Iterative Methods

A LevelAQAEdexcelOCR

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.

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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 2x33x5=02x^3-3x-5=0 can be rearranged to: x=3x+523x=sqrt[3]{dfrac{3x+5}{2}}

 

First add 3x+53x+5 to both sides to give:

2x3=3x+52x^3=3x+5

Next divide both sides by 22:

x3=3x+52x^3=dfrac{3x+5}{2}

Finally take the cube root of both sides:

x=3x+523x=sqrt[3]{dfrac{3x+5}{2}}

Therefore, the iteration formula is: xn+1=3xn+523x_{n+1}=sqrt[3]{dfrac{3x_{n}+5}{2}} to find approximate roots.

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Using Iterations to Draw Diagrams

After you have used an iteration method, you can form a sequence of iterations using xn+1=f(xn)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=xy=x and y=f(x)y=f(x), where f(x)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 x0x_0 draw a vertical line until it meets y=xy=x

3) Next draw a horizontal line from this point to the line y=f(x)y=f(x). This point is the first iteration, x1x_1.

4) After this, draw a vertical line from this point to the line y=xy=x and then a horizontal line to y=f(x)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
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Example 1: Iteration Formula

Starting with x0=1x_0=1, use the iteration formula xn+1=9xn+1x_{n+1}=sqrt{dfrac{9}{x_{n}+1}} to solve x3+x29=0x^3+x^2-9=0 to 11 decimal place.

xnx_n denotes the approximation of the solution at the nnth iteration

Starting with x0=1x_0=1, x1=91+1=2.121320344x_1=sqrt{dfrac{9}{1+1}}=2.121320344

Substitute this value back into the iteration formula, x2=92.121320344+1=1.698056292x_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 11 decimal place.

So,

x3=91.698056292+1=1.826399382x_3=sqrt{dfrac{9}{1.698056292+1}}=1.826399382

Then,

x4=91.826399382+1=1.784450437x_4=sqrt{dfrac{9}{1.826399382+1}}=1.784450437

As we can see, x3x_3 and x4x_4 both round to the same value to 11 decimal place, so the root is x=1.8x=1.8

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Example 2: Drawing Iteration Diagrams

7xx2+12=07x-x^2+12=0 can be rearranged to give the iteration formula: xn=7x+12x_n=sqrt{7x+12}

Starting with x0=8x_0=8, use the iteration formula to find x1x_1, x2x_2 and x3x_3 and hence sketch a diagram to show that the sequence xnx_n converges.

 

Starting with x0=8x_0=8:

x1=7(8)+12=8.246211251x_1=sqrt{7(8)+12}=8.246211251

x2=7(8.246211251)+12=8.350058608x_2=sqrt{7(8.246211251)+12}=8.350058608

x3=7(8.350058608)+12=8.393474266x_3=sqrt{7(8.350058608)+12}=8.393474266

We can now create the diagram for this iteration formula:

Draw the lines y=xy=x and y=7x+12y=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.

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Iterative Methods Example Questions

Example 1: Show that 6x+8x2=06x+8-x^2=0 can be rearranged to x=6x+8x=sqrt{6x+8} and hence start with x0=7x_0=7 to find a root to 11 decimal place.

[3 marks]

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Add x2x^2 to both sides of 6x+8x2=06x+8-x^2=0:

 

x2=6x+8x^2=6x+8

 

Then take the square root of each side:

 

x=6x+8x=sqrt{6x+8}

 

Using the iteration formula xn+1=6xn+8x_{n+1}=sqrt{6x_n+8} and x0=7x_0=7:

 

x1=6(7)+8=7.071067812x_1=sqrt{6(7)+8}=7.071067812

 

x2=6(7.071067812)+8=7.101155319x_2=sqrt{6(7.071067812)+8}=7.101155319

 

x3=6(7.101155319)+8=7.113854927x_3=sqrt{6(7.101155319)+8}=7.113854927

 

Thus, the one approximate root of 6x+8x2=06x+8-x^2=0 is 7.17.1 to 11 decimal place.

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Question 2: Use the formula xn+1=7x+113x_{n+1}=sqrt[3]{7x+11} with x0=3x_0=3 to find a root to the equation 11+7xx3=011+7x-x^3=0 to 33 decimal places.

[2 marks]

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Using x0=3x_0=3:

 

x1=7(3)+113=3.174802104x_1=sqrt[3]{7(3)+11}=3.174802104

 

x2=7(3.174802104)+113=3.214762996x_2=sqrt[3]{7(3.174802104)+11}=3.214762996

 

x3=7(3.214762996)+113=3.223760026x_3=sqrt[3]{7(3.214762996)+11}=3.223760026

 

x4=7(3.223760026)+113=3.225778757x_4=sqrt[3]{7(3.223760026)+11}=3.225778757

 

x5=7(3.225778757)+113=3.226231368x_5=sqrt[3]{7(3.225778757)+11}=3.226231368

 

x4x_4 and x5x_5 both round to the same value to 33 decimal places.

So the approximate root is 3.2263.226 to 33 decimal places.

 

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Question 3: The equation 11x3x2+12=011x-3x^2+12=0 can be rearranged to give the iteration formula:

xn+1=11x+123x_{n+1}=sqrt{dfrac{11x+12}{3}}

a) Use the iteration formula and x0=4x_0=4 to find the root of the equation to 11 decimal place.

 

b) Sketch a diagram to show the convergence of the sequence for x1x_1, x2x_2 and x3x_3.

 

[4 marks]

A Level AQAEdexcelOCR

a) Starting with x0=4x_0=4:

 

x1=11(4)+123=4.320493799x_1=sqrt{dfrac{11(4)+12}{3}}=4.320493799

 

x2=11(4.320493799)+63=4.454414731x_2=sqrt{dfrac{11(4.320493799)+6}{3}}=4.454414731

 

x3=11(4.454414731)+63=4.509196604x_3=sqrt{dfrac{11(4.454414731)+6}{3}}=4.509196604

 

x2x_2 and x3x_3 both round to 4.54.5 to 11 decimal places, so this is our approximate root.

 

b)

 

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Additional Resources

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Exam Tips Cheat Sheet

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Formula Booklet

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Specification Points Covered

I2 – Solve equations approximately using simple iterative methods; be able to draw associated cobweb and staircase diagrams; understand how such methods can fail
I4 – Use numerical methods to solve problems in context

Iterative Methods Worksheet and Example Questions

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Newton-Raphson Method and Other Recurrence Relations

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