SOLUTION OF NONLINEAR EQUATIONS

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SOLUTION OF NONLINEAR EQUATIONS

• Few examples of nonlinear equations follow

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• The primary reason why we solve nonlinear
  equations by using computer method is that
  nonlinear equations have no closed-form solution
  except for very few problem.
• Many researcher had been found the analytic
  solution of polynomial equations for fourth
  order, but there are no closed-form solutions for
  higher order.
• Roots of the nonlinear equations are found by
  computer methods based on iterative procedures.

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METHODS TO FIND ROOTS(we discuss only one root)

• Bisection Method
• False Position Method
• Fixed Point Method
• Newtons Method
• Secant Method
• Note
• The first two methods require a preliminary
  effort to estimate an appropriate interval that
  contains the desired root.
• The last three methods need an initial guess to
  find the root.

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Bisection Algorithm

• Select interval such that
• Compute
• If then else
• When the interval is within a given tolerance or
  the n-th iteration is reached then stopped
  iteration, else repeat step 2

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Application

• Consider the equation
• Find an approximation value of the root within a
  tolerance of
• Repeat problem above with ten iteration
• Answer
• The problem is solved with matlab program

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Error Analysis

• The interval size after n iteration steps becomes
• This also represents the maximum error bound
• Hence, the number of iteration steps required for
  the given error tolerance by is the smallest
  integer satisfying
• or equivalently

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Scheme Analysis

• Advantage
• The bisection method is the simplest, safest, and
  most robust scheme for finding one root in a
  given interval
• Disadvantage
• Slow to converge for a large interval
• Cannot find a pair of double roots
• Does not recognize the difference between root
  and singularity

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False Position Algorithm

• Select interval such that
• Compute
• If then else
• When the interval is within a given tolerance or
  the n-th iteration is reached then stopped
  iteration, else repeat step 2

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Application

• Consider the equation
• Find an approximation value of the root within a
  tolerance of
• Repeat problem above with ten iteration
• Answer
• The problem is solved with matlab program

12
Error Analysis

• The interval size after n iteration steps becomes
• This also represents the maximum error bound
• Hence, the number of iteration steps required for
  the given error tolerance by is the smallest
  integer satisfying
• or equivalently

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Scheme Analysis

• Advantage
• The bisection method is the simplest, safest, and
  most robust scheme for finding one root in a
  given interval
• Disadvantage
• Slow to converge for a large interval
• Cannot find a pair of double roots
• Does not recognize the difference between root
  and singularity
• example