Weierstrass M-Test and Absolutely Uniformly Convergence

In summary: So in your case, since you're testing convergence for a given x, your bound a is allowed to be a function of x.In summary, the author is trying to solve a homework problem. The first post was riddled with errors, so the author is posting again. The author has a solution to the problem that can be proved using the Weierstrass M-test. The problem is that the author does not like constantly bumping his/her threads, so he/she is asking for feedback on whether or not the proof is mathematically sound.
  • #1
end3r7
171
0
I'm posting this again because the other was plagued with errors in the first post. My fault and I apologize. I'll do a better job this time, I hope.

Homework Statement


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] uniformly convergent on [0,1]?



Homework Equations


N/A


The Attempt at a Solution


Alright, a) is simple actually, a simple application of the root/ratio test will do, but here comes the catch.

I know that b) is true, since a power series is uniformly convergent inside its radius of convergence and the endpoint in this case is trivial. It just seems, however, that with a comparison with a geometric series, I can use the Weierstrass M-Test to prove all 3 at once.

For any 'x' in [0,1), I can choose an a such that 0 <= x <= a <1 by the density of R. Therefore
[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]

I have just chosen the function to be uniformally cauchy on [0,1) U {1} (have I?) by a basic comparison test with a series which is uniformly cauchy on (0,1) and by considering the end point separately.
But it also seems that the Weierstrass M-Test gives me "Absolutely Uniformly Convergence" (that is, the absolute value of the function is also uniformly convergent) for free. Am I right? Is that always going to hold true, or is this a property of power series?
 
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  • #2
Alternatively, could I also just make a few arguments reasoning from (a).

I know the series is absolutely convergent on [0,1]. So its absolute value will be uniformly convergent inside its radius of convergence. I proceed to test the endpoint x = 1, which turns it into a series of zeros. So I have just shown its absolute value to be uniformly convergent on [0,1) U {1}.
If the absolute value is uniformly convergent, than the series itself must be uniformly convergent.

Is this an acceptable way to proceed?
 
  • #3
Please, I don't like constantly bumping my threads, but I'm really struggling to understand this concept. =/
 
  • #4
Yeah, I'm pretty sure the Weierstrass M-test tests for uniform convergence by essentially testing for absolute uniform convergence...
 
  • #5
Thank you many times for replying.

Okay, so is my proof mathematically sound?

I also have a quick follow up though regarding the order/flow of the argument.

In my argument, I essentially "fix" my 'x' and choose my 'a'. But it's also true that between 'a' and 1, I can find an 'x'. Now, I'm certain I can write a more rigorous argument showing that the limit of 'a' is 1, but I'm wondering if that's an overkill, since my whole purpose in bringing in teh Weierstrass M Test was to simplify things (as well as solve them, since it's the only method I know).
 
  • #6
Well, I believe that to use the Weierstrass M-test you have to find a bound that works for all x on whatever set you're testing convergence in, not just a given x in the set. That is, your bound a is allowed to be a function of n, but it cannot be a function of x.
 

Related to Weierstrass M-Test and Absolutely Uniformly Convergence

1. What is the Weierstrass M-Test?

The Weierstrass M-Test is a mathematical tool used to determine whether a series of functions converges absolutely and uniformly. It involves finding an upper bound for the absolute value of each function in the series and then comparing it to a convergent series.

2. How does the Weierstrass M-Test work?

The Weierstrass M-Test works by finding an upper bound, M, for the absolute value of each function in the series. If the series of M values converges, then the original series of functions converges absolutely and uniformly. This is because the M values act as a comparison to a known convergent series.

3. What is the difference between absolute and uniform convergence?

Absolute convergence means that the series of functions converges when taking the absolute value of each term. This is important because it guarantees that the series will converge regardless of the order in which the terms are added. Uniform convergence means that the series converges at the same rate at all points in the domain, providing a more precise measure of convergence.

4. When should the Weierstrass M-Test be used?

The Weierstrass M-Test should be used when determining whether a series of functions converges absolutely and uniformly. It is particularly useful for complex series where other convergence tests may not be applicable.

5. What are some applications of the Weierstrass M-Test?

The Weierstrass M-Test has many applications in mathematics, particularly in analysis and calculus. It is commonly used to prove the convergence of power series, which are important in the study of functions and their behavior. It is also used in areas such as differential equations, Fourier series, and complex analysis.

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