What arguments are available to deduce that the radius of convergence of the Taylor series of $$f(x)=\frac{x^3-1}{x-2}$$ centered at $x=0$ is 2, where $f$ is considered as a real function of a real variable?

Note: I am aware that the answer is the distance to the closest singularity, but I do not want an invocation of that fact. What arguments could you use to get to the result if you had forgotten this fact?

Observe that the series for $x^3-1$ (namely itself) has radius $\infty$, while the series for $(x-2)^{-1}$ has radius 2. Since the radius of convergence of the product of two power series is at least the minimum of the radii of the factors, and since Taylor series are multiplicative, we conclude that the radius of convergence of the Taylor series of $f$ at $x=0$ is at least 2.

The harder part is showing it can't be more than 2. It is obvious that $f$ can't equal its Taylor series past 2 because of the discontinuity, but this alone does not imply that the Taylor series for $f$ can't converge past 2. The result follows, though, as soon as we show $f$ equals its Taylor series out to 2.

So far as I can see, there are two possibilities:

1. argue by direct calculation

This approach ignores the observations above; we simply compute. The Taylor series of the product $(x^3-1)(x-2)^{-1}$ is the product of the Taylor series of the factors,


The ratio test shows this converges when $|x|<2$.

2. argue using Cauchy's theorem on the expansion of a function in a power series in its domain of analyticity (theorem 16.7 in Markushevich)

Since the corresponding complex function is differentiable in the disk $|z|<2$, Cauchy's theorem says it equals its Taylor series on that disk; hence the same is true for the real function $f$, which must therefore equal its Taylor series out to $x=2$. Since $f$ has an infinite discontinuity at $2$ but Taylor series are continuous, the series can't converge past 2.

Are there other approaches that don't amount to these two in disguise? I'm curious whether I'm missing something.

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    $\begingroup$ ... nearest non-removable singularity, of course. $\endgroup$ – Robert Israel Aug 16 '16 at 17:02
  • $\begingroup$ @hardmath: I misspoke. What I meant is I want the argument, not just an invocation of the fact. $\endgroup$ – symplectomorphic Aug 16 '16 at 17:05
  • $\begingroup$ Okay, the theorem says if the power series (say about zero, for simplicity) converges at $x=a$, then the function is absolutely convergenet (thus analytic) in the open disk of radius $a$ (about the origin). So there are no non-removable singularities closer than the radius of convergence, ie. radius of convergence is at least the distance to non-removable singularity. You want a proof that absolute convergence of a power series implies analyticity? $\endgroup$ – hardmath Aug 16 '16 at 17:10
  • $\begingroup$ @hardmath: no. I already have an argument that the radius of convergence must be at least 2: see my 3rd paragraph. what I want is the rest of the argument, or another argument entirely, that is not simply an invocation of "a theorem says the answer is 2." as far as I can tell, Robert Israel's answer is just my second argument. $\endgroup$ – symplectomorphic Aug 16 '16 at 17:13
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    $\begingroup$ @hardmath: thanks, that is a slightly different take on the issue and helpful. I feel that my use of the word "heuristic" distracted everyone from the substance of the question. $\endgroup$ – symplectomorphic Aug 16 '16 at 17:38

It's not a "heuristic". The function $f(z) = \dfrac{z^3-1}{z-2}$ is analytic on the complex plane except for the singularity at $z=2$, which is a pole (note that $|f(z)| \to \infty$ as $z \to 2$). Therefore its Taylor series about $z=a$ has radius of convergence $|a-2|$.

  • $\begingroup$ Well, yes, but what I meant is I want the reasoning behind the "therefore." Isn't this just argument #2? $\endgroup$ – symplectomorphic Aug 16 '16 at 17:03
  • $\begingroup$ I am asking for the reasoning behind your similar answer here, which Pete Clark also suggested it would be useful to give, in that context. I believe the reasoning proceeds by way of the theorem I cite in argument #2. $\endgroup$ – symplectomorphic Aug 16 '16 at 17:42
  • $\begingroup$ Yes, it is basically from Cauchy's theorems. $\endgroup$ – Robert Israel Aug 16 '16 at 19:14

Boring approach: $f(x)=P(x)+\frac{C}{x-2}$, where $P(x)$ is a polynomial and $C$ a constant. The conclusion follows.
Another approach without mentioning poles(but uses the same idea): since the power series of $f(x)$ converges to $f(x)$ at least in $(0,2)$, it tends to infinity as $x\to 2$. Hence it cannot converge in an interval containing $2$. This shows its radius of convergence has to be $2$.

  • $\begingroup$ yes, thanks; hardmath mentioned this approach in the comments. I was missing the more elementary way of getting to the fact that $f(x)$ equals its Taylor series on $(0,2)$. this falls out of the argument using complex analysis, but it is also obvious from the real setting if we write the function as your sum. $\endgroup$ – symplectomorphic Aug 17 '16 at 6:23

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