# 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

## 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 Level

## 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 Level
A Level

## 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 $n$th 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 Level

## 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 Level

## 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)

A Level

A Level

A Level

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