Taylor Expansion
Thereom
- $I := [a, b] \subseteq \mathbb{R}$,
- $f: I \rightarrow \mathbb{R}$,
- $k \in \mathbb{N}$,
For all $x \in (a, b)$,
\[\begin{eqnarray} f(x) & = & f(a) + f^{\prime}(a) (x - a) + \frac{1}{2!} f^{\prime\prime}(a) (x - a)^{2} + \cdots + \frac{1}{k!} f^{(k)}(a) (x - a)^{k} + h_{k}(x) (x - a)^{k} \nonumber \\ & = & f(a) + \sum_{i=1}^{k} \frac{1}{k!} f^{(k)}(a) (x - a)^{k} + R_{k}(x) \nonumber . \end{eqnarray}\]$R_{k}(x)$ has many forms
(1) If $f$ is $C^{k+1}((a, b))$ and $f^{(k)}$ is continuous on $I$, then
\[\begin{eqnarray} R_{k}(x) & = & \frac{ f^{(k + 1)}(z) }{ (k + 1)! } (x - a)^{k + 1} \nonumber \end{eqnarray}\]proof
(1)
Let $x \in \mathbb{R}$ be fixed. Let $G: I \rightarrow \mathbb{R}$ be continuous on $I$ and diffierentiable between $(a, x)$.
\[F(t) := f(t) + f^{\prime}(t)(x - t) + \frac{1}{2!} f^{\prime\prime}(t)(x - t)^{2} + \cdots + \frac{1}{k!} f^{(k)}(t)(x - t)^{k} .\] \[\begin{eqnarray} F^{\prime}(t) & = & f^{\prime}(t) + \left( f^{\prime\prime}(t)(x - t) - f^{\prime}(t) \right) + \left( \frac{ f^{(3)}(t) }{ 2! } (x - t)^{2} - \frac{ f^{(2)}(t) }{ 1! } (x - t) \right) + \cdots \left( \frac{ f^{(k+1)}(t) }{ k! } (x - t)^{k} - \frac{ f^{(k)}(t) }{ (k - 1)! } (x - t)^{k - 1} \right) \nonumber \\ & = & \frac{ f^{(k+1)}(t) }{ k! } (x - t)^{k} \nonumber \end{eqnarray}\]By mean-value theorem, for some $z \in (a, x)$,
\[\begin{eqnarray} & & \frac{ F^{\prime}(z) }{ G^{\prime}(z) } = \frac{ F(x) - F(a) }{ G(a) - G(a) } \nonumber \\ & \Leftrightarrow & F(x) - F(a) = F^{\prime}(z) \frac{ G(x) - G(a) }{ G^{\prime}(z) } . \nonumber \end{eqnarray}\]Note that $R_{k}(x) = F(a) - F(x)$. Suppose that $G(t) := (t - x)^{k+1}$.
\[\begin{eqnarray} F(x) - F(a) & = & \frac{ f^{(k+1)}(z) }{ k! } (x - z)^{k} \frac{ G(x) - G(a) }{ G^{\prime}(z) } \nonumber \\ & = & \frac{ f^{(k+1)}(z) }{ k! } (x - z)^{k} \frac{ - (a - x)^{k+1} }{ (k + 1) (z - x)^{k} } \nonumber \\ & = & \frac{ f^{(k+1)}(z) }{ (k + 1)! } (a - x)^{k+1} \end{eqnarray}\]$\Box$
Thereom multiple variables
- $I \subseteq \mathbb{R}^{n}$,
- open
- convex
- $f: I \rightarrow \mathbb{R}$,
- $k \in \mathbb{N}$,
For all $x, a \in I$,
\[\begin{eqnarray} f(x) & = & f(a) + \sum_{\abs{\alpha} \le k} \left( D^{\alpha} f(a) \right) \prod_{j=1}^{n} (x - a)^{\alpha_{j}} + \sum_{\abs{\alpha} = k} h_{\alpha}(x) \prod_{j=1}^{n} (x - a)^{\alpha_{j}} \nonumber . \end{eqnarray}\]$R_{\alpha}(x)$ has many forms
(1) If $f$ is $C^{k+1}(I)$ and $D^{\alpha}f$ is continuous on $I$, then
\[\begin{eqnarray} R_{\alpha}(x) & = & \sum_{\abs{\alpha} = k + 1} \frac{ D^{\alpha}f(x + c(x - a)) }{ (k + 1)! } \prod_{j=1}^{n} (x - a)^{\alpha_{j}} \nonumber \end{eqnarray}\]where $c \in [0, 1]$.
proof
(1)
Let $x, a \in \mathbb{R}$ be fixed.
\[g(t) := f(a + t(x - a)) .\] \[\begin{eqnarray} g(1) & = & g(0) + \sum_{i=1}^{k} \frac{1}{k!} g^{(k)}(0) + \frac{ g^{(k + 1)}(c) }{ (k + 1)! } \end{eqnarray}\]Let $G: I \rightarrow \mathbb{R}$ be continuous on $I$ and diffierentiable between $I$.
\[F(t) := f( \sum_{\abs{\alpha} = k + 1} \frac{ D^{\alpha}f(x + c(x - a)) }{ (k + 1)! } \prod_{j=1}^{n} (x - a)^{\alpha_{j}} .\]$\Box$