Lecture 23

Question 1

Bisection method

Answer 1

• Continuous function interval [a,b] containing root
• evaluate mid-point m=(a+b)/2
• check signs. a=m if sign(f(m))==sign(f(b)) else b=m.
  Interval reduced to half every iteration

Question 2

Convergence iterative method

Answer 2

ek = |xk-x| (error at iteration k) converges at rate r if
lim(kinf) ||ek+1||/||ek||^r = C
r=1 linear
r=2 quadratic
1

Question 3

Bisection method convergence

Answer 3

Linear r=1

Question 4

Bisection method operations

Answer 4

two function evaluations at first iteration then only one needed (f(m))

Question 5

Bisection length interval, # iterations for tolerance

Answer 5

length = b-a/(2^k)

length <= tol then k >= log2(b-a/tol)

Question 6

Newton’s method

Answer 6

Local convergence (sometimes does not converge towards a solution)
x0 initial guess
xk+1 = xk - f(xk)/f'(xk)

Question 7

Newton’s method convergence

Answer 7

Typically quadratic r=2

Question 8

Newton’s method operations

Answer 8

2 evaluations per iteration (function and first derivative)

Question 9

Secant method

Answer 9

Newton’s with first derivative approximation (still local convergence)…
needs two starting guess xk, xk-1.
f’(xk) = (f(xk)-f(xk-1))/(xk - xk-1)

Question 10

Secant method operations

Answer 10

Only one function evaluation per iteration

Question 11

Secant method convergence

Answer 11

superlinear, r= (1+sqrt(5))/2 = 1.618 (golden ratio)

Question 12

Bisection method python

Answer 12

import scipy.optimize as opt
def f(x):
    return x**3 - x - 1

root = opt.bisect(f, a=1, b=2)

Question 13

Newton’s method python

Answer 13

import scipy.optimize as opt

def f(x):
    return x**3 - x - 1
def fprime(x):
    return 3 * x**2 - 1

root = opt.newton(f, x0=1, fprime=fprime)

Question 14

Secant method python

Answer 14

import scipy.optimize as opt

def f(x):
    return x**3 - x - 1

root = opt.newton(f, x0=1)

(same as Newton’s without fprime)

Question 15

A step in Newton’s method can always be carried out for any smooth function and any value of the current guess xk.

Answer 15

False.

Question 16

Python derivative

Answer 16

from sympy import *
x = Symbol('x')
y = x**2 + 1
yp = diff(y, x)
f = lambdify(x, yp, 'numpy')