Really sorry, but I'm still very confused. Perhaps there's an alternate derivation, but that power series is a Maclaurin expansion to me.- you derive it from expanding $exp(x)$ at $x=0$. Sticking in some function instead of ##x## would seem to break that derivation...Otherwise, say I stuck in...
That expansion is the Maclaurin expansion for ##e^{x}##:
$$ exp(x) = \sum_{n=0}^{\infty} \frac{d^{n}}{dx^{n}}\bigg[exp(x)\bigg]_{x=0} \frac{x^{n}}{n!} = \sum_{n=0}^{\infty} \frac{x^{n}}{n!} $$
But when we have ##exp(u) ## where ##u=f(x)##, it's not clear to me why we still get the same...
Sure, I do know what a pole is. But that doesn't help.Obviously the coefficients of the expansion depend on the expansion point... My question is maybe better written as..." From the formal definition of the Taylor expansion, how do we arrive at (1)? "
Thanks
So, I was doing a question on Laurent series. Part of it asked me to work out the pole of the function:
$$ exp \bigg[\frac{1}{z-1}\bigg]$$
The answer is ##1## - since, we can write out a Maclaurin expansion:
(1) $$ exp\bigg[\frac{1}{z-1}\bigg] = 1+\frac{1}{z-1}+\frac{1}{2!}\frac{1}{(z-1)^{2}}...