
Newton's method
Let f(x) = (cos(x))^2  3*x
Show that abs(f''(x)/f'(y))<=B, for x,y exist in [0,1], B is constant>0
Hence or otherwise deduce that Newton's method will converge.
I found that
abs(f''(x)/f'(y)) = abs( (2(sin(x)^2)  2*(cos(x)^2)) / (2cosy*siny  3))
I can deduce that abs(2(sin(x)^2)  2*(cos(x)^2)) is <=2
and that abs(2cosy*siny  3)>=3
so abs( (2(sin(x)^2)  2*(cos(x)^2)) / (2cosy*siny  3))<=2/3
and so B=1? Am I correct??
Thank you very much!!!

Re: Newton's method
Couldn't edit my post, but I can now see that B>0 as a positive constant can be a fraction so is B=2/3?
And after that how can I show that Newton's method converges? If the condition abs(f''(x)/f'(y))<=B, for x,y exist in [0,1] is satisfied then don't we automatically know that the method converges?
Thank you very much!!!

Re: Newton's method
Can someone help me please? :) Thanks again!