Analytic Function and Power Series Expansion In my real analysis class, my professor mentioned about the following that I did not quite follow:
Given $f$ analytic at $x_0$, say $f(x) = \sum_{n=0}^{\infty}\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n$ with $|x-x_0|<\delta$, where $\delta$ is the radius of analyticity.
Now, if we choose $x_1 \in (x_0 - \delta, x_0 + \delta)$, and expand it like $\sum_{n=0}^{\infty}\frac{f^{(n)}(x_1)}{n!}(x-x_1)^n$, he claimed that this power series at $x_1$ should also equal to $f(x)$.
I did not understand two parts.


*

*Why can we expand at $x_1$ like that? The $\delta$ above is defined around $x_0$, but what if the $x$ in the $x_1$ expansion does not satisfy $|x-x_0|<\delta$? 

*Why the two series are equal?

*Is the radius of analyticity simply just radius of convergence of the power series?


Thanks a lot for any explanations or clarifications. Feel free to mention any relevant theorems related.
 A: 
If $f(z) = \sum_{n=0}^\infty c_n z^n$ converges for $|z| < R$ then for any $|z|+|a| < R$, $f(a+z) = \sum_{k=0}^\infty b_k z^k$ where $b_k = \frac{f^{(k)}(a)}{k!}$.

Proof : $f(z) = \sum_{n=0}^\infty c_n z^n$ converges absolutely for $|z|< R$ thus
$$f(a+z) = \sum_{n=0}^\infty c_n (z+a)^n = \sum_{n=0}^\infty c_n \sum_{k=0}^n {n \choose k} z^k a^{n-k}$$ where the latter double sum converges absolutely for $|z|+|a| < R$ as $(|z|+|a|)^n = \sum_{k=0}^n |{n \choose k} z^k a^{n-k}|$ 
thus we can reorganize to get 
$$f(a+z) = \sum_{k=0}^\infty z^k \sum_{n=k}^\infty c_n {n \choose k}a^{n-k} = \sum_{k=0}^\infty z^k \frac{f^{(k)}(a)}{k!}$$
A: $f$ is analytic in $|x-x_1|<\delta -|x_1-x_0|$ because this inequlaity implies that $|x-x_0| <\delta$. Hence $f$ has a power series expansion in the disk with center $x_1$ and radius $\delta -|x-x_0|$. This implies that $f(x)=\sum \frac {f^{k}(x-1)} {k!} (x-x_1)^{k}$ in that disk. [ This theorem can be found in any book on Complex Analysis]. 
Any analytic function has a power series expansion around a point $x$ in the domain valid in the largest disk around $x$ contained in the domain. I hope this ansers part 3). 
