Two questions about trig functions of matrices So the

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Magdalena Warren

Answered question

2022-04-09

Two questions about trig functions of matrices
So the first part of this question asked me to define cosA and sinA where A is a nxn real matrix. I used the taylor series expansions of sin and cos to get

\sin A = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{(2 n + 1)!} A^{2 n + 1}

\cos A = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{(2 n)!} A^{2 n}

For part a. I need to show that

\frac{d}{d t} \sin (A t) = A \cos (A t)

\frac{d}{d t} \cos (A t) = - A \sin (A t)

I was able to do this for sin with the following steps:

\frac{d}{d t} \sin (A t) = \frac{d}{d t} (\sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{(2 n + 1)!} {(A t)}^{2 n + 1}) = \frac{d}{d t} (\sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{(2 n + 1)!} A^{2 n + 1} t^{2 n + 1}) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{(2 n + 1)!} A^{2 n + 1} (2 n + 1) t^{2 n} = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{(2 n)!} A^{2 n + 1} t^{2 n} = A \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{(2 n + 1)!} {(A t)}^{2 n} = A \cos (A t)

However, for cosine I get stuck. I start off the same way:

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Answer & Explanation

firenzesunzc65

Beginner2022-04-10Added 16 answers

Step 1
Instead of focusing on a particular function, consider a generic analytic function and its derivative.
$\begin{aligned} f (x) & = \sum_{k = 0}^{\infty} β_{k} x^{k} ⟹ g (x) = \frac{d f}{d x} = \sum_{k = 0}^{\infty} k β_{k} x^{k - 1} \\ \dot{f} & = (\frac{d f}{d x}) (\frac{d x}{d t}) = g \dot{x} \end{aligned}$
Now apply these functions to a square matrix
$\begin{aligned} F & = f (X) = \sum_{k = 0}^{\infty} β_{k} X^{k} ⟹ G & = g (X) = \sum_{k = 0}^{\infty} k β_{k} X^{k - 1} \\ \dot{F} & = \sum_{k = 0}^{\infty} β_{k} \sum_{j = 0}^{k - 1} X^{j} \dot{X} X^{k - j - 1} \neq G \dot{X} \end{aligned}$
The problem is that (in general) $\dot{X}$ does not commute with $X$ .
Step 2
However, given a constant matrix $A$ , the matrix $X = A t$ does commutes with its time derivative since
$\begin{aligned} \dot{X} & = A ⟹ X \dot{X} & = \dot{X} X = A^{2} t \end{aligned}$
Substituting this X into the expression for $\dot{F}$ allows terms to be collected exactly they way they were in the scalar case, yielding $\dot{F} = G A = A G$ .
Step 3
In this particular problem, $f (x) = \sin (x) and g (x) = \cos (x)$
For part b, if you are given a functional identity
$f_{1} (x) + f_{2} (x) = f_{3} (x)$
it doesn't matter if you evaluate the LHS or the RHS, because they are identical.
You merely need to recognize the RHS as a function, which can be translated into matrix form
$x^{0} = 1 \Rightarrow X^{0} = I$
Or if you prefer, you can think of it as a Taylor series with a single term
$f_{3} (x) = (\sin^{2} (x) + \cos^{2} (x)) = (\sum_{k = 0}^{0} x^{k}) = 1$

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