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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
New contributor
$endgroup$
|
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
New contributor
$endgroup$
$begingroup$
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
New contributor
$endgroup$
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
New contributor
New contributor
edited yesterday
Bernard
122k741116
122k741116
New contributor
asked yesterday
aardvarkaardvark
1
1
New contributor
New contributor
$begingroup$
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).
$endgroup$
$begingroup$
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!.
$endgroup$
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 $a_2=frac12neq0$.
$endgroup$
$begingroup$
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).
$endgroup$
$begingroup$
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).
$endgroup$
$begingroup$
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).
$endgroup$
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
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!.
$endgroup$
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!.
$endgroup$
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!.
$endgroup$
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
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 $a_2=frac12neq0$.
$endgroup$
$begingroup$
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 $a_2=frac12neq0$.
$endgroup$
$begingroup$
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 $a_2=frac12neq0$.
$endgroup$
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
166k22132235
$begingroup$
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 |
$begingroup$
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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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
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– Jyrki Lahtonen
yesterday
1
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FWIW, I asked this at Mathematica.SE.
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– Jyrki Lahtonen
yesterday