Taylor's Theorem

Theorem
Taylor's Theorem states that any infinitely differentiable function (including one where the derivative is $0$) can be approximated by a series of polynomials.

One Variable
Let $f$ be a real function which is continuous on the closed interval $\left[{a..b}\right]$ and $n$ times differentiable on the open interval $\left({a..b}\right)$.

Let $\xi \in \left({a..b}\right)$.

Then, given any $x \in \left({a..b}\right)$:

where $R_n$ (sometimes denoted $E_n$) is known as the error term, and satisfies:
 * $\displaystyle R_n = \frac 1 {\left({n+1}\right)!} \left({x - \xi}\right)^{n+1} f^{\left({n+1}\right)} \left({\eta}\right)$

where $\eta \in \R$ is some (at this point unspecified) real number such that $x \le \eta \le \xi$.

Note that when $n = 1$ Taylor's Theorem reduces to the Mean Value Theorem.

The expression:
 * $\displaystyle f \left({x}\right) = \sum_{n=0}^\infty \frac {\left({x - \xi}\right)^n} {n!} f^{\left({n}\right)} \left({x}\right)$

where $n$ is taken to the limit, is known as the Taylor series expansion of $f$ about $\xi$.

Integral version
We first prove Taylor's theorem with the integral remainder term.

The Fundamental Theorem of Calculus states that


 * $\displaystyle \int_a^x f' \left({t}\right) \ \mathrm d t = f \left({x}\right) - f \left({a}\right)$

which can be rearranged to:


 * $\displaystyle f \left({x}\right) = f \left({a}\right) + \int_a^x f'(t) \ \mathrm d t$

Now we can see that an application of Integration by Parts yields:

Another application yields:
 * $\displaystyle f \left({x}\right) = f \left({a}\right)+(x-a) f' \left({a}\right)+ \frac 1 2 (x-a)^2f \left({a}\right) + \frac 1 2 \int_a^x (x-t)^2 f' \left({t}\right) \ \mathrm d t$

By repeating this process, we may derive Taylor's theorem for higher values of $n$.

This can be formalized by applying the technique of Principle of Mathematical Induction. So, suppose that Taylor's theorem holds for a $n$, that is, suppose that:

We can rewrite the integral using integration by parts. An antiderivative of $(x-t)^n$ as a function $t$ is given by $\dfrac{-(x-t)^{n+1}}{n+1}$, so:

The last integral can be solved immediately, which leads to


 * $\displaystyle R_n = \frac{f^{(n+1)}(\xi)}{(n+1)!} (x-a)^{n+1}$

Mean value theorem
An alternative proof, which holds under milder technical assumptions on the function $f$, can be supplied using the Cauchy Mean Value Theorem.

Let $G$ be a real-valued function continuous on $[a,x]$ and differentiable with non-vanishing derivative on $(a,x)$.

Let:
 * $\displaystyle F(t) = f(t) + \frac{f'(t)}{1!}(x-t) + \cdots + \frac{f^{(n)}(t)}{n!}(x-t)^n$

By the Cauchy Mean Value Theorem:
 * $\displaystyle \frac{F'(\xi)}{G'(\xi)} = \frac{F(x) - F(a)}{G(x) - G(a)} \qquad (1)$

for some $\xi\in(a,x)$.

Note that the numerator $F(x)-F(a)=R_n$ is the remainder of the Taylor polynomial for $f(x)$.

On the other hand, computing $F^{\prime} (t)$:
 * $\displaystyle F'(t) = f'(t) - f'(t) + \frac{f(t)}{1!}(x-t) - \frac{f(t)}{1!}(x-t) + \cdots + \frac{f^{(n+1)}(t)}{n!}(x-t)^n = \frac{f^{(n+1)}(t)}{n!}(x-t)^n$

Putting these two facts together and rearranging the terms of $(1)$ yields:
 * $\displaystyle R_n = \frac{f^{(n+1)}(\xi)}{n!}(x-\xi)^n\cdot\frac{G(x)-G(a)}{G'(\xi)}$

which was to be shown.

Note that the Lagrange Form of the Remainder comes from taking $G \left({t}\right) = \left({x - t}\right)^{n+1}$ and the given Cauchy Form of the Remainder comes from taking $G \left({t}\right) = t - a$.