Mixing
Using Parserval's identity in a Fourier series
$begingroup$
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?
fourier-series
edited Mar 21 at 12:42
Rócherz
3,0013821
asked Jan 30 at 0:11
codingnightcodingnight
927
$endgroup$
add a comment |
$begingroup$
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?
fourier-series
edited Mar 21 at 12:42
Rócherz
3,0013821
asked Jan 30 at 0:11
codingnightcodingnight
927
$endgroup$
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
add a comment |
$begingroup$
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?
fourier-series
edited Mar 21 at 12:42
Rócherz
3,0013821
asked Jan 30 at 0:11
codingnightcodingnight
927
$endgroup$
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
fourier-series
edited Mar 21 at 12:42
Rócherz
3,0013821
asked Jan 30 at 0:11
codingnightcodingnight
927
edited Mar 21 at 12:42
Rócherz
3,0013821
asked Jan 30 at 0:11
codingnightcodingnight
927
edited Mar 21 at 12:42
Rócherz
3,0013821
edited Mar 21 at 12:42
Rócherz
3,0013821
edited Mar 21 at 12:42
Rócherz
3,0013821
3,0013821
asked Jan 30 at 0:11
codingnightcodingnight
927
asked Jan 30 at 0:11
codingnightcodingnight
927
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
add a comment |
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
add a comment |
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$begingroup$
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