The terms of the sequence generated by the iterative formula x n + 1 = 6 7 x n + 1 x n 3 with initial value x 1 = 1 . 5 , converge to α . State an equation satisfied by α , and hence find the exact value of α .

Given iterative formula is x n + 1 = 6 7 x n + 1 x n 3 So, x = 6 7 x + 1 x 3 is the equation satisfied by α . ⇒ 7 x = 6 x + 6 x 3 ⇒ x = 6 x 3 ⇒ x 4 = 6 ⇒ x = 6 1 4 ∴ α = 6 1 4 = 1 . 56508

The terms of the sequence generated by the iterative formula x n + 1 = 6 7 x n + 1 x n 3 with initial value x 1 = 1 . 5 , converge to α . State an equation satisfied by α , and hence find the exact value of α .

Given iterative formula is x n + 1 = 6 7 x n + 1 x n 3 So, x = 6 7 x + 1 x 3 is the equation satisfied by α . ⇒ 7 x = 6 x + 6 x 3 ⇒ x = 6 x 3 ⇒ x 4 = 6 ⇒ x = 6 1 4 ∴ α = 6 1 4 = 1 . 56508

The equation x 3 + x 2 - 8 = 0 has one real root, denoted by α . Use iterative formula x n + 1 = 8 x n − 1 2 to determine α correct to 1 decimal place. Give the result of each iteration to 4 decimal places.

Given, x n + 1 = 8 x n − 1 2 with x 1 = 1.5 x 2 = 8 x 1 − 1 2 = 8 1 . 5 − 1 2 = 5 . 34 - 0 . 5 = 2 . 2 x 3 = 8 x 2 − 1 2 = 8 2 . 2 − 1 2 = 3 . 63 - 0 . 5 = 1 . 77 x 4 = 8 x 3 − 1 2 = 8 1 . 77 − 1 2 = 4 . 52 - 0 . 5 = 2 . 00 x 5 = 8 x 4 − 1 2 = 8 2 . 00 − 1 2 = 4 - 0 . 5 = 1 . 87 x 6 = 8 x 5 − 1 2 = 8 1 . 87 − 1 2 = 4 . 28 - 0 . 5 = 1 . 94 x 7 = 8 x 6 − 1 2 = 8 1 . 94 − 1 2 = 4 . 12 - 0 . 5 = 1 . 90 x 8 = 8 x 7 − 1 2 = 8 1 . 90 − 1 2 = 4 . 21 - 0 . 5 = 1 . 93 x 9 = 8 x 8 − 1 2 = 8 1 . 93 − 1 2 = 4 . 15 - 0 . 5 = 1 . 91 x 10 = 8 x 9 − 1 2 = 8 1 . 91 − 1 2 = 4 . 19 - 0 . 5 = 1 . 92 So, α = 1.9 to one decimal place.

EASY

AS and A Level

IMPORTANT

Earn 100

The terms of the sequence generated by the iterative formula $x_{n + 1} = \frac{6}{7} (x_{n} + \frac{1}{{x_{n}}^{3}})$ with initial value $x_{1} = 1.5,$ converge to $α .$

State an equation satisfied by $α,$ and hence find the exact value of $α .$

Important Questions on Numerical Solutions of Equations

MEDIUM

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 6 - Numerical Solutions of Equations>END-OF-CHAPTER REVIEW EXERCISE 6>Q 2

The equation $x^{3} + \frac{x}{2} - 8 = 0$ has one real root, denoted by $α .$

Find, by calculation, the pair of consecutive integers between which $α$ lies.

EASY

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 6 - Numerical Solutions of Equations>END-OF-CHAPTER REVIEW EXERCISE 6>Q 2

The equation $x^{3} + \frac{x}{2} - 8 = 0$ has one real root, denoted by $α .$

Show that, if a sequence of values given by the iterative formula $x_{n + 1} = \sqrt{\frac{8}{x_{n}} - \frac{1}{2}}$ converges, then it converges to $α .$

MEDIUM

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 6 - Numerical Solutions of Equations>END-OF-CHAPTER REVIEW EXERCISE 6>Q 2

The equation $x^{3} + \frac{x}{2} - 8 = 0$ has one real root, denoted by $α .$

Use iterative formula $x_{n + 1} = \sqrt{\frac{8}{x_{n}} - \frac{1}{2}}$ to determine $α$ correct to $1$ decimal place. Give the result of each iteration to $4$ decimal places.

MEDIUM

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 6 - Numerical Solutions of Equations>END-OF-CHAPTER REVIEW EXERCISE 6>Q 3

By sketching a suitable pair of graphs, show that the equation

e^{2 x + 1} = 14 - x^{3}

has exactly one real root.

MEDIUM

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 6 - Numerical Solutions of Equations>END-OF-CHAPTER REVIEW EXERCISE 6>Q 3

By sketching a suitable pair of graphs, show that the equation

e^{2 x + 1} = 14 - x^{3}

has exactly one real root. Show by calculation that this root lies between

0.5

and

1

MEDIUM

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 6 - Numerical Solutions of Equations>END-OF-CHAPTER REVIEW EXERCISE 6>Q 3

By sketching a suitable pair of graphs, show that the equation

e^{2 x + 1} = 14 - x^{3}

has exactly one real root. Show that this root also satisfies the equation

x = \frac{\ln (14 - x^{3}) - 1}{2}

MEDIUM

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 6 - Numerical Solutions of Equations>END-OF-CHAPTER REVIEW EXERCISE 6>Q 3

Use an iteration process based on the equation

x = \frac{\ln (14 - x^{3}) - 1}{2}

with a suitable starting value, to find the root correct to

4

decimal places. Give the result of each step of the process to

6

decimal places.

MEDIUM

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 6 - Numerical Solutions of Equations>END-OF-CHAPTER REVIEW EXERCISE 6>Q 4

The sequence of values given by the iterative formula

x_{n + 1} = \frac{x_{n} (1 + \sec^{2} x_{n}) - \tan x_{n}}{\sec^{2} x_{n} - 1}

with initial value

x_{1} = 1,

converges to

α .

Use this formula to find

α

correct to

2

decimal places, showing the result of each iteration to

4

decimal places.

The terms of the sequence generated by the iterative formula xn+1=67xn+1xn3 with initial value x1=1.5, converge to α. State an equation satisfied by α, and hence find the exact value of α.

Important Questions on Numerical Solutions of Equations

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The terms of the sequence generated by the iterative formula $x_{n + 1} = \frac{6}{7} (x_{n} + \frac{1}{{x_{n}}^{3}})$ with initial value $x_{1} = 1.5,$ converge to $α .$

State an equation satisfied by $α,$ and hence find the exact value of $α .$