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many with fixed points at : Example: ≔ − ≔ +3 … If has fixed point at , then = − ( ) has a zero at 2 . Why study fixed-point iteration? 3 1. Sometimes easier to analyze 2. Analyzing fixed-point problem can help us find good root-finding methods

Example 1 (From BFB p60) Consider the equation f(x) = x3 +4x2 10. By the Interme-diate Value Theorem there exists a root in the interval [1;2]. There are many ways to change the equation f(x) = 0 to a xed point iteration of the form x= g(x). Here are 5 such examples (which you can verify are equivalent to f(x) = 0 yourself with algebra): a) x= g

Algorithm - Fixed Point Iteration Scheme Given an equation f(x) = 0 Convert f(x) = 0 into the form x = g(x) Let the initial guess be x0 Do xi+1= g(xi) while (none of the convergence criterion C1 or C2 is met) C1. Fixing apriori the total number of iterations N. C2. |(whereiis the iteration number) less than some tolerance

Iteration, Fixed points Paul Seidel 18.01 Lecture Notes, Fall 2011 Take a function f(x). De nition. A xed point is a point x such that f(x) = x : Graphically, these are exactly those points where the graph of f, whose equation is y = f(x), crosses the diagonal, whose equation is y = x. You can often solve for them exactly: Example.

• A number is a fixed point for a given function if = • Root finding =0 is related to fixed-point iteration = –Given a root-finding problem =0, there are many with fixed points at : Example: ≔ − ≔ +3 … If has fixed point at , then = − ( ) has

• A first simple and useful example is the Babylonian method for computing the square root of a>0, which consists in taking , i.e. the mean value of x and a/x, to approach the limit (from whatever starting point ). This is a special case of Newton's method quoted below.• The fixed-point iteration converges to the unique fixed point of the function fo…

The resulting iteration method may or may not converge, though. Page 2. Example. We begin with an example. Consider solving the two equations.

to a xed point iteration x n+1 = g(x n) by doing algebra on f(x) = 0. Newton’s method is also a xed point iteration of the form x n+1 = g(x n), where g(x n) = x n f(xn) f0(xn). But we didn’t get this xed point iteration by algebra like the 5 in the example, we got it …

FIXED POINT ITERATION The idea of the xed point iteration methods is to rst reformulate a equation to an equivalent xed point problem: f(x) = 0 x = g(x) and then to use the iteration: with an initial guess x 0 chosen, compute a sequence x n+1 = g(x n); n 0 in the hope that x n! . There are in nite many ways to introduce an equivalent xed point

The rationale behind that definition is the fact that this sequence will become stationary after some index k, if (and only if) x(k) is a fixed point of Φ.

Fixed point Iteration:The transcendental equation f(x) = 0can be converted algebraically into the form x = g(x)and then using the iterative scheme with the recursive relation xi+1= g(xi), i = 0, 1, 2, . . ., with some initial guessx0 is called the fixed point iterative scheme. Algorithm - Fixed Point Iteration Scheme

However, it is important to ensure that the conversion yields a function g for which fixed-point iteration will converge. Example We use fixed-point iteration ...

Example. To determine the xed points of the function f(x) = x3, we solve x3 = x)x3 x = 0)x(x2 1) = 0 so the xed points are x = 0 and x = 1, x = +1. Example. The function f(x) = x2 + x+ 1 has no xed points. Example. The function f(x) = x 2 + 1 x has a xed point at x = p 2. Example. The function f(x) = cos(x) has a xed point, as one can see by looking at the graph.

Examples[edit] ; x n + 1 = cos ⁡ x n ; f ( x ) = cos ⁡ x ; x 0 . ; | x n − x | ≤ q n 1 − q | x 1 − x 0 | = C q n ; q = 0.85 ...

Sample Problem. Functional (Fixed-Point) Iteration. Fixed-Point Algorithm. To find the fixed point of g in an interval [a,b], given the equation.

1. Algorithm & Example-1 f(x)= ...

Examples Example 1. Consider the equation x = 1 + 0:5 sinx: Here g(x) = 1 + 0:5 sinx: Note that 0:5 g(x) 1:5 for any x 2R. Also, g(x) is a continuous function. Applying the existence lemma, we conclude that the equation x = 1 + 0:5 sinx has a solution in [a;b] with a 0:5 and b 1:5. Example 2. Similarly, the equation x = 3 + 2 sinx

Example 1 (From BFB p60) Consider the equation f(x) = x3 +4x2 −10. By the Interme- diate Value Theorem there exists a root in the interval [1,2]. There are ...