Fixed-Point Iteration in Mathematics: Understanding the Derivative Mean-Value Theorem
Concept of fixed-point iteration in mathematics through examples and explanations, including the derivative mean-value theorem. Learn how to locate roots using iterative methods and understand the importance of fixed points in mathematical functions.
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Sec:6.1 Fixed- Point Iteration
1) Online Computing 2) MATLAB workshop 2) Exam-I and Exam-II Exam-I Date: Oct 19 Exam-II Date: Oct 23 Time: two hours between 5PM-10PM
Sec:6.1 Fixed- Point Iteration The number ? is a fixed point for a given function ? if. ? = ? ? Example: A fixed point for g occurs precisely when the graph of ? = ?(?) intersects the graph of ? = ? Determine any fixed points of the function ?(?) = ?2 2. solution ? = ?(?) ? = ?2 2 0 = ?2 ? 2 ? = 1,? = 2
Sec:6.1 Fixed- Point Iteration The derivative mean-value theorem states that if a function ?(?) and its first derivative are continuous over an interval ? ? ?, then there exists at least one value of ? = ? within the interval such that ? ? ? ? = ? ? (? ?) ?(?) ?(?) ? ? ? (? ) = The right-hand side of this equation is the slope of the line joining ?(?) and ?(?). Slope of the tangent line is ? (? ) Slope of this line is ?(?) ?(?) ? ? ?
Sec:6.1 Fixed- Point Iteration Example1 simple fixed-point iteration ?2 2? + 3 = 0 Step 1 can be simply manipulated to yield ? =?2+ 3 rearranging the function 2 ? = ?(?) ? (?) = ? so that x is on the left-hand side of the equation: ? = ?(?) (*) Example2 sin(?) = 0 can be simply manipulated to yield ? = ? + sin(?) ? = ?(?) Step 2 given an initial guess at the root ??, (*) can be used to compute a new estimate ??+1as expressed by the iterative formula Note Convert the problem from root- finding to finding fixed-point ??+?= ?(??)
Sec:6.1 Fixed- Point Iteration Step 1 Example1 Use simple fixed-point iteration to locate the root of rearranging the function ? = ? ? ? = ?(?) ? ? = ? ? ? ?? ? Step 2 1 0.0000000000000 2 1.0000000000000 3 0.3678794411714 4 0.6922006275553 5 0.5004735005636 6 0.6062435350856 7 0.5453957859750 8 0.5796123355034 9 0.5601154613611 10 0.5711431150802 11 0.5648793473910 12 0.5684287250291 ??+?= ?(??) ??+?= ? ?? Starting with an initial guess of x0 = 0 = ? ?= ? ??= ? ?? = ? ?= 0.367879 ??= ? ?? = ? ?.??????= 0.692201 ??= ? ?? Thus, each iteration brings the estimate closer to the true value of the root: 0.56714329
Sec:6.1 Fixed- Point Iteration ?? ?? ?? clear; clc; format long x(1) = 0; g = @(x) exp(-x); f = @(x) exp(-x) - x; true_root = 0.56714329; for k=1:11 x(k+1) = g( x(k) ); end e_t = 100 * abs( x - true_root ) ./ true_root; res = [[1:12]' x' e_t']; fprintf('%d %16.13f %16.13f\n', res'); ??= ???% ??(%) ?? ? 1 0.0000000000000 100.00 2 1.0000000000000 76.32 3 0.3678794411714 35.13 4 0.6922006275553 22.05 5 0.5004735005636 11.75 6 0.6062435350856 6.89 7 0.5453957859750 3.83 8 0.5796123355034 2.19 9 0.5601154613611 1.23 10 0.5711431150802 0.70 11 0.5648793473910 0.39 12 0.5684287250291 0.22 Notice that the true percent relative error for each iteration is roughly proportional (by a factor of about 0.5 to 0.6) to the error from the previous iteration. This property, called linear convergence
Convergence Sec:6.1 Fixed- Point Iteration Suppose that the true solution is ??,?+?= ? ? ??,? ??= ?(??) < ??,? ( If ? ? the iterative equation is < ? ) the errors decrease with each iteration ??+?= ?(??) Subtracting these equations yields ??,?+? < ??,? ?? ??+?= ? ?? ?(??) The derivative mean-value theorem gives ??,?? < ??,? < < ??,?< ??,? ?? ??+?= ? ? ( ?? ??) Step 1 where ? is somewhere between ??and ?? rearranging the function ? (?) = ? so that x is on the left-hand side of the equation: ? = ?(?) If the true error for iteration n is defined as ??,?= ?? ?? ??,?+?= ? ? ??,? Select ? ? ?? ???? ? ? ??,?+?= ? ? ??,? < ?
