Homework Help: Uniform Convergence question

1. Nov 19, 2013

trap101

I have a question where I am supposed to show that a series does not converge uniformly, I get the majority of the question, but one part in the solution I can't see the rationale or how they decided on the result:

It has to do with the partial sum:

SN= (1 - (-x2)N+1)/ (1+x2)

The interval of consideration is (-1,1).

the reason for the lack of convergence they say: Suppose N is odd. Then it is possible to choose an x' $\in$(-1,1) so that (1 - (-x2)N+1) < 1/4

My question is how do they obtain that bound?

2. Nov 19, 2013

brmath

Is the $S_N$ you define the Nth term of the series or the sum up to N of the series?

3. Nov 19, 2013

trap101

the Nth partial sum, so I suppose that would mean the Nth term of the series

4. Nov 19, 2013

brmath

Just to keep my own head straight let me write m = N + 1 and we have

$S_{m-1}(x) = \frac {1 - (-1)^m(x^{2m)}}{1 + x^2}$

You are writing that that regardless of m we can show that the numerator T is < 1/4 for some x. If m is even, (meaning your N+1 is even and N is odd) you have

T = $1-x^{2m}$. If you could take x = 1 you would have T = 0. But since we are in the open interval (0,1) then take x to be something close to 1. How close x has to be to 1 depends on m. Obviously this can be made to be less than 1/4.

However there is just no way T is less than 1/4 when when m is odd, since it is 1 + (something positive).

With that said, I do believe the sequence $S_N$ converges to 1. For it to converge uniformly, you have to go back to the definition and say: given an $\epsilon$ there exists an N independent of x such that n > N $\Rightarrow |S_N(x) -1| < \epsilon$.

I do think it is not uniform convergence, because for x = 0 you only need N =0; whereas for x close to $\pm$ 1 you may need very large N. This paragraph is not a proof of anything; it is just a thinking exercise to discern reasonableness.

Off the top of my head, I don't see what the computation on T and the 1/4 does to help you. Could you check and see whether you transcribed the problem properly. And were you supposed to prove the convergence is not uniform?

5. Nov 19, 2013

trap101

I provided a photo of the question followed by the solution from the solution manual. It is question 1b) followed by solution b)

With that being said, I see your reasoning for the first portion stating about T = $1-x^{2m}$, but it is drawing the conclusion of being able to make it converge to 1 and being able to draw the fact that if x = 0, you only need N = 0 whereas for x close to ± 1 you may need very large N.

How did you conclude that?

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6. Nov 20, 2013

brmath

One thing your thumbnail shows me is that $S_N \rightarrow 1/(1 + x^2)$ which is not 1; my error.

We already discussed where they got the bound of $S_n(x_0) < 1/4$. Now they say that $1/(x^2 + 1) > 1/2$, which is surely true. After that I'm not quite with them. I have

$S_n(x_0) -1 < - 3/4$ so that $|S_n(x_0) -1| > 3/4$ and $\frac{|S_n(x_0) -1}{1+x^2} >\frac { 3/4}{1/2} = 3/2$ This is not quite the inequality they have, and perhaps you can see where I or they went wrong. Whichever inequality it is, they are now invoking a theorem I don't know to conclude that if this expression is bounded away from 0 the convergence cannot be uniform. Perhaps they presented that theorem to you?

I would tackle it this way. Let x = q/(q+1) (for example 9/10) where q is any positive integer. Then for some m $(q/(q+1))^{2m} < 1/2$ so that $1 - x^{2m} > 1/2$; For a larger m $1-x^{2m} > 3/4$ etc. If you choose an $\epsilon$ there exists an m such that n > m $\Rightarrow |1-x^{2m}| < \epsilon$. So $|1-x^{2m}| \rightarrow 1$ which is what we want.

Now choose y = 100q/(100q + 1). The expression $|1-y^{2m}|$ will also converge to 1, but the m it takes to get within any given distance of 1 is larger than the m it takes when x = q/(q + 1).

I hope you can see intuitively that the larger q gets the larger m has to be, and there is no bound on the m -- it will get larger and larger as q gets larger. This means the convergence cannot be uniform because the m such n > m $\rightarrow |1-x^{2m}|< \epsilon$ depends on x.

What I wrote above is very close to a proof that the convergence is not uniform. To nail it down completely you have to show a little more explicitly that a bigger q means a bigger m. That is something like: if $(q/(q+1))^{2m} < 1/2$ then $(10^kq/(10^kq + 1))^2m$ is not less than 1/2 for the same m;you get to choose your k to make it work.

Since I'm not supposed to do your homework for you, I think you might try to prove the step above. To do it in general show that if $(q/(q+1))^2m$ < 1/2 there will always exist a larger number r > q such that $(r/(r+1))^2m$ > 1/2. If you choose r = $10^kq$ where k is at your discretion, that might be easiest.

Once you have show this much, replace your 1/2 with 1/p where p > 1 and show that the argument for 1/2 works just as well for 1/p. Then you are done.

7. Nov 20, 2013

brmath

I just wanted to add that the proofs provided in textbooks are not always intuitive. Sometimes one can't avoid that. But in this case I think you can avoid your textbook proof and do it along the lines I suggest.

In general with non-uniform convergence the problem is usually at one point -- in the case above x $\rightarrow 1$. If you can figure out what point is messing up the uniformity, then you can usually figure out a proof of the non-uniformity.