- #1

- 113

- 5

$$ exp \bigg[\frac{1}{z-1}\bigg]$$

The answer is ##1## - since, we can write out a Maclaurin expansion:

(1) $$ exp\bigg[\frac{1}{z-1}\bigg] = 1+\frac{1}{z-1}+\frac{1}{2!}\frac{1}{(z-1)^{2}} $$

But, I can't actually justify this expansion -surely if we have an expression like ##exp(f(x))##, then the expansion should be:

(2) $$ exp\bigg[f(x)\bigg] = exp[f(0)]+f'(0)exp[f(0)]x+ \ldots $$

Which definitely doesn't agree with the result stated in (1). Why is this?

Thanks!!