How can i calculate e^A*t

How can i calculate e^A*t without using Markov Chain?

Where e=exp , A is a square matrix, and t is a variable

[Wayne King]

When you say t is a variable, what do you mean? what is the dimension of t?

[Nick]

ok , t>=0 but this question it is not about how to calculate it on Matlab just to see the result because Elad answer is correct (exp(A.*t)) but i want deeper meaning of the solution.

[Walter Roberson]

What deeper meaning? It's a formula presented outside of any context. Are you asking for the deeper meaning of why MATLAB uses exp(x) to represent e^x ?

[Elad]

Using the dot after A, means it will multiply each element in the matrix separately.

[Walter Roberson]

If t is a scalar then using the dot or not using the dot means the same thing. Using the dot makes a difference when neither A nor t are scalar.

[Oleg Komarov]

My crystall ball tells me that Nick is looking for a deeeeper meeaninnggg... deeeeeeeperr..

[Walter Roberson]

Perhaps Nick is data mining? I wonder what he will unearth?

[ABCD]

Dear Nick, do you mean this?

>> a = [1 2 3 ; 2 5 2; 1 4 3]

a =

     1     2     3
     2     5     2
     1     4     3

>> syms t

>> exp(a*t)

ans =

[ exp(t), exp(2*t), exp(3*t)]

[ exp(2*t), exp(5*t), exp(2*t)]

[ exp(t), exp(4*t), exp(3*t)]

[CHENG WEI SHEN]

???

[Walter Roberson]

@CHENG WEI SHEN

Please expand on your question ?

Answer 1

[Elad]

exp(A.*t)

Answer 2

[Kye Taylor]

Use the expm function for computing a matrix exponential

[KJ N]

To help anyone else coming here: ignore all the other answers saying to use exp. Using expm is the right one for this situation. If you want to compute the matrix exponential e^(A t), where A is a n x n square matrix and t is a variable, and you DO NOT want to do simply do the by-element exponential, i.e., you want to compute the equivalent of the inverse Laplace of s*eye(n)-A, which is important in state-space analysis of linear systems, you want to use expm(A*t), not exp(A*t).

>> A = [0 1; -2 -3]
A =
   0     1
  -2    -3
>> syms t;expm(A*t)
ans =
[   2*exp(-t) - exp(-2*t),   exp(-t) - exp(-2*t)]
[ 2*exp(-2*t) - 2*exp(-t), 2*exp(-2*t) - exp(-t)]
>> syms s;ilaplace(inv(s*eye(rank(A))-A))
ans =
[   2*exp(-t) - exp(-2*t),   exp(-t) - exp(-2*t)]
[ 2*exp(-2*t) - 2*exp(-t), 2*exp(-2*t) - exp(-t)]

[Walter Roberson]

Kaleb Nelson, no: the poster said above https://www.mathworks.com/matlabcentral/answers/39854-how-can-i-calculate-e-a-t#comment_82495 that exp(A.*t) is the correct solution.

[KJ N]

exp() only does computes the exponential of A element-by-element, as shown above like this: >> a = [1 2 3 ; 2 5 2; 1 4 3]

a =

     1     2     3
     2     5     2
     1     4     3
>> syms t

>> exp(a*t)

ans =

[ exp(t), exp(2*t), exp(3*t)]

[ exp(2*t), exp(5*t), exp(2*t)]

[ exp(t), exp(4*t), exp(3*t)]

If that's what you're going for, that's great, but not terribly difficult to compute by hand for even somewhat large n x n matrices with integer elements. However, the original poster said they wanted to avoid using the markov chain (a somewhat onerous process, especially when done by hand for large matrices, even with simple integer values as the elements), leading me to understand they were referring to the matrix exponential, not the element-by-element exponential, hence the correct answer in this case would be to use expm(). I had been looking for the same answer, and Kye Taylor was the only post saying use expm instead of exp, so I thought I would try to ensure those in the future looking for the same answer as myself would be helped by a clarification.

[Walter Roberson]

We tried a number of times to get the original poster to clarify, but all we got was that they want the exp() solution and that they are looking for a "deeper reason" for something. The poster effectively defined the exp() solution as being the correct one.

Your analysis might well be what the poster really needed, but it is contrary to what little they defined as being correct for their needs.

Answer 3

[Junsheng SU]

syms t; expm(A*t);

[Lars Nagel]

Worked perfectly for me!

Answer 4

[Shenhai]

I guess it is not always possible to get the close form solution of exp(At)...

Sometimes I can get result with: exp(At) = iL(sI-A)^-1, where iL is the inverse Laplace transformation, like:

syms s t
A = [0 1;0 0];
expAt = ilaplace(inv(s*eye(size(A,1))-A),s,t);

This will give the result as: [1 t;0 1]

Any other ideas?

Answer 5

[Shahram Bekhrad]

As far as I'm aware you probably need it for finding the answer of a state space equation. I myself couldn't find any good function or command yet, so you might have to write a Script file (m-file) and find it. you can use about 3 or 4 way of calculating the said statement. These things are taught in courses like modern control theory. I used the following expression but still have some difficulties. exp(A.t)=I+At+ (At)^2/2! + (At)^3/3!+ (At)^4/4!+. . .

Answer 6

[ABCD]

Dear Nick, do you mean this?

>> a = [1 2 3 ; 2 5 2; 1 4 3]

a =

     1     2     3
     2     5     2
     1     4     3

>> syms t >> exp(a*t)

ans =

[ exp(t), exp(2*t), exp(3*t)] [ exp(2*t), exp(5*t), exp(2*t)] [ exp(t), exp(4*t), exp(3*t)]

[ABCD]

>> a = [1 2 3 ; 2 5 2; 1 4 3]

a =

     1     2     3
     2     5     2
     1     4     3

>> syms t

>> exp(a*t)

ans =

[ exp(t), exp(2*t), exp(3*t)]

[ exp(2*t), exp(5*t), exp(2*t)]

[ exp(t), exp(4*t), exp(3*t)]