Ch6 Roots Eqn

1

Finding Roots of Equations

“Numerical Methods with MATLAB”, Recktenwald, Chapter 6and

“Numerical Methods for Engineers”, Chapra and Canale, 5th Ed., Part Two, Chapters 5, 6 and 7 and

“Applied Numerical Methods with MATLAB”, Chapra, 2nd Ed., Part Two, Chapters 5 and 6

PGE 310: Formulation and Solution in Geosystems Engineering

Dr. Balhoff

1

Martini and Wine Glass

2 2 2R h a R

2 236capV h a h

2 2163

3 3capV h R h

Martini glass filled to 6cm depth contains ~ 216 /3 of wine. How deep must you fill wine glass to ensure that it contains the same amount of wine?

Volume of a spherical cap

Use Pythagorean Theorem2 22a Rh h

32 216

( ) 7 03 3

hf h h

Substitute into formula (R=7)

2

2

Peng-Robinson Equation of State

Ideal gas law: PVi=RT

Many gases not “ideal”: Reservoirs High temperatures

High Pressures

Peng-Robinson accounts for non-ideality

At P= 50 bar and T=473K….

2 22i i i

RT aP

V b V bV b

2 20.4572.3 6

0.077824.7

c

c

c

c

R Ta E

P

RTb

P

2

39325 2.3 6( ) 50 0

24.7 49.4 611ii i i

Ef V

V V V

Methane

3

Falling Parachutist

Newton’s Law:

Force Balance:

Plugging in gravity and drag:

Integrate:

What is “c” if: g = 9.8 m/s2, m = 68.1 kg, v =40 m/s, t = 10s

dvF ma m

dt

( / )( ) 1 c m tgmv t e

c

D U

dvF F m

dt

dvmg cv m

dt

040138.667

)( 146843.0 cec

cf

4

3

Roots of Nonlinear Equations: f(x) = 0

Find the root of the quadratic equation

Many problems aren’t so simple Martini and wine glass

Peng-Robinson Equation of State

Falling Parachutist

Use a “numerical” method to solve

Approximate technique

a

acbbxcbxaxxf

2

4;0)(

22

no “analytical” solution!!!

5

Root Finding Techniques

Bracketing Methods Graphical

Bisection

False Position

Open Methods Fixed Point Iteration

Newton-Raphson

Secant Method

Polynomials Muller’s Method

Bairstows Method

32 216

( ) 7 03 3

hf h h

2

39325 2.3 6( ) 50 0

24.7 49.4 611ii i i

Ef V

V V V

040138.667

)( 146843.0 cec

cf

6

4

Graphical Methods

• Find “c”, but how?

• Maybe I could just plot points?

c = 4; f(c) = 34.115c = 8; f(c) = 17.653 c = 12; f(c) = 6.067c = 16; f(c) = -2.269 c = 20; f(c) = -8.401

040138.667

)( 146843.0 cec

cf

Nonlinear Equation for drag coefficient

root

7

Bisection (Interval Halving)

We saw from the graph The root: 12 < c < 16

Why f(c) went from being positive to negative?

The value of “c”, that gave f(c)= 0 had to be between those numbers

Theorem: If f(x) is real & continuous in the interval xL to xU and f(xL) and f(xU) have opposite signs, then there is at least one root between xL to xU

In English: If “f(x)” goes from positive to negative, it must have been zero somewhere in the middle

0)()( UL xfxf Means positive number multiplied by negative #

8

5

f(c) > 0

f(c) < 09

1. Bracket the root between 12 and 16

2. Bisect in half; x=14?

3. Is root between 14 and 16? Yes.

4. Bisect in half; x=15

5. Is root between 15 and 16? No.

Bisection Method

10

6

Bisection Method: basic algorithm

1. Choose xL and xU so that root is bracketed: f(xL)*f(xU) < 0

2. Divide “x” in half xr = (xL + xU)/2

3. Calculate f(xr): evaluate function at new point

4. Determine if f(xr)* f(xU) < 0 OR if f(xr)* f(xL) < 0

5. Replace either xU or xL with xr

6. Repeat iterative strategy

11

False-Position Method

Bisection is “brute-force” and inefficient

No account is taken for magnitude of f(xU) and f(xL) If f(xU) is closer to zero than f(xL), xU is probably closer to the root

Replace the curve with a straight line to give a “false position”

• Line creates “similar triangles”

• Need a formula to find the x-intercept

• Sounds like a geometry problem

( ) ( )l u

r l r u

f x f x

x x x x

ul

uluur xfxf

xxxfxx

)(

7

Convergence

Bisection guaranteed to converge

“Brute force”

May converge slowly/oscillate

False position also guaranteed to converge Converges linearly and hopefully

quicker

There are pitfalls

ul

uluur xfxf

xxxfxx

)(

vs.

2lu

r

xxx

Pitfalls of False Position

While the false-position WILL converge to the root eventually…

Here is a case where it does very slowly

ul

uluur xfxf

xxxfxx

)(

14

8

Summary of Bracketing Methods

Bracketing methods guarantee convergence! Graphical can be difficult for accurate answers

Bisection is simple and straightforward Bounds the root

Bisects and creates new bounds

False Position takes advantage of f(x) magnitude Uses similar triangles to find new bounds

Converges linearly and usually more efficient

Next time….Open Methods Only require 1 guess

Usually faster, but convergence not guaranteed

15

Fixed Point Iteration

Re-arrange equation to get x = g(x)

4x2-x+3=0 x=4x2+3sin(x)=0 x=sin(x)+x

So, the solution will be the intersection point between these two curves: f1(x)=x and f2(x)=g(x)

Use an “iterative” method to solve for x1. Guess a value of xk

2. Calculate g(xk)3. Update xk+1=g(xk) 4. Iteratively update “x” until the solution

converges

May be several ways to rearrange equation Some arrangements converge faster Some arrangements may DIVERGE

9

Convergence of Fixed Point Iteration : x=g(x)

Fixed Point does not always converge

Converges b/w x = a & b ONLY IF

It means that:

'( ) 1g x (for all a < x< b)

Divergexgorxg 1)(1)(

Convergexgorxg 0)(11)(0

Why? Because of the slopes:g’(x) is the slope of f2(x)

1 is the slope of f1(x)

Look at the next slides

Fixed Point Iteration Convergence for Positive Slopes

f1(x)=x

f2(x)=g(x)

Root

x

y

1st try2nd try

1)(0 xg

1 is the slope of f1(x)g’(x) is the slope of f2(x)

Lower slope of f2(x) leads to convergence of iterations

CONVERGES !Monotonically

f1(x)=xf2(x)=g(x)

Root

x

y

1st try

2nd try

1)( xg

DIVERGES !

Higher slope of f2(x) leads to divergence of iterations

10

Fixed Point Iteration Convergence for Negative Slopes

f1(x)=x

f2(x)=g(x)

x

y

1st try

2nd try

0)(1 xg

CONVERGES !Oscillatory

f1(x)=x

x

y

1st try

2nd try

1)( xg

DIVERGES !

f2(x)=g(x)

Higher slope of f2(x) leads to divergence of iterations

Lower slope of f2(x) leads to convergence of iterations

1 is the slope of f1(x)g’(x) is the slope of f2(x)

Newton’s Method (or Newton-Raphson)

Most widely used method

Often converges much faster (quadratic convergence)

Idea: Use the line tangent to the curve to find the new root

Derive from a Taylor Series

1

( )

'( )k

k kk

f xx x

f x

20

11

Convergence of Newton-Raphson Method

Usually converges quadratically

Example:

Solved with 2 methods:

Newton-Raphson with x0=0

False-Position Method with xl=0 and xu=20

xexf x )( (true solution = 0.567143290409784)

Iterations  true error

x0 = 0 100.000000000%

x1 = 0.500000000000000 11.838858282%

x2 = 0.566311003197218 0.146750782%

x3 = 0.567143165034862 0.000022106%

x4 = 0.567143290409781 0.000000000%

Newton‐Raphson

Iterations  true error

x0 = 0.952380952 67.925984240%

x1 = 0.607944265065116 7.194121018%

x2 = 0.571658116501746 0.796064446%

x3 = 0.567645088312370 0.088478152%

x4 = 0.567199089558233 0.009838633%

False‐Position

Quadratic curve

Newton-Raphson Pitfalls

(a) Inflection point near root; f”(x)=0, causes divergence

(b) Oscillates around local minimum or maximum

(c) Near-zero slopes can jump to location several roots away

(d) Zero slope is true disaster!

• No general convergence criterion!

• Depends on nature of the function

• Best remedy is to have guess “sufficiently” close to root

• Some functions – no guess works!

22

12

The Secant Method

Derivative can be difficult or inconvenient to calculate for some functions

Approximate the derivative Remember the derivative is “slope of the line tangent to the curve”

Calculate the slope of a line that approximates the tangent line

1

1

( ) ( )' k k

kk k

f x f xf x

x x

• 2 points make a line (slope = rise/run)

1

( )

'( )k

k kk

f xx x

f x

11

1

( )( )

( ) ( )k k k

k kk k

f x x xx x

f x f x

• Substitute into Newton’s Formula

23

False Position versus Secant Method

False Position and Secant Method look like the same formula

What is the difference?

( )( )

( ) ( )U L U

r UL U

f x x xx x

f x f x

False Position:

replaces new guess with one of the bounds

Always “bracketed”

Secant Method:

replaces guesses in strict sequence

So xk+1 replaces xk and xk replaces xk-1

No guarantee of bracketing root

11

1

( )( )

( ) ( )k k k

k kk k

f x x xx x

f x f x

24

13

Convergence Comparison

Newton’s method converges fastest Quadratic convergence

sometimes (like many open methods) it may fail

Bracketing Methods slower But convergence is guaranteed

Bisection is the slowest of all

25

Modified Secant Method

Newton’s method is fast (quadratic convergence) but derivative may not be available

Secant method uses two points to approximate the derivative, but approximation may be poor if points are far apart.

Modified Secant method is a much better approximation because it uses one point, and the derivative is found by using another point some small distance, , away

1

( )

'( )k

k kk

f xx x

f x

( )' k k

k

f x f xf x

26

14

“Multiple Roots”

Several equations may have more than root 2nd order polynomials have 2 roots, right?

Start with an initial guess close to the root you want

But “multiple roots” refers to equations with the same root multiple times:

Problems arise with multiple roots Bracketing methods don’t work, b/c no sign change

f(x) and f’(x) both zero exactly at root

Newton’s method because linearly, not quadratically convergent….but an alternative formula is quadratic

3 2( ) 1 1 3 5 5 3f x x x x x x x

1

( ) '( )

' ( ) "( )i i

i ii i i

f x f xx x

f x f x f x

27

Matlab Built-In Functions

fzero uses a hybrid of several methodsr = fzero(fun,x0,options)

r = fzero(fun,x0)

roots function gives multiple roots of polynomial

Consider f(x) = 1x2 - 3x + 2 = 0

>> roots ([1 -3 2])

ans =

2

1

root function “.m” file

Your initial guess

28

15

Review

Need to find roots of nonlinear equations

Martini and wine glass

Peng-Robinson EOS

Parachute Problem

Learned a few different methods Closed – Convergence guaranteed by bracketing the root

Bisection

False Position

Open – No guaranteed Convergence, but much faster

Fixed Point Iteration

Newton’s

Secant and modified secant Method

32 216

( ) 7 03 3

hf h h

2

39325 2.3 6( ) 50 0

24.7 49.4 611ii i i

Ef V

V V V

040138.667

)( 146843.0 cec

cf

29