MGU Mathematics Series — Numerical Methods
Part II: Roots, Interpolation, and Approximation of Functions
Bisection · Fixed-Point · Newton · Secant · Convergence · Error
Chapter 3 of 14
3
MGU Mathematics Series  ·  Numerical Methods  ·  Library Textbook Edition

Chapter 3:
Solving Nonlinear
Equations Numerically

When algebra runs out of formulas — bisection, fixed-point iteration, Newton's method, and the secant method as controlled sequences of approximations converging to a root

■ Part II of V  ·  Roots, Interpolation, and Approximation ■

■ 14 Sections  •  Chapter 3 of 14 ■

Prerequisites: Calculus (derivatives, continuity, IVT) · Basic algebra · Functions

Chapter Introduction and Learning Objectives

Many equations that arise naturally in science, engineering, economics, and mathematics cannot be solved by algebraic formulas. The quadratic formula solves degree-two polynomials, but there is no general formula for degree-five polynomials. There is no formula that isolates \( x \) in an equation like \( e^x = 3x \) or \( x\cos x = 1 \). For these problems, numerical root-finding methods create controlled sequences of approximations that converge to a solution.

This chapter teaches four major methods: the bisection method, fixed-point iteration, Newton's method, and the secant method. Each method approaches root-finding differently, and each carries its own trade-offs of reliability, speed, and required information. By the end of the chapter, you should understand not only how to apply these methods, but why they work, when they fail, and how to judge whether a numerical root is trustworthy.

Opening Question

An engineer designing a cable suspension bridge needs to find the horizontal tension in a cable whose shape follows the equation \( \frac{T}{w}\cosh\!\left(\frac{wx}{T}\right) = H \), where \(T\) is the unknown tension, \(w\) is the weight per unit length, \(x\) is a horizontal distance, and \(H\) is a known height. This equation cannot be solved algebraically for \(T\). A formula does not exist. Yet the engineer must find \(T\) to design the bridge safely. How does mathematics solve a problem like this?

The answer is iteration. Instead of finding an exact symbolic solution, a numerical method creates a sequence of better and better approximations. If the sequence converges, the approximations approach the true answer. The engineer accepts an approximation that is close enough within a specified tolerance. This is numerical root-finding.

14Sections
9Objectives
4Core Methods
14Total Chapters

Learning Objectives

By the end of this chapter, you should be able to:

  1. 1Recognize a root-finding problem and distinguish it from a symbolic equation-solving problem.
  2. 2Explain the role of the Intermediate Value Theorem in bracketing methods.
  3. 3Apply the bisection method, compute error bounds, and determine the number of steps needed for a given tolerance.
  4. 4Apply fixed-point iteration, recognize convergence conditions, and identify diverging cases.
  5. 5Apply Newton's method, interpret it geometrically, and identify cases where it fails.
  6. 6Apply the secant method and explain how it approximates Newton's method without requiring derivatives.
  7. 7Apply appropriate stopping criteria to terminate root-finding iterations.
  8. 8Compare root-finding methods by reliability, speed, and required information.
  9. 9Identify common mistakes in numerical root-finding and interpret reliability warnings.

Sections at a Glance

Vocabulary Before You Begin

Root / Zero
A value \(r\) such that \(f(r)=0\). "Root of the equation" and "zero of the function" are used interchangeably.
Bracketing method
A method maintaining an interval \([a,b]\) known to contain a root, with \(f(a)\cdot f(b) < 0\).
Bisection method
A bracketing method that halves the interval at each step; guaranteed to converge; linear convergence.
Fixed-point
A value \(x\) such that \(g(x)=x\). Roots of \(f\) correspond to fixed points of \(g(x)=x-f(x)\).
Fixed-point iteration
Repeated application of \(p_{n+1}=g(p_n)\); converges when \(|g'(x)|<1\) near the fixed point.
Newton's method
Tangent-line iteration: \(p_{n+1}=p_n - f(p_n)/f'(p_n)\). Quadratic convergence near simple roots.
Secant method
Approximates Newton's method using two previous function values instead of an analytic derivative.
Convergence
The property that a sequence of iterates approaches a limiting value (the root).
Linear convergence
Error decreases by a constant factor each step. Bisection is the canonical example.
Quadratic convergence
Error roughly squares each step; correct digits roughly double per iteration. Newton's method exhibits this near simple roots.
Stopping criterion
A condition used to decide when iteration has produced a sufficiently accurate approximation.
Tolerance
The maximum acceptable error \(\varepsilon\) in a numerical result.
Multiplicity
If \(f(x)=(x-r)^m q(x)\) with \(q(r)\neq 0\), then \(r\) has multiplicity \(m\). Repeated roots slow Newton's method.
Order of convergence
A measure of how quickly error decreases per iteration. Bisection: order 1. Newton: order 2. Secant: order ≈1.618.
Sections 3.1 – 3.2
What Root-Finding Is and Where It Comes From
Nonlinear equations · Abel–Ruffini · Zeros · Multiplicity · Sign changes

What a Root-Finding Problem Is

A root-finding problem begins with a function \(f\) and asks: for what input value \(r\) does \(f(r) = 0\)?

This question is central to mathematics and its applications. Solving an equation like \(g(x)=h(x)\) is equivalent to finding the root of \(f(x)=g(x)-h(x)\). Finding where a curve crosses a horizontal level is a root problem. Computing equilibrium states, break-even points, resonance frequencies, and phase transitions all reduce to root problems.

When the function \(f\) is linear, the root is immediate: if \(f(x)=ax+b\), then \(r=-b/a\). When \(f\) is quadratic, the quadratic formula applies. But the quadratic formula does not generalize. For polynomials of degree five or higher, no general algebraic formula for the roots exists. This is a deep result in algebra known as the Abel–Ruffini theorem. For transcendental functions like \(f(x)=e^x - 3x\) or \(f(x)=\cos x - x\), there is similarly no general algebraic procedure.

Numerical root-finding replaces the missing formula with a systematic approximation process. An algorithm generates a sequence of values \(p_0, p_1, p_2, \ldots\) that, under the right conditions, converges to a root \(r\). Each value is computable, and each is closer to \(r\) than the previous.

The numerical analyst's job is not simply to run the algorithm. It is to understand when the algorithm will converge, how fast it will converge, how much error remains at any step, and when it will fail.

Roots, Zeros, and Solutions

The terms root, zero, and solution are used interchangeably in root-finding, but their usage reflects slightly different framings. A zero of \(f\) is a value \(r\) where \(f(r)=0\) — this is function language. A root of the equation \(f(x)=0\) is the same value described as an equation solution. A solution of \(g(x)=h(x)\) becomes a root of \(f(x)=g(x)-h(x)\).

An important subtlety: a root has a multiplicity. If \(f(x)=(x-r)^m\cdot q(x)\) where \(q(r)\neq 0\), then \(r\) is a root of multiplicity \(m\). Simple roots (multiplicity 1) behave well under most numerical methods. Repeated roots (multiplicity greater than 1) create difficulties that will be noted in later sections.

Example 3.2.1
Locating a root by sign change
  • Function Let \(f(x) = x^3 - 2x - 5\).
  • Evaluate At \(x=2\): \(f(2)=8-4-5=-1 < 0\). At \(x=3\): \(f(3)=27-6-5=16 > 0\).
  • Conclude A sign change occurs on \((2,3)\). By the Intermediate Value Theorem (Section 3.3), a root must lie in that interval.
Sections 3.3 – 3.4
Locating Roots and Establishing Brackets
IVT · Graphical estimates · Bracket condition · Reliability

Graphical Root Estimates

Before applying an algorithm, it helps to sketch or reason about the behavior of \(f\) near its roots. A graph can reveal the approximate location of roots (where the curve crosses the \(x\)-axis), whether roots are simple or repeated, whether there are multiple roots in a region, and whether the function is well-behaved or difficult.

Diagram Instruction

Draw a smooth curve representing \(f(x)\) crossing the \(x\)-axis at approximately \(x=r\). The curve should be clearly negative to the left of \(r\) and positive to the right. Label \(r\) as the root. Also draw a second case where the curve is tangent to the axis (a repeated root), showing that it touches zero without crossing.

Intermediate Value Theorem

If \(f\) is continuous on \([a,b]\) and \(f(a)\) and \(f(b)\) have opposite signs, then there exists at least one \(c\in(a,b)\) such that \(f(c)=0\).

This theorem guarantees existence of a root but does not locate it. Numerical methods provide the location.

Student Warning

Graphical methods can miss roots that are close together, roots where the function nearly touches zero, or roots in regions not shown by the graph. Always supplement graphical analysis with analytical reasoning about the function's behavior.

Bracketing Methods

A bracketing method maintains an interval \([a,b]\) that is known to contain a root. The key requirement is the bracket condition: \(f(a)\) and \(f(b)\) must have opposite signs, i.e.,

\[ f(a)\cdot f(b) < 0 \]

When this condition holds and \(f\) is continuous, the IVT guarantees a root in \((a,b)\). The bracket is gradually narrowed, trapping the root in a smaller and smaller interval.

Bracketing methods are reliable: they always converge to a root as long as the initial bracket is valid and the function is continuous. They do not require derivative information. They cannot jump past the root or diverge. The trade-off is speed — bracketing methods generally converge more slowly than methods that use derivative information.

Strength
Guaranteed Convergence
As long as the bracket condition holds and \(f\) is continuous, bisection never fails. It always finds a root.
Trade-off
Linear Speed
Bisection halves the interval each step. To gain one decimal digit of accuracy, about 3–4 steps are required. For high precision, many steps accumulate.
Setup
Finding the Initial Bracket
Evaluate \(f\) at a sequence of points to find a sign change; use graphical estimates; or use physical knowledge to constrain likely root locations.
Sections 3.5 – 3.6
Bisection Method and Error Bounds
Algorithm · Worked example · Guaranteed error · Step count formula

Bisection Method

The bisection method is the simplest and most reliable bracketing method. At each step, it evaluates \(f\) at the midpoint of the current bracket and replaces one endpoint with the midpoint, keeping the bracket condition satisfied.

Algorithm: Bisection Method Bracketing — Guaranteed Convergence
Purpose:  Find a root of f on [a, b] within tolerance tol
Inputs:   Continuous function f
          Bracket [a, b] with f(a)*f(b) < 0
          Tolerance tol
          Maximum iterations N

Steps:
  1. Verify f(a) * f(b) < 0. If not, stop: bracket condition fails.
  2. For n = 1, 2, ..., N:
       a. Compute midpoint:  p = (a + b) / 2
       b. Compute f(p).
       c. If |b - a| / 2 < tol, or f(p) = 0:
            accept p as the root. Stop.
       d. If f(a) * f(p) < 0:  set b = p   (root in left half)
          Else:                  set a = p   (root in right half)
  3. If max iterations reached: report p with a warning.

Stopping criterion:  |b - a| / 2 < tol  or  f(p) = 0
Output:   Approximation p of the root
Notes:    Guaranteed to converge if f continuous and bracket valid.
          Convergence is linear: each step halves the interval.
Diagram Instruction

Draw a bracket \([a,b]\) on the \(x\)-axis with \(f(a)<0\) and \(f(b)>0\). Mark the midpoint \(p_1\). Show that \(f(p_1)>0\), so the new bracket becomes \([a,p_1]\). Mark \(p_2\) and continue for one more step to visualize the narrowing interval.

Worked Example 3.5.1
Bisection applied to \(f(x)=x^3-2x-5\) on \([2,3]\), tolerance \(0.01\)
  • Think We confirmed \(f(2)=-1<0\) and \(f(3)=16>0\). The bracket condition is satisfied.
  • Method Bisection. Each step halves the interval. We continue until interval width falls below \(0.01\).
  • Compute
    Step\(a\)\(b\)\(p\)\(f(p)\)New bracket
    12.0003.0002.5005.625\([2.000, 2.500]\)
    22.0002.5002.2501.891\([2.000, 2.250]\)
    32.0002.2502.1250.346\([2.000, 2.125]\)
    42.0002.1252.063−0.351\([2.063, 2.125]\)
    52.0632.1252.094−0.009\([2.094, 2.125]\)
    62.0942.1252.1090.167\([2.094, 2.109]\)
    72.0942.1092.1020.079\([2.094, 2.102]\)
    At step 7, interval width \(=0.008 < 0.01\). Accept \(p\approx 2.094\).
  • Check The true root is approximately \(r\approx 2.0946\). Our approximation differs by about \(0.0006\), well within tolerance.
  • Interpret Bisection found the root in seven steps, narrowing a length-1 interval to length \(0.008\) by repeated halving.

Error Bounds for Bisection

One of bisection's most valuable properties is that the error bound at each step is known exactly, without knowing the true root. After \(n\) steps, the midpoint \(p_n\) satisfies

\[ |p_n - r| \leq \frac{b-a}{2^n} \]

This is the bisection error bound. It decreases by a factor of 2 at each step — linear convergence. If the desired tolerance is \(\varepsilon\), we need the minimum number of steps:

\[ n > \log_2\!\left(\frac{b-a}{\varepsilon}\right) \]
Example 3.6.1
How many steps for tolerance \(10^{-4}\) on \([2,3]\)?
  • Compute \[n > \log_2\!\left(\frac{1}{0.0001}\right) = \log_2(10000) \approx 13.3\]
  • Interpret Fourteen steps are sufficient to guarantee error below \(10^{-4}\). This can be planned before running the algorithm — no trial-and-error required.
Why Bisection Matters Despite Slow Convergence

Linear convergence is slow compared to Newton's quadratic convergence. However, bisection never diverges when started with a valid bracket. For problems where safety is paramount and derivative information is unavailable, bisection is the method of choice. Its predictable step count is a genuine advantage.

Section 3.7
Fixed-Point Iteration
Reformulating \(f(x)=0\) · Contraction mapping · Convergence condition · Worked example · Divergence

Fixed-Point Iteration

A different approach to root-finding is to reformulate \(f(x)=0\) as a fixed-point problem: find \(x\) such that \(g(x)=x\). A value \(p\) satisfying \(g(p)=p\) is called a fixed point of \(g\). The connection to root-finding is direct: if we write \(g(x)=x-f(x)\), then \(g(p)=p\) exactly when \(f(p)=0\).

There are many ways to rewrite \(f(x)=0\) in the form \(g(x)=x\), and different rewritings produce different iterations with very different convergence properties. Fixed-point iteration starts from an initial guess \(p_0\) and computes

\[ p_{n+1} = g(p_n) \]

repeatedly. If the sequence converges to \(p\), and if \(g\) is continuous, then \(p = g(p)\), so \(p\) is a fixed point and a root of \(f\).

Convergence Condition: Contraction Mapping

If \(g\) is continuously differentiable and \(|g'(x)| < 1\) for all \(x\) in an interval containing the fixed point \(p\), then fixed-point iteration converges for any starting point in that interval.

Intuitively, \(|g'(x)|<1\) means \(g\) compresses distances: iterates get closer to the fixed point rather than spreading apart.

Algorithm: Fixed-Point Iteration Conditional Convergence
Purpose:  Find fixed point of g(x) = x (i.e., root of f(x) = 0 rewritten as g(x) = x)
Inputs:   Function g, initial guess p_0, tolerance tol, max iterations N

Steps:
  1. For n = 0, 1, 2, ..., N-1:
       a. Compute p_{n+1} = g(p_n).
       b. If |p_{n+1} - p_n| < tol:  accept p_{n+1}. Stop.
  2. If max iterations reached: report best approximation with warning.

Stopping criterion:  |p_{n+1} - p_n| < tol
Output:   Approximation of fixed point p
Notes:    Converges when |g'(x)| < 1 near the fixed point.
          May diverge or converge to wrong value if |g'(x)| >= 1.
          The choice of g dramatically affects convergence.
Worked Example 3.7.1
Fixed-point iteration on \(f(x)=x^3-2x-5\) using \(g(x)=(2x+5)^{1/3}\)
  • Think From \(x^3=2x+5\), take the cube root: \(x=(2x+5)^{1/3}=g(x)\). Start with \(p_0=2\).
  • Check \(g'\) \(g'(x)=\frac{2}{3}(2x+5)^{-2/3}\). Near \(x\approx 2.09\): \(g'(2.09)\approx\frac{2}{3}(9.18)^{-2/3}\approx 0.152\). Since \(|g'|\approx 0.152<1\), convergence is expected.
  • Compute
    \(n\)\(p_n\)\(g(p_n)\)\(|p_{n+1}-p_n|\)
    02.00002.08010.0801
    12.08012.09240.0123
    22.09242.09420.0018
    32.09422.09450.0003
    42.09452.09460.0001
  • Check Accepted value \(p\approx 2.0946\) matches the known root to four decimal places. Differences between successive iterates decreased by roughly a factor of 7 per step — consistent with \(|g'|\approx 0.15\).
  • Interpret Fixed-point iteration converged in four steps from \(p_0=2\), faster than bisection. Convergence depended critically on choosing a \(g\) with small derivative near the root.
Divergence Warning: Choice of g Matters Enormously

The alternative reformulation \(g(x)=(2x+5)/x^2\) gives \(g'(2.09)\approx -1.55\). Since \(|g'|\approx 1.55>1\), that iteration would diverge from the root. Always compute \(g'\) near the expected root before committing to an iteration.

Sections 3.8 – 3.9
Newton's Method and Its Geometric Meaning
Tangent-line derivation · Newton–Raphson formula · Quadratic convergence · Failure modes

Newton's Method

Newton's method is one of the most important and widely used algorithms in all of numerical mathematics. It converges much faster than bisection or fixed-point iteration, typically requiring very few iterations to achieve high accuracy.

The idea comes directly from calculus. Near a root \(r\), the function \(f\) can be approximated by its tangent line at the current estimate \(p_n\):

\[ L(x) = f(p_n) + f'(p_n)(x - p_n) \]

Setting \(L(x)=0\) and solving for \(x\) gives the next estimate:

\[ p_{n+1} = p_n - \frac{f(p_n)}{f'(p_n)} \]
Newton–Raphson Formula

Given \(f\) differentiable and an initial estimate \(p_0\), the Newton–Raphson iteration is:

\(\displaystyle p_{n+1} = p_n - \frac{f(p_n)}{f'(p_n)}\)

Near a simple root, this converges quadratically: the number of correct decimal digits roughly doubles at each step.

Algorithm: Newton's Method (Newton–Raphson) Quadratic Convergence
Purpose:  Find a root of f(x) = 0
Inputs:   Differentiable function f, derivative f'
          Initial guess p_0, tolerance tol, max iterations N

Steps:
  1. For n = 0, 1, 2, ..., N-1:
       a. Compute f(p_n) and f'(p_n).
       b. If f'(p_n) = 0: STOP. Derivative is zero; cannot proceed.
       c. Compute p_{n+1} = p_n - f(p_n) / f'(p_n).
       d. If |p_{n+1} - p_n| < tol or |f(p_{n+1})| < tol: accept p_{n+1}. Stop.
  2. If max iterations reached: report best approximation with warning.

Stopping criterion:  |p_{n+1} - p_n| < tol  or  |f(p_{n+1})| < tol
Output:   Approximation of root r
Notes:    Quadratic convergence near simple roots.
          Requires f'(p_n) ≠ 0.
          Starting point must be sufficiently close to root.
          Slow or fails at repeated roots. May cycle or diverge for bad starts.
Worked Example 3.8.1
Newton's method on \(f(x)=x^3-2x-5\) with \(p_0=2\), tolerance \(10^{-4}\)
  • Think We need \(f'(x)=3x^2-2\). The iteration is \(p_{n+1}=p_n-(p_n^3-2p_n-5)/(3p_n^2-2)\).
  • Compute
    \(n\)\(p_n\)\(f(p_n)\)\(f'(p_n)\)\(p_{n+1}\)
    02.0000−1.000010.00002.1000
    12.10000.261011.23002.0767
    22.07670.006710.93392.0946
    32.0946\(\approx 0\)11.15772.0946
  • Check The known root is \(r\approx 2.09455\). Newton's method achieved four-decimal accuracy in three steps from \(p_0=2\), compared to seven steps by bisection to achieve only \(0.01\) accuracy.
  • Interpret Each step roughly doubled the number of correct decimal digits. This is quadratic convergence — an extraordinary improvement over linear methods.

Tangent Line Interpretation of Newton's Method

The geometric meaning of Newton's method is clear and elegant. At each step: (1) we are at the current estimate \(p_n\); (2) we draw the tangent line to \(y=f(x)\) at the point \((p_n, f(p_n))\); (3) we find where the tangent line crosses the \(x\)-axis; (4) that crossing point is \(p_{n+1}\).

The tangent line is the best local linear approximation to \(f\). It points toward the root. If the curve is well-behaved and the starting point is close enough to the root, the tangent line crossing is much closer to the true root than the starting point.

Diagram Instruction

Draw a smooth curve \(y=f(x)\) crossing the \(x\)-axis at \(r\). Choose a starting point \(p_0\) to the right of \(r\) on the curve. Draw the tangent line at \((p_0, f(p_0))\). Mark where it crosses the \(x\)-axis as \(p_1\). Draw the tangent at \((p_1, f(p_1))\) and mark \(p_2\). Show the iterates converging rapidly toward \(r\).

Quadratic convergence. Near a simple root, there exists a constant \(C\) such that

\[ |p_{n+1} - r| \leq C \cdot |p_n - r|^2 \]

If the error at step \(n\) is \(0.01\), the error at step \(n+1\) is roughly \(C\cdot 0.0001\). Each step doubles the number of correct decimal digits.

Failure Mode 1
Zero Derivative
If \(f'(p_n)=0\), the tangent line is horizontal and never crosses the \(x\)-axis. The iteration cannot proceed.
Failure Mode 2
Poor Starting Point
If \(p_0\) is far from the root, the tangent may cross far away, leading to divergence or convergence to a different root.
Failure Mode 3
Repeated Root
At a repeated root where \(f(r)=0\) and \(f'(r)=0\), convergence slows to linear. Modified Newton methods can restore quadratic convergence.
Failure Mode 4
Oscillation
If the curve oscillates rapidly or has an inflection point near the root, Newton's method can cycle between values without converging.
Section 3.10
Secant Method
Finite-difference derivative · Two starting values · Superlinear convergence

Secant Method

Newton's method requires evaluating both \(f(p_n)\) and \(f'(p_n)\) at each step. In some applications, the derivative is difficult or expensive to compute, or the function may be defined only at data points, not by an explicit formula.

The secant method approximates the derivative using two previous function values:

\[ f'(p_n) \approx \frac{f(p_n) - f(p_{n-1})}{p_n - p_{n-1}} \]

Substituting into the Newton formula:

\[ p_{n+1} = p_n - f(p_n)\cdot\frac{p_n - p_{n-1}}{f(p_n) - f(p_{n-1})} \]

The secant method requires two starting values, \(p_0\) and \(p_1\), rather than one starting value and a derivative formula.

Diagram Instruction

Draw a smooth curve \(y=f(x)\). Choose two starting points \((p_0, f(p_0))\) and \((p_1, f(p_1))\) on the curve. Draw the secant line connecting them. Mark where the secant crosses the \(x\)-axis as \(p_2\). Then draw the secant through \((p_1, f(p_1))\) and \((p_2, f(p_2))\) and mark \(p_3\). Compare with Newton tangent lines to visualize the approximation.

Algorithm: Secant Method Superlinear Convergence — No Derivative Required
Purpose:  Find a root of f(x) = 0 without requiring f'
Inputs:   Function f, two initial guesses p_0 and p_1
          Tolerance tol, max iterations N

Steps:
  1. For n = 1, 2, ..., N:
       a. Compute q_0 = f(p_{n-1}),  q_1 = f(p_n).
       b. If q_1 - q_0 = 0: STOP. Division by zero; method cannot proceed.
       c. Compute p_{n+1} = p_n - q_1 * (p_n - p_{n-1}) / (q_1 - q_0).
       d. If |p_{n+1} - p_n| < tol: accept p_{n+1}. Stop.
  2. If max iterations reached: report best approximation with warning.

Stopping criterion:  |p_{n+1} - p_n| < tol
Output:   Approximation of root r
Notes:    No derivative required; approximates Newton's method.
          Superlinear convergence (order ≈ 1.618, the golden ratio).
          Requires two starting values.
          Does not maintain a bracket; convergence not guaranteed.
Worked Example 3.10.1
Secant method on \(f(x)=x^3-2x-5\) with \(p_0=2\), \(p_1=3\), tolerance \(10^{-4}\)
  • Think No derivative needed. We use the secant formula with the two starting values directly.
  • Compute
    \(n\)\(p_{n-1}\)\(p_n\)\(f(p_{n-1})\)\(f(p_n)\)\(p_{n+1}\)
    12.00003.0000−1.000016.00002.0588
    23.00002.058816.0000−0.39082.0943
    32.05882.0943−0.3908−0.01082.0946
    42.09432.0946−0.01080.00012.0946
  • Check After four steps, \(p_4\approx 2.0946\), accurate to four decimal places. The secant method converged nearly as fast as Newton's method, without requiring the derivative.
  • Interpret The secant method converges at order approximately \(1.618\) — faster than linear but slower than quadratic. In practice it often performs nearly as well as Newton's method, especially when derivatives are expensive to compute.
Sections 3.11 – 3.12
Stopping Criteria and Comparing Methods
When to stop · Combined criteria · Method comparison table · Hybrid strategy

Stopping Criteria

A fundamental question in any iterative algorithm is: when do we stop? Several stopping criteria are commonly used in root-finding. No single criterion is universally best; combining two is safest.

CriterionFormulaWhat It TestsLimitation
Absolute change \(|p_{n+1}-p_n|<\varepsilon\) Successive iterates are close Can be satisfied even when far from root if function is very flat
Relative change \(\dfrac{|p_{n+1}-p_n|}{|p_{n+1}|}<\varepsilon\) Change relative to magnitude of root Better when root is large or small in magnitude
Function value \(|f(p_{n+1})|<\varepsilon\) Whether the equation is nearly satisfied Flat functions can give small \(|f|\) far from the root
Interval width (bisection) \(\dfrac{b-a}{2^n}<\varepsilon\) Rigorous guaranteed error bound Only applies to bisection
Best Practice: Combine Two Criteria

Stop when both the absolute change in the approximation is small and the function value is small. Satisfying only one may be misleading. Always include a maximum iteration limit as a safety measure to prevent infinite loops.

Comparing Root-Finding Methods

MethodConvergence GuaranteeSpeedRequiresMain Failure Mode
Bisection Yes — with valid bracket and continuous \(f\) Linear (halves each step) \(f\), sign-change bracket Slow for high accuracy
Fixed-point Conditional (\(|g'|<1\)) Linear (rate depends on \(|g'|\)) \(f\), clever \(g\) Diverges if \(|g'|\geq 1\)
Newton No — requires good start Quadratic near simple roots \(f\), \(f'\) Bad start; \(f'=0\); repeated roots
Secant No — requires good starts Superlinear (order ≈1.618) \(f\), two starting values Bad starts; \(f(p_n)=f(p_{n-1})\)
Practical Hybrid Strategy

In practice, analysts often combine methods. A few bisection steps locate the root roughly and confirm it lies in a small interval. Newton's method or the secant method then refines the approximation quickly. This hybrid approach combines bisection's reliability with Newton's speed — the dominant strategy in professional scientific computing.

Common Root-Finding Mistakes

Mistake 1
"I can just start bisection anywhere on the function."
Bisection requires a valid bracket — an interval where \(f\) has opposite signs at the endpoints. Without verifying \(f(a)\cdot f(b)<0\), no root is guaranteed in \([a,b]\). Always check the bracket condition before beginning.
Mistake 2
"Newton's method will always find the root quickly from any starting point."
Newton's method can diverge, cycle, or converge to the wrong root when the starting point is far from the target root. Use graphical inspection or a few bisection steps to choose a good starting point before switching to Newton.
Mistake 3
"Any reformulation \(g(x)=x\) will work for fixed-point iteration."
The iteration \(p_{n+1}=g(p_n)\) diverges when \(|g'(x)|\geq 1\) near the root. Checking the derivative of \(g\) near the expected root before iterating can save substantial wasted computation.
Mistake 4
"If \(|f(p_n)|\) is small, the root is found."
A function can be nearly zero over a wide region if it is very flat, so small \(|f(p_n)|\) does not guarantee \(p_n\) is close to the root. Always combine a function-value check with an approximation-change check.
Mistake 5
"Slow convergence means the method is failing."
Bisection converges slowly by design. For tolerance \(10^{-10}\), it requires about 33 steps from a length-1 bracket. This is expected and correct behavior, not a failure mode.
Mistake 6
"Newton's method is always quadratic."
Quadratic convergence applies only near simple roots. At a root of multiplicity \(m>1\), Newton's method converges only linearly. Modified Newton methods (multiplying by \(m\)) can restore quadratic convergence.
Mistake 7
"The secant method is just a worse version of Newton's method."
The secant method eliminates the need for an analytic derivative, which is often unavailable or expensive. It converges at order ≈1.618 and performs nearly as well as Newton in practice. When derivatives are costly, the secant method can actually be more efficient overall.

Preparing for Interpolation and Approximation

Root-finding is our first deep encounter with iterative algorithms, convergence, stopping criteria, and reliability in numerical methods. The ideas introduced here reappear throughout the rest of the book.

In Chapter 4, interpolation will use polynomial functions constructed from data, and the ideas of convergence and error bounds will recur in measuring how well interpolating polynomials approximate a function. In Chapter 9, the Jacobi and Gauss-Seidel methods for linear systems are fixed-point iterations in a higher-dimensional setting. In Chapter 11, Newton's method for optimization is a direct extension applied to \(f'(x)=0\). In Chapter 12, ODE solvers use Newton-like iterations to step forward in time.

The most important lesson of this chapter is not any single formula, but the habit of thinking carefully about convergence, error, reliability, and the conditions under which an algorithm can be trusted.

MGU Library Connection
  • Numerical Methods — Appendix G: Root-Finding Method Reference (formulas and reliability notes)
  • Numerical Methods — Appendix E: Error Analysis Reference
  • Calculus — Chapter on Derivatives (tangent lines, differentiation rules)
  • Calculus — Intermediate Value Theorem and Continuity
  • Linear Algebra — Iterative methods (Chapter 9 of this book uses fixed-point ideas in matrix form)
  • Companion resource: Root-Finding Algorithm Comparison Sheet

Chapter Summary

Key Terms Review

Root / Zero
A value \(r\) where \(f(r)=0\); equivalent to a solution of \(f(x)=0\).
Bracket
An interval \([a,b]\) where \(f(a)\cdot f(b)<0\), guaranteeing a root by the IVT.
Bisection method
Halves the bracket each step; linear convergence; guaranteed by IVT.
Fixed-point iteration
Iterates \(p_{n+1}=g(p_n)\); converges when \(|g'|<1\) near root.
Newton's method
Tangent-line iteration; quadratic convergence; requires \(f'\).
Secant method
Finite-difference Newton; superlinear convergence; no \(f'\) required.
Linear convergence
Error decreases by a constant factor per step (bisection).
Quadratic convergence
Error roughly squares per step; digits double (Newton).
Stopping criterion
Condition for terminating iteration; best to combine two.
Multiplicity
Order \(m\) of a root; multiplicity \(>1\) slows Newton to linear.
ReviewConcept Review Questions
  1. What condition must be satisfied before bisection can begin? What theorem guarantees a root under this condition?
  2. Explain in your own words why bisection always converges but Newton's method may not.
  3. What does \(|g'(x)|<1\) mean geometrically, and why does it ensure convergence of fixed-point iteration?
  4. The Newton formula involves \(f'(p_n)\) in the denominator. What goes wrong when \(f'(p_n)=0\)?
  5. Describe the tangent-line interpretation of Newton's method. Why does the tangent crossing approximate the root better when the starting point is close to the root?
  6. How does the secant method approximate the derivative? What is the geometric interpretation?
  7. A student applies Newton's method and obtains \(p_5=1.99999\) with \(f(p_5)=10^{-6}\). The student declares success. What else should be checked?
  8. Why is it recommended to combine bisection and Newton's method in practice?
  9. What does quadratic convergence mean? Give a numerical illustration comparing it to linear convergence.
  10. A root has multiplicity 2. What happens to Newton's method convergence at such a root?
PracticeMethod Practice

Bisection

  1. Apply the bisection method to \(f(x)=\cos x - x\) on \([0,1]\) for four steps. Compute the error bound after each step.
  2. For \(f(x)=e^x-2\) on \([0,1]\), determine how many bisection steps are required to guarantee error below \(10^{-5}\).
  3. Apply bisection to \(f(x)=x^4-3\) on \([1,2]\) for five steps. Record each midpoint and error bound.

Fixed-Point Iteration

  1. Verify that \(g(x)=\cos x\) has a fixed point near \(x=0.739\). Apply fixed-point iteration from \(p_0=0.7\) for five steps. Compute \(g'(0.739)\) and verify the convergence condition.
  2. For \(f(x)=x^2-3\), write two different fixed-point reformulations. Compute \(g'(x)\) for each near \(x=\sqrt{3}\approx 1.732\) and determine which will converge.
  3. Apply fixed-point iteration to \(f(x)=x^3-x-1\) using \(g(x)=(x+1)^{1/3}\) from \(p_0=1.3\) for five steps.

Newton's Method

  1. Apply Newton's method to \(f(x)=x^3-x-1\) with \(p_0=1.5\), tolerance \(10^{-6}\). Record all iterates.
  2. Apply Newton's method to \(f(x)=\sin x - x/2\) with \(p_0=2\) for four steps.
  3. Apply Newton's method to \(f(x)=x^2-2\) starting from \(p_0=1\). Recognize this as the classical algorithm for computing \(\sqrt{2}\).

Secant Method

  1. Apply the secant method to \(f(x)=x^3-x-1\) with \(p_0=1.0\), \(p_1=1.5\), tolerance \(10^{-4}\). Record all iterates.
  2. Apply the secant method to \(f(x)=e^x-3x\) with \(p_0=0\), \(p_1=1\) for four steps. Compare convergence with bisection on the same function.
PracticeAlgorithm Practice
  1. Write the bisection algorithm in pseudocode from first principles. Identify each step's purpose in plain language.
  2. Write the Newton iteration explicitly for \(f(x)=x^5-x-1\). Compute \(f'(x)\) and write the full iteration formula.
  3. Explain why the secant method requires two starting values while Newton's method requires only one.
  4. Design a stopping criterion that combines absolute change and function value. Explain the rationale for combining both and give an example where each alone would be misleading.
PracticeComputational Interpretation
  1. A root-finding algorithm produces iterates: \(1.5,\ 1.8,\ 2.0,\ 2.09,\ 2.094,\ 2.0946\). Without knowing the method, estimate the order of convergence from the error reductions.
  2. An engineer computes \(f(p_{10})=10^{-8}\) and concludes the root is accurate to eight decimal places. Is this conclusion valid? What additional information is needed?
  3. Two methods are applied to find a root. Method A takes 50 steps but always converges. Method B takes 5 steps but fails 10% of the time. Discuss when each is preferable and what a hybrid strategy might look like.
  4. A function satisfies \(f(1.99)=-0.001\) and \(f(2.01)=0.001\). Discuss whether a root must exist between them and estimate its location. What assumption is required?
ApplicationsReal-World Applications
  1. Projectile range. A projectile is launched at angle \(\theta\) with initial speed \(v_0=50\) m/s. The range is \(R(\theta)=(v_0^2/g)\sin 2\theta\). Find the angle \(\theta\) that achieves \(R=220\) m, given \(g=9.8\) m/s². Reformulate as \(f(\theta)=0\) and solve using Newton's method.
  2. Equilibrium. The equation \(e^{-x}=x\) arises in certain equilibrium problems. Show that it has a root in \((0,1)\) and find it using bisection to four decimal places.
  3. Supply and demand. A supply-demand model gives \(S(p)=p^2+2p\) and \(D(p)=50-3p\). Equilibrium satisfies \(S(p)=D(p)\). Reformulate as a root problem and solve using Newton's method.
  4. Classic example. The function \(f(x)=x-\cos x\) has exactly one root. Find it using both bisection and Newton's method and compare the number of steps required for tolerance \(10^{-6}\).
  5. Finance. Interest rate \(r\) satisfying a present-value equation obeys \(1000(1+r)^{10}-2000=0\). Solve for \(r\) using Newton's method and interpret the result as an annual return rate.
Error AnalysisError Analysis Problems
  1. For bisection applied to \(f(x)=x^3-2x-5\) on \([2,3]\), compute the guaranteed error bound after 5, 10, and 20 steps. How many steps are needed to guarantee accuracy of \(10^{-8}\)?
  2. The fixed-point iteration \(g(x)=(x+3)/4\) is applied near \(x=1\). Compute \(g'(1)\) and estimate the ratio of successive errors. How many steps to reach tolerance \(10^{-6}\) from \(p_0=1.5\)?
  3. Newton's method is applied to \(f(x)=x^2-a\) to compute \(\sqrt{a}\). Derive the Newton formula explicitly and demonstrate quadratic convergence for \(a=2\), \(p_0=1\) by computing three iterates and the error at each step.
  4. Explain why Newton's method converges only linearly at a double root \(f(x)=(x-r)^2\). Compute the derivative ratio that determines the linear convergence constant.
  5. Compare the number of function evaluations required by bisection and Newton's method to achieve error below \(10^{-8}\), starting from an interval of length 1 for bisection or a starting point within distance \(0.5\) of the root for Newton.
ExtensionChallenge and Extension
  1. Newton fractals. Newton's method can be applied in the complex plane to find complex roots of polynomials. Describe what a Newton fractal is and why it arises near the boundaries between basins of attraction. (Conceptual description; no computation required.)
  2. Modified Newton's method. For a root of multiplicity \(m\), the modified iteration is \(p_{n+1}=p_n - m\cdot f(p_n)/f'(p_n)\). Verify that this restores quadratic convergence for \(f(x)=(x-1)^2\) with \(m=2\).
  3. Brent's method. Brent's method combines bisection, secant, and inverse quadratic interpolation. Research its basic idea and describe how it avoids the failure modes of each individual method while maintaining bisection's reliability.
  4. Divergence analysis. Apply Newton's method to \(f(x)=x^3-2x+2\) starting from \(p_0=0\). What happens? Explain the behavior by examining the graph of \(f\) near \(p_0=0\).
Chapter 3 Checkpoint
Complete without reference to worked examples — confirm core skills before moving to Chapter 4
Checkpoint Problems
  • Apply the bisection method to \(f(x)=x^3+x-1\) on \([0,1]\) for five steps. Record \(a\), \(b\), \(p\), \(f(p)\), and error bound at each step.
  • For the function above, determine how many bisection steps are required to guarantee accuracy within \(10^{-6}\).
  • Reformulate \(x^3+x-1=0\) as \(g(x)=x\). Compute \(g'(x)\) and determine whether fixed-point iteration will converge from \(p_0=0.7\).
  • Apply Newton's method to \(f(x)=x^3+x-1\) with \(p_0=0.7\) and tolerance \(10^{-5}\). Record all iterates and the stopping step.
  • Apply the secant method to \(f(x)=x^3+x-1\) with \(p_0=0\) and \(p_1=1\) for four steps.
  • Compare the three methods for this function: which converged fastest? Which is safest? Justify with specific evidence from your computations.
  • A student reports Newton's method gave \(p_3=0.6823\) for a root of \(f(x)=x^3+x-1\). Verify this by computing \(f(0.6823)\) and comparing with the bisection error bound from the second problem above. Is the answer trustworthy?
Concept Mastery
  • I can explain the Intermediate Value Theorem and use it to justify a bracket.
  • I can compute the bisection error bound and determine the step count needed for a given tolerance.
  • I can verify a fixed-point convergence condition using \(g'(x)\) before iterating.
  • I can apply Newton's method and explain the tangent-line geometry.
  • I can apply the secant method and explain why it does not require a derivative formula.
  • I can combine stopping criteria and explain why one criterion alone may be misleading.
  • I can identify at least three failure modes of Newton's method and how to detect them.
Bridge Note — Chapter 4
Interpolation and Polynomial Approximation

Chapter 3 solved the problem of finding where a function is zero. Chapter 4 addresses a different question: given function values at a set of known points, how do we estimate the function at other points? This is the problem of interpolation. We will construct polynomial functions that pass exactly through specified data, study the Lagrange and Newton divided-difference forms, analyze the error in interpolation, and encounter Runge's phenomenon — a cautionary example showing that passing exactly through more points does not always mean better approximation. The ideas of error analysis and convergence studied here will reappear in a new form.

Root-finding ideas also recur later: in Chapter 9, the Jacobi and Gauss-Seidel iterations are fixed-point methods applied to linear systems; in Chapter 11, Newton's method for optimization is the same algorithm applied to \(f'(x)=0\); and in Chapter 12, ODE solvers use Newton-like corrections to propagate solutions forward in time.