What is the e^At Problem for a 4x4 Matrix?

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The discussion focuses on solving the e^{At} problem for a given 4x4 matrix A, where t represents time. The matrix A is structured with specific values and requires proper formatting in LaTeX for clarity. A suggestion is made to represent the matrix as a block matrix, which simplifies finding a general formula for its powers. Participants emphasize using Taylor's series to derive the solution effectively. The conversation highlights the importance of correct notation and mathematical representation in solving matrix exponentiation problems.
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e^At problem!

1. Homework Statement [/b]

Hi
for given 4 by 4 matrix, say A, find the e^{At}
which t is time variable.


Homework Equations



A= {\left(\begin{array}{cccc} 0 & a & 0 & 0 \\ -a& 0 & 0 & 0\\ 0 & 0 & 0 & a\\0 & 0 & -a & 0 \end{array}\right)t}Y(0).


The Attempt at a Solution

 
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jahandideh said:
1. Homework Statement [/b]

Hi
for given 4 by 4 matrix, say A, find the e^{At}
which t is time variable.


Homework Equations



A= {\left(\begin{array}{cc} 0 & a & 0 & 0 \\ -a& 0 & 0 & 0\\ 0 & 0 & 0 & a\\0 & 0 & -a & 0 \end{array}\right)t}Y(0).


The Attempt at a Solution

......
 
The Latex should be

A= {\left(\begin{array}{cccc} 0 & a & 0 & 0 \\ -a& 0 & 0 & 0\\ 0 & 0 & 0 & a\\0 & 0 & -a & 0 \end{array}\right)t}Y(0)
You forgot the ending "[ / tex ]" and, the reason it didn't show up right when I edited your post and in sutupidmath's post is because you also have "{cc}" rather than the "{cccc}" you need for 4 columns.

Now, what have you done about this problem your self? This can be written as a "block" matrix of the form
\left(\begin{array}{cc}A & 0 \\ 0 & A\end{array}\right)
where
A= \left(\begin{array}{cc}0 & a \\ -a & 0\end{array}\right)
so it should be easy to find a general formula for the powers of the matrix and then use the Taylor's series for ex.
 
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I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks

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