Exercise02 Sol
Transcript of Exercise02 Sol
-
- 1 -
AMS 147 Computational Methods and Applications
Exercise #2
1. Consider Newtons method for solving (x) = 0.
Suppose x is a root of f x( ) and f x( ) 0 . Do we expect Newtons method to converge if
the starting point x0 is sufficiently close to x ?
Answer: Yes.
2. Consider the iteration xn+1 =xn + 1
3.
Does the iteration converge for arbitrary starting point x0? And why?
Suppose it converges to x * . What is the value of x *?
What is the order of convergence? And why?
Answer:
It converges for arbitrary starting point x0 because g x( ) =x + 1
3 is a contraction mapping.
x * satisfies x* =x * +1
3 ==> x* = 0.5 .
The convergence is linear (first order) because g x *( ) =1
30 .
3. Suppose Newtons method converges to x and f x( ) 0 (i.e. x is a simple root).
What is the order of convergence?
Answer: quadratic (second order) convergence.
4. Suppose Newtons method converges to x and f x( ) = 0
-
AMS 147 Computational Methods and Applications
- 2 -
(i.e. x is a root of multiplicity > 1).
What is the order of convergence?
Answer: linear (first order) convergence.
5. Consider the golden search method for minimization.
Suppose we start the iteration with interval a, b[ ] . Let N be the number of iterations needed to
reach the error tolerance tol. Express N in terms of a, b and tol.
Answer:
Nlog
tolb a
log g( ), g =
5 1
2