Newton's Method Formula
Textbook Reference
| Primary source | OpenStax Calculus Volume 1, Section 4.9: “Newton’s Method” |
| Direct link | https://openstax.org/books/calculus-volume-1/pages/4-9-newtons-method |
| Textbook used in class | Stewart, Calculus, Section 3.8: “Newton’s Method” |
Opening Scenario
Many equations have no exact algebraic solution. The equation $x^3 = 2$ can be solved: $x = 2^{1/3} = \sqrt[3]{2}$. But can you compute $\sqrt[3]{2}$ exactly as a decimal? No -- it is irrational: $1.2599210...$, going on forever.
Newton’s Method is an algorithm for finding a decimal approximation to a root of $f(x) = 0$ to as many digits of accuracy as you want. It uses the tangent line to the curve as a guide: from a rough starting guess, follow the tangent line to where it crosses the $x$-axis. That crossing point is a better approximation, usually much better. Repeat.
Quick Reference
Newton’s Method. Starting from an initial approximation $x_1$, generate a sequence of better approximations by: $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}, \quad n = 1, 2, 3, \ldots$$
Each $x_n$ is called an iterate. Under favorable conditions, the sequence converges rapidly to a root of $f(x) = 0$.
Key Concepts
1. Where the Formula Comes From
Start at a guess $x_1$. Draw the tangent line to the curve $y = f(x)$ at the point $(x_1, f(x_1))$.
The tangent line has slope $f'(x_1)$ and passes through $(x_1, f(x_1))$: $$y - f(x_1) = f'(x_1)(x - x_1).$$
Find where this tangent line crosses the $x$-axis (set $y = 0$): $$-f(x_1) = f'(x_1)(x - x_1) \Rightarrow x = x_1 - \frac{f(x_1)}{f'(x_1)}.$$
Call this crossing point $x_2$. Now repeat the process from $x_2$: draw the tangent there, find where it crosses the $x$-axis, and call that $x_3$. The general step is: $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}.$$
The idea is that the tangent line is a good local linear approximation to the curve, so where the tangent crosses the axis should be close to where the curve crosses the axis.
2. A First Iteration
Example 1. Use Newton’s Method with $x_1 = 2$ to find one iteration approximating $\sqrt[3]{2}$.
Rewrite as $f(x) = x^3 - 2 = 0$.
$f'(x) = 3x^2$.
$x_2 = x_1 - \dfrac{f(x_1)}{f'(x_1)} = 2 - \dfrac{2^3 - 2}{3 \cdot 2^2} = 2 - \dfrac{6}{12} = 2 - 0.5 = 1.5$.
So after one step, the approximation is $x_2 = 1.5$.
Check: $1.5^3 = 3.375 \neq 2$. The approximation is not exact, but $f(1.5) = 1.375$ is smaller than $f(2) = 6$, so $x_2$ is closer to the root than $x_1$. Applying Newton’s Method again from $x_2 = 1.5$ gives a still better approximation.
Boxed answer: First Newton iterate: $x_2 = 1.5$.
Recall the true value: $\sqrt[3]{2} \approx 1.2599$. So $x_2 = 1.5$ has error about $0.24$. One more iteration will reduce that error dramatically.
3. Choosing a Good Starting Point
The formula works best when $x_1$ is close to the root and the function is well-behaved nearby. A poor choice of $x_1$ can cause the method to fail or converge to the wrong root.
Reasonable starting-point strategies:
- Bracket the root using the Intermediate Value Theorem: if $f(a) < 0 < f(b)$, a root lies in $(a, b)$. Start anywhere in $(a, b)$.
- Use any rough estimate (integer, endpoint, midpoint of the bracket).
- For equations involving square roots or cube roots, the nearest perfect power is usually a good start.
What to avoid:
- Starting at a point where $f'(x_1) = 0$ (the formula is undefined -- division by zero).
- Starting at an inflection point or saddle point where the tangent parallels the $x$-axis.
4. The Tangent Line as a Model
A useful way to understand each step: Newton’s Method replaces the curve $f$ locally with its tangent line (the best linear approximation at $x_n$), then solves the linear equation (finding where the line crosses zero) to get the next approximation.
Since the tangent line agrees with the curve to first order (same value and same slope at $x_n$), the root of the tangent line is usually close to the root of the curve. The approximation improves rapidly -- typically, the number of correct decimal digits roughly doubles at each step.
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Forgetting to compute $f'(x_n)$ | Using the same derivative value for all iterations | Recompute $f(x_n)$ and $f'(x_n)$ at the current iterate $x_n$, not the initial one |
| Starting where $f'(x_1) = 0$ | Choosing $x_1 = 0$ for $f(x) = x^3$ | The formula divides by $f'(x_1)$; if that is zero, the method fails |
| Computing $f'(x_n)$ for the wrong function | Using the derivative of $x^3$ for $f(x) = x^3 - 2$ | $f(x) = x^3 - 2$, so $f'(x) = 3x^2$; the formula requires $f'$, not the derivative of the original equation’s left side separately |
Common Misconceptions
the Newton iteration uses the same derivative value $f'(x_1)$ for every step.
This is the concept-image-conflicts-definition error about what the recurrence formula requires. Each iterate $x_{n+1}$ is computed from the tangent line at the current point $x_n$, so both $f(x_n)$ and $f'(x_n)$ must be recomputed at the updated value of $x_n$ before each step. Using the initial derivative $f'(x_1)$ for all subsequent steps produces a different sequence that generally does not converge to the root and loses the quadratic convergence that makes the method powerful.
Newton’s Method always converges to a root when one exists.
This is the concept-image-conflicts-definition error about convergence conditions. The method fails or diverges when $f'(x_n) = 0$ at any iterate (division by zero in the formula), when the starting point is far from the desired root, or when the function has a root where the derivative is undefined or zero. For $f(x) = x^{1/3}$, starting from $x_1 = 1$ produces the diverging sequence $1, -2, 4, -8, \ldots$ even though the root $x = 0$ exists.
Leveled Practice
Level 1 -- Direct Application
Problem 1. For $f(x) = x^2 - 5$ (a root of this is $\sqrt{5}$), compute $x_2$ starting from $x_1 = 2$.
Show answer
$f'(x) = 2x$.
$x_2 = 2 - \dfrac{4 - 5}{4} = 2 - \dfrac{-1}{4} = 2 + 0.25 = 2.25$.
Check: $2.25^2 = 5.0625$. The true $\sqrt{5} \approx 2.2361$. Error reduced from $0.236$ to about $0.014$.
Boxed answer: $x_2 = 2.25$.
Problem 2. Write the Newton iteration formula for finding a root of $f(x) = x^3 - x - 1$.
Show answer
$f'(x) = 3x^2 - 1$.
$x_{n+1} = x_n - \dfrac{x_n^3 - x_n - 1}{3x_n^2 - 1}$.
Level 2 -- Multiple Steps
Problem 3. The equation $\cos x = x$ has a solution near $x = 0.7$. Write the Newton iteration for $f(x) = \cos x - x$ and compute $x_2$ starting from $x_1 = 0.7$. (Use $\cos(0.7) \approx 0.7648$ and $\sin(0.7) \approx 0.6442$.)
Show answer
$f(x) = \cos x - x$. $f'(x) = -\sin x - 1$.
$f(0.7) = \cos(0.7) - 0.7 \approx 0.7648 - 0.7 = 0.0648$.
$f'(0.7) = -\sin(0.7) - 1 \approx -0.6442 - 1 = -1.6442$.
$x_2 = 0.7 - \dfrac{0.0648}{-1.6442} = 0.7 + 0.0394 \approx 0.7394$.
Boxed answer: $x_2 \approx 0.739$. (The true solution is the Dottie number $\approx 0.7391$.)
Mastery Checklist
Mental Model
Newton’s Method is the follow-the-tangent strategy. At each step, you are standing at a point on the curve. You cannot see the root directly, but you can see the tangent line. You walk along the tangent to where it hits the ground ($y = 0$), then stand at that point and repeat.
Because the tangent line is the best linear approximation to the curve, following it is usually a fast path toward the root. In most cases, the number of accurate decimal digits roughly doubles each step. This is called quadratic convergence -- much faster than trial-and-error.
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