Chapter 3Chapter 3

Roots of EquationsRoots of Equations

ObjectivesObjectives•Understanding what roots problems are and where they occur in engineering and science •Knowing how to determine a root graphically•Knowing how to solve a root problem with bracketing method•Understand the open method and recognize the difference between the open and bracketing methods•Knowing how to use MATLAB built-in function

Content•Introduction•Graphical method•Bracketing method (Bisection and false position)•Open method (Newton Raphson and secant methods)•MATLAB built-in function•Conclusion

IntroductionMany engineering and science problems

predict dependent variable(s) as a function of independent variables

Heat balance : time : time&space

Mass balance : mass concen. : time&space

Kirchoff’s law: current/voltage : time

Introduction (cont’d)See the previous assignment you encountered

1 km

What is the ground velocity ?

tmcec

gmtv )/(1)(

Says that u know the velocity

t=0, v = 0;

Introduction (cont’d)

tmcec

gmtv )/(1)(

from

Find the distant

dtec

gmvdtts

t

tmc 0

)/(1)(

1)( )/( tmce

c

mt

c

gmts

hence

Find t1km that s = 1000 m from

11000 1)/(

1kmtmc

km ec

mt

c

gm

Introduction (cont’d)

11000 1)/(

1kmtmc

km ec

mt

c

gm

So solve t1km from

Can u solve the above equation for t1km ?

Of course, the equation is nonlinear and t1km is an implicit variable !!

Introduction (cont’d)If we are solving a simple nonlinear equation, it’s piece of cake.

a

acbbxcbxax

2

40

22

?0sin

?02345

xxx

xfexdxcxbxax

But what about these

Yes, it’s time for numerical methods !!!

Introduction (cont’d)Nonlinear Equation

Solvers

Bracketing Graphical Open Methods

BisectionFalse Position (Regula-Falsi)

Newton Raphson

Secant

All Iterative

Graphical…Graphical method means that we plot graph fromthe target nonlinear equation and observe the zeroof it !!

Very simple but too iterative.

Here try with parachutist problem…..

Graphical… (cont’d)

11000 1)/(

1kmtmc

km ec

mt

c

gmStart with the distant equation

Find t1km such that

010001)( 1)/(11

kmtmc

kmkm ec

mt

c

gmtf

is satisfied.

Plot f(t1km) from t1km = {0,T} make sure that the zero of f(t1km) falls within this interval.

Graphical… (cont’d)Try with MATLAB to plot the function from t=0,..,50Utilize the command “inline”

0 5 10 15 20 25 30 35 40 45 50-1000

-500

0

500

1000

1500

2000

2500

3000

3500

time (s)

f(x) zero

Around 12.7, more accurate result can attained if u zoom in the figure !!

Bisection

Bracketing…Sometimes called two-point method

Underline principle

Root of f (x) is bounded by{xL,xU} where

f(xL)f(xU) < 0

Bisection (cont’d)How many iterations will it take?Length of the first Interval Δx0=b-aAfter 1 iteration εa(1)=Δx0/2

After 2 iterations εa(2) =Δx0/4

After k iterations εa(n) =Δx0/2n

0

2log ( ) a ss

xn

Bracketing:BisectionFor the arbitrary equation of one variable, f(x)=0

1. Pick xl and xu such that they bound the root of interest, check if f(xl).f(xu) <0.

2. Estimate the root by evaluating f[(xl+xu)/2].

3. Find the pair • If f(xl). f[(xl+xu)/2]<0, root lies in the lower interval, then

xu=(xl+xu)/2 and go to step 2.

Bisection (cont’d)• If f(xl). f[(xl+xu)/2]>0, root

lies in the upper interval, then xl= [(xl+xu)/2, go to step 2.

• If f(xl). f[(xl+xu)/2]=0, then root is (xl+xu)/2 and terminate.

4. Compare s with a

5. If a< s, stop. Otherwise repeat the process.

0

2na n

x

Bisection (cont’d)

Bisection (cont’d)

Bisection (cont’d)Evaluation of Method

ProsEasyAlways find rootNumber of iterations

required to attain an absolute error can be computed a priori.

ConsSlowKnow a and b that bound

rootMultiple rootsNo account is taken of

f(xl) and f(xu), if f(xl) is closer to zero, it is likely that root is closer to xl .

Bisection (cont’d)If the absolute magnitude of the error εs =

1e-4

and Lo=2, how many iterations will you have to do to get the required accuracy in the solution?

0

2na n

x

0

2log ( ) a ss

xn

2 4

2log ( ) 14.3

10 a sn

Try with simple example : f=sin(10x)+cos(3x) [x=3.7,3.9] es = 10-5

Bisection (cont’d)MATLAB: Try with the parachutist problem in Chapter 1

11000 1)/(

1kmtmc

km ec

mt

c

gm

Find t1km for m = 68.1 kg, c = 12.5 kg/s given that L0 = 60s and εs = 10-4

Plot approximation error and true error

False position If a real root is bounded

by xl and xu of f(x)=0, then we can approximate the solution by doing a linear interpolation between the points

[xl, f(xl)] and [xu, f(xu)] to find the xr value such that l(xr)=0, where l(x) is the linear approximation of f(x).

False position (cont’d)

1. Find a pair of values of x, xl and xu such that fl=f(xl) <0 and fu=f(xu) >0.

2. Estimate the value of the root from the following formula

and evaluate f(xr).

Procedure

lu

luulr ff

fxfxx

False position (cont’d)Procedure (cont’d)3. Use the new point to replace one of the original points,

keeping the two points on opposite sides of the x axis.

If f(xr)<0 then xl=xr == > fl=f(xr)

If f(xr)>0 then xu=xr == > fu=f(xr)

If f(xr)=0 then you have found the root and need go no further!

False position (cont’d)Procedure (cont’d)4. See if the new xl and xu are close enough for convergence

to be declared. If they are not go back to step 2.

Now u may wonder how good of this technique is?

Comparing with the bisection method, which one is better ?

I would like u try this method with parachutist problem.

False position… (cont’d) Why this method?

Faster Always converges for a single root.

Pitfalls of the False-Position Method.

Method applied to some curvature functions may slowly converge.

Open methodsOpen methods

are based on formulas that require only a single starting value of x or two starting values that do not necessarily bracket the root.

Fixed-point…(cont’d)

... 2, 1,k ,given )(

)(0)(

1

okk xxgx

xxgxf

•Bracketing methods are “convergent”.

•Fixed-point methods may sometime “diverge”, depending on the stating point (initial guess) and how the function behaves.

•Rearrange the function so that x is on the left side of the equation:

Fixed-point…(cont’d)Convergence

x=g(x) can be expressed as a pair of equations:

y1=x

y2=g(x) (component equations)

Plot them separately.

Fixed-point…(cont’d)Fixed-point iteration converges if

x)f(x) line theof (slope 1)( xg

•When the method converges, the error is roughly proportional to or less than the error of the previous step, therefore it is called “linearly convergent.”

Fixed-point…(cont’d)Example

xxg

or

xxg

or

xxg

xxxxf

21)(

2)(

2)(

02)(2

2

Rearrange x = g(x)

Newton-RaphsonMost widely used method.Based on Taylor series expansion:

)(

)(

)(0

g,Rearrangin

0)f(x when xof value theisroot The!2

)()()()(

1

1

1i1i

32

1

i

iii

iiii

iiii

xf

xfxx

xx)(xf)f(x

xOx

xfxxfxfxf

Newton-Raphson formula

Solve for

Newton-Raphson… (cont’d)A convenient method for

functions whose derivatives can be evaluated analytically. It may not be convenient for functions whose derivatives cannot be evaluated analytically.

Newton-Raphson… (cont’d)

Secant methodA slight variation of Newton’s method for functions

whose derivatives are difficult to evaluate. For these cases the derivative can be approximated by a backward finite divided difference.

,3,2,1)()(

)(

)()()(

1

11

1

1

ixfxf

xxxfxx

xfxf

xxxf

ii

iiiii

ii

iii

Secant method(cont’d)• Requires two initial

estimates of x , e.g, xo, x1. However, because f(x) is not required to change signs between estimates, it is not classified as a “bracketing” method.

• The scant method has the same properties as Newton’s method. Convergence is not guaranteed for all xo, f(x).

Secant method(cont’d)