Question

Are there any ways to find a good initial value for Newton's Method to guarantee convergence?

Answer 1

http://en.wikipedia.org/wiki/Bisection_method

Answer 2

from advance calculus absolute value [f(x)f''(x)/(f'(x))^2]< 1 for all x in an intervalabout a root r, then the method will converge to r. estudier answer is also correct by using the method of bisection

Answer 3

Thanks alot, I will check out more on this!

Answer 4

that is the absolute value must be less than 1in order to guarantee the convergence of the formula for newtons method...goodluck jsaetrum

Answer 5

thanks!

Answer 6

I tested whether an initial value x0=-1.5 makes a convergence on the function f(x) = x - 2sin x.
However I get [f(x)f''(x)/((f'(x))^2] > 1 despite of that Newton's Method converges to a solution.
Does it mean that a guarantee of convergence is not given until you get actually < 1?