Uniform Convergence of Power Series

[markiv]

Given a power series $\sum a_n x^n$ with radius of convergence $R$, it seems that the series converges uniformly on any compact set contained in the disc of radius $R$. This might be a silly question, but what's an example of a power series that doesn't actually also converge uniformly on the whole open disc of radius $R$? I am assuming uniform convergence on all compact subsets does not imply uniform convergence on the whole open disc?

 

[micromass]

Consider the Taylor expansion of $f(x)=\tan(x)$. This has a radius of convergence of $\pi/2$. But the series cannot converge uniformly on the entire disk $]-\pi/2,\pi/2[$. Indeed, it is a theorem that the uniform limit of bounded functions is bounded. Clearly, $\tan(x)$ isn't bounded.

 

[jbunniii]

Another easy example:
$$\sum_{n = 0}^{\infty} x^n = \frac{1}{1 - x}$$
for $|x| < 1$. This is clearly unbounded as $x$ approaches 1.

 

[Mute]

Another easy example:
$$\sum_{n = 0}^{\infty} x^n = \frac{1}{1 - x}$$
for $|x| < 1$. This is clearly unbounded as $x$ approaches 1.

Hm. If I recall correctly from my Complex variables class years ago, the geometric series actually is uniformly convergent on $|z| < 1$. However, it is not uniformly convergent on $|z| \leq 1$.

Do I remember incorrectly?

 

[jbunniii]

Hm. If I recall correctly from my Complex variables class years ago, the geometric series actually is uniformly convergent on $|z| < 1$. However, it is not uniformly convergent on $|z| \leq 1$.

Do I remember incorrectly?

The partial sums are
$$s_N(x) = \sum_{n = 0}^{N-1}x^n = \frac{1 - x^{N}}{1 - x}$$
The limit is
$$s(x) = \sum_{n = 0}^{\infty}x^n = \frac{1}{1 - x}$$
So
$$s(x) - s_N(x) = \frac{x^N}{1 - x}$$
This is the error in approximating the series by the N'th partial sum. For any fixed N, this error is arbitrarily large as $x \rightarrow 1$. If the convergence were uniform, we would be able to uniformly bound the error as small as we like by making N large enough.

I think what you are remembering is that the convergence is uniform on $|z| \leq R$ for any positive $R < 1$.

 

[markiv]

Thank you. This makes a lot of sense.

 

[Mute]

The partial sums are
$$s_N(x) = \sum_{n = 0}^{N-1}x^n = \frac{1 - x^{N}}{1 - x}$$
The limit is
$$s(x) = \sum_{n = 0}^{\infty}x^n = \frac{1}{1 - x}$$
So
$$s(x) - s_N(x) = \frac{x^N}{1 - x}$$
This is the error in approximating the series by the N'th partial sum. For any fixed N, this error is arbitrarily large as $x \rightarrow 1$. If the convergence were uniform, we would be able to uniformly bound the error as small as we like by making N large enough.

I think what you are remembering is that the convergence is uniform on $|z| \leq R$ for any positive $R < 1$.

Yes, that's probably the result I was thinking of.