- #1

- 26

- 0

*open*disc?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter markiv
- Start date

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

- 26

- 0

Physics news on Phys.org

- #2

- 22,183

- 3,321

- #3

- 3,486

- 257

[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

Homework Helper

- 1,388

- 12

jbunniii said:

[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

- 3,486

- 257

The partial sums areMute 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?

[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

- 26

- 0

Thank you. This makes a lot of sense.

- #7

Homework Helper

- 1,388

- 12

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.

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.

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.

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.

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.

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.

Share:

- Replies
- 0

- Views
- 63

- Replies
- 21

- Views
- 1K

- Replies
- 11

- Views
- 1K

- Replies
- 8

- Views
- 2K

- Replies
- 2

- Views
- 2K

- Replies
- 4

- Views
- 2K

- Replies
- 3

- Views
- 1K

- Replies
- 5

- Views
- 802

- Replies
- 2

- Views
- 1K

- Replies
- 9

- Views
- 1K