9 Numerical methods — printable coverage pack

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.

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.

Tier 1 · Easy

1. Let $f(x)=x^2-x-1$ . Use endpoint values to establish that $(1,2)$ brackets a zero of $f$ .[2 marks]
Answer
- $f(1)=-1$ and $f(2)=1$ , so continuity and the change of sign give a root in $(1,2)$ .
Method: Substitute the endpoints: $f(1)=1-1-1=-1$ and $f(2)=4-2-1=1$ . A polynomial is continuous, and the endpoint values have opposite signs, so the intermediate value theorem guarantees at least one root between $1$ and $2$ .

Tier 2 · Standard

1. For $f(x)=x^3+x-5$ , show that there is exactly one root in $(1,2)$ .[4 marks]
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.
Method: Calculate $f(1)=1+1-5=-3$ and $f(2)=8+2-5=5$ . Continuity gives at least one root in $(1,2)$ . 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.

Tier 3 · Hard

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 marks]
Answer
- $h$ changes sign across $x=2$ but has no root because it is discontinuous there.
- $k$ has the repeated root $x=2$ but does not change sign across it.
Method: For $h$ , $h(1)=-1$ and $h(3)=1$ , but $h$ is undefined at $x=2$ , so continuity fails and the sign change occurs across a vertical asymptote rather than a root. For $k$ , $k(2)=0$ , but squaring makes $k(x)\geq0$ on both sides of $2$ , so the graph touches the axis at a repeated root without crossing it.

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

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.

Tier 1 · Easy

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 marks]
Answer
- $x_1=2.5$ and $x_2=2.45$
Method: Substitute $x_0=2$ to obtain $x_1=\frac12(2+6/2)=2.5$ . Then $x_2=\frac12(2.5+6/2.5)=\frac12(4.9)=2.45$ .

Tier 2 · Standard

1. 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.[4 marks]
Answer
- $x_1=2.2361$ , $x_2=2.1826$ and $x_3=2.1948$ .
- A limiting value satisfies $x^2+x-7=0$ .
Method: Repeated substitution gives $x_1=\sqrt5=2.236067\ldots$ , $x_2=\sqrt{7-2.236067\ldots}=2.182643\ldots$ and $x_3=\sqrt{7-2.182643\ldots}=2.194847\ldots$ . If the sequence tends to $L$ , then $L=\sqrt{7-L}$ , so $L^2+L-7=0$ .

Tier 3 · Hard

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 marks]
Answer
- $x_1=2.6$ , $x_2=1.96$ , $x_3=2.216$ and $x_4=2.1136$ .
- The limit is $\frac{15}{7}$ , and the construction is a convergent cobweb because $g'(x)=-0.4$ .
Method: Apply the recurrence successively: $x_1=3-0.4=2.6$ , $x_2=3-0.4(2.6)=1.96$ , $x_3=3-0.4(1.96)=2.216$ and $x_4=3-0.4(2.216)=2.1136$ . A limit $L$ obeys $L=3-0.4L$ , hence $L=3/1.4=15/7$ . Since the constant gradient is negative, the iterates alternate across the fixed point, producing a cobweb; since $|-0.4|<1$ , it converges.

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.

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.

Tier 1 · Easy

1. Use one Newton-Raphson step on $f(x)=x^2-7$ starting from $x_0=2.5$ . Find $x_1$ .[2 marks]
Answer
- $x_1=2.65$
Method: Here $f'(x)=2x$ . Therefore $x_1=2.5-\frac{2.5^2-7}{2(2.5)}=2.5-\frac{-0.75}{5}=2.65$ .

Tier 2 · Standard

1. 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.[5 marks]
Answer
- $x_{n+1}=x_n-\frac{x_n^3+x_n-4}{3x_n^2+1}$ .
- $x_1=1.38287$ and $x_2=1.37881$ .
Method: Set $f(x)=x^3+x-4$ , so $f'(x)=3x^2+1$ . Substitution in the Newton-Raphson formula gives $x_{n+1}=x_n-\frac{x_n^3+x_n-4}{3x_n^2+1}$ . Using $x_0=1.3$ gives $x_1=1.382866\ldots$ , and substituting this unrounded value gives $x_2=1.378806\ldots$ .

Tier 3 · Hard

1. Newton-Raphson is applied to $f(x)=x^3-2x+2$ with $x_0=0$ . Calculate $x_1$ , $x_2$ and $x_3$ , then explain why the method fails from this starting value.[4 marks]
Answer
- $x_1=1$ , $x_2=0$ and $x_3=1$ .
- The iterates form the two-cycle $0,1,0,1,\ldots$ and do not approach a root.
Method: Since $f'(x)=3x^2-2$ , Newton-Raphson gives $x_{n+1}=x_n-\frac{x_n^3-2x_n+2}{3x_n^2-2}$ . From $x_0=0$ , $x_1=0-2/(-2)=1$ . Then $x_2=1-1/1=0$ , so the same two values repeat and $x_3=1$ . The recurrence is trapped in a cycle rather than converging.

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.

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.

Tier 1 · Easy

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 marks]
Answer
- $8$
Method: The strip width is $h=1$ . The trapezium rule gives $\frac12[1+9+2(3)]=\frac12(16)=8$ .

Tier 2 · Standard

1. 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.[4 marks]
Answer
- $4.75$
- It is an overestimate because $f''(x)=2>0$ , so the curve is convex.
Method: Here $h=0.5$ and the ordinates at $x=0,0.5,1,1.5,2$ are $1,1.25,2,3.25,5$ . Thus the estimate is $\frac{0.5}{2}[1+5+2(1.25+2+3.25)]=0.25(19)=4.75$ . Since $f''(x)=2>0$ , each chord lies above the curve, so the result is an overestimate.

Tier 3 · Hard

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 marks]
Answer
- Trapezium estimate $=0.6970$ .
- $0.6345<\int_1^2\frac1x\,dx<0.7595$ .
Method: The width is $h=0.25$ and the ordinates are $1,\frac45,\frac23,\frac47,\frac12$ . The trapezium estimate is $\frac18[1+\frac12+2(\frac45+\frac23+\frac47)]=\frac{1171}{1680}=0.6970$ . Since $1/x$ is decreasing, right endpoints give the lower sum $\frac14(\frac45+\frac23+\frac47+\frac12)=\frac{533}{840}=0.6345$ , while left endpoints give the upper sum $\frac14(1+\frac45+\frac23+\frac47)=\frac{319}{420}=0.7595$ .

9.5 · Use numerical methods to solve problems in context.

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.

Tier 1 · Easy

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 marks]
Answer
- $T(3)-20=1$ and $T(4)-20=-2$ , so the return time lies between $3$ and $4$ hours after noon.
Method: Define $f(t)=T(t)-20=-2+4t-t^2$ . Then $f(3)=-2+12-9=1$ and $f(4)=-2+16-16=-2$ . The model is continuous and the sign changes, so a solution occurs in $(3,4)$ .

Tier 2 · Standard

1. 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.[5 marks]
Answer
- $t_1=5.38550$ and $t_2=5.36486$ .
- The culture reaches $1000$ cells after approximately $5.365$ hours.
Method: Differentiate to get $f'(t)=60e^{0.3t}$ . Use $t_{n+1}=t_n-\frac{200e^{0.3t_n}-1000}{60e^{0.3t_n}}$ . Substitution of $t_0=5$ gives $t_1=5.385502\ldots$ , and the next substitution gives $t_2=5.364857\ldots$ . The positive time is therefore about $5.365$ hours.

Tier 3 · Hard

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 marks]
Answer
- Estimated inflow $=487.5$ litres and estimated final volume $=837.5$ litres.
- The model predicts overflow, with $37.5$ litres in excess of capacity at $30$ minutes.
Method: With spacing $h=5$ , the trapezium estimate is $\frac52[10+11+2(14+18+21+19+15)]=\frac52(195)=487.5$ litres. Adding the initial $350$ litres gives $837.5$ litres, which is $837.5-800=37.5$ litres above capacity, so the tank must have overflowed by the end.