What Is Newtons Method?
Newtons Method (also called the Newton–Raphson method) is an iterative technique for finding a root of a function, a value $r$ where $f(r) = 0$. You start with a guess and repeatedly apply one rule that turns each estimate into a better one:
$$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$
Here $f'(x_n)$ is the derivative of $f$ evaluated at the current guess $x_n$. The rule needs $f$ to be differentiable near the root and needs $f'(x_n) \neq 0$ at every step, since that value sits in the denominator.
The idea behind the formula is geometric. At the point $x_n$ the curve $y = f(x)$ has a tangent line. That straight line is the best local stand-in for the curve, and a straight line is easy to solve. So instead of asking where the curve crosses the x-axis, Newton's Method asks where the tangent crosses it, and takes that crossing as the next guess.
How Do You Use Newtons Method?
The method is a short loop. Given a function $f$, its derivative $f'$, and a starting guess $x_0$:
Compute $f(x_n)$ and $f'(x_n)$.
Apply $x_{n+1} = x_n - \dfrac{f(x_n)}{f'(x_n)}$.
Stop when successive guesses agree to the accuracy you need; otherwise repeat with $x_{n+1}$.
Where does the formula come from? Write the tangent line to $y = f(x)$ at the point $x_n$. A line through $\big(x_n, f(x_n)\big)$ with slope $f'(x_n)$ is:
$$L(x) = f(x_n) + f'(x_n),(x - x_n)$$
The next guess is where this line meets the x-axis, so set $L(x) = 0$ and solve for $x$:
$$ \begin{aligned} 0 &= f(x_n) + f'(x_n),(x - x_n) \[4pt] x - x_n &= -\frac{f(x_n)}{f'(x_n)} \[4pt] x &= x_n - \frac{f(x_n)}{f'(x_n)} \end{aligned} $$
That $x$ is exactly $x_{n+1}$. The algebra and the picture say the same thing: follow the tangent to where it hits the x-axis, and stand there. Because the derivative is the slope of that tangent, the whole method is a piece of applied differential calculus. For the geometry of tangents and the lines perpendicular to them, see tangents and normals.
How Do You Approximate The Square Root Of 2 With Newtons Method?
The square root of $2$ is the positive root of $f(x) = x^2 - 2$, since $f(x) = 0$ means $x^2 = 2$. Its derivative is $f'(x) = 2x$, so the update rule becomes:
$$x_{n+1} = x_n - \frac{x_n^2 - 2}{2x_n} = \frac{1}{2}\left(x_n + \frac{2}{x_n}\right)$$
Start with $x_0 = 1.5$ and apply it.
Step 1.
$$x_1 = \frac{1}{2}\left(1.5 + \frac{2}{1.5}\right) = \frac{1}{2}\big(1.5 + 1.333333\big) = 1.416667$$
Step 2.
$$x_2 = \frac{1}{2}\left(1.416667 + \frac{2}{1.416667}\right) = \frac{1}{2}\big(1.416667 + 1.411765\big) = 1.414216$$
Step 3.
$$x_3 = \frac{1}{2}\left(1.414216 + \frac{2}{1.414216}\right) = 1.41421356$$
The true value is $\sqrt{2} = 1.41421356\ldots$, so three steps from a rough guess already pin down eight digits. Squaring the result gives $1.41421356^2 = 1.99999999$, which confirms it.
Table 1: Each step of Newtons Method for $\sqrt{2}$, with the size of $f(x_n) = x_n^2 - 2$ shrinking toward zero.
Step $n$ | Guess $x_n$ | $f(x_n) = x_n^2 - 2$ | Correct digits |
|---|---|---|---|
0 | 1.500000 | $+0.250000$ | 1 |
1 | 1.416667 | $+0.006945$ | 3 |
2 | 1.414216 | $+0.000007$ | 5 |
3 | 1.41421356 | $\approx 0$ | 8+ |
Look at the last column. The count of correct digits roughly doubles each step (1, then 3, then 5, then 8+). That doubling is the signature of quadratic convergence, and it is why the method is so fast once it gets going.
How Do You Solve A Cubic With Newtons Method?
Newton's Method shines when there is no clean formula to solve $f(x) = 0$. Take the cubic $f(x) = x^3 - 2x - 5$, whose derivative is $f'(x) = 3x^2 - 2$. This is the very equation Isaac Newton used to demonstrate the method. Start at $x_0 = 2$.
Step 1. $f(2) = 8 - 4 - 5 = -1$ and $f'(2) = 12 - 2 = 10$, so
$$x_1 = 2 - \frac{-1}{10} = 2.1$$
Step 2. $f(2.1) = 9.261 - 4.2 - 5 = 0.061$ and $f'(2.1) = 13.23 - 2 = 11.23$, so
$$x_2 = 2.1 - \frac{0.061}{11.23} = 2.094568$$
Step 3. Repeating once more gives
$$x_3 = 2.094551$$
Final answer: the root is $x \approx 2.094551$. Substituting back, $f(2.094551) = (2.094551)^3 - 2(2.094551) - 5 \approx 0$, which checks out. A cubic has no friendly root here, yet three tangent steps land six correct decimals.
Why Does Newtons Method Work?
The method works because near a root a smooth curve looks almost straight, and straight lines are trivial to solve. Replacing the curve by its tangent introduces a small error, and Newton's Method is a machine for shrinking that error fast.
The tangent is the best straight-line copy of the curve. Close to $x_n$, the curve and its tangent nearly coincide, so the tangent's x-intercept sits near the curve's.
The error gets squared each step. If the current error is $e_n = x_n - r$, then for a well-behaved function the next error satisfies $e_{n+1} \approx C,e_n^{,2}$ for some constant $C$. Squaring a small number makes it tiny, which is the digit-doubling seen in the $\sqrt{2}$ table.
A good start matters more than a good function. The squaring only helps once $e_n$ is already small. Far from the root the tangent can point almost anywhere, which is where the failures in the next section come from.
The condition $f'(x_n) \neq 0$ is not a technicality. Where the derivative is zero the tangent is horizontal, a flat line that never meets the x-axis, so there is no next guess to compute.
When Does Newtons Method Fail?
Newton's Method is fast, not foolproof. Being honest about the failures is part of using it well. Four things go wrong in practice.
Table 2: The four ways Newtons Method breaks, each with a verified example.
Failure mode | What goes wrong | Example |
|---|---|---|
Zero derivative | $f'(x_n) = 0$ makes the tangent horizontal; the update divides by zero | Any $f$ at a turning point, e.g. $f(x) = x^2 - 2$ started at $x_0 = 0$ |
Poor starting guess | A far-off $x_0$ sends the tangent toward the wrong root, or nowhere useful | Landing near a local max or min of $f$ |
Cycling | The guesses repeat in a loop and never settle | $f(x) = x^3 - 2x + 2$ from $x_0 = 0$: gives $0, 1, 0, 1, \ldots$ |
Divergence | The guesses grow without bound | $f(x) = x^{1/3}$: the rule becomes $x_{n+1} = -2x_n$, doubling away from $0$ |
The cycling example is worth checking by hand. With $f(x) = x^3 - 2x + 2$ and $f'(x) = 3x^2 - 2$, starting at $x_0 = 0$ gives $f(0) = 2$, $f'(0) = -2$, so $x_1 = 0 - \tfrac{2}{-2} = 1$. Then $f(1) = 1$, $f'(1) = 1$, so $x_2 = 1 - \tfrac{1}{1} = 0$. The estimates bounce between $0$ and $1$ forever and never reach the true root.
The cube-root case is the honest opposite of quadratic convergence. For $f(x) = x^{1/3}$ the update simplifies to $x_{n+1} = x_n - \dfrac{x_n^{1/3}}{\tfrac{1}{3}x_n^{-2/3}} = x_n - 3x_n = -2x_n$, so each guess doubles in size and flips sign, marching away from the root at $0$ no matter how close you start. The lesson is that Newton's Method rewards a good starting guess and a well-behaved function, and punishes neither gently.
Who Invented Newtons Method?
The method carries two names because two people built it, decades apart, and a third gave it the form taught today.
The averaging rule for square roots, $x_{n+1} = \tfrac{1}{2}\big(x_n + \tfrac{a}{x_n}\big)$, is older still. It was known to Babylonian scribes and to Heron of Alexandria almost two thousand years ago, and it is exactly Newton's Method applied to $f(x) = x^2 - a$. The ancient square-root trick turned out to be a special case of a rule nobody would state in general for centuries.
Where Is Newtons Method Used In The Real World?
The method is not a classroom curiosity. It is one of the most heavily used algorithms in computing, because most real equations have no closed-form solution and must be solved by fast iteration.
Calculators and computers taking square roots. The square-root and reciprocal buttons on a calculator run Newton's Method (the Babylonian averaging rule) internally, since it reaches full precision in a handful of steps.
Computer graphics. The famous "fast inverse square root" used in 3D game engines to normalise lighting vectors is a single Newton step refining a clever first guess.
Engineering simulation. Circuit simulators and structural-analysis tools solve large nonlinear systems by Newton's Method at every time step.
Finance. The internal rate of return of an investment is the root of a polynomial in the discount rate, and spreadsheets find it with Newton's Method.
Astronomy. Locating a planet in its orbit means solving Kepler's equation, which has no algebraic solution and is solved by iteration.
One rule for turning a guess into a better guess quietly powers graphics cards, circuit design, financial models, and orbital mechanics. Mathematics built for a single cubic in the 1600s now runs inside the machine in your pocket.
What Are The Most Common Mistakes With Newtons Method?
These errors, drawn from the failure cases documented on The Math Doctors, MIT's calculus notes, and physics-department write-ups on Newton–Raphson breakdown, account for most wrong or non-terminating answers.
Forgetting to check that $f'(x_n)$ is not zero.
Where it slips in:
A student applies the formula mechanically and lands on a point where the tangent is horizontal, then divides by zero or gets a wildly large next guess.
Don't do this:
Do not push a value through the update without looking at the derivative first.
The correct way:
Before each step confirm $f'(x_n) \neq 0$. If the derivative is zero or tiny, nudge the starting guess and restart; the tangent must actually reach the x-axis.
Trusting the method with a careless starting guess.
Where it slips in:
A student picks $x_0$ far from the root, or near a peak or valley of the curve, and the iterates run off, cycle, or converge to a different root than intended.
Don't do this:
Do not assume any starting value converges. Newton's Method is only guaranteed to work close to the root.
The correct way:
Sketch $f$ or test signs to bracket the root, then start $x_0$ inside that bracket, near where the curve crosses the axis.
Stopping at the wrong moment, or never stopping.
Where it slips in:
A student either quits after one step and reports a rough guess as the answer, or loops forever chasing digits that will not settle because the case is diverging.
Don't do this:
Do not treat a single step as the final answer, and do not keep iterating a run that is clearly growing or oscillating.
The correct way:
Stop when consecutive guesses agree to the accuracy you need, for example when $|x_{n+1} - x_n|$ is below a chosen tolerance. If the values are not settling, the method is failing and the starting guess or the function is the problem.
Practice Problems On Newtons Method
Give each answer to four decimal places unless stated otherwise.
One step for $f(x) = x^2 - 5$ from $x_0 = 2$.
(Answer: $x_1 = 2 - \tfrac{-1}{4} = 2.2500$.)Two steps for $\sqrt{5}$ using $f(x) = x^2 - 5$, $x_0 = 2$.
(Answer: $x_1 = 2.2500$, $x_2 = 2.2361$; true $\sqrt{5} = 2.2360679$.)One step for $f(x) = x^2 - 10$ from $x_0 = 3$.
(Answer: $x_1 = 3 - \tfrac{-1}{6} = 3.1667$.)One step for $f(x) = x^3 - 7$ from $x_0 = 2$, using $f'(x) = 3x^2$.
(Answer: $x_1 = 2 - \tfrac{1}{12} = 1.9167$.)Explain why $f(x) = x^2 + 1$ with any real $x_0$ never converges.
(Answer: $f$ has no real root, so the iterates cannot approach one.)Show that $f(x) = x^3 - 2x + 2$ from $x_0 = 0$ cycles.
(Answer: $x_1 = 1$, $x_2 = 0$, repeating; no convergence.)
Where Should You Go Next After Newtons Method?
Newton's Method sits on top of two ideas, the derivative and the root of a function, so the natural next steps deepen those.
Derivative. Every step of the method uses $f'(x_n)$, so a firm grasp of how derivatives are computed makes the whole loop routine.
Zeros of a function. Newton's Method is one tool for finding these; seeing what a zero means and how many a function can have frames when the method applies.
Tangents and normals. The tangent line is the engine of the method, and this is where its geometry is built up carefully.
If your child is meeting root-finding and the calculus behind it, a live Bhanzu trainer teaches Newton's Method from the tangent-line picture first, so the formula is understood rather than memorised, inside the Bhanzu math program.
Was this article helpful?
Your feedback helps us write better content