MHB Calculate Cosine of a Matrix: Solutions to Systems

[jessicamorgan]

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Please help, I know the definition of a cosine of a matrix is cos(a) = I-1/2!A^2+1/4!A^4-...
But I am unsure how this would help me find solutions to these systems.

 

[I like Serena]

Please help, I know the definition of a cosine of a matrix is cos(a) = I-1/2!A^2+1/4!A^4-...
But I am unsure how this would help me find solutions to these systems.

Hi jessicamorgan! Welcome to MHB! (Smile)

That is indeed the definition of the cosine of a matrix.
We won't need it other than knowing it exists and behaves like a regular cosine.

Suppose we solve the equation for the 1-dimensional case, when the cosine is really a regular cosine.
What would the solution be?

 

[jessicamorgan]

Hi jessicamorgan! Welcome to MHB! (Smile)

That is indeed the definition of the cosine of a matrix.
We won't need it other than knowing it exists and behaves like a regular cosine.

Suppose we solve the equation for the 1-dimensional case, when the cosine is really a regular cosine.
What would the solution be?

Hi, thanks for replying.
I'm unsure of the question, I don't understand how to solve or what the cosine has to do with it. :/

 

[HallsofIvy]

It looks to me like there are two different ways to look at this problem.
1) That you have already learned that, for numeric functions, the general solution to [math]y''= -A^2y[/math] is $$y(x)= C cos(\sqrt{A}x)+ Dsin(\sqrt{A}x)$$ and are asked to use the series definitions to write that same solution for A a matrix.

2) That you have already learned that you can differentiate such a series "term by term" so that, given [math]y= cos(Ax)= I-1/2!(Ax)^2+1/4!(Ax)^4-...[/math] you have [math]y'= -A(Ax)+ 1/3!A(Ax)^3- ...= -A(Ax- 1/3! (Ax)^3+ ...)[/math] and then [math]y''= -A(A- 1/2!(Ax)^2+ ...)= -A^2y[/math].