Newton's Method and Antiderivatives

Most equations that arise in practice have no solution formula. A polynomial of degree five or higher has no root formula at all, and a transcendental equation like $cos x = x$ has none either. Newton's method solves such an equation numerically by repeatedly replacing the curve with its tangent line and jumping to the tangent's root. Antidifferentiation runs the derivative backward, recovering a function from its rate of change.

Newton's method

To solve $f (x) = 0$ is to locate an $x$ -intercept of the graph of $f$ . Newton's method starts from a guess $x_{1}$ and improves it by following the tangent. Near $x_{1}$ the tangent line hugs the curve, so the tangent's $x$ -intercept is close to the curve's. The tangent is a line, so its intercept is easy to compute.

The tangent at $(x_{1}, f (x_{1}))$ has slope $f^{'} (x_{1})$ , so its equation is $y - f (x_{1}) = f^{'} (x_{1}) (x - x_{1})$ . Setting $y = 0$ and solving for the intercept $x_{2}$ , provided $f^{'} (x_{1}) \neq = 0$ ,

x_{2} = x_{1} - \frac{f ( x _{1} )}{f ^{'} ( x _{1} )} .

Repeating from $x_{2}$ , then $x_{3}$ , and so on produces a sequence defined by the same rule at every step.

Each step drops from the guess to the curve, rides the tangent down to the axis, and lands closer to the root

r

Newton used $f (x) = x^{3} - 2 x - 5$ to demonstrate the method, starting from $x_{1} = 2$ . With $f^{'} (x) = 3 x^{2} - 2$ ,

x_{2} = 2 - \frac{f ( 2 )}{f ^{'} ( 2 )} = 2 - \frac{- 1}{10} = 2.1, x_{3} = 2.1 - \frac{f ( 2.1 )}{f ^{'} ( 2.1 )} = 2.1 - \frac{0.061}{11.23} \approx 2.0946.

The correct root to four decimals is $2.0946$ , reached in two steps. When it converges, Newton's method roughly doubles the number of correct digits per iteration.

The doubling of correct digits per step is visible when the iterates are laid out.

Worked example. Compute $62$ to eight decimals. The value is the positive root of $f (x) = x^{6} - 2$ , so with $f^{'} (x) = 6 x^{5}$ ,

x_{n + 1} = x_{n} - \frac{x _{n}^{6} - 2}{6 x _{n}^{5}} .

Starting from $x_{1} = 1$ , the iterates are

$n$	$x_{n}$	correct decimals
$1$	$1.00000000$	$0$
$2$	$1.16666667$	$0$
$3$	$1.12644368$	$1$
$4$	$1.12249707$	$3$
$5$	$1.12246205$	$8$
$6$	$1.12246205$	$8$

The count of correct decimals roughly doubles from $x_{3}$ onward, the hallmark of quadratic convergence. Since $x_{5}$ and $x_{6}$ agree, $62 = 1.12246205$ .

Algorithm:

\textsc{Newton}(f, f', x_1, \varepsilon)

— approximate a root of

f(x)=0

1
$x \gets x_1$
2
repeat
3
$x_{\text{new}} \gets x - f(x)/f'(x)$
4
if $|x_{\text{new}} - x| < \varepsilon$ then return $x_{\text{new}}$
5
$x \gets x_{\text{new}}$
6
until a step limit is reached

When Newton's method fails

The tangent shortcut assumes the tangent points back toward the root. A small derivative tilts the tangent nearly flat, throwing its intercept far away, and a badly placed guess can send the iterates outside the domain or into an endless oscillation.

A near-horizontal tangent at

x_{1}

(small

f^{'} (x_{1})

) sends

x_{2}

farther from the root than

x_{1}

was.

A poorly placed start can also trap the iteration in a loop. For $f (x) = x^{3} - 2 x + 2$ starting at $x_{1} = 0$ , the tangent lands at $x_{2} = 1$ , and the tangent at $x_{2}$ lands back at $x_{1} = 0$ ; the iterates cycle between $0$ and $1$ and never approach the real root near $- 1.77$ .

Newton's method can cycle: from

x_{1} = 0

the tangent lands at

x_{2} = 1

, and the tangent there returns to

x_{1} = 0

, so the iterates loop.

Newton's method is a local method: it converges quickly from a good starting point and unreliably from a poor one. A rough sketch or a bracketing of the root supplies the good start.

Antiderivatives

Differentiation sends a function to its rate of change. Recovering the function from the rate reverses that operation. A physicist with a velocity wants position; an engineer with a leak rate wants total volume lost. Each asks for a function whose derivative is known.

Antiderivatives are never unique. If $F (x) = \frac{1}{3} x^{3}$ then $F^{'} (x) = x^{2}$ , but so does $G (x) = \frac{1}{3} x^{3} + 100$ , and so does any $\frac{1}{3} x^{3} + C$ . The corollary to the Mean Value Theorem shows there are no others: two functions with the same derivative on an interval differ by a constant.

The constant is a vertical shift, so the antiderivatives of a function form a family of curves stacked one above another, all with identical slope at each $x$ . Fixing one member requires one extra fact.

The antiderivatives of a function are vertical translates of one curve; every member has the same slope at each

x

, and

C

selects the height.

Reading antiderivatives off differentiation rules

Every differentiation formula, read from right to left, is an antidifferentiation formula. Two structural rules and the reversed power rule cover most cases.

Function $f (x)$	Particular antiderivative $F (x)$
$c f (x)$	$c F (x)$
$f (x) + g (x)$	$F (x) + G (x)$
$x^{n} (n \neq = - 1)$	$\frac{x ^{n + 1}}{n + 1}$
$cos x$	$sin x$
$sin x$	$- cos x$
$sec^{2} x$	$tan x$
$sec x tan x$	$sec x$

The power rule reversed, $x^{n} \mapsto \frac{x ^{n + 1}}{n + 1}$ , fails at $n = - 1$ because the exponent $n + 1$ would be zero; the antiderivative of $x^{- 1}$ is deferred to the natural logarithm. For example, to antidifferentiate $g^{'} (x) = 4 sin x + 2 x^{4} - x^{- 1/2}$ , apply the table term by term:

g (x) = - 4 cos x + \frac{2}{5} x^{5} - 2 x + C .

Initial conditions

An equation involving the derivatives of an unknown function is a differential equation, and its general solution carries the arbitrary constant. An extra condition on the function selects one solution from the family.

For $f^{'} (x) = x x = x^{3/2}$ with $f (1) = 2$ , the general antiderivative is $f (x) = \frac{2}{5} x^{5/2} + C$ . Imposing $f (1) = 2$ gives $\frac{2}{5} + C = 2$ , so $C = \frac{8}{5}$ and $f (x) = \frac{2}{5} x^{5/2} + \frac{8}{5}$ . A second-order equation needs two conditions, one for each integration.

Recovering motion from acceleration

Antidifferentiation reads position from velocity and velocity from acceleration. Given $a (t)$ , one antiderivative recovers $v (t)$ up to a constant fixed by the initial velocity, and a second recovers $s (t)$ up to a constant fixed by the initial position. The geometry is the reverse of the sign chart: the slope of $f$ is known at every point, and the curve is the one threaded through those slopes from the starting height.

Near the surface of the earth, gravity supplies a constant downward acceleration of about $32$ ft/s $^{2}$ , so a projectile's motion follows from antidifferentiating that constant twice.

Height of the thrown ball,

s (t) = - 16 t^{2} + 48 t + 432

: it rises to a peak at

t = 1.5

s, then falls, reaching the ground near

t = 6.9

Reconstructing

f

from its slope: the known

f^{'}

prescribes a tangent direction at each

x

, and one initial point selects the curve through them.

Stewart, §3.8 — Newton's Method: the tangent-line derivation of the iteration $x_{n + 1} = x_{n} - f (x_{n}) / f^{'} (x_{n})$ , the cubic example, and the small-derivative failure case; §3.9 — Antiderivatives: the definition, the general antiderivative $F + C$ from the Mean Value Theorem corollary, the antidifferentiation table, and initial-value problems. ↩