Let $z = x + i y$, and consider the following function:

$$ f(z) = e^{\frac{1}{1 + z^2}} \qquad z \in \mathbb{C} $$

Note that at $z = \pm i$ the function does not converge. We see that on $\mathbb{R}$ when $y = 0$, this becomes:

$$ f(x) = e^{\frac{1}{1 + x^2}} \qquad x \in \mathbb{R}, $$

which converges nicely for all values $x \in \mathbb{R}$. Furthermore, this function is analytic on $\mathbb{R}$, and so there exists some power series of form:

$$ f(x) = \sum_{n = 0}^{\infty} a_n (x - x_0)^n \qquad a_n, x, x_0 \in \mathbb{R} $$

that converges for all $x \in \mathbb{R}$. Does this power series naturally extend to the complex plane if we have $x \mapsto z$, i.e.:

$$ f(z) = \sum_{n = 0}^{\infty} b_n (z - z_0)^n \qquad z \in \mathbb{C}, \quad z_0 = x_0 + i0, \quad b_n = a_n + i 0 $$

In this case, would the function shown above thus demonstrate that a complex-valued power series that converges on all of $\mathbb{R}$ not necessarily converge in $\mathbb{C}$?

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    $\begingroup$ Why do you think that $f(x)$ has a power series with an infinite radius of convergence on $\mathbb{R}$? $\endgroup$ Nov 20 '17 at 22:05
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    $\begingroup$ Your Taylor expansion on $\mathbb R$ will have radius of converge equal to the distance between $x_0$ and $\pm i$. For example, if you expand around $0$ the series has radius of convergence $1$. It starts as $e - e x^2 + (3 e x^4)/2 - (13 e x^6)/6 + (73 e x^8)/24 - (167 e x^{10})/40 + (4051 e x^{12})/720 + O(x^{13})$ . $\endgroup$
    – lulu
    Nov 20 '17 at 22:06
  • $\begingroup$ Ah, that makes sense, thank you for pointing this out! $\endgroup$
    – Anton Xue
    Nov 20 '17 at 22:07

No. If a power series converges on all $\mathbb{R}$, it(s extension) converges for all complex numbers. One quick way to see why this holds is considering the formula for the radius of convergence and recalling that one proves absolute convergence with it, thus holding for $\mathbb{C}$ as well.


The fundamental flaw in your analysis is this (but see @AloizioMacedo's comment):

We see that on $\mathbb{R}$ when $y = 0$, this becomes:

$$ f(x) = e^{\frac{1}{1 + x^2}} \qquad x \in \mathbb{R}, $$

which converges nicely for all values $x \in \mathbb{R}$.

The problem is that the function you've written can't "converge" because it's not a series.

There is a power-series for $e^x$, and there's one for $\frac{1}{1+x^2}$. But that latter series only converges for $x^2 < 1$, so there isn't an everywhere convergent power series for that function on the reals (or at least "the obvious one isn't everywhere convergent").

  • $\begingroup$ I don't think that is the fundamental flaw per se. That block could be easily omitted and the error would still exist. I think that the crux is in "Furthermore, this function is analytic on $\mathbb{R}$, and so (assumes that the function is given globally by a power series)". I know this is exactly what you pointed out in the answer, this comment is just to suggest that the fundamental flaw lies elsewhere. $\endgroup$
    – Aloizio Macedo
    Nov 20 '17 at 22:15
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    $\begingroup$ I guess I agree -- the sentence I quoted seemed to me to be shorthand for the sentence that followed, but if you replace "converges nicely" with "is defined" (which might be what OP meant), then the second is where things go wrong. The first sentence, as written, seems to convey a fundamental misunderstanding...but it could be just a momentary lapse in word-choice. $\endgroup$ Nov 20 '17 at 23:30

No. You might have heard the term "radius of convergence" before - in this case, that isn't an abuse of language. Power series converge within open disks in the complex plane, and the radius of convergence is the distance to the nearest singularity. In this case, since you require your series to converge on $\mathbb{R}$, the radius of convergence would have to be $\infty$, so that the series would converge on all of $\mathbb{C}$


It is easy to prove that if a power series converges for some $z_1\ne 0$, then it converges absolutely for every complex $z$ with $|z|<|z_1|$:

In order for the the power series to converge at $z_1$, the sequence $|a_nz_1^n|$ must go to $0$ -- in particular it must be bounded by some number $K$. We then have that $$ |a_nz^n| = \left| \frac{z}{z_1} \right|^n \cdot |a_nz_1^n| \le K\left|\frac{z}{z_1} \right|^n $$ so the terms of the power series at $z$ are bounded by a geometric sequence with common factor $\left|\frac{z}{z_1}\right|$ which is assumed to be $<1$. This forces the partial sums of the power series at $z$ to be a Cauchy sequence, and therefore it converges.

In particular we can let $z_1$ be real, so if a power series converges for all reals, it will necessarily also converge on all of $\mathbb C$.


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