Lecture 6 Single Variable Problems

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Lecture 6 - Single Variable Problems

• CVEN 302
• September 8, 2001

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Lectures Goals

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

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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.

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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).
• 0 f(xk) (xk1 - xk)f (xk)
• xk1 _at_ xk - f(xk) / f (xk)

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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

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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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Mullers Method

• Mullers method is an interpolation method that
  uses quadratic interpolation rather than linear.
  
• A second degree polynomial is used to fit three
  points in the vicinity of the root.

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Mullers Method

• Do while x2 - x1 gt tolerance value 1
• or f(x3) gt tolerance value 2
• c1 (f(x2)-f(x1))/(x2 - x1)
• c2 (f(x3)-f(x2))/(x3 - x2)
• d1 (c2-c1)/(x3 - x1)
• s c2 d1 (x3 - x2)
• x4 x3 - 2f(x3)/
• s sign(s)sqrt( s2
  -4f(x3)d1
• Set x1 x2
• Set x2 x3
• Set x3 x4
• END loop

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Mullers Method

• This method uses a Newton form of an
  interpolating polynomials.

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Example Problem

• f(x) x5 x3 4x2 - 3x - 2
• The roots are around (-2,-1),(-1,0) and (0,1)
• Look at (1,0) choice three points, 1.5, 0.5, 0.25

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Mullers Method

• x1 f(x1) x2 f(x2) x3
  f(x3) c1 c2 d1
  s1 x4
• 1.5000 13.4688 0.2500 -2.4834 0.5000 -2.3438
  12.7617 0.5586 12.2031 3.6094 0.8145
• 0.2500 -2.4834 0.5000 -2.3438 0.8145 -0.8900
  0.5586 4.6205 7.1938 6.8839 0.9300
• 0.5000 -2.3438 0.8145 -0.8900 0.9300 0.1698
  4.6205 9.1854 10.6157 10.4101 0.9134
• 0.8145 -0.8900 0.9300 0.1698 0.9134 -0.0049
  9.1854 10.5319 13.6309 10.3057 0.9139
• etc. etc. etc. etc. etc. etc.
  etc. etc. etc. etc. etc.

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Summary

• 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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Summary

• Mullers method -need to know the f(x) and 3
  values.

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Homework

• Check the Homework webpage