Function can be represented by a Taylor series

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Discussion Overview

The discussion centers on the conditions under which a function can be considered analytic, particularly in relation to its Taylor series representation at a point with a radius of convergence of zero. The scope includes theoretical aspects of analysis and the definition of analyticity in mathematical contexts.

Discussion Character

  • Debate/contested
  • Conceptual clarification
  • Mathematical reasoning

Main Points Raised

  • Some participants argue that if a function can be represented by a Taylor series at a point with a radius of convergence of zero, it is not considered analytic at that point.
  • One participant emphasizes that having a Taylor series that converges to the function only at the point does not satisfy the conditions for analyticity.
  • Another participant provides an example of a function that has all derivatives equal to zero at a point, resulting in a Taylor series that is not equal to the function elsewhere, thus illustrating that such a function is not analytic.
  • Questions are raised about whether a power series representation that converges in a neighborhood implies analyticity, with responses indicating that convergence must be to the function itself.

Areas of Agreement / Disagreement

Participants generally agree that a function represented by a Taylor series with a radius of convergence of zero is not analytic at that point. However, there are nuanced discussions regarding the implications of power series convergence and the definition of analyticity.

Contextual Notes

Limitations include the dependence on specific definitions of analyticity and the conditions under which a Taylor series is considered representative of a function.

IniquiTrance
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If a function can be represented by a Taylor series at x0, but only at this point, (radius of convergence = 0), is it considered analytic there?
 
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IniquiTrance said:
If a function can be represented by a Taylor series at x0, but only at this point, (radius of convergence = 0), is it considered analytic there?

No. All you have is f(x0)=f(x0)
 


Could you please elaborate on your response? Not sure I follow...
 


Given any function (any function at all, seriously), the "Taylor Series" around x0 with 0 radius of convergence is

f(x)=f(x0). This is pretty pointless
 


In order to be "analytic" at a point, the Taylor's series for the function, around that point, must converge to the function in some neighborhood of the function.

And it depends upon what you mean by "represented by the Taylor's series".

The function
[tex]f(x)= e^{-\frac{1}{x^2}[/tex]
if [itex]x\ne 0[/itex], f(0)= 0, has all derivatives at 0 equal to 0 and so its Taylor's series, about x= 0, exists, has infinite radius of convergence, but is equal to f only at x= 0. That function is NOT "analytic".
 


I have a question about analyticity: Suppose I want to show that a function [itex]f(z)[/itex] is analytic in some open subset [itex]\Omega[/itex] of the complex plane. Is it enough to show that [itex]f[/itex] has a power series representation that converges for every [itex]z[/itex] in [itex]\Omega[/itex]?
 


AxiomOfChoice said:
I have a question about analyticity: Suppose I want to show that a function [itex]f(z)[/itex] is analytic in some open subset [itex]\Omega[/itex] of the complex plane. Is it enough to show that [itex]f[/itex] has a power series representation that converges for every [itex]z[/itex] in [itex]\Omega[/itex]?

Only when the convergence is to f(z) itself. As previously noted the power series for e-1/x2 is all 0, not the function itself.
 


IniquiTrance said:
If a function can be represented by a Taylor series at x0, but only at this point, (radius of convergence = 0), is it considered analytic there?

Nope. Analyticity is a neighborhood property.
 

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