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 &lt; 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 &lt; 1.

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