Taylor series radius of convergence

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Discussion Overview

The discussion revolves around finding the radius of convergence for Taylor series, particularly in cases where a generic nth term is difficult to establish. Participants explore specific examples, such as the series for sqrt(1 - x^2), and question the methods for deriving these series and their convergence properties.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant, Sam, questions how to find the radius of convergence for series without a clear nth term, using sqrt(1 - x^2) as an example.
  • Another participant asserts that without estimating the nth term, determining the radius of convergence is not feasible.
  • A third participant provides a specific series expansion for sqrt(1 - x^2) and suggests methods such as expanding sqrt(1+t) about t=0 and substituting t=-x^2, referencing the generalized binomial theorem.
  • There is a request for clarification on how the substitution affects differentiation and the nth term, indicating confusion about the process and its implications for higher-order terms.
  • Additional examples of functions like log(1-x^2) and exp(1-x^2) are mentioned, suggesting a broader inquiry into similar series expansions.

Areas of Agreement / Disagreement

Participants express differing views on the feasibility of finding a generic nth term for certain series, with some asserting it is impossible while others provide specific expansions. The discussion remains unresolved regarding the best approach to derive these terms and their implications for convergence.

Contextual Notes

Participants highlight limitations in their understanding of the substitution process and its effects on differentiation, indicating that assumptions about the behavior of series may not be fully explored.

saminny
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Hi,

We need a generic expression of a taylor series nth term to find out the radius of convergence of the series. However, there are series where I don't think it is even possible to find a generic term. How do we find the radius of convergence in such cases?

e.g. sqrt (1 - x^2)

There is no generic nth term..

Please help.

thanks,

Sam
 
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One can't find the radius of convergence if one can't estimate the nth term.
 
[tex]\sqrt{1-x^2} = \sum_{n=0}^{\infty} \frac{ (2n)! x^{2n}}{4^n(1-2n)(n!)^2}[/tex]

You may have seen this more easily if you tried to expand [tex]\sqrt{1+t}[/tex] about t=0 then substituted [itex]t=-x^2[/itex]. Or, still easier, used the generalized binomial theorem.
 
Gib Z said:
[tex]\sqrt{1-x^2} = \sum_{n=0}^{\infty} \frac{ (2n)! x^{2n}}{4^n(1-2n)(n!)^2}[/tex]

You may have seen this more easily if you tried to expand [tex]\sqrt{1+t}[/tex] about t=0 then substituted [itex]t=-x^2[/itex]. Or, still easier, used the generalized binomial theorem.

eehhh.. how did you get that result? I've been trying to find the nth term for so many hours with no success. If you expand [itex]\sqrt{1+t}[/itex] about t=0, how can you just substitute t for x^2 since that would change the differentiation of each term and effects of that additional 2x will compound as we go higher order. if you could provide some explanation, that would be really helpful. thanks.

Some more examples
[tex]\log{1-x^2}[/tex]
[tex]\exp(1-x^2)[/tex]
 
saminny said:
eehhh.. how did you get that result? I've been trying to find the nth term for so many hours with no success. If you expand [itex]\sqrt{1+t}[/itex] about t=0, how can you just substitute t for x^2 since that would change the differentiation of each term and effects of that additional 2x will compound as we go higher order. if you could provide some explanation, that would be really helpful. thanks.

Some more examples
[tex]\log{1-x^2}[/tex]
[tex]\exp(1-x^2)[/tex]


I understood it..
 

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