Questions on locating roots and iteration

Question 1

f(x) = x³ + x − 5. Show that f(x) = 0 has a root between x = 1 and x = 2.

Accepted 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. State both the sign change and the continuity of f on the interval; the two values alone do not establish a root.

Question 2

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.

Accepted 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.

Question 3

Show that the equation x³ + x − 5 = 0 can be rearranged into the iterative form x = (5 − x)^1/3.

Accepted 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 is acceptable, though not every rearrangement produces an iteration that converges.

Question 4

f(x) = x³ − 5x + 1. Show that f(x) = 0 has a root in the interval (0, 1) and a second root in the interval (2, 3).

Accepted Answer

f(0) = 1 > 0 and f(1) = 1 − 5 + 1 = −3 < 0, so the sign changes across (0, 1). f(2) = 8 − 10 + 1 = −1 < 0 and f(3) = 27 − 15 + 1 = 13 > 0, so the sign changes again across (2, 3). A polynomial is continuous everywhere, so each sign change traps a root. M1 for evaluating f at the ends of an interval, A1 for the first conclusion, A1 for the second.

Question 5

Using the iteration x_n+1 = (5 − x_n)^1/3 with x₀ = 1.5, find x₁, x₂ and x₃, giving each to 5 decimal places.

Accepted Answer

x₁ = (5 − 1.5)^1/3 = 3.5^1/3 = 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.

Question 6

Show that α = 1.516 is a root of x³ + x − 5 = 0 correct to 3 decimal places.

Accepted 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. The interval tested has to be the one that rounds to 1.516: the interval (1.515, 1.517) is twice as wide and settles nothing about the third decimal place.

Question 7

The cobweb diagram for the iteration x_n+1 = (5 − x_n)^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.

Accepted 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.

Question 8

f(x) = x² − 5x + 5. Show that f(1) and f(4) have the same sign, and show that f(x) = 0 nevertheless has two roots between x = 1 and x = 4.

Accepted Answer

f(1) = 1 − 5 + 5 = 1 and f(4) = 16 − 20 + 5 = 1, both positive, so the endpoints show no sign change. The quadratic formula gives x = (5 ± √5)/2, that is x = 1.382 and x = 3.618, and both lie between 1 and 4. B1 for the two endpoint values, M1 for solving the quadratic, A1 for both roots, A1 for placing them inside the interval. Testing the midpoint finds them: f(2.5) = −1.25, so the sign changes once on either side of it. The ends of a wide interval settle nothing about the number of roots inside it, because an even number of crossings returns the curve to the side it started on.

Question 9

The equation x = cos x, with x in radians, has a root α. Show that α = 0.739 correct to 3 decimal places.

Accepted Answer

Work with f(x) = x − cos x, whose roots are the solutions of the equation. The values that round to 0.739 are those between 0.7385 and 0.7395, so those are the two to test. f(0.7385) = 0.7385 − 0.73948… = −0.00098… < 0 and f(0.7395) = 0.7395 − 0.73881… = 0.00069… > 0. M1 for testing the ends of the rounding interval, A1 for both values, A1 for the sign change together with the continuity of f there, A1 for the conclusion that α = 0.739 to 3 decimal places. Keep the calculator in radians: in degrees cos 0.7385 is 0.99992 and both test values come out negative, so no sign change appears at all.

Question 10

Show that x³ − 3x² + 1 = 0 can be written as x = (3x² − 1)^1/3, and use the iteration x_n+1 = (3x_n² − 1)^1/3 with x₀ = 2.9 to find x₁, x₂ and x₃, giving each to 5 decimal places.

Accepted Answer

Adding 3x² − 1 to both sides gives x³ = 3x² − 1, and taking the cube root of both sides gives the printed form. Then x₁ = (3 × 2.9² − 1)^1/3 = 24.23^1/3 = 2.89368, x₂ = 2.88931 and x₃ = 2.88627. M1 for the rearrangement, A1 for x₁, A1 for x₂, A1 for x₃. Each iterate is smaller than the last and none crosses the root at 2.87939, which is the staircase case: g′ is positive there, about 0.69, so the approach is from one side. That gradient is close to 1, so the sequence closes in slowly and 21 steps are needed before the fifth decimal place stops moving.

Question 11

The equation x³ − 5x + 1 = 0 has a root α = 0.2016 correct to 4 decimal places. It can be rearranged as x = (5x − 1)^1/3 and as x = (x³ + 1)/5. State which of the two iterations converges to α from a nearby starting value, and why.

Accepted Answer

Differentiate each rearrangement and measure it at α. For g(x) = (5x − 1)^1/3, g′(x) = (5/3)(5x − 1)^−2/3, which is about 41 at α, so iterates started near α move away from it. For g(x) = (x³ + 1)/5, g′(x) = 3x²/5, which is about 0.024 at α, well inside 1, so that iteration converges. M1 for differentiating a rearrangement, A1 for both gradients, B1 for naming (x³ + 1)/5.

Question 12

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.

Accepted 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.

Question 13

An iteration x_n+1 = g(x_n) 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.

Accepted Answer

If x_n → α then, g being continuous, taking limits on both sides of x_n+1 = g(x_n) 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.

Question 14

f(x) = x³ + 3x − 5 has one real root, α = 1.1542 correct to 4 decimal places, at which f′(α) = 7.00 correct to 3 significant figures. For a non-zero constant k, the iteration x_n+1 = x_n − k f(x_n) has α among its fixed points. Show that every fixed point of this iteration is a root of f(x) = 0, and find the values of k for which iterates started sufficiently close to α converge to it.

Accepted Answer

A fixed point satisfies x = x − k f(x), so k f(x) = 0, and k is not zero, so f(x) = 0: the fixed points of the iteration are exactly the roots of f. M1 for setting g(x) = x, A1 for the conclusion. Writing g(x) = x − k f(x) gives g′(x) = 1 − k f′(x), so g′(α) = 1 − 7.00k. M1 for differentiating, A1 for 1 − 7.00k. Convergence from nearby starting values needs |g′(α)| < 1, that is −1 < 1 − 7.00k < 1, which rearranges to 0 < k < 2/7.00, so 0 < k < 0.286 to 3 decimal places. M1 for the inequality, A1 for the interval. The upper end is set by the gradient at the root and the lower end by the sign of k, since a negative k moves the iterates the wrong way. Replacing the constant k by 1/f′(x_n), which changes at every step, is the Newton-Raphson method.

Question 15

For y = tan x, with x in radians, tan 1 = 1.5574 > 0 and tan 2 = −2.185 < 0. Explain why the sign change between x = 1 and x = 2 does not locate a root of tan x = 0, and give an interval of width 1 with whole-number ends on which a sign change does locate the root at x = π.

Accepted Answer

tan x = 0 exactly at the whole multiples of π, and no multiple of π lies between 1 and 2, so there is no root there. B1 for the roots of tan x = 0, B1 for none of them lying in the interval. The sign change comes from the asymptote at x = π/2 = 1.5708…, across which tan x passes from large positive values to large negative ones. B1 for the asymptote. The test requires the function to be continuous across the interval, and tan is not continuous on [1, 2], so the sign change establishes nothing. B1 for continuity as the condition that fails. The interval (3, 4) serves: tan 3 = −0.1425 < 0 and tan 4 = 1.1578 > 0, tan is continuous across it because the next asymptote is at 3π/2 = 4.712…, and π = 3.1416… lies inside. B1 for the interval with both values quoted.