Show ž 1 Cosnx Converges Uniformly on R to a Continuous N 1 N2 Function
Show ##\{nx^{n}(1-x)\}_{n = 0}^{\infty}## converges uni
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Homework Statement
show ##\{nx^{n}(1-x)\}## converges uni on [0,1]
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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
Answers and Replies
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).
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.
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 [Broken]
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 [Broken]
i am quite literally picking this up verbatim from my text
it is attached to this post
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Yeah we havent been given that theorem, this was me trying to tackle this thing with the tools availableBut, 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.
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.
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 doesnt converge uni then isnt it weird that
$$\lim_{n\rightarrow \infty}\int_{0}^{1} f_{n} \rightarrow 0?$$
Isn't uniform convergence a necessary condition for that?
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No, uniform convergence is a sufficient condition for the convergence of the integral, but by no means a necessary one - as this example shows.ok, so if it doesnt converge uni then isnt it weird that
$$\lim_{n\rightarrow \infty}\int_{0}^{1} f_{n} \rightarrow 0?$$Isn't uniform convergence a necessary condition for that?
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?)
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]
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.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]
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 theoremAs 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.
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.
I do not see where you are getting that the series converges to 1. thats not even the point wise limit. Maybe i am misunderstanding youYou 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 did not use a theorem in particular, could you be more specific as to what you find unclear in the proof I gave ?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 do not see where you are getting that the series converges to 1. thats not even the point wise limit. Maybe i am misunderstanding you
For ##|x| < 1## we have
[tex]\sum_{n=0}^N x^n = \frac{1-x^{N+1}}{1-x} \to \frac{1}{1-x} [/tex]
as ##N \to \infty##. Thus, ##\sum_n (1-x) x^n \to 1##.
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 use a theorem in particular, could you be more specific as to what you find unclear in the proof I gave ?
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.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?
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 convI 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.
No. I used only the definition of uniform convergence, and you can follow that proof. Try 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
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If so, try ## \epsilon=\frac{1}{2e} ## and check your definition. Do you see why there is a contradiction ?
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
[tex] S_N(x) = \sum_{n=0}^N (1-x) x^n [/tex]
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]##.
To expand this last step, let's assume uniform convergence, and take ## \epsilon=\frac{1}{2e} ## .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.
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.
Wabbit,@Euklidian-Space as you asked for more detail :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.
i am getting ##f_{n}(x_{n}) = n\left(1-\frac{n}{n+1}\right)^{n+1}##
does that still converge to ##\frac{1}{e}?##
No it doesn't, but what you're getting is incorrect - check your calculations again, you must have let a mistake slip by somewhere.Wabbit,
i am getting ##f_{n}(x_{n}) = n\left(1-\frac{n}{n+1}\right)^{n+1}##does that still converge to ##\frac{1}{e}?##
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.
<Snip>.
I don't understand, aren't you saying here that the series converges pointwise to ##0## on ##[0,1]## but ##f_n(x_n) \rightarrow 1/e ##. How is this possible?
To state it in more general terms, if ## f_n\rightarrow f ## and ## x_n\rightarrow x ## we might expect to conclude that ## f_n(x_n)\rightarrow f(x) ## - and while it is easy to prove that this is a theorem if the convergence is uniform, the conclusion does not hold if it is only pointwise.I don't understand, aren't you saying here that the series converges pointwise to ##0## on ##[0,1]## but ##f_n(x_n) \rightarrow 1/e ##. How is this possible?
That sequence was built I imagine precisely to illustrate how, lacking uniform convergence or another suitable assumption, the diagonal sequence may converge to a limit which is not the pointwise limit. Another very similar one which makes this perhaps clearer is the sequence of piecewise linear functions with ##f_n(0)= f_n(1-\frac{2}{n})=0 ; f_n(1-\frac{1}{n})=1 ; f_n(1)=0 ## and linear interpolation between those points : for any given ## x ## the sequence ## f_n(x) ## is constant and equal to zero except for a finite number of terms, yet by definition ## f_n(x_n)=1 ## for ## x_n=1-\frac{1}{n} ## .
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