P3.8 Numerical methods
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A2 content for Pure Mathematics 3 (specification sections 6.1 and 6.2). Curve sketching comes from P1 and the trapezium rule, the other numerical method in the course, from P2.
What this page covers
8 points
- Locating a root of \(f(x) = 0\) by a change of sign of a continuous function
- The situations where the change-of-sign test misleads: asymptotes, two roots close together, repeated roots
- Rearranging \(f(x) = 0\) into the form \(x = g(x)\)
- Carrying out an iteration \(x_{n+1} = g(x_n)\) and giving the iterates to a stated accuracy
- Proving that a root is correct to a given number of decimal places
- Staircase and cobweb diagrams
- Deciding (qualitatively) whether an iteration converges or diverges
- Putting every step together in a full multi-part question
In one minute
6 points
- If \(f\) is continuous on \([a, b]\) and \(f(a)\) and \(f(b)\) have opposite signs, then \(f(x) = 0\) has at least one root between \(a\) and \(b\); this change-of-sign test, together with fixed-point iteration, lets you find roots of equations that have no algebraic solution.
- The test misleads if \(f\) is not continuous on the interval (an asymptote), and it can miss roots if there are two (or a repeated root) inside the interval.
- Iteration: rearrange \(f(x) = 0\) as \(x = g(x)\), choose \(x_0\), and compute \(x_1 = g(x_0)\), \(x_2 = g(x_1)\), … If the values settle down, they settle on a root.
- Proving accuracy: to show \(\alpha = 2.279\) to 3 d.p., show that \(f\) changes sign between 2.2785 and 2.2795, then conclude.
- Diagrams: a staircase approaches the root from one side; a cobweb spirals round it from alternate sides.
- Convergence: near the root, if the curve \(y = g(x)\) is less steep than \(y = x\), the iteration converges; if it is steeper, the iterates move away.
Contents
14 sections
- How does a change of sign show that a root exists?
- When can a change of sign mislead you?
- How do you rearrange f(x) = 0 into the form x = g(x)?
- How do you carry out an iteration?
- How do you show that a root is correct to a given accuracy?
- What do staircase and cobweb diagrams show?
- Why do some iterations fail to converge?
- How does a full numerical methods question fit together?
- Exam tips for P3.8
- Common misconceptions
- How this topic connects
- Key formulae
- Check your understanding
- Specification checklist
How does a change of sign show that a root exists?
Equations such as \(x^3 - 3x - 5 = 0\), \(\ln x + 2x - 4 = 0\) or \(e^{x/2} = 3x - 1\) have no algebraic solution you can write down, yet you may still be asked for a root to three decimal places. Numerical methods give you two tools: the change-of-sign test, which traps a root in an interval, and fixed-point iteration, which homes in on it.
A root of \(f(x) = 0\) is a value of \(x\) where the graph of \(y = f(x)\) meets the \(x\)-axis. If the graph is continuous (an unbroken curve) between \(x = a\) and \(x = b\), and it is below the axis at one end and above it at the other, it must cross the axis somewhere in between. You cannot get from one side of a line to the other without crossing it.
Polynomials, \(e^x\), \(\sin x\) and \(\cos x\) are continuous everywhere; \(\ln x\) is continuous for \(x > 0\); \(\frac{1}{x-k}\) and \(\tan x\) have breaks. Always think about whether the interval contains a break before you use the test.
\(f(x) = x^3 - 3x - 5\). Show that the equation \(f(x) = 0\) has a root \(\alpha\) in the interval \([2.2, 2.3]\).
Evaluate \(f\) at both ends: \(f(2.2) = 10.648 - 6.6 - 5 = -0.952\) and \(f(2.3) = 12.167 - 6.9 - 5 = 0.267\).
\(f(2.2) < 0\) and \(f(2.3) > 0\), so there is a change of sign, and \(f\) is continuous because it is a polynomial. There is a root \(\alpha\) with \(2.2 < \alpha < 2.3\).
Write "\(f(2.2) = -0.952 < 0\), \(f(2.3) = 0.267 > 0\)". One or two significant figures are enough as long as the sign is right, but a bare "negative, positive" cannot be checked by the reader.
When can a change of sign mislead you?
The test has a condition (continuity) and a limitation (it counts crossings in pairs). Each gives a way to be misled.
- A sign change with no root. If \(f\) has a vertical asymptote (or any break) inside the interval, \(f\) can jump from negative to positive without ever being zero. Typical culprits: fractions whose denominator is zero inside the interval, and \(\tan x\) near \(\frac{\pi}{2}\).
- Roots with no sign change. If the interval contains two roots (or any even number), \(f\) goes down through the axis and back up, so \(f(a)\) and \(f(b)\) have the same sign. The same happens with a repeated root, where the curve only touches the axis.
- The interval is too wide. A large interval may contain several roots, so a sign change tells you little. Sketch the curve first, or use smaller steps.
\(f(x) = \frac{2x+1}{x-3}\), \(x \ne 3\). A student finds \(f(2)\) and \(f(4)\) and concludes that \(f(x) = 0\) has a root between 2 and 4. (a) Find \(f(2)\) and \(f(4)\). (b) Explain why the student's conclusion is wrong. (c) Find the root of \(f(x) = 0\).
(a) \(f(2) = 5 \div (-1)\) and \(f(4) = 9 \div 1\). \(f(2) = -5\), \(f(4) = 9\)
(b) Check the condition the test depends on. \(f\) is not continuous on \([2, 4]\): the denominator is zero at \(x = 3\), where the graph has a vertical asymptote. The sign changes by jumping across the asymptote, not by crossing the \(x\)-axis.
(c) A fraction is zero when its numerator is zero: \(2x + 1 = 0\). \(x = -\frac12\), which is not in \([2, 4]\).
Writing "there is a change of sign, so there is a root" without mentioning continuity. The continuity condition is exactly what fails in worked example 2, and it is part of the reasoning a full answer needs.
How do you rearrange f(x) = 0 into the form x = g(x)?
Make \(x\) the subject of the equation in a way that leaves \(x\) on the right as well. Any value that satisfies \(x = g(x)\) is then a root of \(f(x) = 0\): the two equations are equivalent. Geometrically, the root is where the line \(y = x\) meets the curve \(y = g(x)\).
One equation has many rearrangements. Common moves:
- isolate a power and take a root: \(x^3 = 3x + 5 \Rightarrow x = \sqrt[3]{3x+5}\)
- divide through by a power of \(x\): \(x^2 = 3 + \frac5x \Rightarrow x = \sqrt{3 + \frac5x}\)
- isolate a linear term: \(3x = x^3 - 5 \Rightarrow x = \frac{x^3 - 5}{3}\)
- undo an exponential or a logarithm: \(e^{x/2} = 3x - 1 \Rightarrow x = 2\ln(3x - 1)\)
In practice the target form is almost always given ("Show that the equation can be written as …"). Your job is to get there with every step visible.
Show that the equation \(x^3 - 3x - 5 = 0\) can be written as (a) \(x = \sqrt[3]{3x+5}\); (b) \(x = \sqrt{3 + \frac{5}{x}}\), \(x > 0\).
(a) Isolate \(x^3\), then take the cube root: \(x^3 = 3x + 5 \Rightarrow\) \(x = \sqrt[3]{3x+5}\)
(b) From \(x^3 = 3x + 5\), divide by \(x\) (allowed, as \(x \ne 0\)): \(x^2 = 3 + \frac5x\). Taking the positive root as \(x > 0\): \(x = \sqrt{3 + \frac5x}\)
Both rearrangements have exactly the same positive root as the original cubic. Whether an iteration based on them actually finds it is a separate question, answered in the sections on diagrams and convergence.
How do you carry out an iteration?
Why does a limit have to be a root? If \(x_n \to \alpha\), then \(x_{n+1} \to \alpha\) too. Taking the limit of both sides of \(x_{n+1} = g(x_n)\) (for continuous \(g\)) gives \(\alpha = g(\alpha)\), which is the rearranged equation. A good starting value comes from a sign-change interval or a sketch. The iteration is only useful if it converges.
Using your calculator
- Type the starting value and press =. Then type the formula with Ans in place of \(x\) (for example \(\sqrt[3]{3\,\text{Ans} + 5}\)) and press = repeatedly. Each press gives the next iterate, and the calculator keeps full accuracy.
- Write down each iterate to the accuracy asked (usually 4 d.p.), but never round and re-type: rounded values fed back in can change later digits.
- Angles in an iteration with sin, cos or tan are in radians. Check your calculator mode before you start.
Use the iteration formula \(x_{n+1} = \sqrt[3]{3x_n + 5}\) with \(x_0 = 2\) to find the values of \(x_1\), \(x_2\) and \(x_3\), giving each to 4 decimal places.
First iterate: \(x_1 = \sqrt[3]{3 \times 2 + 5} = \sqrt[3]{11} = 2.223980\ldots\)
Feed back the unrounded value (Ans): \(x_2 = \sqrt[3]{3 \times 2.223980\ldots + 5} = 2.2684\ldots\), then \(x_3 = 2.2770\ldots\) \(x_1 = 2.2240\), \(x_2 = 2.2684\), \(x_3 = 2.2770\)
The values are increasing towards the root \(\alpha \approx 2.2790\): each step closes most of the remaining gap. Writing \(x_1\) as 2.224 (3 d.p.) instead of 2.2240 does not meet the "4 decimal places" instruction; trailing zeros count.
How do you show that a root is correct to a given accuracy?
Iterates that agree to 3 d.p. strongly suggest the answer but do not prove it. The proof is a change of sign over the right interval.
"\(\alpha = 2.279\) correct to 3 d.p." means \(2.2785 \le \alpha < 2.2795\). So you evaluate \(f\) at the bounds 2.2785 and 2.2795, half a unit of the last decimal place either side. If \(f\) changes sign between the bounds (and \(f\) is continuous there), the root lies in \([2.2785, 2.2795]\), and every number in that interval rounds to 2.279. For 2 d.p. the bounds are \(\pm 0.005\); for 4 d.p., \(\pm 0.00005\).
Show that \(\alpha = 2.279\), correct to 3 decimal places, where \(\alpha\) is the root of \(x^3 - 3x - 5 = 0\) in \([2.2, 2.3]\).
Choose the interval \([2.2785, 2.2795]\). Evaluate \(f\) at both bounds: \(f(2.2785) = -0.00653\) and \(f(2.2795) = 0.00606\).
There is a change of sign and \(f\) is continuous, so \(2.2785 < \alpha < 2.2795\). Hence \(\alpha = 2.279\) to 3 d.p.
The values of \(f\) at the bounds are tiny, so give them to at least 1 or 2 significant figures; "−0.0065 and 0.0061" is fine, "0.00 and 0.01" is not.
Testing \(f(2.278)\) and \(f(2.280)\). A sign change there shows the root is between 2.278 and 2.280, which includes numbers such as 2.2782 that round to 2.278. Only the bounds 2.2785 and 2.2795 prove the 3 d.p. value.
What do staircase and cobweb diagrams show?
Draw \(y = g(x)\) and \(y = x\) on the same axes; the root is where they cross. From \(x_0\) on the \(x\)-axis go vertically to the curve: the height there is \(g(x_0) = x_1\). Then go horizontally to the line \(y = x\): you are now directly above \(x_1\) on the \(x\)-axis. Repeat: vertical to the curve, horizontal to the line.
- Staircase: the path climbs (or descends) in steps on one side of the root. The iterates approach the root from one direction only. This happens when \(g\) is increasing near the root.
- Cobweb: the path spirals round the crossing point. The iterates land alternately above and below the root. This happens when \(g\) is decreasing near the root.
The equation \(\ln x + 2x - 4 = 0\) has a single root \(\alpha\). (a) Show that \(\alpha\) lies between 1.7 and 1.8. (b) Show that the equation can be written as \(x = 2 - \frac12\ln x\). (c) Using \(x_{n+1} = 2 - \frac12\ln x_n\) with \(x_0 = 1\), find \(x_1\), \(x_2\), \(x_3\) and \(x_4\) to 4 decimal places, and state whether a staircase or a cobweb diagram would show these iterates.
(a) Let \(f(x) = \ln x + 2x - 4\), continuous for \(x > 0\). \(f(1.7) = -0.0694 < 0\) and \(f(1.8) = 0.188 > 0\): change of sign and \(f\) continuous. \(1.7 < \alpha < 1.8\)
(b) Isolate the \(2x\) term and divide by 2: \(2x = 4 - \ln x \Rightarrow\) \(x = 2 - \frac12\ln x\)
(c) \(\ln 1 = 0\) makes the first step exact. \(x_1 = 2.0000\), \(x_2 = 1.6534\), \(x_3 = 1.7486\), \(x_4 = 1.7206\) The iterates are alternately above and below \(\alpha\) (\(2.0000 > \alpha\), \(1.6534 < \alpha\), \(1.7486 > \alpha\), …), so the diagram is a cobweb.
Why do some iterations fail to converge?
Different rearrangements of the same equation behave differently. What matters is how steep the curve \(y = g(x)\) is near the root, compared with the line \(y = x\), which has gradient 1.
- Less steep than \(y = x\) (gradient between −1 and 1): each step moves the iterate closer to the root, so the iteration converges (as long as \(x_0\) is close enough). The flatter the curve, the faster it converges.
- Steeper than \(y = x\) (gradient greater than 1, or less than −1): each step throws the iterate further away, so it diverges from that root, as a staircase going outwards or a cobweb unwinding.
With P3 differentiation you can check this: work out \(g'(x)\) near the root and see whether its size is less than 1. For \(\sqrt[3]{3x+5}\) the gradient at the root is about 0.19; for \(\frac{x^3-5}{3}\) it is \(x^2 \approx 5.2\). A diverging iteration can also wander off to a different root, or stop working because \(g\) is undefined (a negative number inside a log or a square root). A good answer describes what actually happens to the numbers.
The equation \(\ln x + 2x - 4 = 0\) from worked example 6, with root \(\alpha \approx 1.727\), can also be written as \(x = e^{4-2x}\). (a) Show this. (b) Use \(x_{n+1} = e^{4-2x_n}\) with \(x_0 = 1.7\) to find \(x_1\), \(x_2\) and \(x_3\) to 4 decimal places. (c) Explain why this iteration cannot be used to find \(\alpha\).
(a) Isolate \(\ln x\), then exponentiate: \(\ln x = 4 - 2x \Rightarrow\) \(x = e^{4-2x}\)
(b) \(x_1 = e^{0.6} = 1.8221\), then \(x_2 = e^{4 - 2(1.8221\ldots)} = 1.4273\), then \(x_3 = 3.1439\). \(x_1 = 1.8221\), \(x_2 = 1.4273\), \(x_3 = 3.1439\)
(c) The iterates jump from one side of \(\alpha\) to the other and get further from \(\alpha\) each time (\(x_0\) is 0.027 from \(\alpha\), \(x_3\) is more than 1.4 away). The sequence diverges: the curve \(y = e^{4-2x}\) is much steeper than \(y = x\) near the root (its gradient there is \(-2\alpha \approx -3.45\)), giving an unwinding cobweb.
Compare \(x = 2 - \frac12\ln x\), whose gradient near the root is \(-\frac{1}{2\alpha} \approx -0.29\): a gentle slope, hence the converging cobweb in worked example 6.
When asked to "comment on" or "explain" the behaviour of an iteration, say what the numbers do ("the values move further from 1.727, alternately above and below it") and give the conclusion ("the iteration diverges, so it cannot be used to find \(\alpha\)"). A word such as "wrong" or "bad" is not an explanation.
How does a full numerical methods question fit together?
The last example strings every skill on this page together, at the length of a full question. Notice how each part feeds the next: the sign change gives an interval, the interval gives a starting value, the iteration gives an approximate value, and a second sign change proves it.
\(f(x) = e^{x/2} - 3x + 1\). The equation \(f(x) = 0\) has two roots, \(\alpha\) and \(\beta\), where \(\alpha < \beta\).
(a) Show that \(\beta\) lies in the interval \([5, 6]\).
(b) Show that \(f(x) = 0\) can be written as \(x = 2\ln(3x - 1)\), \(x > \frac13\).
(c) Use the iteration formula \(x_{n+1} = 2\ln(3x_n - 1)\) with \(x_0 = 5\) to find \(x_1\), \(x_2\) and \(x_3\), giving your answers to 4 decimal places.
(d) Show that \(\beta = 5.470\) correct to 3 decimal places.
(e) A student tries instead \(x_{n+1} = \frac13\left(e^{x_n/2} + 1\right)\) with \(x_0 = 5.5\). Find \(x_1\) and \(x_2\) to 4 decimal places and explain why this iteration will not find \(\beta\).
(a) \(f(5) = e^{2.5} - 14 = -1.82 < 0\) and \(f(6) = e^3 - 17 = 3.09 > 0\). Change of sign and \(f\) is continuous, so \(5 < \beta < 6\).
(b) Isolate the exponential, then take logarithms: \(e^{x/2} = 3x - 1 \Rightarrow \frac{x}{2} = \ln(3x - 1) \Rightarrow\) \(x = 2\ln(3x - 1)\) (needs \(3x - 1 > 0\), i.e. \(x > \frac13\)).
(c) Iterate with Ans: \(x_1 = 2\ln 14 = 5.2781\), then \(x_2 = 5.3939\), \(x_3 = 5.4402\). \(x_1 = 5.2781\), \(x_2 = 5.3939\), \(x_3 = 5.4402\)
(d) Bounds 5.4695 and 5.4705: \(f(5.4695) = -0.00261 < 0\) and \(f(5.4705) = 0.00210 > 0\). Change of sign, \(f\) continuous, so \(5.4695 < \beta < 5.4705\). \(\beta = 5.470\) to 3 d.p.
(e) \(x_1 = 5.5475\), \(x_2 = 5.6730\) \(x_0 = 5.5\) is already above \(\beta \approx 5.470\), and the iterates are increasing, moving away from \(\beta\): the sequence diverges from \(\beta\) (a staircase going outwards), because near \(\beta\) the curve \(y = \frac13\left(e^{x/2} + 1\right)\) is much steeper than \(y = x\) (gradient about 2.57).
The iterates in (c) increased steadily towards \(\beta\): a converging staircase, since \(2\ln(3x - 1)\) is increasing but shallow near \(\beta\) (gradient about 0.39). The formula in (e) is not useless: from a small start such as \(x_0 = 0\) it converges to the other root, \(\alpha \approx 0.8409\), where its curve is shallow.
Exam tips for P3.8
Values (with signs), "change of sign and \(f\) is continuous", and a conclusion naming the interval. Learn it as one sentence and write it the same way every time; it covers both "show that there is a root" and "show that the root is correct to 3 d.p.".
The answer is printed in the question, so what counts is the steps in between. Write each operation on its own line (isolate, divide, take a root, take logs) and finish with the given form exactly.
For "comment on" an iteration, say whether the values converge or diverge, whether they approach from one side (staircase) or alternate (cobweb), and, if it fails, link it to the steepness of \(y = g(x)\) near the root.
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| A change of sign always means there is a root. | Only for a continuous function; across an asymptote the sign can change with no root. | Check \(f\) is continuous on the interval and say so. |
| No change of sign means no root. | Two roots (or a repeated root) in the interval leave \(f\) with the same sign at both ends. | Use a sketch or smaller intervals when roots may be close together. |
| To show 2.279 to 3 d.p., test 2.278 and 2.280. | That interval contains values that round to 2.278 or 2.280. | Test the bounds 2.2785 and 2.2795. |
| Every correct rearrangement \(x = g(x)\) will find the root. | The rearrangement is equivalent, but the iteration converges only if \(g\) is shallow enough near the root. | Near the root, gradient of \(g\) between −1 and 1: converges; steeper than \(y = x\): diverges. |
| Iterates that agree to 3 d.p. prove the root to 3 d.p. | Agreement is evidence, not proof; the sequence could still drift. | Prove it with a change of sign over the bounds. |
| Rounding each iterate before feeding it back makes no difference. | Rounding errors carry forward and can change the 4th decimal place. | Keep full accuracy with Ans; round only what you write down. |
How this topic connects
P1 Curve sketching and intersections
The number of roots of an equation such as \(e^{x/2} = 3x - 1\) is the number of times two graphs cross; a quick sketch tells you how many roots to look for and roughly where.
P3 Exponentials, logarithms, trigonometry and differentiation
Most equations on this page mix \(e^x\), \(\ln x\) or trigonometric functions with powers of \(x\), which is exactly why they cannot be solved algebraically; the rearrangements use the laws of logarithms and inverse functions. Setting \(\frac{dy}{dx} = 0\) for curves such as \(y = x^2 - 4\sin x\) often produces an equation that needs iteration, and \(g'(x)\) near the root tells you whether an iteration will converge.
P2 The trapezium rule, and IAL Physics
The trapezium rule is the other numerical method in the course: it estimates an area that cannot be found exactly. In physics, equations from models that cannot be rearranged for the unknown are solved numerically in the same spirit.
Key formulae
| Formula | Use | In the formulae list? |
|---|---|---|
| \(f\) continuous on \([a, b]\) and \(f(a)f(b) < 0 \Rightarrow\) root in \((a, b)\) | locating a root by a change of sign | No — learn it |
| \(x_{n+1} = g(x_n)\) | fixed-point iteration from a rearrangement \(x = g(x)\) | No — learn it |
| \(\alpha = c\) to \(n\) d.p.: test \(c \pm 0.5 \times 10^{-n}\) | bounds for proving the accuracy of a root | No — learn it |
| gradient of \(g\) near \(\alpha\) between −1 and 1 \(\Rightarrow\) converges | judging whether an iteration will work (qualitative) | No — learn it |
| \(g\) increasing near \(\alpha\): staircase; \(g\) decreasing: cobweb | predicting the shape of the diagram | No — learn it |
| \(e^{\ln x} = x\), \(\ln(e^x) = x\) | rearranging equations with exponentials and logarithms | No — learn it |
Check your understanding
1. Show that \(\sqrt5\) lies between 2.2 and 2.3 using a change of sign.
Let \(f(x) = x^2 - 5\). \(f(2.2) = -0.16 < 0\) and \(f(2.3) = 0.29 > 0\). \(f\) is continuous and changes sign, so the positive root of \(x^2 = 5\), which is \(\sqrt5\), lies between 2.2 and 2.3.
2. \(\tan 1.5 \approx 14.1\) and \(\tan 1.7 \approx -7.7\). Does \(\tan x = 0\) have a root between 1.5 and 1.7?
No. \(\tan x\) has a vertical asymptote at \(x = \frac{\pi}{2} \approx 1.571\), inside the interval, so it is not continuous there. The sign change comes from the asymptote; the roots of \(\tan x = 0\) are \(0, \pi, 2\pi, \ldots\)
3. An iteration gives 2.31, 2.27, 2.29, 2.28, 2.285, … Staircase or cobweb?
Cobweb: the values land alternately above and below the limit (about 2.28), so they converge from alternate sides.
4. Which two values of \(x\) would you test to show that a root is 0.74 correct to 2 decimal places?
0.735 and 0.745: half a unit of the second decimal place either side of 0.74.
5. Why can \(x = \frac{x^3-5}{3}\) not be used to find the root 2.279 of \(x^3 - 3x - 5 = 0\)?
Near the root the curve \(y = \frac{x^3-5}{3}\) has gradient \(x^2 \approx 5.2\), much steeper than \(y = x\), so each iteration moves further from the root: the sequence diverges whatever starting value near the root you choose (except the root itself).
6. Using \(x_{n+1} = \sqrt[3]{3x_n + 5}\) with \(x_0 = 1\), find \(x_1\) and \(x_2\) and say what kind of diagram they give.
\(x_1 = \sqrt[3]{8} = 2\) exactly, and \(x_2 = \sqrt[3]{11} = 2.2240\) (4 d.p.). Both are below the root 2.279 and increasing, as \(g\) is increasing near the root, so the diagram is a converging staircase.
Specification checklist
- (P3 6.1) I can show that an equation has a root in a given interval by a change of sign, stating that the function is continuous.
- (P3 6.1) I can explain why a change of sign may fail to locate a root: a discontinuity such as an asymptote, two or more roots close together, or a repeated root.
- (P3 6.2) I can rearrange an equation \(f(x) = 0\) into a given form \(x = g(x)\), showing every step.
- (P3 6.2) I can use an iteration \(x_{n+1} = g(x_n)\) to find iterates and a root to a stated accuracy, keeping full calculator accuracy and working in radians where needed.
- (P3 6.2) I can prove that a root is correct to a given number of decimal places by a change of sign over the appropriate bounds.
- (P3 6.2) I can draw and interpret staircase and cobweb diagrams on a graph of \(y = g(x)\) and \(y = x\).
- (P3 6.2) I can describe, without formal proof, when an iteration converges or diverges, and explain why a particular rearrangement fails.