Hey there! Sign in to join this conversationNew here? Join for free
    • PS Helper
    • Thread Starter
    Offline

    14
    PS Helper
    So I've just proved the Weierstrass M-test; that is, if a series of functions u_n : E \to \mathbb{R} is such that for all x \in E, |u_n(x)| \le M_n where \displaystyle \sum_{n \ge 0} M_n converges, then \displaystyle \sum_{n \ge 0} u_n(x) converges uniformly on E

    I'm given that E=\mathbb{R} \smallsetminus \mathbb{Z}, let u_0(x) = \dfrac{1}{x^2} and let u_n(x) = \dfrac{1}{(x-n)^2} + \dfrac{1}{(x+n)^2} for n>0. I have to show that \displaystyle f(x) = \sum_{n \ge 0} u_n(x) converges uniformly on E_R = \{ x \in E\, :\, |x|<R \} for each R \in \mathbb{R}.

    Now I'm a bit confused here. If I'm supposed to use the Weierstrass M-test on this then each of the u_i should be bounded; but u_0(x) is unbounded on any E_R, for example; and indeed u_n(x) is unbounded on any E_R with R \ge n.

    Showing uniform convergence of the series probably isn't too difficult just using my bare hands; I suppose I'm just being thrown off by the Weierstrass M-test in the first half of the question. So if someone could verify whether or not I'm going mad, I'd appreciate it :p:
    Offline

    17
    ReputationRep:
    We can ignore the first R (or 2R if it makes life easier, which it might) terms of the series since that's a finite sum. The rest of the terms don't have the boundedness problem.
    • PS Helper
    • Thread Starter
    Offline

    14
    PS Helper
    (Original post by DFranklin)
    We can ignore the first R (or 2R if it makes life easier, which it might) terms of the series since that's a finite sum. The rest of the terms don't have the boundedness problem.
    Duh, I'm a moron. Thanks
    • PS Helper
    • Thread Starter
    Offline

    14
    PS Helper
    Just whilst I'm here, I might as well just check that I did the last part right.

    I have to determine whether \displaystyle \sum_{n \ge 0} u_n(x) converges uniformly on E = \mathbb{R} \smallsetminus \mathbb{Z}. I've said it doesn't:

    I've noted that u_n(x) > 0 for all n and that each u_n is unbounded for |x|<n and bounded on |x| \ge n+1. Suppose that f_k(x) = \displaystyle \sum_{0 \le n \le k} u_n(x) and f(x) = \displaystyle \sum_{n \ge 0} u_n(x). Then I note that f_k(x) is bounded/unbounded on the same regions as u_k(x).

    The series converges uniformly on E iff (f_k) \to f uniformly on E iff \displaystyle \sup_{x \in E} |f_k(x) - f(x)| \to 0 as k \to \infty. If f is bounded on |x|>R for some R then choosing any k \ge R + 1, we must have that f_k is unbounded in a neighbourhood of x=k and so the supremum is infinite. Similarly, if f is unbounded on |x|>R for all R \in \mathbb{R} then since each f_n is bounded on |x|>n+1, the supremum is again infinite. In either case, the supremum certainly doesn't converge to zero, and so \displaystyle \sum_{n \ge 0} u_n(x) doesn't converge uniformly on E.

    Once again I'm fairly sure this is correct, but is it a bit clumsy? Is there a neater way of saying this?

    Thanks to anyone for their help.
    Offline

    17
    ReputationRep:
    I think you can do it more painlessly: suppose the sum uniformly converges over the reals, take epsilon = 1. By assumption of unif conv we can find N s.t. n > N => |f_n(x) - f(x)| < 1 for all real x. But f_n+1(n+1.1) > f_n(n+1.1) + 100, and so f(n+1.1) > f_n(n+1.1) + 100. Contradiction.
    • PS Helper
    • Thread Starter
    Offline

    14
    PS Helper
    (Original post by DFranklin)
    I think you can do it more painlessly: suppose the sum uniformly converges over the reals, take epsilon = 1. By assumption of unif conv we can find N s.t. n > N => |f_n(x) - f(x)| < 1 for all real x. But f_n+1(n+1.1) > f_n(n+1.1) + 100, and so f(n+1.1) > f_n(n+1.1) + 100. Contradiction.
    That's certainly less awkward; thanks.
 
 
 
Poll
Do you agree with the PM's proposal to cut tuition fees for some courses?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.