I need to compute e^A for the matrix
A= 0 ∏
....-∏ 0
where the diagonal are zeros and the other diagonal has pi on top and negative pi on the bottom.
I'm not quite sure where to start.
Thanks
$\displaystyle e^A $ is defined as
$\displaystyle *\sum_{i=0}^{\infty} \frac{A^n}{n!} $
You can find that
if $\displaystyle B=e^A $
then $\displaystyle B(1,1) = B(2,2) =1 -\frac{\pi^2}{2!}+ \frac{\pi^4}{4!} - \frac{\pi^6}{6!} + \frac{\pi^8}{8!} - ... =-1$
$\displaystyle B(1,2) = \frac{\pi}{1!}- \frac{\pi^3}{3!} + \frac{\pi^5}{5!} - \frac{\pi^7}{7!} + ... = 0 $
$\displaystyle B(2,1) = -\frac{\pi}{1!}+ \frac{\pi^3}{3!} - \frac{\pi^5}{5!} + \frac{\pi^7}{7!} - ... = 0 $
If your prof wants you to evaluate those series then I don't know (I used Maple). Well I don't have the time... or both.
I don't think so because it is
$\displaystyle \sum_{i=1}^{\infty} (-1)^{n}\cdot\frac{\pi^{2n}}{(2n)!} $ because the first term is 0 not 1 we must integrate from 1 to infinity not 0 to infinity. But I may be mistaking.
Here are the 10 first terms
-4.934802202
-0.876090073
-2.211352843
-1.976022212
-2.001829103
-1.999899529
-2.000004167
-1.999999864
-2.000000003
-1.999999999
Yes you're right because we have to add the identity matrix at the beginning. SOrry.
$\displaystyle B=e^A $
then $\displaystyle B(1,1) = B(2,2) =1 -\frac{\pi^2}{2!}+ \frac{\pi^4}{4!} - \frac{\pi^6}{6!} + \frac{\pi^8}{8!} - ... =-1$
$\displaystyle B(1,2) = \frac{\pi}{1!}- \frac{\pi^3}{3!} + \frac{\pi^5}{5!} - \frac{\pi^7}{7!} + ... = 0 $
$\displaystyle B(2,1) = -\frac{\pi}{1!}+ \frac{\pi^3}{3!} - \frac{\pi^5}{5!} + \frac{\pi^7}{7!} - ... = 0 $
Somebody knows how to calculate these sum by hand?
You might consider starting by diagonalizing A--
Notice that $\displaystyle A = P D P^{-1}$
where
$\displaystyle P = \begin{pmatrix}1 &1 \\
i &-i \end{pmatrix}$
$\displaystyle D =\begin{pmatrix}\pi i &0 \\
0 &-\pi i \end{pmatrix}$
$\displaystyle P^{-1} = \frac{1}{2} \begin{pmatrix}1 &-i \\
1 &i \end{pmatrix}$
(I'm assuming you know how to go about diagonalizing a matrix; if not, there is an article on Wikipedia:
Diagonalizable matrix - Wikipedia, the free encyclopedia.)