Uniform Convergence: Explaining f_n(x)=1-x^n all over 1-x

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Discussion Overview

The discussion centers around the uniform convergence of the function f_n(x) = (1 - x^n) / (1 - x) as n approaches infinity. Participants explore the conditions under which this convergence occurs, particularly focusing on the behavior of the function across different intervals of x.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested

Main Points Raised

  • One participant asks whether f_n(x) is uniformly convergent and requests clarification on its independence from x and the relevant intervals.
  • Another participant points out that the original question lacks clarity regarding the variable n and its limit.
  • It is noted that if x is between -1 and 1, then x^n converges to 0, leading to 1 - x^n converging to 1. However, for x > 1 or x < -1, x^n does not converge, and thus 1 - x^n does not converge either.
  • For x = 1, the function is constantly 0, while for x = -1, the behavior alternates between 0 for even n and 2 for odd n, indicating non-convergence.
  • One participant claims that a sequence of functions converges uniformly on a compact set, stating that 1 - x^n converges uniformly on [ε, 1] for any ε > 0 but does not converge uniformly on (0, 1].
  • Another participant suggests that the interval of convergence should be [ε - 1, 1].
  • A separate question is raised about proving the uniform convergence of (1 + z/n)^n on closed intervals containing zero, with a mention of Cauchy's convergence criterion.

Areas of Agreement / Disagreement

Participants express differing views on the uniform convergence of f_n(x), with some agreeing on specific intervals of convergence while others propose alternative interpretations. The discussion remains unresolved regarding the overall uniform convergence of the function.

Contextual Notes

There are limitations regarding the assumptions made about the intervals and the nature of convergence, particularly concerning the behavior of x^n as n approaches infinity.

jack5322
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could someone please explain if f_n(x)= 1-x^n all over 1-x is uniformly convergent? can someone show why its independent of x if it is and on what intervals? Also, can someone explain if it isn't uniform convergent then can they show a proof and explain. Thanks any help will be greatly appreciated!
 
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Your question is incomplete. What variable (n?) is changing and what is it going to?
 
n is changing, and its going to infinity
 
If x lies between -1 and 1, then xn converges to 0 so 1- xn converges to 1. If x> 1 or x< -1 then xn does not converge at all so 1- xn does not converge. If x= 1, 1- xn= 0 for all x while if x= -1, 1- xn= 0 for even n and 2 for odd n and so 1- (-1)n does not converge. If a sequence of functions converges on a compact (closed and bounded) set, then it converges uniformly on that set so 1- xn converges uniformly on [itex][\epsilon, 1][/itex] for any [itex]\epsilon> 0[/itex] but does NOT converge uniformly on (0, 1].
 
thank you
 
HallsofIvy said:
If x lies between -1 and 1, then xn converges to 0 so 1- xn converges to 1. If x> 1 or x< -1 then xn does not converge at all so 1- xn does not converge. If x= 1, 1- xn= 0 for all x while if x= -1, 1- xn= 0 for even n and 2 for odd n and so 1- (-1)n does not converge. If a sequence of functions converges on a compact (closed and bounded) set, then it converges uniformly on that set so 1- xn converges uniformly on [itex][\epsilon, 1][/itex] for any [itex]\epsilon> 0[/itex] but does NOT converge uniformly on (0, 1].

I believe the interval of convergence should be[itex][\epsilon-1, 1][/itex]
 
Last edited:
how would we rigorously prove that (1+z/n)^n converges uniformly on any closed interval containing zero and R where R is a number between zero and positive infinity? I was thinking of using cauchy's convergence criterion but I always get stuck. Any suggestions?
 

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