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For $|x| < 1$ we have the identity $$ \frac{1}{1 + x^2} = 1 - x^2 + x^4 -x^6 \dots$$

I read that this holds because the left-hand side has singularities at $\pm i$ and by requiring $|x|<1$ we exclude them.

But: if we think of $\mathbb C$ as $\mathbb R^2$ then there are more points than $|x|<1$ where the left-hand side is defined, for example $2+2i$.

Does the identity above not hold for all $x \in \mathbb C \setminus \{i, -i\}$? And if yes, why do we require $|x|<1$ and in general, power series' domain of convergence to be disk shaped? (like $|z|<K$ instead of giving a precise domain like $\mathbb C \setminus \{ \text{ points where it goes wrong} \}$)

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I think it is a theorem that the converge area must be a disk. –  user46090 Oct 29 '12 at 10:49
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The RHS is not defined for complex numbers $x$ such that $|x|\geqslant1$ hence the identity LHS = RHS has no meaning outside the unit disk of equation $|x|\lt1$. –  Did Oct 29 '12 at 10:50
    
@did Alright, for this example the RHS is not defined (infinite) outside the unit disk but what about other power series like eg the power series of $\sin (x)$? –  Frosty Oct 29 '12 at 11:09
    
What about them? The power series of the sine has infinite radius of convergence hence one is rarely outside the disk of convergence... –  Did Oct 29 '12 at 11:11

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We do not require that the domain of convergence for a power series will be disk shaped, it just happens to be the case. This is quite a difference between general series of functions and power series.

There is a theorem that states that every power series $\sum_{i=1}^{\infty} a_n (z - z_0)^n$ comes with an associated number $0 \leq R \leq \infty $ called the radius of convergence. It satisfies the following properties:

  1. The power series converges for all $z$ such that $|z - z_0| < R$ (all $x$'s whose distance from $z_0$ is less than $R$).
  2. The power series diverges for all $z$ such that $|z - z_0| > R$.
  3. When $|z - z_0| = R$, the series might converge or might diverge. This must be checked on a case-to-case basis.

You can write various explicit formulas for the radius of converge in terms of the coefficients $a_n$. For example, the Cauchy-Hadamard formula for the radius of convergence is $$ R = \frac{1}{\mathrm{lim \;sup}_{n \rightarrow \infty} |a_n|^{\frac{1}{n}}} $$

So, given the power series $\sum_{i=0}^{\infty} (-1)^i z^{2i}$ around $0$, if you plug $z = 2$, you can see that the terms of the series are $(-1)^i 2^{2i}$, and so they don't even tend to $0$ and the series diverges. If you plug $z=1$, you see that the series $1 - 1 + 1 - 1 + ...$ also diverges. This means that the radius of convergence must be $1$ or less. You can prove directly, or using the formula, that $R$ is exactly $1$.


Everything said above is independent of whether you know to what function the power series converges. We didn't care above that the power series converged to $1/(1 + z^2)$. Denote by $B = \{ z \; | \; |z| < 1 \}$ the unit disc. What happens is that while the power series converges only when $|z| < 1$, the function it converges to, $u(z) = \frac{1}{1 + z^2} : B \rightarrow \mathbb{C}$, defined a priori only on $B$, can be extended to a function $\tilde{u}(z) = \frac{1}{1 + z^2} : \mathbb{C} \setminus \{ \pm i \} \rightarrow \mathbb{C}$ defined on a larger (maximal) domain of definition. But the power series won't converge to that function - it won't converge to anything when $|z| > 1$.

If you start the other way, from $\tilde{u}$, a function defined on $\mathbb{C} \setminus \{ \pm i \}$, and try to develop it into a Taylor series around $z = 0$, then what happens is that you get your series, the series has radius of convergence $R = 1$, and it converges on $|z| < 1$ to $\tilde{u}\restriction_B = u$.

Its not always the case that the Taylor expansion of a function converges to the function itself. If you knew in advance that the Taylor series must converge to $\tilde{u}$, you could deduce that its radius of convergence can't be more than $R = 1$, as $\tilde{u}$ is not defined at $z = \pm i$. And indeed, because the function $u$ is rational (more generally, holomorphic on its domain of definition), there is a theorem that states the Taylor series of $\tilde{u}$ around $z_0$ must converge to the function $\tilde{u}$ on the maximal possible (open) disc around $z_0$. For $z_0 = 0$, we have $|z| < 1$.


I hope I didn't confuse you too much, but somehow behind such "innocent" claims lies a lot of difficult / interesting / subtle mathematics which is worth being familiar with.

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You did not confuse me, your answer is beautiful and I now understand it. I am now in the process of becoming familiar with these subtle mathematics!! –  Frosty Jan 9 at 21:15
    
Just to be clear: $\widetilde{u}$ is also holomorphic at for example $z_0=2$, right? So then its Taylor series at $z_0=2$ converges to it on for example a disk of radius $1/2$ around $z_0=2$? –  Frosty Apr 8 at 13:17
    
Yep. In fact, it will even converge to $\tilde{u}$ on an open disc of radius $\sqrt{5}$ (the maximal possible open disc around $z_0 = 2$ in the domain of definition of $\tilde{u}$). –  levap Apr 8 at 13:47

To address your question of why a power series' domain of convergence must be a disk, you can obtain a formal proof in any complex analysis textbook, but perhaps let me try to motivate why this should be. When you consider a power series $\sum_n a_n z^n$, even though it may have different values for different values of $z$ on the circle $|z| = R$, one of the most features of analytic functions is that they somehow have some form of "radial" growth that relates all the values of the function along a circle; this is made precise by the Cauchy integral formulas. If a singularity occurs at one point in a power series representation, then this leads to "messing up" the convergence at points beyond.

For an even simpler example, consider the geometric series formula, which holds for $|z| <1$: $$ \frac{1}{1 - z} = 1 + z + z^2 + \ldots $$ The function $1/(1 - z)$ is only undefined at $z = 1$, but try plugging in $z = 2$ into the above series representation.

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Regarding the geometric series, I understand that it converges for $|z|<1$ and diverges for $|z|>1$ but what about all the complex numbers with $|z|=1$? –  Pranasas Nov 21 at 9:56

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