Numerical Methods

What you'll learn

How to locate roots using a change of sign.
How to use fixed-point iteration of the form $x_{n+1}=g(x_n)$ .
How to prove a numerical answer is correct to a given number of decimal places.
How to apply the Newton-Raphson method, including to stationary points.

Why numerical methods matter

Some equations cannot be solved neatly by factorising or rearranging. For example, equations mixing powers, exponentials, logarithms or trigonometric functions often need approximate solutions.

A numerical method is a repeated calculation that gets you closer to the answer. In A-Level questions, you are usually asked to show enough working to justify that your decimal answer is reliable.

Roots and sign changes

Definition

Roots and continuity

A root of $f (x) = 0$ is a value of $x$ that makes the function equal to zero.
A continuous function has no jumps or breaks. Polynomials are continuous everywhere; logarithms and roots are continuous only on their valid domains.

Key Idea

The sign-change test

If $f (x)$ is continuous on $[a, b]$ and $f (a)$ and $f (b)$ have opposite signs, then there is at least one root between $a$ and $b$ .

A continuous curve crossing the x-axis between endpoints with opposite signs.

This is often worth 2 marks: one for evaluating both ends, and one for stating the sign change and continuity.

Example

Showing a root lies in an interval

Show that $f(x)=x^3+x^2-2x-1$ has a root between 1 and 1.5.

Check that $f (x)$ is continuous. Since it is a polynomial, it is continuous for all real $x$ .
Evaluate the function at both ends of the interval:
$f(1)=13+12−2(1)−1=−1f(1.5)=1.53+1.52−2(1.5)−1=1.625\begin{aligned} f(1)&=1^3+1^2-2(1)-1=-1\\ f(1.5)&=1.5^3+1.5^2-2(1.5)-1=1.625 \end{aligned}$
The values have opposite signs, since $f (1) < 0$ and $f (1.5) > 0$ .
Therefore, by continuity, there is a root between 1 and 1.5.

Common Mistake

Overclaiming from a sign change

A sign change proves at least one root in the interval, not necessarily exactly one. Also, no sign change does not prove there is no root.

Two sketches showing why the sign-change test gives existence, not uniqueness or non-existence.

Turning a curve problem into an equation

Many numerical methods questions first ask you to differentiate a curve and find a stationary point.

Definition

Stationary point

A stationary point is a point on a curve where the gradient is zero, so you solve $dydx=0\frac{dy}{dx}=0$ .

You may need the product rule, chain rule, and standard derivatives such as:

$ddxex=ex\frac{d}{dx}e^x=e^x$
$ddxln⁡x=1x\frac{d}{dx}\ln x=\frac{1}{x}$
$ddxx=12x\frac{d}{dx}\sqrt{x}=\frac{1}{2\sqrt{x}}$

Example

Forming an equation for a stationary point

The curve has equation $y=(10−x)ln⁡xy=(10-x)\ln x$ , where $x > 0$ . Show that the $x$ coordinate of a stationary point satisfies $x=101+ln⁡xx=\frac{10}{1+\ln x}$ .

Use the product rule with $10 - x$ and $ln⁡x\ln x$ .
Differentiate:
$dydx=−1⋅ln⁡x+(10−x)⋅1x=10x−1−ln⁡x\begin{aligned} \frac{dy}{dx} &= -1\cdot \ln x+(10-x)\cdot \frac{1}{x}\\ &= \frac{10}{x}-1-\ln x \end{aligned}$
At a stationary point, set the derivative equal to zero and rearrange:
$10x−1−ln⁡x=010x=1+ln⁡xx=101+ln⁡x\begin{aligned} \frac{10}{x}-1-\ln x&=0\\ \frac{10}{x}&=1+\ln x\\ x&=\frac{10}{1+\ln x} \end{aligned}$
So the stationary point’s $x$ coordinate satisfies $x=101+ln⁡xx=\frac{10}{1+\ln x}$ .

Fixed-point iteration

Definition

Fixed-point iteration

If an equation can be written as $x = g (x)$ , then a fixed-point iteration uses $x_{n+1}=g(x_n)$ , starting from an initial value $x_0$ or $x_1$ .

The idea is simple: put your current estimate into the right-hand side, and the output becomes the next estimate.

Example

Using an iteration formula

Show that $e^{x+1}+x-6=0$ can be written as $x=ln⁡(6−x)−1x=\ln(6-x)-1$ . Then use $x_0=0.5$ to find $x_1$ , $x_2$ and $x_3$ to 3 decimal places.

Rearrange the equation:
$ex+1+x−6=0ex+1=6−xx+1=ln⁡(6−x)x=ln⁡(6−x)−1\begin{aligned} e^{x+1}+x-6&=0\\ e^{x+1}&=6-x\\ x+1&=\ln(6-x)\\ x&=\ln(6-x)-1 \end{aligned}$
Therefore the iteration formula is $x_{n+1}=\ln(6-x_n)-1$ .
Substitute repeatedly, keeping full calculator accuracy between lines:
$x1=ln⁡(6−0.5)−1=0.704748…x2=ln⁡(6−0.704748…)−1=0.666811…x3=ln⁡(6−0.666811…)−1=0.673949…\begin{aligned} x_1&=\ln(6-0.5)-1=0.704748\ldots\\ x_2&=\ln(6-0.704748\ldots)-1=0.666811\ldots\\ x_3&=\ln(6-0.666811\ldots)-1=0.673949\ldots \end{aligned}$
So, to 3 decimal places, $x_1=0.705$ , $x_2=0.667$ and $x_3=0.674$ .

Tip

Keep full accuracy

Do not type the rounded value back into your calculator unless the question explicitly tells you to. Store or reuse the full displayed value.

Proving an answer to a given accuracy

Getting an iteration value is not always enough. If asked to prove a root is correct to 3 decimal places, choose a tiny interval that rounds to that value.

For example, to prove $α=0.673\alpha=0.673$ to 3 decimal places, test the endpoints 0.6725 and 0.6735.

The rounding interval for proving that a root is 0.673 to three decimal places.

Example

Proving a root to 3 decimal places

Let $f(x)=e^{x+1}+x-6$ . Prove that the root $α\alpha$ near the iterates is 0.673 to 3 decimal places.

To prove 0.673 to 3 decimal places, use the interval $[0.6725, 0.6735]$ .
Evaluate $f (x)$ at both endpoints:
$f(0.6725)=e1.6725+0.6725−6=−0.00204…f(0.6735)=e1.6735+0.6735−6=0.00429…\begin{aligned} f(0.6725)&=e^{1.6725}+0.6725-6=-0.00204\ldots\\ f(0.6735)&=e^{1.6735}+0.6735-6=0.00429\ldots \end{aligned}$
Since the signs are different and $f (x)$ is continuous, there is a root between 0.6725 and 0.6735.
Every number in this interval rounds to 0.673 to 3 decimal places, so $α=0.673\alpha=0.673$ to 3 decimal places.

Convergence of fixed-point iteration

Definition

Convergence

An iteration converges if the sequence of values gets closer and closer to a fixed point. It diverges if the values move away or fail to settle.

Graphically, fixed points are intersections of $y = g (x)$ and $y = x$ . A cobweb or staircase diagram shows whether repeated vertical and horizontal moves approach the intersection.

A cobweb diagram showing iteration values moving towards the fixed point where y=g(x) meets y=x.

A useful test near the root is: if $∣g′(α)∣<1|g'(\alpha)|<1$ , the iteration usually converges; if $∣g′(α)∣>1|g'(\alpha)|>1$ , it usually diverges.

Example

Deciding whether an iteration can be used

Consider $x_{n+1}=\ln(6-x_n)$ , starting with $x_1=2$ . Decide whether the iteration is likely to find the fixed point.

The fixed point is where $y=ln⁡(6−x)y=\ln(6-x)$ meets $y = x$ .
Here $g(x)=ln⁡(6−x)g(x)=\ln(6-x)$ , so $g′(x)=−16−xg'(x)=-\frac{1}{6-x}$ . Near the fixed point, the gradient has magnitude less than 1.
Calculate the first few terms:
$x2=ln⁡(4)=1.386…x3=ln⁡(6−1.386…)=1.529…x4=ln⁡(6−1.529…)=1.497…\begin{aligned} x_2&=\ln(4)=1.386\ldots\\ x_3&=\ln(6-1.386\ldots)=1.529\ldots\\ x_4&=\ln(6-1.529\ldots)=1.497\ldots \end{aligned}$
The values bounce either side but move closer, so the iteration can be used.

Common Mistake

Domain checks

Iteration can fail immediately if a value makes the next calculation undefined, such as taking $ln⁡\ln$ of a negative number or dividing by zero.

Newton-Raphson method

Newton-Raphson is usually faster than fixed-point iteration, but it needs a derivative.

Key Idea

Newton-Raphson formula

To solve $f (x) = 0$ , use $xn+1=xn−f(xn)f′(xn)x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$ . This comes from drawing a tangent to the curve and seeing where it meets the $x$ -axis.

Newton-Raphson uses the tangent at the current estimate to find the next estimate on the x-axis.

Example

Applying Newton-Raphson

Use Newton-Raphson with $x_0=1.5$ to solve $x^3+2x-5=0$ to 3 decimal places.

Let $f(x)=x^3+2x-5$ .
Differentiate:
$f'(x)=3x^2+2$
Substitute into the Newton-Raphson formula:
$xn+1=xn−xn3+2xn−53xn2+2x_{n+1}=x_n-\frac{x_n^3+2x_n-5}{3x_n^2+2}$
Starting with $x_0=1.5$ , calculate:
$x1=1.342857…x2=1.328384…x3=1.328269…\begin{aligned} x_1&=1.342857\ldots\\ x_2&=1.328384\ldots\\ x_3&=1.328269\ldots \end{aligned}$
Since $x_2$ and $x_3$ agree to 3 decimal places, the root is $x = 1.328$ .

Common Mistake

Forgetting the derivative

Newton-Raphson always needs both $f (x)$ and $f^{'} (x)$ . A common error is to substitute into the original function twice instead of using the derivative in the denominator.

Newton-Raphson for stationary points

For a stationary point, solve $dydx=0\frac{dy}{dx}=0$ . You can do this by applying Newton-Raphson to the derivative.

If $F(x)=dydxF(x)=\frac{dy}{dx}$ , then use:

xn+1=xn−F(xn)F′(xn)x_{n+1}=x_n-\frac{F(x_n)}{F'(x_n)}

Example

Approximating a stationary point

The curve $y=xsin⁡xy=x\sin x$ has a maximum for $π2<x<π\frac{\pi}{2}<x<\pi$ . Use Newton-Raphson starting with $x_0=2$ to approximate the $x$ coordinate.

The curve y=x sin x has a maximum between π/2 and π where the tangent is horizontal.

Differentiate:
$dydx=sin⁡x+xcos⁡x\frac{dy}{dx}=\sin x+x\cos x$
Set the derivative equal to zero. Since $cos⁡x≠0\cos x\neq 0$ in this interval, divide by $cos⁡x\cos x$ :
$tan⁡x+x=0\tan x+x=0$
Let $F(x)=tan⁡x+xF(x)=\tan x+x$ , so $F'(x)=\sec^2 x+1$ .
Apply Newton-Raphson:
$xn+1=xn−tan⁡xn+xnsec⁡2xn+1x_{n+1}=x_n-\frac{\tan x_n+x_n}{\sec^2 x_n+1}$
Starting with $x_0=2$ :
$x1=2.0273…x2=2.0288…x3=2.0288…\begin{aligned} x_1&=2.0273\ldots\\ x_2&=2.0288\ldots\\ x_3&=2.0288\ldots \end{aligned}$
The $x$ coordinate is approximately 2.0288 to 4 decimal places.

Tip

Radians mode

For calculus with trigonometric functions, your calculator must be in radians unless a question explicitly says degrees.

Common Mistake

When Newton-Raphson breaks

Newton-Raphson cannot be used if $f'(x_n)=0$ or if $f'(x_n)$ is undefined. It may also fail if the starting value is too far from the root.

Exam technique

In the exam

For interval questions, write both endpoint values, state the sign change, and mention continuity.
For iteration, show the formula used and keep full calculator accuracy between terms.
To prove 3 decimal places, test the two half-way endpoints around the rounded value.
For Newton-Raphson, clearly write $f^{'} (x)$ and the substitution formula before giving iterates.
Check domains carefully, especially with $ln⁡x\ln x$ , square roots and fractions.

Self review

Check yourself

What interval would you test to prove that a root is 1.426 to 3 decimal places?
Why does a sign change prove a root only when the function is continuous?
What goes wrong in Newton-Raphson if $f'(x_n)=0$ ?

Numerical Methods

What you'll learn

How to locate roots using a change of sign.
How to use fixed-point iteration of the form $x_{n+1}=g(x_n)$ .
How to prove a numerical answer is correct to a given number of decimal places.
How to apply the Newton-Raphson method, including to stationary points.

Why numerical methods matter

Some equations cannot be solved neatly by factorising or rearranging. For example, equations mixing powers, exponentials, logarithms or trigonometric functions often need approximate solutions.

Roots and sign changes

Definition

Roots and continuity

A root of $f (x) = 0$ is a value of $x$ that makes the function equal to zero.
A continuous function has no jumps or breaks. Polynomials are continuous everywhere; logarithms and roots are continuous only on their valid domains.

Key Idea

The sign-change test

If $f (x)$ is continuous on $[a, b]$ and $f (a)$ and $f (b)$ have opposite signs, then there is at least one root between $a$ and $b$ .

A continuous curve crossing the x-axis between endpoints with opposite signs.

This is often worth 2 marks: one for evaluating both ends, and one for stating the sign change and continuity.

Example

Showing a root lies in an interval

Show that $f(x)=x^3+x^2-2x-1$ has a root between 1 and 1.5.

Check that $f (x)$ is continuous. Since it is a polynomial, it is continuous for all real $x$ .
Evaluate the function at both ends of the interval:
$f(1)=13+12−2(1)−1=−1f(1.5)=1.53+1.52−2(1.5)−1=1.625\begin{aligned} f(1)&=1^3+1^2-2(1)-1=-1\\ f(1.5)&=1.5^3+1.5^2-2(1.5)-1=1.625 \end{aligned}$
The values have opposite signs, since $f (1) < 0$ and $f (1.5) > 0$ .
Therefore, by continuity, there is a root between 1 and 1.5.

Common Mistake

Overclaiming from a sign change

A sign change proves at least one root in the interval, not necessarily exactly one. Also, no sign change does not prove there is no root.

Two sketches showing why the sign-change test gives existence, not uniqueness or non-existence.

Turning a curve problem into an equation

Many numerical methods questions first ask you to differentiate a curve and find a stationary point.

Definition

Stationary point

A stationary point is a point on a curve where the gradient is zero, so you solve $dydx=0\frac{dy}{dx}=0$ .

You may need the product rule, chain rule, and standard derivatives such as:

$ddxex=ex\frac{d}{dx}e^x=e^x$
$ddxln⁡x=1x\frac{d}{dx}\ln x=\frac{1}{x}$
$ddxx=12x\frac{d}{dx}\sqrt{x}=\frac{1}{2\sqrt{x}}$

Example

Forming an equation for a stationary point

The curve has equation $y=(10−x)ln⁡xy=(10-x)\ln x$ , where $x > 0$ . Show that the $x$ coordinate of a stationary point satisfies $x=101+ln⁡xx=\frac{10}{1+\ln x}$ .

Use the product rule with $10 - x$ and $ln⁡x\ln x$ .
Differentiate:
$dydx=−1⋅ln⁡x+(10−x)⋅1x=10x−1−ln⁡x\begin{aligned} \frac{dy}{dx} &= -1\cdot \ln x+(10-x)\cdot \frac{1}{x}\\ &= \frac{10}{x}-1-\ln x \end{aligned}$
At a stationary point, set the derivative equal to zero and rearrange:
$10x−1−ln⁡x=010x=1+ln⁡xx=101+ln⁡x\begin{aligned} \frac{10}{x}-1-\ln x&=0\\ \frac{10}{x}&=1+\ln x\\ x&=\frac{10}{1+\ln x} \end{aligned}$
So the stationary point’s $x$ coordinate satisfies $x=101+ln⁡xx=\frac{10}{1+\ln x}$ .

Fixed-point iteration

Definition

Fixed-point iteration

If an equation can be written as $x = g (x)$ , then a fixed-point iteration uses $x_{n+1}=g(x_n)$ , starting from an initial value $x_0$ or $x_1$ .

The idea is simple: put your current estimate into the right-hand side, and the output becomes the next estimate.

Example

Using an iteration formula

Show that $e^{x+1}+x-6=0$ can be written as $x=ln⁡(6−x)−1x=\ln(6-x)-1$ . Then use $x_0=0.5$ to find $x_1$ , $x_2$ and $x_3$ to 3 decimal places.

Rearrange the equation:
$ex+1+x−6=0ex+1=6−xx+1=ln⁡(6−x)x=ln⁡(6−x)−1\begin{aligned} e^{x+1}+x-6&=0\\ e^{x+1}&=6-x\\ x+1&=\ln(6-x)\\ x&=\ln(6-x)-1 \end{aligned}$
Therefore the iteration formula is $x_{n+1}=\ln(6-x_n)-1$ .
Substitute repeatedly, keeping full calculator accuracy between lines:
$x1=ln⁡(6−0.5)−1=0.704748…x2=ln⁡(6−0.704748…)−1=0.666811…x3=ln⁡(6−0.666811…)−1=0.673949…\begin{aligned} x_1&=\ln(6-0.5)-1=0.704748\ldots\\ x_2&=\ln(6-0.704748\ldots)-1=0.666811\ldots\\ x_3&=\ln(6-0.666811\ldots)-1=0.673949\ldots \end{aligned}$
So, to 3 decimal places, $x_1=0.705$ , $x_2=0.667$ and $x_3=0.674$ .

Tip

Keep full accuracy

Do not type the rounded value back into your calculator unless the question explicitly tells you to. Store or reuse the full displayed value.

Proving an answer to a given accuracy

Getting an iteration value is not always enough. If asked to prove a root is correct to 3 decimal places, choose a tiny interval that rounds to that value.

For example, to prove $α=0.673\alpha=0.673$ to 3 decimal places, test the endpoints 0.6725 and 0.6735.

The rounding interval for proving that a root is 0.673 to three decimal places.

Example

Proving a root to 3 decimal places

Let $f(x)=e^{x+1}+x-6$ . Prove that the root $α\alpha$ near the iterates is 0.673 to 3 decimal places.

To prove 0.673 to 3 decimal places, use the interval $[0.6725, 0.6735]$ .
Evaluate $f (x)$ at both endpoints:
$f(0.6725)=e1.6725+0.6725−6=−0.00204…f(0.6735)=e1.6735+0.6735−6=0.00429…\begin{aligned} f(0.6725)&=e^{1.6725}+0.6725-6=-0.00204\ldots\\ f(0.6735)&=e^{1.6735}+0.6735-6=0.00429\ldots \end{aligned}$
Since the signs are different and $f (x)$ is continuous, there is a root between 0.6725 and 0.6735.
Every number in this interval rounds to 0.673 to 3 decimal places, so $α=0.673\alpha=0.673$ to 3 decimal places.

Convergence of fixed-point iteration

Definition

Convergence

An iteration converges if the sequence of values gets closer and closer to a fixed point. It diverges if the values move away or fail to settle.

Graphically, fixed points are intersections of $y = g (x)$ and $y = x$ . A cobweb or staircase diagram shows whether repeated vertical and horizontal moves approach the intersection.

A cobweb diagram showing iteration values moving towards the fixed point where y=g(x) meets y=x.

A useful test near the root is: if $∣g′(α)∣<1|g'(\alpha)|<1$ , the iteration usually converges; if $∣g′(α)∣>1|g'(\alpha)|>1$ , it usually diverges.

Example

Deciding whether an iteration can be used

Consider $x_{n+1}=\ln(6-x_n)$ , starting with $x_1=2$ . Decide whether the iteration is likely to find the fixed point.

The fixed point is where $y=ln⁡(6−x)y=\ln(6-x)$ meets $y = x$ .
Here $g(x)=ln⁡(6−x)g(x)=\ln(6-x)$ , so $g′(x)=−16−xg'(x)=-\frac{1}{6-x}$ . Near the fixed point, the gradient has magnitude less than 1.
Calculate the first few terms:
$x2=ln⁡(4)=1.386…x3=ln⁡(6−1.386…)=1.529…x4=ln⁡(6−1.529…)=1.497…\begin{aligned} x_2&=\ln(4)=1.386\ldots\\ x_3&=\ln(6-1.386\ldots)=1.529\ldots\\ x_4&=\ln(6-1.529\ldots)=1.497\ldots \end{aligned}$
The values bounce either side but move closer, so the iteration can be used.

Common Mistake

Domain checks

Iteration can fail immediately if a value makes the next calculation undefined, such as taking $ln⁡\ln$ of a negative number or dividing by zero.

Newton-Raphson method

Newton-Raphson is usually faster than fixed-point iteration, but it needs a derivative.

Key Idea

Newton-Raphson formula

To solve $f (x) = 0$ , use $xn+1=xn−f(xn)f′(xn)x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$ . This comes from drawing a tangent to the curve and seeing where it meets the $x$ -axis.

Newton-Raphson uses the tangent at the current estimate to find the next estimate on the x-axis.

Example

Applying Newton-Raphson

Use Newton-Raphson with $x_0=1.5$ to solve $x^3+2x-5=0$ to 3 decimal places.

Let $f(x)=x^3+2x-5$ .
Differentiate:
$f'(x)=3x^2+2$
Substitute into the Newton-Raphson formula:
$xn+1=xn−xn3+2xn−53xn2+2x_{n+1}=x_n-\frac{x_n^3+2x_n-5}{3x_n^2+2}$
Starting with $x_0=1.5$ , calculate:
$x1=1.342857…x2=1.328384…x3=1.328269…\begin{aligned} x_1&=1.342857\ldots\\ x_2&=1.328384\ldots\\ x_3&=1.328269\ldots \end{aligned}$
Since $x_2$ and $x_3$ agree to 3 decimal places, the root is $x = 1.328$ .

Common Mistake

Forgetting the derivative

Newton-Raphson always needs both $f (x)$ and $f^{'} (x)$ . A common error is to substitute into the original function twice instead of using the derivative in the denominator.

Newton-Raphson for stationary points

For a stationary point, solve $dydx=0\frac{dy}{dx}=0$ . You can do this by applying Newton-Raphson to the derivative.

If $F(x)=dydxF(x)=\frac{dy}{dx}$ , then use:

xn+1=xn−F(xn)F′(xn)x_{n+1}=x_n-\frac{F(x_n)}{F'(x_n)}

Example

Approximating a stationary point

The curve $y=xsin⁡xy=x\sin x$ has a maximum for $π2<x<π\frac{\pi}{2}<x<\pi$ . Use Newton-Raphson starting with $x_0=2$ to approximate the $x$ coordinate.

The curve y=x sin x has a maximum between π/2 and π where the tangent is horizontal.

Differentiate:
$dydx=sin⁡x+xcos⁡x\frac{dy}{dx}=\sin x+x\cos x$
Set the derivative equal to zero. Since $cos⁡x≠0\cos x\neq 0$ in this interval, divide by $cos⁡x\cos x$ :
$tan⁡x+x=0\tan x+x=0$
Let $F(x)=tan⁡x+xF(x)=\tan x+x$ , so $F'(x)=\sec^2 x+1$ .
Apply Newton-Raphson:
$xn+1=xn−tan⁡xn+xnsec⁡2xn+1x_{n+1}=x_n-\frac{\tan x_n+x_n}{\sec^2 x_n+1}$
Starting with $x_0=2$ :
$x1=2.0273…x2=2.0288…x3=2.0288…\begin{aligned} x_1&=2.0273\ldots\\ x_2&=2.0288\ldots\\ x_3&=2.0288\ldots \end{aligned}$
The $x$ coordinate is approximately 2.0288 to 4 decimal places.

Tip

Radians mode

For calculus with trigonometric functions, your calculator must be in radians unless a question explicitly says degrees.

Common Mistake

When Newton-Raphson breaks

Newton-Raphson cannot be used if $f'(x_n)=0$ or if $f'(x_n)$ is undefined. It may also fail if the starting value is too far from the root.

Exam technique

In the exam

For interval questions, write both endpoint values, state the sign change, and mention continuity.
For iteration, show the formula used and keep full calculator accuracy between terms.
To prove 3 decimal places, test the two half-way endpoints around the rounded value.
For Newton-Raphson, clearly write $f^{'} (x)$ and the substitution formula before giving iterates.
Check domains carefully, especially with $ln⁡x\ln x$ , square roots and fractions.

Self review

Check yourself

What interval would you test to prove that a root is 1.426 to 3 decimal places?
Why does a sign change prove a root only when the function is continuous?
What goes wrong in Newton-Raphson if $f'(x_n)=0$ ?

Numerical Methods

What you'll learn

Why numerical methods matter

Roots and sign changes

Turning a curve problem into an equation

Fixed-point iteration

Proving an answer to a given accuracy

Convergence of fixed-point iteration

Newton-Raphson method

Newton-Raphson for stationary points

Recap questions

You've reached the end

Numerical Methods

What you'll learn

Why numerical methods matter

Roots and sign changes

Turning a curve problem into an equation

Fixed-point iteration

Proving an answer to a given accuracy

Convergence of fixed-point iteration

Newton-Raphson method

Newton-Raphson for stationary points

Recap questions

You've reached the end

Numerical Methods

What you'll learn

Why numerical methods matter

Roots and sign changes

Turning a curve problem into an equation

Fixed-point iteration

Proving an answer to a given accuracy

Convergence of fixed-point iteration

Newton-Raphson method

Newton-Raphson for stationary points

Recap questions

You've reached the end

Numerical Methods Revision Guide

Numerical Methods

What you'll learn

Why numerical methods matter

Roots and sign changes

Turning a curve problem into an equation

Fixed-point iteration

Proving an answer to a given accuracy

Convergence of fixed-point iteration

Newton-Raphson method

Newton-Raphson for stationary points

Recap questions

You've reached the end

Numerical Methods Revision Guide