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 $\size {z - \tuple {x_0, 0} } < R$
where $\tuple {x_0, 0}$ 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 $\tuple {x, 0}$ 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 $\size {x - x_0} < R$
=== Corollary: Taylor Series reaches closest Singularity ===
=== Corollary: Taylor Series of Analytic Function has infinite Radius of Convergence ===
== Proof ==
=== 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 $\size {x - x_0} < R$.
Category:Taylor Series
Category:Real Analysis