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
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:
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 $\size {x - x_0} < R$
Corollary: Taylor Series of Analytic Function has infinite Radius of Convergence
Let $F$ be a complex function.
Let $F$ be analytic everywhere.
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 $0$.
Let $x_0$ be a point in $\R$.
Then:
- the Taylor series of $f$ about $x_0$ converges to $f$ at every point in $\R$.
Proof
Lemma
Let $y > 1$.
Then:
- $\ds \lim_{n \mathop \to \infty} \frac n {y^n} = 0$
$\Box$
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:
- $\map f x = \ds \sum_{i \mathop = 0}^n \frac {\paren {x - x_0}^i} {i!} \map {f^{\paren i} } {x_0} + \map {R_n} x$
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:
\(\ds \size {\map {R_n} x}\) | \(=\) | \(\ds \size {\frac 1 {n!} \int_{x_0}^x \paren {x - t}^n \map {f^{\paren {n \mathop + 1} } } t \rd t}\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \frac 1 {n!} \int_{x_0}^x \size {x - t}^n \size {\map {f^{\paren {n \mathop + 1} } } t} \size {\d t}\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \frac 1 {n!} \int_{x_0}^x \size {x - t}^n \size {\frac {M r \paren {n + 1}!} {\paren {r - \size {t - x_0} }^{\paren {n \mathop + 2} } } } \size {\d t}\) | where $M \in \R_{\ge 0}$ , by Bound for Analytic Function and Derivatives | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {n!} \int_{x_0}^x \size {x - t}^n \frac {M r \paren {n + 1}!} {\paren {r - \size {t - x_0} }^{\paren {n \mathop + 2} } } \size {\d t}\) | as $r - \size {t - x_0} > 0$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {n!} \int_{x_0}^x \size {x - t}^n \frac {M r \paren {n + 1}!} {\paren {r - \size {t - x_0} }^2} \frac 1 {\paren {r - \size {t - x_0} }^n} \size {\d t}\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \frac 1 {n!} \int_{x_0}^x \size {x - t}^n \frac {M r \paren {n + 1}!} {\paren {r - \size {x - x_0} }^2} \frac 1 {\paren {r - \size {t - x_0} }^n} \size {\d t}\) | as $0 < \size {r - \size {x - x_0} } \le \size {r - \size {t - x_0} }$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {n!} \frac {M r \paren {n + 1}!} {\paren {r - \size {x - x_0} }^2} \int_{x_0}^x \frac {\size {x - t}^n} {\paren {r - \size {t - x_0} }^n} \size {\d t}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {M r \paren {n + 1} } {\paren {r - \size {x - x_0} }^2} \int_{x_0}^x \paren {\frac {\size {x - t} } {\paren {r - \size {t - x_0} } } }^n \size {\d t}\) |
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 $\size {\map {R_n} x}$:
\(\ds \size {x - t}\) | \(=\) | \(\ds \size {x - y + y - t}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \size {y - t - \paren {y - x} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \size {\size {y - t} - \size {y - x} }\) | as $\paren {y - t}$ and $\paren {y - x}$ have the same sign because either $t \le x < y$ or $y < x \le t$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \size {y - t} - \size {y - x}\) | as $\size {y - t} \ge \size {y - x}$ because either $t \le x < y$ or $y < x \le t$ |
Also, we have:
\(\ds r - \size {t - x_0}\) | \(=\) | \(\ds \size {r - \size {t - x_0} }\) | as $r - \size {t - x_0} > 0$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \size {\size {y - x_0} - \size {t - x_0} }\) | as $\size {y - x_0} = r$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \size {y - x_0 - \paren {t - x_0} }\) | as $\paren {y - x_0}$ and $\paren {t - x_0}$ have the same sign because either $x_0 \le t < y$ or $y < t \le x_0$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \size {y - t}\) |
We combine these two results to get:
\(\ds \frac {\size {x - t} } {r - \size {t - x_0} }\) | \(=\) | \(\ds \frac {\size {y - t} - \size {y - x} } {\size {y - t} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1 - \frac {\size {y - x} } {\size {y - t} }\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds 1 - \frac {\size {y - x} } r\) | as $r \ge \size {y - t}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {r - \size {y - x} } r\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\size {y - x_0} - \size {y - x} } r\) | as $\size {y - x_0} = r$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\size {\size {y - x_0} - \size {y - x} } } r\) | as $\size {y - x_0} \ge \size {y - x}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\size {y - x_0 - \paren {y - x} } } r\) | as $\paren {y - x_0}$ and $\paren {y - x}$ have the same sign because either $x_0 < x < y$ or $y < x < x_0$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\size {x - x_0} } r\) |
We use this result in the expression for the bound for $\size {\map {R_n} x}$:
\(\ds \size {\map {R_n} x}\) | \(\le\) | \(\ds \frac {M r \paren {n + 1} } {\paren {r - \size {x - x_0} }^2} \int_{x_0}^x \paren {\frac {\size {x - t} } {\paren {r - \size {t - x_0} } } }^n \size {\d t}\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \frac {M r \paren {n + 1} } {\paren {r - \size {x - x_0} }^2} \int_{x_0}^x \paren {\frac {\size {x - x_0} } r}^n \size {\d t}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {M r \paren {n + 1} } {\paren {r - \size {x - x_0} }^2} \paren {\frac {\size {x - x_0} } r}^n \int_{x_0}^x \size {\d t}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {M r \paren {n + 1} } {\paren {r - \size {x - x_0} }^2} \paren {\frac {\size {x - x_0} } r}^n \size {x - x_0}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {M r \size {x - x_0} } {\paren {r - \size {x - x_0} }^2} \frac {\paren {n + 1} } {\paren {\frac r {\size {x - x_0} } }^n}\) |
We have:
- $\ds \frac r {\size {x - x_0} } > 1$ 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:
\(\ds \lim_{n \mathop \to \infty} \size {\map {R_n} x}\) | \(\le\) | \(\ds \lim_{n \mathop \to \infty} \frac {M r \size {x - x_0} } {\paren {r - \size {x - x_0} }^2} \frac {n + 1} {\paren {\frac r {\size {x - x_0} } }^n}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \frac {M r \size {x - x_0} } {\paren {r - \size {x - x_0} }^2} \frac {n + 1} n \frac n {\paren {\frac r {\size {x - x_0} } }^n}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {M r \size {x - x_0} } {\paren {r - \size {x - x_0} }^2} \lim_{n \mathop \to \infty} \frac {n + 1} n \frac n {\paren {\frac r {\size {x - x_0} } }^n}\) | Multiple Rule for Real Sequences | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {M r \size {x - x_0} } {\paren {r - \size {x - x_0} }^2} \lim_{n \mathop \to \infty} \frac {n + 1} n \lim_{n \mathop \to \infty} \frac n {\paren {\frac r {\size {x - x_0} } }^n}\) | Product Rule for Real Sequences | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {M r \size {x - x_0} } {\paren {r - \size {x - x_0} }^2} 1 \lim_{n \mathop \to \infty} \frac n {\paren {\frac r {\size {x - x_0} } }^n}\) | as $\ds \lim_{n \mathop \to \infty} \frac {n + 1} n = 1$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {M r \size {x - x_0} } {\paren {r - \size {x - x_0} }^2} 0\) | as $\ds \lim_{n \mathop \to \infty} \frac n {\paren {\frac r {\size {x - x_0} } }^n} = 0$ | |||||||||||
\(\ds \) | \(=\) | \(\ds 0\) |
So:
\(\ds \lim_{n \mathop \to \infty} \size {\map {R_n} x}\) | \(\le\) | \(\ds 0\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \lim_{n \mathop \to \infty} \size {\map {R_n} x}\) | \(=\) | \(\ds 0\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \lim_{n \mathop \to \infty} \map {R_n} x\) | \(=\) | \(\ds 0\) |
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 $\size {x - x_0} < R$.
$\blacksquare$