Solution to the Basel Problem in complex analysis pole issue.












3














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:



square contour



How would we do this?






calculus real-analysis integration sequences-and-series complex-analysis







share|cite|improve this question


















  • 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
















3














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:



square contour



How would we do this?






calculus real-analysis integration sequences-and-series complex-analysis







share|cite|improve this question


















  • 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














3












3








3







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:



square contour



How would we do this?






calculus real-analysis integration sequences-and-series complex-analysis







share|cite|improve this question













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:



square contour



How would we do this?






calculus real-analysis integration sequences-and-series complex-analysis




calculus real-analysis integration sequences-and-series complex-analysis






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 12 '15 at 14:38









Amad27

5,32321753




5,32321753








  • 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




    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










1 Answer
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oldest

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2














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
$$
sum_{n=1}^inftyfrac1{n^2}=frac{pi^2}6
$$






share|cite|improve this answer























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    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
    $$
    sum_{n=1}^inftyfrac1{n^2}=frac{pi^2}6
    $$






    share|cite|improve this answer




























      2














      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
      $$
      sum_{n=1}^inftyfrac1{n^2}=frac{pi^2}6
      $$






      share|cite|improve this answer


























        2












        2








        2






        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
        $$
        sum_{n=1}^inftyfrac1{n^2}=frac{pi^2}6
        $$






        share|cite|improve this answer














        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
        $$
        sum_{n=1}^inftyfrac1{n^2}=frac{pi^2}6
        $$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 15 '15 at 11:05

























        answered Jan 15 '15 at 1:16









        robjohn♦

        264k27303623




        264k27303623






























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