Proving that $frac{1}{2pi}int_{0}^{2pi}|f(re^{it})|^2dt=sum_{n=0}^{infty}|a_n|^2r^{2n}$












2












$begingroup$


Let $f(z)=sum_{n=0}^{infty}a_nz^n$ with radius of convergence equals to $R$. Show that for every $r<R$:



$$frac{1}{2pi}int_{0}^{2pi}|f(re^{it})|^2dt=sum_{n=0}^{infty}|a_n|^2r^{2n}$$



I tried this:



$$int_{0}^{2pi}|f(re^{it})|^2dt=int_{0}^{2pi}f(re^{it})overline{f(re^{it})}dt=int_{0}^{2pi}sum_{n=0}^{infty}a_nr^ne^{int}sum_{n=0}^{infty}overline{a_n}r^ne^{-int}dt=
int_{0}^{2pi}{sum_{n=0}^{infty}sum_{k=0}^{n}}a_koverline{a_{n-k}}r^ne^{int}$$



But this is about it. Any ideas?



Thanks!






complex-analysis







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    Are you aware of the relationship that $int_{0}^{2pi} e^{i(n-m)t} = 2pidelta_{nm}$, where $delta_{nm}$ is the Kronecker delta? Also you really need to use different indices for your sums. Make one an index over $n$ and the other an index over $m$. The two sums are separate from each other so you should demonstrate this with different indices. It doesn't make sense otherwise.
    $endgroup$
    – Cameron Williams
    Jul 24 '13 at 19:19












  • $begingroup$
    This is a restatement of Parseval's theorem for holomorphic functions on the disk if you're familiar with Fourier theory.
    $endgroup$
    – Cameron Williams
    Jul 24 '13 at 19:28






  • 1




    $begingroup$
    Your last line is not correct. Using different indices in the line before (as suggested by Cameron), say $n$ and $m$, and permuting $Sigma$ and $int$, you should get $sum_{m,n=0}^infty a_noverline{a_m} r^{n+m} int_0^{2pi} e^{i(n-m)t}dt$.
    $endgroup$
    – Etienne
    Jul 24 '13 at 19:37


















2












$begingroup$


Let $f(z)=sum_{n=0}^{infty}a_nz^n$ with radius of convergence equals to $R$. Show that for every $r<R$:



$$frac{1}{2pi}int_{0}^{2pi}|f(re^{it})|^2dt=sum_{n=0}^{infty}|a_n|^2r^{2n}$$



I tried this:



$$int_{0}^{2pi}|f(re^{it})|^2dt=int_{0}^{2pi}f(re^{it})overline{f(re^{it})}dt=int_{0}^{2pi}sum_{n=0}^{infty}a_nr^ne^{int}sum_{n=0}^{infty}overline{a_n}r^ne^{-int}dt=
int_{0}^{2pi}{sum_{n=0}^{infty}sum_{k=0}^{n}}a_koverline{a_{n-k}}r^ne^{int}$$



But this is about it. Any ideas?



Thanks!






complex-analysis







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    Are you aware of the relationship that $int_{0}^{2pi} e^{i(n-m)t} = 2pidelta_{nm}$, where $delta_{nm}$ is the Kronecker delta? Also you really need to use different indices for your sums. Make one an index over $n$ and the other an index over $m$. The two sums are separate from each other so you should demonstrate this with different indices. It doesn't make sense otherwise.
    $endgroup$
    – Cameron Williams
    Jul 24 '13 at 19:19












  • $begingroup$
    This is a restatement of Parseval's theorem for holomorphic functions on the disk if you're familiar with Fourier theory.
    $endgroup$
    – Cameron Williams
    Jul 24 '13 at 19:28






  • 1




    $begingroup$
    Your last line is not correct. Using different indices in the line before (as suggested by Cameron), say $n$ and $m$, and permuting $Sigma$ and $int$, you should get $sum_{m,n=0}^infty a_noverline{a_m} r^{n+m} int_0^{2pi} e^{i(n-m)t}dt$.
    $endgroup$
    – Etienne
    Jul 24 '13 at 19:37
















2












2








2


4



$begingroup$


Let $f(z)=sum_{n=0}^{infty}a_nz^n$ with radius of convergence equals to $R$. Show that for every $r<R$:



$$frac{1}{2pi}int_{0}^{2pi}|f(re^{it})|^2dt=sum_{n=0}^{infty}|a_n|^2r^{2n}$$



I tried this:



$$int_{0}^{2pi}|f(re^{it})|^2dt=int_{0}^{2pi}f(re^{it})overline{f(re^{it})}dt=int_{0}^{2pi}sum_{n=0}^{infty}a_nr^ne^{int}sum_{n=0}^{infty}overline{a_n}r^ne^{-int}dt=
int_{0}^{2pi}{sum_{n=0}^{infty}sum_{k=0}^{n}}a_koverline{a_{n-k}}r^ne^{int}$$



But this is about it. Any ideas?



Thanks!






complex-analysis







share|cite|improve this question









$endgroup$




Let $f(z)=sum_{n=0}^{infty}a_nz^n$ with radius of convergence equals to $R$. Show that for every $r<R$:



$$frac{1}{2pi}int_{0}^{2pi}|f(re^{it})|^2dt=sum_{n=0}^{infty}|a_n|^2r^{2n}$$



I tried this:



$$int_{0}^{2pi}|f(re^{it})|^2dt=int_{0}^{2pi}f(re^{it})overline{f(re^{it})}dt=int_{0}^{2pi}sum_{n=0}^{infty}a_nr^ne^{int}sum_{n=0}^{infty}overline{a_n}r^ne^{-int}dt=
int_{0}^{2pi}{sum_{n=0}^{infty}sum_{k=0}^{n}}a_koverline{a_{n-k}}r^ne^{int}$$



But this is about it. Any ideas?



Thanks!






complex-analysis




complex-analysis






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jul 24 '13 at 19:17









catch22catch22

1,3411123




1,3411123








  • 3




    $begingroup$
    Are you aware of the relationship that $int_{0}^{2pi} e^{i(n-m)t} = 2pidelta_{nm}$, where $delta_{nm}$ is the Kronecker delta? Also you really need to use different indices for your sums. Make one an index over $n$ and the other an index over $m$. The two sums are separate from each other so you should demonstrate this with different indices. It doesn't make sense otherwise.
    $endgroup$
    – Cameron Williams
    Jul 24 '13 at 19:19












  • $begingroup$
    This is a restatement of Parseval's theorem for holomorphic functions on the disk if you're familiar with Fourier theory.
    $endgroup$
    – Cameron Williams
    Jul 24 '13 at 19:28






  • 1




    $begingroup$
    Your last line is not correct. Using different indices in the line before (as suggested by Cameron), say $n$ and $m$, and permuting $Sigma$ and $int$, you should get $sum_{m,n=0}^infty a_noverline{a_m} r^{n+m} int_0^{2pi} e^{i(n-m)t}dt$.
    $endgroup$
    – Etienne
    Jul 24 '13 at 19:37
















  • 3




    $begingroup$
    Are you aware of the relationship that $int_{0}^{2pi} e^{i(n-m)t} = 2pidelta_{nm}$, where $delta_{nm}$ is the Kronecker delta? Also you really need to use different indices for your sums. Make one an index over $n$ and the other an index over $m$. The two sums are separate from each other so you should demonstrate this with different indices. It doesn't make sense otherwise.
    $endgroup$
    – Cameron Williams
    Jul 24 '13 at 19:19












  • $begingroup$
    This is a restatement of Parseval's theorem for holomorphic functions on the disk if you're familiar with Fourier theory.
    $endgroup$
    – Cameron Williams
    Jul 24 '13 at 19:28






  • 1




    $begingroup$
    Your last line is not correct. Using different indices in the line before (as suggested by Cameron), say $n$ and $m$, and permuting $Sigma$ and $int$, you should get $sum_{m,n=0}^infty a_noverline{a_m} r^{n+m} int_0^{2pi} e^{i(n-m)t}dt$.
    $endgroup$
    – Etienne
    Jul 24 '13 at 19:37










3




3




$begingroup$
Are you aware of the relationship that $int_{0}^{2pi} e^{i(n-m)t} = 2pidelta_{nm}$, where $delta_{nm}$ is the Kronecker delta? Also you really need to use different indices for your sums. Make one an index over $n$ and the other an index over $m$. The two sums are separate from each other so you should demonstrate this with different indices. It doesn't make sense otherwise.
$endgroup$
– Cameron Williams
Jul 24 '13 at 19:19






$begingroup$
Are you aware of the relationship that $int_{0}^{2pi} e^{i(n-m)t} = 2pidelta_{nm}$, where $delta_{nm}$ is the Kronecker delta? Also you really need to use different indices for your sums. Make one an index over $n$ and the other an index over $m$. The two sums are separate from each other so you should demonstrate this with different indices. It doesn't make sense otherwise.
$endgroup$
– Cameron Williams
Jul 24 '13 at 19:19














$begingroup$
This is a restatement of Parseval's theorem for holomorphic functions on the disk if you're familiar with Fourier theory.
$endgroup$
– Cameron Williams
Jul 24 '13 at 19:28




$begingroup$
This is a restatement of Parseval's theorem for holomorphic functions on the disk if you're familiar with Fourier theory.
$endgroup$
– Cameron Williams
Jul 24 '13 at 19:28




1




1




$begingroup$
Your last line is not correct. Using different indices in the line before (as suggested by Cameron), say $n$ and $m$, and permuting $Sigma$ and $int$, you should get $sum_{m,n=0}^infty a_noverline{a_m} r^{n+m} int_0^{2pi} e^{i(n-m)t}dt$.
$endgroup$
– Etienne
Jul 24 '13 at 19:37






$begingroup$
Your last line is not correct. Using different indices in the line before (as suggested by Cameron), say $n$ and $m$, and permuting $Sigma$ and $int$, you should get $sum_{m,n=0}^infty a_noverline{a_m} r^{n+m} int_0^{2pi} e^{i(n-m)t}dt$.
$endgroup$
– Etienne
Jul 24 '13 at 19:37












2 Answers
2






active

oldest

votes


















3












$begingroup$

$$ int_0^{2pi}| f(re^{it})|^2 dt = int_0^{2pi}f(re^{it})overline{f(re^{it})}dt = int_0^{2pi}sum_{n=0}^{infty}a_nr^ne^{int}sum_{m=0}^{infty}overline{a_m}r^me^{-imt} dt $$



Assuming we can commute the sums (which I guess you could argue from uniform convergence of the power series) we have



$$ frac{1}{2pi}int_0^{2pi}| f(re^{it})|^2 dt = sum_{n=0}^{infty}sum_{m=0}^{infty}a_noverline{a_m}r^{n+m}int_0^{2pi}e^{i(n-m)t}dt = frac{1}{2pi}sum_{n=0}^{infty}sum_{m=0}^{infty}a_noverline{a_m}r^{n+m} 2pidelta_{nm} = sum_{n=0}^{infty}|a_n|^2r^{2n}.$$



As for how to so that $int_0^{2pi}e^{i(n-m)t}dt = 2pidelta_{nm}$, the proof is as follows:



Suppose $n = m$ then we wish to evaluate $int_0^{2pi}e^{i(n-m)t}dt = int_0^{2pi}e^{0}dt = int_0^{2pi} 1 dt = 2pi$.



Suppose $nneq m$ then we have $int_0^{2pi}e^{i(n-m)t} dt = frac{1}{i(n-m)}e^{i(n-m)t}|_0^{2pi} = frac{1}{i(n-m)}left(e^{2pi i(n-m)} - 1right)$. But since $2pi i(n-m)$ is a multiple of $2pi$, $e^{2pi i(n-m)} = 1$ and so it evaluates to $0$. QED.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Thanks. For completeness I will write the answer:



    $$int_{0}^{2pi}|f(re^{it})|^2dt=int_{0}^{2pi}f(re^{it})overline{f(re^{it})}dt=int_{0}^{2pi}left( sum_{n=0}^{infty}a_nr^ne^{int}right) left(sum_{n=0}^{infty}overline{a_n}r^ne^{-int}right) dt=
    int_{0}^{2pi}{sum_{n=0}^{infty}sum_{k=0}^{n}}a_koverline{a_{n-k}}r^ne^{i(2k-n)t}dt=
    {sum_{n=0}^{infty}sum_{k=0}^{n}}a_koverline{a_{n-k}}r^nint_{0}^{2pi}e^{i(2k-n)t}dt=
    2pisum_{n=0}^{infty}a_{n/2}overline{a_{n/2}}r^n=2pisum_{n=0}^{infty}|a_{n}|^2r^{2n}$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Again, you really need to use different indices for your summations. What you have written doesn't really make sense. I will write it out an an answer since you kind of have the idea.
      $endgroup$
      – Cameron Williams
      Jul 24 '13 at 19:48












    • $begingroup$
      @CameronWilliams: Thanks a lot. About the indices - I get what you're saying but I'm used to this notation. I've added brackets - does it make it clearer?
      $endgroup$
      – catch22
      Jul 24 '13 at 20:10










    • $begingroup$
      I understand but you cannot repeat indices like that because it can lead to erroneous results. What you have done in your fourth term in the series of equalities is effectively discount all of the values of $k > n$ so you have considered only about half of your overall state space for $(n,k)$. If you want to look at your state space of $(n,k)$ like a matrix, you basically only looked at the lower triangular part of the matrix. In this case it worked out because the only values that mattered lied along the diagonal but in general this is not true.
      $endgroup$
      – Cameron Williams
      Jul 24 '13 at 20:17










    • $begingroup$
      @CameronWilliams: How is what I did different from here: en.wikipedia.org/wiki/Cauchy_product ?
      $endgroup$
      – catch22
      Jul 24 '13 at 20:39










    • $begingroup$
      I didn't realize you were using the Cauchy product. Then what you have is fine. But I am going to make an edit to the Wikipedia article because they shouldn't be reusing indices like they are. It is not proper mathematical form.
      $endgroup$
      – Cameron Williams
      Jul 24 '13 at 20:42











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    2 Answers
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    2 Answers
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    3












    $begingroup$

    $$ int_0^{2pi}| f(re^{it})|^2 dt = int_0^{2pi}f(re^{it})overline{f(re^{it})}dt = int_0^{2pi}sum_{n=0}^{infty}a_nr^ne^{int}sum_{m=0}^{infty}overline{a_m}r^me^{-imt} dt $$



    Assuming we can commute the sums (which I guess you could argue from uniform convergence of the power series) we have



    $$ frac{1}{2pi}int_0^{2pi}| f(re^{it})|^2 dt = sum_{n=0}^{infty}sum_{m=0}^{infty}a_noverline{a_m}r^{n+m}int_0^{2pi}e^{i(n-m)t}dt = frac{1}{2pi}sum_{n=0}^{infty}sum_{m=0}^{infty}a_noverline{a_m}r^{n+m} 2pidelta_{nm} = sum_{n=0}^{infty}|a_n|^2r^{2n}.$$



    As for how to so that $int_0^{2pi}e^{i(n-m)t}dt = 2pidelta_{nm}$, the proof is as follows:



    Suppose $n = m$ then we wish to evaluate $int_0^{2pi}e^{i(n-m)t}dt = int_0^{2pi}e^{0}dt = int_0^{2pi} 1 dt = 2pi$.



    Suppose $nneq m$ then we have $int_0^{2pi}e^{i(n-m)t} dt = frac{1}{i(n-m)}e^{i(n-m)t}|_0^{2pi} = frac{1}{i(n-m)}left(e^{2pi i(n-m)} - 1right)$. But since $2pi i(n-m)$ is a multiple of $2pi$, $e^{2pi i(n-m)} = 1$ and so it evaluates to $0$. QED.






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      $$ int_0^{2pi}| f(re^{it})|^2 dt = int_0^{2pi}f(re^{it})overline{f(re^{it})}dt = int_0^{2pi}sum_{n=0}^{infty}a_nr^ne^{int}sum_{m=0}^{infty}overline{a_m}r^me^{-imt} dt $$



      Assuming we can commute the sums (which I guess you could argue from uniform convergence of the power series) we have



      $$ frac{1}{2pi}int_0^{2pi}| f(re^{it})|^2 dt = sum_{n=0}^{infty}sum_{m=0}^{infty}a_noverline{a_m}r^{n+m}int_0^{2pi}e^{i(n-m)t}dt = frac{1}{2pi}sum_{n=0}^{infty}sum_{m=0}^{infty}a_noverline{a_m}r^{n+m} 2pidelta_{nm} = sum_{n=0}^{infty}|a_n|^2r^{2n}.$$



      As for how to so that $int_0^{2pi}e^{i(n-m)t}dt = 2pidelta_{nm}$, the proof is as follows:



      Suppose $n = m$ then we wish to evaluate $int_0^{2pi}e^{i(n-m)t}dt = int_0^{2pi}e^{0}dt = int_0^{2pi} 1 dt = 2pi$.



      Suppose $nneq m$ then we have $int_0^{2pi}e^{i(n-m)t} dt = frac{1}{i(n-m)}e^{i(n-m)t}|_0^{2pi} = frac{1}{i(n-m)}left(e^{2pi i(n-m)} - 1right)$. But since $2pi i(n-m)$ is a multiple of $2pi$, $e^{2pi i(n-m)} = 1$ and so it evaluates to $0$. QED.






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        $$ int_0^{2pi}| f(re^{it})|^2 dt = int_0^{2pi}f(re^{it})overline{f(re^{it})}dt = int_0^{2pi}sum_{n=0}^{infty}a_nr^ne^{int}sum_{m=0}^{infty}overline{a_m}r^me^{-imt} dt $$



        Assuming we can commute the sums (which I guess you could argue from uniform convergence of the power series) we have



        $$ frac{1}{2pi}int_0^{2pi}| f(re^{it})|^2 dt = sum_{n=0}^{infty}sum_{m=0}^{infty}a_noverline{a_m}r^{n+m}int_0^{2pi}e^{i(n-m)t}dt = frac{1}{2pi}sum_{n=0}^{infty}sum_{m=0}^{infty}a_noverline{a_m}r^{n+m} 2pidelta_{nm} = sum_{n=0}^{infty}|a_n|^2r^{2n}.$$



        As for how to so that $int_0^{2pi}e^{i(n-m)t}dt = 2pidelta_{nm}$, the proof is as follows:



        Suppose $n = m$ then we wish to evaluate $int_0^{2pi}e^{i(n-m)t}dt = int_0^{2pi}e^{0}dt = int_0^{2pi} 1 dt = 2pi$.



        Suppose $nneq m$ then we have $int_0^{2pi}e^{i(n-m)t} dt = frac{1}{i(n-m)}e^{i(n-m)t}|_0^{2pi} = frac{1}{i(n-m)}left(e^{2pi i(n-m)} - 1right)$. But since $2pi i(n-m)$ is a multiple of $2pi$, $e^{2pi i(n-m)} = 1$ and so it evaluates to $0$. QED.






        share|cite|improve this answer









        $endgroup$



        $$ int_0^{2pi}| f(re^{it})|^2 dt = int_0^{2pi}f(re^{it})overline{f(re^{it})}dt = int_0^{2pi}sum_{n=0}^{infty}a_nr^ne^{int}sum_{m=0}^{infty}overline{a_m}r^me^{-imt} dt $$



        Assuming we can commute the sums (which I guess you could argue from uniform convergence of the power series) we have



        $$ frac{1}{2pi}int_0^{2pi}| f(re^{it})|^2 dt = sum_{n=0}^{infty}sum_{m=0}^{infty}a_noverline{a_m}r^{n+m}int_0^{2pi}e^{i(n-m)t}dt = frac{1}{2pi}sum_{n=0}^{infty}sum_{m=0}^{infty}a_noverline{a_m}r^{n+m} 2pidelta_{nm} = sum_{n=0}^{infty}|a_n|^2r^{2n}.$$



        As for how to so that $int_0^{2pi}e^{i(n-m)t}dt = 2pidelta_{nm}$, the proof is as follows:



        Suppose $n = m$ then we wish to evaluate $int_0^{2pi}e^{i(n-m)t}dt = int_0^{2pi}e^{0}dt = int_0^{2pi} 1 dt = 2pi$.



        Suppose $nneq m$ then we have $int_0^{2pi}e^{i(n-m)t} dt = frac{1}{i(n-m)}e^{i(n-m)t}|_0^{2pi} = frac{1}{i(n-m)}left(e^{2pi i(n-m)} - 1right)$. But since $2pi i(n-m)$ is a multiple of $2pi$, $e^{2pi i(n-m)} = 1$ and so it evaluates to $0$. QED.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jul 24 '13 at 20:01









        Cameron WilliamsCameron Williams

        22.5k43680




        22.5k43680























            0












            $begingroup$

            Thanks. For completeness I will write the answer:



            $$int_{0}^{2pi}|f(re^{it})|^2dt=int_{0}^{2pi}f(re^{it})overline{f(re^{it})}dt=int_{0}^{2pi}left( sum_{n=0}^{infty}a_nr^ne^{int}right) left(sum_{n=0}^{infty}overline{a_n}r^ne^{-int}right) dt=
            int_{0}^{2pi}{sum_{n=0}^{infty}sum_{k=0}^{n}}a_koverline{a_{n-k}}r^ne^{i(2k-n)t}dt=
            {sum_{n=0}^{infty}sum_{k=0}^{n}}a_koverline{a_{n-k}}r^nint_{0}^{2pi}e^{i(2k-n)t}dt=
            2pisum_{n=0}^{infty}a_{n/2}overline{a_{n/2}}r^n=2pisum_{n=0}^{infty}|a_{n}|^2r^{2n}$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Again, you really need to use different indices for your summations. What you have written doesn't really make sense. I will write it out an an answer since you kind of have the idea.
              $endgroup$
              – Cameron Williams
              Jul 24 '13 at 19:48












            • $begingroup$
              @CameronWilliams: Thanks a lot. About the indices - I get what you're saying but I'm used to this notation. I've added brackets - does it make it clearer?
              $endgroup$
              – catch22
              Jul 24 '13 at 20:10










            • $begingroup$
              I understand but you cannot repeat indices like that because it can lead to erroneous results. What you have done in your fourth term in the series of equalities is effectively discount all of the values of $k > n$ so you have considered only about half of your overall state space for $(n,k)$. If you want to look at your state space of $(n,k)$ like a matrix, you basically only looked at the lower triangular part of the matrix. In this case it worked out because the only values that mattered lied along the diagonal but in general this is not true.
              $endgroup$
              – Cameron Williams
              Jul 24 '13 at 20:17










            • $begingroup$
              @CameronWilliams: How is what I did different from here: en.wikipedia.org/wiki/Cauchy_product ?
              $endgroup$
              – catch22
              Jul 24 '13 at 20:39










            • $begingroup$
              I didn't realize you were using the Cauchy product. Then what you have is fine. But I am going to make an edit to the Wikipedia article because they shouldn't be reusing indices like they are. It is not proper mathematical form.
              $endgroup$
              – Cameron Williams
              Jul 24 '13 at 20:42
















            0












            $begingroup$

            Thanks. For completeness I will write the answer:



            $$int_{0}^{2pi}|f(re^{it})|^2dt=int_{0}^{2pi}f(re^{it})overline{f(re^{it})}dt=int_{0}^{2pi}left( sum_{n=0}^{infty}a_nr^ne^{int}right) left(sum_{n=0}^{infty}overline{a_n}r^ne^{-int}right) dt=
            int_{0}^{2pi}{sum_{n=0}^{infty}sum_{k=0}^{n}}a_koverline{a_{n-k}}r^ne^{i(2k-n)t}dt=
            {sum_{n=0}^{infty}sum_{k=0}^{n}}a_koverline{a_{n-k}}r^nint_{0}^{2pi}e^{i(2k-n)t}dt=
            2pisum_{n=0}^{infty}a_{n/2}overline{a_{n/2}}r^n=2pisum_{n=0}^{infty}|a_{n}|^2r^{2n}$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Again, you really need to use different indices for your summations. What you have written doesn't really make sense. I will write it out an an answer since you kind of have the idea.
              $endgroup$
              – Cameron Williams
              Jul 24 '13 at 19:48












            • $begingroup$
              @CameronWilliams: Thanks a lot. About the indices - I get what you're saying but I'm used to this notation. I've added brackets - does it make it clearer?
              $endgroup$
              – catch22
              Jul 24 '13 at 20:10










            • $begingroup$
              I understand but you cannot repeat indices like that because it can lead to erroneous results. What you have done in your fourth term in the series of equalities is effectively discount all of the values of $k > n$ so you have considered only about half of your overall state space for $(n,k)$. If you want to look at your state space of $(n,k)$ like a matrix, you basically only looked at the lower triangular part of the matrix. In this case it worked out because the only values that mattered lied along the diagonal but in general this is not true.
              $endgroup$
              – Cameron Williams
              Jul 24 '13 at 20:17










            • $begingroup$
              @CameronWilliams: How is what I did different from here: en.wikipedia.org/wiki/Cauchy_product ?
              $endgroup$
              – catch22
              Jul 24 '13 at 20:39










            • $begingroup$
              I didn't realize you were using the Cauchy product. Then what you have is fine. But I am going to make an edit to the Wikipedia article because they shouldn't be reusing indices like they are. It is not proper mathematical form.
              $endgroup$
              – Cameron Williams
              Jul 24 '13 at 20:42














            0












            0








            0





            $begingroup$

            Thanks. For completeness I will write the answer:



            $$int_{0}^{2pi}|f(re^{it})|^2dt=int_{0}^{2pi}f(re^{it})overline{f(re^{it})}dt=int_{0}^{2pi}left( sum_{n=0}^{infty}a_nr^ne^{int}right) left(sum_{n=0}^{infty}overline{a_n}r^ne^{-int}right) dt=
            int_{0}^{2pi}{sum_{n=0}^{infty}sum_{k=0}^{n}}a_koverline{a_{n-k}}r^ne^{i(2k-n)t}dt=
            {sum_{n=0}^{infty}sum_{k=0}^{n}}a_koverline{a_{n-k}}r^nint_{0}^{2pi}e^{i(2k-n)t}dt=
            2pisum_{n=0}^{infty}a_{n/2}overline{a_{n/2}}r^n=2pisum_{n=0}^{infty}|a_{n}|^2r^{2n}$$






            share|cite|improve this answer











            $endgroup$



            Thanks. For completeness I will write the answer:



            $$int_{0}^{2pi}|f(re^{it})|^2dt=int_{0}^{2pi}f(re^{it})overline{f(re^{it})}dt=int_{0}^{2pi}left( sum_{n=0}^{infty}a_nr^ne^{int}right) left(sum_{n=0}^{infty}overline{a_n}r^ne^{-int}right) dt=
            int_{0}^{2pi}{sum_{n=0}^{infty}sum_{k=0}^{n}}a_koverline{a_{n-k}}r^ne^{i(2k-n)t}dt=
            {sum_{n=0}^{infty}sum_{k=0}^{n}}a_koverline{a_{n-k}}r^nint_{0}^{2pi}e^{i(2k-n)t}dt=
            2pisum_{n=0}^{infty}a_{n/2}overline{a_{n/2}}r^n=2pisum_{n=0}^{infty}|a_{n}|^2r^{2n}$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jul 24 '13 at 20:06

























            answered Jul 24 '13 at 19:45









            catch22catch22

            1,3411123




            1,3411123












            • $begingroup$
              Again, you really need to use different indices for your summations. What you have written doesn't really make sense. I will write it out an an answer since you kind of have the idea.
              $endgroup$
              – Cameron Williams
              Jul 24 '13 at 19:48












            • $begingroup$
              @CameronWilliams: Thanks a lot. About the indices - I get what you're saying but I'm used to this notation. I've added brackets - does it make it clearer?
              $endgroup$
              – catch22
              Jul 24 '13 at 20:10










            • $begingroup$
              I understand but you cannot repeat indices like that because it can lead to erroneous results. What you have done in your fourth term in the series of equalities is effectively discount all of the values of $k > n$ so you have considered only about half of your overall state space for $(n,k)$. If you want to look at your state space of $(n,k)$ like a matrix, you basically only looked at the lower triangular part of the matrix. In this case it worked out because the only values that mattered lied along the diagonal but in general this is not true.
              $endgroup$
              – Cameron Williams
              Jul 24 '13 at 20:17










            • $begingroup$
              @CameronWilliams: How is what I did different from here: en.wikipedia.org/wiki/Cauchy_product ?
              $endgroup$
              – catch22
              Jul 24 '13 at 20:39










            • $begingroup$
              I didn't realize you were using the Cauchy product. Then what you have is fine. But I am going to make an edit to the Wikipedia article because they shouldn't be reusing indices like they are. It is not proper mathematical form.
              $endgroup$
              – Cameron Williams
              Jul 24 '13 at 20:42


















            • $begingroup$
              Again, you really need to use different indices for your summations. What you have written doesn't really make sense. I will write it out an an answer since you kind of have the idea.
              $endgroup$
              – Cameron Williams
              Jul 24 '13 at 19:48












            • $begingroup$
              @CameronWilliams: Thanks a lot. About the indices - I get what you're saying but I'm used to this notation. I've added brackets - does it make it clearer?
              $endgroup$
              – catch22
              Jul 24 '13 at 20:10










            • $begingroup$
              I understand but you cannot repeat indices like that because it can lead to erroneous results. What you have done in your fourth term in the series of equalities is effectively discount all of the values of $k > n$ so you have considered only about half of your overall state space for $(n,k)$. If you want to look at your state space of $(n,k)$ like a matrix, you basically only looked at the lower triangular part of the matrix. In this case it worked out because the only values that mattered lied along the diagonal but in general this is not true.
              $endgroup$
              – Cameron Williams
              Jul 24 '13 at 20:17










            • $begingroup$
              @CameronWilliams: How is what I did different from here: en.wikipedia.org/wiki/Cauchy_product ?
              $endgroup$
              – catch22
              Jul 24 '13 at 20:39










            • $begingroup$
              I didn't realize you were using the Cauchy product. Then what you have is fine. But I am going to make an edit to the Wikipedia article because they shouldn't be reusing indices like they are. It is not proper mathematical form.
              $endgroup$
              – Cameron Williams
              Jul 24 '13 at 20:42
















            $begingroup$
            Again, you really need to use different indices for your summations. What you have written doesn't really make sense. I will write it out an an answer since you kind of have the idea.
            $endgroup$
            – Cameron Williams
            Jul 24 '13 at 19:48






            $begingroup$
            Again, you really need to use different indices for your summations. What you have written doesn't really make sense. I will write it out an an answer since you kind of have the idea.
            $endgroup$
            – Cameron Williams
            Jul 24 '13 at 19:48














            $begingroup$
            @CameronWilliams: Thanks a lot. About the indices - I get what you're saying but I'm used to this notation. I've added brackets - does it make it clearer?
            $endgroup$
            – catch22
            Jul 24 '13 at 20:10




            $begingroup$
            @CameronWilliams: Thanks a lot. About the indices - I get what you're saying but I'm used to this notation. I've added brackets - does it make it clearer?
            $endgroup$
            – catch22
            Jul 24 '13 at 20:10












            $begingroup$
            I understand but you cannot repeat indices like that because it can lead to erroneous results. What you have done in your fourth term in the series of equalities is effectively discount all of the values of $k > n$ so you have considered only about half of your overall state space for $(n,k)$. If you want to look at your state space of $(n,k)$ like a matrix, you basically only looked at the lower triangular part of the matrix. In this case it worked out because the only values that mattered lied along the diagonal but in general this is not true.
            $endgroup$
            – Cameron Williams
            Jul 24 '13 at 20:17




            $begingroup$
            I understand but you cannot repeat indices like that because it can lead to erroneous results. What you have done in your fourth term in the series of equalities is effectively discount all of the values of $k > n$ so you have considered only about half of your overall state space for $(n,k)$. If you want to look at your state space of $(n,k)$ like a matrix, you basically only looked at the lower triangular part of the matrix. In this case it worked out because the only values that mattered lied along the diagonal but in general this is not true.
            $endgroup$
            – Cameron Williams
            Jul 24 '13 at 20:17












            $begingroup$
            @CameronWilliams: How is what I did different from here: en.wikipedia.org/wiki/Cauchy_product ?
            $endgroup$
            – catch22
            Jul 24 '13 at 20:39




            $begingroup$
            @CameronWilliams: How is what I did different from here: en.wikipedia.org/wiki/Cauchy_product ?
            $endgroup$
            – catch22
            Jul 24 '13 at 20:39












            $begingroup$
            I didn't realize you were using the Cauchy product. Then what you have is fine. But I am going to make an edit to the Wikipedia article because they shouldn't be reusing indices like they are. It is not proper mathematical form.
            $endgroup$
            – Cameron Williams
            Jul 24 '13 at 20:42




            $begingroup$
            I didn't realize you were using the Cauchy product. Then what you have is fine. But I am going to make an edit to the Wikipedia article because they shouldn't be reusing indices like they are. It is not proper mathematical form.
            $endgroup$
            – Cameron Williams
            Jul 24 '13 at 20:42


















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