Lecture 5 - Single Variable Problems CVEN 302 June 12, 2002.

Lecture 5 - Single Variable Lecture 5 - Single Variable ProblemsProblems

CVEN 302

June 12, 2002

Lecture’s GoalsLecture’s Goals

• Introduction of Solving Equations of 1 Variable– Bisection Method – Linear interpolation & Secant Method– Newton’s Method– Muller’s Method– Fixed Point Iteration

General ConsiderationsGeneral Considerations

• Is the function to be evaluated often?

• How much precision is needed?

– If so you may be able to create a custom algorithm for the problem.

– Engineering calculations are often required for a few significant figures. A simple root finding procedure may be adequate.


• How fast and robust must the method be?

– If the root finding is embedded in another program that automatically changes the parameter f(x), it is important that the root finder is robust. A robust procedure is relatively immune to initial guess and it converges quickly without being overly sensitive to the behavior of the function.


• Is the function a polynomial?

– There are special procedures for finding roots of polynomials. These should be used instead of general procedures such as bisection, Newton’s method or secant method.

Does the function have singularities?Does the function have singularities?

Some root finding procedure will converge to a singularity, as well as converge to a root. This must be guarded against.

Basic Root Finding TechniquesBasic Root Finding Techniques

The initial procedure is to find the roots of a function. The main concept of each technique is to bracket the root and do a series of iteration until the method converges on a solution.

Root Finding TechniquesRoot Finding Techniques

• Bracket the roots• Determine the method• Iterate to find the solution

Convergence CriteriaConvergence Criteria

• The algorithm must decide on how close to the root the guess should be before stopping.

• Two criteria can be applied in testing.

– Magnitude by which estimate of root changes.

– Magnitude by which the function will change.

Bisection MethodBisection Method

• The method is known as the Bolzano method and can be called interval halving technique.

• The method is simple and straight-forward.

• Given a bracketed root, the method repeatedly halves the interval while continuing to bracket the root and it will converge on the solution.

Bisection Method (Algorithm)Bisection Method (Algorithm)

Do while 0.5*|x1 - x2| >= tolerance value

Set x3 =(x1 + x2)/2

IF f(x3) of opposite sign of f(x1);

Set x2 = x3;

ELSE

Set x1 = x3;

ENDIF

END loop


DemoBisect: the example program does a simple bisection for a cubic polynomial equation.


• Example Problemf(x) = a5 x5 + a4 x4 + a3 x3 + a2 x2 + a1 x + a0

• Let the functiona0 = -2, a1 = -3, a2 = 4, a3 = 1, a4 = 0, a5 = 1

f(x) = x5 + x3 + 4x2 - 3x - 2

Bisection MethodBisection MethodFunctional Plot

-1200

-1000

-800

-600

-400

-200

0

200

400

600

800

-6 -4 -2 0 2 4

x value

f(x

)

Functional Plot

-6

-4

-2

0

2

4

6

8

-2 -1.5 -1 -0.5 0 0.5 1 1.5

x value

f(x

)

There are 3 roots

(a) -2 < x < -1

(b) -1 < x < 0

(c) 0.5 < x <1.5


x1 f(x1) x2 f(x2) x3 f(x3)

-2.0000 -20.000 -1.0000 3.00000 1.0000 -1.5000 0.53125

-2.0000 -20.000 -1.5000 0.53125 0.5000 -1.7500 -6.27246

-1.7500 -6.2725 -1.5000 0.53125 0.2500 -1.6250 -2.18448

-1.6250 -2.1848 -1.5000 0.53125 0.1250 -1.5625 -0.67480

-1.5625 -0.6748 -1.5000 0.53125 0.0625 -1.53125 -0.03612

etc. etc. etc. etc. etc. etc. etc.

Linear InterpolationLinear Interpolation

• The bisection method is not very efficient. We would like to converge to the root at a faster rate with a different algorithm.

• One of these method is linear interpolation method or the method of false position (Latinized version Regula Falsi )

Method of Linear InterpolationMethod of Linear InterpolationRegula FalsiRegula Falsi

Do while |x2 - x1| >= tolerance value 1

or |f(x3)|>= tolerance value 2

Set x3 = x2 - f(x2)*(x2 - x1)/(f(x2)-f(x1))

IF f(x3) of opposite sign of f(x1);

Set x2 = x3;

ELSE

Set x1 = x3;

ENDIF

END loop

Regula Flasi or Regula Flasi or Linear InterpolationLinear Interpolation

The program uses the slope of the two points to find the intersection. However, the upper bound is kept constant.

The program uses a similar triangle to estimate the location of the root.

Linear InterpolationLinear Interpolation

• Same example problem

f(x) = x5 + x3 + 4x2 - 3x - 2

• roots are between (-1.7,-1.3), (-1,0), & (0.5,1.5)

Regula Falsi MethodRegula Falsi Method

x1 f(x1) x2 f(x2) x2-x1 f(x2)-f(x1) x3 f(x3)

-1.7000 -4.4516 -1.3000 2.75007 0.4000 7.2016 -1.45275 1.2635

-1.7000 -4.4516 -1.4528 1.2635 0.2472 5.7143 -1.50743 0.40265

-1.7000 -4.4516 -1.5074 0.40265 0.1926 4.8547 -1.52339 0.11307

-1.7000 -4.4516 -1.5234 0.11293 0.1766 4.5645 -1.52777 0.03052

etc. etc. etc. etc. etc. etc. etc. etc.

Secant MethodSecant Method

• The algorithm is similar to the linear interpolation method but it oscillates from one side to the next.

• The method converges quickly with well behaved functions.


Do while |x2 - x1| >= tolerance value 1


Set x3 = x2 - f(x2)*(x2 - x1)/(f(x2)-f(x1))

Set x1 = x2;

Set x2 = x3;

END loop

Comparison between Linear Comparison between Linear Interpolation and Secant MethodInterpolation and Secant Method

The secant can be faster method, because it does not have a large slope at the root.

Problems with the Secant MethodProblems with the Secant Method

The convergence condition is :

x3 = x2 - f(x2)*

(x2-x1)/(f(x2)-f(x1))

It is tempting to rewrite the

convergence condition.

x3 = (f(x2)*x1-f(x1)* x2 )

/(f(x2)-f(x1))

Problems with the Secant MethodProblems with the Secant Method

This method can be

catastrophic if the points

are near a point where the

first derivative is zero.

Example: SecantProb

Try for the interval (1,3)


• The Secant method is the same as the linear interpolation method, but you do not do a comparison between +/- values and do the substitution.

• Problem if the points are on opposite ends of a peak or trough.

Secant Method x1 f(x1) x2 f(x2) x2x1 f(x2)-f(x1) x3 f(x3)

-1.7000 -4.4516 -1.3000 2.7501 0.4000 7.2016 -1.45275 1.2635

-1.4528 1.2635 -1.7000 -4.4516 -0.2472 -5.7143 -1.50743 0.40265

-1.5074 0.4031 -1.4528 1.2635 0.0546 0.8596 -1.53300 -0.07000

-1.5330 -0.0700 -1.5074 0.4031 0.0256 0.4731 -1.52921 -0.00297

etc. etc. etc. etc. etc. etc. etc. etc.

Newton’s MethodNewton’s Method

• This method is one of the most widely used methods to solve equations also known as the Newton-Raphson.

• The idea for the method comes from a Taylor series expansion, where you know the function and its’ first derivative.

f(xk+1) = f(xk) + (xk+1 - xk)*f ‘(xk) + ...

Newton’s MethodNewton’s Method• The goal of the calculations is to find a f(x)=0, so

set f(xk+1) = 0 and rearrange the equation. [ f ‘(xk) is the first derivative of f(x). ]

dx

xdfxf

xx

dx

xdfxxxf

k

kk1k

kk1kk0

Newton-Raphson MethodNewton-Raphson Method

The method uses the

slope of the line to

project to the x axis and

find the root. The

method converges on the

solution quickly.

Newton’s MethodNewton’s MethodDo while |x2 - x1| >= tolerance value 1


or f’(x1) ~= 0

Set x2 = x1 - f(x1)/f’(x1)

Set x1 = x2;

END loop

Newton’s MethodNewton’s Method

• Same example problem

f(x) = x5 + x3 + 4x2 - 3x - 2

and

f’(x) = 5x4 + 3x2 + 8x - 3

• roots are between (-1.7,-1.3), (-1,0), & (0.5,1.5)

Newton’s MethodNewton’s Method x1 f(x1) f’(x1) f(x1)/f’(x1) x2

-2.0000 -20.0000 73.0000 -0.27397 -1.7260

-1.7260 -5.3656 36.5037 -0.14699 -1.5790

-1.5790 -1.0423 22.9290 -0.04546 -1.5335

-1.5335 -0.0797 19.4275 -0.00410 -1.5294

-1.5294 -0.0005 19.1381 -0.00003 -1.5294

etc. etc. etc. etc. etc.

SummarySummary

• Bisection - need to know f(x) and two bounds.• Regula Flasi (linear interpolation) - need to know

the function and two bounds.• Secant - need to know f(x) and two bounds.• Newton’s - need to know f(x) and f’(x) and an

initial guess.

HomeworkHomework

• Check the Homework webpage

Lecture 5 - Single Variable Problems CVEN 302 June 12, 2002.

Documents

Transcript of Lecture 5 - Single Variable Problems CVEN 302 June 12, 2002.