Higher-Degree Approximations and Taylor Polynomials

The linearisation $L(x)=f(a)+f'(a)(x-a)$ is the best straight line at $a$: it gets the value and the slope right and everything else wrong. A Taylor polynomial simply keeps going — match the second derivative too, then the third, then the $n$-th — and the approximation improves dramatically near $a$. This topic supplies the formula, the standard Maclaurin polynomials worth knowing cold, and the piece that turns an approximation into mathematics rather than optimism: the Lagrange remainder, which converts "close" into a guaranteed numerical error bound.

Why go beyond linear

Fix $a$ and write $h=x-a$. Linearisation has error of size roughly $Ch^{2}$; the degree-$n$ Taylor polynomial has error of size roughly $Ch^{n+1}$. So halving the distance to the centre divides the error of $T_n$ by about $2^{\,n+1}$, not by $2$. Concretely, for $f(x)=e^{x}$ at $a=0$, approximating $e^{0.5}=1.6487212707\dots$:

Each extra degree buys close to another decimal digit of accuracy here. That is the whole motivation.

Definition

Definition. Suppose $f$ is $n$ times differentiable at $a$. The $n$-th degree Taylor polynomial of $f$ centred at $a$ is

$T_n(x)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^{2}+\cdots+\frac{f^{(n)}(a)}{n!}(x-a)^{n},$

i.e. $T_n(x)=c_0+c_1(x-a)+\cdots+c_n(x-a)^n$ with $c_k=\dfrac{f^{(k)}(a)}{k!}$. When $a=0$ it is called the Maclaurin polynomial.

Three immediate remarks. (i) $T_0$ is the constant $f(a)$ and $T_1$ is exactly the linearisation. (ii) $T_n$ has degree at most $n$ — it can be lower, since $f^{(n)}(a)$ may be $0$. (iii) The variable in the powers is $x-a$, never $x$, unless $a=0$.

The matching-derivatives characterisation

Theorem. If $f$ is $n$ times differentiable at $a$, then $T_n$ is the unique polynomial $P$ of degree at most $n$ satisfying

$P^{(k)}(a)=f^{(k)}(a)\quad\text{for } k=0,1,2,\dots,n.$

Why. Write $P(x)=c_0+c_1(x-a)+\cdots+c_n(x-a)^n$. Differentiating $k$ times kills every term of lower degree and, at $x=a$, kills every term of higher degree, leaving $P^{(k)}(a)=k!\,c_k$. Setting $k!\,c_k=f^{(k)}(a)$ forces $c_k=f^{(k)}(a)/k!$, and the coefficients are therefore determined.

Run this backwards to read derivatives off coefficients: if you know $T_n$, then $f^{(k)}(a)=k!\,c_k$.

Theorem (order of contact). With the same hypothesis,

$\lim_{x\to a}\frac{f(x)-T_n(x)}{(x-a)^{n}}=0,$

and $T_n$ is the only polynomial of degree at most $n$ with this property. This is what licenses the shortcuts below: if any legitimate algebra produces a polynomial $P$ of degree $\le n$ whose difference from $f$ vanishes faster than $(x-a)^n$, then $P$ is $T_n$ — no need to differentiate $n$ times.

Standard Maclaurin polynomials

Note $\ln x$ has no Maclaurin polynomial: it is not defined at $0$. Use $\ln(1+x)$, or centre at $a=1$.

Building new ones from old. Substituting $u=g(x)$, where $g$ is a polynomial with $g(0)=0$ (if $g$ is not already one, replace it by its own Maclaurin polynomial first), into a known Maclaurin polynomial, or multiplying two known ones together — and in both cases discarding every power above $x^{n}$ — produces $T_n$ of the composite/product, by the uniqueness theorem above. Keep enough terms of the outer polynomial that nothing of degree $\le n$ can be lost. Example: from $e^{u}\approx 1+u+\frac{u^{2}}{2}+\frac{u^{3}}{6}$ with $u=-x^{2}$,

$e^{-x^{2}}:\quad T_6(x)=1-x^{2}+\frac{x^{4}}{2}-\frac{x^{6}}{6}.$

Reading off the coefficient of $x^{4}$: $\dfrac{f^{(4)}(0)}{4!}=\dfrac12$, so $f^{(4)}(0)=12$ — obtained without ever differentiating four times.

Worked Example 1 — approximate and bound

Estimate $e^{0.1}$ with $T_3$ and give a rigorous error bound.

For $f(x)=e^{x}$, every derivative is $e^{x}$, so $f^{(k)}(0)=1$ and $T_3(x)=1+x+\frac{x^{2}}{2}+\frac{x^{3}}{6}$. Then

$$\begin{aligned} T_3(0.1)&=1+0.1+\frac{0.01}{2}+\frac{0.001}{6}\\ &=1+0.1+0.005+0.00016667\\ &=1.10516667. \end{aligned}$$

Bound: $f^{(4)}(t)=e^{t}$, which is increasing, so on $[0,0.1]$ we have $|f^{(4)}(t)|\le e^{0.1}<1.2$. Take $M=1.2$:

$|R_3(0.1)|\le \frac{M\,|0.1-0|^{4}}{4!}=\frac{1.2\times 10^{-4}}{24}=5\times 10^{-6}.$

So $e^{0.1}=1.105167\pm 0.000005$, i.e. $1.105161<e^{0.1}<1.105172$. (True value $1.1051709\dots$; true error $4.25\times10^{-6}$ — the bound holds and is realistic.)

Taylor's Theorem — the Lagrange remainder

Define the remainder $R_n(x)=f(x)-T_n(x)$.

Theorem (Taylor, Lagrange form). Let $n\ge 0$. Suppose $f^{(n)}$ is continuous on the closed interval with endpoints $a$ and $x$, and $f^{(n+1)}$ exists at every point of the corresponding open interval. Then there exists a number $c$ strictly between $a$ and $x$ with

$R_n(x)=\frac{f^{(n+1)}(c)}{(n+1)!}\,(x-a)^{n+1}.$

The case $n=0$ is exactly the Mean Value Theorem: $f(x)=f(a)+f'(c)(x-a)$. The number $c$ depends on both $x$ and $n$ and is essentially never computable — which is fine, because we only ever bound $f^{(n+1)}(c)$.

Corollary (Taylor's Inequality). If in addition $|f^{(n+1)}(t)|\le M$ for all $t$ in an interval $I$ containing $a$, then for every $x$ in $I$,

$|R_n(x)|\le \frac{M}{(n+1)!}\,|x-a|^{n+1}.$

Procedure: (1) fix $n$ and the interval $I$; (2) compute $f^{(n+1)}$; (3) find $M=\max_{t\in I}|f^{(n+1)}(t)|$ — over the whole interval, not just at $a$; (4) insert the largest $|x-a|$ occurring in $I$ for a bound valid throughout $I$.

Sign information is free too. Since $c$ lies between $a$ and $x$, the sign of $R_n$ is often decidable. E.g. for $x>0$ there is a $c$ with $0<c<x$ and $e^{x}=1+x+\frac{x^{2}}{2}+\frac{e^{c}x^{3}}{6}$; here $e^{c}x^{3}/6>0$, so $e^{x}>1+x+\frac{x^{2}}{2}$. (At $x=0$ no such $c$ exists — Taylor's Theorem needs $x\ne a$ — but both sides equal $1$ there, so $e^{x}\ge 1+x+\frac{x^{2}}{2}$ for all $x\ge0$.)

Worked Example 2 — bounding the error over an interval

How accurate is $\cos x\approx 1-\frac{x^{2}}{2}+\frac{x^{4}}{24}$ on $[-0.5,0.5]$?

Naive route, $n=4$: $f^{(5)}(t)=-\sin t$, so $M=\max_{|t|\le 0.5}|\sin t|=\sin 0.5<0.48$, and

$|R_4(x)|\le \frac{0.48\,(0.5)^{5}}{5!}=\frac{0.48\times 0.03125}{120}=1.25\times 10^{-4}.$

Better route: for $\cos$, $T_4=T_5$ (the $x^{5}$ coefficient is $0$), so the same polynomial is also the degree-$5$ Taylor polynomial and we may use $n=5$. Now $f^{(6)}(t)=-\cos t$ and $M=1$:

$|R_5(x)|\le \frac{1\cdot (0.5)^{6}}{6!}=\frac{0.015625}{720}=2.17\times 10^{-5}.$

Nearly six times sharper, for free. (Check at the worst point: $T_4(0.5)=0.87760417$, $\cos 0.5=0.8775825619\dots$, true error $2.160\times10^{-5}$ — just under the bound.)

Common mistakes