Applying Newton's Method
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
The previous lesson derived the Newton iteration formula $x_{n+1} = x_n - \dfrac{f(x_n)}{f'(x_n)}$ and computed one step. Running multiple steps produces a sequence of iterates. That sequence sometimes converges to a root, sometimes does not, and a stopping test tells you when you have enough accuracy.
In practice, Newton’s Method is fast when it works (a few steps give many decimal digits of accuracy) but can fail spectacularly when it does not.
Quick Reference
Newton’s Method recurrence: $x_{n+1} = x_n - \dfrac{f(x_n)}{f'(x_n)}$.
Stopping criterion: Stop when $|x_{n+1} - x_n| < \varepsilon$ (two successive iterates agree to within the desired tolerance $\varepsilon$).
Convergence: Under favorable conditions, the number of correct decimal digits roughly doubles with each step (quadratic convergence).
Failure: The method can fail if $f'(x_n) = 0$ (undefined step), if the starting value is too far from the root, or if the function has pathological behavior near the root.
Key Concepts
1. Running Multiple Iterations
After computing $x_2$ from $x_1$, plug $x_2$ back in to get $x_3$: $$x_3 = x_2 - \frac{f(x_2)}{f'(x_2)}.$$
Repeat until successive iterates agree to the desired number of decimal places. On a calculator, keep all digits in the memory -- rounding intermediate results can prevent the method from converging.
Table format for tracking:
| $n$ | $x_n$ | $f(x_n)$ | $f'(x_n)$ | $x_{n+1}$ |
|---|---|---|---|---|
| 1 | initial guess | ... | ... | ... |
| 2 | ... | ... | ... | ... |
2. Convergence: Why It Is Usually Fast
Near a simple root $r$ (where $f(r) = 0$ and $f'(r) \neq 0$), the error $e_n = x_n - r$ satisfies approximately: $$|e_{n+1}| \approx \frac{|f''(r)|}{2|f'(r)|} e_n^2.$$
This is quadratic convergence: each step squares the error. If $x_3$ has error $10^{-4}$, then $x_4$ has error $\approx 10^{-8}$ and $x_5$ has error $\approx 10^{-16}$. This is much faster than linear convergence (where each step multiplies the error by a fixed factor less than 1).
3. When Newton’s Method Fails
Case 1 -- Division by zero. If $f'(x_n) = 0$, the next iterate is undefined. This happens if the current iterate is at a local extremum or at a point where $f'$ accidentally equals zero.
Case 2 -- Divergence. If the starting value is far from the root or the function has a large second derivative relative to the first derivative, the iterates can diverge (grow without bound) or cycle.
Classic example of cycling: For $f(x) = \sqrt[3]{x}$ (cube root), if $x_1 = 1$: $f'(x) = \frac{1}{3}x^{-2/3}$, and the iteration produces $x_2 = -2$, $x_3 = 4$, alternating in sign with increasing magnitude. The method diverges even though the root $x=0$ exists.
Case 3 -- Converging to the wrong root. If there are multiple roots and the starting point is not close to the desired one, the method may converge to a different root. Always check that $f(x_n) \approx 0$ at the limiting value.
4. Stopping Criterion
Two practical criteria:
- Absolute change: Stop when $|x_{n+1} - x_n| < \varepsilon$.
- Function value: Stop when $|f(x_n)| < \delta$.
Use both for reliability. The absolute change criterion can fail for nearly flat functions; the function value criterion can fail near points where $f$ is very small but $f' = 0$ (saddle).
Worked Example
Use Newton’s Method to find $\sqrt{2}$ to 4 decimal places, starting from $x_1 = 1$.
Set $f(x) = x^2 - 2 = 0$. $f'(x) = 2x$.
$x_{n+1} = x_n - \dfrac{x_n^2-2}{2x_n} = \dfrac{x_n^2+2}{2x_n}$ (after simplifying).
| $n$ | $x_n$ | $x_{n+1}$ |
|---|---|---|
| 1 | 1 | $(1+2)/2 = 1.5$ |
| 2 | 1.5 | $(2.25+2)/3 = 1.4167$ |
| 3 | 1.4167 | $(2.0069+2)/2.8334 \approx 1.41422$ |
| 4 | 1.41422 | $\approx 1.41421$ |
After 4 steps, $x_4 \approx 1.41421$. The true value is $\sqrt{2} \approx 1.41421356\ldots$
$|x_4 - x_3| = |1.41421 - 1.41422| < 0.00001$, so 4 decimal places are confirmed.
Boxed answer: $\sqrt{2} \approx 1.41421$ (4 decimal places, reached in 4 iterations from $x_1 = 1$).
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Rounding intermediate results | Rounding $x_2 = 1.4167$ to $1.4$ before computing $x_3$ | Carry full precision through all iterations; round only at the end |
| Stopping too early | Stopping after one iteration | The stopping criterion is $|x_{n+1}-x_n| < \varepsilon$; check this each step |
| Starting at a critical point of $f$ | Choosing $x_1$ where $f'(x_1) = 0$ | Check that $f'(x_1) \neq 0$ before starting; pick a different initial value if needed |
| Not verifying convergence to the right root | Reporting the limit without checking $f(x_n) \approx 0$ | Confirm $|f(x_n)|$ is small at the reported root |
Common Misconceptions
once two successive iterates agree to several decimal places, those digits are guaranteed to be correct.
This is the concept-image-conflicts-definition error about what the stopping criterion certifies. The criterion $|x_{n+1} - x_n| < \varepsilon$ confirms that the iterates have stabilized, but it does not rule out convergence to the wrong root or to a spurious fixed point. The final iterate should also be checked by computing $|f(x_n)|$; if that value is not small, the sequence has not converged to a root of $f$. On problems with multiple roots, verifying $f(x_n) \approx 0$ is essential.
intermediate results can be rounded to fewer digits during the iteration to save work.
This is the concept-image-conflicts-definition error about how precision accumulates. Newton’s Method achieves quadratic convergence only when full precision is carried through each step. Rounding $x_2 = 1.41667$ to $1.4$ before computing $x_3$ introduces an error that the method then amplifies rather than reduces, potentially preventing convergence to the desired accuracy. All intermediate values should be kept at full calculator or computer precision; rounding occurs only in the final reported answer.
Leveled Practice
Level 1 -- Run Two Steps
Problem 1. Use Newton’s Method with $f(x) = x^3 - 3$ (root at $\sqrt[3]{3}$), starting from $x_1 = 1.5$. Compute $x_2$ and $x_3$.
Show answer
$f'(x) = 3x^2$.
$x_2 = 1.5 - \dfrac{1.5^3-3}{3\cdot 1.5^2} = 1.5 - \dfrac{0.375}{6.75} = 1.5 - 0.0556 \approx 1.4444$.
$f(1.4444) = 1.4444^3 - 3 \approx 3.0135 - 3 = 0.0135$. $f'(1.4444) = 3(1.4444)^2 \approx 6.263$.
$x_3 = 1.4444 - 0.0135/6.263 \approx 1.4444 - 0.00216 \approx 1.4422$.
True value: $\sqrt[3]{3} \approx 1.4422$. Confirmed.
Boxed answers: $x_2 \approx 1.4444$; $x_3 \approx 1.4422$.
Level 2 -- Identify a Failure
Problem 2. Explain why Newton’s Method with $f(x) = x^{1/3}$ and $x_1 = 1$ fails to converge.
Show answer
$f'(x) = \frac{1}{3}x^{-2/3}$.
$x_2 = 1 - \dfrac{1^{1/3}}{\frac{1}{3}\cdot 1^{-2/3}} = 1 - \dfrac{1}{1/3} = 1 - 3 = -2$.
$x_3 = -2 - \dfrac{(-2)^{1/3}}{\frac{1}{3}(-2)^{-2/3}} = -2 - \dfrac{-2^{1/3}}{2^{-2/3}/3} = -2 - (-3\cdot 2^{1/3}\cdot 2^{2/3}) = -2 + 3\cdot 2 = 4$.
The iterates alternate in sign ($1, -2, 4, -8, \ldots$) and diverge. The root is at $x = 0$, but the tangent lines near $x = 0$ for $f(x) = x^{1/3}$ have nearly vertical slope (not zero slope) yet the iteration jumps past the root each time due to the concavity structure.
This is the classic failure mode: the function $f(x) = x^{1/3}$ has an inflection point at the root, violating the condition that $f'(r) \neq 0$ (here $f'(0)$ is undefined/infinite). Newton’s Method should not be applied to this function at the root.
Mastery Checklist
Mental Model
Newton’s Method is a self-correcting process. Each iteration replaces the current approximation with a better one, using the tangent line to guess where the function crosses zero. When it works, the errors shrink quadratically -- each step roughly doubles the number of correct digits. The method fails when the tangent line points in the wrong direction (bad starting point) or does not exist (zero derivative). The key discipline is to run the iteration systematically in a table, carry full precision, and stop only when the successive-iterate criterion is satisfied.
Connections
Looking back
- Newton’s Method formula (Section 3.8): Derived the iteration $x_{n+1} = x_n - f(x_n)/f'(x_n)$; this lesson applies it repeatedly.
Looking ahead
This is the final topic in Chapter 3. Newton’s Method is revisited in numerical analysis courses and is the basis of most root-finding algorithms in scientific computing.
Back to Newton’s Method Formula | Back to Chapter 3 Overview