Iteration Equations: Comprehensive Guide with Worksheet 

Iteration

  • Iteration is a mathematical process of repeatedly refining a solution or estimate to a problem.
  • It is a valuable tool for finding numerical solutions to equations that cannot be solved analytically.

In this article, we will discuss: 

  1. What is Iteration?
  2. Steps for Solving an equation using iteration
  3. Steps to locate the roots of an equation between two points

Here is one more link to practice a few extra questions: Maths Genie Iteration Questions

What is Iteration?

  • Iteration is a mathematical process that involves repeating a specific sequence of steps until a satisfactory solution or estimate is achieved.
  • The main idea behind iteration is to use the previous solution to generate a new one, which should be closer to the actual answer.
  • This process is repeated until the desired level of accuracy is achieved.
  • Iteration is commonly used to solve nonlinear equations, such as transcendental or polynomial equations, that cannot be solved analytically.
  • It can also be used to find the roots of a function or estimate the value of an integral.

Steps for Solving an equation using iteration

  • The following steps can be used to solve an equation using iteration:

Step #1: Rearrange the equation to get it into the form f(x) = 0, where f(x) is a continuous function.

Step #2: Select an initial guess, x₀, for the solution.

Step #3: Apply the iteration formula xn+1 = g(xn), where g(x) is a function that generates a new solution using the previous solution.

Step #4: Repeat step #3 until the desired level of accuracy is achieved.

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Solved Example: 

Question 1: Solve the equation 3x – x3 = -11 using iteration.

Solution: 

  • Step #1: Rearrange the equation to get it into the form f(x) = 0. We get

f(x) = 3x – x3 + 11.

  • Step #2: Select an initial guess, x₀, for the solution. Let x₀ = 3.
  • Step #3: Apply the iteration formula:

Iteration Formula

which generates a new solution using the previous solution.

  • Step #4: Repeat step #3 until the desired level of accuracy is achieved.

x1 = (11 + 3x0)1/3 = (11 + 3x32)1/3 = 2.7144

x2 = (11 + 3x1)1/3 = (11 + 3x2.7144)1/3 = 2.6751

x3 = (11 + 3x2)1/3 = (11 + 3x2.6751)1/3 = 2.6696

x4 = (11 + 3x3)1/3 = (11 + 3x2.6696)1/3 = 2.6688

x5 = (11 + 3x4)1/3 = (11 + 3x2.6688)1/3 = 2.6687

After several iterations, we get x5 ≈ 2.6687

 

Question 2: Solve the equation ex – 4x = 0 using iteration.

Solution: 

  • Step #1: Rearrange the equation to get it into the form f(x) = 0. We get

f(x) = ex – 4x.

  • Step #2: Select an initial guess, x₀, for the solution. Let x₀ = 1.
  • Step #3: Apply the iteration formula

xᵢ = ln(4xᵢ),

which generates a new solution using the previous solution.

  • Step #4: Repeat step #3 until the desired level of accuracy is achieved.

Iteration 1: x₁ = ln(4 x 1) = ln(4) = 1.386294361119891

Iteration 2: x₂ = ln(4 x 13863) = 1.712932688682084

Iteration 3: x₃ = ln(4 x 17129) = 1.924482201641693

Iteration 4: x₄ = ln(4 x 19245) = 2.040960554861554

Iteration 5: x5 = ln(4 x 20410) = 2.099734244947598

Iteration 6: x6 = ln(4 x 20997) = 2.128088838501357

At this point, we have achieved the desired level of accuracy and can stop iterating.

After several iterations, we get x6 ≈ 2.128

Steps to locate the roots of an equation between two points

  • The following steps can be used to locate the roots of an equation between two points:

Step #1: Find the value of the function at the two intervals a and b.

Step #2: Check if there is a sign change between the two values obtained in step #2.

If the signs are different, there is a root between the two intervals.

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Solved Example: 

Question: Find the root of the equation f(x) = x3 – 4x2 + 5x – 2 between the intervals [0, 2].

Solution: 

  • Step #1: Find the value of the function at the two intervals a = 0 and b = 2.

    f(0) = -2 and f(2) = 2

  • Step #2: Check if there is a sign change between the two values obtained in step #2.

Since there is a sign change between -2 and 2, there is a root between the intervals [0, 2].

Practice Questions

Question 1: Solve the equation x3 – 2x – 5 = 0. Take the initial value as x = 2.

Question 2: Solve the equation 2x2 – 5x + 3 = 0. Take the initial value as x = 1.

Question 3: Solve the equation ex – x – 2 = 0. Take the initial value as x = 1.

Question 4: Solve the equation 4cos(x) – x = 0. Take the initial value as x = 1.

Question 5: Solve the equation ln(x) – x + 2 = 0. Take the initial value as x = 2.

Question 6: Solve the equation x2 – x – 1 = 0. Take the initial value as x = 2.

Question 7: Solve the equation 3x – 2x = 0. Take the initial value as x = 1.

Question 8: Solve the equation sin(x) + x – 2 = 0. Take the initial value as x = 1.

Question 9: Solve the equation x3 + x + 1 = 0. Take the initial value as x = 0.

Question 10: Solve the equation 5x2 – 6x – 2 = 0. Take the initial value as x = 1.

Solutions:

Question 1: Solve the equation x3 – 2x – 5 = 0. Take the initial value as x = 2.

Solution:

Using the method of iteration, we can start with an initial value of x = 2 and iterate using the formula:

xₙ₊₁ = ∛(xₙ² + 5/2)

Iteration 1:

x₁ = ∛(2² + 5/2) ≈ ∛(4 + 5/2) ≈ ∛(8.5) ≈ 2.0801

Iteration 2:

x₂ = ∛(2.0801² + 5/2) ≈ ∛(4.3264 + 5/2) ≈ ∛(7.3264) ≈ 2.1259

Continuing the iterations, we can approach the solution of the equation.

Question 2: Solve the equation 2x2 – 5x + 3 = 0. Take the initial value as x = 1.

Solution:

Using the method of iteration, we can start with an initial value of x = 1 and iterate using the formula:

xₙ₊₁ = (5xₙ – 3)/(2xₙ)

Iteration 1:

x₁ = (5(1) – 3)/(2(1)) = (5 – 3)/2 = 2/2 = 1

Iteration 2:

x₂ = (5(1) – 3)/(2(1)) = (5 – 3)/2 = 2/2 = 1

Continuing the iterations, we find that x remains 1, indicating that it is the solution to the equation.

Question 3: Solve the equation ex – x – 2 = 0. Take the initial value as x = 1.

Solution:

Using the method of iteration, we can start with an initial value of x = 1 and iterate using the formula:

xₙ₊₁ = ln(xₙ + 2)

Iteration 1:

x₁ = ln(1 + 2) = ln(3) ≈ 1.0986

Iteration 2:

x₂ = ln(1.0986 + 2) ≈ ln(3.0986) ≈ 1.1314

Continuing the iterations, we can approach the solution of the equation.

Question 4: Solve the equation 4cos(x) – x = 0. Take the initial value as x = 1.

Solution:

Using the method of iteration, we can start with an initial value of x = 1 and iterate using the formula:

xₙ₊₁ = arccos(xₙ/4)

Iteration 1:

x₁ = arccos(1/4) ≈ 1.3181

Iteration 2:

x₂ = arccos(1.3181/4) ≈ 1.2863

Continuing the iterations, we can approach the solution of the equation.

Question 5: Solve the equation ln(x) – x + 2 = 0. Take the initial value as x = 2.

Solution:

Using the method, we can start with an initial value of x = 2 and iterate using the formula:

xₙ₊₁ = e(xₙ – 2)

Iteration 1:

x₁ = e(2 – 2) = e0 = 1

Iteration 2:

x₂ = e(1 – 2) = e-1 ≈ 0.3679

Continuing the iterations, we can approach the solution of the equation.

Question 6: Solve the equation x2 – x – 1 = 0. Take the initial value as x = 2.

Solution:

Using the method, we can start with an initial value of x = 2 and iterate using the formula:

xₙ₊₁ = 1 + 1/xₙ

Iteration 1:

x₁ = 1 + 1/2 = 1.5

Iteration 2:

x₂ = 1 + 1/1.5 = 1.6667

Continuing the iterations, we can approach the solution of the equation.

Question 7: Solve the equation 3x – 2x = 0. Take the initial value as x = 1.

Solution:

Using the method, we can start with an initial value of x = 1 and iterate using the formula:

xₙ₊₁ = log₂(3xₙ)

Iteration 1:

x₁ = log₂(3(1)) = log₂(3) ≈ 1.585

Iteration 2:

x₂ = log₂(3(1.585)) ≈ log₂(4.755) ≈ 2.246

Continuing the iterations, we can approach the solution of the equation.

Question 8: Solve the equation sin(x) + x – 2 = 0. Take the initial value as x = 1.

Solution:

Using the method of iteration, we can start with an initial value of x = 1 and iterate using the formula:

xₙ₊₁ = 2 – sin(xₙ)

Iteration 1:

x₁ = 2 – sin(1) ≈ 1.158

Iteration 2:

x₂ = 2 – sin(1.158) ≈ 1.603

Continuing the iterations, we can approach the solution of the equation.

Question 9: Solve the equation x3 + x + 1 = 0. Take the initial value as x = 0.

Solution:

Using the method of iteration, we can start with an initial value of x = 0 and iterate using the formula:

xₙ₊₁ = -1/(xₙ2 + 1)

Iteration 1:

x₁ = -1/(02 + 1) = -1/1 = -1

Iteration 2:

x₂ = -1/((-1)2 + 1) = -1/2

Continuing the iterations, we can approach the solution of the equation.

Question 10: Solve the equation 5x2 – 6x – 2 = 0. Take the initial value as x = 1.

Solution:

Using the method of iteration, we can start with an initial value of x = 1 and iterate using the formula:

xₙ₊₁ = (5xₙ2 – 2) / 6

Iteration 1:

x₁ = (5(1)2 – 2) / 6 = (5 – 2) / 6 = 3/6 = 0.5

Iteration 2:

x₂ = (5(0.5)2 – 2) / 6 ≈ (5(0.25) – 2) / 6 ≈ (1.25 – 2) / 6 ≈ -0.125

Continuing the iterations, we can approach the solution of the equation.