Convergence of Taylor series in a point implies analyticity

In summary, the conversation discusses the convergence of a Taylor series for a function f(x) in a given interval (a,b), centered at a point x_0 in that interval. The question is whether choosing a different point x_1 in the same interval would allow for convergence of the Taylor series at that point, and if so, how to prove it. It is suggested that the proof may require knowledge of the absolute convergence of the series for f(x) and its derivatives. The conversation also raises the question of whether the equality of f(x) and its Taylor series at a point implies the same for its derivatives.
  • #1
crick
43
4
Suppose that the Taylor series of a function ##f: (a,b) \subset \mathbb{R} \to \mathbb{R}## (with ##f \in C^{\infty}##), centered in a point ##x_0 \in (a,b)## converges to ##f(x)## ##\forall x \in (x_0-r, x_0+r)## with ##r >0##. That is
$$f(x)=\sum_{n \geq 0} \frac{f^{(n)}(x_0)}{n!} (x-x_0)^n \,\,\,\,\,\,\,\,\,\,\,\,\,\forall x \in (x_0-r,x_0+r) $$
(In particular, I do not know in advance anything about the analyticity of ##f(x)## in ##(a,b)##, but only the convergence of the Taylor series in ##x_0##.)

Does this imply that, chosen a ##x_1 \in (x_0-r,x_0+r)##, the Taylor series of ##f(x)## centered in ##x_1## converges at least in ##(x_1-r',x_1+r')## with ##r'=r-|x_0-x_1|##? That is
$$f(x)=\sum_{n \geq 0} \frac{f^{(n)}(x_1)}{n!} (x-x_1)^n \,\,\,\,\,\,\,\,\,\,\,\,\,\forall x \in (x_1-r',x_1+r') $$
If this is true, how can I prove it?
 
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  • #2
Fooling around with the power series, I see that the conclusion is easily provable if you additionally know the following:
  1. The series for [itex]f(x)[/itex] is absolutely convergent (so that you can reorder terms and get the same answer).
  2. The series for [itex]\frac{df}{dx}, \frac{d^2 f}{dx^2}, ...[/itex] are also absolutely convergent.
As to point two--can you prove that if [itex]f(x)[/itex] is equal to its Taylor series at a point, then [itex]f^{(n)}(x)[/itex] is equal to its Taylor series at that point?

As to point one--I'm not sure whether you can get away without assuming that, or not.
 

Related to Convergence of Taylor series in a point implies analyticity

1. What is the definition of analyticity in terms of Taylor series convergence?

Analyticity refers to the property of a function being represented by a convergent Taylor series expansion around a point. In other words, if a function is analytic at a point, its derivatives at that point can be computed by the coefficients of its Taylor series.

2. How does the convergence of a Taylor series at a point imply analyticity?

If a Taylor series converges at a point, it means that the function is equal to its Taylor series expansion at that point. This implies that the function can be represented by an infinite series of polynomial terms, which is a characteristic of analytic functions.

3. Is the converse true - does analyticity always imply convergence of the Taylor series at a point?

No, analyticity does not always imply the convergence of the Taylor series at a point. A function can be analytic at a point but its Taylor series may not converge at that point. However, if a function is analytic on a closed domain, then its Taylor series will converge at any point within that domain.

4. Can a function be analytic at a point but not analytic on a larger interval?

Yes, a function can be analytic at a point but not analytic on a larger interval. This is because analyticity only requires the function to have a Taylor series expansion at a single point, but it may not converge on a larger interval due to singularities or other discontinuities.

5. What is the significance of analyticity in mathematics and science?

Analyticity is a fundamental concept in mathematics and science, particularly in complex analysis and differential equations. It allows us to represent and manipulate functions in terms of their Taylor series, making it easier to solve equations and analyze their behavior. It also has applications in fields such as physics, engineering, and economics.

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