Removable_singularity Knowpia

In complex analysis, a removable singularity of a holomorphic function is a point at which the function is undefined, but it is possible to redefine the function at that point in such a way that the resulting function is regular in a neighbourhood of that point.

For instance, the (unnormalized) sinc function, as defined by

{\text{sinc}}(z)={\frac {\sin z}{z}}

has a singularity at $z = 0$ . This singularity can be removed by defining ${\text{sinc}}(0):=1,$ which is the limit of $sinc$ as $z$ tends to 0. The resulting function is holomorphic. In this case the problem was caused by $sinc$ being given an indeterminate form. Taking a power series expansion for ${\textstyle {\frac {\sin(z)}{z}}}$ around the singular point shows that

{\text{sinc}}(z)={\frac {1}{z}}\left(\sum _{k=0}^{\infty }{\frac {(-1)^{k}z^{2k+1}}{(2k+1)!}}\right)=\sum _{k=0}^{\infty }{\frac {(-1)^{k}z^{2k}}{(2k+1)!}}=1-{\frac {z^{2}}{3!}}+{\frac {z^{4}}{5!}}-{\frac {z^{6}}{7!}}+\cdots .

Formally, if $U\subset \mathbb {C}$ is an open subset of the complex plane $\mathbb {C}$ , $a\in U$ a point of $U$ , and $f:U\setminus \{a\}\rightarrow \mathbb {C}$ is a holomorphic function, then $a$ is called a removable singularity for $f$ if there exists a holomorphic function $g:U\rightarrow \mathbb {C}$ which coincides with $f$ on $U\setminus \{a\}$ . We say $f$ is holomorphically extendable over $U$ if such a $g$ exists.

Riemann's theorem edit

Riemann's theorem on removable singularities is as follows:

Theorem — Let $D\subset \mathbb {C}$ be an open subset of the complex plane, $a\in D$ a point of $D$ and $f$ a holomorphic function defined on the set $D\setminus \{a\}$ . The following are equivalent:

$f$ is holomorphically extendable over $a$ .
$f$ is continuously extendable over $a$ .
There exists a neighborhood of $a$ on which $f$ is bounded.
$\lim _{z\to a}(z-a)f(z)=0$ .

The implications 1 ⇒ 2 ⇒ 3 ⇒ 4 are trivial. To prove 4 ⇒ 1, we first recall that the holomorphy of a function at $a$ is equivalent to it being analytic at $a$ (proof), i.e. having a power series representation. Define

h(z)={\begin{cases}(z-a)^{2}f(z)&z\neq a,\\0&z=a.\end{cases}}

Clearly, h is holomorphic on $D\setminus \{a\}$ , and there exists

h'(a)=\lim _{z\to a}{\frac {(z-a)^{2}f(z)-0}{z-a}}=\lim _{z\to a}(z-a)f(z)=0

by 4, hence h is holomorphic on D and has a Taylor series about a:

h(z)=c_{0}+c_{1}(z-a)+c_{2}(z-a)^{2}+c_{3}(z-a)^{3}+\cdots \,.

We have c₀ = h(a) = 0 and c₁ = h'(a) = 0; therefore

h(z)=c_{2}(z-a)^{2}+c_{3}(z-a)^{3}+\cdots \,.

Hence, where $z\neq a$ , we have:

f(z)={\frac {h(z)}{(z-a)^{2}}}=c_{2}+c_{3}(z-a)+\cdots \,.

However,

g(z)=c_{2}+c_{3}(z-a)+\cdots \,.

is holomorphic on D, thus an extension of $f$ .

Other kinds of singularities edit

Unlike functions of a real variable, holomorphic functions are sufficiently rigid that their isolated singularities can be completely classified. A holomorphic function's singularity is either not really a singularity at all, i.e. a removable singularity, or one of the following two types:

In light of Riemann's theorem, given a non-removable singularity, one might ask whether there exists a natural number $m$ such that $\lim _{z\rightarrow a}(z-a)^{m+1}f(z)=0$ . If so, $a$ is called a pole of $f$ and the smallest such $m$ is the order of $a$ . So removable singularities are precisely the poles of order 0. A holomorphic function blows up uniformly near its other poles.
If an isolated singularity $a$ of $f$ is neither removable nor a pole, it is called an essential singularity. The Great Picard Theorem shows that such an $f$ maps every punctured open neighborhood $U\setminus \{a\}$ to the entire complex plane, with the possible exception of at most one point.

External links edit

Removable singular point at Encyclopedia of Mathematics

Summary

Riemann's theorem edit

Other kinds of singularities edit

See also edit

External links edit