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 $e^{0.5}=1.6487212707\dots$ :

$n$	$T_n(0.5)$	error
$1$	$1.5$	$1.49\times 10^{-1}$
$2$	$1.625$	$2.37\times 10^{-2}$
$3$	$1.6458333\dots$	$2.89\times 10^{-3}$
$4$	$1.6484375$	$2.84\times 10^{-4}$
$5$	$1.6486979\dots$	$2.34\times 10^{-5}$

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

$f(x)$	Maclaurin polynomial (centre $0$ )	general term	notes
$e^{x}$	$1+x+\dfrac{x^{2}}{2!}+\cdots+\dfrac{x^{n}}{n!}$	$\dfrac{x^{k}}{k!}$ , $k\ge 0$	every derivative is $1$ at $0$
$\sin x$	$x-\dfrac{x^{3}}{3!}+\dfrac{x^{5}}{5!}-\cdots$	$\dfrac{(-1)^{m}x^{2m+1}}{(2m+1)!}$ , $m\ge 0$	odd powers only; $T_{2m+1}=T_{2m+2}$
$\cos x$	$1-\dfrac{x^{2}}{2!}+\dfrac{x^{4}}{4!}-\cdots$	$\dfrac{(-1)^{m}x^{2m}}{(2m)!}$ , $m\ge 0$	even powers only; $T_{2m}=T_{2m+1}$
$\ln(1+x)$	$x-\dfrac{x^{2}}{2}+\dfrac{x^{3}}{3}-\cdots+\dfrac{(-1)^{n-1}x^{n}}{n}$	$\dfrac{(-1)^{k-1}x^{k}}{k}$ , $k\ge 1$	needs $x>-1$ ; no constant term
$\dfrac{1}{1-x}$	$1+x+x^{2}+\cdots+x^{n}$	$x^{k}$ , $k\ge 0$	error is exactly $\dfrac{x^{n+1}}{1-x}$ for $x\ne 1$
$(1+x)^{k}$	$1+kx+\dfrac{k(k-1)}{2!}x^{2}+\dfrac{k(k-1)(k-2)}{3!}x^{3}+\cdots$	$\dfrac{k(k-1)\cdots(k-j+1)}{j!}x^{j}$ , $j\ge 1$ (here $k$ is the exponent, $j$ the term index)	needs $1+x>0$ if $k$ is not a whole number; terminates if $k$ is a non-negative integer

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

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 | Wrong | Right | |---|---| | $T_3(x)=f(a)+f'(a)(x-a)+f''(a)(x-a)^{2}+f'''(a)(x-a)^{3}$ | Divide by $k!$: the coefficients are $f''(a)/2$ and $f'''(a)/6$. | | Centring at $a=2$ but writing powers of $x$. | Every power must be $(x-2)^{k}$. Do **not** expand and re-collect in powers of $x$ unless asked. | | Leaving $x$ in the coefficients: "$T_2(x)=e^{x}+e^{x}x+\frac{e^{x}}{2}x^{2}$". | Derivatives are evaluated **at the centre**: $T_2(x)=1+x+\frac{x^{2}}{2}$ for $a=0$. Coefficients are numbers. | | Given $T_3(x)=2-5(x-1)+4(x-1)^{2}$, concluding $f''(1)=4$. | $f''(1)=2!\cdot 4=8$. Multiply the coefficient by $k!$. | | Taking $M=\lvert f^{(n+1)}(a)\rvert$ at the centre only. | $M$ must dominate on the *entire* interval. For $\ln(1+x)$, $n=3$, on $[-0.2,0.2]$: $M=6$ gives the bound $4.0\times10^{-4}$, but the true error at $x=-0.2$ is $4.77\times 10^{-4}$ — the "bound" is false. Correct $M=6/(0.8)^{4}=14.65$. | | "The error equals the next term $\dfrac{f^{(n+1)}(a)}{(n+1)!}(x-a)^{n+1}$." | The next term is a *heuristic*. Lagrange has $f^{(n+1)}(c)$ with $c$ unknown between $a$ and $x$; only that gives a proof. | | Expecting $T_4$ of $\sin x$ to have an $x^{4}$ term. | $\sin$ is odd: $T_3=T_4=x-\frac{x^{3}}{6}$. Exploit this to get a better remainder bound. | | Writing a Maclaurin polynomial for $\ln x$ or $\sqrt{x}$. | Neither is differentiable at $0$ ($\ln x$ is not even defined there). Use $\ln(1+x)$, $(1+x)^{1/2}$, or a different centre. | | Assuming more terms always means a better answer at a fixed $x$. | For $f(x)=\frac{1}{1-x}$ at $x=2$: $f(2)=-1$ but $T_n(2)=1+2+\cdots+2^{n}\to+\infty$. Accuracy is guaranteed only where the remainder tends to $0$. | | Using $c=x$ or $c=a$ "to be safe". | $c$ is strictly between them and unknown; bound $\lvert f^{(n+1)}\rvert$ over the closed interval instead. |

$n$	$(n+1)!$	$3/(n+1)!$
$6$	$5040$	$5.95\times10^{-4}$
$7$	$40\,320$	$7.44\times10^{-5}$
$8$	$362\,880$	$8.27\times10^{-6}$

Navigation

Higher-Degree Approximations and Taylor Polynomials

Why go beyond linear

Definition

The matching-derivatives characterisation

Standard Maclaurin polynomials

Worked Example 1 — approximate and bound

Key terms

Practice Problems