2
votes
3answers
66 views

Does the Taylor series of an infinitely differentiable function converge; and if yes, does it converge to the function? [duplicate]

I have googled it, but I am not satisfied with those. So my questions are: Let $D$ be an open set in $\mathbb{R}$. Let $f:D\rightarrow \mathbb{R}$ be a infinitely differentiable ...
0
votes
1answer
16 views

Convergence of Taylor series about centre of open disc for analytic function.

I define a function on an open set of the complex plane to be analytic if about any point $z_0$ in that set it can be expanded as a power series in $(z - z_0)$ that converges in some neighbourhood of ...
3
votes
2answers
112 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
42 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 ...
0
votes
1answer
44 views

Suppose $f \in C^{\infty}(\mathbb R)$ and $\lim_{n \to \infty} \frac{1}{n!} \int_0^a x^n f^{(n+1)}(a-x)dx=0.$ Show $f$ is analytic on $\mathbb{R}$.

Suppose that $f \in C^{\infty} (-\infty , \infty)$ and that $$\lim_{n \to \infty} \frac{1}{n!} \int_0^a x^n f^{(n+1)}(a-x)dx=0$$ for all $a\in \mathbb{R}$. Prove that $f$ is analytic on ...
3
votes
2answers
360 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
323 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
342 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
1k 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
895 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 ...