Nov 10, 2020 CO 372: Assignment 3 DUE: Nov 19, 10PM (Wat. Time) Assignment Guidelines: The current TA, the current course instructor, the course notes, and the lectures, are your sources for help. THIS ASSIGNMENT IS TO BE DONE ALONE, WITHOUT DISCUSSION WITH OTHER STUDENTS. 1. Let ( ) : n nF x be a continuously differentiable mapping and suppose *x is a local minimizer to the problem: 21 2 2 min { ( )} where ( ) ( ) .x f x f x F x True or false: Either *( ) 0F x or the rows of *( )J x are linearly dependent. Justify your answer. [6 pts] 2. Assume 1: nf is twice continuously differentiable and assume we are using a descent direction algorithm . a. Show that if ( ) 0kf x and the direction kd is obtained by solving ( )k k kB d f x where n n kB is SPD, then kd is a descent direction for f at kx [ 3 pts] b. Show that if kB is symmetric, but not positive definite, then kd may not be a descent direction even if ( )k k kB d f x . [3 pts] 3. a. True or False: If kx is a sequence of n-vectors converging to a point *x at a quadratic rate then kx converges to *x at a linear rate with a (positive) linear contraction factor 1 2 L . Explain. [3 pts] b. Let : nf be twice continuously differentiable. True or False: The Newton process for minimization generates descent directions. That is, if 2 ( ) ( )k kf x s f x , and ( ) 0,kf x then s is (necessarily) a descent direction for f at kx . Explain. [5 pts] 4. Let 1 2 ( ) , : 0 .T Tq x g x x Hx x x Suppose satisfies 0 .x x Further, let d be a vector such that ( ) 0, 0T Tq x d d Hd . What’s a formula for computing * argmin{ ( ) : 0, }q x d x d ? [10 pts] Nov 10, 2020 CO 372: Assignment 3 DUE: Nov 19, 10PM (Wat. Time) 5. Function msine2 is available to you on LEARN. function [f,g,h] = msine2(x,varargin) %MSINE2. continuously diff function joining a sine fcn with a Gaussian %(joined at x = -12). % % INPUT % x: the single variable % varargin:a cell array.Problem variables(just 2 cells used in msine2) %OUTPUT % f: the function value at x % g: the derivative value at x % h: the 2nd derivative value at x NOTE 1: In this problem assign problem parameters: varargin{1}=1, varargin{2}=.95 NOTE 2: Output parameter h is not required by either your bisection method (part a.), or Newton method (part b.). a. Implement your own bisection method and use it to search for a zero of f on [a,b] – i.e., determine a point *x where 10. *| ( ) | 10f x . For each problem record your final values * *, ( )x f x , the number of iterations required to get to *x , and a flag indicating one of three things: a zero was found, there is no zero in [a,b], no zero found but there may or may not be a zero in [a,b]. Record these results in a very NEAT and CLEAR table. Try with [ , ] [ 18,20], [ 18,13], [ 18,2]a b . Observations on your results? [5 pts] b. Implement your own pure Newton method (without safeguards) and attempt to solve this problem starting from the 4 starting points : 10, 8, 12, 17.378925x Be sure to restrict the total number of iterations to less than 50, and also terminate if 10 10 * *| ( ) | 10 or | ( ) | 10f x f x .Record the final value of the function, its derivative, the number of iterations, and a flag indicating if a zero was successfully located or not. Observations? [7 pts] c. There are actually 5 distinct zeros of this function on [-18,20] . Using either or both of your zero finders developed above, perhaps in coordination with a 1-D minimizer such as the MATLAB function FMINBND, or another, locate as many zeros as you can .List for each found zero * ,x the computed function value * *( ) along with f x x and succinctly describe (in English) your strategy (combining your bisection function, your Newton function, and possibly a minimizer) . [8 pts] Please include your documented MATLAB code with your answer. [Total: 50 points]
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