# Homework Help: A proof in Numerical Analysis

1. Sep 25, 2005

### pinodk

Hello there!
yet another proof, that i need help on
I am supposed to prove that the following statement holds for the secant method
dk+1/ek -> -1 for k->Infinity
where
dk+1 is the next change and ek is the error.

I have this idea, but i want to hear whether its a valid proof.

i use the expression for the secant method

xk+1 = xk - f(xk) * ( xk-xk-1/f(xk)-f(xk-1) )

and derive that
dk+1 = xk+1 - xk = - f(xk) * ( xk-xk-1/f(xk)-f(xk-1) ) (1)

I then use an expression in the lecture book, saying that
f(xk) = ek* ( f(xk)-f(xk-1)/xk-xk-1 ) - (ek-1*ek * f''(xa)/2 )

My argument is then that for k->Infinity, i will get that - (ek-1*ek * f''(xa)/2 ) goes towards zero. xa is in the interval between the exact solution and the current x, xk.
This is the part that im not sure if im right about, can i argue like this?

I then get the following expression

f(xk) = ek* ( f(xk)-f(xk-1)/xk-xk-1 )

Where I use the expression (1) and get
f(xk) = ek* (- f(xk) /dk+1)
Ánd from this I get
dk+1/ek = -1

Cheers
-Daniel

PS: How do you make those javascript math expressions ive seen in some of the posts?