Uniform convergence

  • Thread starter 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 isnt uniform convergent then can they show a proof and explain. Thanks any help will be greatly appreciated!
 

Answers and Replies

  • #2
mathman
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Your question is incomplete. What variable (n???) is changing and what is it going to?
 
  • #3
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n is changing, and its going to infinity
 
  • #4
HallsofIvy
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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].
 
  • #5
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thank you
 
  • #6
mathman
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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]
 
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  • #7
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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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