Taylor Series reaches closest Singularity

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:
 * a pole
 * an essential singularity
 * a branch point

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.

Also see

 * Taylor Series of Analytic Function has infinite Radius of Convergence