Show ##\{nx^{n}(1-x)\}_{n = 0}^{\infty}## converges uni

[Euklidian-Space]

Homework Statement

show $\{nx^{n}(1-x)\}$ converges uni on [0,1]

Homework Equations

The Attempt at a Solution

Note, that $\sum nx^n(1-x)$ converges uni on [0,1] by Abels' Theorem. Therefore the series follows the cauchy criterion,
$\forall \epsilon > 0$, $\exists N > 0$ such that if n,n > N then

$$\left|(m+1)x^{m+1}(1 - x) + ... + nx^{n}(1- x)\right| < \epsilon\,\,\, \forall x \in [0,1]$$,

Now since $nx^n(1 - x) \geq 0$ for all n we can say

$$\left|nx^{n}(1 - x) - mx^{m}(1 - x)\right| < \epsilon\,\,\, \forall x \in [0,1].$$

Which implies that $\{nx^n(1 - x)\}$ is uniformly cauchy $\forall x \in [0,1]$ and therefore uniformly convergent for $x \in [0,1]$

Dont know if this is right

 

[wabbit]

Note, that $\sum nx^n(1−x)$ converges uni on [0,1] by Abels' Theorem

Care to elaborate ?

Note that once you've said that, you're done - if the series converges uniformly so does the sequence of course, there is no need for a Cauchy argument. But your first step is mysterious (and unnecessary).

 

[Euklidian-Space]

Care to elaborate ?

Note that once you've said that, you're done - if the series converges uniformly so does the sequence of course, there is no need for a Cauchy argument. But your first step is mysterious (and unnecessary).

Well my version of Abel's theorem states
Let $g(x) = \sum a_{n}x^{n}$ that converges at some point R > 0. Then the series converges uniformly on [0,R]. I used the cauchy argument since we never proved series uniformly convergent implies the sequence is uniformly convergent.

 

[micromass]

Well my version of Abel's theorem states
Let $g(x) = \sum a_{n}x^{n}$ that converges at some point R > 0. Then the series converges uniformly on [0,R].

That is:
1) Incorrect, if $g(x)$ converges at some point $R>0$, then you are only guaranteed convergence on $[0,r]$ for $r<R$.
2) Not what Abel's theorem says.

See http://jlms.oxfordjournals.org/content/s1-39/1/81.extract

 

[Euklidian-Space]

That is:
1) Incorrect, if $g(x)$ converges at some point $R>0$, then you are only guaranteed convergence on $[0,r]$ for $r<R$.
2) Not what Abel's theorem says.

See http://jlms.oxfordjournals.org/content/s1-39/1/81.extract

i am quite literally picking this up verbatim from my text
it is attached to this post

 

[micromass]

Which book is this? The flaw in the proof seems to be assuming that $\left( \frac{x}{R} \right)^n$ is monotone decreasing, which is not the case for $x=R$.

 

[Euklidian-Space]

understanding analysis, Abbot

 

[micromass]

OK, never mind. The theorem and proof is correct. Your proof is correct too then.

 

[micromass]

But, your proof is unnecessarily complicated by the way. It is much easier to find the maximum of the function on each interval, and to prove those maxima converge to $0$. There is no real reason to transform this into a power series.

 

[Euklidian-Space]

But, your proof is unnecessarily complicated by the way. It is much easier to find the maximum of the function on each interval, and to prove those maxima converge to $0$. There is no real reason to transform this into a power series.

Yeah we haven't been given that theorem, this was me trying to tackle this thing with the tools available

 

[wabbit]

Actually, your sequence $f_n(x)=nx^n(1-x)$ does not converge uniformly on $[0,1]$.
It converges pointwise to zero on that interval. However, if you set $x_n=\frac{n}{n+1}$ which is the value at which $f_n$ reaches its maximum over $[0,1]$ , you find that $f_n(x_n)=(\frac{n}{n+1})^{n+1}=(1-\frac{1}{n+1})^{n+1} \ _{\overrightarrow{n\rightarrow\infty}}\frac{1}{e}>0$ and this proves that the convergence is not uniform.

Edit : the reason your proof is incorrect, and the theorem you quote is not applicable here, is that $\sum nx^n(1-x)$ is not a power series. You can of course write it as $(1-x)\sum nx^n$ but then you will note that the power series $\sum n x^n$ does not converge for $x=1$ so you cannot use that theorem.

 

[Euklidian-Space]

Actually, your sequence $f_n(x)=nx^n(1-x)$ does not converge uniformly on $[0,1]$.
It converges pointwise to zero on that interval. However, if you set $x_n=\frac{n}{n+1}$ which is the value at which $f_n$ reaches its maximum over $[0,1]$ , you find that $f_n(x_n)=(\frac{n}{n+1})^{n+1}=(1-\frac{1}{n+1})^{n+1} \ _{\overrightarrow{n\rightarrow\infty}}\frac{1}{e}>0$ and this proves that the convergence is not uniform.

Edit : the reason your proof is incorrect, and the theorem you quote is not applicable here, is that $\sum nx^n(1-x)$ is not a power series. You can of course write it as $(1-x)\sum nx^n$ but then you will note that the power series $\sum n x^n$ does not converge for $x=1$ so you cannot use that theorem.

ok, so if it doesn't converge uni then isn't it weird that
$$\lim_{n\rightarrow \infty}\int_{0}^{1} f_{n} \rightarrow 0?$$

Isn't uniform convergence a necessary condition for that?

 

[wabbit]

ok, so if it doesn't converge uni then isn't it weird that
$$\lim_{n\rightarrow \infty}\int_{0}^{1} f_{n} \rightarrow 0?$$

Isn't uniform convergence a necessary condition for that?

No, uniform convergence is a sufficient condition for the convergence of the integral, but by no means a necessary one - as this example shows.

 

[Ray Vickson]

Well my version of Abel's theorem states
Let $g(x) = \sum a_{n}x^{n}$ that converges at some point R > 0. Then the series converges uniformly on [0,R]. I used the cauchy argument since we never proved series uniformly convergent implies the sequence is uniformly convergent.

This just implies that the series $\sum_{n=0}^N x^n$ converges uniformly on $[0, 1 - \epsilon]$ for any small $\epsilon > 0$. Thus, $\sum_{n=0}^N (1-x)x^n$ converges to 1 uniformly on $[0, 1-\epsilon]$. This does not say what happens right at $x = 1$, because then you have a series of terms that all = 0 (and, somehow, you want that series to converge to 1?)

 

[Euklidian-Space]

This just implies that the series $\sum_{n=0}^N x^n$ converges uniformly on $[0, 1 - \epsilon]$ for any small $\epsilon > 0$. Thus, $\sum_{n=0}^N (1-x)x^n$ converges to 1 uniformly on $[0, 1-\epsilon]$. This does not say what happens right at $x = 1$, because then you have a series of terms that all = 0 (and, somehow, you want that series to converge to 1?)

hmmm, never said anything about converging to 1...I said that the theorem implies that it converges uni on [0,1]. It does say what happens at x = 1. The original series equals zero at x = 1. Therefore it converges at some point x = R = 1 > 0. Therefore it converges for [0,R] = [0,1]

 

[wabbit]

hmmm, never said anything about converging to 1...I said that the theorem implies that it converges uni on [0,1]. It does say what happens at x = 1. The original series equals zero at x = 1. Therefore it converges at some point R = 1. Therefore it converges for [0,R] = [0,1]

As I mentionned in my previous post, the original series you wrote is not a power series so you cannot apply Abel's theorem to it.

 

[Euklidian-Space]

As I mentionned in my previous post, the original series you wrote is not a power series so you cannot apply Abel's theorem to it.

my only issue wabbit is that in my class we did not get the theorem you applied to show it is not uni conv. i do not know how to approach it without that theorem

 

[Ray Vickson]

hmmm, never said anything about converging to 1...I said that the theorem implies that it converges uni on [0,1]. It does say what happens at x = 1. The original series equals zero at x = 1. Therefore it converges at some point x = R = 1 > 0. Therefore it converges for [0,R] = [0,1]

You did not mention convergence to 1, but I did.

You have a series that converges uniformly to 1 for $0 \leq x \leq 1 - \epsilon$ (for any small $\epsilon > 0$), but which converges to 0 when $x = 1$ exactly. You may call that uniform convergence on [0,1], but I---for one---would not.

 

[Euklidian-Space]

You did not mention convergence to 1, but I did.

You have a series that converges uniformly to 1 for $0 \leq x \leq 1 - \epsilon$ (for any small $\epsilon > 0$), but which converges to 0 when $x = 1$ exactly. You may call that uniform convergence on [0,1], but I---for one---would not.

I do not see where you are getting that the series converges to 1. that's not even the point wise limit. Maybe i am misunderstanding you

 

[wabbit]

my only issue wabbit is that in my class we did not get the theorem you applied to show it is not uni conv. i do not know how to approach it without that theorem

I did not use a theorem in particular, could you be more specific as to what you find unclear in the proof I gave ?

 

[Ray Vickson]

I do not see where you are getting that the series converges to 1. that's not even the point wise limit. Maybe i am misunderstanding you

For $|x| < 1$ we have
\sum_{n=0}^N x^n = \frac{1-x^{N+1}}{1-x} \to \frac{1}{1-x}
as $N \to \infty$. Thus, $\sum_n (1-x) x^n \to 1$.

 

[Euklidian-Space]

I did not use a theorem in particular, could you be more specific as to what you find unclear in the proof I gave ?

you used the lim sup right? Finding the maximum of $f_{n}$ and showing that the limit of $f_{n}(x_{max})$ does not go to zero?

 

[wabbit]

you used the lim sup right? Finding the maximum of $f_{n}$ and showing that the limit of $f_{n}(x_{max})$ does not go to zero?

I did not exactly use a limit-sup (and certainly no theorem about lim sup) but yes for the rest. It would be easier if you quoted the step you are having difficulty with so I could elaborate on it.

 

[Euklidian-Space]

I did not exactly use a limit-sup but yes for the rest. It would be easier if you quoted the step you are having difficulty with so I could elaborate on it.

well, what you used to show it is not uni conv we have not been taught, therefore i must find another way to show that the sequence is not uni conv

 

[wabbit]

well, what you used to show it is not uni conv we have not been taught, therefore i must find another way to show that the sequence is not uni conv

No. I used only the definition of uniform convergence, and you can follow that proof. Try it.

 

[Euklidian-Space]

We are using different definitions then. here is mine...

 

[wabbit]

OK. So you understand why $f_n(x_n)\rightarrow \frac{1}{e}>0$ and the only thing you miss is why this contradicts the definition of uniform convergence $f_n\rightarrow_u 0$ , is that right ?

If so, try $\epsilon=\frac{1}{2e}$ and check your definition. Do you see why there is a contradiction ?

 

[Ray Vickson]

We are using different definitions then. here is mine...

Right, so sticking with YOUR definition of uniform convergence, we have that the sequence of functions
S_N(x) = \sum_{n=0}^N (1-x) x^n
converges uniformly (to 1) on the interval $[0,1-\epsilon]$. (You need to do a bit of work and use some previous theorems to establish uniformity of convergence, but it is not very hard.) However, $S_N(1) = 0$ for all $N$, so by your very own definition, the convergence of $S_N(x)$ cannot be uniform (because the limit function $\lim_N S_N(x)$ is not continuous on $[0,1]$.

 

[wabbit]

OK I'll let Ray explain, there's no point in doing it in two voices other than creating confusion, this thread is already getting unreadable. I'll move aside, you can ping me if you need.

 

[wabbit]

@Euklidian-Space as you asked for more detail :

Actually, your sequence $f_n(x)=nx^n(1-x)$ does not converge uniformly on $[0,1]$.
It converges pointwise to zero on that interval. However, if you set $x_n=\frac{n}{n+1}$ which is the value at which $f_n$ reaches its maximum over $[0,1]$ , you find that $f_n(x_n)=(\frac{n}{n+1})^{n+1}=(1-\frac{1}{n+1})^{n+1} \ _{\overrightarrow{n\rightarrow\infty}}\frac{1}{e}>0$ and this proves that the convergence is not uniform.

To expand this last step, let's assume uniform convergence, and take $\epsilon=\frac{1}{2e}$ .
From the convergence of $f_n(x_n) \rightarrow \frac{1}{e}$ we know that
(1) $\exists n_0,\forall n\geq n_0, f_n(x_n)>\frac{1}{2e}$
But uniform convergence $f_n\rightarrow_u 0$ tells us that
(2) $\exists N, \forall n\geq N, \forall x \in [0,1], f(x)<\frac{1}{2e}$
Now take any $n\geq\max(n_0,N)$ and set $x=x_n$ .
By (1) we have $f_n(x)>\frac{1}{2e}$ , but (2) yields $f_n(x)<\frac{1}{2e}$ so we have a contradiction.