Lecture 5 Single Variable Problems

1
Lecture 5 - Single Variable Problems

• CVEN 302
• June 12, 2002

2
Lectures Goals

• Introduction of Solving Equations of 1 Variable
• Bisection Method
• Linear interpolation Secant Method
• Newtons Method
• Mullers Method
• Fixed Point Iteration

3
General Considerations

• Is the function to be evaluated often?

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

• How much precision is needed?

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

4
General Considerations

• 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.

5
General Considerations

• 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, Newtons
  method or secant method.

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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.
7
Basic 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.
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Root Finding Techniques

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

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Convergence 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.

10
Bisection 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.

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Bisection Method (Algorithm)

• Do while 0.5x1 - x2 gt 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

12
Bisection Method
DemoBisect the example program does a simple
bisection for a cubic polynomial equation.
13
Bisection Method

• Example Problem
• f(x) a5 x5 a4 x4 a3 x3 a2 x2 a1 x a0
  
• Let the function
• a0 -2, a1 -3, a2 4, a3 1, a4 0, a5
  1
• f(x) x5 x3 4x2 - 3x - 2

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Bisection Method

• There are 3 roots
• (a) -2 lt x lt -1
• (b) -1 lt x lt 0
• (c) 0.5 lt x lt1.5

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Bisection Method

• x1 f(x1) x2 f(x2) d
  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.

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Linear 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 )

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Method of Linear InterpolationRegula Falsi

• Do while x2 - x1 gt tolerance value 1
• or f(x3)gt
  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

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Regula Flasi or Linear 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.
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Linear Interpolation

• Same example problem
• f(x) x5 x3 4x2 - 3x - 2
• roots are between (-1.7,-1.3), (-1,0), (0.5,1.5)

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Regula 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.

21
Secant 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.

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Secant Method

• Do while x2 - x1 gt tolerance value 1
• or f(x3)gt
  tolerance value 2
• Set x3 x2 - f(x2)(x2 - x1)/(f(x2)-f(x1))
• Set x1 x2
• Set x2 x3
• END loop

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Comparison between Linear Interpolation and
Secant Method
The secant can be faster method, because it does
not have a large slope at the root.
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Problems 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))

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Problems 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)

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Secant Method

• 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.

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Secant Method

• x1 f(x1) x2 f(x2)
  x2-x1 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.

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Newtons 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(xk1) f(xk) (xk1 - xk)f (xk) ...

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Newtons Method

• The goal of the calculations is to find a f(x)0,
  so set f(xk1) 0 and rearrange the equation.
  f (xk) is the first derivative of f(x).

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Newton-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.

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Newtons Method

• Do while x2 - x1 gt tolerance value 1
• or f(x2)gt tolerance
  value 2
• or f(x1) 0
• Set x2 x1 - f(x1)/f(x1)
• Set x1 x2
• END loop

32
Newtons 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)

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Newtons 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.

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Summary

• 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.
• Newtons - need to know f(x) and f(x) and an
  initial guess.

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Homework

• Check the Homework webpage