1. Rearranging the equation f(x)=0 into the form x=g(x)

2. If I want to solve the equation
x3-x2-3x+0.5=0
we can draw a graph and read off the points where it crosses the x axis.
But we can also solve it by rearranging the equation and finding where the 2 graphs cross.
Both ways find the root.

3. x3-x2-3x+0.5=0 3x=x3-x2+0.5 x=(x3-x2+0.5)/3
So we can now plot the graphs of the 2
functions y=x and y=(x3-x2+0.5)/3
Iterative formula used to find the root.

4. y=x3-2x2-3x+0.5
All the same roots
y=x
y=(x3-2x2+0.5)/3

5. Using the iterative formula
From the graph you can tell that there is root in the interval [0,1].
I will let my starting value x1=1 and then find x2 by substitution. x
g(x) 1 2 3 4 5 6 7 8

6. What does this look like on the graph?
What to do
Choose a value of x, x1.
Find corresponding value of g(x1).
Take value g(x1) as new x, x2.
Find value of g(x2).
What this looks like
Take starting point on x axis.
Move vertically to curve.
Move across to line.

7. What it looks like on a graph.
STAIRCASE

8. COBWEB

9. Failure
This method fails if at the root of y=g(x) and y=x the gradient is>1 or <-1.
You need to show failure using excel too

10. If g ‘(x) is small then we get rapid convergence.
If g ‘(x) is close to ±1 then we get slow convergence.
If g ‘(x) is > ±1 then we get failure. ie divergence.
You can also get failure when you find another root from what you wanted to find.