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Calculating Fourier series of $cos^2(t)$ gives unexpected result
Use WolframAlpha to compute the real Fourier series of a functionIs it possible for cosine functions to have Fourier sine series expressions or sine functions to have Fourier cosine series expressions?Finding Fourier series of $sin^2 x$ (STILL not clear - read comments)The coefficients of the Fourier series of the product of two real valued functionsFourier Series; odd and even half-range expansionAfter calculating Fourier series coefficients for $x(t)=2 cos(4t) + 4 sin(10t)$, why am I getting all zeroes for all coefficients?Equivalence of two definitions of Fourier Series(Trigonometric) Fourier series of sawtooth integralEven and odd functions with Fourier seriesFourier series of a fourier series.
$begingroup$
As I understand it:
$cos^2(t)$ is even because it is a product of two even functions $cos(t)$.
The Fourier series and Fourier cosine series of an even function is the same link.
So in the fourier series expansion $cos^2(t)=a_0 over 2+sum_n=1^infty(a_ncos(n omega t) + b_n sin(n omega t))$, I expect $a_nneq 0$ and $b_n=0$.
I try to get the the coefficients with wolfram alpha like this:
$a_n$ with FourierCosCoefficient[$cos(t)^2,t,n$] gives zero link
$b_n$ with FourierSinCoefficient[$cos(t)^2,t,n$] gives non-zero link
Which of my assumptions are wrong?
(I also got $a_n=0$ when calculating by hand so the question is not primarily about wolfram alpha but about the relation between Fourier series, Fourier cosine series, and even functions)
fourier-series wolfram-alpha
edited yesterday
Bernard
122k741116
asked yesterday
aardvarkaardvark
1
New contributor
aardvark is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
|
show 2 more comments
$begingroup$
As I understand it:
$cos^2(t)$ is even because it is a product of two even functions $cos(t)$.
The Fourier series and Fourier cosine series of an even function is the same link.
So in the fourier series expansion $cos^2(t)=a_0 over 2+sum_n=1^infty(a_ncos(n omega t) + b_n sin(n omega t))$, I expect $a_nneq 0$ and $b_n=0$.
I try to get the the coefficients with wolfram alpha like this:
$a_n$ with FourierCosCoefficient[$cos(t)^2,t,n$] gives zero link
$b_n$ with FourierSinCoefficient[$cos(t)^2,t,n$] gives non-zero link
Which of my assumptions are wrong?
(I also got $a_n=0$ when calculating by hand so the question is not primarily about wolfram alpha but about the relation between Fourier series, Fourier cosine series, and even functions)
fourier-series wolfram-alpha
edited yesterday
Bernard
122k741116
asked yesterday
aardvarkaardvark
1
New contributor
aardvark is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Looks to me that already $a_0 = frac12 ne 0$... Can you detail your computations ?
$endgroup$
– Gâteau-Gallois
yesterday
1
$begingroup$
I'm not familiar enough with WolframAlpha to say, but it seems to me this is likely to be a misunderstanding of what the function does. For example, where's the $omega$ in that call? Maybe it's trying to project the function on a different period? As it stands, $cos^2(t) = 1/2+1/2cos(2t)$, so that's all your Fourier series.
$endgroup$
– Okarin
yesterday
$begingroup$
The correct answer is $a_0=a_2=frac 1 2$ and all other coefficients $0$.
$endgroup$
– Kavi Rama Murthy
yesterday
$begingroup$
Surprisingly Mathematica gives the same answer. OTOH to the integralFourierCosCoefficient[(1+Cos[2t])/2,t,n]it gives the expected(DiscreteDelta[n-2]+2 DiscreteDelta[n])/2
$endgroup$
– Jyrki Lahtonen
yesterday
1
$begingroup$
FWIW, I asked this at Mathematica.SE.
$endgroup$
– Jyrki Lahtonen
yesterday
|
show 2 more comments
$begingroup$
As I understand it:
$cos^2(t)$ is even because it is a product of two even functions $cos(t)$.
The Fourier series and Fourier cosine series of an even function is the same link.
So in the fourier series expansion $cos^2(t)=a_0 over 2+sum_n=1^infty(a_ncos(n omega t) + b_n sin(n omega t))$, I expect $a_nneq 0$ and $b_n=0$.
I try to get the the coefficients with wolfram alpha like this:
$a_n$ with FourierCosCoefficient[$cos(t)^2,t,n$] gives zero link
$b_n$ with FourierSinCoefficient[$cos(t)^2,t,n$] gives non-zero link
Which of my assumptions are wrong?
(I also got $a_n=0$ when calculating by hand so the question is not primarily about wolfram alpha but about the relation between Fourier series, Fourier cosine series, and even functions)
fourier-series wolfram-alpha
edited yesterday
Bernard
122k741116
asked yesterday
aardvarkaardvark
1
New contributor
aardvark is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
As I understand it:
$cos^2(t)$ is even because it is a product of two even functions $cos(t)$.
The Fourier series and Fourier cosine series of an even function is the same link.
So in the fourier series expansion $cos^2(t)=a_0 over 2+sum_n=1^infty(a_ncos(n omega t) + b_n sin(n omega t))$, I expect $a_nneq 0$ and $b_n=0$.
I try to get the the coefficients with wolfram alpha like this:
$a_n$ with FourierCosCoefficient[$cos(t)^2,t,n$] gives zero link
$b_n$ with FourierSinCoefficient[$cos(t)^2,t,n$] gives non-zero link
Which of my assumptions are wrong?
(I also got $a_n=0$ when calculating by hand so the question is not primarily about wolfram alpha but about the relation between Fourier series, Fourier cosine series, and even functions)
fourier-series wolfram-alpha
fourier-series wolfram-alpha
edited yesterday
Bernard
122k741116
asked yesterday
aardvarkaardvark
1
New contributor
aardvark is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Bernard
122k741116
asked yesterday
aardvarkaardvark
1
New contributor
aardvark is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Bernard
122k741116
edited yesterday
Bernard
122k741116
edited yesterday
Bernard
122k741116
122k741116
asked yesterday
aardvarkaardvark
1
New contributor
aardvark is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
aardvarkaardvark
1
asked yesterday
aardvarkaardvark
1
1
New contributor
aardvark is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
aardvark is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
aardvark is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
Looks to me that already $a_0 = frac12 ne 0$... Can you detail your computations ?
$endgroup$
– Gâteau-Gallois
yesterday
1
$begingroup$
I'm not familiar enough with WolframAlpha to say, but it seems to me this is likely to be a misunderstanding of what the function does. For example, where's the $omega$ in that call? Maybe it's trying to project the function on a different period? As it stands, $cos^2(t) = 1/2+1/2cos(2t)$, so that's all your Fourier series.
$endgroup$
– Okarin
yesterday
$begingroup$
The correct answer is $a_0=a_2=frac 1 2$ and all other coefficients $0$.
$endgroup$
– Kavi Rama Murthy
yesterday
$begingroup$
Surprisingly Mathematica gives the same answer. OTOH to the integralFourierCosCoefficient[(1+Cos[2t])/2,t,n]it gives the expected(DiscreteDelta[n-2]+2 DiscreteDelta[n])/2
$endgroup$
– Jyrki Lahtonen
yesterday
1
$begingroup$
FWIW, I asked this at Mathematica.SE.
$endgroup$
– Jyrki Lahtonen
yesterday
|
show 2 more comments
$begingroup$
Looks to me that already $a_0 = frac12 ne 0$... Can you detail your computations ?
$endgroup$
– Gâteau-Gallois
yesterday
1
$begingroup$
I'm not familiar enough with WolframAlpha to say, but it seems to me this is likely to be a misunderstanding of what the function does. For example, where's the $omega$ in that call? Maybe it's trying to project the function on a different period? As it stands, $cos^2(t) = 1/2+1/2cos(2t)$, so that's all your Fourier series.
$endgroup$
– Okarin
yesterday
$begingroup$
The correct answer is $a_0=a_2=frac 1 2$ and all other coefficients $0$.
$endgroup$
– Kavi Rama Murthy
yesterday
$begingroup$
Surprisingly Mathematica gives the same answer. OTOH to the integralFourierCosCoefficient[(1+Cos[2t])/2,t,n]it gives the expected(DiscreteDelta[n-2]+2 DiscreteDelta[n])/2
$endgroup$
– Jyrki Lahtonen
yesterday
1
$begingroup$
FWIW, I asked this at Mathematica.SE.
$endgroup$
– Jyrki Lahtonen
yesterday
$begingroup$
Looks to me that already $a_0 = frac12 ne 0$... Can you detail your computations ?
$endgroup$
– Gâteau-Gallois
yesterday
$begingroup$
Looks to me that already $a_0 = frac12 ne 0$... Can you detail your computations ?
$endgroup$
– Gâteau-Gallois
yesterday
1
1
$begingroup$
I'm not familiar enough with WolframAlpha to say, but it seems to me this is likely to be a misunderstanding of what the function does. For example, where's the $omega$ in that call? Maybe it's trying to project the function on a different period? As it stands, $cos^2(t) = 1/2+1/2cos(2t)$, so that's all your Fourier series.
$endgroup$
– Okarin
yesterday
$begingroup$
I'm not familiar enough with WolframAlpha to say, but it seems to me this is likely to be a misunderstanding of what the function does. For example, where's the $omega$ in that call? Maybe it's trying to project the function on a different period? As it stands, $cos^2(t) = 1/2+1/2cos(2t)$, so that's all your Fourier series.
$endgroup$
– Okarin
yesterday
$begingroup$
The correct answer is $a_0=a_2=frac 1 2$ and all other coefficients $0$.
$endgroup$
– Kavi Rama Murthy
yesterday
$begingroup$
The correct answer is $a_0=a_2=frac 1 2$ and all other coefficients $0$.
$endgroup$
– Kavi Rama Murthy
yesterday
$begingroup$
Surprisingly Mathematica gives the same answer. OTOH to the integral
FourierCosCoefficient[(1+Cos[2t])/2,t,n]it gives the expected (DiscreteDelta[n-2]+2 DiscreteDelta[n])/2$endgroup$
– Jyrki Lahtonen
yesterday
$begingroup$
Surprisingly Mathematica gives the same answer. OTOH to the integral
FourierCosCoefficient[(1+Cos[2t])/2,t,n]it gives the expected (DiscreteDelta[n-2]+2 DiscreteDelta[n])/2$endgroup$
– Jyrki Lahtonen
yesterday
1
1
$begingroup$
FWIW, I asked this at Mathematica.SE.
$endgroup$
– Jyrki Lahtonen
yesterday
$begingroup$
FWIW, I asked this at Mathematica.SE.
$endgroup$
– Jyrki Lahtonen
yesterday
|
show 2 more comments
3 Answers
3
active
oldest
votes
$begingroup$
Just read the documentation:
https://reference.wolfram.com/language/ref/FourierSinCoefficient.html
The command FourierSinCoefficient does not compute the coefficients $b_n$ in the full Fourier series, but the coefficients in the Fourier sine series (= the full Fourier series of the odd extension of the function in question).
answered yesterday
Hans LundmarkHans Lundmark
35.8k564115
$endgroup$
$begingroup$
Thanks, this shows why I was wrong to assume I could calculate $b_n$ with FourierSinCoefficient. I still think I should be able to calculate $a_n$ with FourierCosCoefficient.
$endgroup$
– aardvark
yesterday
add a comment |
$begingroup$
The simplest way to find the Fourier series of this function is to write it as $frac 1+cos(2t) 2$. This expression is in fact the Fourier series!.
answered yesterday
Kavi Rama MurthyKavi Rama Murthy
66k42867
$endgroup$
2
$begingroup$
I'm sure we all know this. The question is why WolframAlpha gives a different answer.
$endgroup$
– Jyrki Lahtonen
yesterday
add a comment |
$begingroup$
It is clear that, for each $ninmathbb N$,$$int_-pi^picos^2(t)sin(nt),mathrm dt=0,$$since $tmapstocos^2(t)sin(nt)$ is an odd function.
On the other hand,$$a_0=frac1piint_-pi^picos^2(t),mathrm dt=1neq0.$$It turns out that $a_1=0$, but $a_2=frac12neq0$.
answered yesterday
José Carlos SantosJosé Carlos Santos
166k22132235
$endgroup$
$begingroup$
I'm sure we all know this. The question is why WolframAlpha gives a different answer.
$endgroup$
– Jyrki Lahtonen
yesterday
1
$begingroup$
No, we don't all know this, since the OP claimed to have obtained $a_n=0$ while calculating by hand.
$endgroup$
– José Carlos Santos
yesterday
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Just read the documentation:
https://reference.wolfram.com/language/ref/FourierSinCoefficient.html
The command FourierSinCoefficient does not compute the coefficients $b_n$ in the full Fourier series, but the coefficients in the Fourier sine series (= the full Fourier series of the odd extension of the function in question).
answered yesterday
Hans LundmarkHans Lundmark
35.8k564115
$endgroup$
$begingroup$
Thanks, this shows why I was wrong to assume I could calculate $b_n$ with FourierSinCoefficient. I still think I should be able to calculate $a_n$ with FourierCosCoefficient.
$endgroup$
– aardvark
yesterday
add a comment |
$begingroup$
Just read the documentation:
https://reference.wolfram.com/language/ref/FourierSinCoefficient.html
The command FourierSinCoefficient does not compute the coefficients $b_n$ in the full Fourier series, but the coefficients in the Fourier sine series (= the full Fourier series of the odd extension of the function in question).
answered yesterday
Hans LundmarkHans Lundmark
35.8k564115
$endgroup$
$begingroup$
Thanks, this shows why I was wrong to assume I could calculate $b_n$ with FourierSinCoefficient. I still think I should be able to calculate $a_n$ with FourierCosCoefficient.
$endgroup$
– aardvark
yesterday
add a comment |
$begingroup$
Just read the documentation:
https://reference.wolfram.com/language/ref/FourierSinCoefficient.html
The command FourierSinCoefficient does not compute the coefficients $b_n$ in the full Fourier series, but the coefficients in the Fourier sine series (= the full Fourier series of the odd extension of the function in question).
answered yesterday
Hans LundmarkHans Lundmark
35.8k564115
$endgroup$
Just read the documentation:
https://reference.wolfram.com/language/ref/FourierSinCoefficient.html
The command FourierSinCoefficient does not compute the coefficients $b_n$ in the full Fourier series, but the coefficients in the Fourier sine series (= the full Fourier series of the odd extension of the function in question).
answered yesterday
Hans LundmarkHans Lundmark
35.8k564115
answered yesterday
Hans LundmarkHans Lundmark
35.8k564115
answered yesterday
Hans LundmarkHans Lundmark
35.8k564115
answered yesterday
Hans LundmarkHans Lundmark
35.8k564115
35.8k564115
$begingroup$
Thanks, this shows why I was wrong to assume I could calculate $b_n$ with FourierSinCoefficient. I still think I should be able to calculate $a_n$ with FourierCosCoefficient.
$endgroup$
– aardvark
yesterday
add a comment |
$begingroup$
Thanks, this shows why I was wrong to assume I could calculate $b_n$ with FourierSinCoefficient. I still think I should be able to calculate $a_n$ with FourierCosCoefficient.
$endgroup$
– aardvark
yesterday
$begingroup$
Thanks, this shows why I was wrong to assume I could calculate $b_n$ with FourierSinCoefficient. I still think I should be able to calculate $a_n$ with FourierCosCoefficient.
$endgroup$
– aardvark
yesterday
$begingroup$
Thanks, this shows why I was wrong to assume I could calculate $b_n$ with FourierSinCoefficient. I still think I should be able to calculate $a_n$ with FourierCosCoefficient.
$endgroup$
– aardvark
yesterday
add a comment |
$begingroup$
The simplest way to find the Fourier series of this function is to write it as $frac 1+cos(2t) 2$. This expression is in fact the Fourier series!.
answered yesterday
Kavi Rama MurthyKavi Rama Murthy
66k42867
$endgroup$
2
$begingroup$
I'm sure we all know this. The question is why WolframAlpha gives a different answer.
$endgroup$
– Jyrki Lahtonen
yesterday
add a comment |
$begingroup$
The simplest way to find the Fourier series of this function is to write it as $frac 1+cos(2t) 2$. This expression is in fact the Fourier series!.
answered yesterday
Kavi Rama MurthyKavi Rama Murthy
66k42867
$endgroup$
2
$begingroup$
I'm sure we all know this. The question is why WolframAlpha gives a different answer.
$endgroup$
– Jyrki Lahtonen
yesterday
add a comment |
$begingroup$
The simplest way to find the Fourier series of this function is to write it as $frac 1+cos(2t) 2$. This expression is in fact the Fourier series!.
answered yesterday
Kavi Rama MurthyKavi Rama Murthy
66k42867
$endgroup$
The simplest way to find the Fourier series of this function is to write it as $frac 1+cos(2t) 2$. This expression is in fact the Fourier series!.
answered yesterday
Kavi Rama MurthyKavi Rama Murthy
66k42867
answered yesterday
Kavi Rama MurthyKavi Rama Murthy
66k42867
answered yesterday
Kavi Rama MurthyKavi Rama Murthy
66k42867
answered yesterday
Kavi Rama MurthyKavi Rama Murthy
66k42867
66k42867
2
$begingroup$
I'm sure we all know this. The question is why WolframAlpha gives a different answer.
$endgroup$
– Jyrki Lahtonen
yesterday
add a comment |
2
$begingroup$
I'm sure we all know this. The question is why WolframAlpha gives a different answer.
$endgroup$
– Jyrki Lahtonen
yesterday
2
2
$begingroup$
I'm sure we all know this. The question is why WolframAlpha gives a different answer.
$endgroup$
– Jyrki Lahtonen
yesterday
$begingroup$
I'm sure we all know this. The question is why WolframAlpha gives a different answer.
$endgroup$
– Jyrki Lahtonen
yesterday
add a comment |
$begingroup$
It is clear that, for each $ninmathbb N$,$$int_-pi^picos^2(t)sin(nt),mathrm dt=0,$$since $tmapstocos^2(t)sin(nt)$ is an odd function.
On the other hand,$$a_0=frac1piint_-pi^picos^2(t),mathrm dt=1neq0.$$It turns out that $a_1=0$, but $a_2=frac12neq0$.
answered yesterday
José Carlos SantosJosé Carlos Santos
166k22132235
$endgroup$
$begingroup$
I'm sure we all know this. The question is why WolframAlpha gives a different answer.
$endgroup$
– Jyrki Lahtonen
yesterday
1
$begingroup$
No, we don't all know this, since the OP claimed to have obtained $a_n=0$ while calculating by hand.
$endgroup$
– José Carlos Santos
yesterday
add a comment |
$begingroup$
It is clear that, for each $ninmathbb N$,$$int_-pi^picos^2(t)sin(nt),mathrm dt=0,$$since $tmapstocos^2(t)sin(nt)$ is an odd function.
On the other hand,$$a_0=frac1piint_-pi^picos^2(t),mathrm dt=1neq0.$$It turns out that $a_1=0$, but $a_2=frac12neq0$.
answered yesterday
José Carlos SantosJosé Carlos Santos
166k22132235
$endgroup$
$begingroup$
I'm sure we all know this. The question is why WolframAlpha gives a different answer.
$endgroup$
– Jyrki Lahtonen
yesterday
1
$begingroup$
No, we don't all know this, since the OP claimed to have obtained $a_n=0$ while calculating by hand.
$endgroup$
– José Carlos Santos
yesterday
add a comment |
$begingroup$
It is clear that, for each $ninmathbb N$,$$int_-pi^picos^2(t)sin(nt),mathrm dt=0,$$since $tmapstocos^2(t)sin(nt)$ is an odd function.
On the other hand,$$a_0=frac1piint_-pi^picos^2(t),mathrm dt=1neq0.$$It turns out that $a_1=0$, but $a_2=frac12neq0$.
answered yesterday
José Carlos SantosJosé Carlos Santos
166k22132235
$endgroup$
It is clear that, for each $ninmathbb N$,$$int_-pi^picos^2(t)sin(nt),mathrm dt=0,$$since $tmapstocos^2(t)sin(nt)$ is an odd function.
On the other hand,$$a_0=frac1piint_-pi^picos^2(t),mathrm dt=1neq0.$$It turns out that $a_1=0$, but $a_2=frac12neq0$.
answered yesterday
José Carlos SantosJosé Carlos Santos
166k22132235
answered yesterday
José Carlos SantosJosé Carlos Santos
166k22132235
answered yesterday
José Carlos SantosJosé Carlos Santos
166k22132235
answered yesterday
José Carlos SantosJosé Carlos Santos
166k22132235
166k22132235
$begingroup$
I'm sure we all know this. The question is why WolframAlpha gives a different answer.
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– Jyrki Lahtonen
yesterday
1
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No, we don't all know this, since the OP claimed to have obtained $a_n=0$ while calculating by hand.
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– José Carlos Santos
yesterday
add a comment |
$begingroup$
I'm sure we all know this. The question is why WolframAlpha gives a different answer.
$endgroup$
– Jyrki Lahtonen
yesterday
1
$begingroup$
No, we don't all know this, since the OP claimed to have obtained $a_n=0$ while calculating by hand.
$endgroup$
– José Carlos Santos
yesterday
$begingroup$
I'm sure we all know this. The question is why WolframAlpha gives a different answer.
$endgroup$
– Jyrki Lahtonen
yesterday
$begingroup$
I'm sure we all know this. The question is why WolframAlpha gives a different answer.
$endgroup$
– Jyrki Lahtonen
yesterday
1
1
$begingroup$
No, we don't all know this, since the OP claimed to have obtained $a_n=0$ while calculating by hand.
$endgroup$
– José Carlos Santos
yesterday
$begingroup$
No, we don't all know this, since the OP claimed to have obtained $a_n=0$ while calculating by hand.
$endgroup$
– José Carlos Santos
yesterday
add a comment |
aardvark is a new contributor. Be nice, and check out our Code of Conduct.
aardvark is a new contributor. Be nice, and check out our Code of Conduct.
aardvark is a new contributor. Be nice, and check out our Code of Conduct.
aardvark is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
Looks to me that already $a_0 = frac12 ne 0$... Can you detail your computations ?
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– Gâteau-Gallois
yesterday
1
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I'm not familiar enough with WolframAlpha to say, but it seems to me this is likely to be a misunderstanding of what the function does. For example, where's the $omega$ in that call? Maybe it's trying to project the function on a different period? As it stands, $cos^2(t) = 1/2+1/2cos(2t)$, so that's all your Fourier series.
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– Okarin
yesterday
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The correct answer is $a_0=a_2=frac 1 2$ and all other coefficients $0$.
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– Kavi Rama Murthy
yesterday
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Surprisingly Mathematica gives the same answer. OTOH to the integral
FourierCosCoefficient[(1+Cos[2t])/2,t,n]it gives the expected(DiscreteDelta[n-2]+2 DiscreteDelta[n])/2$endgroup$
– Jyrki Lahtonen
yesterday
1
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FWIW, I asked this at Mathematica.SE.
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– Jyrki Lahtonen
yesterday