1
$\begingroup$

Suppose I have a power series approximation $y$ to an invertible function $f(x)$, and I know that $y$ convergences around $x$ on an interval $(-R,R)$, $R$ being the radius of convergence.

How are the radii of convergence of the power series $y$ and its reversion (i.e. inverse) related?

Problem:

I have a power series approximation to a function which I cannot invert, so I need the reversion of the power series $y^{-1}$ to approximate the function inverse $f^{-1}$.

If I have the function $f(x)$ on an interval $a \leq x_{0} \leq b$ in the domain $D(f)$, and a power series which converges on that same interval, how can I determine on which interval in the range $R(f)$ the inverse power series will converge?

I obviously want to have a radius of convergence of the inverse power series such that it covers the range $f(a) \leq f(x) \leq f(b)$.

$\endgroup$
  • 1
    $\begingroup$ Without further restriction, there are no relation. Consider the function $f(x) = x^3$. It is invertible over whole $\mathbb{R}$ and yet its inverse function doesn't have a power series expansion at $x = 0$. $\endgroup$ – achille hui Jul 26 '14 at 13:20
  • $\begingroup$ So I would have to check both the series and its inverse for convergence separetely? And determine the amount of terms of the series required to obtain a prescribred radius of convergence of the inverse in an iterative manner? $\endgroup$ – jjack Jul 26 '14 at 15:02
2
$\begingroup$

This is an expansion and follow up of my comment.

As mentioned in comment.

$\hspace0.5in$ Without further restriction, there are no relation.

An example is the function $f(x) = x^3$ which is invertible over the real axis and yet its inverse function doesn't have a power series expansion at $x = 0$.

In general, if your function is invertible only over real axis (or part of it), there isn't much one can do but check the power expansions of the function and its inverse function separately.

However, if your function is analytic and injective over some open subset of $\mathbb{C}$ (such a function is known as an univalent function), there is a theorem by Koebe which can help you.

Koebe quarter theorem The image of any univalent function $\varphi : B(0,1) \to \mathbb{C}$ from the unit disk $B(0,1)$ onto a subset of the complex plane contains the disk $\displaystyle\;B\left( \varphi(0), \frac{|\varphi'(0)|}{4}\right)\;$.

Let's say your function $f$ is univalent over the open disk $B(z_0,R)$. Since $f$ is analytic and locally injective at $z = z_0$, $f'(z_0) \ne 0$. Now define a function $\varphi : B(0,1) \to \mathbb{C}$ by

$$B(0,1) \ni \omega \mapsto \varphi(\omega) = f(z_0 + R \omega ) \in \mathbb{C}$$

We have $\varphi(0) = f(z_0)$ and $\varphi'(0) = R f'(z_0)$. By Koebe quarter theorem, $$f(B(z_0,R)) = \varphi(B(0,1)) \supset B\left( f(z_0), \frac{R|f'(z_0)|}{4} \right)$$

This implies the nearest singularity of the inverse function $\varphi^{-1}$ in $\mathbb{C}$ is at least at a distance $\displaystyle\;\frac{R|f'(z_0)|}{4}$ from $f(z_0)$. As a result, the radius of convergence of the power series of $\varphi^{-1}$ and hence that of $f^{-1}$ at $f(z_0)$ is at least $\displaystyle\;\frac{R|f'(z_0)|}{4}$.

$\endgroup$
  • $\begingroup$ This is a useful theorem. In genral, what are my options if I cannot invert a certain function? Is there only the reversion of power series approximations at a number of points in the domain? $\endgroup$ – jjack Jul 27 '14 at 7:46
  • $\begingroup$ @jjack I'm not an expert on any of this. The only other thing that comes to my mind is Lagrange inversion theorem which sometimes allow you to derive the "formal" power series of the inverse function more easily. $\endgroup$ – achille hui Jul 27 '14 at 8:38
  • $\begingroup$ Thank you. I think I will post a separate question on this. $\endgroup$ – jjack Jul 27 '14 at 8:45

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.