Prove about radius of convergence

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Homework Help Overview

The discussion revolves around proving the radius of convergence of a power series, specifically addressing the relationship between the coefficients and the limit superior of their absolute values. The original poster expresses urgency in completing the task and seeks assistance in understanding the concepts involved.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to relate the limit superior of the coefficients to the radius of convergence and expresses confusion about the ratio test. Some participants question the correctness of the expression involving the coefficients, while others suggest that the ratio test might be a useful approach.

Discussion Status

Participants are exploring different methods to tackle the problem, with some offering insights into the ratio test and others reflecting on their course background. There is acknowledgment of the original poster's struggle and a mix of suggestions and clarifications regarding the mathematical expressions involved.

Contextual Notes

There is mention of course prerequisites and topics covered, indicating that the participants have varying levels of familiarity with the concepts of power series and convergence. The original poster notes a lack of exposure to the ratio test in their current course, which may affect their approach.

jacobrhcp
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Homework Statement



Prove that the radius of convergence [tex]\rho[/tex] of the power series [tex]\sum[/tex]ck (z-a)^k over all k, equals 1/R when ck is not 0 and you know that:

|[tex]\frac{ck+1}{ck}[/tex]|=R>0

Homework Equations



I was planning on using that the radius of convergence is in this case:
[tex]\rho[/tex]= 1/limsup(|ck|^1/k) ( and k->infinity)

The Attempt at a Solution



I tried to make it sensible that

limsup(|ck|^1/k)=|[tex]\frac{ck+1}{ck}[/tex]|=R

I've been staring at it for quite some hours now (it's 3 in the morning and it's got to be done by 9 o'clock this morning... so any help would be greatly appreciated, though I understand if you think it's my own fault)
 
Last edited:
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I'm pretty sure a quick ratio test does the trick
 
I'm not so sure what a ratio test is (I'm reading about on wikipedia right now) but we haven't handeled it yet, and so I can't use it.

edit: unless I prove it, ofcourse.

edit2: taking another look at it, it looks like I'm trying to prove this ratio test for general power series here, but they didn't tell me it's name <_<

edit3: and thanks for your help, ofcourse =)
 
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it's 3:30 and it's looking hopeless; I'm going to go to bed now. Thanks all.
 
I find it hard to imagine anyone working with power series if they have not already learned how to handle numerical series. And I find it hard to believe that you could have already learned numerical series without learning the ratio test!


I am a bit concerned about your [tex]\frac{ck+1}{ck}[/tex]
I would think it ought to be [tex]\frac{c(k+1)}{ck}[/tex].
 
well, I did the assignment now, without the ratio test. I used the epsilon-definition of the limsup, and it followed quite easily.

The ratio test they did in an other course (on Fourier analysis), but we skipped the use of it, but proved the theorem by ourselves.



if you'd like to know: here are the course specifics (translated to english by me)

course: analysis 2

foreknowledge: analysis 1, linear algebra, calc1, foundations of higher mathematics

description:

After a short repeat of definitions and basics about functions in more variables, we handle theorems about exchanging. These are about exchanging limits, order of integration, differential operators, limits under the integral. After that we introduce line integrals and gradient vector fields. Complex line integrals are used to research complex differentiable functions of complex variables.

In the second part we study power series and Fourier series. We show that a function is complex differentiable if and only if it's equal to the sum of convergent power series.






and you're right about what it ought to be.
 

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