Using Parserval's identity in a Fourier series












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I have the function $$f(x)=begin{cases}1, space |x|leq a \ 0, space a<|x|leq 1/2end{cases}$$
I have calculated the Fourier series of this even function: $$Sf(x)=2a+2sum_{n=0}^infty frac{sin(2pi na)}{pi n}cos(2pi nx)$$
Now I need to calculate $$sum_{n=0}^inftyfrac{sin^2(2pi na)}{n^2}$$
How do I use Parserval's Identity to calculate this last sum? I have tried $$2a=int_{-a}^a1^2=2a^2+sum_{n=0}^inftyfrac{sin^2(2pi na)}{pi^2 n^2}$$ And therefore $$(2a-2a^2)pi^2=sum_{n=0}^inftyfrac{sin^2(2pi na)}{n^2}$$
Is this correct? If not, what is the correct way to do it?










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  • 1




    $begingroup$
    Those sums should probably run from $1$ to $infty$; you don't want a literal $frac00$ term in there, and the zero term $2a$ has a different formula anyway.
    $endgroup$
    – jmerry
    Jan 30 at 0:17










  • $begingroup$
    The leading constant should be part of the first fourier coefficient to have the form $Sf = sum_n c_n cos(2pi nx)$ for which we can apply Parserval's identity.
    $endgroup$
    – Nicholas Parris
    Jan 30 at 0:25










  • $begingroup$
    @NicholasParris How would you plug in the leading constant into the sum? I'm also interested in this problem
    $endgroup$
    – John Keeper
    Feb 8 at 1:55
















0












$begingroup$


I have the function $$f(x)=begin{cases}1, space |x|leq a \ 0, space a<|x|leq 1/2end{cases}$$
I have calculated the Fourier series of this even function: $$Sf(x)=2a+2sum_{n=0}^infty frac{sin(2pi na)}{pi n}cos(2pi nx)$$
Now I need to calculate $$sum_{n=0}^inftyfrac{sin^2(2pi na)}{n^2}$$
How do I use Parserval's Identity to calculate this last sum? I have tried $$2a=int_{-a}^a1^2=2a^2+sum_{n=0}^inftyfrac{sin^2(2pi na)}{pi^2 n^2}$$ And therefore $$(2a-2a^2)pi^2=sum_{n=0}^inftyfrac{sin^2(2pi na)}{n^2}$$
Is this correct? If not, what is the correct way to do it?










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Those sums should probably run from $1$ to $infty$; you don't want a literal $frac00$ term in there, and the zero term $2a$ has a different formula anyway.
    $endgroup$
    – jmerry
    Jan 30 at 0:17










  • $begingroup$
    The leading constant should be part of the first fourier coefficient to have the form $Sf = sum_n c_n cos(2pi nx)$ for which we can apply Parserval's identity.
    $endgroup$
    – Nicholas Parris
    Jan 30 at 0:25










  • $begingroup$
    @NicholasParris How would you plug in the leading constant into the sum? I'm also interested in this problem
    $endgroup$
    – John Keeper
    Feb 8 at 1:55














0












0








0





$begingroup$


I have the function $$f(x)=begin{cases}1, space |x|leq a \ 0, space a<|x|leq 1/2end{cases}$$
I have calculated the Fourier series of this even function: $$Sf(x)=2a+2sum_{n=0}^infty frac{sin(2pi na)}{pi n}cos(2pi nx)$$
Now I need to calculate $$sum_{n=0}^inftyfrac{sin^2(2pi na)}{n^2}$$
How do I use Parserval's Identity to calculate this last sum? I have tried $$2a=int_{-a}^a1^2=2a^2+sum_{n=0}^inftyfrac{sin^2(2pi na)}{pi^2 n^2}$$ And therefore $$(2a-2a^2)pi^2=sum_{n=0}^inftyfrac{sin^2(2pi na)}{n^2}$$
Is this correct? If not, what is the correct way to do it?










share|cite|improve this question











$endgroup$




I have the function $$f(x)=begin{cases}1, space |x|leq a \ 0, space a<|x|leq 1/2end{cases}$$
I have calculated the Fourier series of this even function: $$Sf(x)=2a+2sum_{n=0}^infty frac{sin(2pi na)}{pi n}cos(2pi nx)$$
Now I need to calculate $$sum_{n=0}^inftyfrac{sin^2(2pi na)}{n^2}$$
How do I use Parserval's Identity to calculate this last sum? I have tried $$2a=int_{-a}^a1^2=2a^2+sum_{n=0}^inftyfrac{sin^2(2pi na)}{pi^2 n^2}$$ And therefore $$(2a-2a^2)pi^2=sum_{n=0}^inftyfrac{sin^2(2pi na)}{n^2}$$
Is this correct? If not, what is the correct way to do it?







fourier-series






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share|cite|improve this question













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edited Mar 21 at 12:42









Rócherz

3,0013821




3,0013821










asked Jan 30 at 0:11









codingnightcodingnight

927




927








  • 1




    $begingroup$
    Those sums should probably run from $1$ to $infty$; you don't want a literal $frac00$ term in there, and the zero term $2a$ has a different formula anyway.
    $endgroup$
    – jmerry
    Jan 30 at 0:17










  • $begingroup$
    The leading constant should be part of the first fourier coefficient to have the form $Sf = sum_n c_n cos(2pi nx)$ for which we can apply Parserval's identity.
    $endgroup$
    – Nicholas Parris
    Jan 30 at 0:25










  • $begingroup$
    @NicholasParris How would you plug in the leading constant into the sum? I'm also interested in this problem
    $endgroup$
    – John Keeper
    Feb 8 at 1:55














  • 1




    $begingroup$
    Those sums should probably run from $1$ to $infty$; you don't want a literal $frac00$ term in there, and the zero term $2a$ has a different formula anyway.
    $endgroup$
    – jmerry
    Jan 30 at 0:17










  • $begingroup$
    The leading constant should be part of the first fourier coefficient to have the form $Sf = sum_n c_n cos(2pi nx)$ for which we can apply Parserval's identity.
    $endgroup$
    – Nicholas Parris
    Jan 30 at 0:25










  • $begingroup$
    @NicholasParris How would you plug in the leading constant into the sum? I'm also interested in this problem
    $endgroup$
    – John Keeper
    Feb 8 at 1:55








1




1




$begingroup$
Those sums should probably run from $1$ to $infty$; you don't want a literal $frac00$ term in there, and the zero term $2a$ has a different formula anyway.
$endgroup$
– jmerry
Jan 30 at 0:17




$begingroup$
Those sums should probably run from $1$ to $infty$; you don't want a literal $frac00$ term in there, and the zero term $2a$ has a different formula anyway.
$endgroup$
– jmerry
Jan 30 at 0:17












$begingroup$
The leading constant should be part of the first fourier coefficient to have the form $Sf = sum_n c_n cos(2pi nx)$ for which we can apply Parserval's identity.
$endgroup$
– Nicholas Parris
Jan 30 at 0:25




$begingroup$
The leading constant should be part of the first fourier coefficient to have the form $Sf = sum_n c_n cos(2pi nx)$ for which we can apply Parserval's identity.
$endgroup$
– Nicholas Parris
Jan 30 at 0:25












$begingroup$
@NicholasParris How would you plug in the leading constant into the sum? I'm also interested in this problem
$endgroup$
– John Keeper
Feb 8 at 1:55




$begingroup$
@NicholasParris How would you plug in the leading constant into the sum? I'm also interested in this problem
$endgroup$
– John Keeper
Feb 8 at 1:55










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