09 chap-2a

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~ Roots of Equations ~

Bracketing Methods

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• Easy

• But, not easy

a

acbb xcbxax

2

40

22

? 0)3sin()10cos(? 0sin

? 02345

x x x x x x

x f exdxcxbxax

Roots of Equations

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Graphical Approach

• Make a plot of the function f(x)and observe where it crosses the

x-axis, i.e. f(x) = 0

• Not very practical but can be usedto obtain rough estimates for roots

• These estimates can be used as

initial guesses for numericalmethods that we’ll study here.

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Fig 5.2

Different cases:

Odd and evennumber of roots

Fig 5.3

exceptions

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Bisection Method

Termination criteria: e < Epsilon OR Max.Iteration is reached

%100 :estimateerror Relativenew

r

old

r

new

r

x

x x

e

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Fig 5.10:

Pseudocode toimplement the

Bisection Method

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Bisection Method

Pros

• Easy

• Always finds a root• Number of iterations

required to attain an

absolute error can be

computed a priori.

How?

Cons

• Slow

• Need to find initialguesses for xl and xu

• Multiple roots

• No account is taken

of the fact that if f(xl)is closer to zero, it is

likely that root is

closer to xl .

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How Many Iterations will It Take?

• Length of the first Interval Lo= xu-xl • After 1 iteration L1=Lo/2

• After 2 iterations L2=Lo/4

• After k iterations Lk =Lo/2k

• If the root (solution) is close to xl then we can write

k error relativedesired x L

error relativedesired

x

L

l

l

k

k

for solve ) _ _ (*2

_ _

0

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Fig 5.11

Minimize functionevaluations in the code.

Why?

Because they are costly

(takes more time)

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The False-Position Method(Regula-Falsi)

• If the root is between xland xu, then we canapproximate the solution

by doing a linear interpolation betweenthe 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).

• Derive xr

using similar triangles (Box 5.1):

Fig. 5.12

l u

l uul r

f f

f x f x x

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The False-Position Method

Good! but be careful Here is a pitfall

Interesting!

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Modified False-Position

One way to mitigate the “one-sided”

nature of the false position (i.e. the pitfall case) is to have the algorithm

detect when one of the bounds is

stuck.

If this occurs, the function value at

the stagnant bound can be divided in

half (original Bisection method).

The resultant code

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How to find good initial guesses?

• Start at one end of the region of interest (xa) and evaluate

f(xa), f(x

a+Dx), f(x

a+2Dx), f(x

a+3Dx), ........

• Continue until the sign of the result changes.

If that happens between f(xa+k*Dx) and f(xa+(k+1)*Dx)

then pick xl= xa+k*Dx and xu= xa+(k+1)*Dx

Problem:

if Dx is too small search is very time consuming

if Dx is too large could miss two closely spaced roots

even worse, if there is a multiple root .

partial solution: if the 1.derivative, f’(.), changes sign in an interval, it suggests a localminima/maxima and the vicinity should be examined more carefully for closely spaced roots.

Ultimate solution:

Know the application and plot the function to see the location of the roots

And pick x l and x u accordingly to start the iterations.

09 chap-2a

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