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Collect Fourier series terms


CoefficientList for exponentials termsAssign Coefficients of Exponential SeriesConstruct Matrices from Coefficient of f[x] in Finite Fields?collect a multivariate polynomial, and simplify all coefficientsCollect terms with same-kind coefficients and factor exponential based on thatCollect terms in Fourier TransformExtracting coefficients from sumExtract terms containing a specific symbol from an expressionPlotting double seriesRearranging similar terms in an equation













3












$begingroup$


Is there a way to make mathematica automatically collect terms like $mathrmCos[n x]$ with $n=1,2,3...$?
I have a long expression coming out of a Fourier series where I see a lot of $Cos[x],Cos[2 x],Cos[3 x]$ and so on, but Simplify is very inefficient. I would like mathematica to automatically collect all coefficients of the various Cos[n x] without having to do it manually for each one of them.










share|improve this question









$endgroup$











  • $begingroup$
    can you show your initial code, so we try to help!
    $endgroup$
    – Alrubaie
    Apr 6 at 17:00










  • $begingroup$
    Try Collect[exp /. Cos[x n_.] -> x^n, x]/. x^n_. -> Cos[n x].
    $endgroup$
    – Somos
    Apr 7 at 2:29
















3












$begingroup$


Is there a way to make mathematica automatically collect terms like $mathrmCos[n x]$ with $n=1,2,3...$?
I have a long expression coming out of a Fourier series where I see a lot of $Cos[x],Cos[2 x],Cos[3 x]$ and so on, but Simplify is very inefficient. I would like mathematica to automatically collect all coefficients of the various Cos[n x] without having to do it manually for each one of them.










share|improve this question









$endgroup$











  • $begingroup$
    can you show your initial code, so we try to help!
    $endgroup$
    – Alrubaie
    Apr 6 at 17:00










  • $begingroup$
    Try Collect[exp /. Cos[x n_.] -> x^n, x]/. x^n_. -> Cos[n x].
    $endgroup$
    – Somos
    Apr 7 at 2:29














3












3








3





$begingroup$


Is there a way to make mathematica automatically collect terms like $mathrmCos[n x]$ with $n=1,2,3...$?
I have a long expression coming out of a Fourier series where I see a lot of $Cos[x],Cos[2 x],Cos[3 x]$ and so on, but Simplify is very inefficient. I would like mathematica to automatically collect all coefficients of the various Cos[n x] without having to do it manually for each one of them.










share|improve this question









$endgroup$




Is there a way to make mathematica automatically collect terms like $mathrmCos[n x]$ with $n=1,2,3...$?
I have a long expression coming out of a Fourier series where I see a lot of $Cos[x],Cos[2 x],Cos[3 x]$ and so on, but Simplify is very inefficient. I would like mathematica to automatically collect all coefficients of the various Cos[n x] without having to do it manually for each one of them.







coefficients






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Apr 6 at 16:49









RiccardoRiccardo

505




505











  • $begingroup$
    can you show your initial code, so we try to help!
    $endgroup$
    – Alrubaie
    Apr 6 at 17:00










  • $begingroup$
    Try Collect[exp /. Cos[x n_.] -> x^n, x]/. x^n_. -> Cos[n x].
    $endgroup$
    – Somos
    Apr 7 at 2:29

















  • $begingroup$
    can you show your initial code, so we try to help!
    $endgroup$
    – Alrubaie
    Apr 6 at 17:00










  • $begingroup$
    Try Collect[exp /. Cos[x n_.] -> x^n, x]/. x^n_. -> Cos[n x].
    $endgroup$
    – Somos
    Apr 7 at 2:29
















$begingroup$
can you show your initial code, so we try to help!
$endgroup$
– Alrubaie
Apr 6 at 17:00




$begingroup$
can you show your initial code, so we try to help!
$endgroup$
– Alrubaie
Apr 6 at 17:00












$begingroup$
Try Collect[exp /. Cos[x n_.] -> x^n, x]/. x^n_. -> Cos[n x].
$endgroup$
– Somos
Apr 7 at 2:29





$begingroup$
Try Collect[exp /. Cos[x n_.] -> x^n, x]/. x^n_. -> Cos[n x].
$endgroup$
– Somos
Apr 7 at 2:29











1 Answer
1






active

oldest

votes


















5












$begingroup$

is this what you are asking?



exp=3Cos[x]+5Cos[2x]+12Cos[3x];

Coefficient[exp,Cos[x #]]&/@Range@3



3, 5, 12







share|improve this answer











$endgroup$












  • $begingroup$
    Yes, thanks this is it!
    $endgroup$
    – Riccardo
    Apr 6 at 17:28










  • $begingroup$
    you are welcome!
    $endgroup$
    – J42161217
    Apr 6 at 17:35











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









5












$begingroup$

is this what you are asking?



exp=3Cos[x]+5Cos[2x]+12Cos[3x];

Coefficient[exp,Cos[x #]]&/@Range@3



3, 5, 12







share|improve this answer











$endgroup$












  • $begingroup$
    Yes, thanks this is it!
    $endgroup$
    – Riccardo
    Apr 6 at 17:28










  • $begingroup$
    you are welcome!
    $endgroup$
    – J42161217
    Apr 6 at 17:35















5












$begingroup$

is this what you are asking?



exp=3Cos[x]+5Cos[2x]+12Cos[3x];

Coefficient[exp,Cos[x #]]&/@Range@3



3, 5, 12







share|improve this answer











$endgroup$












  • $begingroup$
    Yes, thanks this is it!
    $endgroup$
    – Riccardo
    Apr 6 at 17:28










  • $begingroup$
    you are welcome!
    $endgroup$
    – J42161217
    Apr 6 at 17:35













5












5








5





$begingroup$

is this what you are asking?



exp=3Cos[x]+5Cos[2x]+12Cos[3x];

Coefficient[exp,Cos[x #]]&/@Range@3



3, 5, 12







share|improve this answer











$endgroup$



is this what you are asking?



exp=3Cos[x]+5Cos[2x]+12Cos[3x];

Coefficient[exp,Cos[x #]]&/@Range@3



3, 5, 12








share|improve this answer














share|improve this answer



share|improve this answer








edited Apr 6 at 17:34

























answered Apr 6 at 17:24









J42161217J42161217

4,733324




4,733324











  • $begingroup$
    Yes, thanks this is it!
    $endgroup$
    – Riccardo
    Apr 6 at 17:28










  • $begingroup$
    you are welcome!
    $endgroup$
    – J42161217
    Apr 6 at 17:35
















  • $begingroup$
    Yes, thanks this is it!
    $endgroup$
    – Riccardo
    Apr 6 at 17:28










  • $begingroup$
    you are welcome!
    $endgroup$
    – J42161217
    Apr 6 at 17:35















$begingroup$
Yes, thanks this is it!
$endgroup$
– Riccardo
Apr 6 at 17:28




$begingroup$
Yes, thanks this is it!
$endgroup$
– Riccardo
Apr 6 at 17:28












$begingroup$
you are welcome!
$endgroup$
– J42161217
Apr 6 at 17:35




$begingroup$
you are welcome!
$endgroup$
– J42161217
Apr 6 at 17:35

















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