Looking for "an easy to understand" proof for following Power series

Question

I'm looking for proof for the following Power series

$exp(X) = \sum_{k=0}^{n} \frac{X^{k}}{k!}$

If X is $A_{nxn}$ matrix, then prove the series is converge

Given

$\exp(X) = \sum_{k=0}^{n} \frac{X^{k}}{k!}$

$\sin(X) = \sum_{k=0}^{n} (-1)^{k}\frac{X^{2k}}{2k!}$

$\cos(X) = \sum_{k=0}^{n} (-1)^{k}\frac{X^{2k+1}}{(2k+1)!}$

$\cosh(X) = \sum_{k=0}^{n} \frac{X^{2k}}{(2k)!}$

$\sinh(X) = \sum_{k=0}^{n} \frac{X^{2k+1}}{(2k+1)!}$

Here is what I try so far, and I'm looking for "easy to understand" proof

Let $ X = \begin{bmatrix} 0 & -\beta \\ \beta & 0 \end{bmatrix} $

$ \exp(X) = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \begin{bmatrix} 0 & -\beta \\ \beta & 0 \end{bmatrix} + \frac{1}{2!} \begin{bmatrix} -\beta^{2} & 0 \\ 0 & \beta^{2} \end{bmatrix} + \frac{1}{3!} \begin{bmatrix} 0 & \beta^{3} \\ -\beta^{3} & 0 \end{bmatrix} + \frac{1}{4!} \begin{bmatrix} \beta^{4} & 0 \\ 0 & \beta^{4} \end{bmatrix} + \frac{1}{5!} \begin{bmatrix} 0 & -\beta^{5}\\ \beta^{5} & 0 \end{bmatrix} $

$ \exp(X) = \begin{bmatrix} 1 + \frac{1}{2!}(-\beta^{2}) + \frac{1}{4!}\beta^{4} + ... & -\beta + \frac{1}{3!}\beta^{3} + ...\\ \beta + \frac{1}{3!}-\beta^{3} + ... & 1 + \frac{1}{2!}\beta^{2} + \frac{1}{4!}\beta^{4} + ... \end{bmatrix} = \begin{bmatrix} \sin(\beta) & -\cos(\beta)\\ \cos(\beta) & \sin(\beta) \end{bmatrix} $

Answer 1

Better to proceed as follows.

Consider for example the series $\sum_{k=0}^\infty\frac1{k!}t^kA^k$, say with $t$ complex or if you prefer take $t\in\mathbb R$. You can verify that the all entries of the series are power series.

So, all that matters for convergence is the radius of convergence. It is easy to show that the radius of convergence of the series of numbers $\sum_{k=0}^\infty\frac1{k!}t^k\|A^k\|$ is infinity, and so the same goes for each entry of the original series.

Conclusion: each entry of $\sum_{k=0}^\infty\frac1{k!}t^kA^k$ converges, for all values of $t$, and you can take say $t=1$.

Answer 2

If you assume $X$ to be diagonalizable, then you immediately see that the series converges. For the general case, with a little extra work using the Jordan form, you can still achieve the result., but here I'm going to show just the diagonalizable case.

So, let $X=B\Lambda B^{-1}$. Then we have $X^k=B\Lambda^k B^{-1}$. Then, writing out the partial sums,

$$ S_n = \sum_{k=0}^n \frac{X^k}{k!} = \sum_{k=0}^n \frac{B\Lambda^kB^{-1}}{k!} = B\left(\sum_{k=0}^n \frac{\Lambda^k}{k!}\right)B^{-1} $$

Now the right hand side is a diagonal matrix (sum of diagonal matrices), whose entries converge to $e^{\lambda_i}$, $\lambda_i$ being the eigenvalues of $A$. Clearly, the candidate limit is the matrix $exp(X):=Be^\Lambda B^{-1}$, where the exponential of a diagonal matrix is intended as the diagonal matrix whose entries are the exponential of the diagonal entries of the input matrix. Now, let's check the norm $\|S_n-exp(X)\|$ (I will assume the norm is submultiplicative):

$$ \|S_n-exp(X)\| = \left\|B\left(\sum_{k=0}^n \frac{\Lambda^k}{k!}\right)B^{-1}-Be^\Lambda B^{-1}\right\|\leq\|B\|\|B^{-1}\|\|D\| $$

where $D$ is a diagonal matrix with entries $D_{ii}=e^{\lambda_i}-\sum_{k=0}^n\frac{\lambda_i^k}{k!}$. Clearly $\|D\|$ approaches 0, in whatever norm you choose.

Edit: just to give an idea, in case of non-diagonalizable matrices, you have $X=B J^k B^{-1}$, where $J$ is the Jordan form of $X$. The diagonal of $J^k$ looks exactly as in the diagonal case, but the Jordan blocks fill up. Fortunately, though, they fill up with terms of the form $p(k)\lambda^s$, where $p$ is a polynomial, and $s\in\{k-r+1,\ldots,k-1\}$, $r$ being the order of the block. Hence, all the extra diagonal terms approach zero.