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Which function is given by a power series whose convergence radius is infinite?

$$A. \ \ \ e^{-\frac{1}{x^2}}$$ $$B. \ \ \ \sin{\left(\frac{1}{x}\right)}$$ $$C. \ \ \ \cos{\left(\frac{1}{1+x^2}\right)}$$ $$D. \ \ \ 1+x+x^3$$

$$$$

When we have the power series $\sum_{ n=0}^{\infty} a_n (x−ξ)^n$ the radius of convergence is infinite when $p=0$ , where $p=\lim \sup \sqrt[n]{|a_n |} $. Does this stand? Do I have to find the power series of all these functions?

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    Can you share what you've tried, and explain what you're having trouble with? Do you know a relationship between the domain of convergence and the poles of the function? –  Jun 25 '14 at 23:39
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    When we have the power series $$\sum_{n=0}^{\infty} a_n (x-\xi)^n$$ the radius of convergence is infinite when $p=0$, where $p=\lim \sup \sqrt[n]{|a_n|}$. Does this stand? Do I have to find the power series of all these functions? –  Jun 25 '14 at 23:55
  • By the way, you should add your comment to the question - otherwise, it might be put on hold as missing context / details / efforts. –  Jun 26 '14 at 00:11
  • Ok, I added it to the question. –  Jun 26 '14 at 00:14
  • A related problem. – Mhenni Benghorbal Jun 26 '14 at 03:34

2 Answers2

1

You have the right formula for the radius of convergence, but it's rather hard to use in this case. A general fact is that

The radius of convergence of a power series centered at a point is the distance to the closest singularity, be it a pole or an essential singularity.

So figure out which of these functions are entire, and you're done. Polynomials are might nice, though.

0

For D it should be clear, that the convergence radius is $\infty$ (why?).

For A, B and C. U can use the power series extensions of the functions:

$e^x=\sum\limits_{n=0}^{\infty} \frac{x^n}{n!} $

$sin(x)=\sum\limits_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!}$

$cos(x)=\sum\limits_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{(2n)!}$

(With this way you are also able to classify the pole(s) if u have one)

I give an example:

Given: $f(z)=\frac{e^z}{z}$ The we obtain: $\sum\limits_{n=0}^{\infty} \frac{z^n}{n!}\frac{1}{z}=\sum\limits_{n=0}^{\infty} \frac{z^{n-1}}{n!}=\frac{1}{z}+1+\frac{z}{2!}+\frac{z^2}{3!}+.....$

As we can see we got problems in 0. In this case 0 is called a pole (pole of order 1), because $\lim_{z \to 0} f(z)=\infty $. So our function can not have a $\infty$ convergence radius.

If you face the situation that there is no addend in the form $\frac{1}{(z-z_0)^n}$, after you have shorten your term, then your function is holomorph on $\mathbb C$ i.e. the convergence radius is $\infty$

Can u now solve your tasks?

Marm
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  • "If you face the situation that there is no addend in the form $\frac{1}{(z-z_0)^n}$, after you have shorten your term, then your function is holomorph on $\mathbb{C}$ i.e. the convergence radius is $\infty$" Is this true? Consider $f(z) = z+z^2+...+z^n+...z^\infty$ whose radius of convergence is 1 but has not principal part. – Mathstudent Mar 02 '20 at 06:31
  • If you have an entire function, then the radius of convergence of the Laurent series at any point is $\infty$, since there is no pole or essential singularity to be encountered. But you cannot conclude a series is entire by simply looking at the Laurent expansion at a point, while what you can say around that point it is analytic. – Mathstudent Mar 02 '20 at 06:35