The title could be: Uniform Convergence of f_n(x) = nx^n(1-x) on [0,1]?

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Homework Statement



Consider f_n(x) = nx^n(1-x) for x in [0,1].

a) What is the limit of f_n(x)?

b) Does f_n \rightarrow f uniformly on [0,1]?

Homework Equations


The Attempt at a Solution



a) 0

b) Yes...

We know that sup|f_n(x) - f(x)| = |n{\frac{1}{2}}^n(1-\frac{1}{2})|...

and

lim_{n \rightarrow \infty} [sup\{ |f_n(x) - f(x)|: x \in [0,1]\}] = 0

So it must be uniformly convergent on [0,1].

Do you think my answer is correct?Thanks in advance
 
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Artusartos said:

Homework Statement



Consider f_n(x) = nx^n(1-x) for x in [0,1].

a) What is the limit of f_n(x)?

b) Does f_n \rightarrow f uniformly on [0,1]?



Homework Equations





The Attempt at a Solution



a) 0

b) Yes...

We know that sup|f_n(x) - f(x)| = |n{\frac{1}{2}}^n(1-\frac{1}{2})|...

Why? How do you know they will always obtain their maximum in 1/2??
 
If we let y = 1-x, then we may write
f_n(1-y) = n y (1 - y)^n
Now what happens if you choose y = 1/n?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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