Showing $f(x) = \frac{x^2}{\sin(x)}$ is analytic near $0$

Question

Problem

Show the function

$$ f(x) = \frac{x^2}{\sin(x)} $$

is an analytic about $x=0$.

Try

We have

$$ f(x) = \frac{x^2}{x - x^3/3! + x^5/5! - \cdots } $$

Letting $f(x) = \sum_{n=0}^\infty a_n x^n $, we have

$$ a_0 = 0, a_1 = 1, a_2 = 0, a_3 = 1/6, a_4 = 0, a_5 = 7/360, \cdots $$

thus $f(x) = x - x^3/6 + \cdots$.

Now we have to decide if $\sum_{n=0}^\infty a_n x^n$ is convergent near zero.

But I cannot see the rules of $a_0, a_1, a_2, \cdots$, so how should I proceed?

It is known that

Assume $f: = \sum_{n=0}^\infty b_n x^n$ and $g:= \sum_{n=0}^\infty c_n x^n$ be convergent near zero with radius of convergence $\rho>0$.

If $g(0) \neq 0$, we have

$$ \frac{f}{g} = \sum_{n=0}^\infty d_n $$

for some radius of convergence $\le \rho$. So relating this fact to my problem,

$$ f(x) = \frac{x}{1 - x^2/3! + x^4/5! - \cdots } $$

may be convergent with radius of convergence $\le \infty$, which is the radius of $x^2$(poylnomial) and $\sin(x)$

What exactly do you men by 'some radius $\leq \rho$'? Do you know that the radius is $>0$? — Kavi Rama Murthy, Jul 15 '19 at 09:03
@KaviRamaMurthy I meant "radius of convergence". Are they still equivocal?. Edited. Thanks — Moreblue, Jul 15 '19 at 09:08
Note that $0$ is a removable singularity and numerator and denominator are both holomorphic, then done. — Qi Zhu, Jul 15 '19 at 09:45

score 1 · Accepted Answer · answered Jul 15 '19 at 09:10

1

We have $f(x) = \frac{x^2}{x - x^3/3! + x^5/5! - \cdots }=\frac{x}{1 - x^2/3! + x^4/5! - \cdots }.$

The nominator and the denominator are power series with radius of convergenc $= \infty$. The power series in the denominator is $ \ne 0$ at $x=0.$ Hence there is $ r>0$ such that

$f(x)= \sum_{n=0}^\infty d_nx^n$ for $|x|<r.$

We have $r= \pi$. Why ?

answered Jul 15 '19 at 09:10

Fred

77,394

The point is whether $x$ can be cancelled in numerator and denominator for all $x\in\mathbb R$? – Nitin Uniyal Jul 15 '19 at 10:10

score 1 · Answer 2 · answered Jul 15 '19 at 09:15

1

$f(x)=\frac x {g(x)}$ where $g(x)=\frac {\sin\, x} x$ (for $x \neq 0$, $g(0)=1$). So what you have written as 'It is known that ....' immediately gives you the result.

answered Jul 15 '19 at 09:15

Kavi Rama Murthy

311,013

score 1 · Answer 3 · 2019-07-15T09:33:39.197

1

$f$ is not defined where $\sin x=0$. Elsewhere, it is the ratio of two analytic functions, hence it is analytic in $\mathbb R\setminus\{k\pi\}$.

Note that discussing analyticity at $x=0$ would be irrelevant.

edited Jul 15 '19 at 09:33

answered Jul 15 '19 at 09:24

score 1 · Answer 4 · answered Jul 15 '19 at 10:34

A real function $f(x)$ is $R-analytic$ at some point $x_0$ if there exists an interval $(x_0-\epsilon,x_0+\epsilon)$ s.t. $f(x)=\sum_{n=0}^\infty a_n(x-x_0)^n$ valid in $(x_0-\epsilon,x_0+\epsilon)$.

It is not so hard to see that $a_n=\frac{f^{(n)}(x_0)}{n!}$ is the Taylor's coefficient. Since $lim_{x\rightarrow 0}f(x)=1$ which assures a removable singularity at $x=0$ (though $a_0=f(0)$ is not explicitly defined in the question). On defining $f(0)=1$ you will get a nice series in non-negative powers of $x$ valid in$(-π,π)$, which shows $f(x)$ is $R-analytic$ at $x=0$.

Showing $f(x) = \frac{x^2}{\sin(x)}$ is analytic near $0$

Problem

Try

4 Answers4