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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 $x_0$.


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?


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.

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    $\begingroup$ I don't get it. The function $f(z)=\frac1z$ is analytic on the open set $D=\{z:z\ne0\}$ according to the "former and correct" definition, but how does it comply with the latter definition? What is $c$? $\endgroup$ – bof Jun 6 '18 at 20:39
  • $\begingroup$ What is c in the second “definition?” $\endgroup$ – Aurel Jun 6 '18 at 20:40
  • $\begingroup$ You are right, I might have been confused, I will edit my question. $\endgroup$ – Leo Jun 6 '18 at 20:42
  • $\begingroup$ In the first definition $b_n$ may depend on $x_0$, i.e. it does not have to be the same series for all $x_0\in D$. $\endgroup$ – A.Γ. Jun 6 '18 at 20:50
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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 $1/(x^2+1)$.

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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 $D$.

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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.

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