Convergence of Taylor Series of Function Analytic on Disk

Theorem
Let $F$ be a complex function.

Let $x_0$ be a point in $\R$.

Let $R$ be an extended real number greater than zero.

Let $F$ be analytic at every point $z \in \C$ satisfying $\size {z - \paren {x_0, 0}} < R$

where $\paren {x_0, 0}$ denotes the complex number with real part $x_0$ and imaginary part $0$.

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 $\paren {x, 0}$ 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 $\size {x - x_0} < R$

Corollary: Taylor Series reaches closest Singularity
Let the singularities of a function be the points at which the function is not analytic.

Let $F$ be analytic everywhere except at a finite number of singularities.

Let $R \in \R_{>0}$ be the distance from $x_0$ to the closest singularity of $F$.

Then:
 * the Taylor series of $f$ about $x_0$ converges to $f$ at every point $x \in \R$ satisfying $\size {x - x_0} < R$

Corollary: Taylor Series of Analytic Function has infinite Radius of Convergence
Let $F$ be analytic everywhere.

Then:
 * the Taylor series of $f$ about $x_0$ converges to $f$ at every point in $\R$

Lemma
Let $r$ be a real number satisfying:
 * $0 < r < R$

Let $x$ be a real number satisfying:
 * $\size {x - x_0} < r$

$f$ has a Taylor series expansion about $x_0$ with radius of convergence greater than zero as $f$ is analytic at $x_0$.

The Taylor's formula with remainder for $f$ about $x_0$ is:
 * $\map f x = \ds \sum_{i \mathop = 0}^n \frac {\paren {x - x_0}^i} {i!} \map {f^{\paren i} } {x_0} + \map {R_n} x$

where
 * $\map {R_n} x = \dfrac 1 {n!} \ds \int_{x_0}^x \paren {x - t}^n \map {f^{\paren {n \mathop + 1} } } t \rd t$

Our first aim is to prove:
 * $\ds \lim_{n \mathop \to \infty} \map {R_n} x = 0$

For the case $x = x_0$, the interval of integration in the expression for $\map {R_n} x$ has zero length.

Therefore, $\map {R_n} x = 0$.

Accordingly, $\ds \lim_{n \mathop \to \infty} \map {R_n} x = 0$ is true for this case.

Now we consider the case $x \ne x_0$.

We have:
 * $0 < r - \size {x - x_0}$ as $\size {x - x_0} < r$

Observe that:
 * $\size {x - x_0} \ge \size {t - x_0}$

Therefore:
 * $r - \size {x - x_0} \le r - \size {t - x_0}$
 * $0 < r - \size {x - x_0} \le r - \size {t - x_0}$
 * $0 < \size {r - \size {x - x_0} } \le \size {r - \size {t - x_0} }$

We have:

Let $y \in \R$ be equal to $x_0 + r$ if $x > x_0$ and $x_0 - r$ if $x < x_0$.

Note that $y > x$ if $x > x_0$ and $y < x$ if $x < x_0$.

The general situation is:
 * $x_0 \le t \le x < y$ if $x > x_0$
 * $y < x \le t \le x_0$ if $x < x_0$

Let us study $\size {x - t}$ in the expression above for the bound for $\size {\map {R_n} x}$:

Also, we have:

We combine these two results to get:

We use this result in the expression for the bound for $\size {\map {R_n} x}$:

We have:
 * $\ds \frac r {\size {x - x_0} } > 1$ as $\size {x - x_0} < r$ and $x \ne x_0$

Therefore:
 * $\ds \lim_{n \mathop \to \infty} \frac n {\paren {\frac r {\size {x - x_0} } }^n} = 0$ by the lemma

Letting $n$ approach $\infty$ in the expression for the bound for $\size {\map {R_n} x}$, we get:

So:

Accordingly, $\ds \lim_{n \mathop \to \infty} \map {R_n} x = 0$ is true for the case $x \ne x_0$.

Thus, $\ds \lim_{n \mathop \to \infty} \map {R_n} x = 0$ holds for every $x$ satisfying $\size {x - x_0} < r$ where $r < R$.

Since we can choose $r$ as close to $R$ as we like, we conclude that $\ds \lim_{n \mathop \to \infty} \map {R_n} x = 0$ holds for every $x$ that satisfies $\size {x - x_0} < R$.

Therefore, the Taylor series expansion of $\map f x$ about $x_0$ converges to $\map f x$ for every $x$ that satisfies $\size {x - x_0} < R$.