show that $e^P \approx I + 1.718*P$
I know that : Let ${P}\in R^{nxn}$ $$e^P = \sum_{k=0}^\infty {\frac{P^k}{k!}}$$
My question is : Let $P \in R^{nxn}$ be a projection. Show that $$e^P \approx I + 1.718*P$$
I have no idea about number 1.718 ?
$\endgroup$ 22 Answers
$\begingroup$Using the fact that for a projection matrix, $P^k=P$ for positive $k$, $$ \begin{align} \sum_{k=0}^\infty {\frac{P^k}{k!}} &=I+\sum_{k=1}^\infty {\frac{P}{k!}}\\ &=I+P\left(\sum_{k=0}^\infty {\frac{1}{k!}}-1\right)\\[9pt] &=I+(e-1)P \end{align} $$
$\endgroup$ 1 $\begingroup$hint
If $P $ is a projection then
$$P^2=P^3=...=P $$
$\endgroup$ 1
