3
votes
2answers
99 views

If it converges, how to show that power series converges to $f(x)$?

I had a very basic question. Suppose $f(x)$ is a function. And let us say it has a power series :- $$f(x) = \sum_{n=0}^\infty a_nx^n.$$ Suppose we are operating inside the region of convergence. ...
0
votes
1answer
41 views

Convergence of an analytic function

Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a smooth function. Let $R$ be the radius of convergence of the Taylor series centered at $a.$ For each $n \in \mathbb{N},$ let $M_n= \sup\{f^{n}(t) : t \in ...
2
votes
2answers
283 views

Analytic functions of a real variable which do not extend to an open complex neighborhood

Do such functions exist? If not, is it appropriate to think of real analytic functions as "slices" of holomorphic functions?
2
votes
1answer
297 views

Radius of convergence of Maclaurin series for $\frac1{\sin z}-1/z+\frac{2z}{z^2-\pi^2}$

What is the radius of convergence of the Taylor series about $z=0$ for $h(z)=\frac1{\sin z}-1/z+\frac{2z}{z^2-\pi^2}$? Here's a plot ...
1
vote
1answer
288 views

Composition Taylor Series

Is there any theorem that specifies when we are allowed to compose the taylor series of two functions? Does it have a name? Thanks.
7
votes
4answers
935 views

How many smooth functions are non-analytic?

We know from example that not all smooth (infinitely differentiable) functions are analytic (equal to their Taylor expansion at all points). However, the examples on the linked page seem rather ...
6
votes
3answers
868 views

Explanation of Maclaurin Series of $x^\pi$

I am reviewing Calc $2$ material and I came across a problem which asked me to explain why $x^\pi$ does not have a Taylor Series expansion around $x=0$. To me it seems that it would have an expansion ...