Convergence of Taylor Series for Various Functions

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SUMMARY

The discussion focuses on determining the convergence of Taylor Series for specific functions: \(\sqrt{x^{2}-x-2}\) around \(x = \frac{1}{3}\), \(\sin(1-\theta^{2})\) around \(\theta = 0\), and \(\tanh(u)\) around \(u = 1\). The key concept is the radius of convergence, defined by the distance to the nearest singularity from the point of expansion. For \(\sqrt{x^{2}-x-2}\), the series converges only within its defined domain, while the convergence of the other two functions is contingent on their respective singularities.

PREREQUISITES
  • Understanding of Taylor Series and their properties
  • Knowledge of singularities in functions
  • Familiarity with the concept of radius of convergence
  • Basic calculus skills for evaluating functions
NEXT STEPS
  • Research the concept of singularities in functions
  • Learn how to calculate the radius of convergence for Taylor Series
  • Study the Taylor Series expansion for \(\sin(x)\) and \(\tanh(x)\)
  • Explore the implications of convergence in complex analysis
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Students studying calculus, mathematicians interested in series convergence, and educators teaching Taylor Series concepts.

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Homework Statement



For what values of x (or [tex]\theta[/tex] or u as appropriate) do you expect the following Taylor Series to converge? DO NOT work out the series.

[tex]\sqrt{x^{2}-x-2}[/tex] about x = 1/3

[tex]sin(1-\theta^{2})[/tex] about [tex]\theta = 0[/tex]


[tex]tanh (u)[/tex] about u =1


Homework Equations





The Attempt at a Solution



I'm not to sure where to begin. Taylor series have a radius of convergence where |x-a|< R, wher a is the nearest singularity, so I suppose that's a starting point?
 
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For what domain is [tex]\sqrt{x^{2}-x-2}[/tex] defined? It can't converge beyond that.
 

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