# Lagrange inversion formula example unclear

The following example is from De Bruijn's Asymptotic methods in analysis (page 24). The considered equation is

$$x^t = e^{-x}$$

The author wants to transform the equation into the form: $$w=z/f(z)$$, in order to use the Lagrange inversion formula. So he sets $$x=1+z$$ and $$t^{-1}=w$$. And then obtains the equation for $$f(z)$$ as: $$f(z)=-z(1+z)/(\log{(1+z}))$$. And then says:

The function $$f(z)$$ is analytic at $$z=0$$: $$f(z)=-1+c_1z+...$$.

It follows that $$x=1-t^{-1}-c_1t^{-2}+...$$

My question is: This function $$f(z)$$ at $$z=0$$ will be $$0/0$$. i.e. undefined! So how come he says the function is analytic at $$z=0$$?

It's a removable singularity; we commonly identify functions with removable singularities with the function which agrees with them on the original domain and with the singularities removed. It's a removable singularity because the numerator and denominator are analytic in a neighborhood of $$z=0$$ and both scale as $$O(z)$$ as $$z \to 0$$.
• But in this case shouldn't the writer define the new function's value at the points of singularity? He didn't mention what will be the value of $f$ at 0. And the use of the Lagrange inversion theorem depends on that value. – Dorgham Sep 21 '18 at 16:31
• @Dorgham It's somewhat redundant to do so, because we almost always select the continuous extension, i.e. we define $f(0)$ to be $\lim_{z \to 0} f(z)$. That's what it means to "remove the singularities"; selecting a discontinuous extension leaves the singularities intact. – Ian Sep 21 '18 at 17:13