A Question concerning the definition of Analytic functions

Question

A power series is an infinite series of the form

$$\sum_{n=1}^∞ a_n(x-c)^n.$$

And a function $f$ is analytic if it is locally given by a power series. Yet the formal definition of an analytic function is the following:

A function $f$ is analytic on an open set $D$ of the real line if for any $x_0∈D$ one can write

$$f(x) = \sum_{n=1}^∞ b_n(x-x_0)^n.$$

for coefficients $b_i∈ℝ$ and the series is convergent to $f(x)$ on a neighborhood of $x_0$.

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Why isn't the definition simply

"$f$ is analytic on an open set $D$ of the real line if one can write

$$f(x) = \sum_{n=1}^∞ a_n(x-c)^n.$$

for some constant $c∈ℝ$, coefficients $a_i∈ℝ$ and all $x∈D$" ?

Since an analytic function is one that is "locally given by a power series", then I would think that the proper way to define an analytic function would be as shown in the example above. Why is this not the case?

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Also, wouldn't it then be possible for a function to be given by a power series $\sum_{n=1}^∞ a_n(x-c)^n$ on an open set $D$, yet for that function not to comply with the former and correct definition of analytic?

I would really appreciate any help/thoughts.

Answer 1

With the actual definition ,for different $x_0$ in $D$ we might get different neighborhoods on which the function is represented by a power series centered at $x_0$.

In your simplified version $$ f(x) = \sum_{n=1}^∞ a_n(x-c)^n.$$ the role of $c$ is not clear.

Thus there is more to the definition of analytic functions than just being represented by a power series about one point $c$ of the region $D$.

Answer 2

Also for the edited version consider the function $f(x)=\frac{1}{x}$ on the positive reals.

It is analytic on this set but you cannot find one power-series that represents it on the full set.

For example around $1$, one has $f(x) = \sum_{n=0}^{\infty}(-1)^n(x-1)^n$, but the radius of convergence is only $1$. Thus, you cannot express $f(7)$ in this way. If you develop around any other point you will face the same problem.

Answer 3

A function given by a power series $f(x) = \sum_{n=0}^\infty a_n (x-c)^n$ convergent in an open neighbourhood $D$ of $c$ is in fact analytic in $D$ according to the "formal and correct" definition: this is a theorem.

On the other hand, there might not be any single $c$ such that the region of convergence of the series in powers of $x-c$ is the whole domain of the analytic function. For example, this is the case for the function $1/(x^2+1)$.