Taylor Series reaches closest Singularity
This page has been identified as a candidate for refactoring of medium complexity. In particular: Separate page for Real and Complex versions Until this has been finished, please leave {{Refactor}} in the code.
New contributors: Refactoring is a task which is expected to be undertaken by experienced editors only. Because of the underlying complexity of the work needed, it is recommended that you do not embark on a refactoring task until you have become familiar with the structural nature of pages of $\mathsf{Pr} \infty \mathsf{fWiki}$.To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Refactor}} from the code. |
Theorem
Let $F$ be a complex function.
Let $F$ be analytic everywhere except at a finite number of singularities.
Let a singularity of $F$ be one of the following:
In the latter case $F$ is a restriction of a multifunction to one of its branches.
Let $x_0$ be a real number.
Let $F$ be analytic at the complex number $\tuple {x_0, 0}$.
Let $R \in \R_{>0}$ be the distance from the complex number $\tuple {x_0, 0}$ to the closest singularity of $F$.
Let the restriction of $F$ to $\R \to \C$ be a real function $f$.
This means:
- $\forall x \in \R: \map f x = \map \Re {\map F {x, 0} }, 0 = \map \Im {\map F {x, 0} }$
where $\tuple {x, 0}$ denotes the complex number with real part $x$ and imaginary part $0$.
Then:
- the Taylor series of $f$ about $x_0$ converges to $f$ at every point $x \in \R$ satisfying $\size {x - x_0} < R$
Proof
We have that $F$ is analytic everywhere except at its singularities.
Also, the distance from the complex number $\tuple {x_0, 0}$ to the closest singularity of $F$ is $R$.
Therefore:
- $F$ is analytic at every point $z \in \C$ satisfying $\size {z - \tuple {x_0, 0} } < R$
where $\tuple {x_0 , 0}$ denotes the complex number with real part $x_0$ and imaginary part $0$.
The result follows by Convergence of Taylor Series of Function Analytic on Disk.
$\blacksquare$