Collect Fourier series termsCoefficientList 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

Collect Fourier series terms

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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













2












$begingroup$


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.






coefficients







share|improve this question









$endgroup$











  • $begingroup$
    can you show your initial code, so we try to help!
    $endgroup$
    – Alrubaie
    5 hours ago















2












$begingroup$


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.






coefficients







share|improve this question









$endgroup$











  • $begingroup$
    can you show your initial code, so we try to help!
    $endgroup$
    – Alrubaie
    5 hours ago













2












2








2





$begingroup$


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.






coefficients







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




coefficients






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked 5 hours ago









RiccardoRiccardo

455




455











  • $begingroup$
    can you show your initial code, so we try to help!
    $endgroup$
    – Alrubaie
    5 hours ago
















  • $begingroup$
    can you show your initial code, so we try to help!
    $endgroup$
    – Alrubaie
    5 hours ago















$begingroup$
can you show your initial code, so we try to help!
$endgroup$
– Alrubaie
5 hours ago




$begingroup$
can you show your initial code, so we try to help!
$endgroup$
– Alrubaie
5 hours ago










1 Answer
1






active

oldest

votes


















3












$begingroup$

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
    5 hours ago










  • $begingroup$
    you are welcome!
    $endgroup$
    – J42161217
    4 hours ago











Your Answer





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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

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
    5 hours ago










  • $begingroup$
    you are welcome!
    $endgroup$
    – J42161217
    4 hours ago















3












$begingroup$

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
    5 hours ago










  • $begingroup$
    you are welcome!
    $endgroup$
    – J42161217
    4 hours ago













3












3








3





$begingroup$

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 4 hours ago

























answered 5 hours ago









J42161217J42161217

4,248324




4,248324











  • $begingroup$
    Yes, thanks this is it!
    $endgroup$
    – Riccardo
    5 hours ago










  • $begingroup$
    you are welcome!
    $endgroup$
    – J42161217
    4 hours ago
















  • $begingroup$
    Yes, thanks this is it!
    $endgroup$
    – Riccardo
    5 hours ago










  • $begingroup$
    you are welcome!
    $endgroup$
    – J42161217
    4 hours ago















$begingroup$
Yes, thanks this is it!
$endgroup$
– Riccardo
5 hours ago




$begingroup$
Yes, thanks this is it!
$endgroup$
– Riccardo
5 hours ago












$begingroup$
you are welcome!
$endgroup$
– J42161217
4 hours ago




$begingroup$
you are welcome!
$endgroup$
– J42161217
4 hours ago

















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