# Domain of convergence of $f^{-1}: \mathbb R ^N \mapsto \mathbb R^N$ taylor series

In another question, I ask about the topology of the singular manifold of the Jacobian. What i want to ask in here is about the radius of convergence of a Taylor series expansion of the inverse function. In particular;

• is the domain of covergence always a ball? (GEdgar answer points that this is not so in general with a very simple counterexample)

• what determines the shape of the domain of convergence in general? is it only bounded by the singular manifold of the Jacobian?

Edit I did an inexcusable mistake in this question; i completely forgot to mention the very important detail that what i am asking here is for what valid extensions exists of the inverse function theorem; in this regard i wanted to gather if the domain of convergence is going to be only bounded by the singular jacobian manifold, or if in general it can be smaller (so the original theorem is all it can be said with generality)

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Investigate the domain of convergence for the Taylor series of $$f(x,y) = \frac{1}{1-x} + \frac{1}{2-y}$$ near the point $(0,0)$. Is that domain a ball?

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good point; they are entirely separable, so the domain of convergence is the cartesian product of the domains for each axis. So the domain is a square patch –  lurscher Nov 17 '11 at 3:00
ok, so i take that in general is hard to say the shape of the convergence domain. –  lurscher Nov 17 '11 at 3:01
ok, so the question would have to change to 'why then speak of a convergence radii'? or more like 'is the domain of convergence only bounded by the singular manifold of the Jacobian'? –  lurscher Nov 17 '11 at 3:04

The domain of convergence can be much smaller than the regions bounded by the singular manifold of the Jacobian. For example, consider this slight modification of GEdgar's example:

$$f(x,y) = \arctan(x) + \arctan(y)$$

It's not a ball, and the Jacobian is everywhere non-singular.

The domain of convergence can also be much larger than the region bounded by the singular manifold of the Jacobian. Let $f$ be any polynomial function. So the domain of convergence is all of $\mathbb R^n$. But polynomials tend to have many degenerate Jacobian matrices. $f(x) = 1-x^2$ is a good one.

So generally-speaking there's no relation at all between the singular set of the Jacobian and the domain of convergence of the Taylor expansion.

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sorry, i messed up really bad in this question; i was thinking in extensions of the inverse function theorem, and i completely forgot to mention that –  lurscher Nov 24 '11 at 11:57