Unit+Standard+5264+Numerical+Solutions

Numerical methods: For solving non- linear equations.

The two methods covered in 13 MAG are: The __bisection method__ and The __Newton-Raphson__ method

**The bisection method:**
The first point is to find an interval that will contain a root or solution to the equation.

We can either draw the graph or make up a table of values.

If we draw the graph, it is easy to tell the interval in which the solution or root lies in. The solution or root is at the point where the function crosses the x axis. Although we cannot see the precise number (x value) at times, we can easily read off the x values on either side to get our interval.

If we draw a table, our solution lies between the two x values the have a sign change (goes from positive to negative or negative to positive).

**Example: let our function be, f(x) = x^3 - x^2 - 3. Find a root for this function between 1 and 2 accurate to 2 dp.**
In this case, our Xo = 1 and our X1 = 2 f(Xo) = -2, this is what our function is worth when we replace every X with our first value, 1. Our f(Xo) is negative f(X1) = 1, our f(X1) value is positive

Since the function changes signs between these two x values, we now know that there is indeed a solution (root) between these two. Our aim now is to find a X value which will make our function = 0.

We do this by taking the mid - point between the two values. In our example our first midpoint is x = 1.5. We then replace every x in our function with our new x. If the result is positive our new X, called Xm replaces the old X value that gave us a positive result, and vice versa.
 * Xo || f(0) || X1 || f(1) || Xm || f(m) (3dp) ||
 * 1 || Neg || 2 || Pos || 1.5 || -1.875 ||
 * 1.5 || Neg || 2 || Pos || 1.75 || -0.703 ||
 * 1.75 || Neg || 2 || Pos || 1.875 || 0.076 ||
 * 1.75 || Neg || 1.875 || Pos || 1.8125 || -0.0331 ||
 * 1.8125 || Neg || 1.875 || Pos || 1.844 || -0.130 ||
 * 1.844 || Neg || 1.875 || Pos || 1.860 || -0.025 ||
 * 1.860 || Neg || 1.875 || Pos || 1.868 || 0.029 ||
 * 1.860 || Neg || 1.868 || Pos || 1.864 || 0.002 ||
 * 1.860 || Neg || 1.864 || Pos || 1.862 ||  ||

From the table, we can now see that our Xm has stabilised at 2 decimal places, and hence the solution to the Question is:
 * __The root of X^3 - X^2 - 3 which lies between 1 and 2 is 1.86 (2 dp).__**

X0 is our fist x value from our interval X1 is our second x value from the interval f0 is the value of the function using x0 f1 is the value of the function using x1