9 Numerical methods — 9MA0 revision notes

9.1

Locate roots of f(x) = 0 by considering changes of sign of f(x) in an interval on which f(x) is sufficiently well behaved; understand how change of sign methods can fail.

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

If a continuous function has values of opposite sign at the endpoints of an interval, the intermediate value theorem guarantees at least one root inside that interval.
Evaluate $f(a)$ and $f(b)$ accurately, record their signs, and conclude that a root lies in $(a,b)$ only after checking that $f$ is continuous there.
For example, $f(1)<0$ and $f(2)>0$ for a continuous $f$ , so the graph must cross the $x$ -axis at least once between $1$ and $2$ .
A sign change across a discontinuity need not contain a root, while a repeated root can touch the axis without changing sign; a common error is to treat the sign test as an equivalence.

Worked example

For $f(x)=x^3+x-5$ , show that there is exactly one root in $(1,2)$ .

1.Calculate $f(1)=1+1-5=-3$ and $f(2)=8+2-5=5$ .
2.Continuity gives at least one root in $(1,2)$ .
3.Also $f'(x)=3x^2+1$ , which is positive for every real $x$ , so $f$ is strictly increasing and cannot cross the axis more than once.

Answer: $f(1)=-3$ and $f(2)=5$ , so a root lies in $(1,2)$ .; Since $f'(x)=3x^2+1>0$ , $f$ is strictly increasing and the root is unique.

Common mistakes

Don't conclude that an even-multiplicity root cannot exist because endpoint values have the same sign.
Don't claim a root from a sign change without checking continuity, or claim uniqueness without a monotonicity argument.

Exam tip

To establish exactly one root, combine a sign change on a continuous interval with a separate uniqueness argument.

Tier 1 · Easy

ORIGINAL

1.

Let

f(x)=x^2-x-1

. Use endpoint values to establish that

(1,2)

brackets a zero of

f

.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1.

The function

f(x)=(x-1)^2

has a root in the interval

[0,2]

. Explain why a search based only on a change of sign between the endpoints does not detect this root.

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

1.

Two sign tests are proposed. For

h(x)=\frac{1}{x-2}

, the values

h(1)

and

h(3)

have opposite signs. For

k(x)=(x-2)^2

, the values

k(1.9)

and

k(2.1)

have the same sign. Explain why neither test gives the correct conclusion about roots in the stated intervals.

(4)

(Total for Question 1 is 4 marks)

9.2

Solve equations approximately using simple iterative methods; be able to draw associated cobweb and staircase diagrams.

Notes

Worked answers & exam appearances →

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

A fixed-point iteration rewrites an equation as $x=g(x)$ and generates approximations using $x_{n+1}=g(x_n)$ from a chosen starting value.
Calculate successive values without premature rounding; on a graph of $y=g(x)$ and $y=x$ , move vertically to $y=g(x)$ and horizontally to $y=x$ to display each iteration.
Near a fixed point $\alpha$ , iterations usually converge when $|g'(\alpha)|<1$ : a positive gradient gives a staircase pattern and a negative gradient gives a cobweb pattern.
Different rearrangements of the same equation can converge at different rates or diverge; a common error is to assume that obtaining $x=g(x)$ automatically produces a useful recurrence.

A staircase diagram alternates between

y=g(x)

and

y=x

to display successive fixed-point iterates.

Worked example

Use $x_{n+1}=\sqrt{7-x_n}$ with $x_0=2$ to calculate $x_1$ , $x_2$ and $x_3$ to $4$ decimal places. State the equation satisfied by any limiting value.

1.Repeated substitution gives $x_1=\sqrt5=2.236067\ldots$ , $x_2=\sqrt{7-2.236067\ldots}=2.182643\ldots$ and $x_3=\sqrt{7-2.182643\ldots}=2.194847\ldots$ .
2.If the sequence tends to $L$ , then $L=\sqrt{7-L}$ , so $L^2+L-7=0$ .

Answer: $x_1=2.2361$ , $x_2=2.1826$ and $x_3=2.1948$ .; A limiting value satisfies $x^2+x-7=0$ .

Common mistakes

Don't draw horizontal moves to $y=g(x)$ and vertical moves to $y=x$ , reversing the cobweb construction.
Don't calculate iterates and fail to state the fixed-point equation or assess convergence from the iteration graph.

Exam tip

Show each substituted iterate to the requested accuracy, then set the limiting value equal to the iteration function.

Tier 1 · Easy

ORIGINAL

1.

The recurrence

x_{n+1}=\frac{1}{2}\left(x_n+\frac{6}{x_n}\right)

is used with

x_0=2

. Find

x_1

and

x_2

.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1.

On a pair of axes, draw

y=x

and

y=2+\dfrac14x

. Starting at

x_0=0

on the

x

-axis, draw the first three steps of the fixed-point construction. State whether the construction is a staircase or a cobweb, and find its limiting value.

(4)

(Total for Question 1 is 4 marks)

Tier 3 · Hard

ORIGINAL

1.

For

x_{n+1}=3-0.4x_n

with

x_0=1

, find

x_1

to

x_4

. Determine the limiting value and state whether the graphical construction is a cobweb or a staircase, giving a reason.

(5)

(Total for Question 1 is 5 marks)

9.3

Solve equations using the Newton-Raphson method and other recurrence relations of the form xₙ₊₁ = g(xₙ); understand how such methods can fail.

Notes

Worked answers & exam appearances →

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

Newton-Raphson replaces the curve locally by its tangent, giving $x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$ for a root of $f(x)=0$ .
Differentiate first, substitute the current approximation into both $f$ and $f'$ , and retain extra calculator digits until the requested final accuracy.
For instance, applying Newton-Raphson to $f(x)=x^2-a$ gives $x_{n+1}=\frac12\left(x_n+\frac{a}{x_n}\right)$ .
The method can fail when $f'(x_n)=0$ , when a tangent sends the next value away from the desired root, or when the values enter a cycle; a common error is to continue without checking the iterates.
Set $f(x)=x^3+x-4$ , so $f'(x)=3x^2+1$ .

Newton–Raphson uses the tangent at

x_n

; its intercept with the

x

-axis is the next iterate

x_{n+1}

.

Worked example

Derive a Newton-Raphson recurrence for $x^3+x-4=0$ . Starting with $x_0=1.3$ , find $x_1$ and $x_2$ to $5$ decimal places.

1.Set $f(x)=x^3+x-4$ , so $f'(x)=3x^2+1$ .
2.Substitution in the Newton-Raphson formula gives $x_{n+1}=x_n-\frac{x_n^3+x_n-4}{3x_n^2+1}$ .
3.Using $x_0=1.3$ gives $x_1=1.382866\ldots$ , and substituting this unrounded value gives $x_2=1.378806\ldots$ .

Answer: $x_{n+1}=x_n-\frac{x_n^3+x_n-4}{3x_n^2+1}$ .; $x_1=1.38287$ and $x_2=1.37881$ .

Common mistakes

Don't choose a starting value where $f'(x_0)=0$ , making the first Newton step undefined.
Don't use the Newton formula with the derivative evaluated at the new iterate rather than the current one.

Exam tip

Write the Newton recurrence explicitly, retain guard digits between iterations and round only the reported values.

Tier 1 · Easy

ORIGINAL

1.

Use one Newton-Raphson step on

f(x)=x^2-7

starting from

x_0=2.5

. Find

x_1

.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1.

Newton-Raphson is applied to

f(x)=x^3-2x+2

starting from

x_0=0

. Calculate

x_1

,

x_2

and

x_3

, and explain why the method fails to converge to a root from this starting value.

(4)

(Total for Question 1 is 4 marks)

Tier 3 · Hard

ORIGINAL

1.

Newton-Raphson is applied to

f(x)=x^2-a

, where

a>0

, starting from

x_0=4

. The first iterate is

x_1=\dfrac{13}{4}

. Find

a

. Then calculate

x_2

exactly and

x_3

to six decimal places, and hence give the positive root of

f(x)=0

correct to four decimal places.

(5)

(Total for Question 1 is 5 marks)

9.4

Understand and use numerical integration of functions, including the use of the trapezium rule and estimating the approximate area under a curve and limits that it must lie between.

Notes

Worked answers & exam appearances →

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

The trapezium rule approximates an integral by replacing the curve over each strip with a straight chord and adding the resulting trapezium areas.
For ordinates $y_0,\ldots,y_n$ at equal spacing $h$ , use $\frac{h}{2}[y_0+y_n+2(y_1+\cdots+y_{n-1})]$ and include each interior ordinate twice.
A convex curve with $f''(x)>0$ lies below its chords, so the trapezium rule overestimates its integral; more, narrower strips usually improve the estimate.
Numerical area is not automatically the signed integral: a common error is to leave contributions below the $x$ -axis negative when the question asks for total area.

The trapezium rule joins consecutive ordinates with straight segments; convexity indicates whether this polygonal boundary lies above or below the curve.

Worked example

Use four equal strips and the trapezium rule to estimate $\int_0^2(x^2+1)\,dx$ . State whether the estimate is an overestimate or an underestimate, with a reason.

1.Here $h=0.5$ and the ordinates at $x=0,0.5,1,1.5,2$ are $1,1.25,2,3.25,5$ .
2.Thus the estimate is $\frac{0.5}{2}[1+5+2(1.25+2+3.25)]=0.25(19)=4.75$ .
3.Since $f''(x)=2>0$ , each chord lies above the curve, so the result is an overestimate.

Answer: $4.75$ ; It is an overestimate because $f''(x)=2>0$ , so the curve is convex.

Common mistakes

Don't apply weight two to the endpoints and weight one to the interior ordinates, reversing the trapezium-rule coefficients.
Don't use the number of ordinates as the number of strips, giving the wrong strip width and endpoint weights.

Exam tip

For the trapezium rule, calculate strip width from the interval and use half weight only on the two endpoints.

Tier 1 · Easy

ORIGINAL

1.

The values of a function are

y=1,3,9

at

x=0,1,2

respectively. Use two strips and the trapezium rule to estimate

\int_0^2 y\,dx

.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1.

The values of a function at

x=0,1,2,3,4

are respectively

2,2.5,4,6.5,10

. Use the trapezium rule to estimate

\displaystyle\int_0^4 f(x)\,\mathrm dx

.

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

1.

For

f(x)=\frac1x

on

1\leq x\leq2

, use four equal strips to find a trapezium-rule estimate. By using left- and right-endpoint rectangles, also give lower and upper bounds for

\int_1^2\frac1x\,dx

. Give decimals to

4

decimal places.

(6)

(Total for Question 1 is 6 marks)

9.5

Use numerical methods to solve problems in context.

Notes

Worked answers & exam appearances →

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

A contextual numerical model translates a physical condition, such as a target value or accumulated quantity, into a root, recurrence or numerical integral.
Define the function and units before applying the method, choose an interval or starting value that is meaningful in the context, and state the requested accuracy.
For a threshold $M(t)=C$ , solve $M(t)-C=0$ numerically; for a varying rate $q(t)$ , approximate the accumulated change with $\int q(t)\,dt$ .
A calculator value is not a complete contextual answer: a common error is to omit units, ignore the model's domain or round in a way that makes a safety decision invalid.

Worked example

A culture is modelled by $P(t)=200e^{0.3t}$ , where $t$ is in hours. Use Newton-Raphson on $f(t)=200e^{0.3t}-1000$ with $t_0=5$ to find $t_1$ and $t_2$ . Hence estimate when the culture reaches $1000$ cells.

1.Differentiate to get $f'(t)=60e^{0.3t}$ .
2.Use $t_{n+1}=t_n-\frac{200e^{0.3t_n}-1000}{60e^{0.3t_n}}$ .
3.Substitution of $t_0=5$ gives $t_1=5.385502\ldots$ , and the next substitution gives $t_2=5.364857\ldots$ .
4.The positive time is therefore about $5.365$ hours.

Answer: $t_1=5.38550$ and $t_2=5.36486$ .; The culture reaches $1000$ cells after approximately $5.365$ hours.

Common mistakes

Don't select a root that satisfies the equation but lies outside the interval allowed by the context.
Don't carry out a numerical iteration but give no contextual time, units or check that the root is meaningful.

Exam tip

After numerical iteration, translate the root back into the context and report a sensible accuracy with units.

Tier 1 · Easy

ORIGINAL

1.

A greenhouse temperature is modelled by

T(t)=18+4t-t^2

degrees Celsius, where

t

is hours after noon. Show that the temperature returns to

20^\circ\mathrm{C}

at a time between

3

and

4

hours after noon.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1.

A particle's height is modelled by

h(t)=20t-4.9t^2

metres. Show that the positive time when the particle returns to the ground lies in

(4.0,4.1)

seconds and, by considering the sign of

h(4.05)

, show that it lies in

(4.05,4.10)

seconds.

(4)

(Total for Question 1 is 4 marks)

Tier 3 · Hard

ORIGINAL

1.

Water enters a tank at rates

10,14,18,21,19,15,11

litres per minute at times

0,5,10,15,20,25,30

minutes. The tank initially contains

350

litres and has capacity

800

litres. Use the trapezium rule to estimate whether the tank overflows during the

30

minutes, and estimate any excess at the end.

(5)

(Total for Question 1 is 5 marks)

Locate roots of f(x) = 0 by considering changes of sign of f(x) in an interval on which f(x) is sufficiently well behaved; understand how change of sign methods can fail.

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Solve equations approximately using simple iterative methods; be able to draw associated cobweb and staircase diagrams.

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Solve equations using the Newton-Raphson method and other recurrence relations of the form xₙ₊₁ = g(xₙ); understand how such methods can fail.

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Understand and use numerical integration of functions, including the use of the trapezium rule and estimating the approximate area under a curve and limits that it must lie between.

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Use numerical methods to solve problems in context.

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

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