Newton’s Method: Formula, Worked Example and When It Fails

Newton’s method (also known as the Newton–Raphson method) is a fast way to approximate a root of \( f(x) = 0 \) when you can’t solve it with algebra. Start with an initial guess \( x_0 \) and repeat the iterative formula:

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

Each pass through the formula is called an iteration. You keep iterating until two successive values agree to the number of decimal places you need. Provided the function \( f(x) \) is differentiable and your initial guess \( x_0 \) is sufficiently close to a simple root, the number of correct decimal places approximately doubles with each iteration.

Why it works: the tangent line picture

Near any point, a smooth curve looks almost like its tangent line. That is the whole idea behind linear approximation. A line is easy to solve: you can always find exactly where it crosses the x-axis. So Newton’s method replaces the hard problem (where does the curve cross the axis?) with an easy one (where does the tangent line cross?), and uses the answer as a better guess.

If your guess is already close to the root, the tangent line hugs the curve tightly in that region, so its x-intercept lands very close to the true root. That is why the method speeds up dramatically as it closes in.

Geometrically, at your current estimate \( x_n \), you evaluate the function height \( f(x_n) \) and slope \( f'(x_n) \). You then construct the tangent line at \( (x_n, f(x_n)) \). Sliding down this tangent line to the x-axis gives your next point, \( x_{n+1} \). If the function is curving gently (small second derivative \( f'' \)) and the slope is steep (large first derivative \( f' \)), the step lands almost directly on top of the actual root.

Where the formula comes from

At the current guess \( x_n \), replace the curve by its tangent line:

$$y - f(x_n) = f'(x_n)(x - x_n)$$

Set \( y = 0 \) to find where this tangent line intersects the horizontal axis, and call this x-intercept \( x_{n+1} \):

$$0 - f(x_n) = f'(x_n)(x_{n+1} - x_n)$$

Here is the algebra, one step at a time. Isolate the term \( (x_{n+1} - x_n) \) by dividing both sides by \( f'(x_n) \), which is allowed as long as the slope is not zero:

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

Add \( x_n \) to both sides. As a result, that gives the standard Newton formula:

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

Notice the division by \( f'(x_n) \): this is where things break when the tangent is flat. A horizontal tangent line (\( f'(x_n) = 0 \)) is parallel to the x-axis and never crosses it, making it impossible to calculate a next step.

Derivation of quadratic convergence using Taylor series

To understand why Newton’s method converges so rapidly near a root, we analyze how the error shrinks from step to step using a second-order Taylor series expansion.

Let \( r \) be the true root, so \( f(r) = 0 \). Define the approximation error at step \( n \) as \( e_n = x_n - r \). Expanding \( f(r) \) around \( x_n \) gives:

$$f(r) = f(x_n) + f'(x_n)(r - x_n) + \frac{f''(\xi_n)}{2}(r - x_n)^2$$

where \( \xi_n \) lies between \( x_n \) and \( r \). Since \( f(r) = 0 \) and \( r - x_n = -e_n \), we rewrite the expansion as:

$$0 = f(x_n) - f'(x_n) e_n + \frac{f''(\xi_n)}{2} e_n^2$$

Divide the entire equation by \( f'(x_n) \):

$$0 = \frac{f(x_n)}{f'(x_n)} - e_n + \frac{f''(\xi_n)}{2 f'(x_n)} e_n^2$$

Rearranging terms reveals the error at the next step, \( e_{n+1} = x_{n+1} - r \):

$$e_n - \frac{f(x_n)}{f'(x_n)} = \frac{f''(\xi_n)}{2 f'(x_n)} e_n^2 \implies e_{n+1} = \frac{f''(\xi_n)}{2 f'(x_n)} e_n^2$$

Quadratic convergence and its speed

Taking the absolute value as \( x_n \) approaches \( r \) shows that:

$$|e_{n+1}| \approx M |e_n|^2 \quad \text{where} \quad M = \left| \frac{f''(r)}{2 f'(r)} \right|$$

Because the error at step \( n+1 \) is proportional to the square of the error at step \( n \), Newton’s method exhibits quadratic convergence. If your error is \( 10^{-2} \), the next error is roughly \( 10^{-4} \), followed by \( 10^{-8} \) and \( 10^{-16} \). Every step approximately doubles the number of correct decimal places.

When the root is a repeated root (for example \( f(x) = (x - 1)^2 \)), \( f'(r) = 0 \), violating the requirement that \( f'(r) \neq 0 \). In that case, the speed drops to linear convergence, meaning the error shrinks only by a constant fraction each step. To restore quadratic convergence at a root of known multiplicity \( m \), use the modified formula \( x_{n+1} = x_n - m \frac{f(x_n)}{f'(x_n)} \).

How to apply Newton’s method step by step

1. Rewrite the equation in the form \( f(x) = 0 \): Move all terms to one side. For example, \( \cos x = x \) becomes \( f(x) = \cos x - x = 0 \).
2. Find \( f'(x) \): Compute the derivative analytically. A derivative calculator is handy for checking this step.
3. Pick a starting value \( x_0 \) close to the root: Check for a sign change using the Intermediate Value Theorem. If \( f(a) \) and \( f(b) \) have opposite signs, a root lies between \( a \) and \( b \). Also, a quick plot in the visual lab helps confirm there are no turning points nearby.
4. Apply the formula repeatedly: Calculate \( x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \). Keep plenty of decimal places during intermediate steps to prevent rounding drift.
5. Check stopping criteria: Stop when two consecutive approximations agree to your desired precision (\( |x_{n+1} - x_n| < \epsilon \)) or when the function residual is negligible \(( |f(x_{n+1})| < \delta )\). Plug your final value into \( f(x) \) to confirm it is nearly zero.

Worked example 1: x³ − 2x − 5 = 0

This is the equation Newton himself used in his 1671 treatise. Here, \( f(x) = x^3 - 2x - 5 \) and \( f'(x) = 3x^2 - 2 \). Testing integer values gives \( f(2) = 2^3 - 2(2) - 5 = -1 \) and \( f(3) = 3^3 - 2(3) - 5 = 16 \). Since \( f(2) 0 \), a root lies in the interval ( [2, 3] ). We choose \( x_0 = 2 \).

n	\( x_n \)	\( f(x_n) \)	\( f'(x_n) \)	Step size \( |x_{n+1} - x_n| \)
0	2.0	-1.0	10.0	0.1
1	2.1	0.061	11.23	0.00543188
2	2.09456812	0.00018573	11.161528	0.00001664
3	2.09455148	\( 1.85 \times 10^{-9} \)	11.161353	\( 1.66 \times 10^{-10} \)

Step 1 in detail: At \( x_0 = 2 \), \( f(2) = -1 \) and \( f'(2) = 3(2^2) - 2 = 10 \). Applying the formula:

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

Step 2: At \( x_1 = 2.1 \), evaluate \( f(2.1) = (2.1)^3 - 2(2.1) - 5 = 9.261 - 4.2 - 5 = 0.061 \). Evaluate \( f'(2.1) = 3(2.1)^2 - 2 = 13.23 - 2 = 11.23 \). Then:

$$x_2 = 2.1 - \frac{0.061}{11.23} = 2.1 - 0.00543187889 = 2.094568121$$

Step 3: At \( x_2 = 2.094568121 \), evaluate \( f(x_2) \approx 0.00018573 \) and \( f'(x_2) \approx 11.161528 \):

$$x_3 = 2.094568121 - \frac{0.00018573}{11.161528} = 2.094551482$$

Evaluating \( f(2.094551482) \) yields approximately \( 1.85 \times 10^{-9} \), confirming the root is \( x \approx 2.09455148 \) to 8 correct decimal places after just three iterations.

Worked example 2: computing √2

Calculating a square root is equivalent to finding the positive zero of \( f(x) = x^2 - 2 \). Also, the derivative is \( f'(x) = 2x \). Substituting these into Newton’s formula yields:

$$x_{n+1} = x_n - \frac{x_n^2 - 2}{2x_n} = \frac{2x_n^2 - (x_n^2 - 2)}{2x_n} = \frac{1}{2}\left(x_n + \frac{2}{x_n}\right)$$

This simplified iterative relation is known as the ancient Babylonian method or Heron’s method. Let’s start with an initial guess of \( x_0 = 1.5 \):

$$x_1 = \frac{1}{2}\left(1.5 + \frac{2}{1.5}\right) = \frac{1}{2}(1.5 + 1.333333333) = 1.416666667$$

$$x_2 = \frac{1}{2}\left(1.416666667 + \frac{2}{1.416666667}\right) = \frac{1}{2}(1.416666667 + 1.411764706) = 1.414215686$$

$$x_3 = \frac{1}{2}\left(1.414215686 + \frac{2}{1.414215686}\right) = \frac{1}{2}(1.414215686 + 1.414211438) = 1.414213562$$

Comparing \( x_3 = 1.414213562 \) with the true value \( \sqrt{2} \approx 1.41421356237 \), we see 9 correct decimal places after 3 iterations. The simplified form has an intuitive geometric interpretation: if your guess \( x_n \) is an overestimate, then \( 2/x_n \) is an underestimate. Averaging the two pulls the next estimate directly into the balance between them.

Worked example 3: solving cos x = x

Transcendental equations cannot be isolated using basic algebra. So, rewrite \( \cos x = x \) into standard form \( f(x) = \cos x - x = 0 \). The derivative is \( f'(x) = -\sin x - 1 \).

Testing initial inputs in radians gives \( f(0) = \cos(0) - 0 = 1 \) and \( f(1) = \cos(1) - 1 \approx 0.5403023 - 1 = -0.4596977 \). Because \( f(0) > 0 \) and \( f(1) < 0 \), a root exists between 0 and 1. We choose \( x_0 = 1.0 \).

$$x_1 = 1 - \frac{\cos(1) - 1}{-\sin(1) - 1} = 1 - \frac{-0.45969769}{-1.84147098} = 1 - 0.24963612 = 0.75036387$$

$$x_2 = 0.75036387 - \frac{\cos(0.75036387) - 0.75036387}{-\sin(0.75036387) - 1} = 0.75036387 - \frac{-0.01892307}{-1.68190006} = 0.73911289$$

$$x_3 = 0.73911289 - \frac{\cos(0.73911289) - 0.73911289}{-\sin(0.73911289) - 1} = 0.73911289 - \frac{-0.00004647}{-1.67363254} = 0.73908513$$

Computing \( x_4 \) gives \( 0.73908513 \), matching \( x_3 \) to all 8 decimal places shown. As a result, three numerical steps transformed a crude starting guess of 1.0 into a high-precision solution.

Worked example 4: an exam-style question

A typical test question states: “Use Newton’s method with \( x_0 = 1 \) to find \( x_2 \) for \( x^3 - x - 1 = 0 \).”

First, identify \( f(x) = x^3 - x - 1 \) and compute \( f'(x) = 3x^2 - 1 \).

• First iteration (\( x_1 \)):

  \( f(1) = 1^3 - 1 - 1 = -1 \)

  \( f'(1) = 3(1^2) - 1 = 2 \)

  $$x_1 = 1 - \frac{-1}{2} = 1 + 0.5 = 1.5$$

• Second iteration (\( x_2 \)):

  \( f(1.5) = (1.5)^3 - 1.5 - 1 = 3.375 - 2.5 = 0.875 \)

  \( f'(1.5) = 3(1.5^2) - 1 = 3(2.25) - 1 = 5.75 \)

  $$x_2 = 1.5 - \frac{0.875}{5.75} = 1.5 - 0.15217391 = 1.34782609$$

On exams, explicitly write down the values for \( f(x_n) \) and \( f'(x_n) \) for every step. If you make an arithmetic error in division, showing these intermediate function evaluations ensures you retain partial credit.

Worked example 5: cube root extraction

To compute \( \sqrt[3]{a} \) without built-in radical functions, solve \( f(x) = x^3 - a = 0 \). The derivative is \( f'(x) = 3x^2 \). Substituting into Newton’s formula gives:

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

Let’s approximate \( \sqrt[3]{5} \), setting \( a = 5 \), starting from \( x_0 = 1.7 \) (since \( 1.7^3 = 4.913 \)):

$$x_1 = \frac{2(1.7)^3 + 5}{3(1.7)^2} = \frac{2(4.913) + 5}{3(2.89)} = \frac{14.826}{8.67} \approx 1.70991926$$

$$x_2 = \frac{2(1.70991926)^3 + 5}{3(1.70991926)^2} = \frac{2(4.9992918) + 5}{3(2.9238238)} = \frac{14.9985836}{8.7714714} \approx 1.70997595$$

Since \( (1.70997595)^3 = 5.00000001 \), the estimate \( x_2 \) is accurate to 7 decimal places in two iterations.

When Newton’s method fails: breakdown modes and remedies

While Newton’s method is remarkably fast, it is not globally convergent and can break down in specific geometric scenarios.

1. Stationary points (zero derivative)

If \( f'(x_n) = 0 \), the formula attempts to divide by zero. Geometrically, the tangent line is horizontal and runs parallel to the x-axis, so it never intersects it.

For \( f(x) = x^3 - 3x + 2 \), the derivative is \( f'(x) = 3x^2 - 3 \). If you choose \( x_0 = 1 \), \( f'(1) = 0 \), and so calculation halts immediately.

Fix: Perturb your guess slightly away from the stationary point (for instance, set \( x_0 = 1.1 \)) or take a single bisection step to move away from the local extremum.

2. Infinite cycles and oscillations

If the tangent line at point A lands exactly on point B, and the tangent line at point B lands back on point A, the method becomes trapped in an infinite loop \(( x_0 \to x_1 \to x_0 \to x_1 \dots )\).

Consider \( f(x) = x^3 - 2x + 2 \) with starting guess \( x_0 = 0 \). Here, \( f'(x) = 3x^2 - 2 \).

• Step 1: \( f(0) = 2 \) and \( f'(0) = -2 \). \( x_1 = 0 - \frac{2}{-2} = 1 \).
• Step 2: \( f(1) = 1 - 2 + 2 = 1 \) and \( f'(1) = 3(1) - 2 = 1 \). \( x_2 = 1 - \frac{1}{1} = 0 \).

The sequence endlessly alternates between 0 and 1 without reaching the true root near \( x \approx -1.7693 \).

Fix: Detect repeating values in your iteration log. For example, change your initial guess or introduce a damping factor \( \gamma \in (0, 1) \) to use a line-search update: \( x_{n+1} = x_n - \gamma \frac{f(x_n)}{f'(x_n)} \).

3. Local divergence and overshooting

When a function has an inflection point near a root or an infinite derivative slope, the tangent line can shoot the next iterate extremely far from the target root.

Example A (Infinite Slope): Consider \( f(x) = x^{1/3} \), which has a root at \( x = 0 \). The derivative is \( f'(x) = \frac{1}{3}x^{-2/3} \). Applying the formula gives:

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

Starting at \( x_0 = 1 \), the iterates become \( x_1 = -2 \), \( x_2 = 4 \), \( x_3 = -8 \). As a result, the distance from the root doubles every step, diverging to infinity.

Example B (Inflection Point Overshoot): For \( f(x) = \arctan(x) \), if your starting guess satisfies \( |x_0| > 1.39177 \), the tangent line projects the next guess further away from zero than the current estimate, causing infinite divergence.

Fix: Bracket the root within a known interval ( [a, b] ) where \( f(a) f(b) < 0 \) and force iterates to stay within bounds.

4. Jumps to distant roots

If your initial guess is close to a local maximum or minimum, the derivative \( f'(x_n) \) is very small. Dividing by a near-zero number produces a massive step that skips nearby roots and lands in a completely different basin of attraction.

For \( f(x) = x^3 - x \), roots exist at \( x = -1, 0, 1 \). Choosing \( x_0 = \frac{1}{\sqrt{5}} \approx 0.4472 \) places the guess near the local minimum at \( x = 1/\sqrt{3} \approx 0.577 \). The step shoots across the origin and converges to \( x = -1 \) instead of the nearest root \( x = 0 \).

Fix: Choose \( x_0 \) closer to the specific root of interest, ensuring no local extrema (\( f'(x) = 0 \)) lie between \( x_0 \) and the target root.

Failure Mode	Root Cause	Symptom	Recommended Remedy
Zero Derivative	\( f'(x_n) = 0 \) at guess	Division by zero; iteration halts	Shift \( x_0 \) by small \( \delta \); use bisection step
Infinite Oscillation	Symmetric tangent intersections	Iterates alternate endlessly between values	Change \( x_0 \); use damped Newton step
Divergence / Overshoot	Inflection points or non-smooth roots	Iterates grow infinitely larger	Bracket root; apply line search / backtracking
Wrong Root Jump	Guess placed near local extrema	Converges to far-away root unexpectedly	Plot graph; restrict initial guess range
Slowed Convergence	Root has multiplicity \( m > 1 \)	Linear convergence instead of quadratic	Use modified formula \( x_{n+1} = x_n - m \frac{f(x_n)}{f'(x_n)} \)

Comparison of numerical root-finding algorithms

No single numerical root-finding method is best for every function. Selecting the optimal method depends on whether derivatives are available, how expensive function evaluations are, and whether guaranteed convergence is required.

Method	Order of Convergence	Evaluations per Step	Guaranteed Convergence?	Requires Derivative?	Primary Strengths & Weaknesses
Newton’s Method	Quadratic (\( p = 2 \))	1 function, 1 derivative	No (requires good \( x_0 \))	Yes (\( f' \))	Extremely fast near simple roots; fails on zero derivatives.
Bisection Method	Linear (\( p = 1 \))	1 function	Yes (if sign change exists)	No	100% reliable; very slow (gains ~1 decimal bit per step).
Secant Method	Superlinear \(( p \approx 1.618 )\)	1 function	No	No	Fast convergence without analytical derivatives; needs 2 initial points.
Halley’s Method	Cubic (\( p = 3 \))	1 function, 1 first derivative, 1 second derivative	No	Yes (\( f', f'' \))	Ultra-fast convergence; evaluating \( f'' \) can be computationally heavy.
Regula Falsi	Linear (\( p = 1 \))	1 function	Yes (if sign change exists)	No	Guaranteed convergence like bisection; can stall at one endpoint.

Compared to the Bisection Method, Newton’s method is infinitely faster when close to a root, requiring typically 3 to 5 steps to reach machine precision where bisection might require 30 to 50 iterations. However, Bisection is completely robust and never diverges when provided a valid bracket. In professional software engineering, hybrid algorithms like Brent’s Method combine the speed of Newton/Secant steps with the absolute guarantee of Bisection.

Newton’s method in optimization and division-free reciprocals

Newton’s method extends beyond simple root finding into several critical areas of scientific computing.

1. Optimization (finding local extrema)

To find a local minimum or maximum of a function \( g(x) \), calculus requires finding where its derivative is zero (critical points), meaning \( g'(x) = 0 \). Therefore, setting \( f(x) = g'(x) \) in Newton’s formula yields the optimization update rule:

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

This update drives the derivative to zero, identifying local peaks or valleys. It requires the second derivative \( g''(x_n) \neq 0 \).

2. Computing division-free reciprocals in hardware

Computer processors often compute floating-point division \( 1/a \) without using slow hardware division circuits. Instead, they apply Newton’s method to solve \( f(x) = \frac{1}{x} - a = 0 \).

The derivative is \( f'(x) = -\frac{1}{x^2} \). Substituting into Newton’s formula gives:

$$x_{n+1} = x_n - \frac{\frac{1}{x_n} - a}{-\frac{1}{x_n^2}} = x_n + x_n^2\left(\frac{1}{x_n} - a\right) = x_n + x_n - a x_n^2 = x_n(2 - a x_n)$$

This remarkable formula calculates \( 1/a \) using only multiplication and subtraction. For example, to find \( 1/3 \approx 0.33333333 \) starting from \( x_0 = 0.3 \):

• \( x_1 = 0.3(2 - 3(0.3)) = 0.3(1.1) = 0.33 \)
• \( x_2 = 0.33(2 - 3(0.33)) = 0.33(1.01) = 0.3333 \)
• \( x_3 = 0.3333(2 - 3(0.3333)) = 0.3333(1.0001) = 0.33333333 \)

3. Relationship to differential equations

Newton’s method shares conceptual roots with Euler’s method. Both techniques utilize tangent lines to project future values. While Euler’s method follows tangent slopes sequentially to approximate the solution curve of a differential equation, Newton’s method uses tangent slopes to find the zero-crossing of an algebraic curve.

Who Newton’s method suits (and who it does not)

Newton’s method is ideal for:

• Students and practitioners who can easily compute the analytical derivative \( f'(x) \).
• Applications requiring extreme numerical precision (10 to 100+ decimal digits) in very few computational steps.
• Situations where a reliable initial guess \( x_0 \) is known from physical constraints or prior steps in a simulation.
• Hardware and low-level software design where divisions or roots must be implemented using basic arithmetic gates.

Newton’s method is NOT suitable for:

• Functions whose derivatives are non-existent, discontinuous, or extremely expensive to evaluate numerically. In these cases, the Secant Method is preferred.
• “Black-box” functions where only output values are returned and no derivative information is exposed.
• Functions with high noise or frequent local fluctuations, where tangent lines bounce wildly. Use the Bisection Method or Brent’s Method instead.

Common mistakes to avoid

• Not setting the equation to zero (\( f(x) = 0 \)): If given \( e^x = 3x \), you must define \( f(x) = e^x - 3x \). Applying Newton’s formula directly to \( e^x \) yields completely wrong results.
• Sign errors in the update formula: The formula subtracts the correction term: \( x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \). Adding \( f/f' \) moves away from the root.
• Using degree mode for trigonometric functions: Calculus derivatives for \( \sin x \) and \( \cos x \) require inputs in radians. Ensure your calculator is set to radian mode.
• Rounding intermediate values too early: Rounding intermediate iterations (such as keeping only 2 decimal places in \( x_1 \)) destroys quadratic convergence and causes calculation drift. Keep at least 4 more decimal places than required in intermediate steps.
• Evaluating \( f' \) at the fixed initial point \( x_0 \): Both \( f \) and \( f' \) must be re-evaluated at the new point \( x_n \) during every single iteration. Using \( f'(x_0) \) repeatedly turns the algorithm into a slow, linear approximation method.

Newton’s method calculator

Enter \( f(x) \), a starting guess, and the number of iterations to see every step computed in real time:

For a full-page tool with a complete step table and graphical tangent visualizations, open the Newton’s method calculator.

Practice problems

Work through these problems to master Newton’s method. Complete solutions are provided below.

1. Calculate one Newton step for \( x^2 - 5 = 0 \) starting from \( x_0 = 2 \).
2. Calculate the next iteration step (\( x_2 \)) from \( x_1 = 2.25 \) for the same function.
3. Derive the simplified Newton iteration formula for finding the cube root \( \sqrt[3]{a} \).
4. Perform one Newton step for \( \cos x - x = 0 \) starting from \( x_0 = 1 \).
5. Find \( x_2 \) for \( x^3 - x - 1 = 0 \) starting from \( x_0 = 1 \).
6. Use the reciprocal iteration \( x_{n+1} = x_n(2 - 3x_n) \) twice starting from \( x_0 = 0.3 \) to approximate \( 1/3 \).

Step-by-step solutions

1. Problem 1 Solution:

   \( f(x) = x^2 - 5 \), \( f'(x) = 2x \). At \( x_0 = 2 \):

   \( f(2) = 2^2 - 5 = -1 \), \( f'(2) = 2(2) = 4 \).

   $$x_1 = 2 - \frac{-1}{4} = 2 + 0.25 = 2.25$$

2. Problem 2 Solution:

   At \( x_1 = 2.25 \):

   \( f(2.25) = 2.25^2 - 5 = 5.0625 - 5 = 0.0625 \).

   \( f'(2.25) = 2(2.25) = 4.5 \).

   $$x_2 = 2.25 - \frac{0.0625}{4.5} = 2.25 - 0.01388889 = 2.23611111$$

3. Next, Problem 3 Solution:

   Set \( f(x) = x^3 - a = 0 \), so \( f'(x) = 3x^2 \).

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

4. Problem 4 Solution:

   \( f(x) = \cos x - x \), \( f'(x) = -\sin x - 1 \). At \( x_0 = 1 \):

   \( f(1) = \cos(1) - 1 \approx -0.45969769 \), \( f'(1) = -\sin(1) - 1 \approx -1.84147098 \).

   $$x_1 = 1 - \frac{-0.45969769}{-1.84147098} \approx 0.75036387$$

5. Problem 5 Solution:

   \( f(x) = x^3 - x - 1 \), \( f'(x) = 3x^2 - 1 \). At \( x_0 = 1 \), \( x_1 = 1.5 \).

   At \( x_1 = 1.5 \): \( f(1.5) = 0.875 \), \( f'(1.5) = 5.75 \).

   $$x_2 = 1.5 - \frac{0.875}{5.75} \approx 1.34782609$$

6. Problem 6 Solution:

   Iteration 1: \( x_1 = 0.3(2 - 3(0.3)) = 0.3(1.1) = 0.33 \).

   Iteration 2: \( x_2 = 0.33(2 - 3(0.33)) = 0.33(1.01) = 0.3333 \).

Frequently asked questions

How many iterations do I need?

Usually between 3 and 6 iterations are enough to reach maximum double-precision accuracy (15–17 decimal places), provided your initial guess is close to a simple root. Stop when successive iterates agree to your target precision or when \( |f(x_n)| \) is within acceptable error tolerances.

Is Newton’s method always faster than bisection?

Near a simple root where \( f'(r) \neq 0 \), Newton’s method is exponentially faster because it converges quadratically rather than linearly. However, Bisection is globally guaranteed to converge whenever you have an initial sign change, whereas Newton’s method can diverge or oscillate if given a poor starting guess.

What happens if f'(x) = 0 at my guess?

The formula divides by zero because the tangent line is completely horizontal and never intersects the x-axis. Shift your starting value slightly away from the stationary point or take a single bisection step to establish a valid non-zero derivative.

How do I choose a good starting value?

Apply the Intermediate Value Theorem by identifying a sign change: if \( f(a) \) and \( f(b) \) have opposite signs, a root lies between them. Choose a value near the midpoint or near the endpoint where \( |f(x)| \) is smaller. Sketching the curve or checking a graph ensures no turning points lie between your guess and the root.

Can Newton’s method find more than one root?

Yes, but each individual run converges to only one root. To find multiple roots, execute the algorithm separately starting from different initial guesses near each target root. Near boundaries between basins of attraction, small changes in \( x_0 \) can change which root the method reaches.

Further reading

• Newton’s Method (Paul’s Online Math Notes) — Worked calculus examples and detailed failure mode illustrations.
• Continuity and the Intermediate Value Theorem (Paul’s Online Math Notes) — Mathematical proof and examples for finding root intervals using sign changes.