Bisection Iteration method & Newton Raphson Method

Both methods help to find the approximate roots of an equation
In bisection iteration

Let me use an example to illustrate
Suppose f(x)=x^4 + 3*x^3 - 1 and we know that f(0)=-1, f(1)=3, so, there is a root lies between 0 and 1.
The method of bisection is to evaluate the average of two numbers.
For example.
x=0.5 f(x)=-0.5625
so, now, we should choose x = (1+0.5)/2 rather than (0+0.5)/2 as there is root between x = 0.5 and x = 1 as the curve passes through the x-axis
x=0.75 f(x)=0.582031
then, we choose (0.75+0.5)/2 as there is root between x = 0.5 and x = 0.75
x=0.625 f(x)=-0.11499
x=0.675 f(x)=0.198257
This method continues until you got -10^-5 < f(x) < 10^-5
______________

For Newton Raphson Method
A formula is required, that's
x = x0 - f(x0)/f'(x0), where x is evaluated from x0 and x0 is the initial condition
f(x)=x^4 + 3*x^3 - 1
f'(x)=4x^3+9x^2
As root lies between 0 and 1, we let x = 1 as initial condition
n is the number of times we have done
for n = 1, x = 1 - f(1)/f'(1),
n = 1 x=0.769231 f(x)=0.715628
for n = 2, x = 0.769231 - f(0.769231)/f'(0.769231)
n=2 x=0.669089 f(x)=0.099031
n=3 x=0.650144 f(x)=0.0030872
n=4 x=0.649514 f(x)=0.000001343

Hope i can help you

2006-11-07 21:39:46 補充：
The number of times each method have to go depends on how accurate your answer needed~