Convergence of Taylor Series of Function Analytic on Disk

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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 $\cmod {z - x_0} < R$.


Let $f = F {\restriction_{\R}}$ be a real function.


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 $\left\lvert{x - x_0}\right\rvert < R$


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 $\R$


Proof

Lemma

Let $y > 1$.


Then:

$\displaystyle \lim_{n \mathop \to \infty} \frac n {y^n} = 0$


Let $r$ be a real number satisfying:

$0 < r < R$

Let $x$ be a real number satisfying:

$\left\lvert{x - x_0}\right\rvert < r$

$f$ 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:

$f \left({x}\right) = \displaystyle \sum_{i \mathop = 0}^n \frac {\left({x - x_0}\right)^i} {i!} f^{\left({i}\right)} \left({x_0}\right) + R_n \left({x}\right)$

where

$R_n \left({x}\right) = \dfrac 1 {n!} \displaystyle \int_{x_0}^x \left({x - t}\right)^n f^{\left({n \mathop + 1}\right)} \left({t}\right) \mathrm d t$

Our first aim is to prove:

$\displaystyle \lim_{n \mathop \to \infty} R_n \left({x}\right) = 0$


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

Therefore, $R_n \left({x}\right) = 0$.

Accordingly, $\displaystyle \lim_{n \mathop \to \infty} R_n \left({x}\right) = 0$ is true for this case.


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


We have:

$0 < r - \left\lvert{x - x_0}\right\rvert$ as $\left\lvert{x - x_0}\right\rvert < r$

Observe that:

$\left\lvert{x - x_0}\right\rvert \ge \left\lvert{t - x_0}\right\rvert$

Therefore:

$r - \left\lvert{x - x_0}\right\rvert \le r - \left\lvert{t - x_0}\right\rvert$
$0 < r - \left\lvert{x - x_0}\right\rvert \le r - \left\lvert{t - x_0}\right\rvert$
$0 < \left\lvert{r - \left\lvert{x - x_0}\right\rvert}\right\rvert\le \left\lvert{r - \left\lvert{t - x_0}\right\rvert}\right\rvert$


We have:

\(\displaystyle \left\lvert{R_n \left({x}\right)}\right\rvert\) \(=\) \(\displaystyle \left\lvert{\frac 1 {n!} \int_{x_0}^x \left({x - t}\right)^n f^{\left({n \mathop + 1}\right)} \left({t}\right) \mathrm d t}\right\rvert\) $\quad$ $\quad$
\(\displaystyle \) \(\le\) \(\displaystyle \frac 1 {n!} \int_{x_0}^x \left\lvert{x - t}\right\rvert^n \left\lvert{f^{\left({n \mathop + 1}\right)} \left({t}\right)}\right\rvert \left\lvert{\mathrm d t}\right\rvert\) $\quad$ $\quad$
\(\displaystyle \) \(\le\) \(\displaystyle \frac 1 {n!} \int_{x_0}^x \left\lvert{x - t}\right\rvert^n \left\lvert{\frac {M r \left({n + 1}\right)!} {\left({r - \left\lvert{t - x_0}\right\rvert}\right)^{\left({n \mathop + 2}\right)} } }\right\rvert \left\lvert{\mathrm d t}\right\rvert\) $\quad$ where $M \in \R_{>=0}$ , by Bound for Analytic Function and Derivatives $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {n!} \int_{x_0}^x \left\lvert{x - t}\right\rvert^n \frac {M r \left({n + 1}\right)!} {\left({r - \left\lvert{t - x_0}\right\rvert}\right)^{\left({n \mathop + 2}\right)} } \left\lvert{\mathrm d t}\right\rvert\) $\quad$ as $r - \left\lvert{t - x_0}\right\rvert > 0$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {n!} \int_{x_0}^x \left\lvert{x - t}\right\rvert^n \frac {M r \left({n + 1}\right)!} {\left({r - \left\lvert{t - x_0}\right\rvert}\right)^2} \frac 1 {\left({r - \left\lvert{t - x_0}\right\rvert}\right)^n} \left\lvert{\mathrm d t}\right\rvert\) $\quad$ $\quad$
\(\displaystyle \) \(\le\) \(\displaystyle \frac 1 {n!} \int_{x_0}^x \left\lvert{x - t}\right\rvert^n \frac {M r \left({n + 1}\right)!} {\left({r - \left\lvert{x - x_0}\right\rvert}\right)^2} \frac 1 {\left({r - \left\lvert{t - x_0}\right\rvert}\right)^n} \left\lvert{\mathrm d t}\right\rvert\) $\quad$ as $0 < \left\lvert{r - \left\lvert{x - x_0}\right\rvert}\right\rvert\le \left\lvert{r - \left\lvert{t - x_0}\right\rvert}\right\rvert$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {n!} \frac {M r \left({n + 1}\right)!} {\left({r - \left\lvert{x - x_0}\right\rvert}\right)^2} \int_{x_0}^x \frac {\left\lvert{x - t}\right\rvert^n} {\left({r - \left\lvert{t - x_0}\right\rvert}\right)^n} \left\lvert{\mathrm d t}\right\rvert\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {M r \left({n + 1}\right)} {\left({r - \left\lvert{x - x_0}\right\rvert}\right)^2} \int_{x_0}^x \left({\frac {\left\lvert{x - t}\right\rvert} {\left({r - \left\lvert{t - x_0}\right\rvert}\right)} }\right)^n \left\lvert{\mathrm d t}\right\rvert\) $\quad$ $\quad$


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 $\left\lvert{x - t}\right\rvert$ in the expression above for the bound for $\left\lvert{R_n \left({x}\right)}\right\rvert$:

\(\displaystyle \left\lvert{x - t}\right\rvert\) \(=\) \(\displaystyle \left\lvert{x - y + y - t}\right\rvert\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left\lvert{y - t - \left({y - x}\right)}\right\rvert\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left\lvert{\left\lvert{y - t}\right\rvert - \left\lvert{y - x}\right\rvert}\right\rvert\) $\quad$ as $\left({y - t}\right)$ and $\left({y-x}\right)$ have the same sign because either $t \le x < y$ or $y < x \le t$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left\lvert{y - t}\right\rvert - \left\lvert{y - x}\right\rvert\) $\quad$ as $\left\lvert{y - t}\right\rvert \ge \left\lvert{y - x}\right\rvert$ because either $t \le x < y$ or $y < x \le t$ $\quad$

Also, we have:

\(\displaystyle r - \left\lvert{t - x_0}\right\rvert\) \(=\) \(\displaystyle \left\lvert{r - \left\lvert{t - x_0}\right\rvert}\right\rvert\) $\quad$ as $r - \left\lvert{t - x_0}\right\rvert > 0$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left\lvert{\left\lvert{y - x_0}\right\rvert - \left\lvert{t - x_0}\right\rvert}\right\rvert\) $\quad$ as $\left\lvert{y - x_0}\right\rvert = r$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left\lvert{y - x_0 - \left({t - x_0}\right)}\right\rvert\) $\quad$ since $\left({y - x_0}\right)$ and $\left({t - x_0}\right)$ have the same sign because either $x_0 \le t < y$ or $y < t \le x_0$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left\lvert{y - t}\right\rvert\) $\quad$ $\quad$

We combine these two results to get:

\(\displaystyle \frac {\left\lvert{x - t}\right\rvert} {r - \left\lvert{t - x_0}\right\rvert}\) \(=\) \(\displaystyle \frac {\left\lvert{y - t}\right\rvert - \left\lvert{y - x}\right\rvert} {\left\lvert{y - t}\right\rvert}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle 1 - \frac {\left\lvert{y - x}\right\rvert} {\left\lvert{y - t}\right\rvert}\) $\quad$ $\quad$
\(\displaystyle \) \(\le\) \(\displaystyle 1 - \frac {\left\lvert{y - x}\right\rvert} r\) $\quad$ as $r \ge \left\lvert{y - t}\right\rvert$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {r - \left\lvert{y - x}\right\rvert} r\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {\left\lvert{y - x_0}\right\rvert - \left\lvert{y - x}\right\rvert} r\) $\quad$ as $\left\lvert{y - x_0}\right\rvert = r$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {\left\lvert{\left\lvert{y - x_0}\right\rvert - \left\lvert{y - x}\right\rvert}\right\rvert} r\) $\quad$ as $\left\lvert{y - x_0}\right\rvert \ge \left\lvert{y - x}\right\rvert$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {\left\lvert{y - x_0 - \left({y - x}\right)}\right\rvert} r\) $\quad$ since $\left({y - x_0}\right)$ and $\left({y - x}\right)$ have the same sign because either $x_0 < x < y$ or $y < x < x_0$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {\left\lvert{x - x_0}\right\rvert} r\) $\quad$ $\quad$

We use this result in the expression for the bound for $\left\lvert{R_n \left({x}\right)}\right\rvert$:

\(\displaystyle \left\lvert{R_n \left({x}\right)}\right\rvert\) \(\le\) \(\displaystyle \frac {M r \left({n + 1}\right)} {\left({r - \left\lvert{x - x_0}\right\rvert}\right)^2} \int_{x_0}^x \left({\frac {\left\lvert{x - t}\right\rvert} {\left({r - \left\lvert{t - x_0}\right\rvert}\right)} }\right)^n \left\lvert{\mathrm d t}\right\rvert\) $\quad$ $\quad$
\(\displaystyle \) \(\le\) \(\displaystyle \frac {M r \left({n + 1}\right)} {\left({r - \left\lvert{x - x_0}\right\rvert}\right)^2} \int_{x_0}^x \left({\frac {\left\lvert{x - x_0}\right\rvert} r}\right)^n \left\lvert{\mathrm d t}\right\rvert\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {M r \left({n + 1}\right)} {\left({r - \left\lvert{x - x_0}\right\rvert}\right)^2} \left({\frac {\left\lvert{x - x_0}\right\rvert} r}\right)^n \int_{x_0}^x \left\lvert{\mathrm d t}\right\rvert\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {M r \left({n + 1}\right)} {\left({r - \left\lvert{x - x_0}\right\rvert}\right)^2} \left({\frac {\left\lvert{x - x_0}\right\rvert} r}\right)^n \left\lvert{x - x_0}\right\rvert\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {M r \left\lvert{x - x_0}\right\rvert} {\left({r - \left\lvert{x - x_0}\right\rvert}\right)^2} \frac {\left({n + 1}\right)} {\left({\frac r {\left\lvert{x - x_0}\right\rvert} }\right)^n}\) $\quad$ $\quad$

We have:

$\displaystyle \frac r {\left\lvert{x - x_0}\right\rvert} > 1$ as $\left\lvert{x - x_0}\right\rvert < r$ and $x \ne x_0$

Therefore:

$\displaystyle \lim_{n \mathop \to \infty} \frac n {\left({\frac r {\left\lvert{x - x_0}\right\rvert} }\right)^n} = 0$ by the lemma

Letting $n$ approach $\infty$ in the expression for the bound for $\left\lvert{R_n \left({x}\right)}\right\rvert$, we get:

\(\displaystyle \lim_{n \mathop \to \infty} \left\lvert{R_n \left({x}\right)}\right\rvert\) \(\le\) \(\displaystyle \lim_{n \mathop \to \infty} \frac {M r \left\lvert{x - x_0}\right\rvert} {\left({r - \left\lvert{x - x_0}\right\rvert}\right)^2} \frac {n + 1} {\left({\frac r {\left\lvert{x - x_0}\right\rvert} }\right)^n}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \lim_{n \mathop \to \infty} \frac {M r \left\lvert{x - x_0}\right\rvert} {\left({r - \left\lvert{x - x_0}\right\rvert}\right)^2} \frac {n + 1} n \frac n {\left({\frac r {\left\lvert{x - x_0}\right\rvert} }\right)^n}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {M r \left\lvert{x - x_0}\right\rvert} {\left({r - \left\lvert{x - x_0}\right\rvert}\right)^2} \lim_{n \mathop \to \infty} \frac {n + 1} n \frac n {\left({\frac r {\left\lvert{x - x_0}\right\rvert} }\right)^n}\) $\quad$ Multiple Rule for Real Sequences $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {M r \left\lvert{x - x_0}\right\rvert} {\left({r - \left\lvert{x - x_0}\right\rvert}\right)^2} \lim_{n \mathop \to \infty} \frac {n + 1} n \lim_{n \mathop \to \infty} \frac n {\left({\frac r {\left\lvert{x - x_0}\right\rvert} }\right)^n}\) $\quad$ Product Rule for Real Sequences $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {M r \left\lvert{x - x_0}\right\rvert} {\left({r - \left\lvert{x - x_0}\right\rvert}\right)^2} 1 \lim_{n \mathop \to \infty} \frac n {\left({\frac r {\left\lvert{x - x_0}\right\rvert} }\right)^n}\) $\quad$ as $\displaystyle \lim_{n \mathop \to \infty} \frac {n + 1} n = 1$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {M r \left\lvert{x - x_0}\right\rvert} {\left({r - \left\lvert{x - x_0}\right\rvert}\right)^2} 0\) $\quad$ as $\displaystyle \lim_{n \mathop \to \infty} \frac n {\left({\frac r {\left\lvert{x - x_0}\right\rvert} }\right)^n} = 0$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle 0\) $\quad$ $\quad$

So:

\(\displaystyle \lim_{n \mathop \to \infty} \left\lvert{R_n \left({x}\right)}\right\rvert\) \(\le\) \(\displaystyle 0\) $\quad$ $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle \lim_{n \mathop \to \infty} \left\lvert{R_n \left({x}\right)}\right\rvert\) \(=\) \(\displaystyle 0\) $\quad$ $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle \lim_{n \mathop \to \infty} R_n \left({x}\right)\) \(=\) \(\displaystyle 0\) $\quad$ $\quad$

Accordingly, $\displaystyle \lim_{n \mathop \to \infty} R_n \left({x}\right) = 0$ is true for the case $x \ne x_0$.


Thus, $\displaystyle \lim_{n \mathop \to \infty} R_n \left({x}\right) = 0$ holds for every $x$ satisfying $\left\lvert{x - x_0}\right\rvert < r$ where $r < R$.

Since we can choose $r$ as close to $R$ as we like, we conclude that $\displaystyle \lim_{n \mathop \to \infty} R_n \left({x}\right) = 0$ holds for every $x$ that satisfies $\left\lvert{x - x_0}\right\rvert < R$.

Therefore, the Taylor series expansion of $f \left({x}\right)$ about $x_0$ converges to $f \left({x}\right)$ for every $x$ that satisfies $\left\lvert{x - x_0}\right\rvert < R$.

$\blacksquare$