You know Newton's Method: $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$. Using the language of order of convergence, we can now say exactly how fast it is near a root, and why it sometimes fails.
Key idea
Near a simple root $r$ (one with $f'(r) \ne 0$), Newton's method converges quadratically: $$e_{n+1} \approx \left|\frac{f''(r)}{2f'(r)}\right|e_n^{2}.$$ Far from a root it can wander, cycle or blow up, and at a multiple root it slows to linear convergence.
Why it works
Expand $f$ about $x_n$ and evaluate at the root: $0 = f(r) = f(x_n) + f'(x_n)(r - x_n) + \frac{f''(\xi)}{2}(r - x_n)^{2}$. Divide by $f'(x_n)$ and use $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$: $$x_{n+1} - r = \frac{f''(\xi)}{2f'(x_n)}(x_n - r)^{2}.$$ The square is the quadratic convergence. The derivation needs $f'(x_n)$ away from $0$ and $x_n$ close enough to $r$ that the error term is small.
Worked example: Digits doubling
Run Newton's method on $f(x) = x^{2} - 2$ from $x_0 = 1$.
- $x_1 = 1.5$, $x_2 = 1.41666\ldots$, $x_3 = 1.4142157$, $x_4 = 1.41421356237$.
- Errors: $8.6 \times 10^{-2}$, $2.5 \times 10^{-3}$, $2.1 \times 10^{-6}$, $1.6 \times 10^{-12}$.
- $C = \frac{f''(\sqrt{2})}{2f'(\sqrt{2})} = \frac{2}{4\sqrt{2}} \approx 0.354$, and $0.354 \times (2.5 \times 10^{-3})^{2} \approx 2.2 \times 10^{-6}$: the formula predicts the next error.
Answer: the correct digits go 1, 2, 5, 11
Worked example: A cycle
Run Newton's method on $f(x) = x^{3} - 2x + 2$ from $x_0 = 0$.
- $f'(x) = 3x^{2} - 2$. $x_1 = 0 - \frac{2}{-2} = 1$.
- $x_2 = 1 - \frac{f(1)}{f'(1)} = 1 - \frac{1}{1} = 0$.
- Back at the start: the iterates cycle $0, 1, 0, 1, \ldots$ forever. The only real root is near $-1.77$, and from $x_0 = -2$ the method finds it in a few steps.
Answer: a 2-cycle, never converging
Worked example: A double root
Run Newton's method on $f(x) = (x - 1)^{2}$ from $x_0 = 2$.
- $\frac{f(x)}{f'(x)} = \frac{(x - 1)^{2}}{2(x - 1)} = \frac{x - 1}{2}$, so $x_{n+1} - 1 = \frac{1}{2}(x_n - 1)$.
- Iterates: $1.5$, $1.25$, $1.125, \ldots$ The error only halves: linear, like bisection.
- At a root of multiplicity $m$ the ratio is $\frac{m - 1}{m}$; the step $x - m\frac{f(x)}{f'(x)}$ restores quadratic convergence when $m$ is known.
Answer: linear, with ratio ½
The local promise matters: quadratic convergence starts only once the error is small enough that $Ce_n < 1$. Before that, Newton's method can behave in many ways. The common failures:
- A start near a point where $f'(x) \approx 0$: the nearly flat tangent throws the next iterate far away.
- A start in the wrong basin: the iterates may cycle or head to a different root.
- A multiple root: convergence slows to linear.
- A function like $\arctan x$: from $|x_0| \gtrsim 1.39$ the iterates grow in size and diverge.
Common mistake
Assuming Newton's method always converges. Its fast convergence is a local promise: only from a good enough start near a simple root. In practice, check that $|f(x_n)|$ decreases, cap the number of steps, and fall back to a safe method such as The Bisection Method when it does not.