Newton’s Method for MATH 140

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Newton's Method appears occasionally on MATH 140 finals. Know the iteration formula x_{n+1} = x_n - f(x_n)/f'(x_n).

Lesson

What is Newton's Method?

Newton's Method (also called Newton-Raphson) is a technique for finding roots of equations — that is, finding where $f(x) = 0$.

Isaac Newton developed this iterative approach: start with a guess, then repeatedly improve it using calculus until you're close enough to the actual root.

The Big Idea: Use the tangent line at your current guess to find a better guess. Repeat until you converge to the root.


Why Do We Need This?

Some equations can't be solved algebraically. For example:

  • $x^5 - x - 1 = 0$ — no formula exists for 5th-degree polynomials!
  • $\cos(x) = x$ — mixing trig and polynomials
  • $e^x = 3x$ — mixing exponentials and polynomials

Newton's Method gives us a way to find approximate solutions to any precision we want.


The Formula

Starting from an initial guess $x_0$, each new approximation is:

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

What This Means

  1. Evaluate $f(x_n)$ — how far is your current guess from being a root?
  2. Evaluate $f'(x_n)$ — what's the slope of the tangent line there?
  3. Compute the ratio and subtract — this gives you the next (better) guess

Key Variables

  • $x_n$ — your current approximation (the $n$-th guess)
  • $x_{n+1}$ — your next approximation (hopefully closer to the root)
  • $f(x_n)$ — the function value at your current guess
  • $f'(x_n)$ — the derivative (slope) at your current guess

The Geometric Intuition

Here's what's actually happening:

  1. You're at a point $(x_n, f(x_n))$ on the curve
  2. Draw the tangent line at that point
  3. Find where the tangent line crosses the x-axis
  4. That x-intercept becomes your new guess $x_{n+1}$

The tangent line is a good local approximation of the curve, so its root is close to the curve's root!

Newton's method
Newton's method

Starting from $x_0 = b$, watch how each tangent line leads us closer to the root $a$.


Example 1: Finding $\sqrt{2}$

Use Newton's Method to approximate $\sqrt{2}$, starting with $x_0 = 1$.

Step 1: Set up the equation

We want $x = \sqrt{2}$, which means $x^2 = 2$, so $x^2 - 2 = 0$.

Let $f(x) = x^2 - 2$, so $f'(x) = 2x$.

Step 2: Write the iteration formula

$$x_{n+1} = x_n - \frac{x_n^2 - 2}{2x_n}$$

Step 3: Iterate!

$x_0 = 1$

$x_1 = 1 - \frac{1^2 - 2}{2(1)} = 1 - \frac{-1}{2} = 1.5$

$x_2 = 1.5 - \frac{1.5^2 - 2}{2(1.5)} = 1.5 - \frac{0.25}{3} = 1.5 - 0.0833... = 1.4167$

$x_3 = 1.4167 - \frac{1.4167^2 - 2}{2(1.4167)} \approx 1.4142$

After just 3 iterations, we have $\sqrt{2} \approx 1.4142$ (actual: $1.41421356...$)

$$\boxed{x_3 \approx 1.4142}$$


Example 2: Solving $\cos(x) = x$

Find a solution to $\cos(x) = x$ using Newton's Method with $x_0 = 0.5$.

Step 1: Rewrite as $f(x) = 0$

$\cos(x) = x \implies \cos(x) - x = 0$

Let $f(x) = \cos(x) - x$, so $f'(x) = -\sin(x) - 1$.

Step 2: Write the iteration formula

$$x_{n+1} = x_n - \frac{\cos(x_n) - x_n}{-\sin(x_n) - 1}$$

Step 3: Iterate!

$x_0 = 0.5$

$f(0.5) = \cos(0.5) - 0.5 \approx 0.8776 - 0.5 = 0.3776$

$f'(0.5) = -\sin(0.5) - 1 \approx -0.4794 - 1 = -1.4794$

$x_1 = 0.5 - \frac{0.3776}{-1.4794} = 0.5 + 0.2552 = 0.7552$

$x_2 \approx 0.7391$

$x_3 \approx 0.7391$ (converged!)

$$\boxed{x \approx 0.7391}$$

This is the Dottie number — a fun mathematical constant!


Example 3: Finding a Root of a Polynomial

Use Newton's Method to find a root of $f(x) = x^3 - 2x - 5$, starting with $x_0 = 2$.

Step 1: Find the derivative

$f(x) = x^3 - 2x - 5$

$f'(x) = 3x^2 - 2$

Step 2: Write the iteration formula

$$x_{n+1} = x_n - \frac{x_n^3 - 2x_n - 5}{3x_n^2 - 2}$$

Step 3: Iterate!

$x_0 = 2$

$f(2) = 8 - 4 - 5 = -1$

$f'(2) = 12 - 2 = 10$

$x_1 = 2 - \frac{-1}{10} = 2 + 0.1 = 2.1$

$x_2 = 2.1 - \frac{2.1^3 - 2(2.1) - 5}{3(2.1)^2 - 2} = 2.1 - \frac{0.061}{11.23} \approx 2.0946$

$x_3 \approx 2.0946$ (converged!)

$$\boxed{x \approx 2.0946}$$


Example 4: One Iteration Only

Given $f(x) = e^x - 4x$ and $x_0 = 2$, find $x_1$ using Newton's Method.

Step 1: Find $f(x_0)$ and $f'(x_0)$

$f(x) = e^x - 4x \implies f'(x) = e^x - 4$

$f(2) = e^2 - 8 \approx 7.389 - 8 = -0.611$

$f'(2) = e^2 - 4 \approx 7.389 - 4 = 3.389$

Step 2: Apply the formula

$$x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 2 - \frac{-0.611}{3.389} = 2 + 0.180 = 2.180$$

$$\boxed{x_1 \approx 2.180}$$


When Newton's Method Fails

Newton's Method doesn't always work! Watch out for:

  1. Bad starting point — if $x_0$ is too far from the root, you might converge to a different root or not converge at all

  2. Derivative is zero — if $f'(x_n) = 0$, you're dividing by zero!

  3. Oscillation — sometimes the method bounces back and forth without converging

  4. Divergence — the approximations can get further from the root instead of closer

Pro tip: Choose $x_0$ by sketching the graph or using a calculator to find roughly where the function crosses zero.


Tips for Newton's Method Problems

  1. Always find $f'(x)$ first — you'll need it for every iteration
  2. Set up the iteration formula before plugging in numbers
  3. Keep track of your iterations — organize as $x_0, x_1, x_2, ...$
  4. Watch for convergence — when successive values agree to the required decimal places, you're done
  5. Use a calculator — the arithmetic can get messy, and that's okay!

Common Mistakes and Misunderstandings

❌ Mistake: Forgetting the minus sign

Wrong: $x_{n+1} = x_n + \frac{f(x_n)}{f'(x_n)}$

Why it's wrong: The formula has a subtraction, not addition.

Correct: $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$


❌ Mistake: Not rewriting the equation as $f(x) = 0$

Wrong: Using $f(x) = \cos(x)$ when solving $\cos(x) = x$.

Why it's wrong: Newton's Method finds roots (where $f(x) = 0$). You need $f(x) = \cos(x) - x$.

Correct: Rearrange to $\cos(x) - x = 0$, then use $f(x) = \cos(x) - x$.


❌ Mistake: Using the wrong derivative

Wrong: Forgetting chain rule or making arithmetic errors with $f'(x)$.

Why it's wrong: An incorrect derivative will give you wrong iterations.

Correct: Double-check your derivative before starting iterations.

Formulas & Reference

Newton's Method Iteration

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

Starting from an initial guess x₀, each iteration uses the tangent line to find a better approximation of the root. Repeat until successive values agree to the desired precision.

Variables:
$x_n$:
Your current approximation (the n-th guess)
$x_{n+1}$:
Your next approximation (hopefully closer to the root)
$f(x_n)$:
The function value at your current guess
$f'(x_n)$:
The derivative (slope) at your current guess
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