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:

What is Iteration?
Steps for Solving an equation using iteration
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 x_n+1 = g(x_n), 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.

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:

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 + 3x_2.7144)^1/3 = 2.6751

x₃ = (11 + 3x₂)^1/3 = (11 + 3x_2.6751)^1/3 = 2.6696

x₄ = (11 + 3x₃)^1/3 = (11 + 3x_2.6696)^1/3 = 2.6688

x₅ = (11 + 3x₄)^1/3 = (11 + 3x_2.6688)^1/3 = 2.6687

After several iterations, we get x₅ ≈ 2.6687

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

Solution:

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

f(x) = e^x – 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.

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 e^x – 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 – 2^x = 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 e^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 = 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)

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^(x^{ₙ – 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 – 2^x = 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.

Struggling with a topic?

Iteration Equations: Comprehensive Guide with Worksheet

What is Iteration?

Steps for Solving an equation using iteration

Steps to locate the roots of an equation between two points

Practice Questions

Solutions:

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