# Iterative Methods

## Iterative Methods Revision

**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.

**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.

**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.

**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

**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**.

## Iterative Methods Example Questions

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

**[3 marks]**

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.

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

**[2 marks]**

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.

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

a) Use the iteration formula and x_0=4 to find the root of the equation to 1 decimal place.

b) Sketch a diagram to show the convergence of the sequence for x_1, x_2 and x_3.

**[4 marks]**

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)

## Iterative Methods Worksheet and Example Questions

### Newton-Raphson Method and Other Recurrence Relations

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