Numerical solution of equations Banker

The equation x³ − 2x − 5 = 0 has one real root, α, which lies between 2 and 2.5.

1. (a) Use the iterative formula xₙ₊₁ = ∛(2xₙ + 5), with x₁ = 2, to find α correct to 3 decimal places. Give the result of each iteration to 5 decimal places. [3]

1. (a) Use the iterative formula xₙ₊₁ = 1 + ln(4 − xₙ), with x₁ = 2, to find the root of the equation x = 1 + ln(4 − x) correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

1. (a) Use the iterative formula xₙ₊₁ = e^(−xₙ/2), with x₁ = 0.5, to find the root of the equation x = e^(−x/2) correct to 3 decimal places. Give the result of each iteration to 4 decimal places. [3]
2. (b) Show by a sign-change argument that your answer is correct to 3 decimal places. [2]

1. (a) Use the iterative formula xₙ₊₁ = √(10 − 2xₙ), with x₁ = 2, to find the positive root of an equation correct to 3 decimal places. Give each iteration to 4 decimal places. [3]
2. (b) Show that this root is a root of x² + 2x − 10 = 0, and find its exact value. [2]

The equation x³ − 3x − 1 = 0 has a root α between 1.5 and 2.

1. (a) Show that α satisfies the equation x = √(3 + 1/x). [2]
2. (b) Use the iterative formula xₙ₊₁ = √(3 + 1/xₙ), with x₁ = 2, to find α correct to 3 decimal places. Give each iteration to 4 decimal places. [3]

The sequence defined by xₙ₊₁ = (2xₙ³ + 7)/(3xₙ²), with x₁ = 2, converges to α.

1. (a) Use the formula to find α correct to 3 decimal places, showing the result of each iteration to 4 decimal places. [2]
2. (b) State an equation satisfied by α and hence find the exact value of α. [2]

The equation x³ − 2x² − 3 = 0 has exactly one real root, α.

1. (a) Show by calculation that α lies between 2 and 3. [2]
2. (b) Show that the equation can be written as x = 2 + 3/x². [1]
3. (c) Use the iterative formula xₙ₊₁ = 2 + 3/xₙ², with x₁ = 2.5, to find α correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

1. (a) Show that the equation 2x − sin x = 2, where x is in radians, has a root between 1 and 2. [2]
2. (b) Show that the equation can be rearranged as x = 1 + ½ sin x. [1]
3. (c) Use the iterative formula xₙ₊₁ = 1 + ½ sin xₙ, with x₁ = 1, to find the root correct to 3 decimal places. Give the result of each iteration to 4 decimal places. [3]

The equation e^x + x² = 12 has a positive root α.

1. (a) Show by calculation that α lies between 2 and 2.5. [2]
2. (b) Show that α satisfies x = ln(12 − x²). [1]
3. (c) Use the iterative formula xₙ₊₁ = ln(12 − xₙ²), with x₁ = 2, to find α correct to 2 decimal places. Give each iteration to 4 decimal places. [3]

The equation x⁴ − x − 5 = 0 has a positive root α.

1. (a) Show that α lies between 1 and 2. [2]
2. (b) Use the iterative formula xₙ₊₁ = (5 + xₙ)^(1/4), with x₁ = 1.5, to find α correct to 3 decimal places. Give each iteration to 4 decimal places. [3]

In a model of a chemical reaction in a laboratory in Islamabad, the steady concentration c (in mol/dm³) satisfies the equation c = 0.4 + 0.6e^(−c).

1. (a) Show that the equation has a root between 0.5 and 1. [2]
2. (b) Use the iterative formula cₙ₊₁ = 0.4 + 0.6e^(−cₙ), with c₁ = 0.5, to find c correct to 3 significant figures. Give each iteration to 4 decimal places. [3]

The positive constant a is such that ∫₁ᵃ ln x dx = 2.

1. (a) Show that a = e^(1 + 1/a). [4]
2. (b) Show by calculation that a lies between 3 and 4. [2]
3. (c) Use an iterative formula based on the equation in part (a), with a₁ = 3.5, to find a correct to 3 decimal places. Give the result of each iteration to 4 decimal places. [3]

The curve y = x² cos x has a stationary point at x = α, where 0 < α < ½π.

1. (a) Show that α satisfies the equation α = tan⁻¹(2/α). [3]
2. (b) Use the iterative formula xₙ₊₁ = tan⁻¹(2/xₙ), with x₁ = 1, to find α correct to 3 decimal places. Give the result of each iteration to 4 decimal places. [3]

The equation x³ − 6x + 2 = 0 has a root α between 2 and 2.5. Two rearrangements are suggested:A: xₙ₊₁ = (xₙ³ + 2)/6B: xₙ₊₁ = ∛(6xₙ − 2)

1. (a) Starting with x₁ = 2.3, find x₂, x₃, x₄ and x₅ using formula A, giving answers to 4 decimal places, and describe what happens. [2]
2. (b) Use formula B with x₁ = 2.3 to find α correct to 2 decimal places, giving each iteration to 4 decimal places. [3]
3. (c) By considering the gradient of each iteration function near x = 2.26, explain why B converges but A does not. [2]

The equation eˣ = 2x + 3 has two roots, α > 0 and β < 0.

1. (a) Use the iterative formula xₙ₊₁ = ln(2xₙ + 3), with x₁ = 2, to find α correct to 2 decimal places. Give each iteration to 4 decimal places. [3]
2. (b) Explain why the formula in part (a) cannot be used to find β, which is close to −1.37. [2]
3. (c) Use the iterative formula xₙ₊₁ = ½(e^(xₙ) − 3), with x₁ = −1, to find β correct to 2 decimal places. Give each iteration to 4 decimal places. [3]

The sequence defined by xₙ₊₁ = (2xₙ³ + 5)/(3xₙ² + 1), with x₁ = 1.5, converges to α.

1. (a) Show that α is a root of the equation x³ + x − 5 = 0. [2]
2. (b) Show that x³ + x − 5 = 0 has only one real root. [2]
3. (c) Use the iteration to find α correct to 4 decimal places, giving each iteration to 5 decimal places. [3]

f(x) = x³ − 4x − 2.

1. (a) Show that the equation f(x) = 0 has a root between 2 and 3. [2]
2. (b) Show that the equation also has a root between −1 and 0. [2]

1. (a) Show by calculation that the equation eˣ = 5 − x has a root between 1 and 2. [2]
2. (b) By considering the graphs of y = eˣ and y = 5 − x, explain why this is the only root. [2]

1. (a) By sketching the graphs of y = ln x and y = 3 − x on the same axes, state the number of roots of the equation ln x = 3 − x. [2]
2. (b) Show by calculation that this root lies between 2 and 3. [2]

1. (a) By sketching the graphs of y = cos x and y = x², where x is in radians, state the number of real roots of the equation cos x = x². [2]
2. (b) Show by calculation that the positive root lies between 0.8 and 0.9. [2]

1. (a) By sketching the graphs of y = eˣ and y = 2/x, show that the equation eˣ = 2/x has exactly one real root. [3]
2. (b) Show that the root lies between 0.8 and 0.9. [2]

1. (a) By sketching the graphs of y = ln(x + 1) and y = 4 − x² for x > −1, show that the equation ln(x + 1) = 4 − x² has exactly one root. [3]
2. (b) Show by calculation that the root lies between 1.7 and 1.8. [2]

The curve y = x sin x, for 0 < x < π, has a maximum point at x = α.

1. (a) Show that α satisfies the equation tan α = −α. [3]
2. (b) Show by calculation that α lies between 2 and 2.1. [2]

1. (a) Show that the equation x³ + 2x − 7 = 0 has exactly one real root. [2]
2. (b) Show that the root lies between 1.5 and 1.6. [2]

f(x) = 2x³ − 3x² − 1.

1. (a) Show that the equation f(x) = 0 has a root α between 1.6 and 1.7. [2]
2. (b) By calculating f(1.65), state whether α is closer to 1.6 or to 1.7. [2]

1. (a) By considering the graphs of y = sec x and y = 3 − x² for 0 ≤ x < ½π, show that the equation sec x = 3 − x² has exactly one root in this interval. [3]
2. (b) Show that this root lies between 1 and 1.1. [2]

The curves y = 4 − x² and y = e^(−x) meet at two points, with x-coordinates α and β, where α < 0 < β.

1. (a) Show that β lies between 1.9 and 2. [2]
2. (b) Find an interval of width 0.1, with ends that are multiples of 0.1, which contains α. Justify your answer. [3]

f(x) = tan x − 2, where x is in radians.

1. (a) A student finds that f(1.5) > 0 and f(1.6) < 0 and says that the equation f(x) = 0 has a root between 1.5 and 1.6. Explain why the student is wrong. [2]
2. (b) Show by calculation that the equation has a root between 1 and 1.2, and state its exact value. [3]

1. (a) By sketching the graphs of y = ln x and y = x − 2, show that the equation ln x = x − 2 has two roots. [2]
2. (b) Show that one root lies between 0.1 and 0.2 and the other between 3 and 4. [4]

A circle has centre O and radius r cm. Points A and B lie on the circle, and angle AOB = x radians, where 0 < x < π. The area of triangle AOB is two thirds of the area of the sector OAB.

1. (a) Show that x satisfies the equation x = 1.5 sin x. [2]
2. (b) Show by calculation that the root lies between 1.4 and 1.6. [2]
3. (c) Use the iterative formula xₙ₊₁ = 1.5 sin xₙ, with x₁ = 1.5, to find x correct to 3 decimal places. Give each iteration to 4 decimal places. [3]