# 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.