Linear Approximation and Newton’s Method

A derivative makes a local model

Near an input $a$ where $f$ is differentiable, the tangent-line approximation is

L(x)=f(a)+f'(a)(x-a).

Differentiability means the approximation error divided by $|x-a|$ tends to zero. It does not make the error zero. A useful base point has an easy function value and derivative and is close to the desired input.

Visual guide

VISUAL GUIDEThe tangent predicts the next root estimate

x² − 2
Tangent at x₀ = 2

For f(x) = x² − 2, start at x₀ = 2. The tangent y = 4x − 6 crosses the axis at x₁ = 1.5, closer to √2. Newton’s method repeats this tangent-intercept construction.

Worked example: estimate a square root

For $f(x)=\sqrt{x}$ , choose $a=4$ . Then $f(4)=2$ and $f'(4)=1/4$ , so $\sqrt{4.1}\approx2+0.1/4=2.025$ . Since the square-root function is concave down, its tangent lies above the graph on the positive domain, so this estimate is an overestimate.

If $|f''|\le M$ between $a$ and $x$ , Taylor's theorem bounds the linearization error by $M|x-a|^2/2$ . This additional smoothness gives a quantitative guarantee. A picture of a nearby tangent alone does not provide an error bound.

From a tangent to a better root guess

At a current estimate $x_n$ , intersect the tangent with the horizontal axis. Solving $0=f(x_n)+f'(x_n)(x_{n+1}-x_n)$ gives Newton's iteration:

x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}.

This needs a nonzero derivative and a function that can be evaluated at the next point. Under suitable smoothness and a sufficiently close starting value near a simple root, convergence is often rapid. Those conditions are local, not automatic for an arbitrary starting guess.

Worked example: compute a square root

To solve $x^2-2=0$ , begin at $x_0=1.5$ . Newton's formula becomes $x_{n+1}=(x_n+2/x_n)/2$ . The first step gives $17/12\approx1.416667$ and the next gives $577/408\approx1.414216$ . Substituting back into $x^2-2$ checks the residual.

A small residual does not always imply a small root error, especially when the function is nearly flat. When possible, combine Newton steps with a verified bracket, or use bisection when a Newton step leaves the allowed interval.

Failure is informative

For $f(x)=x^3-2x+2$ , starting at zero produces $x_1=1$ , and starting from $1$ returns to zero. The iteration cycles rather than converges. A stopping rule should limit iterations and check both changes and residuals; it must not label an arbitrary last iterate a root.

Practice

Approximate $1/1.02$ using a tangent at $1$ .
Take one Newton step for $x^2-3=0$ from $x_0=2$ .
Why is $x_0=0$ invalid for Newton applied to $x^2-2$ ?

Show worked solutions

For $1/x$ , the value and derivative at $1$ are $1$ and $-1$ , so the estimate is $0.98$ .
$x_1=2-(4-3)/4=7/4$ .
The derivative is $2x$ , so the formula divides by zero at the starting point.