Collect Fourier series terms The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)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
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Collect Fourier series terms
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)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
$begingroup$
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
asked Apr 6 at 16:49
RiccardoRiccardo
505
$endgroup$
add a comment |
$begingroup$
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
asked Apr 6 at 16:49
RiccardoRiccardo
505
$endgroup$
$begingroup$
can you show your initial code, so we try to help!
$endgroup$
– Alrubaie
Apr 6 at 17:00
$begingroup$
TryCollect[exp /. Cos[x n_.] -> x^n, x]/. x^n_. -> Cos[n x].
$endgroup$
– Somos
Apr 7 at 2:29
add a comment |
$begingroup$
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
asked Apr 6 at 16:49
RiccardoRiccardo
505
$endgroup$
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
asked Apr 6 at 16:49
RiccardoRiccardo
505
asked Apr 6 at 16:49
RiccardoRiccardo
505
asked Apr 6 at 16:49
RiccardoRiccardo
505
asked Apr 6 at 16:49
RiccardoRiccardo
505
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
$begingroup$
TryCollect[exp /. Cos[x n_.] -> x^n, x]/. x^n_. -> Cos[n x].
$endgroup$
– Somos
Apr 7 at 2:29
add a comment |
$begingroup$
can you show your initial code, so we try to help!
$endgroup$
– Alrubaie
Apr 6 at 17:00
$begingroup$
TryCollect[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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
is this what you are asking?
exp=3Cos[x]+5Cos[2x]+12Cos[3x];
Coefficient[exp,Cos[x #]]&/@Range@3
3, 5, 12
answered Apr 6 at 17:24
J42161217J42161217
4,473324
$endgroup$
$begingroup$
Yes, thanks this is it!
$endgroup$
– Riccardo
Apr 6 at 17:28
$begingroup$
you are welcome!
$endgroup$
– J42161217
Apr 6 at 17:35
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
is this what you are asking?
exp=3Cos[x]+5Cos[2x]+12Cos[3x];
Coefficient[exp,Cos[x #]]&/@Range@3
3, 5, 12
answered Apr 6 at 17:24
J42161217J42161217
4,473324
$endgroup$
$begingroup$
Yes, thanks this is it!
$endgroup$
– Riccardo
Apr 6 at 17:28
$begingroup$
you are welcome!
$endgroup$
– J42161217
Apr 6 at 17:35
add a comment |
$begingroup$
is this what you are asking?
exp=3Cos[x]+5Cos[2x]+12Cos[3x];
Coefficient[exp,Cos[x #]]&/@Range@3
3, 5, 12
answered Apr 6 at 17:24
J42161217J42161217
4,473324
$endgroup$
$begingroup$
Yes, thanks this is it!
$endgroup$
– Riccardo
Apr 6 at 17:28
$begingroup$
you are welcome!
$endgroup$
– J42161217
Apr 6 at 17:35
add a comment |
$begingroup$
is this what you are asking?
exp=3Cos[x]+5Cos[2x]+12Cos[3x];
Coefficient[exp,Cos[x #]]&/@Range@3
3, 5, 12
answered Apr 6 at 17:24
J42161217J42161217
4,473324
$endgroup$
is this what you are asking?
exp=3Cos[x]+5Cos[2x]+12Cos[3x];
Coefficient[exp,Cos[x #]]&/@Range@3
3, 5, 12
answered Apr 6 at 17:24
J42161217J42161217
4,473324
edited Apr 6 at 17:34
answered Apr 6 at 17:24
J42161217J42161217
4,473324
answered Apr 6 at 17:24
J42161217J42161217
4,473324
answered Apr 6 at 17:24
J42161217J42161217
4,473324
4,473324
$begingroup$
Yes, thanks this is it!
$endgroup$
– Riccardo
Apr 6 at 17:28
$begingroup$
you are welcome!
$endgroup$
– J42161217
Apr 6 at 17:35
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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