Use the iteration formula xn+1 = 3√(9xn + 2) with x0 = 2 to find x1 correct to 3 decimal places.
Worked Solution & Answer
x1 = 3√(9(2) + 2) = 3√(20) = 2.714 B1
Answer: x1 = 2.714
Iteration — Notes & Worked Examples Show / Hide
Explanation:
To show that an equation f(x) = 0 has a root between x = a and x = b, evaluate f(a) and f(b), state that there is a change of sign (one positive, one negative) and f(x) is continuous. An iterative formula xn+1 = g(xn) is formed by rearranging f(x) = 0 to isolate one x term.
To iterate efficiently on a calculator, type the starting value x0 and press =, then enter the formula using the ANS key in place of xn. Press = repeatedly to generate x1, x2, x3, writing down full calculator displays at each step.
Example 1: Show x3 + 2x − 6 = 0 has a root between 1 and 2, and rearrange to x = ∛(6 − 2x).
Solution:
Let f(x) = x3 + 2x − 6:
f(1) = 13 + 2(1) − 6 = −3 (< 0)
f(2) = 23 + 2(2) − 6 = +6 (> 0)
Change of sign shows a root lies in (1, 2).
Rearrange: x3 = 6 − 2x ⇒ x = ∛(6 − 2x)
Example 2: Use xn+1 = ∛(6 − 2xn) with x0 = 1.5 to find x1, x2, x3 (3 d.p.).
Solution:
Type 1.5 =, then ∛(6 − 2 × ANS):
x1 = ∛(6 − 2(1.5)) = ∛3 = 1.4422...
x2 = ∛(6 − 2(1.4422...)) = 1.4605...
x3 = ∛(6 − 2(1.4605...)) = 1.4548...
Answer: x1 = 1.442, x2 = 1.461, x3 = 1.455
Use the iteration formula xn+1 = 3√(10xn + 5) with x0 = 2 to find x1 correct to 3 decimal places.
Worked Solution & Answer
x1 = 3√(10(2) + 5) = 3√(25) = 2.924 B1
Answer: x1 = 2.924
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Use x1 = 1 to work out an approximate solution to x = 5 - 1x
Use x1 = 1 to work out an approximate solution to x = 5 - 1x
Give your answer to 4 significant figures.
x =
Worked Solution & Answer
x₂ = 5 – 1/1 = 4. M1
x₃ = 5 – 1/4 = 4.75.
x₄ = 5 – 1/4.75 = 4.78947... M1
x₅ = 5 – 1/4.78947... = 4.79120...
x₆ = 4.79128... → 4.791 (to 4 s.f.). A1
x₃ = 5 – 1/4 = 4.75.
x₄ = 5 – 1/4.75 = 4.78947... M1
x₅ = 5 – 1/4.78947... = 4.79120...
x₆ = 4.79128... → 4.791 (to 4 s.f.). A1
Answer: x = 4.791
xn+1 = 3√(3xn + 7)
Use a starting value of x1 = 2 to work out a solution to x = 3√(3x + 7).
Give your answer to 3 decimal places.
Give your answer to 3 decimal places.
Answer
Worked Solution & Answer
Iterative calculations: M1
x1 = 2
x2 = 3√(3(2) + 7) = 3√(13) ≈ 2.35133
x3 = 3√(3(2.35133) + 7) ≈ 3√(14.0540) ≈ 2.41316
x4 = 3√(3(2.41316) + 7) ≈ 3√(14.2395) ≈ 2.42375 M1
x5 = 3√(3(2.42375) + 7) ≈ 3√(14.2712) ≈ 2.42555
x6 = 3√(3(2.42555) + 7) ≈ 2.42586
x7 ≈ 2.42591
Rounded to 3 decimal places, x = 2.426. A1
x1 = 2
x2 = 3√(3(2) + 7) = 3√(13) ≈ 2.35133
x3 = 3√(3(2.35133) + 7) ≈ 3√(14.0540) ≈ 2.41316
x4 = 3√(3(2.41316) + 7) ≈ 3√(14.2395) ≈ 2.42375 M1
x5 = 3√(3(2.42375) + 7) ≈ 3√(14.2712) ≈ 2.42555
x6 = 3√(3(2.42555) + 7) ≈ 2.42586
x7 ≈ 2.42591
Rounded to 3 decimal places, x = 2.426. A1
Answer: Answer 2.426
An approximate solution to an equation is found using the iterative formula
xn+1 = (xn)3 - 210 with x1 = -1
xn+1 = (xn)3 - 210 with x1 = -1
(a) Work out the values of x2 and x3 [2 marks]
(b) Work out the solution to 5 decimal places. [1 mark]
x =
Worked Solution & Answer
(a)
• x2 = (-1)3 - 210 = -1 - 210 = -0.3 M1
• x3 = (-0.3)3 - 210 = -0.027 - 210 = -2.02710 = -0.2027 A1
• x3 = (-0.3)3 - 210 = -0.027 - 210 = -2.02710 = -0.2027 A1
Answer (a): –0.3, –0.2027
(b)
Continue iterating:
• x4 = (-0.2027)3 - 210 ≈ -0.20083
• x5 = (-0.20083)3 - 210 ≈ -0.20081
• x6 = (-0.20081)3 - 210 ≈ -0.20081
To 5 decimal places, x = -0.20081. B1
• x4 = (-0.2027)3 - 210 ≈ -0.20083
• x5 = (-0.20083)3 - 210 ≈ -0.20081
• x6 = (-0.20081)3 - 210 ≈ -0.20081
To 5 decimal places, x = -0.20081. B1
Answer (b): x = –0.20081
(a) Show that x⁴ – x² – 5 = 0 can be written as x = ⁴√(x² + 5). (1)
(b) Starting with x₀ = 1.5, use the iteration formula three times to find x₃. (3)
(b) Starting with x₀ = 1.5, use the iteration formula three times to find x₃. (3)
(a) Show that the equation x4 - x2 - 5 = 0 can be written in the form x = 4√(x2 + 5) [1 mark]
(b) Starting with x0 = 1.5
use the iteration formula xn+1 = 4√(xn2 + 5) three times to find an estimate for a solution of x4 - x2 - 5 = 0 [3 marks]
use the iteration formula xn+1 = 4√(xn2 + 5) three times to find an estimate for a solution of x4 - x2 - 5 = 0 [3 marks]
Worked Solution & Answer
(a)
x4 = x2 + 5 ⇒ x = 4√(x2 + 5)
Answer (a): x4 = x2 + 5 ⇒ x = 4√(x2 + 5)
(b)
x₁ = ⁴√(1.5² + 5) = 1.640909... M1
x₂ = ⁴√(x₁² + 5) = 1.665418... M1
x₃ = ⁴√(x₂² + 5) = 1.669763... A1
x₂ = ⁴√(x₁² + 5) = 1.665418... M1
x₃ = ⁴√(x₂² + 5) = 1.669763... A1
Answer (b): 1.66976
(a) Show that x = 13 - 9x is a rearrangement of x2 - 13x + 9 = 0.
You must show each stage of your working.
(b) Use the iteration formula
xn+1 = 13 - 9xn and x1 = 12
to find a solution of x2 - 13x + 9 = 0 correct to 2 decimal places.
You must give all your calculated values of xn+1.
You must show each stage of your working.
(b) Use the iteration formula
xn+1 = 13 - 9xn and x1 = 12
to find a solution of x2 - 13x + 9 = 0 correct to 2 decimal places.
You must give all your calculated values of xn+1.
(a) Show that x = 13 - 9x is a rearrangement of x2 - 13x + 9 = 0.
You must show each stage of your working. [1 mark]
You must show each stage of your working. [1 mark]
(b) Use the iteration formula
xn+1 = 13 - 9xn and x1 = 12
to find a solution of x2 - 13x + 9 = 0 correct to 2 decimal places.
You must give all your calculated values of xn+1. [3 marks]
xn+1 = 13 - 9xn and x1 = 12
to find a solution of x2 - 13x + 9 = 0 correct to 2 decimal places.
You must give all your calculated values of xn+1. [3 marks]
Worked Solution & Answer
(a)
x2 - 13x + 9 = 0 ⇒ x2 = 13x - 9
Divide by x:
x = 13 - 9x. C1
Divide by x:
x = 13 - 9x. C1
Answer (a): x2 = 13x - 9 => x = 13 - 9/x
(b)
x1 = 12 M1
x2 = 13 - 912 = 12.25
x3 = 13 - 912.25 ≈ 12.2653 M1
x4 = 13 - 912.2653 ≈ 12.2662
To 2 decimal places, x = 12.27. A1
x2 = 13 - 912 = 12.25
x3 = 13 - 912.25 ≈ 12.2653 M1
x4 = 13 - 912.2653 ≈ 12.2662
To 2 decimal places, x = 12.27. A1
Answer (b): 12.27
Using xn+1 = -2 - 4xn2
with x0 = -2.5
with x0 = -2.5
(a) find the values of x1, x2 and x3 [3 marks]
(b) Explain the relationship between the values of x1, x2 and x3 and the equation x3 + 2x2 + 4 = 0 [2 marks]
Worked Solution & Answer
(a)
x₁ = –2 – 4/(–2.5)² = –2.64 M1
x₂ = –2 – 4/(–2.64)² = –2.57392... M1
x₃ = –2 – 4/(–2.57392...)² = –2.603767 A1
x₂ = –2 – 4/(–2.64)² = –2.57392... M1
x₃ = –2 – 4/(–2.57392...)² = –2.603767 A1
Answer (a): x₁ = -2.64, x₂ = -2.57392, x₃ = -2.60377
(b)
x = –2 – 4/x² ⇔ x³ + 2x² + 4 = 0, so the iterates converge to the root M1 C1
Answer (b): The iterative formula is a rearrangement of x³ + 2x² + 4 = 0 (dividing by x² gives x + 2 + 4/x² = 0 ⇒ x = –2 – 4/x²), so x₁, x₂ and x₃ are successive estimates of the solution to the equation.
(a) Show that x³ + 2x – 6 = 0 has a solution between x = 1 and x = 2 (2) (b) Show that x³ + 2x – 6 = 0 rearranges to x = 6/(x² + 2) (1) (c) Starting with x₀ = 1.45, use the iteration twice to 4 d.p. (3)
(a) Show that the equation x3 + 2x - 6 = 0 has a solution between x = 1 and x = 2 [2 marks]
(b) Show that the equation x3 + 2x - 6 = 0 can be rearranged to give x = 6x2 + 2 [1 mark]
(c) Starting with x0 = 1.45
use the iteration formula xn+1 = 6xn2 + 2 twice to find an estimate for the solution of x3 + 2x - 6 = 0
Give your answer correct to 4 decimal places. [3 marks]
use the iteration formula xn+1 = 6xn2 + 2 twice to find an estimate for the solution of x3 + 2x - 6 = 0
Give your answer correct to 4 decimal places. [3 marks]
Worked Solution & Answer
(a)
f(1) = 1 + 2 – 6 = –3 < 0 and f(2) = 8 + 4 – 6 = 6 > 0 M1
sign change so root between 1 and 2 C1
sign change so root between 1 and 2 C1
Answer (a): f(1) = 1 + 2 – 6 = –3 < 0 and f(2) = 8 + 4 – 6 = 6 > 0; sign change so root between 1 and 2
(b)
x³ + 2x = 6 ⇒ x(x² + 2) = 6 ⇒ x = 6 / (x² + 2)
Answer (b): x³ + 2x = 6 ⇒ x(x² + 2) = 6 ⇒ x = 6 / (x² + 2)
(c)
x₁ = 6 / (1.45² + 2) = 6 / 4.1025 = 1.46252285... M2
x₂ = 6 / (1.46252285² + 2) = 1.4496349... ≈ 1.4496 A1
x₂ = 6 / (1.46252285² + 2) = 1.4496349... ≈ 1.4496 A1
Answer (c): 1.4496
(a) Show x³ + 7x – 5 = 0 has a solution between 0 and 1. (2) (b) Rearrange to x = 5/(x² + 7). (2) (c) Start with x₀ = 1 and iterate 3 times. (3)
(a) Show that the equation x3 + 7x - 5 = 0 has a solution between x = 0 and x = 1 [2 marks]
(b) Show that the equation x3 + 7x - 5 = 0 can be arranged to give x = 5x2 + 7 [2 marks]
(c) Starting with x0 = 1, use the iteration formula x_{n+1} = 5xn2 + 7 three times to find an estimate for the solution of x3 + 7x - 5 = 0 [3 marks]
Worked Solution & Answer
(a)
f(0) = –5, f(1) = 3, change of sign M1 C1
Answer (a): Let f(x) = x3 + 7x - 5: f(0) = -5 < 0 and f(1) = 1 + 7 - 5 = 3 > 0. Since there is a change of sign and f(x) is continuous, there is a root between x = 0 and x = 1.
(b)
x³ + 7x = 5 M1
⇒ x(x² + 7) = 5
⇒ x = 5/(x² + 7) C1
⇒ x(x² + 7) = 5
⇒ x = 5/(x² + 7) C1
Answer (b): x3 + 7x = 5 ⇒ x(x2 + 7) = 5 ⇒ x = 5x2 + 7
(c)
x₁ = 5/8 = 0.625 M1
x₂ = 5/(0.625² + 7) = 0.67653277 M1
x₃ = 5/(0.67653277² + 7) = 0.6704483 A1
x₂ = 5/(0.625² + 7) = 0.67653277 M1
x₃ = 5/(0.67653277² + 7) = 0.6704483 A1
Answer (c): 0.6704483