Function can be represented by a Taylor series

[IniquiTrance]

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?

 

[mathman]

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)

 

[IniquiTrance]

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

 

[Office_Shredder]

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

 

[HallsofIvy]

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
$$f(x)= e^{-\frac{1}{x^2}$$
if $x\ne 0$, 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".

 

[AxiomOfChoice]

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

 

[mathman]

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

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.

 

[edwinksl]

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.