Iterative Methods

Iterative Methods

A LevelAQAEdexcelOCRAQA 2022Edexcel 2022OCR 2022

Iterative Methods

Using iteration allows you to find approximate roots to a given level of accuracy. When using iterative methods, you substitute an approximate value of the root into an iteration formula, and then you substitute this new approximate root back in until you get a root that is to the desired accuracy.

A Level AQA Edexcel OCR

Forming Iteration Formulas by Rearranging Equations

Iteration formulas are formed by rearranging equations and isolating a single variable.

There are many ways to rearrange equations, and not all will give you an iteration formula that converges to give a root. So an exam question will most likely ask you to show that an equation can be rearranged into a specific form.

Example: 

Show that 2x^3-3x-5=0 can be rearranged to: x=\sqrt[3]{\dfrac{3x+5}{2}}

 

First add 3x+5 to both sides to give:

2x^3=3x+5

Next divide both sides by 2:

x^3=\dfrac{3x+5}{2}

Finally take the cube root of both sides:

x=\sqrt[3]{\dfrac{3x+5}{2}}

Therefore, the iteration formula is: x_{n+1}=\sqrt[3]{\dfrac{3x_{n}+5}{2}} to find approximate roots.

A LevelAQAEdexcelOCR

Using Iterations to Draw Diagrams

After you have used an iteration method, you can form a sequence of iterations using x_{n+1}=f(x_n) and then plot these points on a diagram to show whether the sequence converges or diverges.

Forming iteration diagrams

1) Sketch the graphs of y=x and y=f(x), where f(x) is the iterative formula. The root of the original equation is the point of intersection of the two graphs.

2) From your starting point x_0 draw a vertical line until it meets y=x

3) Next draw a horizontal line from this point to the line y=f(x). This point is the first iteration, x_1.

4) After this, draw a vertical line from this point to the line y=x and then a horizontal line to y=f(x). Repeat this step for the remaining iterations.

5) If after each step points are getting closer to the roots, the sequence is converging. If after each step the points are getting further away from the root, the sequence is diverging.

 

There are two types of diagrams – staircase diagrams and cobweb diagrams.

In convergent staircase diagrams, the iterations increasingly get closer to the root.

In convergent cobweb diagrams, the iterations alternate between going above and below the root, progressively getting closer.

 

A LevelAQAEdexcelOCR
A Level AQA Edexcel OCR

Example 1: Iteration Formula

Starting with x_0=1, use the iteration formula x_{n+1}=\sqrt{\dfrac{9}{x_{n}+1}} to solve x^3+x^2-9=0 to 1 decimal place.

x_n denotes the approximation of the solution at the nth iteration

Starting with x_0=1, x_1=\sqrt{\dfrac{9}{1+1}}=2.121320344

Substitute this value back into the iteration formula, x_2=\sqrt{\dfrac{9}{2.121320344+1}}=1.698056292

Repeat the previous step until you get consecutive answers that are the same when rounded to 1 decimal place.

So,

x_3=\sqrt{\dfrac{9}{1.698056292+1}}=1.826399382

Then,

x_4=\sqrt{\dfrac{9}{1.826399382+1}}=1.784450437

As we can see, x_3 and x_4 both round to the same value to 1 decimal place, so the root is x=1.8

A LevelAQAEdexcelOCR

Example 2: Drawing Iteration Diagrams

7x-x^2+12=0 can be rearranged to give the iteration formula: x_n=\sqrt{7x+12}

Starting with x_0=8, use the iteration formula to find x_1, x_2 and x_3 and hence sketch a diagram to show that the sequence x_n converges.

 

Starting with x_0=8:

x_1=\sqrt{7(8)+12}=8.246211251

x_2=\sqrt{7(8.246211251)+12}=8.350058608

x_3=\sqrt{7(8.350058608)+12}=8.393474266

We can now create the diagram for this iteration formula:

Draw the lines y=x and y=\sqrt{7x+12} on the same set of axis.

Then draw on the lines corresponding with the iterations.

We can see that the sequence is a convergent staircase.

A LevelAQAEdexcelOCR

Example Questions

Add x^2 to both sides of 6x+8-x^2=0:

 

x^2=6x+8

 

Then take the square root of each side:

 

x=\sqrt{6x+8}

 

Using the iteration formula x_{n+1}=\sqrt{6x_n+8} and x_0=7:

 

x_1=\sqrt{6(7)+8}=7.071067812

 

x_2=\sqrt{6(7.071067812)+8}=7.101155319

 

x_3=\sqrt{6(7.101155319)+8}=7.113854927

 

Thus, the one approximate root of 6x+8-x^2=0 is 7.1 to 1 decimal place.

Using x_0=3:

 

x_1=\sqrt[3]{7(3)+11}=3.174802104

 

x_2=\sqrt[3]{7(3.174802104)+11}=3.214762996

 

x_3=\sqrt[3]{7(3.214762996)+11}=3.223760026

 

x_4=\sqrt[3]{7(3.223760026)+11}=3.225778757

 

x_5=\sqrt[3]{7(3.225778757)+11}=3.226231368

 

x_4 and x_5 both round to the same value to 3 decimal places.

So the approximate root is 3.226 to 3 decimal places.

 

a) Starting with x_0=4:

 

x_1=\sqrt{\dfrac{11(4)+12}{3}}=4.320493799

 

x_2=\sqrt{\dfrac{11(4.320493799)+6}{3}}=4.454414731

 

x_3=\sqrt{\dfrac{11(4.454414731)+6}{3}}=4.509196604

 

x_2 and x_3 both round to 4.5 to 1 decimal places, so this is our approximate root.

 

b)

 

Additional Resources

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Formula Booklet

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Worksheet and Example Questions

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Newton-Raphson Method and Other Recurrence Relations

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