Taylor's Theorem/One Variable

Theorem

Let $f$ be a real function which is continuous on the closed interval $\left[{a \,.\,.\, b}\right]$ and $n + 1$ times differentiable on the open interval $\left({a \,.\,.\, b}\right)$.

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

Then, given any $x \in \left({a \,.\,.\, b}\right)$, there exists some $\eta \in \R: x \le \eta \le \xi$ or $\xi \le \eta \le x$ such that:

 $\displaystyle f \left({x}\right)$ $=$ $\displaystyle \frac 1 {0!} f \left({\xi}\right)$ $\quad$ $\quad$ $\displaystyle$  $\, \displaystyle + \,$ $\displaystyle \frac 1 {1!} \left({x - \xi}\right) f^{\prime} \left({\xi}\right)$ $\quad$ $\quad$ $\displaystyle$  $\, \displaystyle + \,$ $\displaystyle \frac 1 {2!} \left({x - \xi}\right)^2 f^{\prime \prime} \left({\xi}\right)$ $\quad$ $\quad$ $\displaystyle$  $\, \displaystyle + \,$ $\displaystyle \cdots$ $\quad$ $\quad$ $\displaystyle$  $\, \displaystyle + \,$ $\displaystyle \frac 1 {n!} \left({x - \xi}\right)^n f^{\left({n}\right)} \left({\xi}\right)$ $\quad$ $\quad$ $\displaystyle$  $\, \displaystyle + \,$ $\displaystyle R_n$ $\quad$ $\quad$

where $R_n$ (sometimes denoted $E_n$) is known as the error term, and satisfies:

$R_n = \dfrac 1 {\left({n + 1}\right)!} \left({x - \xi}\right)^{n + 1} f^{\left({n + 1}\right)} \left({\eta}\right)$

Note that when $n = 0$ Taylor's Theorem reduces to the Mean Value Theorem.

The expression:

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

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

Proof

Integral Version

This proof requires $f^{\left({n}\right)}$ to be absolutely continuous on $\left[{a \,.\,.\,x}\right]$, so that the Fundamental Theorem of Calculus holds.

Except at the end when the Mean Value Theorem is invoked, differentiability of $f^{\left({n}\right)}$ need not be assumed, since absolute continuity implies:

differentiability almost everywhere
the validity of the Fundamental Theorem of Calculus

provided the integrals involved are understood as Lebesgue integrals.

Consequently, the integral form of the remainder holds with this particular weakening of the assumptions on $f$.

We first prove Taylor's Theorem with the integral remainder term.

The Fundamental Theorem of Calculus states that:

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

which can be rearranged to:

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

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

 $\displaystyle f \left({x}\right)$ $=$ $\displaystyle f \left({a}\right) + x f' \left({x}\right) - a f' \left({a}\right) - \int_a^x t f'' \left({t}\right) \rd t$ $\quad$ $u = f' \left({t}\right)$ and $\d v = \d t$ $\quad$ $\displaystyle$ $=$ $\displaystyle f \left({a}\right) + \int_a^x x f'' \left({t}\right) \rd t + x f' \left({a}\right) - a f' \left({a}\right) - \int_a^x \, t f'' \left({t}\right) \rd t$ $\quad$ $\displaystyle \int_a^x x f'' \left({t}\right) \rd t = x f' \left({x}\right)-x f' \left({a}\right)$ $\quad$ $\displaystyle$ $=$ $\displaystyle f \left({a}\right) + \left({x - a}\right) f' \left({a}\right) + \int_a^x \left({x - t}\right) f'' \left({t}\right) \rd t$ $\quad$ factoring out some common terms $\quad$

Another application yields:

$\displaystyle f \left({x}\right) = f \left({a}\right) + \left({x - a}\right) f' \left({a}\right) + \frac 1 2 \left({x - a}\right)^2 f'' \left({a}\right) + \frac 1 2 \int_a^x \left({x - t}\right)^2 f''' \left({t}\right) \rd t$

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:

 $\displaystyle f \left({x}\right)$ $=$ $\displaystyle f \left({a}\right)$ $\quad$ $\quad$ $\displaystyle$ $+$ $\displaystyle \frac{f' \left({a}\right)} {1!} \left({x - a}\right)$ $\quad$ $\quad$ $\displaystyle$ $+$ $\displaystyle \cdots$ $\quad$ $\quad$ $\displaystyle$ $+$ $\displaystyle \frac{f^{\left({n}\right)} \left({a}\right)} {n!} \left({x - a}\right)^n$ $\quad$ $\quad$ $\displaystyle$ $+$ $\displaystyle \int_a^x \frac{f^{\left({n + 1}\right)} \left({t}\right)} {n!} \left({x - t}\right)^n \rd t$ $\quad$ $*$ $\quad$

We can rewrite the integral using Integration by Parts.

A primitive of $\left({x - t}\right)^n$ as a function $t$ is given by $\dfrac {-\left({x - t}\right)^{n + 1} } {n + 1}$.

So:

 $\displaystyle$  $\displaystyle \int_a^x \frac{f^{\left({n + 1}\right)} \left({t}\right)} {n!} \left({x - t}\right)^n \rd t$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle - \left[{\frac {f^{\left({n + 1}\right)} \left({t}\right)} {\left({n + 1}\right) n!} \left({x - t}\right)^{n + 1} }\right]_a^x + \int_a^x \frac {f^{\left({n + 2}\right)} \left({t}\right)} {\left({n + 1}\right) n!} \left({x - t}\right)^{n + 1} \rd t$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \frac {f^{\left({n + 1}\right)} \left({a}\right)} {\left({n + 1}\right)!} \left({x - a}\right)^{n + 1} + \int_a^x \frac {f^{\left({n + 2}\right)} \left({t}\right)} {\left({n + 1}\right)!} \left({x - t}\right)^{n + 1} \rd t$ $\quad$ $\quad$

The last integral can be solved immediately, which leads to

$R_n = \dfrac {f^{\left({n + 1}\right)} \left({\xi}\right)} {\left({n + 1}\right)!} \left({x - a}\right)^{n + 1}$

$\blacksquare$

Proof using Cauchy 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 $\left[{a \,.\,.\, x}\right]$ and differentiable with non-vanishing derivative on $\left({a \,.\,.\, x}\right)$.

Let:

$F \left({t}\right) = f \left({t}\right) + \dfrac {f' \left({t}\right)} {1!} \left({x - t}\right) + \dotsb + \dfrac {f^{\left({n}\right)} \left({t}\right)} {n!} \left({x - t}\right)^n$

By the Cauchy Mean Value Theorem:

$(1): \quad \dfrac {F' \left({\xi}\right)} {G' \left({\xi}\right)} = \dfrac {F \left({x}\right) - F \left({a}\right)} {G \left({x}\right) - G \left({a}\right)}$

for some $\xi \in \left({a \,.\,.\, x}\right)$.

Note that the numerator:

$F \left({x}\right) - F \left({a}\right) = R_n$

is the remainder of the Taylor polynomial for $f \left({x}\right)$.

On the other hand, computing $F' \left({\xi}\right)$:

$F' \left({\xi}\right) = f' \left({\xi}\right) - f' \left({\xi}\right) + \dfrac {f'' \left({\xi}\right)} {1!} \left({x - \xi}\right) - \dfrac {f'' \left({\xi}\right)} {1!} \left({x - \xi}\right) + \dotsb + \dfrac {f^{\left({n + 1}\right)} \left({t}\right)} {n!} \left({x - \xi}\right)^n = \dfrac {f^{\left({n + 1}\right)} \left({\xi}\right)} {n!} \left({x - \xi}\right)^n$

Putting these two facts together and rearranging the terms of $(1)$ yields:

$R_n = \dfrac {f^{\left({n + 1}\right)} \left({\xi}\right)} {n!} \left({x - \xi}\right)^n \dfrac {G \left({x}\right) - G \left({a}\right)} {G' \left({\xi}\right)}$

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 $G \left({t}\right) = t - a$.

$\blacksquare$

Proof using Rolle's Theorem directly

Yet another proof for Lagrange Form of the Remainder can be constructed applying Rolle's theorem directly $n$ times; this proof might be easier to visualize geometrically.

Let the function $g$ be defined as:

$g \left({t}\right) = R_n \left({t}\right) - \dfrac {\left({t - a}\right)^{n + 1} } {\left({x - a}\right)^{n + 1} } R_n \left({x}\right)$

Then:

$g^{\left({k}\right)} \left({a}\right) = 0$

for $k = 0, \dotsc, n$, and $g \left({x}\right) = 0$.

Apply Rolle's Theorem successively to $g, g', \dotsc, g^{\left({n}\right)}$.

Then there exist:

$\xi_1, \ldots, \xi_{n + 1}$

between $a$ and $x$ such that:

$g' \left({\xi_1}\right) = 0, g'' \left({\xi_2}\right) = 0, \ldots, g^{\left({n + 1}\right)} \left({\xi_{n + 1} }\right) = 0$

Let $\xi = \xi_{n + 1}$.

Then:

$0 = g^{\left({n + 1}\right)} \left({\xi}\right) = f^{\left({n + 1}\right)} \left({\xi}\right) - \dfrac {\left({n + 1}\right)!} {\left({x - a}\right)^{n + 1} } R_n \left({x}\right)$

and the formula for $R_n \left({x}\right)$ follows.

$\blacksquare$

Source of Name

This entry was named for Brook Taylor.