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tutorials/tutorial4_root_finding.ipynb
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tutorials/tutorial4_root_finding.ipynb
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"cells": [
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{
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"cell_type": "markdown",
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"id": "3c51341a",
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"metadata": {},
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"source": [
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"# Tutorial 4: Root-Finding Methods\n",
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"---\n",
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"In this tutorial, we study classical **root-finding algorithms** for nonlinear equations. We will:\n",
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"\n",
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"- Define the root-finding problem mathematically\n",
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"- Derive several algorithms (bisection, fixed-point, Newton, secant)\n",
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"- Discuss convergence conditions and error behavior\n",
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"- Compare methods with worked examples using the `numethods` package.\n"
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]
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},
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{
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"cell_type": "markdown",
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"id": "2999aa4f",
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"metadata": {},
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"source": [
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"## 1. Problem Setup and Notation\n",
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"\n",
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"We seek to solve a nonlinear scalar equation:\n",
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"$$\n",
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"f(x) = 0, \\quad f: \\mathbb{R} \\to \\mathbb{R}.\n",
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"$$\n",
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"with a continuously differentiable function.\n",
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"\n",
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"\n",
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"### Root, residual, and error\n",
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"- A **root** $r$ satisfies $f(r)=0$.\n",
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"- **Absolute error:** $(e_k = |x_k - x^\\star|)$.\n",
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"- **Residual:** $(r_k = |f(x_k)|)$. \n",
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"Note that small residual does not always imply small error.\n",
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"\n",
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"### Multiplicity\n",
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"A root $r$ has **multiplicity** $m$ if\n",
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"$$\n",
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"f(r) = f'(r) = \\dots = f^{(m-1)}(r) = 0, \\quad f^{(m)}(r) \\neq 0.\n",
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"$$\n",
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"If $x^\\star$ satisfies $f(x^\\star)=0$ and $f'(x^\\star)\\ne 0$, we say the root is **simple** (multiplicity 1).\n",
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"\n",
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"If $f'(x^\\star)=\\cdots=f^{(m-1)}(x^\\star)=0$ and $f^{(m)}(x^\\star)\\ne 0$, we say the root has **multiplicity** (m).\n",
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"\n",
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"- **Simple roots** ($m=1$): most methods converge rapidly.\n",
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"- **Multiple roots** ($m>1$): convergence often slows.\n"
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]
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},
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{
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"cell_type": "markdown",
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"id": "b610cea2",
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"metadata": {},
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"source": [
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"## 2. Bisection Method\n",
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"\n",
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"**Assumption (Intermediate Value Theorem):** If f is continuous on ([a,b]) and (f(a),f(b) < 0),\n",
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"then there exists $x^\\star$ in (a,b) with $f(x^\\star)=0$.\n",
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"\n",
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"- Assumes $f$ is continuous on $[a,b]$ with $f(a)f(b)<0$.\n",
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"- Repeatedly bisect interval and select subinterval containing the root.\n",
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"\n",
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"**Iteration:**\n",
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"$$\n",
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"c_k = \\frac{a_k+b_k}{2}, \\quad f(c_k).\n",
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"$$\n",
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"\n",
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"**Error bound:** interval length halves each step:\n",
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"$$\n",
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"|c_k-r| \\le \\frac{b-a}{2^k}.\n",
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"$$\n",
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"- Convergence: **linear**, guaranteed.\n"
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]
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},
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{
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"cell_type": "markdown",
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"id": "3be4a510",
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"metadata": {},
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"source": [
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"## 3. Fixed-Point Iteration\n",
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"- Rewrite equation as $x=g(x)$.\n",
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"- Iterate:\n",
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"$$\n",
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"x_{k+1} = g(x_k).\n",
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"$$\n",
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"\n",
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"**Convergence theorem (Banach fixed-point):** If g is continuously differentiable near $(x^\\star)$ and\n",
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"$$\n",
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"|g'(x_\\star)| < 1,\n",
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"$$\n",
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"then for initial guesses $x_0$ sufficiently close to $x^\\star$, the iterates converge **linearly** to $x^\\star$ with asymptotic rate $|g'(x^\\star)|$.\n",
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"\n",
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"**Choice of g.** Different rearrangements yield different g's with different convergence properties.\n",
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"A poor choice (with $(|g'|\\ge 1))$ can diverge.\n",
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"\n",
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"- Rate: linear with factor $|g'(r)|$.\n"
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]
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},
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{
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"cell_type": "markdown",
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"id": "40e66a30",
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"metadata": {},
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"source": [
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"## 4. Newton’s Method\n",
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"From Taylor expansion:\n",
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"$$\n",
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"f(x) \\approx f(x_k) + f'(x_k)(x-x_k).\n",
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"$$\n",
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"Set $f(x)=0$ to solve for next iterate:\n",
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"$$\n",
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"x_{k+1} = x_k - \\frac{f(x_k)}{f'(x_k)}.\n",
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"$$\n",
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"\n",
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"- **Quadratic convergence** for simple roots.\n",
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"- For multiple roots: drops to linear.\n",
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"- Requires derivative, sensitive to initial guess.\n"
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]
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},
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{
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"cell_type": "markdown",
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"id": "84888305",
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"metadata": {},
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"source": [
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"## 5. Secant Method\n",
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"- Avoids derivative by approximating slope with finite difference:\n",
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"$$\n",
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"x_{k+1} = x_k - f(x_k) \\frac{x_k - x_{k-1}}{f(x_k)-f(x_{k-1})}.\n",
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"$$\n",
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"\n",
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"- Convergence order: $\\approx 1.618$ (superlinear).\n",
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"- More efficient than Newton if derivative expensive.\n"
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]
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},
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{
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"cell_type": "markdown",
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"id": "7ed8a0b5",
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"metadata": {},
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"source": [
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"## 6. Stopping Criteria\n",
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"We stop iteration when:\n",
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"- $|f(x_k)| \\le \\varepsilon$ (residual small), or\n",
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"- $|x_{k+1}-x_k| \\le \\varepsilon(1+|x_{k+1}|)$ (relative step small).\n"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 1,
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"id": "4c9f93d9",
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"metadata": {},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"Bisection root: 1.4142135605216026\n",
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"Newton root: 1.4142135623730951\n",
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"Secant root: 1.414213562373095\n",
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"Fixed-point failed: Fixed-point iteration did not converge\n"
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]
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}
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],
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"source": [
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"from numethods import Bisection, FixedPoint, NewtonRoot, Secant\n",
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"import math\n",
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"\n",
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"# Example function: f(x) = x^2 - 2\n",
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"f = lambda x: x**2 - 2\n",
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"df = lambda x: 2*x\n",
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"\n",
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"# Bisection\n",
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"bisect = Bisection(f, 0, 2, tol=1e-8)\n",
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"root_b = bisect.solve()\n",
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"print('Bisection root:', root_b)\n",
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"\n",
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"# Newton\n",
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"newton = NewtonRoot(f, df, 1.0, tol=1e-12)\n",
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"root_n = newton.solve()\n",
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"print('Newton root:', root_n)\n",
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"\n",
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"# Secant\n",
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"sec = Secant(f, 0, 2, tol=1e-12)\n",
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"root_s = sec.solve()\n",
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"print('Secant root:', root_s)\n",
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"\n",
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"# Fixed point: g(x)=sqrt(2) ~ rewriting\n",
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"g = lambda x: (2/x) # not always convergent, but demonstrates\n",
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"try:\n",
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" fp = FixedPoint(g, 1.0, tol=1e-8)\n",
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" root_fp = fp.solve()\n",
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" print('Fixed-point root:', root_fp)\n",
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"except Exception as e:\n",
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" print('Fixed-point failed:', e)\n"
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]
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},
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{
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"cell_type": "markdown",
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"id": "833b538c",
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"metadata": {},
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"source": [
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"## 7. Comparison of Methods\n",
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"| Method | Requires derivative | Convergence rate | Guarantee? |\n",
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"|--------|---------------------|------------------|------------|\n",
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"| Bisection | No | Linear | Yes (if sign change) |\n",
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"| Fixed-Point | No | Linear | Not always |\n",
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"| Newton | Yes | Quadratic | Locally (good guess) |\n",
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"| Secant | No | ~1.618 (superlinear) | Locally (good guess) |\n"
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]
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},
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{
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"cell_type": "markdown",
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"id": "0d372aa0",
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"metadata": {},
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"source": [
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"## 8. Exercises\n",
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"1. Apply all four methods to $f(x)=\\cos x - x$.\n",
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"2. Try Newton’s method on $f(x)=(x-1)^2$ and compare convergence rate with $f(x)=x^2-2$.\n",
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"3. Modify Secant to stop if denominator too small.\n"
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]
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}
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],
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