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Solutions of Equations in One Variable. +. -. Suppose f(x) is a continuous function of x within interval [a, b]. f(a) = - ive and f(b) = + ive. There exist at least a number p in [a, b] with f(p) = 0. Meaning, p is a root of the equation f(x) = 0. +. -. Bisection Method
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Solutions of Equations in One Variable
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Suppose f(x) is a continuous function of x within interval [a, b]. f(a) = - ive and f(b) = + ive There exist at least a number p in [a, b] with f(p) = 0. Meaning, p is a root of the equation f(x) = 0
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Bisection Method (Binary Search) The Bisection Method calls for a repeated halving of subintervals of [a, b] each time locating the half containing p.
Seta1 = aandb1 = b. Find the midpoint between a1 and b1. Midpoint,
If f(p1) = 0, then p1 is the root of the equation within [a, b]. If f(p1) ¹ 0, then what? Then find if f(p1) has the same sign as either f(a1) or f(b1).
IF f(p1) has the same sign as f(a1) , then the root is in [p1, b1]. Set a2 = p1 and b2 = b1. IF f(p1) has the same sign as f(b1) , then the root is in [a1, p1]. Set a2 = a1 and b2 = p1.
The rootis in the interval [a2, b2].Divide the interval in two halves and repeat the process.
has a root in [1, 2].
The Method of False Position The method is based on bracketing the root between two points. At the beginning choose two points, so that Now draw a line joining The x-intercept of the line is
Now bracket the root between either Which pair to choose? On the other hand Let us assume that This means that the root is between
and the process continues … Now draw a line joining The x-intercept of the line is
has a root in [1, 2].
has a root in [1, 2]. Fixed-Point Iteration Rewrite f(x) = 0 in the form of x = g(x) and iterate.
We can rewrite f(x) in the form of x = g(x) in the following ways.
Why some expressions failed to deliver the root? To deliver the root, g(x) for all x in [a, b] must stay within [a, b].
(p1, f(p1)) (p2, 0) (p1, 0) Consider the triangle (p2, 0), (p1, 0) and (p1, f(p1)).
(p1, f(p1)) (p2, 0) (p1, 0) The slope,