2
$\begingroup$

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 $z=0$?

$\endgroup$
2
$\begingroup$

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 $z \to 0$.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Dorgham Sep 21 '18 at 16:31
  • $\begingroup$ @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. $\endgroup$ – Ian Sep 21 '18 at 17:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.