Transcript FOUNDATIONS - Korea University

Graphics
2. Solving Equations
of One Variable
Korea University Computer Graphics Lab.
Lee Seung Ho / Shin Seung Ho
Roh Byeong Seok / Jeong So Hyeon
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Graphics Lab @ Korea University
Contents






KUCG
Bisection Method
Regula Falsi and Secant Method
Newton’s Method
Muller’s Method
Fixed-Point Iteration
Matlab’s Method
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Graphics
Bisection Method
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Bisection Method
KUCG
1
0.5
0
ab
m
2
-0.5
y  x 3
-1
2
-1.5
-2
1
1.1
1.2
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1.3
1.4
1.5
1.6
1.7
1.8
1.9
2
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Finding the Square Root of 3 Using
Bisection
How can we get
y  x 3
2
KUCG
3?
y 0: x  3
3
b1  a1
error is at most
2k
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Approximating the Floating Depth
for a Cork Ball by Bisection(1/2)
KUCG
Cork ball
Radius : 1
Density : 0.25
y  x  3x  1
3
2
if y  0,
we can get floatingdepth of cork ball by x [0,1]
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Approximating the Floating Depth
for a Cork Ball by Bisection(2/2)
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KUCG
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Discussion of Bisection Method
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KUCG
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Fixed-Point Iteration
KUCG
x  g (x)

Solution of equation

Convergence Theorem of fixed-point iteration
if  1  g ( x)  1 / 2,
 xk  x   xk  xk 1
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Fixed-Point Iteration to Find a Zero
of a Cubic Function
KUCG
f ( x)  5 x 3  10x  3  0,
 10x  5 x 3  3

x  0.5 x 3  0.3  g ( x)
 g ( x)  1.5 x 2
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Matlab’s Methods(1/2)

roots(p)


KUCG
p : vector
Example
3
2
p
(
x
)

x

7
x
 14x  7

EDU> r = roots(p); (p=[1 -7 14 -7])
r = 3.8019
2.445
0.75302
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Matlab’s Methods(2/2)

fzero( ‘function name’,x0 )



KUCG
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
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Graphics
Regular Falsi and
Secant Methods
2005. 3. 23
Byungseok Roh
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Regula Falsi Method
KUCG

The regula falsi method start with two point, (a, f(a)) and
(b,f(b)), satisfying the condition that f(a)f(b)<0.

The straight line through the two points (a, f(a)), (b, f(b)) is
y  f (a) 

f (b)  f (a )
( x  a)
ba
The next approximation to the zero is the value of x where the
straight line through the initial points crosses the x-axis.
x a
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ba
af (b)  bf (a)
f (a) 
f (b)  f (a)
f (b)  f (a)
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Regula Falsi Method (cont.)
KUCG

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.
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Example

KUCG
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 a=1 and b=2.
 Our first approximation to the zero is
ba
2 1
( f (b))  2 
(6)
f (b)  f (a)
6 1
 2  6 / 7  8 / 7  1.1429
x b

We then find the value of the function:
y  f ( x)  (8 / 7)3  2  0.5073

Since f(a) and y are both negative, but y
and f(b) have opposite signs
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Example (cont.)

Calculation of
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3
KUCG
2 using regula falsi.
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Secant Method
KUCG

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:
x1  x0
x2  x1 
y1
y1  y0

At the k-th stage, the new approximation
to the zero is
xk 1  xk 

xk  xk 1
yk
yk  yk 1
The secant method has converged with a tolerance of 104.
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Example

KUCG
Finding the Square Root of 3 by Secant Method

To find a numerical approximation to
y  f ( x)  x 2  3.
3 , we seek the zero of

Since f(1)=-2 and f(2)=1, we take as our starting bounds on the
zero x0  1 and x1  2 .
 Our first approximation to the zero is
x2  x1 

x1  x0
2 1
5
y1  2 
(1)   1.667
y1  y0
1  (2)
3
Calculation of
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3 using secant method.
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Graphics
NEWTON’S METHOD
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Newton’s Method
KUCG

Newton’s method uses straight-line approximation which is the
tangent to curve.

.
f ( x)  f ' ( x0 )(x  x0 )  f ( x0 )

Intersection point
f ( x0 )
x1  x0 
f ' ( x0 )
xk 1  xk 
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f ( xk )
f ' ( xk )
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Example
KUCG

Finding Square Root of ¾

3
4 approximate the zero of
f ( x)  4 x 2  3 using the fact
that f ' ( x)  8 x .
x1  x0 

f ( x0 )
2
 0 .5 
1
f ' ( x0 )
4
Continuing for one more step
f ( x1 )
1
x2  x1 
 1.0   7 / 8  0.875
f ' ( x1 )
8
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Finding Floating Depth for a
Wooden Ball

KUCG
Volume of submerged
segment of the Sphere
V  x(3r 2  x 2 ) / 6

To find depth at which the ball
float, volume of submerged
segment is  time.
x(3r 2 x 2 ) / 6   (4 / 3)

Simplifies to
x3  3x 2  4  0
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Finding Floating Depth for a
Wooden Ball (cont.)

KUCG
To find depth a ball, density is one-third of water float.
f ( x)  x3  3x 2  4 / 3
Calculation f(x) using Newton’s Method
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Oscillations in Newton Method

Ex)
KUCG
Newton’s method give Oscillatory result for some funtions & initial
estimates.
f ( x)  x3  3x 2  x  3
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Graphics
Muller’s Method
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Muller’s Method
KUCG

based on a quadratic approximation
 procedure
1.
2.
3.

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.
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Muller’s Method (Cont’)
KUCG
y  y3  c2 ( x  x3 )  d1 ( x  x3 )(x  x2 )
y3  y 2
c2  c1
y2  y1
c1 
c2 
d1 
x3  x1
x2  x1
x3  x2
x4  x3 
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2 y3
s  sign(s) s 2  4 y3d1
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Example

KUCG
Finding the sixth root of 2 using Muller’s method
6
y

f
(
x
)

x
 2, x1  0.5, x2  1.5 , x3  1.0

 y1  1.9844, y2  9.3906, y3  1
c1 
y2  y1
 11.375
x2  x1
c2 
y3  y2
 20.781
x3  x2
D
c2  c1
 18.812
x3  x1
s  c2  d1 ( x3  x2 )  20.781 18.812(0.5)  11.375
x4  x3 
 1
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2 y3
s  sign( s ) s 2  4 y3d1
2(1)
11.375 11.375  4(1)18.812
2
 1.0779
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Example (Cont’)
i
x
KUCG
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
Calculation of
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6
converge
2 using Muller’s method
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Another Challenging Problem

y  x10  0.5
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KUCG
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
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MATLAB function for Muller’s
Method

KUCG
P.65~66 code
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