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Locating roots and iteration questions
Plenty of equations have no solution you can reach by rearrangement, and this unit stops trying. A sign change traps a root inside an interval, a rearranged equation iterated from a starting value walks towards it, and staircase and cobweb diagrams show whether the walk is closing in or running away.
8 original questions · 22 marks · the locating roots and iteration notes · Numerical methods
Every question here is written for this library rather than taken from a past paper. Write your answer out before opening the worked one: the answers award marks point by point, and the marks are easier to see when you have something of your own to compare against.
f(x) = x3 + x − 5. Show that f(x) = 0 has a root between x = 1 and x = 2.
Worked answer
f(1) = 1 + 1 − 5 = −3 < 0 and f(2) = 8 + 2 − 5 = 5 > 0. There is a change of sign, and f is continuous on [1, 2], so a root lies in the interval. M1 for evaluating f at both ends, A1 for the sign change, continuity and conclusion. Both halves of that sentence carry marks. Candidates who compute the two values but never state the sign change or the continuity lose the conclusion mark.Explain why a change of sign of f between a and b does not always mean a root of f(x) = 0 lies in (a, b), and give the condition that repairs the argument.
Worked answer
The function might be discontinuous. Take f(x) = 1/x, which is negative at x = −1 and positive at x = 1 yet never equals zero, because of the asymptote at x = 0. The argument needs f to be continuous across the whole interval before the sign change proves anything. B1 for a counterexample with a discontinuity, B1 for continuity as the missing condition.Show that the equation x3 + x − 5 = 0 can be rearranged into the iterative form x = (5 − x)1/3.
Worked answer
From x3 + x − 5 = 0, isolate the cube to get x3 = 5 − x, then take the cube root of both sides to get x = (5 − x)1/3. M1 for isolating the cube, A1 for the printed form. Any correct isolation earns the marks, though not every rearrangement produces an iteration that converges.Using the iteration xn+1 = (5 − xn)1/3 with x0 = 1.5, find x1, x2 and x3, giving each to 5 decimal places.
Worked answer
x1 = (5 − 1.5)1/3 = 3.51/3 = 1.51829. x2 = (5 − 1.51829)1/3 = 1.51564. x3 = (5 − 1.51564)1/3 = 1.51603. The iterates close in on each other while alternating either side of the root. M1 for a correct first iterate, A1 for x2, A1 for x3. Feed the calculator's full unrounded value into the next step rather than the printed 5 decimal places, or the last digit will drift.Show that α = 1.516 is a root of x3 + x − 5 = 0 correct to 3 decimal places.
Worked answer
Test the ends of the interval that rounds to 1.516. With f(x) = x3 + x − 5, f(1.5155) = −0.0038… < 0 and f(1.5165) = 0.0041… > 0. The sign change, together with the continuity of f, traps the root in (1.5155, 1.5165), and every number in that interval rounds to 1.516. M1 for testing the ends of the rounding interval, A1 for both values, A1 for the sign change and conclusion. Testing 1.515 and 1.517 instead is the standard mistake, since that interval is twice as wide and proves nothing about the third decimal place.The cobweb diagram for the iteration xn+1 = (5 − xn)1/3 spirals inwards towards the root α ≈ 1.516. State the feature of g(x) = (5 − x)1/3 near the root that guarantees convergence.
Worked answer
The gradient of g is small there. g′(α) ≈ −0.145, and an iteration of the form x = g(x) converges near a root whenever |g′| < 1. B1 for the gradient condition, B1 for its value at the root. The negative sign is what makes the path spiral in from alternate sides rather than climb a staircase from one side.The equation x3 + x − 5 = 0 also rearranges to x = 5 − x3. Starting from x0 = 1.5, calculate x1 and x2, and explain what happens to this iteration.
Worked answer
x1 = 5 − 1.53 = 5 − 3.375 = 1.625, and x2 = 5 − 1.6253 = 0.70898…. The values then swing ever further from the root, reaching 4.64 and then −95. Here g(x) = 5 − x3 has g′(x) = −3x2, which is about −6.9 near the root, and |g′| > 1 throws the iterates away. M1 for a correct iterate, A1 for both values, M1 for differentiating g, A1 for the size of the gradient explaining the divergence. A valid rearrangement is not always a usable one, and the size of g′ decides which.An iteration xn+1 = g(xn) converges to a value α. Explain why α must satisfy the equation x = g(x), and why a sequence that appears to settle on a calculator display is not by itself a proof that α is accurate to the digits shown.
Worked answer
If xn → α then, g being continuous, taking limits on both sides of xn+1 = g(xn) gives α = g(α), which rearranges back to f(α) = 0. Settling digits show only that successive iterates agree with each other, and a slowly converging iteration can hold four decimal places steady for several steps while still being wrong in the fourth. The accuracy claim needs the sign-change test on the rounding interval of the claimed value. M1 for taking limits on both sides, A1 for α = g(α), B1 for successive iterates only agreeing with each other, B1 for the sign-change test being needed.
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