Iterative methods

Use the substitution u = cos x to show that

Fig. 1 shows part of the curve y = The shaded region is bounded by the curve, the x-axis, and the line x = a.

You are given that the area of the shaded region is a square units.

Show that the value of a satisfies the equation a - cos^-1(1 - 2e^-a) = 0.

Show by calculation that the value of a lies between 1.1 and 1.2.

Use the iterative formula

a_n+1 = cos^-1(1 - 2e^-a_n),

with starting value a₁ = 1.2, to find the value of a correct to 2 decimal places. Show the

result of each step of the iterative process.

Fig. 2 shows the curve y = cos^-1(1 - 2e^-x).

Explain why the gradient of the curve y = cos^-1(1 - 2e^-x) at the point where x = a, where a

is the value found in part (d), lies in the interval (- 1, 0).

The diagram shows the curve with equation y = (x³ - 2x²)ln x. The curve has a point of inflection at the point M.

Show that the x-coordinate of M satisfies the equation

Use an iterative formula, based on the equation in part (a)(i), to determine the x-coordinate of M correct to 2 decimal places. Use an initial value of 1.1 and show the

result of each step of the iterative process.

Determine the exact area of the shaded region, giving your answer in the form p ln q - r,

where p and r are positive rational numbers and q is a positive integer.

The diagram shows part of the curve f(x) = + 2. The equation f(x) = 0 has a positive root α close to x = 0.3.

Explain why using the sign change method with x = 0 and x = 1 will fail to locate α.

Show that the equation f(x) = 0 can be written as x =

An alternative iterative formula is x_n+1 = F(x_n), where F(x_n) = ln(2 - 8x_n²).

By considering F ′ (0.3) explain why this iterative formula will not find α.

Show that θ = 1.25 sin θ.

The equation θ = 1.25 sin θ has only one root for θ > 0.

This root can be found by using the iterative formula θ_n+1 = 1.25 sin θ_n with a starting value of θ₁ = 0.5.

• Write down the values of θ₂, θ₃ and θ₄.

• Hence find the value of this root correct to 3 significant figures.

Draw a suitable diagram to show why using an iterative process with the formula

θ_n+1 = sin^-1(0.8θ_n) does not converge to the root found in (b).

In this question you must show detailed reasoning.

The diagram shows the curve with equation y = . The tangent to the curve at the point P has gradient 2.

Show that the x-coordinate of P satisfies the equation

4x³ + 3x - 3 = 0.

Show by calculation that the x-coordinate of P lies between 0.5 and 1.

Show that the iteration

cannot converge to the x-coordinate of P whatever starting value is used.

Use the Newton-Raphson method, with initial value 0.5, to determine the coordinates of P

correct to 5 decimal places.

The equation 6 arcsin(2x - 1) - x² = 0 has exactly one real root.

Show by calculation that the root lies between 0.5 and 0.6.

In order to find the root, the iterative formula

x_n+1 = p + q sin(rx_n²),

with initial value x₀ = 0.5, is to be used.

Determine the values of the constants p, q and r.

Hence find the root correct to 4 significant figures. Show the result of each step of the iteration

process.

Show that θ = 0.4 + sin θ.

Let F (θ) = 0.4 + sin θ.

By considering the value of F ′ (θ) where θ = 1.2, explain why using an iterative method based on the equation in part (a) will converge to the root, assuming that 1.2 is sufficiently close to

the root.

Use the iterative formula θ_n+1 = 0.4 + sin θ_n with a starting value of 1.2 to find the value of θ correct to 4 significant figures.

You should show the result of each iteration.

Use a change of sign method to show that the value of θ found in part (c) is correct to 4

significant figures.