0
$\begingroup$

Given $A = \begin{bmatrix} 1 & 1 \\ & 2 \end{bmatrix}$,

$e^A$ is computed by the formula

$e^A= \begin{bmatrix} 1 & \\ & 1 \end{bmatrix} + \frac{1}{1!}\begin{bmatrix} 1 & 1 \\ & 2 \end{bmatrix} + \frac{1}{2!}\begin{bmatrix} 1 & 3 \\ & 4 \end{bmatrix} + \dots = \begin{bmatrix} e & * \\ & e^2 \end{bmatrix}$.

I am trying to figure out what $*$ is in the above formula.

By calculating $A^n$,

$A=\begin{bmatrix} 1 & 1 \\ & 2 \end{bmatrix}$,

$A^2=\begin{bmatrix} 1 & 1 \\ & 2 \end{bmatrix}\begin{bmatrix} 1 & 1 \\ & 2 \end{bmatrix}=\begin{bmatrix} 1 & 1+2 \\ & 2^2 \end{bmatrix}$,

$A^n=\begin{bmatrix} 1 & 1+2+2^2+\dots+2^{n-1} \\ & 2^n \end{bmatrix}=\begin{bmatrix} 1 & \frac{2^n-1}{2-1} \\ & 2^n \end{bmatrix}$.

Thus $* = \Sigma_{n=1}^{\infty}\frac{2^n-1}{n!}$.

My question is (1) whether I came to the right place until now, (2) if the sum $*$ exists and (3) if so, what it is(does it have a explicit formula).

$\endgroup$
11
  • $\begingroup$ For series look to exponent en.wikipedia.org/wiki/Taylor_series#Exponential_function $\endgroup$ – zkutch Jul 17 '20 at 11:24
  • $\begingroup$ What is the lower left entry? Zero? Same as upper entry? $\endgroup$ – Oscar Lanzi Jul 17 '20 at 11:30
  • $\begingroup$ @OscarLanzi Zero it is. $\endgroup$ – Henry Choi Jul 17 '20 at 11:32
  • 1
    $\begingroup$ But $A^0\ne\begin{bmatrix} 1 & 1 \\ & 1 \end{bmatrix}$, and $A^2\ne\begin{bmatrix} 1 & 3 \\ & 3 \end{bmatrix}$. $\endgroup$ – TonyK Jul 17 '20 at 11:54
  • 1
    $\begingroup$ And finally, $* = \Sigma_{n=1}^{\infty}\frac{2^n-1}{n!}$, not $1+\Sigma_{n=1}^{\infty}\frac{2^n-1}{n!}$. I think I'm done now. $\endgroup$ – TonyK Jul 17 '20 at 12:03
5
$\begingroup$

You might want to use the fact that $$ e^x = \sum_{n\geq 0} \dfrac{x^n}{n!}. $$ Hence $$ \sum_{n=1}^{\infty} \dfrac{2^n - 1}{n!} = \sum_{n\geq 0} \dfrac{2^n - 1^n}{n!} = e^2 - e^1 = e^2 - e. $$

$\endgroup$
1
$\begingroup$

It does converge, for example by the ratio test.

You can find the limit by splitting the sum into two: it's $\sum \frac{2^n}{n!} - \sum \frac{1}{n!}$. Both of these you can recognise from $e^x = \sum \frac{x^n}{n!}$.

$\endgroup$
1
$\begingroup$

$\Sigma_{n=1}^{\infty}\frac{x^n}{n!}$ is a powerseries with radius of convergence $R=+\infty$. Thus both $\Sigma_{n=1}^{\infty}\frac{2^n}{n!}$ and $\Sigma_{n=1}^{\infty}\frac{-1}{n!}=-\Sigma_{n=1}^{\infty}\frac{1}{n!}=-\Sigma_{n=1}^{\infty}\frac{1^n}{n!}$ are convergent. Take the sum of two.

$\endgroup$
1
$\begingroup$

By the ratio test

$$\frac{2-2^{-n}}{n+1}\le\frac34$$ as soon as $n>1$.

(Of course, you can split the two terms and use the well-known Taylor development.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.