# Function can be represented by a Taylor series

1. Oct 30, 2009

### 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?

2. Oct 30, 2009

### mathman

Re: Analyticity

No. All you have is f(x0)=f(x0)

3. Oct 30, 2009

### IniquiTrance

Re: Analyticity

4. Oct 31, 2009

### Office_Shredder

Staff Emeritus
Re: Analyticity

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

5. Oct 31, 2009

### HallsofIvy

Staff Emeritus
Re: Analyticity

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".

6. Oct 31, 2009

### AxiomOfChoice

Re: Analyticity

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 $\Omega$?

7. Oct 31, 2009

### mathman

Re: Analyticity

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.

8. Nov 2, 2009

### edwinksl

Re: Analyticity

Nope. Analyticity is a neighborhood property.