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1. ### I Expansion of $e^{f(x)}$

Sorry for the slow reply all - many thanks for the help - that actualy makes a lot more sense :)
2. ### I Expansion of $e^{f(x)}$

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...
3. ### I Expansion of $e^{f(x)}$

$$\frac{d}{dx}\bigg( e^{f(x)} \bigg) = f'(x) e^{f(x)} \neq e^{f(x)}$$ and etc. for higher differentials in the series... ??
4. ### I Expansion of $e^{f(x)}$

But surely the fact that this is now a function and not a simple number must go into the differential? I.e we need to use some form of chain rule...
5. ### I Expansion of $e^{f(x)}$

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...
6. ### I Expansion of $e^{f(x)}$

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
7. ### I Expansion of $e^{f(x)}$

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}}...