Uniform Convergence of Power Series

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Given a power series [itex]\sum a_n x^n[/itex] with radius of convergence [itex]R[/itex], it seems that the series converges uniformly on any compact set contained in the disc of radius [itex]R[/itex]. 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 [itex]R[/itex]? I am assuming uniform convergence on all compact subsets does not imply uniform convergence on the whole open disc?
 

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  • #2
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Consider the Taylor expansion of [itex]f(x)=\tan(x)[/itex]. This has a radius of convergence of [itex]\pi/2[/itex]. But the series cannot converge uniformly on the entire disk [itex]]-\pi/2,\pi/2[[/itex]. Indeed, it is a theorem that the uniform limit of bounded functions is bounded. Clearly, [itex]\tan(x)[/itex] isn't bounded.
 
  • #3
jbunniii
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Another easy example:
[tex]\sum_{n = 0}^{\infty} x^n = \frac{1}{1 - x}[/tex]
for [itex]|x| < 1[/itex]. This is clearly unbounded as [itex]x[/itex] approaches 1.
 
  • #4
Mute
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Another easy example:
[tex]\sum_{n = 0}^{\infty} x^n = \frac{1}{1 - x}[/tex]
for [itex]|x| < 1[/itex]. This is clearly unbounded as [itex]x[/itex] 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?
 
  • #5
jbunniii
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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
[tex]s_N(x) = \sum_{n = 0}^{N-1}x^n = \frac{1 - x^{N}}{1 - x}[/tex]
The limit is
[tex]s(x) = \sum_{n = 0}^{\infty}x^n = \frac{1}{1 - x}[/tex]
So
[tex]s(x) - s_N(x) = \frac{x^N}{1 - x}[/tex]
This is the error in approximating the series by the N'th partial sum. For any fixed N, this error is arbitrarily large as [itex]x \rightarrow 1[/itex]. 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 [itex]|z| \leq R[/itex] for any positive [itex]R < 1[/itex].
 
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Thank you. This makes a lot of sense.
 
  • #7
Mute
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The partial sums are
[tex]s_N(x) = \sum_{n = 0}^{N-1}x^n = \frac{1 - x^{N}}{1 - x}[/tex]
The limit is
[tex]s(x) = \sum_{n = 0}^{\infty}x^n = \frac{1}{1 - x}[/tex]
So
[tex]s(x) - s_N(x) = \frac{x^N}{1 - x}[/tex]
This is the error in approximating the series by the N'th partial sum. For any fixed N, this error is arbitrarily large as [itex]x \rightarrow 1[/itex]. 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 [itex]|z| \leq R[/itex] for any positive [itex]R < 1[/itex].
Yes, that's probably the result I was thinking of.
 

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