1. Some Comments on Root finding

2. Avoid subtraction of nearly equal numbers
Some Comments on Root finding
Consider the following two numbers
a= ;
b= ;
c = a - b
ans =
Avoid subtraction of nearly equal numbers
a=1e-16; b=1; c=-1;
a+(b+c)
(a+b)+c
Associative property
𝑎+ 𝑏+𝑐 = 𝑎+𝑏 +𝑐
ans =
e-16
why

3. Avoid subtraction of nearly equal numbers
Some Comments on Root finding
Avoid subtraction of nearly equal numbers
% solve x^2 + b*x + 1 = 0
f x*(a*x+b)+c;
xeps = 1e-8;
b = (2/xeps) + xeps; a=1; c=1;
f a*x^2 + b*x +c
x1 = (-b+sqrt(b^2-4*a*c))/(2*a)
y1=(-2*c)/(b+sqrt(b^2-4*a*c))
f(x1)
f(y1)
abs((x1-y1)/y1)*100
Solve:
𝑥 2 +𝑏 𝑥+1=0
𝑏=2× 10 8
𝑥 1 = −𝑏+ 𝑏 2 −4𝑎𝑐 2𝑎
rationalizing the numerator
x1 =
y1 =
e-09
ans =
100
𝑥 1 = −2𝑐 𝑏+ 𝑏 2 −4𝑎𝑐

4. Avoid subtraction of nearly equal numbers
Some Comments on Root finding
Avoid subtraction of nearly equal numbers
The Secant Method
Algebraically equivalent formula
𝑥 𝑛+1 = 𝑥 𝑛 − 𝑓( 𝑥 𝑛 )( 𝑥 𝑛 − 𝑥 𝑛−1 ) 𝑓 𝑥 𝑛 −𝑓( 𝑥 𝑛−1 )
𝑥 𝑛+1 = 𝑥 𝑛−1 𝑓 𝑥 𝑛 − 𝑥 𝑛 𝑓 𝑥 𝑛−1 𝑓 𝑥 𝑛 −𝑓( 𝑥 𝑛−1 )
in general, this iteration equation is likely to be less accurate than the one given in left box.
why
Here, we subtract two nearly equal numbers in both numerator and denominator.

5. Formula Error Equation guess
Some Comments on Root finding
Formula
Error Equation
guess
x0 suff close to x*
Any
𝑥 𝑛+1 = 𝑥 𝑛 − 𝑓( 𝑥 𝑛 ) 𝑓′( 𝑥 𝑛 )
𝑥 ∗ − 𝑥 𝑛+1 = 𝑓 2 𝜉 2 𝑓 ′ 𝑥 𝑛 𝑥 ∗ − 𝑥 𝑛 2
Newton’s
Secant
𝑥 𝑛+1 = 𝑥 𝑛 − 𝑓( 𝑥 𝑛 )( 𝑥 𝑛 − 𝑥 𝑛−1 ) 𝑓 𝑥 𝑛 −𝑓( 𝑥 𝑛−1 )
𝑥 𝑛+1 = 𝑎 𝑛 + 𝑏 𝑛 2
𝑥 ∗ − 𝑥 𝑛 ≤ 𝑏−𝑎 2 𝑛
Bisection
Fixed-point
𝑥 𝑛+1 =𝑓( 𝑥 𝑛 )
𝑥 ∗ − 𝑥 𝑛 ≤ 𝑘 𝑛 𝑥 ∗ − 𝑥 𝑛−1
𝑔′(𝑥) ≤𝑘 ∀𝑥
False position
Secant +test

6. Some Comments on Root finding
Cq1 solution
Cq2 solution

7. Apply Newton’s method to find a solution to 𝑥−6𝑐𝑜𝑠𝑥=0, 𝑥 0 =2
Some Comments on Root finding
Apply Newton’s method to find a solution to
𝑥−6𝑐𝑜𝑠𝑥=0, 𝑥 0 =2
In the interval [0,𝜋] that is accurate to within 10 −3
𝑥 𝑛+1 = 𝑥 𝑛+1 − 𝑥−6𝑐𝑜𝑠𝑥 1+6sin⁡(𝑥)
Newton’s method: 𝑛
𝑥 𝑛
𝑥 𝑛+1 − 𝑥 𝑛
e-01
e-02
e-04
e-09

8. Multiple Root Some Comments on Root finding
𝑥=1 is a zero of multiplicity 2 of the the function 𝑓 𝑥 = (𝑥−1) 2 (𝑥−3) 2
Note that no change of sign close to the root. (bisection ,FP ??)
Definition:
Theorem
p of f (x) = 0 is a zero of multiplicity m of f if for 𝑥=𝑝, we can write
𝑓 𝑥 = (𝑥−𝑝) 𝑚 𝑞 𝑥
where 𝑞(𝑝)≠0
p is a zero of multiplicity m of f if and only if
𝑓 𝑝 = 𝑓 (1) 𝑝 =⋯= 𝑓 𝑚−1 𝑝 =0
and 𝑓 (𝑚) 𝑝 ≠0

9. Multiple Root Some Comments on Root finding Example:
Show that f has a zero of multiplicity 2 at x = 0.
𝑓 𝑥 = 𝑒 𝑥 −𝑥−1
Newton’s method are shown in Table
The sequence is clearly converging to 0, but not quadratically.

10. modification of Newton’s method
Some Comments on Root finding
modification of Newton’s method
Example:
Show that f has a zero of multiplicity 2 at x = 0.
𝑓 𝑥 = 𝑒 𝑥 −𝑥−1
If p is a zero of f of multiplicity m
𝑓 𝑥 = (𝑥−𝑝) 𝑚 𝑞 𝑥
then p is a zero of 𝑓′ of multiplicity m-1
Hence p is a simple zero of μ 𝑥 = 𝑓(𝑥) 𝑓′(𝑥)
Newton’s method can then be applied to μ 𝑥
The only drawback to this method is the additional calculation of

11. modification of Newton’s method
Some Comments on Root finding
modification of Newton’s method
Example:
Show that f has a zero of multiplicity 2 at x = 0.
𝑓 𝑥 = 𝑒 𝑥 −𝑥−1
If p is a zero of f of multiplicity m
𝑓 𝑥 = (𝑥−𝑝) 𝑚 𝑞 𝑥
then p is also a fixed point of
𝑔 𝑥 =𝑥−𝑚 𝑓(𝑥) 𝑓′(𝑥)
𝑥 𝑛+1 = 𝑥 𝑛 −𝑚 𝑓( 𝑥 𝑛 ) 𝑓′( 𝑥 𝑛 )

12. Nonlinear system of equations:
Some Comments on Root finding
Nonlinear system of equations:
𝑓 1 =𝑥 𝑥 2 + 𝑥 3 2 −8
𝑓 2 =2𝑥 𝑥 2 2 − 𝑥 3 −11
𝑓 3 =𝑥 𝑥 𝑥 3 2 −14
𝑥 𝑥 2 + 𝑥 3 2 =8
2𝑥 𝑥 2 2 − 𝑥 3 =11
𝑥 𝑥 𝑥 3 2 =14
Convert into 𝐹 𝑋 =0
𝐹 𝑋 = 𝑥 𝑥 2 + 𝑥 3 2 −8 2𝑥 𝑥 2 2 − 𝑥 3 −11 𝑥 𝑥 𝑥 3 2 −14
𝑥 𝑥 2 + 𝑥 3 2 −8=0
2𝑥 𝑥 2 2 − 𝑥 3 −11=0
𝑥 𝑥 𝑥 3 2 −14=0
𝑋= 𝑥 1 𝑥 2 𝑥 3

13. Some Comments on Root finding
Def: Jacobian Matrix
𝐹 𝑋 = 𝑥 𝑥 2 + 𝑥 3 2 −8 2𝑥 𝑥 2 2 − 𝑥 3 −11 𝑥 𝑥 𝑥 3 2 −14
𝐽 𝑋 = 𝜕 𝑓 1 𝑥 1 𝜕 𝑓 1 𝑥 2 𝜕 𝑓 1 𝑥 3 𝜕 𝑓 2 𝑥 1 𝜕 𝑓 2 𝑥 2 𝜕 𝑓 2 𝑥 3 𝜕 𝑓 3 𝑥 1 𝜕 𝑓 3 𝑥 2 𝜕 𝑓 3 𝑥 3
𝑋= 𝑥 1 𝑥 2 𝑥 3
Example:
𝐽 𝑋 = 2 𝑥 𝑥 𝑥 2 2 −1 2 𝑥 1 2 𝑥 2 2 𝑥 3
Example:
𝑋 (0) = 1 −1 1
𝐽 𝑋 (0) = −1 2 −2 2

14. 𝑋 (𝑛+1) = 𝑋 (𝑛) − 𝐽 ( 𝑋 (𝑛) ) −𝟏 𝑓( 𝑋 (𝑛) )
Some Comments on Root finding
Newton’s method for single variable
𝑥 𝑛+1 = 𝑥 𝑛 − 𝑓( 𝑥 𝑛 ) 𝑓′( 𝑥 𝑛 )
𝑥 𝑛+1 = 𝑥 𝑛 − 𝑓′( 𝑥 𝑛 ) −𝟏 𝑓( 𝑥 𝑛 )
Newton’s method for system of nonlinear equations
𝐹 𝑋 =0
𝑋 (𝑛+1) = 𝑋 (𝑛) − 𝐽 ( 𝑋 (𝑛) ) −𝟏 𝑓( 𝑋 (𝑛) )
𝑋 (𝑛) = 𝑥 1 (𝑛) 𝑥 2 (𝑛) 𝑥 3 (𝑛)

15. 𝑨𝑌 𝑛 =𝑏 Newton’s method for system of nonlinear equations 𝐹 𝑋 =0
Some Comments on Root finding
Newton’s method for system of nonlinear equations
𝐹 𝑋 =0
𝑋 (𝑛+1) = 𝑋 (𝑛) − 𝐽 ( 𝑋 (𝑛) ) −𝟏 𝑓( 𝑋 (𝑛) )
𝑋 (𝑛) = 𝑥 1 (𝑛) 𝑥 2 (𝑛) 𝑥 3 (𝑛)
𝑋 (𝑛+1) − 𝑋 𝑛 =− 𝐽 ( 𝑋 (𝑛) ) −𝟏 𝑓( 𝑋 (𝑛) )
𝑋 (𝑛+1) − 𝑋 𝑛 = 𝐽 𝑋 𝑛 −𝟏 (−𝑓 𝑋 𝑛 )
𝑌 𝑛 = 𝐽 𝑋 𝑛 −𝟏 (−𝑓 𝑋 𝑛 )
𝐽 𝑋 𝑛 𝑌 𝑛 = −𝑓 𝑋 𝑛
𝑨𝑌 𝑛 =𝑏
This is a linear system of n-equations in n-unknowns
where 𝐴=𝐽 𝑋 𝑛 𝒂𝒏𝒅 𝑏= −𝑓 𝑋 𝑛

16. ALGORITHM: (Newton’s method for system of nonlinear equations)
Some Comments on Root finding
ALGORITHM: (Newton’s method for system of nonlinear equations)
𝐹= 𝑥 𝑥 2 + 𝑥 3 2 −8 2𝑥 𝑥 2 2 − 𝑥 3 −11 𝑥 𝑥 𝑥 3 2 −14
𝐽 𝑋 = 2 𝑥 𝑥 𝑥 2 2 −1 2 𝑥 1 2 𝑥 2 2 𝑥 3
𝑿 (𝟎) = 𝒙 𝟏 (𝟎) 𝒙 𝟐 (𝟎) 𝒙 𝟑 (𝟎)
𝑋 (0) = 1 −1 1
Given initial guess
𝐽 𝑋 (0) = −1 2 −2 2
1) Calculate Jacobian Matrix
𝑨= 𝑱 𝑿 (𝟎)
𝑏= −7 −7 −13
2) Calculate right-hand-side vector
𝒃= −𝒇 𝑿 𝟎
−1 2 − 𝑦 1 (0) 𝑦 2 (0) 𝑦 3 (0) = −7 −7 −13
→ 𝑦 1 (0) 𝑦 2 (0) 𝑦 3 (0) = 19/18 −4/3 28/9
3) Solve the linear system:
𝑨𝒀 𝟎 =𝒃
4) update
𝑋 (1) = −
𝑿 (𝟎) = 𝑿 (𝟎) + 𝒀 (𝟎)
Repeat (1 - 4)

17. Some Comments on Root finding
clear; clc
f [ x(1)^2+x(2)+…
x(3)^2-8;2*x(1)+3*x(2)^2-x(3)-11; x(1)^2+x(2)^2+x(3)^2-14];
x=[1;-1;1];
format short
for n=1:10
[J] = jacob(x);
F = - f(x);
y = J\F
x = x + y
end x
function [J] = jacob(x);
J=[ 2*x(1) *x(3); ...
*x(2) ; …
2*x(1) 2*x(2) *x(3)]; 1 2 3 4 n
𝒙 𝟏 (𝒏)
𝒙 𝟐 (𝒏)
𝒙 𝟑 (𝒏)
𝑥 𝑥 2 + 𝑥 3 2 =8
2𝑥 𝑥 2 2 − 𝑥 3 =11
𝑥 𝑥 𝑥 3 2 =14