Complex Analysis Radius of Convergence.

In summary, the conversation is discussing the radius of convergence and the conditions for convergence. The first question asks about what to do at the interval (R-eps, R), and the second question is about the necessity and sufficiency of the condition for convergence. The author uses (L+eps)|Z|<1 to prove the radius of convergence, but there is concern about whether this condition holds true for all |Z|<R. The conversation also mentions a possible mistake and the shrinking of the interval due to the small value of ε. There is also confusion about what is being proven in the text.
  • #1
kidsasd987
143
4
Hello, I have two questions regarding the Radius of convergence.

1. What should we do at the interval (R-eps, R)
2. It used definition to prove radius of convergence, but I am not sure if it is necessary-sufficient condition of convergence. I get that this can be a sufficient condition but not sure of the necessity
 

Attachments

  • 스크린샷 2016-06-21 오후 4.01.05.png
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  • RadiusOfConv.png
    RadiusOfConv.png
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  • #2
If I understand the question, this may help -- The ε was picked arbitrarily small so the proof shows convergence all the way to ε = 0. So there really is no specific interval (R-ε, R) to worry about.
 
  • #3
kidsasd987 said:
what I am concerend about is,
hm the author used (L+eps)|Z|<1 to argue that L^n*|Z|^n < (L+eps)^n|Z|^n < r^n whereas r<1

but the thing is as I showed at the uploaded png file (if the derivation is correct), it is clear that |Z|= R-eps satisfies the condition |Z| < R and if |Z|= R-eps, r = 1.

which implies (L+eps)|Z|<1 cannot be true. for all |Z|<R
Ah, also I made a mistake. It should be R+R^2*eps-eps-R*eps^2
hmm do you agree?
 
Last edited:
  • #4
FactChecker said:
If I understand the question, this may help -- The ε was picked arbitrarily small so the proof shows convergence all the way to ε = 0. So there really is no specific interval (R-ε, R) to worry about.
maybe I screwed up at somewhere.
 

Attachments

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  • #5
kidsasd987 said:
maybe I screwed up at somewhere.
Maybe not. But the point is that ε >0 is arbitrarily small (or r is arbitrarily near 1) so the the interval you are worried about shrinks down to nothing.
 
  • #6
FactChecker said:
Maybe not. But the point is that ε >0 is arbitrarily small (or r is arbitrarily near 1) so the the interval you are worried about shrinks down to nothing.

Although we ignore the interval, the argument that (L+eps)|Z|<1 does not hold true if we take |Z| = R - eps. (I guess) It just bugs me and I am stuck at that page ;(

and my last conclusion R<1 doesn't seem right at all too.
 
  • #7
It isn't clear to me what is to be proven. We've only been shown the last part of the text that states "the theorem".
 
  • #8
Stephen Tashi said:
It isn't clear to me what is to be proven. We've only been shown the last part of the text that states "the theorem".

I am sorry.
 

Attachments

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1. What is the radius of convergence in complex analysis?

The radius of convergence in complex analysis is a measure of the distance from the center of a power series to the nearest point where the series converges. It is usually denoted as R and can be calculated using the ratio test or the Cauchy-Hadamard formula.

2. How is the radius of convergence determined?

The radius of convergence is determined by analyzing the behavior of the coefficients of the power series. If the series converges for all values of the variable within a certain distance from the center, then the radius of convergence will be that distance. If the series diverges for all values of the variable outside that distance, then the radius of convergence will be 0.

3. What is the significance of the radius of convergence?

The radius of convergence is significant because it tells us which values of the variable will result in a convergent power series. It also allows us to determine the interval of convergence, which is the range of values for the variable where the series will converge. In addition, the radius of convergence is used in many applications of complex analysis, such as in the study of analytic functions and in solving differential equations.

4. Can the radius of convergence be infinite?

Yes, the radius of convergence can be infinite. This means that the power series will converge for all values of the variable. In other words, the series will have an infinite interval of convergence. This is often the case for simple, well-behaved functions such as polynomials and exponential functions.

5. How does changing the center of a power series affect the radius of convergence?

Changing the center of a power series does not affect the radius of convergence. The radius of convergence only depends on the behavior of the coefficients of the series and not on the location of the center. However, changing the center can change the interval of convergence, as the distance from the center to the nearest point of convergence may vary.

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