at 1:16
robjohn♦
263k27301622
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
We have the Laurent expansion at $z=0$:
$$
begin{align}
frac{picot(pi z)}{z^2}
&=fracpi{z^2}frac{cos(pi z)}{sin(pi z)}\
&=fracpi{z^2}frac{1-frac{(pi z)^2}2+frac{(pi z)^4}{24}-dots}{pi zleft(1-frac{(pi z)^2}6+frac{(pi z)^4}{120}-dotsright)}\
&=frac1{z^3}left(1-frac{(pi z)^2}3-frac{(pi z)^4}{45}-dotsright)\[6pt]
&=frac1{z^3}bbox[5px,border:2px solid #C00000]{-frac{pi^2}{3z}}-frac{pi^4z}{45}-dots
end{align}
$$
Thus, the residue of $frac{picot(pi z)}{z^2}$ at $z=0$ is $-frac{pi^2}3$.
Since $picot(pi z)$ has residue $1$ at each integer, at $z=n$, where $n$ is a non-zero integer, the residue of $frac{picot(pi z)}{z^2}$ is $frac1{n^2}$.
Consider
$$
frac1{2pi i}int_gamma frac{picot(pi z)}{z^2},mathrm{d}z
$$
along the contour
$$
gamma=Re^{i[0,2pi]}cup re^{-i[0,2pi]}
$$
where $R=left(n+frac12right)pitoinfty$ where $ninmathbb{Z}$ and $rto0$.
The integral along $gamma$ is the sum of the residues of the singularities inside. The singularities inside the contour are those at the non-zero integers, zero being excluded by the small clockwise circle. That is,
$$
frac1{2pi i}int_gamma frac{picot(pi z)}{z^2},mathrm{d}z=2sum_{n=1}^inftyfrac1{n^2}
$$
The integral along the large circle tends to $0$ because $picot(pi z)$ is bounded on the whole of the large circle and $left|frac1{z^2}right|=frac1{R^2}$ on a circle of length $2pi R$.
The integral along the small clockwise circle is the negative of the residue of the singularity inside; that is, $frac{pi^2}3$. Thus,
$$
frac1{2pi i}int_gamma frac{picot(pi z)}{z^2},mathrm{d}z=frac{pi^2}3
$$
Putting these two together, we get
$$
sum_{n=1}^inftyfrac1{n^2}=frac{pi^2}6
$$
answered Jan 15 '15 at 1:16
robjohn♦
263k27301622
We have the Laurent expansion at $z=0$:
$$
begin{align}
frac{picot(pi z)}{z^2}
&=fracpi{z^2}frac{cos(pi z)}{sin(pi z)}\
&=fracpi{z^2}frac{1-frac{(pi z)^2}2+frac{(pi z)^4}{24}-dots}{pi zleft(1-frac{(pi z)^2}6+frac{(pi z)^4}{120}-dotsright)}\
&=frac1{z^3}left(1-frac{(pi z)^2}3-frac{(pi z)^4}{45}-dotsright)\[6pt]
&=frac1{z^3}bbox[5px,border:2px solid #C00000]{-frac{pi^2}{3z}}-frac{pi^4z}{45}-dots
end{align}
$$
Thus, the residue of $frac{picot(pi z)}{z^2}$ at $z=0$ is $-frac{pi^2}3$.
Since $picot(pi z)$ has residue $1$ at each integer, at $z=n$, where $n$ is a non-zero integer, the residue of $frac{picot(pi z)}{z^2}$ is $frac1{n^2}$.
Consider
$$
frac1{2pi i}int_gamma frac{picot(pi z)}{z^2},mathrm{d}z
$$
along the contour
$$
gamma=Re^{i[0,2pi]}cup re^{-i[0,2pi]}
$$
where $R=left(n+frac12right)pitoinfty$ where $ninmathbb{Z}$ and $rto0$.
The integral along $gamma$ is the sum of the residues of the singularities inside. The singularities inside the contour are those at the non-zero integers, zero being excluded by the small clockwise circle. That is,
$$
frac1{2pi i}int_gamma frac{picot(pi z)}{z^2},mathrm{d}z=2sum_{n=1}^inftyfrac1{n^2}
$$
The integral along the large circle tends to $0$ because $picot(pi z)$ is bounded on the whole of the large circle and $left|frac1{z^2}right|=frac1{R^2}$ on a circle of length $2pi R$.
The integral along the small clockwise circle is the negative of the residue of the singularity inside; that is, $frac{pi^2}3$. Thus,
$$
frac1{2pi i}int_gamma frac{picot(pi z)}{z^2},mathrm{d}z=frac{pi^2}3
$$
Putting these two together, we get
$$
sum_{n=1}^inftyfrac1{n^2}=frac{pi^2}6
$$
answered Jan 15 '15 at 1:16
robjohn♦
263k27301622
edited Jan 15 '15 at 11:05
answered Jan 15 '15 at 1:16
robjohn♦
263k27301622
answered Jan 15 '15 at 1:16
robjohn♦
263k27301622
answered Jan 15 '15 at 1:16
robjohn♦
263k27301622
263k27301622
add a comment |
add a comment |
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5
I don't know what you entered into WolframAlpha, but yes, $(picotpi z)/z^2$ has simple poles at all $ninmathbb{Z}setminus{0}$. The residue at $z=n$ is $1/n^2$ for $nneq 0$. (This is why one uses that trick with the cotangent.)
– mickep
Jan 12 '15 at 14:40
Is this account yours? If so, you should probably try to merge that account with this one by following the instructions on this page.
– robjohn♦
Jan 15 '15 at 0:26