Taylor and Maclaurin Series

Manipulating the geometric series produces a power series for a function without ever asking the direct question: given $f$ , what coefficients make $f (x) = \sum c_{n} (x - a)^{n}$ ? The coefficients are determined by the derivatives of $f$ at the center.

The coefficient formula

Suppose $f$ has a power series representation about $a$ ,

f (x) = c_{0} + c_{1} (x - a) + c_{2} (x - a)^{2} + c_{3} (x - a)^{3} + \dots, ∣ x - a ∣ < R .

Setting $x = a$ gives $f (a) = c_{0}$ . Differentiating term by term (legal inside $R$ ) and setting $x = a$ each time isolates one coefficient per step:

f^{'} (x) = c_{1} + 2 c_{2} (x - a) + 3 c_{3} (x - a)^{2} + \dots \Rightarrow f^{'} (a) = c_{1},

f^{''} (x) = 2 c_{2} + 2 \cdot 3 c_{3} (x - a) + \dots \Rightarrow f^{''} (a) = 2 c_{2},

f^{'''} (x) = 2 \cdot 3 c_{3} + \dots \Rightarrow f^{'''} (a) = 3! c_{3} .

The pattern is $f^{(n)} (a) = n! c_{n}$ , so $c_{n} = f^{(n)} (a) / n!$ .

The coefficients are unique: however a power series for $f$ is obtained, its coefficients must be these. Substituting them back names the series.

Taylor polynomials and the remainder

Having a Taylor series is not the same as equalling it: the series might converge to something other than $f$ , or not converge at all. To check, look at its partial sums.

$T_{1}$ is the tangent-line linearization; $T_{n}$ agrees with $f$ in value and in the first $n$ derivatives at $a$ . The function equals its Taylor series exactly when $T_{n} (x) \to f (x)$ . Writing the gap as $R_{n} (x) = f (x) - T_{n} (x)$ , the remainder, the condition is $lim_{n \to \infty} R_{n} (x) = 0$ .

For

f (x) = sin x

, the Taylor polynomials

T_{1}, T_{3}, T_{5}

match the curve over ever-wider intervals about the origin; the remainder

R_{n} (x)

is the vertical gap between

sin x

and the polynomial.

To show $R_{n} \to 0$ , bound the remainder by the size of the next derivative.

Paired with the fact that $x^{n} / n! \to 0$ for every real $x$ , this drives the remainder to zero whenever the derivatives stay uniformly bounded.

The exponential, sine, and cosine

Exponential. For $f (x) = e^{x}$ , every derivative is $e^{x}$ , so $f^{(n)} (0) = 1$ and the Maclaurin series is $\sum x^{n} / n!$ . On any interval $∣ x ∣ \leq d$ the derivatives satisfy $∣ f^{(n + 1)} (x) ∣ = e^{x} \leq e^{d}$ , so Taylor's Inequality with $M = e^{d}$ gives $∣ R_{n} (x) ∣ \leq e^{d} ∣ x ∣^{n + 1} / (n + 1)! \to 0$ . Hence

e^{x} = n = 0 \sum \infty \frac{x ^{n}}{n !} = 1 + x + \frac{x ^{2}}{2 !} + \frac{x ^{3}}{3 !} + \dots for all x .

Setting $x = 1$ recovers $e = \sum 1/ n!$ .

The Taylor polynomials of

e^{x}

at the origin. Each added term lifts the polynomial closer to

e^{x}

over a wider interval;

T_{3}

already tracks the curve well past the center.

Sine. For $f (x) = sin x$ the derivatives cycle through $cos x, - sin x, - cos x, sin x$ , giving $f (0), f^{'} (0), \dots = 0, 1, 0, - 1, 0, 1, \dots$ . Every derivative is $\pm sin x$ or $\pm cos x$ , so $∣ f^{(n + 1)} (x) ∣ \leq 1$ and $M = 1$ works for all $x$ . Thus

sin x = n = 0 \sum \infty (- 1)^{n} \frac{x ^{2 n + 1}}{( 2 n + 1 )!} = x - \frac{x ^{3}}{3 !} + \frac{x ^{5}}{5 !} - \dots for all x .

Cosine. Rather than repeat the computation, differentiate the sine series term by term (legal inside the interval of convergence, here all of $R$ ):

cos x = n = 0 \sum \infty (- 1)^{n} \frac{x ^{2 n}}{( 2 n )!} = 1 - \frac{x ^{2}}{2 !} + \frac{x ^{4}}{4 !} - \dots for all x .

Existing series combine without any new differentiation. Multiplying the cosine series by $x$ gives the series for $x cos x$ ; substituting into the exponential series gives series for $e^{- x^{2}}$ and the like.

The binomial series

The last standard series generalizes the Binomial Theorem to any real exponent. For $f (x) = (1 + x)^{k}$ with $k$ real, $f^{(n)} (0) = k (k - 1) \dots (k - n + 1)$ , so the Maclaurin coefficients are the binomial coefficients

(n k) = \frac{k ( k - 1 ) ( k - 2 ) \dots ( k - n + 1 )}{n !} .

The Ratio Test gives radius of convergence $1$ for non-integer $k$ .

If $k$ is a nonnegative integer the coefficients eventually vanish and the series terminates, reducing to the ordinary Binomial Theorem. Otherwise it is infinite.

Example. To expand $1/ 4 - x$ , factor: $1/ 4 - x = \frac{1}{2} (1 - x /4)^{- 1/2}$ , then apply the binomial series with $k = - \frac{1}{2}$ and $x$ replaced by $- x /4$ . The result converges for $∣ x ∣ < 4$ .

A reference table

The standard Maclaurin series, collected in one place, recur constantly.

Function	Maclaurin series	Radius
$\frac{1}{1 - x}$	$n = 0 \sum \infty x^{n}$	$1$
$e^{x}$	$n = 0 \sum \infty \frac{x ^{n}}{n !}$	$\infty$
$sin x$	$n = 0 \sum \infty (- 1)^{n} \frac{x ^{2 n + 1}}{( 2 n + 1 )!}$	$\infty$
$cos x$	$n = 0 \sum \infty (- 1)^{n} \frac{x ^{2 n}}{( 2 n )!}$	$\infty$
$tan^{- 1} x$	$n = 0 \sum \infty (- 1)^{n} \frac{x ^{2 n + 1}}{2 n + 1}$	$1$
$ln (1 + x)$	$n = 1 \sum \infty (- 1)^{n - 1} \frac{x ^{n}}{n}$	$1$
$(1 + x)^{k}$	$n = 0 \sum \infty (n k) x^{n}$	$1$

Approximating with a controlled error

In practice, a low-degree Taylor polynomial approximates $f$ near $a$ , and Taylor's Inequality (or, for alternating series, the first omitted term) certifies the accuracy over an interval.

Over

∣ x - a ∣ \leq d

the graph of

T_{n}

stays inside an error band of half-width

M d^{n + 1} / (n + 1)!

around

f

; the band widens as

x

leaves the center.

Integrals and limits from series

A Maclaurin series turns two otherwise hard computations into arithmetic on a convergent series: integrating a function with no elementary antiderivative, and evaluating an indeterminate limit without repeated l'Hospital differentiation.

The area

\int_{0}^{1} e^{- x^{2}} d x

(shaded) has no elementary antiderivative; expanding

e^{- x^{2}}

as a series and integrating term by term evaluates it to

0.7475

with error below

0.001

The Newtonian limit of relativistic energy

Taylor polynomials formalize the physicist's habit of keeping only the leading correction. In special relativity the kinetic energy of a mass $m$ moving at speed $v$ is

K = m c^{2} (\frac{1}{1 - v ^{2} / c ^{2}} - 1) .

Expanding $(1 - v^{2} / c^{2})^{- 1/2}$ by the binomial series and keeping terms through $(v / c)^{2}$ ,

K = m c^{2} (\frac{1}{2} \frac{v ^{2}}{c ^{2}} + \frac{3}{8} \frac{v ^{4}}{c ^{4}} + \dots) = \frac{1}{2} m v^{2} + \frac{3}{8} \frac{m v ^{4}}{c ^{2}} + \dots .

At speeds small compared with $c$ the higher terms are negligible and the relativistic energy reduces to the Newtonian $\frac{1}{2} m v^{2}$ . The Taylor series makes precise the sense in which one theory contains another as a low-order approximation, and quantifies the correction the next term supplies.

Relativistic kinetic energy (dark) against the Newtonian

\frac{1}{2} m v^{2}

(light) as functions of speed. The two agree closely at low speed, where the leading Taylor term dominates, and separate as

v

approaches the speed of light, where the energy grows without bound.