Mixing
$int_0^t e^{sA}cos(omega s)ds$ with $A$ matrix
$begingroup$
Let $A$ be a singular square matrix and $omega,tinmathbb{R}^{*+}$. How to compute the following integral?
$$I = int_0^t e^{sA}cos(omega s),mathrm{d}s$$
Since I am looking for a numerical solutions, I am open to approximations. For example,
$$ I = sum_{k=0}^infty int_0^t dfrac{(sA)^k}{k!} cos(omega s),mathrm{d}s = sum_{k=0}^infty dfrac{A^k}{k!}int_0^t s^k cos(omega s),mathrm{d}s$$
so I can truncate the above sum but I'm wondering if there is a more ingenious approach.
calculus integration matrix-equations matrix-exponential
asked Jan 30 at 10:46
anderstoodanderstood
2,1791030
$endgroup$
add a comment |
$begingroup$
Let $A$ be a singular square matrix and $omega,tinmathbb{R}^{*+}$. How to compute the following integral?
$$I = int_0^t e^{sA}cos(omega s),mathrm{d}s$$
Since I am looking for a numerical solutions, I am open to approximations. For example,
$$ I = sum_{k=0}^infty int_0^t dfrac{(sA)^k}{k!} cos(omega s),mathrm{d}s = sum_{k=0}^infty dfrac{A^k}{k!}int_0^t s^k cos(omega s),mathrm{d}s$$
so I can truncate the above sum but I'm wondering if there is a more ingenious approach.
calculus integration matrix-equations matrix-exponential
asked Jan 30 at 10:46
anderstoodanderstood
2,1791030
$endgroup$
$begingroup$
Complex-exponential form of $cos$?..
$endgroup$
– metamorphy
Jan 30 at 10:49
$begingroup$
@metamorphy I thought of it but I am stuck with $e^{sA}e^{iomega s}$ since it does not equal $e^{sA+iomega s}$.
$endgroup$
– anderstood
Jan 30 at 10:55
1
$begingroup$
Well, $e^X e^Y=e^{X+Y}$ for commuting matrices $X, Y$. So if your $omega$ is just a scalar, then $e^{sA} e^{iomega s}=e^{s(A+iomega I)}$.
$endgroup$
– metamorphy
Jan 30 at 11:02
add a comment |
$begingroup$
Let $A$ be a singular square matrix and $omega,tinmathbb{R}^{*+}$. How to compute the following integral?
$$I = int_0^t e^{sA}cos(omega s),mathrm{d}s$$
Since I am looking for a numerical solutions, I am open to approximations. For example,
$$ I = sum_{k=0}^infty int_0^t dfrac{(sA)^k}{k!} cos(omega s),mathrm{d}s = sum_{k=0}^infty dfrac{A^k}{k!}int_0^t s^k cos(omega s),mathrm{d}s$$
so I can truncate the above sum but I'm wondering if there is a more ingenious approach.
calculus integration matrix-equations matrix-exponential
asked Jan 30 at 10:46
anderstoodanderstood
2,1791030
$endgroup$
Let $A$ be a singular square matrix and $omega,tinmathbb{R}^{*+}$. How to compute the following integral?
$$I = int_0^t e^{sA}cos(omega s),mathrm{d}s$$
Since I am looking for a numerical solutions, I am open to approximations. For example,
$$ I = sum_{k=0}^infty int_0^t dfrac{(sA)^k}{k!} cos(omega s),mathrm{d}s = sum_{k=0}^infty dfrac{A^k}{k!}int_0^t s^k cos(omega s),mathrm{d}s$$
so I can truncate the above sum but I'm wondering if there is a more ingenious approach.
calculus integration matrix-equations matrix-exponential
calculus integration matrix-equations matrix-exponential
asked Jan 30 at 10:46
anderstoodanderstood
2,1791030
asked Jan 30 at 10:46
anderstoodanderstood
2,1791030
asked Jan 30 at 10:46
anderstoodanderstood
2,1791030
asked Jan 30 at 10:46
anderstoodanderstood
2,1791030
asked Jan 30 at 10:46
anderstoodanderstood
2,1791030
2,1791030
$begingroup$
Complex-exponential form of $cos$?..
$endgroup$
– metamorphy
Jan 30 at 10:49
$begingroup$
@metamorphy I thought of it but I am stuck with $e^{sA}e^{iomega s}$ since it does not equal $e^{sA+iomega s}$.
$endgroup$
– anderstood
Jan 30 at 10:55
1
$begingroup$
Well, $e^X e^Y=e^{X+Y}$ for commuting matrices $X, Y$. So if your $omega$ is just a scalar, then $e^{sA} e^{iomega s}=e^{s(A+iomega I)}$.
$endgroup$
– metamorphy
Jan 30 at 11:02
add a comment |
$begingroup$
Complex-exponential form of $cos$?..
$endgroup$
– metamorphy
Jan 30 at 10:49
$begingroup$
@metamorphy I thought of it but I am stuck with $e^{sA}e^{iomega s}$ since it does not equal $e^{sA+iomega s}$.
$endgroup$
– anderstood
Jan 30 at 10:55
1
$begingroup$
Well, $e^X e^Y=e^{X+Y}$ for commuting matrices $X, Y$. So if your $omega$ is just a scalar, then $e^{sA} e^{iomega s}=e^{s(A+iomega I)}$.
$endgroup$
– metamorphy
Jan 30 at 11:02
$begingroup$
Complex-exponential form of $cos$?..
$endgroup$
– metamorphy
Jan 30 at 10:49
$begingroup$
Complex-exponential form of $cos$?..
$endgroup$
– metamorphy
Jan 30 at 10:49
$begingroup$
@metamorphy I thought of it but I am stuck with $e^{sA}e^{iomega s}$ since it does not equal $e^{sA+iomega s}$.
$endgroup$
– anderstood
Jan 30 at 10:55
$begingroup$
@metamorphy I thought of it but I am stuck with $e^{sA}e^{iomega s}$ since it does not equal $e^{sA+iomega s}$.
$endgroup$
– anderstood
Jan 30 at 10:55
1
1
$begingroup$
Well, $e^X e^Y=e^{X+Y}$ for commuting matrices $X, Y$. So if your $omega$ is just a scalar, then $e^{sA} e^{iomega s}=e^{s(A+iomega I)}$.
$endgroup$
– metamorphy
Jan 30 at 11:02
$begingroup$
Well, $e^X e^Y=e^{X+Y}$ for commuting matrices $X, Y$. So if your $omega$ is just a scalar, then $e^{sA} e^{iomega s}=e^{s(A+iomega I)}$.
$endgroup$
– metamorphy
Jan 30 at 11:02
add a comment |
1 Answer
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$begingroup$
As suggested by @metamorphy in the comments, it suffices to write
$$ I = operatorname{Re}left(int_0^t e^{s(A+iomega I_n)},mathrm{d}sright)$$
and the computation boils down to the integral of a matrix exponential as addressed in this post.
answered Jan 30 at 13:37
anderstoodanderstood
2,1791030
$endgroup$
add a comment |
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$begingroup$
As suggested by @metamorphy in the comments, it suffices to write
$$ I = operatorname{Re}left(int_0^t e^{s(A+iomega I_n)},mathrm{d}sright)$$
and the computation boils down to the integral of a matrix exponential as addressed in this post.
answered Jan 30 at 13:37
anderstoodanderstood
2,1791030
$endgroup$
add a comment |
$begingroup$
As suggested by @metamorphy in the comments, it suffices to write
$$ I = operatorname{Re}left(int_0^t e^{s(A+iomega I_n)},mathrm{d}sright)$$
and the computation boils down to the integral of a matrix exponential as addressed in this post.
answered Jan 30 at 13:37
anderstoodanderstood
2,1791030
$endgroup$
add a comment |
$begingroup$
As suggested by @metamorphy in the comments, it suffices to write
$$ I = operatorname{Re}left(int_0^t e^{s(A+iomega I_n)},mathrm{d}sright)$$
and the computation boils down to the integral of a matrix exponential as addressed in this post.
answered Jan 30 at 13:37
anderstoodanderstood
2,1791030
$endgroup$
As suggested by @metamorphy in the comments, it suffices to write
$$ I = operatorname{Re}left(int_0^t e^{s(A+iomega I_n)},mathrm{d}sright)$$
and the computation boils down to the integral of a matrix exponential as addressed in this post.
answered Jan 30 at 13:37
anderstoodanderstood
2,1791030
answered Jan 30 at 13:37
anderstoodanderstood
2,1791030
answered Jan 30 at 13:37
anderstoodanderstood
2,1791030
answered Jan 30 at 13:37
anderstoodanderstood
2,1791030
2,1791030
add a comment |
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$begingroup$
Complex-exponential form of $cos$?..
$endgroup$
– metamorphy
Jan 30 at 10:49
$begingroup$
@metamorphy I thought of it but I am stuck with $e^{sA}e^{iomega s}$ since it does not equal $e^{sA+iomega s}$.
$endgroup$
– anderstood
Jan 30 at 10:55
1
$begingroup$
Well, $e^X e^Y=e^{X+Y}$ for commuting matrices $X, Y$. So if your $omega$ is just a scalar, then $e^{sA} e^{iomega s}=e^{s(A+iomega I)}$.
$endgroup$
– metamorphy
Jan 30 at 11:02