Lecture 3II Root Finding

數值方法 2008, Applied Mathematics NDHU 1

An iterative approach for root finding Newton method

Lecture 3II Root Finding


Problem statement

• Find x such that f(x)=0 for a given f• f denotes an arbitrary single-variable function

Root finding


1102)(2

xxf x >> x=linspace(-1,3);>> plot(x,2.^x.^2-10*x+1)


Root finding

Input: an expressionPlot given functionFind a root


Tangent line

xxn

y=f(x)

yn=f(xn)

xn+1

)(')(

)('

)('

1

1

n

nn

n

nnn

nn

nn

xfxfx

xfyxx

xxyxfslope

Taylor series


)('/)()('/)()(0))((')(

)( ))((')()(

at

nnn

nnn

nnn

nnn

n

xfxfxxxfxfxx

xxxfxfxfzero toSet

xxxfxfxfxxExpand


An iterative approach

Start at some guessRefine current guess by

Find a tangent line that passes (xn,f(xn)) Set xn to a point at intersection of the

tangent line to horizontal axis


Updating rule

)(')(

1n

nnn xf

xfxx


Newton method

An Iterative approach

Random guess Refine current guess by an updating rule If a halting condition holds, go to step 2

otherwise exit



n=0 ; Set xn

Determine xn+1

Increase n by one

Halting condition

T

F



Set x randomly

Apply an updating rule

to refine x

~( halting condition)exit


Halting condition

Let x denote the current guessThe absolute value of f(x) is expected as

close as possible to zeroHalting Condition

abs(f(x)) < epslonepslon denotes a predefined small positive

number


Initialization

Random initialization A guess by the user



x = rand*2-1; s=sym(ss); ep=10^-6f=inline(s); s1=diff(s); f1=inline(s1)

x=x-f(x)/f1(x);fprintf('%f\n' ,x)

~( abs(f(x)) < ep)exit

input an expression, ss


Demo_newton

source code demo_newtonfstr=input('input a function:','s');x_ini=input('guess its zero:');x_zero=newton(fstr,x_ini);

newton.m


Main Program


Newton method


>> demo_newtoninput a function:cos(x)guess its zero:2 iter=1 x=1.542342 fx=0.028450 iter=2 x=1.570804 fx=-0.000008 iter=3 x=1.570796 fx=0.000000>>


Example

demo_newtoninput a function:2.^x.^2-10*x+1guess its zero:1.5


Example

>> demo_newtoninput a function:2.^x.^2-10*x+1guess its zero:0.5

Second order expansion


2

22

2

2

!2)('')(')(

),(')('',!2

)(''2

4 ,0

0)(!2

)(''))((')(

)(

)(!2

)(''))((')()(

at

nn

nnn

nnn

nn

nnn

nn

nnn

n

xxfxfxxfc

xfxfbxfa

aacbbxcbxax

xxxfxxxfxf

xfzero toSet

xxxfxxxfxfxf

xxExpand

Second order method

Update rule


2

2

1

!2)('')(')(

),(')('',!2

)(''2

4

nn

nnn

nnn

n

xxfxfxxfc

xfxfbxfa

aacbbx


An iterative approachx = rand*2-1; s=sym(ss); ep=10^-6f=inline(s); s1=diff(s); f1=inline(s1)

s2=diff(s1); f2=inline(s2)

a=f2(x)/2;b=-f2(x)+f1(x);c=f(x)-x*f1(x)+f2(x)*x^2/2;

x=…fprintf('%f\n' ,x)



2

2

1

!2)('')(')(

),(')('',!2

)(''2

4

nn

nnn

nnn

n

xxfxfxxfc

xfxfbxfa

aacbbx


Unconstrained Optimization

Problem statement Given a differentiable function, y=g(x),

unconstrained optimization aims to find the minimum of g(x)

Let x denote a minimum and dxxdgxf )()(

0 that suchx Then

f(x)find


Optimization by Newton method

Use symbolic differentiation to find the first derivative of a given function

Apply the Newton method to find zeros of the first derivative


An iterative approachx = rand*2-1; s=sym(ss); ep=10^-6g=inline(s); s1=diff(s); f=inline(s1);

s2=diff(s1); f1=inline(s2)

x=x-f(x)/f1(x);fprintf('%f gx=%f\n' ,x, g(x))




First derivative ss=input('input a function:','s');s=sym(ss);f=inline(s);s1=diff(s);f1=inline(s1);x=linspace(-5,5);plot(x,f(x));hold on;plot(x,f1(x));


>> x=linspace(-2*pi,2*pi);>> plot(x,(x-tanh(2*x+10)).^2)


>> demo_mininput a function:(x-tanh(2*x+10)).^2guess its minimum:4

Local minimum


>> demo_mininput a function:(x-tanh(2*x+10)).^2guess its minimum:-4

Local Minimum


Global minimum


>> demo_mininput a function:0.2*x.^3-3*x+cos(x)guess its minimum:2


Exercise

ex3II.pdf

Documents

Lecture 3II Root Finding