Algebra

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Iteration

Iteration is another method of find approximate solutions to complicated quadratic equations.

Iteration is used to find an approximate numerical solution to an equation. This is particularly useful for equations that are complicated to solve or do not have whole number solutions.

Look at $x^3 + x = 5$ .

We don't have an easy algebraic way to solve this, but we can repeatedly use an iterative formula to find more accurate values of $x$ . However, the value of $x$ may not ever be exact.

What is an iterative formula?

An iterative formula is in the form $x=$ , and allows us to work out values of $x$ .

Let's try making an iterative formula for $x^3 + x = 5$

Rearrange to make $x$ the subject

We need to find an equation with $x$ as the subject in its singular form. Since the two $x$ terms are raised to different powers, we do not collect them together.

First, make $x^3$ the subject

$x^3 = 5 - x$

Then cube root both sides to isolate singular $x$

$x =\sqrt[3](5 - x)$

Nice!

$x =\sqrt[3](5 - x)$ is our iterative formula which we can use to find increasingly accurate values of $x$ .

Let's try another one! Make an iterative formula for $x^2 + x = 56$

Make $x^2$ the subject of the equation.

Square root both sides

Write an iterative formula for $x^2 - 6x = 7$

Write an iterative formula for $x^2 + 5x -24 = 0$

Once we have an iterative formula, we can use it to find values for $x$ .

We will call the $x$ on the left of the formula $X_{n+1}$ and the $x$ on the right $X_n$ .

$x = \sqrt[3]{5 - x}$ $\rightarrow$

$X_{n+1} = \sqrt[3]{5 - X_n}$

This just means that we will start with a value of $X1$ on the right and use the formula to find the second value, $X2$ . We then repeat this process to find $X3$ , and so on.

Let's use the iterative formula to find some values : $X_{n+1} = \sqrt[3]{5 - X_n}$

The first value of $n$ is 1

It is the first time using the formula

We can choose any value to try for $X_1$

We will use $2$

Put these numbers into the formula

$X_{{\color{#21affb}1}+1} = \sqrt[3]{5 - {\color{#21affb}2}}$

Find the value of $X_2$

$X_2=\sqrt[3]{5 - 2}$ so $X_2=1.44224957$

Use $X_2$ to find $X_3$

$X_{\color{#21affb}2}+1 = \sqrt[3]{5 - {\color{#21affb}1.44224957}}$

Solve, to 4 s.f., $X_{{\color{#21affb}2}+1} = \sqrt[3]{5 - {\color{#21affb}1.44224957}}$

$X3=1.526599655$ . What is $X4$ if $X_{{\color{#21affb}3}+1} = \sqrt[3]{5 - {\color{#21affb}1.526599655}}$

$X4=1.514438405$ . What is $X5$ if $X_{{\color{#21affb}4}+1} = \sqrt[3]{5 - {\color{#21affb}1.514438405}}$

Notice the difference between $X$ values

The difference between successive values is getting smaller. This means that we are getting more accurate values.

So, we can round our number

$X_5$ and values after it all round to $1.52$

This is our answer!

Approximately, $x\approx1.52$

To solve $\sqrt{x} + 1 - x = 0$ , we can use the iterative formula: $X_n+1 = \sqrt{X_n} + 1$ . If $X1 = 2$ , what is $X2$ ?

To solve $\sqrt{x} + 1 - x = 0$ , we can use the iterative formula: $X_n+1 = \sqrt{X_n} + 1$ . If $X_2 = 2.414213562$ what is $X_3$ ?

If you keep iterating using the formula $X_n+1 = \sqrt{X_n} + 1$ , what is the approximate solution to $\sqrt{x} + 1 - x = 0$ to 3 significant figures?