Applications of Taylor and Maclaurin Series, calculus lesson

Nobody can write the integral

\int_0^{0.5} e^{-t^2} \, dt

in terms of familiar functions, yet a handful of terms of a series give it to five decimal places. The same series explain why $\sin \theta \approx \theta$ is good enough for a swinging pendulum, and they turn many hard limits into one line of algebra. In each case the method replaces a function by a polynomial and keeps track of what was thrown away.

Key terms

Polynomial approximation

Definition If $f$ is of class $C^{n+1}$ on an open interval containing $a$ , then near $a$ we approximate $f(x) \approx T_n(x)$ , the Taylor polynomial of degree $n$ . By Taylor’s theorem there are $C > 0$ and $\delta > 0$ with

\lvert f(x) - T_n(x) \rvert \le C \lvert x - a \rvert^{n+1} \qquad \text{for } \lvert x - a \rvert < \delta.

In plain words Close to the center, a Taylor polynomial is an excellent stand-in for the function, and its error shrinks like a power of the distance.

Example Use $\cos x \approx 1 - \frac{x^2}{2} + \frac{x^4}{24}$ at $x = 0.2$ . It gives

1 - 0.02 + 0.0000667 = 0.9800667,

while $\cos 0.2 = 0.9800666\ldots$ .

Scale of neglected terms

Definition Write $g(x) = O\bigl(h(x)\bigr)$ as $x \to a$ if there are constants $C > 0$ and $\delta > 0$ with $\lvert g(x) \rvert \le C \lvert h(x) \rvert$ for $0 < \lvert x - a \rvert < \delta$ . With this notation the polynomial approximation above reads

f(x) = T_n(x) + O\bigl((x - a)^{n+1}\bigr) \quad \text{as } x \to a.

For a convergent power series the tail after the term in $(x - a)^n$ is $O\bigl((x - a)^{n+1}\bigr)$ as $x \to a$ .

In plain words The part you drop has roughly the size of the first term you dropped. Halving $x$ cuts an $O(x^3)$ error by about a factor of $8$ .

Example As $x \to 0$ ,

\sin x = x - \frac{x^3}{6} + O(x^5),

so the error of $\sin x \approx x$ is close to $-\frac{x^3}{6}$ .

$x$	$\sin x - x$	$-\frac{x^3}{6}$
$0.4$	$-0.010582$	$-0.010667$
$0.2$	$-0.001331$	$-0.001333$
$0.1$	$-0.000167$	$-0.000167$

Limits of indeterminate forms

Definition Suppose that as $x \to a$ ,

f(x) = c\,(x - a)^m + O\bigl((x - a)^{m+1}\bigr), \qquad g(x) = d\,(x - a)^m + O\bigl((x - a)^{m+1}\bigr),

with $d \neq 0$ . Then

\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{c}{d}.

To use this, expand numerator and denominator until the first nonzero terms appear, divide by the common power, and let $x \to a$ .

In plain words Expand both top and bottom, cancel what cancels, and the leading terms decide the limit.

Example Since $\sin x - x = -\frac{x^3}{6} + O(x^5)$ ,

\lim_{x \to 0} \frac{\sin x - x}{x^3} = \lim_{x \to 0} \left(-\frac{1}{6} + O(x^2)\right) = -\frac{1}{6}.

Applying l’Hôpital’s rule would take three rounds of differentiation.

Functions defined by integrals

Definition If $f(t) = \sum c_n (t - a)^n$ for $\lvert t - a \rvert < R$ , then for every $x$ with $\lvert x - a \rvert < R$

\int_a^x f(t) \, dt = \sum_{n=0}^{\infty} c_n \frac{(x - a)^{n+1}}{n + 1},

by term-by-term integration. This gives power series for functions defined as integrals whose integrands have no elementary antiderivative.

In plain words Expand the integrand, integrate each power, and you have a series for the integral.

Example From $e^{-t^2} = \sum \frac{(-1)^n t^{2n}}{n!}$ , valid for all $t$ ,

\int_0^x e^{-t^2} \, dt = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n + 1) \, n!} = x - \frac{x^3}{3} + \frac{x^5}{10} - \frac{x^7}{42} + \cdots

for every real $x$ . At $x = 0.5$ these four terms give

0.5 - 0.0416667 + 0.003125 - 0.0001860 = 0.4612723.

Error estimates

Definition Two standard bounds control the error of a truncated series. If the series is alternating, with terms whose absolute values decrease to $0$ , the error is at most the absolute value of the first omitted term and has its sign. In general, when $f$ is $n + 1$ times differentiable and $\lvert f^{(n+1)} \rvert \le M$ on the closed interval between $a$ and $x$ , Taylor’s inequality bounds the error of $T_n(x)$ by

\lvert f(x) - T_n(x) \rvert \le \frac{M}{(n + 1)!} \lvert x - a \rvert^{n+1}.

In plain words Always pair an approximation with a guaranteed bound on how wrong it can be.

Example The series for the integral of $e^{-t^2}$ from $0$ to $0.5$ is alternating, with terms that decrease to $0$ . The first omitted term is

\frac{0.5^9}{9 \cdot 4!} \approx 9.04 \times 10^{-6},

0.4612723 < \int_0^{0.5} e^{-t^2} \, dt < 0.4612723 + 0.0000091.

The true value is $0.4612810\ldots$ , an error of $8.7 \times 10^{-6}$ .

Asymptotic expansions

Definition A sequence of coefficients $c_0, c_1, c_2, \ldots$ gives an asymptotic expansion of $f$ as $x \to a$ , written $f(x) \sim \sum c_k (x - a)^k$ , if for every $N \ge 0$

f(x) - \sum_{k=0}^{N} c_k (x - a)^k = O\bigl((x - a)^{N+1}\bigr) \quad \text{as } x \to a.

Expansions as $x \to \infty$ are defined in the same way with powers of $\frac{1}{x}$ . The series need not converge, and even when it converges it need not equal $f$ . What matters is that each truncation is accurate to the stated order.

In plain words An asymptotic expansion promises that each extra term makes the approximation better as you get close to the limit point, not that the full series adds up to $f$ .

Example For large $x > 0$ , writing $\sqrt{x^2 + 1} = x \sqrt{1 + x^{-2}}$ and expanding gives

\sqrt{x^2 + 1} = x + \frac{1}{2x} - \frac{1}{8x^3} + O\left(\frac{1}{x^5}\right) \quad \text{as } x \to \infty.

At $x = 10$ the two leading correction terms give

\frac{1}{20} - \frac{1}{8000} = 0.049875,

while $\sqrt{101} - 10 = 0.0498756\ldots$ .

Applications to physics

Definition Physical formulas are often replaced by the leading terms of their Taylor series when a variable is small. For an ideal pendulum of length $L$ , with the angle $\theta$ in radians and $g$ the acceleration due to gravity, the motion obeys

\theta'' = -\frac{g}{L} \sin \theta.

Using $\sin \theta = \theta + O(\theta^3)$ gives the small angle model

\theta'' = -\frac{g}{L} \theta.

For $0 < \theta \le \frac{\pi}{2}$ the Lagrange remainder of $T_2(\theta) = \theta$ gives

\theta - \sin \theta = \frac{\cos z}{6} \theta^3

for some $z$ between $0$ and $\theta$ , and $0 < \cos z < 1$ there, so

0 \le \theta - \sin \theta \le \frac{\theta^3}{6},

and the error of replacing $\sin \theta$ by $\theta$ , as a fraction of $\theta$ , is at most $\frac{\theta^2}{6}$ .

In plain words For small swings the restoring force is almost exactly proportional to the angle, and the series says how small is small enough.

Example At $\theta = 0.2$ radians, about $11.5$ degrees, $\sin 0.2 = 0.198669$ . As a fraction of $\theta$ the error is

\frac{0.2 - 0.198669}{0.2} \approx 0.00665,

just within the bound $\frac{0.04}{6} \approx 0.00667$ . That is under one percent, which is why the small angle model works so well for gentle swings.

Check yourself

Use series to find $\lim_{x \to 0} \frac{e^x - 1 - x}{x^2}$ .

Show the answer

The exponential series gives

e^x - 1 - x = \frac{x^2}{2} + O(x^3),

so the quotient is $\frac{1}{2} + O(x)$ and the limit is $\frac{1}{2}$ .

Estimate $\int_0^1 \sin(x^2) \, dx$ with two terms of a series and bound the error.

Show the answer

Substituting $x^2$ into the sine series gives

\sin(x^2) = x^2 - \frac{x^6}{6} + \frac{x^{10}}{120} - \cdots,

so the integral is

\int_0^1 \sin(x^2) \, dx = \frac{1}{3} - \frac{1}{42} + \frac{1}{1320} - \cdots.

Two terms give $0.309524$ . The series alternates with decreasing terms, so the error is at most $\frac{1}{1320} \approx 0.00076$ . The true value is $0.310268$ .

For which $x \ge 0$ does $\sin x \approx x$ have error less than $0.001$ , according to the alternating series bound?

Show the answer

The error is at most $\frac{x^3}{6}$ , and $\frac{x^3}{6} < 0.001$ when $x^3 < 0.006$ , that is $x < 0.1817$ approximately.

Applications of Taylor and Maclaurin Series

Key terms

Polynomial approximation

Scale of neglected terms

Limits of indeterminate forms

Functions defined by integrals

Error estimates

Asymptotic expansions

Applications to physics

Check yourself

How this lesson connects

Builds on

Resources for this lesson

Taylor polynomial explorer