Title: 2. Solving Equations of One Variable
12. Solving Equations of One Variable
- Korea University Computer Graphics Lab.
- Lee Seung Ho / Shin Seung Ho Roh Byeong Seok /
Jeong So Hyeon
2Contents
- Bisection Method
- Regula Falsi and Secant Method
- Newtons Method
- Mullers Method
- Fixed-Point Iteration
- Matlabs Method
3Bisection Method
4Bisection Method
5Finding the Square Root of 3 Using Bisection
How can we get ?
6Approximating the Floating Depth for a Cork Ball
by Bisection(1/2)
Cork ball Radius 1 Density 0.25
7Approximating the Floating Depth for a Cork Ball
by Bisection(2/2)
8Discussion of Bisection Method
9Fixed-Point Iteration
- Solution of equation
-
- Convergence Theorem of fixed-point iteration
10Fixed-Point Iteration to Find a Zero of a Cubic
Function
11Matlabs Methods(1/2)
- roots(p)
- p vector
- Example
-
-
- EDUgt r roots(p) (p1 -7 14 -7)
- r 3.8019
- 2.445
- 0.75302
12Matlabs Methods(2/2)
- fzero( function name,x0 )
- function name string
- x0 initial estimate of the root
- Example
- function y flat10(x)
- y x.10 0.5
- z fzero(flat10,0.5)
- z 0.93303
13Regular Falsi and Secant Methods
- 2005. 3. 23
- Byungseok Roh
14Regula Falsi Method
- The regula falsi method start with two point, (a,
f(a)) and (b,f(b)), satisfying the condition that
f(a)f(b)lt0. - The straight line through the two points (a,
f(a)), (b, f(b)) is - The next approximation to the zero is the value
of x where the straight line through the initial
points crosses the x-axis.
15Regula Falsi Method (cont.)
- If there is a zero in the interval a, c, we
leave the value of a unchanged and set b c. - On the other hand, if there is no zero in a, c,
the zero must be in the interval c, b so we
set a c and leave b unchanged.
- The stopping condition may test the size of y,
the amount by which the approximate solution x
has changed on the last iteration, or whether the
process has continued too long. - Typically, a combination of these conditions is
used.
16Example
- Finding the Cube Root of 2 Using Regula Falsi
- Since f(1) -1, f(2)6, we take as our starting
bounds on the zero a1 and b2. - Our first approximation to the zero is
- We then find the value of the function
- Since f(a) and y are both negative, but y and
f(b) have opposite signs
17Example (cont.)
- Calculation of using regula falsi.
18Secant Method
- The secant method, closely related to the regula
falsi method, results from a slight modification
of the latter.
- Instead of choosing the subinterval that must
contain the zero, we form the next approximation
from the two most recently generated points - At the k-th stage, the new approximation to the
zero is
- The secant method has converged with a tolerance
of .
19Example
- Finding the Square Root of 3 by Secant Method
- To find a numerical approximation to , we
seek the zero of - .
- Since f(1)-2 and f(2)1, we take as our starting
bounds on the zero and . - Our first approximation to the zero is
- Calculation of using secant method.
20NEWTONS METHOD
21Newtons Method
- Newtons method uses straight-line approximation
which is the tangent to curve. - .
- Intersection point
22Example
- Finding Square Root of ¾
- approximate the zero of
using the fact that . - Continuing for one more step
-
23Finding Floating Depth for a Wooden Ball
- Volume of submerged segment of the Sphere
- To find depth at which the ball float, volume of
submerged segment is time. - Simplifies to
24Finding Floating Depth for a Wooden Ball (cont.)
- To find depth a ball, density is one-third of
water float.
Calculation f(x) using Newtons Method
25Oscillations in Newton Method
- Newtons method give Oscillatory result for some
funtions initial estimates. - Ex)
26Mullers Method
27Mullers Method
- based on a quadratic approximation
- procedure
- Decide the parabola passing through (x1,y1), (x2,
y2) and (x3,y3) - Solve the zero(x4) that is closest to x3
- Repeat 1,2 until x converge to predefined
tolerance - advantage
- Requires only function values
- Derivative need not be calculated
- X can be an imaginary number.
28Mullers Method (Cont)
29Example
- Finding the sixth root of 2 using Mullers method
- , ,
, -
30Example (Cont)
i x y
1 0.5 -1.9844
2 1.5 9.3906
3 1 -1
4 1.0779 -0.43172
5 1.117 -0.05635
6 1.1255 0.00076162
7 1.1255 -4.7432e-07
converge
Calculation of using Mullers method
31Another Challenging Problem
Tolerance 0.0001
step x Y
1 0 -0.5
2 1 0.5
3 0.5 -0.49902
4 0.80875 -0.38029
5 0.9081 -0.11862
6 0.94325 0.057542
7 0.93269 -0.0018478
8 0.93303 -6.3021e-06
9 0.93303 -3.1235e-10
32MATLAB function for Mullers Method