The Taylor polynomials 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.

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

If $f$ is differentiable $k$ times at $c$, then the **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 $k$.

It is straightforward (by differentiating a polynomial ansatz?) to find an explicit formula (in the variable $x$):

$T^k_{f,c}(x) = \sum_{n = 0}^k \frac{f^{(n)}(c) (x-c)^n}{n!} .$

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.

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 $k = -1$):

If $f$ is continuous at $c$, then

$\lim_{x \to c} \frac{{f(x) - T^k_{f,c}(x)}}{{\|x - c\|}^k} = 0 .$

(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:

If $f^{(k)}$ is absolutely continuous on $[\min(x,c),\max(x,c)]$, then

$f(x) = T^k_{f,c}(x) + \int_{t=a}^x \frac{f^{(k+1)}(t) (x-t)^{k+1} \,\mathrm{d}t}{(k + 1)!} .$

(Note that $f^{(k+1)}$ is defined almost everywhere and 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:

If $f^{(k)}$ is absolutely continuous on $[\min(x,c),\max(x,c)]$, then

${|f(x) - T^k_{f,c}(x)}| \leq \frac{M {|x-c|}^{k+1}}{(k + 1)!} ,$

where $M$ is any essential upper bound of $f^{(k+1)}$ on $[\min(x,c),\max(x,c)]$.

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:

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

$f(x) = T^k_{f,c}(x) + \frac{f^{(k+1)}(t) (x-c)^{k+1}}{(k + 1)!} .$

All of these can be generalized in a fairly straightforward way to functions of several variables.

Last revised on November 21, 2023 at 04:56:47. See the history of this page for a list of all contributions to it.