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Uniform and pointwise convergence?
Convergence and uniform convergence?
Let (n=1 -> infinity) x(1-x)^n.
a) Show that this series is convergent on [0,1] to f, where f(0) = 0 and f(x) = 1 for 0<x<=1.
By using the ratio test,...
x(1-x)^n+1/x(1-x)^n = 1-x
So, i find that |1-x|<1, so x is convergent on 0 to 2, so it is obviously convergent on [0,1].
What I don't understand is how to find the proper function where f(0) = 0 and f(x) = 1 on 0<x<=1.
b) Show that this series is not uniformly convergent on [0,1].
I know that there is a theorem for sequences that states that if f isn't continuous, then the sequence isn't uniformly convergent. Does the same hold here?
1 Answer
- SuleimanLv 610 years agoFavorite Answer
Observe that Σ(1-x)^n is just the ordinary geometric series for 0 < x ≤ 1,
so on this interval
Σx(1-x)^n = xΣ(1-x)^n = x/(1 - (1 - x)) = 1.
And of course Σ0.(1-0)^n = 0, so the series converges pointwise to
f(x) = 1 (0 < x ≤ 1)
f(0) = 0.
f is discontinuous, so convergence is not uniform.
