=== Lemma === {{:Convergence of Taylor Series of Function Analytic on Disk/Lemma}}{{qed|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: :$\map f x = \displaystyle \sum_{i \mathop = 0}^n \frac {\paren {x - x_0}^i} {i!} \map {f^{\paren i} } {x_0} + \map {R_n} x$ 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: {{begin-eqn}} {{eqn | l = \size {\map {R_n} x}       | r = \size {\frac 1 {n!} \int_{x_0}^x \paren {x - t}^n \map {f^{\paren {n \mathop + 1} } } t \rd t} }} {{eqn | o = \le       | r = \frac 1 {n!} \int_{x_0}^x \size {x - t}^n \size {\map {f^{\paren {n \mathop + 1} } } t} \size {\d t} }} {{eqn | o = \le       | r = \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}       | c = where $M \in \R_{\ge 0}$ , by Bound for Analytic Function and Derivatives }} {{eqn | r = \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}       | c =  as $r - \size {t - x_0} > 0$ }} {{eqn | r = \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} }} {{eqn | o = \le       | r = \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}       | c =  as $0 < \size {r - \size {x - x_0} } \le \size {r - \size {t - x_0} }$ }} {{eqn | r = \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} }} {{eqn | r = \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} }} {{end-eqn}} 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 $\size {\map {R_n} x}$: {{begin-eqn}} {{eqn | l = \size {x - t}       | r = \size {x - y + y - t} }} {{eqn | r = \size {y - t - \paren {y - x} } }} {{eqn | r = \size {\size {y - t} - \size {y - x} }       | c = as $\paren {y - t}$ and $\paren {y - x}$ have the same sign because either $t \le x < y$ or $y < x \le t$ }} {{eqn | r = \size {y - t} - \size {y - x}       | c = as $\size {y - t} \ge \size {y - x}$ because either $t \le x < y$ or $y < x \le t$ }} {{end-eqn}} Also, we have: {{begin-eqn}} {{eqn | l = r - \size {t - x_0}       | r = \size {r - \size {t - x_0} }       | c = as $r - \size {t - x_0} > 0$ }} {{eqn | r = \size {\size {y - x_0} - \size {t - x_0} }       | c = as $\size {y - x_0} = r$ }} {{eqn | r = \size {y - x_0 - \paren {t - x_0} }       | c = as $\paren {y - x_0}$ and $\paren {t - x_0}$ have the same sign because either $x_0 \le t < y$ or $y < t \le x_0$ }} {{eqn | r = \size {y - t} }} {{end-eqn}} We combine these two results to get: {{begin-eqn}} {{eqn | l = \frac {\size {x - t} } {r - \size {t - x_0} }       | r = \frac {\size {y - t} - \size {y - x} } {\size {y - t} } }} {{eqn | r = 1 - \frac {\size {y - x} } {\size {y - t} } }} {{eqn | o = \le       | r = 1 - \frac {\size {y - x} } r       | c = as $r \ge \size {y - t}$ }} {{eqn | r = \frac {r - \size {y - x} } r }} {{eqn | r = \frac {\size {y - x_0} - \size {y - x} } r       | c = as $\size {y - x_0} = r$ }} {{eqn | r = \frac {\size {\size {y - x_0} - \size {y - x} } } r       | c = as $\size {y - x_0} \ge \size {y - x}$ }} {{eqn | r = \frac {\size {y - x_0 - \paren {y - x} } } r       | c = as $\paren {y - x_0}$ and $\paren {y - x}$ have the same sign because either $x_0 < x < y$ or $y < x < x_0$ }} {{eqn | r = \frac {\size {x - x_0} } r }} {{end-eqn}} We use this result in the expression for the bound for $\size {\map {R_n} x}$: {{begin-eqn}} {{eqn | l = \size {\map {R_n} x}       | o = \le       | r = \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} }} {{eqn | o = \le       | r = \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} }} {{eqn | r = \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} }} {{eqn | r = \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} }} {{eqn | r = \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} }} {{end-eqn}} 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: {{begin-eqn}} {{eqn | l = \lim_{n \mathop \to \infty} \size {\map {R_n} x}       | o = \le       | r = \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} }} {{eqn | r = \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} }} {{eqn | r = \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}       | c = Multiple Rule for Real Sequences }} {{eqn | r = \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}       | c = Product Rule for Real Sequences }} {{eqn | r = \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}       | c = as $\displaystyle \lim_{n \mathop \to \infty} \frac {n + 1} n = 1$ }} {{eqn | r = \frac {M r \size {x - x_0} } {\paren {r - \size {x - x_0} }^2} 0       | c = as $\displaystyle \lim_{n \mathop \to \infty} \frac n {\paren {\frac r {\size {x - x_0} } }^n} = 0$ }} {{eqn | r = 0 }} {{end-eqn}} So: {{begin-eqn}} {{eqn | l = \lim_{n \mathop \to \infty} \size {\map {R_n} x}       | o = \le       | r = 0 }} {{eqn | ll= \leadsto       | l = \lim_{n \mathop \to \infty} \size {\map {R_n} x}       | r = 0 }} {{eqn | ll= \leadsto       | l = \lim_{n \mathop \to \infty} \map {R_n} x       | r = 0 }} {{end-eqn}} 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 $\size {x - x_0} < R$. {{qed}} Category:Taylor Series Category:Real Analysis ovjqg4qs3tpb8besvlr31u2a1cox9ua