In this tutorial, you will learn:

- How to find a range of the variable, x, where a root - or a solution - of a function could exist in
- How to turn a function into an
*iterative*form - How to use
**iteration**to find a specific root - or a solution - How to find the square root of a number by iteration

If f(x) is a function and it changes its sign from +ve to -ve or vice versa in a certain range of x, then a root of the function exists within the same range. The following image shows it:

**E.g.1**Find a range of x in which a root for f(x) = x^{2} -x -3 exists.

f(x) = x^{2} -x -3

f(2) = 4 - 2 - 3 = -1 => -ve

f(3) = 9 - 2 - 3 = 4 => +ve

Since there is a change in sign in the range, 2≤ x ≤3, a root of the function exists in the same range.

Sometimes, there is not a change in sign, yet there will be a root in a particular region. On the other hand, a change in sign does not necessarily indicate the existance of a root within a range. The following images illustrate this:

If f(x) = 0 and it can be rearranged in such a way that x = g(x), where both f(x) and g(x) are functions of x, **x = g(x)** is said to be in **iterative** form.

**E.g.**

x^{2} - x - 6 = 0

Let's make 'x' the subject of this equation; there are three possibilities:

- x = √(6 + x)
- x = x
^{2}- 6 - x = 1 + 6 / x

In the above iterations, g(x) = √(6 + x) or g(x) = x^{2}- 6 or g(x) = 1 + 6 / x.

When the above equation - x^{2} -x - 6 = 0 - is written in one of the above forms
with 'x' being the subject, they are said to be in an iterative form. That means, when the a value is substituted
for 'x' on the right hand side the value of 'x' on the left can be obtained.
Then the latter is put back in the equation to generate the other value. This
goes on until, x **approaches** a **constant** value. This
value is taken as a solution. Therefore, the actual iterative formula takes the following form, depending on the
rearrangement.

- x
_{n+1}= √(6 + x_{n}) - x
_{n+1}= x_{n2}- 6 - x
_{n+1}= 1 + 6/x_{n}

The first value of x - x_{0} - is taken from a range of possibilities - guesses. Now let's solve one of the
above equations using iteration.

**x _{n+1} = √(6 + x_{n})**

First, guess a range where a solution could exist. Press the button and it will provide you with one.

Now fill in the value of 'x_{0}' from one of the value you get and press the iterate button to find the solution:

x_{0}=

Have you noticed the way 'x' values approach the solution?; the longer you go the better.You can now see the beauty of iteration; it helps us find the root of an a function; since we have different forms of iteration, we can use some of them to find all the roots.

Have you noticed the
way 'x' values approach the solution? The longer you go the better. You can now see the
**beauty of iteration**; it helps us to find the solution of an equation; since we
have different forms of iteration, we can use each one of them to find some of the roots, if not all.

**E.g.1**

Let's look at another simple example - a quadratic function.

x^{2} - 10x + 5 = 0

x^{2} = 10x - 5

:- x => x = 10 -5/x

x_{n+1} = 10 -5/x_{n}

You can practise this interactively with the following applet: chose x_{0} as 7, 8, 9 and 10 and see how it works; you will be amazed at the power and beauty of iteration.

Of course, in this simple example, a quadratic function was used for illustration. Iteration is generally used for higher polynomials such as cubic functions and more complex functions.

**E.g.2**

Find a root of the function, x^{3} + 4x - 3 by iteration.

f(x) = x^{3} + 4x - 3

f(0) = -3; f(1) = 2 => There is a change in sign between x = 0 and x = 1; therefore, a root could exist in the range.

Now, let's make the iterative formula.

x^{3} + 4x - 3 = 0

x = (3 - x^{3})/4

x_{n+1} = (3 - x_{n}^{3})/4

Let x_{0}, the initial value = 0.

x_{1} = 0.5

x_{2} = 0.71875

x_{3} = 0.65717

x_{4} = 0.67904

x_{5} = 0.67172

x_{6} = 0.67422

x_{7} = 0.673377...

So, the root is approximately, x = 0.673 (3 d.p.).

The following animation shows how to use Microsoft Excel in finding a root of a function by iteration.

**E.g.**

Find a root of the function, f(x) = x^{3} + 3x - 5 by iteration.

x_{n+1} = (5 - x_{n}^{3})/3

f(1) = -1 => -ve; f(2) = 9 => +ve;

So, there must be a root between x = 1 and x = 2. The iteration will find it.

Let x_{0} = 1.

As you can see, x = 1.67 is a root to 2 d.p.

You can experiment with iteration here: change the value of **x** in **x** and the **number of iterations**n and experiment with it.

**E.g.**

Solve x^{2} - x - 3 = 0.

x^{2} - x - 3 = 0.

x^{2} = x + 3

x = 1 + 3/x

x_{n+1} = 1 + 3/x_{n}

You may use x_{0} = 2 or 4 for the practice.

Iteration can be used to find the square root of number. This is how it works:

Let x^{2} = N

:-x => x = N/x

+x => 2x = N/x + x

:-2 => x = N/2x + x/2

x = 1/2[N/x + x]

x_{n+1} = 1/2[N/x_{n} + x_{n}]

Take x_{0} = 1 and iterate.

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