Numerical methods – Year 13 A level maths revision notes to download as PDF

Worked example 1

Show that \(x^3+x-3=0\) has a root in \([1,2]\).

\(f(1)=-1<0\) and \(f(2)=7>0\), and \(f\) is continuous, so there is a root in \([1,2]\).

Worked example 2

Show that \(e^x-3x=0\) has a root near 0.6, and that the root of \(x^3+x-3=0\) is 1.213 to 3 decimal places.

\(f(0.6)=0.022>0\) and \(f(0.7)=-0.086<0\), so there is a root in \([0.6,0.7]\). For the cubic: \(f(1.2125)=-0.0049<0\) and \(f(1.2135)=0.0005>0\), so \(1.2125\lt \alpha<1.2135\).

Worked example 3

Use \(x_{n+1}=\sqrt[3]{3-x_n}\) with \(x_0=1\) to solve \(x^3+x-3=0\).

\(x_1=1.2599\), \(x_2=1.2028\), \(x_3=1.2158\), \(x_4=1.2129\), \(x_5=1.2135\), \(x_6=1.2134\). The values alternate about the root and close in on \(\alpha\approx1.2134\).

Worked example 4

Use Newton-Raphson on \(f(x)=x^3+x-3\) from \(x_0=1.2\).

\(f'(x)=3x^2+1\). \(x_1=1.2-\dfrac{-0.072}{5.32}=1.2135\), and the next step gives \(1.21341\).

Worked example 5

Use Newton-Raphson to find \(\sqrt5\).

With \(f(x)=x^2-5\): \(x_{n+1}=\dfrac12\left(x_n+\dfrac5{x_n}\right)\). From \(x_0=2\): \(2.25\), then \(2.2361=\sqrt5\).

Worked example 6

Estimate \(\int_1^3\ln x\,dx\) with 4 strips.

\(h=0.5\); the ordinates are \(0, 0.4055, 0.6931, 0.9163, 1.0986\), so the estimate is \(\dfrac{0.5}{2}\left[0+1.0986+2(0.4055+0.6931+0.9163)\right]=1.2821\). The exact value is \(3\ln3-2=1.2958\), so the estimate is too small by about 1.1%.

Worked example 7

A lake is measured every 10 m with depths \(0,2.1,3.4,3.8,2.9,1.5,0\) metres. Estimate the cross-section area.

\(\dfrac{10}{2}\left[0+2(2.1+3.4+3.8+2.9+1.5)+0\right]=10\times13.7=137\ \mathrm{m^2}\).

Worked example 8

Show that \(x=2\sin x\) has a positive root near 1.895, and use \(x_{n+1}=2\sin x_n\).

\(f(x)=x-2\sin x\): \(f(1.8)=-0.148\) and \(f(2)=0.181\), so there is a root in \([1.8,2]\). The iteration gives \(1.948,1.860,1.917,1.881,1.904,\dots\), a cobweb pattern.

Worked example 9: Showing a root to a given accuracy

Show that the root of \(x^3+2x-7=0\) is \(1.569\) to 3 decimal places.

Let \(f(x)=x^3+2x-7\). Evaluate at the two ends of the interval that round to 1.569: \(f(1.5685)=3.8588+3.1370-7=-0.0042<0\) and \(f(1.5695)=3.8662+3.1390-7=0.0052>0\). There is a change of sign and \(f\) is continuous, so the root lies in \([1.5685,1.5695]\), and every number in that interval rounds to 1.569.

Worked example 10: Using the ANS key

Explain how to calculate \(x_{n+1}=\sqrt{3x_n+1}\) with \(x_0=3\) using a calculator.

Type 3, press equals. Then type \(\sqrt{3\,\text{ANS}+1}\) and press equals repeatedly: the display gives \(x_1=3.1623\), \(x_2=3.2383\), \(x_3=3.2734\) and so on, slowly converging to the root \(3.3028\) of \(x^2-3x-1=0\). Keep all the digits on the calculator and only round at the end; rounding the iterates early can change the final digit.

Worked example 11: Newton-Raphson for a reciprocal

Use Newton-Raphson on \(f(x)=\dfrac1x-3\) to derive an iteration for \(\dfrac13\) that needs no division. Perform two steps from \(x_0=0.3\).

\(f\prime(x)=-\dfrac1{x^2}\). So \(x_{n+1}=x_n-\dfrac{1/x_n-3}{-1/x_n^2}=x_n+x_n^2\left(\dfrac1{x_n}-3\right)=2x_n-3x_n^2\). From \(x_0=0.3\): \(x_1=0.6-0.27=0.33\); \(x_2=0.66-0.3267=0.3333\). Early calculators used this trick to divide using only multiplication.

Worked example 12: Trapezium rule with data

A cyclist records her speed (m/s) every 5 s: 0, 4, 7, 9, 10, 10 for \(t=0,5,10,15,20,25\). Estimate the distance travelled in 25 s.

Distance \(=\int v\,dt\approx\dfrac52\left[0+10+2(4+7+9+10)\right]=2.5\times70=175\) m. The speed levels off so the speed-time curve is concave, and the trapezium rule underestimates slightly. The average speed is \(\dfrac{175}{25}=7\) m/s.

Worked example 13: Converting an equation for iteration

Rearrange \(x^3=5x-1\) into the form \(x=g(x)\) in two ways and decide which converges to the root near 0.2.

(A) \(x=\dfrac{x^3+1}5\) gives \(g\prime(x)=\dfrac{3x^2}5\), which at \(x=0.2\) is \(0.024\), very small, so the iteration converges quickly: from \(x_0=0\): \(0.2\), \(0.2016\), \(0.20164\). (B) \(x=\sqrt[3]{5x-1}\) gives \(g\prime(x)=\dfrac{5}{3(5x-1)^{2/3}}\), which at the root near 0.2 involves the cube root of a tiny number, so \(|g\prime|\) is large and the iteration fails. (B) finds the larger root near 2.13 instead.

Worked example 14: Newton-Raphson for a trigonometric equation

Use Newton-Raphson with \(x_0=1\) to solve \(x=2\sin x\) (radians) and find the positive root to 4 decimal places.

Let \(f(x)=x-2\sin x\). Then \(f\prime(x)=1-2\cos x\). \(x_1=1-\dfrac{1-2\sin1}{1-2\cos1}=1-\dfrac{-0.68294}{-0.08060}=-7.47\) which is nowhere near the root: the derivative is almost zero at \(x_0=1\) because \(\cos1\approx0.5\). This is the typical failure. A better start from the graph is \(x_0=2\): \(f(2)=0.18141\), \(f\prime(2)=1.83229\), so \(x_1=1.90099\), \(x_2=1.89551\), \(x_3=1.89549\). The root is 1.8955 to 4 d.p. Always sketch the graph to choose a start where the gradient is not close to zero.

Worked example 15: Is the estimate an over- or under-estimate?

The trapezium rule is used on \(\displaystyle\int_0^3\ln(x+1)\,dx\) with 3 strips. State whether the estimate is too large or too small, then calculate it.

\(y=\ln(x+1)\) has second derivative \(-\dfrac1{(x+1)^2}<0\), so the curve is concave and the chords lie below it: the estimate is too small. Ordinates for \(h=1\): \(0,0.6931,1.0986,1.3863\). Estimate \(=\dfrac12\left[0+1.3863+2(0.6931+1.0986)\right]=\dfrac12\times4.9697=2.4849\). The exact value is \(4\ln4-3=2.5452\), so the estimate is \(2.4\%\) too small.

Worked example 16: Using bounds to show accuracy

Show that the root of \(e^x=3x\) near 1.5 is \(1.512\) to 3 decimal places.

Let \(f(x)=e^x-3x\). \(f(1.5115)=4.5335-4.5345=-0.0010<0\) and \(f(1.5125)=4.5381-4.5375=0.0006>0\). There is a change of sign on \([1.5115,1.5125]\), so the root is 1.512 to 3 d.p.