# Thread: you're a god if you can prove this

1. ## you're a god if you can prove this

Let f be some given function for which we wish to find a such that
f(alpha)= 0. Suppose that the equation f(x)= 0 may be arranged into
the form x=g(x), such that for some interval (a,b) alpha is in (a,b) and
g(x) belongs to (a,b) . Further, suppose that g is differentiable with |g'(x)|</= C
for x belonging to (a,b), where C is some positive number. (Note that x=g(x)
implies that alpha=g(alpha) If we have the iteration method
Xn+1=g(Xn)
where X0 belonging to (a,b) is some given value, prove that
|alpha - Xn+1| </= C|alpha - Xn|

please note that when i say "belonging to" i mean that symbol that looks a bit like an 'E'

2. ## Re: you're a god if you can prove this

Originally Posted by limpfisch
Let f be some given function for which we wish to find a such that
f(alpha)= 0. Suppose that the equation f(x)= 0 may be arranged into
the form x=g(x), such that for some interval (a,b) alpha is in (a,b) and
g(x) belongs to (a,b) . Further, suppose that g is differentiable with |g'(x)|</= C
for x belonging to (a,b), where C is some positive number. (Note that x=g(x)
implies that alpha=g(alpha) If we have the iteration method
Xn+1=g(Xn)
where X0 belonging to (a,b) is some given value, prove that
|alpha - Xn+1| </= C|alpha - Xn|

please note that when i say "belonging to" i mean that symbol that looks a bit like an 'E'
What have you been able to do so far?

-Dan