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Math 175: Numerical Analysis II. Lecturer: Jomar Fajardo Rabajante IMSP, UPLB 2 nd Sem AY 2012-2013. SOLVING AND SOLVING AGAIN…. . Solutions of Linear Equations [high-school/ Math 17] Solutions of Nonlinear Equations (finding roots of nonlinear equations) [Math 17/Math 175]
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Math 175: Numerical Analysis II Lecturer: Jomar Fajardo Rabajante IMSP, UPLB 2ndSem AY 2012-2013
SOLVING AND SOLVING AGAIN… • Solutions of Linear Equations [high-school/ Math 17] • Solutions of Nonlinear Equations (finding roots of nonlinear equations) [Math 17/Math 175] • Solutions of Linear Systems (Systems of Linear Equations) [high school/Math 17/Math 120/ then later in Math 175] • Solutions of Nonlinear Systems (Systems of Nonlinear Equations) [Math 17/Math175]
SOLVING AND SOLVING AGAIN… • We will DELAY the discussion of Numerical Solutions to LINEAR SYSTEMS. • Your LABORATORY instructors will discuss Numerical Solutions to NONLINEAR SYSTEMS, specifically applying Multivariate Newton’s Method (similar to our Newton-Raphson but here we will use the concept of Jacobian).
NOW, • We will discuss an optional topic which is NUMERICAL OPTIMIZATION (unconstrained) • We will consider two methods: • Golden-section Search (univariate) (lecture class) • Newton’s Method (laboratory class) • Univariate (Math 36) • Multivariate (Math 38) Of course Exhaustive Search is still applicable but not encouraged. Another famous method is the Steepest Descent.
GOLDEN SECTION SEARCH • We will only consider MINIMIZATION. Why? • Similar to bisection method. It uses a bracket. It is also linearly but globally convergent. • The function should be UNIMODAL on the interval in consideration (i.e. there is only ONE dip ).
GOLDEN SECTION SEARCH It uses the golden ratio or the divine proportion φ: Theorem: After k steps of Golden Section Search with starting interval [a,b], the midpoint of the final interval is within of the minimum, where The theorem says that (b-a) is cut approx (38.2%)^k times, then after that get the midpoint.
GOLDEN SECTION SEARCH Algorithm: Given funimodal with minimum in [a,b] while (0.5*(g^k)*(b-a)>tol) if f(a+(1-g)*(b-a))<f(a+g*(b-a)) b=a+g*(b-a) else a=a+(1-g)*(b-a) end end approxmin=(a+b)/2