Calculating infinite sum using Parseval's theorem The 2019 Stack Overflow Developer Survey Results Are InI have a problem with my proof of the Basel problem with Fourier seriesInner product of function of period $2pi$ with exponentialFourier series to calculate infinite seriesComplex Fourier series for $f(x)=cos(3x)$Evaluate $sum_n=-infty^infty hatg(n)$Proving linearity of an operator using boundedness.Convergence of the series $sum_n|langle f,e_nrangle|$Proof of Parseval's TheoremFourier series of $psi (x) = cos(2x)$ using the basis $e_n = dfrace^inxsqrt2pi$Avg Area of triangle
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Calculating infinite sum using Parseval's theorem
The 2019 Stack Overflow Developer Survey Results Are InI have a problem with my proof of the Basel problem with Fourier seriesInner product of function of period $2pi$ with exponentialFourier series to calculate infinite seriesComplex Fourier series for $f(x)=cos(3x)$Evaluate $sum_n=-infty^infty hatg(n)$Proving linearity of an operator using boundedness.Convergence of the series $sum_n|langle f,e_nrangle|$Proof of Parseval's TheoremFourier series of $psi (x) = cos(2x)$ using the basis $e_n = dfrace^inxsqrt2pi$Avg Area of triangle
$begingroup$
For$alpha in mathbbR backslash mathbbZ$, consider the fnunction $[0,2pi] to mathbbC : x mapsto fracpisin pi alpha e^i(pi - x)alpha$, and prove that $sum_n=-infty^infty frac1(n+alpha)^2 = fracpi^2(sin pi alpha)^2$. Recall that $x mapsto (2pi)^-1/2 e^inx : n in mathbbZ$ is an orthonormal basis for $L^2_mathbbC[0,2pi]$
My Attempt:
We use Parseval's Theorem, the fact that $|f|^2 = sum_n in mathbbZ |langle f,e_n rangle|^2$, with the inner product norm on $L^2_mathbbC[0,2pi]$. Firstly, we compute the left-hand side. We have
beginalign*
|f|^2 & = int_0^2pi f(x)^2 dx = fracpi^2(sin pi alpha)^2 int_0^2pi e^2i(pi - x)alpha dx \
& = fracpi^2(sin pi alpha)^2 e^2ipialpha int_0^2pi e^-2ix alpha dx
= fracpi^2(sin pi alpha)^2 e^2ipialpha left[ -frac12ialphae^-2ix alpha right]_0^2pi \
& = fracpi^2(sin pi alpha)^2 e^2ipialpha left( -frac12ialphae^-4ipi alpha + frac12ialpha right) =
fracpi^2(sin pi alpha)^2 e^2ipialpha left( frac1 - e^-4ipi alpha2ialpha right) \
& = fracpi^2(sin pi alpha)^2 left( frace^2ipialpha - e^-2ipi alpha2ialpha right)
= fracpi^2(sin pi alpha)^2 left( fracsin (2pi alpha)alpha right)
= frac2 pi^2 cos (pi alpha)sin (pi alpha) alpha.
endalign*
Here we use the trigonometric identities
$$
sin beta = frace^ibeta - e^-ibeta2i, textrm and sin(2beta) = 2sin(beta)cos(beta).
$$
On the left-hand side we have
beginalign*
langle f,e_n rangle & = int_0^2pi frac1sqrt2pi e^inx fracpisin pi alpha e^i(pi - x)alpha dx = fracpisin (pi alpha) sqrt2pi int_0^2pi e^inx + i(pi - x)alpha dx \
& = fracpisin (pi alpha) sqrt2pi e^i pi alpha int_0^2pi e^(in- ialpha)x dx
= fracpisin (pi alpha) sqrt2pi e^i pi alpha left[ frac1in - ialpha e^(in- ialpha)x right]_0^2pi \
& = fracpisin (pi alpha) sqrt2pi e^i pi alpha left( frac1in - ialpha e^(in- ialpha)2pi - frac1in - ialpha right) \
& = fracpisin (pi alpha) sqrt2pi e^i pi alpha left( frace^i(n-alpha)2pi - 1i(n - alpha) right) = fracpisin (pi alpha) sqrt2pi left( frac-(e^i pi alpha - e^-ipialpha)i(n - alpha) right) \
& = fracpisin (pi alpha) sqrt2pi frac-2 sin(pi alpha)n - alpha
= frac-2pi(n-alpha)sqrt2pi
endalign*
Hence, we have that
$$
|langle f,e_n rangle|^2 = frac2pi(n-alpha)^2.
$$
This is what I tried, things don't seem to add up to the desired outcome and I can't find any mistakes. Anyone has suggestions? I find the computations I made rather laborious, perhaps there are some tricks to do this simpler? I'm not so familiar with trigonometry and complex integration. Thanks in advance.
integration functional-analysis analysis fourier-series complex-integration
asked Mar 23 at 19:06
SigurdSigurd
555212
$endgroup$
|
show 5 more comments
$begingroup$
For$alpha in mathbbR backslash mathbbZ$, consider the fnunction $[0,2pi] to mathbbC : x mapsto fracpisin pi alpha e^i(pi - x)alpha$, and prove that $sum_n=-infty^infty frac1(n+alpha)^2 = fracpi^2(sin pi alpha)^2$. Recall that $x mapsto (2pi)^-1/2 e^inx : n in mathbbZ$ is an orthonormal basis for $L^2_mathbbC[0,2pi]$
My Attempt:
We use Parseval's Theorem, the fact that $|f|^2 = sum_n in mathbbZ |langle f,e_n rangle|^2$, with the inner product norm on $L^2_mathbbC[0,2pi]$. Firstly, we compute the left-hand side. We have
beginalign*
|f|^2 & = int_0^2pi f(x)^2 dx = fracpi^2(sin pi alpha)^2 int_0^2pi e^2i(pi - x)alpha dx \
& = fracpi^2(sin pi alpha)^2 e^2ipialpha int_0^2pi e^-2ix alpha dx
= fracpi^2(sin pi alpha)^2 e^2ipialpha left[ -frac12ialphae^-2ix alpha right]_0^2pi \
& = fracpi^2(sin pi alpha)^2 e^2ipialpha left( -frac12ialphae^-4ipi alpha + frac12ialpha right) =
fracpi^2(sin pi alpha)^2 e^2ipialpha left( frac1 - e^-4ipi alpha2ialpha right) \
& = fracpi^2(sin pi alpha)^2 left( frace^2ipialpha - e^-2ipi alpha2ialpha right)
= fracpi^2(sin pi alpha)^2 left( fracsin (2pi alpha)alpha right)
= frac2 pi^2 cos (pi alpha)sin (pi alpha) alpha.
endalign*
Here we use the trigonometric identities
$$
sin beta = frace^ibeta - e^-ibeta2i, textrm and sin(2beta) = 2sin(beta)cos(beta).
$$
On the left-hand side we have
beginalign*
langle f,e_n rangle & = int_0^2pi frac1sqrt2pi e^inx fracpisin pi alpha e^i(pi - x)alpha dx = fracpisin (pi alpha) sqrt2pi int_0^2pi e^inx + i(pi - x)alpha dx \
& = fracpisin (pi alpha) sqrt2pi e^i pi alpha int_0^2pi e^(in- ialpha)x dx
= fracpisin (pi alpha) sqrt2pi e^i pi alpha left[ frac1in - ialpha e^(in- ialpha)x right]_0^2pi \
& = fracpisin (pi alpha) sqrt2pi e^i pi alpha left( frac1in - ialpha e^(in- ialpha)2pi - frac1in - ialpha right) \
& = fracpisin (pi alpha) sqrt2pi e^i pi alpha left( frace^i(n-alpha)2pi - 1i(n - alpha) right) = fracpisin (pi alpha) sqrt2pi left( frac-(e^i pi alpha - e^-ipialpha)i(n - alpha) right) \
& = fracpisin (pi alpha) sqrt2pi frac-2 sin(pi alpha)n - alpha
= frac-2pi(n-alpha)sqrt2pi
endalign*
Hence, we have that
$$
|langle f,e_n rangle|^2 = frac2pi(n-alpha)^2.
$$
This is what I tried, things don't seem to add up to the desired outcome and I can't find any mistakes. Anyone has suggestions? I find the computations I made rather laborious, perhaps there are some tricks to do this simpler? I'm not so familiar with trigonometry and complex integration. Thanks in advance.
integration functional-analysis analysis fourier-series complex-integration
asked Mar 23 at 19:06
SigurdSigurd
555212
$endgroup$
$begingroup$
I believe the norm of $f$ should be an integral of $|f|^2$, and not merely $f^2$
$endgroup$
– bobcliffe
Mar 23 at 19:11
$begingroup$
Isn't the inner product on $L^2[a,b]$ defined as $langle f,g rangle = int_a^b f(x)g(x)dx$? Also, what would be $|e^i(pi-x)alpha|^2$?
$endgroup$
– Sigurd
Mar 23 at 19:21
$begingroup$
That is the inner product when the functions are real-valued. If they are complex-valued, it is $langle f,grangle =int_a^b f(x)overlineg(x), dx$.
$endgroup$
– Minus One-Twelfth
Mar 23 at 19:24
$begingroup$
Oh yeah you're right. How to compute absolute value of complex e-powers though?
$endgroup$
– Sigurd
Mar 23 at 19:25
$begingroup$
Use the fact that if $t$ is any real number, then $boxedcolorblueleft$. This can be shown by using Euler's formula ($e^it = cos t +i sin t$) and the identity $cos^2 t + sin^2 t =1$.
$endgroup$
– Minus One-Twelfth
Mar 23 at 19:26
|
show 5 more comments
$begingroup$
For$alpha in mathbbR backslash mathbbZ$, consider the fnunction $[0,2pi] to mathbbC : x mapsto fracpisin pi alpha e^i(pi - x)alpha$, and prove that $sum_n=-infty^infty frac1(n+alpha)^2 = fracpi^2(sin pi alpha)^2$. Recall that $x mapsto (2pi)^-1/2 e^inx : n in mathbbZ$ is an orthonormal basis for $L^2_mathbbC[0,2pi]$
My Attempt:
We use Parseval's Theorem, the fact that $|f|^2 = sum_n in mathbbZ |langle f,e_n rangle|^2$, with the inner product norm on $L^2_mathbbC[0,2pi]$. Firstly, we compute the left-hand side. We have
beginalign*
|f|^2 & = int_0^2pi f(x)^2 dx = fracpi^2(sin pi alpha)^2 int_0^2pi e^2i(pi - x)alpha dx \
& = fracpi^2(sin pi alpha)^2 e^2ipialpha int_0^2pi e^-2ix alpha dx
= fracpi^2(sin pi alpha)^2 e^2ipialpha left[ -frac12ialphae^-2ix alpha right]_0^2pi \
& = fracpi^2(sin pi alpha)^2 e^2ipialpha left( -frac12ialphae^-4ipi alpha + frac12ialpha right) =
fracpi^2(sin pi alpha)^2 e^2ipialpha left( frac1 - e^-4ipi alpha2ialpha right) \
& = fracpi^2(sin pi alpha)^2 left( frace^2ipialpha - e^-2ipi alpha2ialpha right)
= fracpi^2(sin pi alpha)^2 left( fracsin (2pi alpha)alpha right)
= frac2 pi^2 cos (pi alpha)sin (pi alpha) alpha.
endalign*
Here we use the trigonometric identities
$$
sin beta = frace^ibeta - e^-ibeta2i, textrm and sin(2beta) = 2sin(beta)cos(beta).
$$
On the left-hand side we have
beginalign*
langle f,e_n rangle & = int_0^2pi frac1sqrt2pi e^inx fracpisin pi alpha e^i(pi - x)alpha dx = fracpisin (pi alpha) sqrt2pi int_0^2pi e^inx + i(pi - x)alpha dx \
& = fracpisin (pi alpha) sqrt2pi e^i pi alpha int_0^2pi e^(in- ialpha)x dx
= fracpisin (pi alpha) sqrt2pi e^i pi alpha left[ frac1in - ialpha e^(in- ialpha)x right]_0^2pi \
& = fracpisin (pi alpha) sqrt2pi e^i pi alpha left( frac1in - ialpha e^(in- ialpha)2pi - frac1in - ialpha right) \
& = fracpisin (pi alpha) sqrt2pi e^i pi alpha left( frace^i(n-alpha)2pi - 1i(n - alpha) right) = fracpisin (pi alpha) sqrt2pi left( frac-(e^i pi alpha - e^-ipialpha)i(n - alpha) right) \
& = fracpisin (pi alpha) sqrt2pi frac-2 sin(pi alpha)n - alpha
= frac-2pi(n-alpha)sqrt2pi
endalign*
Hence, we have that
$$
|langle f,e_n rangle|^2 = frac2pi(n-alpha)^2.
$$
This is what I tried, things don't seem to add up to the desired outcome and I can't find any mistakes. Anyone has suggestions? I find the computations I made rather laborious, perhaps there are some tricks to do this simpler? I'm not so familiar with trigonometry and complex integration. Thanks in advance.
integration functional-analysis analysis fourier-series complex-integration
asked Mar 23 at 19:06
SigurdSigurd
555212
$endgroup$
For$alpha in mathbbR backslash mathbbZ$, consider the fnunction $[0,2pi] to mathbbC : x mapsto fracpisin pi alpha e^i(pi - x)alpha$, and prove that $sum_n=-infty^infty frac1(n+alpha)^2 = fracpi^2(sin pi alpha)^2$. Recall that $x mapsto (2pi)^-1/2 e^inx : n in mathbbZ$ is an orthonormal basis for $L^2_mathbbC[0,2pi]$
My Attempt:
We use Parseval's Theorem, the fact that $|f|^2 = sum_n in mathbbZ |langle f,e_n rangle|^2$, with the inner product norm on $L^2_mathbbC[0,2pi]$. Firstly, we compute the left-hand side. We have
beginalign*
|f|^2 & = int_0^2pi f(x)^2 dx = fracpi^2(sin pi alpha)^2 int_0^2pi e^2i(pi - x)alpha dx \
& = fracpi^2(sin pi alpha)^2 e^2ipialpha int_0^2pi e^-2ix alpha dx
= fracpi^2(sin pi alpha)^2 e^2ipialpha left[ -frac12ialphae^-2ix alpha right]_0^2pi \
& = fracpi^2(sin pi alpha)^2 e^2ipialpha left( -frac12ialphae^-4ipi alpha + frac12ialpha right) =
fracpi^2(sin pi alpha)^2 e^2ipialpha left( frac1 - e^-4ipi alpha2ialpha right) \
& = fracpi^2(sin pi alpha)^2 left( frace^2ipialpha - e^-2ipi alpha2ialpha right)
= fracpi^2(sin pi alpha)^2 left( fracsin (2pi alpha)alpha right)
= frac2 pi^2 cos (pi alpha)sin (pi alpha) alpha.
endalign*
Here we use the trigonometric identities
$$
sin beta = frace^ibeta - e^-ibeta2i, textrm and sin(2beta) = 2sin(beta)cos(beta).
$$
On the left-hand side we have
beginalign*
langle f,e_n rangle & = int_0^2pi frac1sqrt2pi e^inx fracpisin pi alpha e^i(pi - x)alpha dx = fracpisin (pi alpha) sqrt2pi int_0^2pi e^inx + i(pi - x)alpha dx \
& = fracpisin (pi alpha) sqrt2pi e^i pi alpha int_0^2pi e^(in- ialpha)x dx
= fracpisin (pi alpha) sqrt2pi e^i pi alpha left[ frac1in - ialpha e^(in- ialpha)x right]_0^2pi \
& = fracpisin (pi alpha) sqrt2pi e^i pi alpha left( frac1in - ialpha e^(in- ialpha)2pi - frac1in - ialpha right) \
& = fracpisin (pi alpha) sqrt2pi e^i pi alpha left( frace^i(n-alpha)2pi - 1i(n - alpha) right) = fracpisin (pi alpha) sqrt2pi left( frac-(e^i pi alpha - e^-ipialpha)i(n - alpha) right) \
& = fracpisin (pi alpha) sqrt2pi frac-2 sin(pi alpha)n - alpha
= frac-2pi(n-alpha)sqrt2pi
endalign*
Hence, we have that
$$
|langle f,e_n rangle|^2 = frac2pi(n-alpha)^2.
$$
This is what I tried, things don't seem to add up to the desired outcome and I can't find any mistakes. Anyone has suggestions? I find the computations I made rather laborious, perhaps there are some tricks to do this simpler? I'm not so familiar with trigonometry and complex integration. Thanks in advance.
integration functional-analysis analysis fourier-series complex-integration
integration functional-analysis analysis fourier-series complex-integration
asked Mar 23 at 19:06
SigurdSigurd
555212
asked Mar 23 at 19:06
SigurdSigurd
555212
asked Mar 23 at 19:06
SigurdSigurd
555212
asked Mar 23 at 19:06
SigurdSigurd
555212
asked Mar 23 at 19:06
SigurdSigurd
555212
555212
$begingroup$
I believe the norm of $f$ should be an integral of $|f|^2$, and not merely $f^2$
$endgroup$
– bobcliffe
Mar 23 at 19:11
$begingroup$
Isn't the inner product on $L^2[a,b]$ defined as $langle f,g rangle = int_a^b f(x)g(x)dx$? Also, what would be $|e^i(pi-x)alpha|^2$?
$endgroup$
– Sigurd
Mar 23 at 19:21
$begingroup$
That is the inner product when the functions are real-valued. If they are complex-valued, it is $langle f,grangle =int_a^b f(x)overlineg(x), dx$.
$endgroup$
– Minus One-Twelfth
Mar 23 at 19:24
$begingroup$
Oh yeah you're right. How to compute absolute value of complex e-powers though?
$endgroup$
– Sigurd
Mar 23 at 19:25
$begingroup$
Use the fact that if $t$ is any real number, then $boxedcolorblueleft$. This can be shown by using Euler's formula ($e^it = cos t +i sin t$) and the identity $cos^2 t + sin^2 t =1$.
$endgroup$
– Minus One-Twelfth
Mar 23 at 19:26
|
show 5 more comments
$begingroup$
I believe the norm of $f$ should be an integral of $|f|^2$, and not merely $f^2$
$endgroup$
– bobcliffe
Mar 23 at 19:11
$begingroup$
Isn't the inner product on $L^2[a,b]$ defined as $langle f,g rangle = int_a^b f(x)g(x)dx$? Also, what would be $|e^i(pi-x)alpha|^2$?
$endgroup$
– Sigurd
Mar 23 at 19:21
$begingroup$
That is the inner product when the functions are real-valued. If they are complex-valued, it is $langle f,grangle =int_a^b f(x)overlineg(x), dx$.
$endgroup$
– Minus One-Twelfth
Mar 23 at 19:24
$begingroup$
Oh yeah you're right. How to compute absolute value of complex e-powers though?
$endgroup$
– Sigurd
Mar 23 at 19:25
$begingroup$
Use the fact that if $t$ is any real number, then $boxedcolorblueleft$. This can be shown by using Euler's formula ($e^it = cos t +i sin t$) and the identity $cos^2 t + sin^2 t =1$.
$endgroup$
– Minus One-Twelfth
Mar 23 at 19:26
$begingroup$
I believe the norm of $f$ should be an integral of $|f|^2$, and not merely $f^2$
$endgroup$
– bobcliffe
Mar 23 at 19:11
$begingroup$
I believe the norm of $f$ should be an integral of $|f|^2$, and not merely $f^2$
$endgroup$
– bobcliffe
Mar 23 at 19:11
$begingroup$
Isn't the inner product on $L^2[a,b]$ defined as $langle f,g rangle = int_a^b f(x)g(x)dx$? Also, what would be $|e^i(pi-x)alpha|^2$?
$endgroup$
– Sigurd
Mar 23 at 19:21
$begingroup$
Isn't the inner product on $L^2[a,b]$ defined as $langle f,g rangle = int_a^b f(x)g(x)dx$? Also, what would be $|e^i(pi-x)alpha|^2$?
$endgroup$
– Sigurd
Mar 23 at 19:21
$begingroup$
That is the inner product when the functions are real-valued. If they are complex-valued, it is $langle f,grangle =int_a^b f(x)overlineg(x), dx$.
$endgroup$
– Minus One-Twelfth
Mar 23 at 19:24
$begingroup$
That is the inner product when the functions are real-valued. If they are complex-valued, it is $langle f,grangle =int_a^b f(x)overlineg(x), dx$.
$endgroup$
– Minus One-Twelfth
Mar 23 at 19:24
$begingroup$
Oh yeah you're right. How to compute absolute value of complex e-powers though?
$endgroup$
– Sigurd
Mar 23 at 19:25
$begingroup$
Oh yeah you're right. How to compute absolute value of complex e-powers though?
$endgroup$
– Sigurd
Mar 23 at 19:25
$begingroup$
Use the fact that if $t$ is any real number, then $boxedcolorblueleft$. This can be shown by using Euler's formula ($e^it = cos t +i sin t$) and the identity $cos^2 t + sin^2 t =1$.
$endgroup$
– Minus One-Twelfth
Mar 23 at 19:26
$begingroup$
Use the fact that if $t$ is any real number, then $boxedcolorblueleft$. This can be shown by using Euler's formula ($e^it = cos t +i sin t$) and the identity $cos^2 t + sin^2 t =1$.
$endgroup$
– Minus One-Twelfth
Mar 23 at 19:26
|
show 5 more comments
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$begingroup$
I believe the norm of $f$ should be an integral of $|f|^2$, and not merely $f^2$
$endgroup$
– bobcliffe
Mar 23 at 19:11
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Isn't the inner product on $L^2[a,b]$ defined as $langle f,g rangle = int_a^b f(x)g(x)dx$? Also, what would be $|e^i(pi-x)alpha|^2$?
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– Sigurd
Mar 23 at 19:21
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That is the inner product when the functions are real-valued. If they are complex-valued, it is $langle f,grangle =int_a^b f(x)overlineg(x), dx$.
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– Minus One-Twelfth
Mar 23 at 19:24
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Oh yeah you're right. How to compute absolute value of complex e-powers though?
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– Sigurd
Mar 23 at 19:25
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Use the fact that if $t$ is any real number, then $boxedcolorblueleft$. This can be shown by using Euler's formula ($e^it = cos t +i sin t$) and the identity $cos^2 t + sin^2 t =1$.
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– Minus One-Twelfth
Mar 23 at 19:26