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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.













0












$begingroup$


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)










share|cite|improve this question









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 integral FourierCosCoefficient[(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















0












$begingroup$


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)










share|cite|improve this question









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 integral FourierCosCoefficient[(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













0












0








0





$begingroup$


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)










share|cite|improve this question









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






share|cite|improve this question









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.











share|cite|improve this question









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.









share|cite|improve this question




share|cite|improve this question








edited yesterday









Bernard

122k741116




122k741116






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




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 integral FourierCosCoefficient[(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






  • 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 integral FourierCosCoefficient[(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










3 Answers
3






active

oldest

votes


















3












$begingroup$

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).






share|cite|improve this answer









$endgroup$












  • $begingroup$
    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


















1












$begingroup$

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!.






share|cite|improve this answer









$endgroup$








  • 2




    $begingroup$
    I'm sure we all know this. The question is why WolframAlpha gives a different answer.
    $endgroup$
    – Jyrki Lahtonen
    yesterday


















0












$begingroup$

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 $a_2=frac12neq0$.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    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










Your Answer





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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

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).






share|cite|improve this answer









$endgroup$












  • $begingroup$
    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















3












$begingroup$

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).






share|cite|improve this answer









$endgroup$












  • $begingroup$
    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













3












3








3





$begingroup$

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).






share|cite|improve this answer









$endgroup$



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).







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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
















  • $begingroup$
    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











1












$begingroup$

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!.






share|cite|improve this answer









$endgroup$








  • 2




    $begingroup$
    I'm sure we all know this. The question is why WolframAlpha gives a different answer.
    $endgroup$
    – Jyrki Lahtonen
    yesterday















1












$begingroup$

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!.






share|cite|improve this answer









$endgroup$








  • 2




    $begingroup$
    I'm sure we all know this. The question is why WolframAlpha gives a different answer.
    $endgroup$
    – Jyrki Lahtonen
    yesterday













1












1








1





$begingroup$

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!.






share|cite|improve this answer









$endgroup$



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!.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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












  • 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











0












$begingroup$

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 $a_2=frac12neq0$.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    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















0












$begingroup$

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 $a_2=frac12neq0$.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    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













0












0








0





$begingroup$

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 $a_2=frac12neq0$.






share|cite|improve this answer









$endgroup$



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 $a_2=frac12neq0$.







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answered yesterday









José Carlos SantosJosé Carlos Santos

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  • $begingroup$
    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






  • 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










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