Uniform Convergence of Power Series

In summary, we discussed the convergence of power series with a radius of convergence R and how it may converge uniformly on compact sets contained in the disc of radius R, but not necessarily on the entire open disc. We also gave examples of such power series, including the Taylor expansion of f(x) = tan(x) and the geometric series. We concluded that the convergence is uniform on |z| < R for any positive R < 1, but not for |z| \leq R.
  • #1
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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
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
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
jbunniii said:
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
Mute said:
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].
 
  • #6
Thank you. This makes a lot of sense.
 
  • #7
jbunniii said:
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.
 

What is uniform convergence of power series?

Uniform convergence of power series is a mathematical concept that refers to the behavior of a sequence of functions, where each function is a power series. It describes the convergence of the series of functions to a single function, as the number of terms in the series increases.

How is uniform convergence different from pointwise convergence?

Pointwise convergence refers to the convergence of a sequence of functions at each individual point in the domain. Uniform convergence, on the other hand, requires that the convergence holds uniformly across the entire domain, rather than just at individual points.

What is the significance of uniform convergence in analysis?

Uniform convergence is an important concept in analysis because it allows for the manipulation and integration of infinite series of functions. It also guarantees the continuity of the limit function, and allows for the interchange of limits and integrals.

What are the conditions for uniform convergence of a power series?

The power series must converge pointwise on a closed interval, and the convergence must be uniform on any compact subset of that interval. Additionally, the series must have a radius of convergence greater than or equal to the length of the interval.

How is the uniform convergence of a power series tested?

The Weierstrass M-test is a commonly used method for testing the uniform convergence of a power series. It states that if the absolute values of the terms in the series can be bounded by a convergent series, then the original series will also converge uniformly.

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