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\begin{document}

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\section*{Taylor&#39;s theorem}

\hypertarget{taylors_theorem}{}\section*{{Taylor's Theorem}}\label{taylors_theorem}

\noindent\hyperlink{idea}{Idea}\dotfill \pageref*{idea} \linebreak
\noindent\hyperlink{preliminary_definitions}{Preliminary definitions}\dotfill \pageref*{preliminary_definitions} \linebreak
\noindent\hyperlink{statements}{Statements}\dotfill \pageref*{statements} \linebreak


\hypertarget{idea}{}\subsection*{{Idea}}\label{idea}

The [[Taylor polynomial]]s of a [[differentiable function]] approximate it by [[polynomial]] functions; the various versions of Taylor's Theorem describe how good this approximation is. The limiting case of this is the [[Taylor series]].

\hypertarget{preliminary_definitions}{}\subsection*{{Preliminary definitions}}\label{preliminary_definitions}

Let $f$ be a [[partial function]] on a [[cartesian space]] $\mathbb{R}^d$ (or $\mathbb{C}^d$), let $c$ be a [[concrete point|point]] in the cartesian space, and let $k$ be a [[natural number]] (actually we can allow $k \geq -1$ also, using [[negative thinking]]).

\begin{defn}
\label{}\hypertarget{}{}
If $f$ is [[differentiable function|differentiable]] $k$ times at $c$, then the \textbf{Taylor polynomial} of $f$ at $c$ with order $k$ is the unique [[polynomial]] in $d$ variables of degree at most $k$ whose derivatives at $c$ match those of $f$ up to order $k$.

\end{defn}
It is straightforward (by differentiating a polynomial [[ansatz]]) to find an explicit formula (in the variable $x$):

\begin{displaymath}
T^k_{f@c}(x) = \sum_{n = 0}^k \frac{f^{(n)}(c) (x-c)^n}{n!} .
\end{displaymath}
We have written this as if $f$ is a function of one variable; but interpret $n$ as a [[multi-index]] whose length $\ell$ satisfies $0 \leq \ell \leq k$, with $f^{(n)}$ the mixed [[partial derivative]] given by that multi-index, $n!$ the [[factorial]] of $\ell$, and $(x-c)^n$ a product of $\ell$ factors of the form $x_i - c_i$ (where $i$ is an index appearing in the multi-index $n$). Then this works in any cartesian space. (Note that we include all possible orderings of a multi-index as separate terms; this leads to repeated terms that, when combined, will cancel some of the factors in the factorial.)

In the case of a function of several variables, we can also manage the maximum degrees in the various variables separately, although nobody seems to bother with this.

\hypertarget{statements}{}\subsection*{{Statements}}\label{statements}

There are several different versions of Taylor's Theorem, all stating an extent to which a Taylor polynomial of $f$ at $c$, when evaluated at $x$, approximates $f(x)$.

Here is the simplest statement, which requires only continuity of $f$ (which we really only need for $k = 0$, since it's automatic for $k \geq 1$ and not actually necessary for $k = -1$):

\begin{theorem}
\label{limit}\hypertarget{limit}{}
If $f$ is [[continuous function|continuous]] at $c$, then

\begin{displaymath}
\lim_{x \to c} \frac{{f(x) - T^k_{f@c}(x)}}{{\|x - c\|}^k} = 0 .
\end{displaymath}
\end{theorem}
(The norm ${\|\cdot\|}$ can be left out in one variable, or placed in the numerator to handle all components of a vector-valued function at once.)

If $f^{(k)}$ has some continuity, then we get a version of Taylor's Theorem with an integral:

\begin{theorem}
\label{integral}\hypertarget{integral}{}
If $f^{(k)}$ is [[absolutely continuous function|absolutely continuous]] on $[\min(x,c),\max(x,c)]$, then

\begin{displaymath}
f(x) = T^k_{f@c}(x) + \int_{t=a}^x \frac{f^{(k+1)}(t) (x-t)^{k+1} \,\mathrm{d}t}{(k + 1)!} .
\end{displaymath}
\end{theorem}
(Note that $f^{(k+1)}$ is [[almost function|defined almost everywhere]] and [[integrable function|Lebesgue integrable]], because $f^{(k)}$ is absolutely continuous.)

This result is not directly very useful is one is using Taylor polynomials to approximate $f$ where one doesn't know its behaviour, but we have a corollary which can often be used:

\begin{theorem}
\label{estimate}\hypertarget{estimate}{}
If $f^{(k)}$ is [[absolutely continuous function|absolutely continuous]] on $[\min(x,c),\max(x,c)]$, then

\begin{displaymath}
{|f(x) - T^k_{f@c}(x)}| \leq \frac{M {|x-c|}^{k+1}}{(k + 1)!} ,
\end{displaymath}
where $M$ is any [[essentially bounded function|essential upper bound]] of $f^{(k+1)}$ on $[\min(x,c),\max(x,c)]$.

\end{theorem}
In many cases, finding a good upper bound of $f^{(k+1)}$ can be reduced to solving $f^{(k+2)}(t) = 0$.

There are also versions that generalize the [[mean value theorem]]:

\begin{theorem}
\label{MVT}\hypertarget{MVT}{}
If $f^{(k)}$ is differentiable on $[\min(x,c),\max(x,c)]$, then, for some $t \in [\min(x,c),\max(x,c)]$,

\begin{displaymath}
f(x) = T^k_{f@c}(x) + \frac{f^{(k+1)}(t) (x-c)^{k+1}}{(k + 1)!} .
\end{displaymath}
\end{theorem}
All of these can be generalized in a fairly straightforward way to functions of several variables.

[[!redirects Taylor's theorem]] [[!redirects Taylor's theorems]] [[!redirects Taylor's theorem]] [[!redirects Taylor's theorems]] [[!redirects Taylor's theorem]] [[!redirects Taylor's theorems]]

[[!redirects Taylor polynomial]] [[!redirects Taylor polynomials]]



\end{document}