Convergence of Taylor Series of Function Analytic on Disk

From ProofWiki
Jump to navigation Jump to search


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 $\size {x - x_0} < R$


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 $\R$


Proof

Lemma

Let $y > 1$.


Then:

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

$\Box$


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:

$\map f x = \displaystyle \sum_{i \mathop = 0}^n \frac {\paren {x - x_0}^i} {i!} \map {f^{\paren i} } {x_0} + \map {R_n} x$

where

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

Our first aim is to prove:

$\displaystyle \lim_{n \mathop \to \infty} \map {R_n} x = 0$


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, $\displaystyle \lim_{n \mathop \to \infty} \map {R_n} x = 0$ 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:

\(\displaystyle \size {\map {R_n} x}\) \(=\) \(\displaystyle \size {\frac 1 {n!} \int_{x_0}^x \paren {x - t}^n \map {f^{\paren {n \mathop + 1} } } t \rd t}\)
\(\displaystyle \) \(\le\) \(\displaystyle \frac 1 {n!} \int_{x_0}^x \size {x - t}^n \size {\map {f^{\paren {n \mathop + 1} } } t} \size {\d t}\)
\(\displaystyle \) \(\le\) \(\displaystyle \frac 1 {n!} \int_{x_0}^x \size {x - t}^n \size {\frac {M r \paren {n + 1}!} {\paren {r - \size {t - x_0} }^{\paren {n \mathop + 2} } } } \size {\d t}\) where $M \in \R_{\ge 0}$ , by Bound for Analytic Function and Derivatives
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {n!} \int_{x_0}^x \size {x - t}^n \frac {M r \paren {n + 1}!} {\paren {r - \size {t - x_0} }^{\paren {n \mathop + 2} } } \size {\d t}\) as $r - \size {t - x_0} > 0$
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {n!} \int_{x_0}^x \size {x - t}^n \frac {M r \paren {n + 1}!} {\paren {r - \size {t - x_0} }^2} \frac 1 {\paren {r - \size {t - x_0} }^n} \size {\d t}\)
\(\displaystyle \) \(\le\) \(\displaystyle \frac 1 {n!} \int_{x_0}^x \size {x - t}^n \frac {M r \paren {n + 1}!} {\paren {r - \size {x - x_0} }^2} \frac 1 {\paren {r - \size {t - x_0} }^n} \size {\d t}\) as $0 < \size {r - \size {x - x_0} } \le \size {r - \size {t - x_0} }$
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {n!} \frac {M r \paren {n + 1}!} {\paren {r - \size {x - x_0} }^2} \int_{x_0}^x \frac {\size {x - t}^n} {\paren {r - \size {t - x_0} }^n} \size {\d t}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {M r \paren {n + 1} } {\paren {r - \size {x - x_0} }^2} \int_{x_0}^x \paren {\frac {\size {x - t} } {\paren {r - \size {t - x_0} } } }^n \size {\d t}\)


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 $\size {\map {R_n} x}$:

\(\displaystyle \size {x - t}\) \(=\) \(\displaystyle \size {x - y + y - t}\)
\(\displaystyle \) \(=\) \(\displaystyle \size {y - t - \paren {y - x} }\)
\(\displaystyle \) \(=\) \(\displaystyle \size {\size {y - t} - \size {y - x} }\) as $\paren {y - t}$ and $\paren {y - x}$ have the same sign because either $t \le x < y$ or $y < x \le t$
\(\displaystyle \) \(=\) \(\displaystyle \size {y - t} - \size {y - x}\) as $\size {y - t} \ge \size {y - x}$ because either $t \le x < y$ or $y < x \le t$

Also, we have:

\(\displaystyle r - \size {t - x_0}\) \(=\) \(\displaystyle \size {r - \size {t - x_0} }\) as $r - \size {t - x_0} > 0$
\(\displaystyle \) \(=\) \(\displaystyle \size {\size {y - x_0} - \size {t - x_0} }\) as $\size {y - x_0} = r$
\(\displaystyle \) \(=\) \(\displaystyle \size {y - x_0 - \paren {t - x_0} }\) as $\paren {y - x_0}$ and $\paren {t - x_0}$ have the same sign because either $x_0 \le t < y$ or $y < t \le x_0$
\(\displaystyle \) \(=\) \(\displaystyle \size {y - t}\)

We combine these two results to get:

\(\displaystyle \frac {\size {x - t} } {r - \size {t - x_0} }\) \(=\) \(\displaystyle \frac {\size {y - t} - \size {y - x} } {\size {y - t} }\)
\(\displaystyle \) \(=\) \(\displaystyle 1 - \frac {\size {y - x} } {\size {y - t} }\)
\(\displaystyle \) \(\le\) \(\displaystyle 1 - \frac {\size {y - x} } r\) as $r \ge \size {y - t}$
\(\displaystyle \) \(=\) \(\displaystyle \frac {r - \size {y - x} } r\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {\size {y - x_0} - \size {y - x} } r\) as $\size {y - x_0} = r$
\(\displaystyle \) \(=\) \(\displaystyle \frac {\size {\size {y - x_0} - \size {y - x} } } r\) as $\size {y - x_0} \ge \size {y - x}$
\(\displaystyle \) \(=\) \(\displaystyle \frac {\size {y - x_0 - \paren {y - x} } } r\) as $\paren {y - x_0}$ and $\paren {y - x}$ have the same sign because either $x_0 < x < y$ or $y < x < x_0$
\(\displaystyle \) \(=\) \(\displaystyle \frac {\size {x - x_0} } r\)

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

\(\displaystyle \size {\map {R_n} x}\) \(\le\) \(\displaystyle \frac {M r \paren {n + 1} } {\paren {r - \size {x - x_0} }^2} \int_{x_0}^x \paren {\frac {\size {x - t} } {\paren {r - \size {t - x_0} } } }^n \size {\d t}\)
\(\displaystyle \) \(\le\) \(\displaystyle \frac {M r \paren {n + 1} } {\paren {r - \size {x - x_0} }^2} \int_{x_0}^x \paren {\frac {size {x - x_0} } r}^n \size {\d t}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {M r \paren {n + 1} } {\paren {r - \size {x - x_0} }^2} \paren {\frac {\size {x - x_0} } r}^n \int_{x_0}^x \size {\d t}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {M r \paren {n + 1} } {\paren {r - \size {x - x_0} }^2} \paren {\frac {\size {x - x_0} } r}^n \size {x - x_0}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {M r \size {x - x_0} } {\paren {r - \size {x - x_0} }^2} \frac {\paren {n + 1} } {\paren {\frac r {\size {x - x_0} } }^n}\)

We have:

$\displaystyle \frac r {\size {x - x_0} } > 1$ as $\size {x - x_0} < r$ and $x \ne x_0$

Therefore:

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

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

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

So:

\(\displaystyle \lim_{n \mathop \to \infty} \size {\map {R_n} x}\) \(\le\) \(\displaystyle 0\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \lim_{n \mathop \to \infty} \size {\map {R_n} x}\) \(=\) \(\displaystyle 0\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \lim_{n \mathop \to \infty} \map {R_n} x\) \(=\) \(\displaystyle 0\)

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


Thus, $\displaystyle \lim_{n \mathop \to \infty} \map {R_n} x = 0$ 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 $\displaystyle \lim_{n \mathop \to \infty} \map {R_n} x = 0$ 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 $\size {x - x_0} < R$.

$\blacksquare$