Finding Real roots of the Algebraic and Transcendental equations. Bisection Method

Jun 18th, 2015
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Analytically, we can usually choose any point in an interval where a change of sign takes place. However, this is subject to certain conditions that vary from method to method. And f(a) and f(b) have opposite signs. In this case a and b are said to bracket a root. the f must have at least one root in the interval (a, b). At each step the method divides the interval in two by computing the midpoint X1 = (a+b) / 2 of the interval and the value of the function f(X1) at that point. Unless X1 is itself a root (which is very unlikely, but possible) there are now two possibilities: either f(a) and f(X1) have opposite signs and bracket a root, or f(X1) and f(b) have opposite signs and bracket a root. The method selects the subinterval that is a bracket as a new interval to be used in the next step.

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Unit 2 Finding Real roots of the Algebraic and Transcendental equations.Bisection MethodInitial ApproximationAnalytically, we can usually choose any point in an interval where a change of sign takes place. However,this is subject to certain conditions that vary from method to method.And f(a) and f(b) have opposite signs. In this case a and b are said to bracket a root. the f must have atleast one root in the interval (a, b).At each step the method divides the interval in two by computing the midpoint X1 = (a+b) / 2 of theinterval and the value of the function f(X1) at that point.Unless X1 is itself a root (which is very unlikely, but possible) there are now two possibilities:either f(a) and f(X1) have opposite signs and bracket a root, or f(X1) and f(b) have opposite signs andbracket a root.The method selects the subinterval that is a bracket as a new interval to be used in the next step.In this way the interval that contains a zero of f is reduced in width by 50% at each step. The process iscontinued until the interval is sufficiently small.Explicitly, if f(a) and f(X1) are opposite signs, then the method sets X1as the new value for b, and if f(b)and f(X1) are opposite signs then the method sets X1 as the new a.(If f(X1)=0 then X1may be taken as the solution and the process stops.)In both cases, the new f(a) and f(b) have opposite signs, so the method is applicable to this smallerintervalX3 9x + 1 = 0Regula Falsi MethodLike the bisection method,

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