Two problems which have been giving me problems

  • Thread starter end3r7
  • Start date
In summary, the first series does not converge uniformly on the interval [0,1] as the terms of the series do not go to zero uniformly on its domain. For the second series, it converges absolutely on [0,1] due to the comparison test. However, it does not converge uniformly on [0,1] as the alternating series test only guarantees convergence for a fixed constant x, not for every x in the interval. Therefore, the sum of absolute values of the second series is not absolutely convergent on [0,1].
  • #1
end3r7
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0

Homework Statement


1) Test the following series for Uniform Convergence
[tex]
\sum\limits_{n = 1}^{\inf } {\frac{{( - 1)^n }}{{n^{x}\ln (x)}}}
[/tex]

2) Let f(n,x) = [tex]
\sum\limits_{n = 1}^{\inf } {( - 1)^n (1-x^{2})x^{n}}
[/tex]
a) Test for absolutely convergence on [0,1]
b) Test for uniformly convergence on [0,1]
c) Is [tex]
\sum\limits_{n = 1}^{\inf } {|f(n,x)|}
[/tex] absolutely convergent on [0,1]?

Homework Equations


The Attempt at a Solution



For the first, I'm utterly lost. Is there an easy way to deal with such series?

For the second, could I just argue that for all 0<=x<1, there exists a, s.t. x < a <1
and thus
[tex]
|\sum\limits_{n = 1}^{\inf } {|f(n,x)|} | <= \sum\limits_{n = 1}^{\inf } {|f(n,x)|} < \sum\limits_{n = 1}^{\inf } {(a)^n} = \frac{a}{1-a}
[/tex]
and for x = 1 and any a > 0
[tex]
|\sum\limits_{n = 1}^{\inf } {|f(n,x)|} | <= \sum\limits_{n = 1}^{\inf } {|f(n,x)|} = 0 < \sum\limits_{n = 1}^{\inf } {(a)^n} = \frac{a}{1-a}[/tex]

This would prove all 3 right? But can I argue taht way? Can I fix my 'x' ahead of time, or does my argument have to work for all x simultaneously? Cuz if it does, then all I would have to do is choose x between a and 1 and the argument would break down.
 
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  • #2
I think I figured out two (will post my work later).

But for 1, could I argue that the sequences of partial sums cannot be uniformly cauchy since they are unbounded near x = 1.
 
  • #3
I confess I might have jumped the gun in saying that I figure out number two. Any help on either would still be greatly appreciated (also, if I could get a mod to give this thread a more decriptive title... I don't think I can change the title).
 
  • #4
For the first one, how about splitting the sum into two: one for n even, one for n odd. Then you can use the p-series test on each.

The argument you use when 0 <= x < 1 works. When x = 1, what's inside the sum is 0 so the whole sum is 0. I find it unnecessary to consider the absolute value of the sum.
 
  • #5
Call me stupid (I'd rather you don't though =P), but I'm not sure I understand your approach for the first.

What I did was show that I can get x arbitrarily close to 1, and since log(x) is continuous, I can get arbitrarily close 1. Basically, I showed that for any 'n', I can make [tex]
log(x) < \frac{1}{ne}
[/tex]
so for x sufficiently close to 1

[tex]
|\frac{1}{{n^{x}\ln (x)}}| > \frac{1}{|{n||\ln (x)|}} > |\frac{ne}{n}| = e
[/tex]

where first inequality holds because x is between 0 and 1.

Therefore the terms of the series do not go to zero uniformly on its domain, so it can't converge.

Is that a valid argument?
 
  • #6
What I wrote is for testing convergence, not uniform convergence. Sorry about that. I'm not familiar with uniform convergence.
 
  • #7
e(ho0n3 said:
What I wrote is for testing convergence, not uniform convergence. Sorry about that. I'm not familiar with uniform convergence.

No problemo. =)

(Btw, uniform convergence is essentially the same, but it has to work for arbitrary x in the interval).
 
  • #8
If x is a fixed constant greater than 0, by the alternating series test, the sum in 1) converges. Does this mean that the sum converges uniformly in the interval [itex](0, \infty)[/itex]?
 
  • #9
Oops, I forgot to say that x belongs to [0,1].

Note that log(1) = 0, so it can't converge.

To converge uniformly, it has to converge for every x in the interval.
 
  • #10
Please, if a mod could, edit the first one so x belongs to the closed interval [0,1].

Thanks =)
 

Related to Two problems which have been giving me problems

1. What are the two main problems you are facing?

The two problems I am currently dealing with are finding a cure for a rare disease and developing renewable energy sources.

2. How long have these problems been giving you trouble?

I have been researching and working on these problems for the past 5 years.

3. What motivated you to pursue these problems?

I have always been passionate about finding solutions to real-world problems and making a positive impact in people's lives. These two problems align with my interests and skills, so I decided to focus on them.

4. Have you made any progress in solving these problems?

Yes, I have made significant progress in both areas. I have identified potential treatments for the rare disease and have also developed a prototype for a new renewable energy source.

5. What are the biggest challenges you face in solving these problems?

The biggest challenge is securing funding and resources to continue my research and bring these solutions to fruition. Additionally, there are ethical and logistical considerations that need to be carefully addressed in both areas.

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