# Problem

I need help with simplifying following sum:

$$1 + \sum_{i=1}^{\infty}{\frac{1}{i!} * (-1)^i * a * (a + b)^{i-1}}$$

and can get the $a$ out to get

$$1 + a*\sum_{i=1}^{\infty}{\frac{1}{i!} * (-1)^i * (a + b)^{i-1}}$$

but really don't know where to go from there. Any help will be appreciated.

## Disclamer

If you suspect that this is homework, you are correct. This is part of HW assignment from my linear algebra course. The assignment was to express

$$exp(A = \begin{pmatrix} -a & b \\ a &-b \end{pmatrix})$$

in the most simple form. I determined that

$$A^n = (a+b)^{n-1} * \begin{pmatrix} (-1)^n*a & (-1)^{n-1}*b \\ (-1)^{n-1}*a & (-1)^n*b \end{pmatrix}$$

from that I got the above mentioned sum as value of $exp(A)_{00}$, but don't know how to simplify it further.

As this is a HW, I would appreciate, if you mentioned how you figure out how to simplify it in your answers. Thanks.

Computing $\exp(A)$, where $A=\pmatrix{-a&b\\a&-b}$, is easy once $A$ is diagonalized. To do this, we need to find the eigenvectors of $A$. The characteristic polynomial \begin{align} \chi_A(\lambda) &= \det(A-\lambda E) = \left|\matrix{-a-\lambda & b \\ a & -b-\lambda}\right|= (-a-\lambda)(-b-\lambda)-ab \\&= \lambda^2+(a+b)\lambda \end{align} has roots at $\lambda=0$ and $\lambda=-(a+b)$, so we have $2$ real eigenvalues whenever $a+b\neq 0$. The eigenvectors are $\pmatrix{b\\a}$ and $\pmatrix{-1\\1}$, respectively. Thus, letting $P=\pmatrix{b&-1\\a&1}$ we have $P^{-1} =\frac{1}{a+b}\pmatrix{1&1\\-a&b}$ and $$P^{-1}AP = \frac{1}{a+b}\pmatrix{1&1\\-a&b} \pmatrix{-a&b\\a&-b} \pmatrix{b&-1\\a&1} = \pmatrix{0&0\\0&-(a+b)}=:D.$$ Now \begin{align} \exp(A) &= \exp(PDP^{-1}) = P\exp(A)P^{-1} = P \pmatrix{1 & 0\\0&e^{-(a+b)}} P^{-1}\\ &= \frac{e^{-(a+b)}}{a+b} \pmatrix{a+be^{a+b}&be^{a+b}-b\\ae^{a+b}-a&ae^{a+b}+b}. \end{align} When $a+b=0$ you have $A^2=0$ so $$\exp(A) = \sum_{n=0}^\infty \frac{1}{n!} A^n = E + A^1 = \pmatrix{-a+1&b\\a&-b+1}.$$