finding roots of algebraic and transcendental equations
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Finding Roots of Algebraic and Transcendental
Equations
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Equations like this are called transcendental equations
Solutions to these equations are always obtained iteratively.Starting point is really important for obtaining the proper solution.Lot of insight can be obtained from geometry and pictures.
Euler-buckling load for a fixed-pinned beam
Y = a Cosh (x/c), equation for a catenary
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Example: Natural Freqencies of cantilever
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There are different ways of approaching a non-linear problem
m is called the order of convergence.Note that for most of these methods the upper and lower bound for the root [a,b] has to be given it is called as bracketingfor obvious reasons.
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If we want to find the solution to the equation f(x) = 0 re-write it in the form:
x = g(x).we then iterate according the following rule:
x(i+1) = g(x(i)), where x(i+i) is the value Obtained after ith iteration. For functions with certain properties we Will always get a convergence to a uunique point no matter what initial point we start with
We will demonstrate with a simple example.
Fixed Point Iteration method
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Error is obtainedas relative error
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Graphical Representation
Solution of x^2 + 0.2 = xStarting point:0.65
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Most basic method to bracket root: Iterative search method
It uses the property that when a function hits its Root, the sign changes From +ve to -ve or vice-versa.So there has to be at least one root within that interval.
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Problems with Iterative search besides being slow
Dx > 0.1: root will not be captured
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contd...
Double-roots at x = 1.0 will not be obtainableThe equation is (x-1)^2 = 0
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contd..
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Matlab code for incremental search
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Matlab codefor incrementalsearch
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Example
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Bisection Method
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Example
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Solution to the following equation when b = 4.0
Geometric approach to finding roots of equations
A good iterative scheme should find all roots in a given bracket irrespective of initial guesss. This is a very tough task because different equations have very unpredictible behavior. This is especially true for higher dimensional equations.
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Geometric and pictorial argument for solving the equation
re-written
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What if we had used
We would have happily walked away from the real solution because the flatness criteria isabsolutely not obeyed.
What if we had used
Computationally a bit cumbersome