Mini-Lesson

Numerical methods

2 missions. Guess, learn, then check. Your place is saved.

Mission 1Radian mode: type 1, then cos(Ans) again and again. Where does it end?
Mission 2How can four quick steps pin down √2 to 10 decimal places?

The specification points and sources are listed under the lesson: see Sources.

Mission 1

Radian mode: type 1, then cos(Ans) again and again. Where does it end?

Make a guess: Radian mode: type 1, then cos(Ans) again and again. Where does it end?

Numerical methods · roots and iteration

Change of sign (A-level only)

If $f$ f is continuous on $[a, b]$ [a, b] and $f (a)$ f(a) and $f (b)$ f(b) have opposite signs, then $f (x) = 0$ f(x) = 0 has a root between $a$ a and $b$ b.
To show a root is 2.09 correct to 2 decimal places, check the signs at the bounds 2.085 and 2.095.
Always finish with a conclusion: "sign change and $f$ f is continuous, so there is a root in the interval."

Worked example Numerical methods · roots and iteration

Showing a root to 2 decimal places

Question

f (x) = x^{3} - 2 x - 5

. Show that

f (x) = 0

has a root

α

between 2 and 3, and that

α = 2.09

correct to 2 decimal places.

1
Evaluate at the ends: $f (2) = - 1 < 0$ f(2) = −1 < 0 and $f (3) = 16 > 0$ f(3) = 16 > 0
2
Conclude: the sign changes and $f$ f is continuous, so $2 < α < 3$ 2 < α < 3
3
Evaluate at the bounds for 2.09: $f (2.085) = - 0.106 < 0$ f(2.085) = −0.106 < 0 and $f (2.095) = 0.005 > 0$ f(2.095) = 0.005 > 0
4
Conclude: the sign changes, so $2.085 < α < 2.095$ 2.085 < α < 2.095

Answer

every number in

(2.085, 2.095)

rounds to 2.09, so

α = 2.09

to 2 decimal places

⚠️ Watch out

testing 2.09 and 2.10 only shows

α

is between them; it could still round to 2.10. Use the bounds 2.085 and 2.095.

Numerical methods · roots and iteration

When change of sign misleads

Two roots (or a touching point) in the interval: $f (x) = (x - 1) (x - 2)$ f(x) = (x − 1)(x − 2) has $f (0) = 2$ f(0) = 2 and $f (3) = 2$ f(3) = 2. No sign change, yet two roots. Use a smaller interval.
A discontinuity: $f (x) = \frac{1}{x - 2}$ f(x) = 1/(x − 2) has $f (1) = - 1$ f(1) = −1 and $f (3) = 1$ f(3) = 1. A sign change, but no root: there is a vertical asymptote at $x = 2$ x = 2.
So the method needs $f$ f to be continuous on the interval, and the interval to be small.

Quick check

?

g (x) = \frac{3}{x - 4} + 1

. A student finds

g (3) = - 2

and

g (5) = 4

and says there must be a root between 3 and 5. What is wrong?

Numerical methods · roots and iteration

Iteration

Rearrange $f (x) = 0$ f(x) = 0 into the form $x = g (x)$ x = g(x). For example $x^{3} - 2 x - 5 = 0$ x³ − 2x − 5 = 0 becomes $x = 3 2 x + 5$ x = ∛(2x + 5).
Pick a starting value $x_{0}$ x₀ and use the iterative formula $x_{n + 1} = g (x_{n})$ xₙ₊₁ = g(xₙ) to get $x_{1}, x_{2}, x_{3}, \dots$ x₁, x₂, x₃, …
If the values converge to $α$ α, then $α = g (α)$ α = g(α), so $α$ α is a root of $f (x) = 0$ f(x) = 0.
On a calculator: type $x_{0}$ x₀, press =, then type the formula with Ans in place of $x_{n}$ xₙ and keep pressing =.

Calculate

?Use the iterative formula

x_{n + 1} = 3 2 x_{n} + 5

with

x_{0} = 2

to find

x_{3}

. Give your answer to 4 decimal places.

Numerical methods · roots and iteration

Staircases, cobwebs and convergence

Near the root $α$ α, the iteration converges if $∣ g^{'} (α) ∣ < 1$ |g'(α)| < 1: the graph of $y = g (x)$ y = g(x) is less steep than $y = x$ y = x. If $∣ g^{'} (α) ∣ > 1$ |g'(α)| > 1 it diverges, even from a start close to $α$ α.
Different rearrangements of one equation can behave differently.
Draw $y = x$ y = x and $y = g (x)$ y = g(x). From $x_{0}$ x₀ go to the curve, across to $y = x$ y = x, back to the curve, and repeat.
$g^{'} (α) > 0$ g'(α) > 0 gives a staircase; $g^{'} (α) < 0$ g'(α) < 0 gives a cobweb, with values either side of $α$ α.

Sort it

?Each formula is a rearrangement of

x^{3} - 2 x - 5 = 0

or of

cos x = x

. At its root,

g^{'} (α)

is given. Will

x_{n + 1} = g (x_{n})

converge to that root from a start nearby?

Tap a statement, then tap the group it belongs in. Then press Check.

Calculate

?The equation

x = e^{- x}

has one root. Use

x_{n + 1} = e^{- x_{n}}

with

x_{0} = 0.5

and iterate until the values agree to 3 decimal places. Give the root to 3 decimal places.

Mission check

?Calculator in radians: type 1, then cos(Ans) again and again. Where does it end up?

Mission 2

How can four quick steps pin down √2 to 10 decimal places?

Make a guess: How can four quick steps pin down √2 to 10 decimal places?

Numerical methods · Newton–Raphson and trapezium rule

The Newton–Raphson method (A-level only)

To solve $f (x) = 0$ f(x) = 0, start from an estimate $x_{0}$ x₀ near the root.
Draw the tangent at $x_{0}$ x₀. Where it crosses the $x$ x-axis is a better estimate $x_{1}$ x₁.
The tangent has gradient $f^{'} (x_{0})$ f'(x₀), which gives the Newton–Raphson formula:

x_{n + 1} = x_{n} - \frac{f ( x _{n} )}{f ^{'} ( x _{n} )}

Repeat until successive estimates agree to the accuracy you need.
When it works, the method converges very fast: the number of correct digits roughly doubles each step.

Worked example Numerical methods · Newton–Raphson and trapezium rule

One root of a cubic

Question

Use the Newton–Raphson method with

x_{0} = 2

to find a root of

x^{3} - 2 x - 5 = 0

correct to 4 decimal places.

1
Differentiate: $f (x) = x^{3} - 2 x - 5$ f(x) = x³ − 2x − 5, so $f^{'} (x) = 3 x^{2} - 2$ f'(x) = 3x² − 2
2
First step: $f (2) = - 1$ f(2) = −1, $f^{'} (2) = 10$ f'(2) = 10, so $x_{1} = 2 - \frac{- 1}{10} = 2.1$ x₁ = 2 − (−1)/10 = 2.1
3
Second step: $f (2.1) = 0.061$ f(2.1) = 0.061, $f^{'} (2.1) = 11.23$ f'(2.1) = 11.23, so $x_{2} = 2.1 - \frac{0.061}{11.23} = 2.094568 \dots$ x₂ = 2.1 − 0.061/11.23 = 2.094568…
4
Third step: $x_{3} = 2.094551 \dots$ x₃ = 2.094551…
5
$x_{2}$ x₂ and $x_{3}$ x₃ agree to 4 decimal places

Answer

x = 2.0946

(4 d.p.)

⚠️ Watch out

keep the minus sign in the formula. Adding

\frac{f ( x _{0} )}{f ^{'} ( x _{0} )}

gives

x_{1} = 1.9

, which moves away from the root.

Calculate

?Use the Newton–Raphson method on

f (x) = x^{3} + x - 3

with

x_{1} = 1

to find

x_{3}

. Give your answer to 4 decimal places.

Numerical methods · Newton–Raphson and trapezium rule

When Newton–Raphson fails

If $f^{'} (x_{n}) = 0$ f'(xₙ) = 0 the tangent is horizontal and never meets the $x$ x-axis: you cannot divide by zero. This happens when $x_{n}$ xₙ is at a stationary point.
If $f^{'} (x_{n})$ f'(xₙ) is small (near a stationary point), the tangent is nearly flat and the next estimate can jump a long way off.
The estimates may then converge to a different root, or not settle at all.
Fix: start close to the root, away from turning points.

Quick check

?Someone applies the Newton–Raphson method to

f (x) = x^{3} - 3 x + 1

with

x_{0} = 1

. What happens?

Numerical methods · Newton–Raphson and trapezium rule

The trapezium rule (A-level only)

Split $[a, b]$ [a, b] into $n$ n strips of equal width $h = \frac{b - a}{n}$ h = (b − a)/n. That needs $n + 1$ n + 1 ordinates $y_{0}, y_{1}, \dots, y_{n}$ y₀, y₁, …, yₙ.

\int_{a b} y d x \approx \frac{h}{2} [y_{0} + y_{n} + 2 (y_{1} + y_{2} + \dots + y_{n - 1})]

If the curve is convex (bends upwards, $f^{''} (x) > 0$ f''(x) > 0) the trapezia sit above it: an overestimate.
If the curve is concave ( $f^{''} (x) < 0$ f''(x) < 0) the result is an underestimate.
More strips give a better estimate.
To trap the area between bounds, use rectangles: on an increasing curve, left-hand heights give a lower bound and right-hand heights an upper bound.

Worked example Numerical methods · Newton–Raphson and trapezium rule

The trapezium rule with four strips

Question

Use the trapezium rule with the values of

2 x + 1

at

x = 0, 0.25, 0.5, 0.75, 1

to estimate

\int_{01} 2 x + 1 d x

. State whether it is an over- or underestimate.

1
Five ordinates give 4 strips, so $h = 0.25$ h = 0.25
2
Ordinates (4 d.p.): 1, 1.2247, 1.4142, 1.5811, 1.7321
3
Apply the rule: $\frac{0.25}{2} [1 + 1.7321 + 2 (1.2247 + 1.4142 + 1.5811)] = 1.3965$ 0.25/2[1 + 1.7321 + 2(1.2247 + 1.4142 + 1.5811)] = 1.3965
4
$\frac{d ^{2} y}{d x ^{2}} = - (2 x + 1)^{- 3/2} < 0$ d²y/dx² = −(2x+1)^(−3/2) < 0, so the curve is concave

Answer

1.3965 (4 d.p.), an underestimate (the exact value is

\frac{3 3 - 1}{3} = 1.3987

)

⚠️ Watch out

h

is the strip width, not

\frac{1}{number of ordinates}

. Using

h = 0.2

gives 1.1172.

Calculate

?A car starts from rest. Its velocity

v

m/s is recorded every 5 seconds.

$t$ t (s)	0	5	10	15	20
$v$ v (m/s)	0	8	13	16	17

Use the trapezium rule with all the data to estimate the distance travelled in the 20 seconds.

m

Sort it

?Will the trapezium rule over- or underestimate the area under each curve on the interval given?

Tap a statement, then tap the group it belongs in. Then press Check.

Mission check

?Starting from

x_{0} = 1

, how can four quick steps pin down

2

to 10 decimal places?

Recap

Big ideas

Sign change of a continuous $f$ f on a small interval shows a root; use the bounds (e.g. 2.085 and 2.095) to prove a rounded value.
Change of sign fails with repeated or paired roots (no change) and with asymptotes (change, but no root).
$x_{n + 1} = g (x_{n})$ xₙ₊₁ = g(xₙ) converges to $α$ α when $∣ g^{'} (α) ∣ < 1$ |g'(α)| < 1; the limit solves $x = g (x)$ x = g(x).
$g^{'} > 0$ g' > 0 gives a staircase diagram, $g^{'} < 0$ g' < 0 a cobweb.
Newton–Raphson: $x_{n + 1} = x_{n} - \frac{f ( x _{n} )}{f ^{'} ( x _{n} )}$ xₙ₊₁ = xₙ − (f(xₙ))/(f'(xₙ)). It fails at a stationary point and can go astray where the gradient is small.
Trapezium rule: $\frac{h}{2} [y_{0} + y_{n} + 2 (y_{1} + \dots + y_{n - 1})]$ h/2[y₀ + yₙ + 2(y₁ + … + yₙ₋₁)] with $h = \frac{b - a}{n}$ h = (b−a)/n for $n$ n strips.
Convex curve: overestimate. Concave curve: underestimate.

🏆

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You've worked through Numerical methods for Edexcel A-level Mathematics (9MA0). 🎉

Mission 1Radian mode: type 1, then cos(Ans) again and again. Where does it end?
Mission 2How can four quick steps pin down √2 to 10 decimal places?

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Specification points and sources for this lesson are listed at the bottom of the page.