Newton s Method for Unconstrained Optimization

Transcription

1 Newton s Method for Unconstrained Optimization Robert M. Freund February, Massachusetts Institute of Technology.

2 Newton s Method Suppose we want to solve: (P:) min f (x) At x = x, f (x) can be approximated by: n x R. f (x) h(x) := f ( x)+ f ( x) T (x x)+ (x x) t H ( x)(x x), 2 which is the quadratic Taylor expansion of f (x) at x = x. Here f (x) is the gradient of f (x) and H (x) is the Hessian of f (x). Notice that h(x) is a quadratic function, which is minimized by solving h(x) =. Since the gradient of h(x) is: we therefore are motivated to solve: which yields h(x) = f ( x)+ H( x)(x x), f ( x)+ H ( x)(x x) =, x x = H( x) f ( x). The direction H ( x) f ( x) is called the Newton direction, or the Newton step at x = x. This leads to the following algorithm for solving (P): Step Given x,set k Newton s Method: Step d k = H(x k ) f (x k ). If d k =, then stop. Step 2 Choose step-size α k =. 2

3 Step 3 Set x k+ x k + α k d k, k k +. Go to Step. Note the following: The method assumes H(x k ) is nonsingular at each iteration. There is no guarantee that f(x k+ ) f(x k ). Step 2 could be augmented by a line-search of f(x k + αd k )tofind an optimal value of the step-size parameter α. Recall that we call a matrix SPD if it is symmetric and positive definite. Proposition. If H(x) is SPD and d := H(x) f(x), then d is a descent direction, i.e., f(x + αd) < f(x) for all sufficiently small values of α. Proof: It is sufficient to show that f(x) t d = f(x) t H(x) f(x) <. This will clearly be the case if H(x) is SPD. Since H(x) is SPD, if v =, < (H(x) v) t H(x)(H(x) v) = v t H(x) v, thereby showing that H(x) is SPD.. Rates of convergence A sequence of numbers {s i } exhibits linear convergence if lim i s i = s and s i+ s lim = δ <. i s i s If δ = in the above expression, the sequence exhibits superlinear convergence. A sequence of numbers {s i } exhibits quadratic convergence if lim i s i = s and s i+ s lim = δ <. i s i s 2 3

4 .. Examples of Rates of Convergence ( ) i Linear convergence: s i = :.,.,., etc. s =. s i+ s =.. s i s Superlinear convergence: s i = i! :, 2, 6, 24, 25,etc. s =. s i+ s i! = = as i. s i s (i +)! i + ( ) (2 i ) Quadratic convergence: s i = :.,.,.,., etc. s =. s i+ s ( 2i ) 2 = =. s i s 2 2i+.2 Quadratic Convergence of Newton s Method We have the following quadratic convergence theorem. In the theorem, we use the operator norm of a matrix M: M := max { Mx x =}. x Theorem. (Quadratic Convergence Theorem) Suppose f(x) is twice continuously differentiable and x is a point for which f(x )=. Suppose that H(x) satisfies the following conditions: there exists a scalar h > for which [H(x )] h there exists scalars β> and L> for which H(x) H(x ) L x x for all x satisfying x x β { } Let x satisfy x x <γ:= min β, 2h 3L,and let x N := x H(x) f(x). Then: ( ) L (i) x N x x x 2 2(h L x x ) (ii) x N x < x x <γ 4

5 ( ) (iii) x 2 3L N x x x 2h Example : Let f (x) =7x ln(x). Then f (x) = f (x) =7 x and H(x) = f (x) = 2. It is not hard to check that x = 7 = is x the unique global minimum. The Newton direction at x is ( ) 2 2 d = H (x) f ( x) f (x) = = x 7 = x 7x. f ( x) x k Newton s method will generate the sequence of iterates {x } satisfying: k+ k ) 2 x = x k +(x k 7(x k ) 2 )=2x k 7(x. Below are some examples of the sequences generated by this method for different starting points. k x k x k x k x k , By the way, the range of quadratic convergence for Newton s method for this function happens to be x (., ). 5

6 Example 2: f(x) = ln( x x 2 ) ln x ln x 2. f(x) = x x 2 x x x 2 x 2 ) 2 ( x, ( ) 2 ( ) 2 x x 2 + x x 2 H(x) =. ( ) 2 ( ) 2 ( ) 2 x x 2 x x 2 + x 2 ( ) 3, f(x )= x = 3, k x k x k 2 x k x e e e 6 Comments: The convergence rate is quadratic: x N x x x 2 3L 2h We typically never know β, h, or L. However, there are some amazing n exceptions, for example f(x) = ln(x j ), as we will soon see. j= The constants β, h, and L depend on the choice of norm used, but the method does not. This is a drawback in the concept. But we do not know β, h, or L anyway. x x x 2 In the limit we obtain x N 6 L 2h

7 We did not assume convexity, only that H(x ) is nonsingular and not badly behaved near x. One can view Newton s method as trying successively to solve f(x) = by successive linear approximations. Note from the statement of the convergence theorem that the iterates of Newton s method are equally attracted to local minima and local maxima. Indeed, the method is just trying to solve f(x) =. What if H(x k ) becomes increasingly singular (or not positive definite)? In this case, one way to fix this is to use H(x k )+ ɛi. () Comparison with steepest-descent. One can think of steepest-descent as ɛ + in () above. The work per iteration of Newton s method is O(n 3 ) So-called quasi-newton methods use approximations of H(x k )at each iteration in an attempt to do less work per iteration. 2 Proof of Theorem. The proof relies on the following two elementary facts. For the first fact, let v denote the usual Euclidian norm of a vector, namely v := v T v. The operator norm of a matrix M is defined as follows: M := max { Mx x =}. x Proposition 2. Suppose that M is a symmetric matrix. Then the following are equivalent:. h> satisfies M h 2. h> satisfies Mv h v for any vector v 7

8 You are asked to prove this as part of your homework for the class. Proposition 2.2 Suppose that f (x) is twice differentiable. Then f (z) f (x) = [H(x + t(z x))] (z x)dt. Proof: Let φ(t) := f (x+t(z x)). Then φ() = f (x) and φ() = f (z), and φ (t) =[H(x + t(z x))] (z x). From the fundamental theorem of calculus, we have: f (z) f (x) = φ() φ() = φ (t)dt ProofofTheorem.: We have: = [H(x + t(z x))] (z x)dt. x N x = x H(x) f (x) x = x x + H (x) ( f (x ) f (x)) = x x + H (x) [H (x + t(x x))] (x x)dt (from Proposition 2.2) = H(x) [H(x + t(x x)) H(x)] (x x)dt 8

9 Therefore x N x H(x) [H(x + t(x x)) H(x)] (x x) dt x x H (x) L t (x x) dt = x x 2 H(x) L tdt = x x 2 H(x) L 2 We now bound H(x).Let v be any vector. Then H(x)v = H (x )v +(H (x) H(x ))v H (x )v (H(x) H(x ))v h v H(x) H (x ) v (from Proposition 2.) h v L x x v = (h L x x ) v. Invoking Proposition 2. again, we see that this implies that Combining this with the above yields H(x). h L x x L x N x x x 2 2(h L x x ), which is (i) of the theorem. Because L x x < 3 we have: x N x x x 2h 3 2h L x x < ( 2(h L x x ) 2 h 2h 3 9 ) x x = x x,

10 which establishes (ii) of the theorem. Finally, we have L L 3L x N x x x 2 2(h L x x ) x x 2 ( ) 2 h = x 2h x 2 2h, which establishes (iii) of the theorem. 3 3 Newton s Method Exercises. (Newton s Method) Suppose we want to minimize the following function: f(x) =9x 4ln(x 7) over the domain X = {x x>7} using Newton s method. (a) Give an exact formula for the Newton iterate for a given value of x. (b) Using a calculator (or a computer, if you wish), compute five iterations of Newton s method starting at each of the following points, and record your answers: x =7.4 x =7.2 x =7. x =7.8 x =7.88 (c) Verify empirically that Newton s method will converge to the optimal solution for all starting values of x in the range (7, ). What behavior does Newton s method exhibit outside of this range? 2. (Newton s Method) Suppose we want to minimize the following function: f(x) =6x 4ln(x 2) 3 ln(25 x) over the domain X = {x 2 < x<25} using Newton s method. (a) Using a calculator (or a computer, if you wish), compute five iterations of Newton s method starting at each of the following points, and record your answers:

11 x = 2.6 x = 2.7 x = 2.4 x = 2.8 x = 3. (b) Verify empirically that Newton s method will converge to the optimal solution for all starting values of x in the range (2, 3.5). What behavior does Newton s method exhibit outside of this range? 3. (Newton s Method) Suppose that we seek to minimize the following function: f(x,x 2 )= 9x x 2 +θ( ln( x x 2 ) ln(x ) ln(x 2 ) ln(5 x +x 2 )), where θ is a given parameter, on the domain X = {(x,x 2 ) x >,x 2 >,x + x 2 <,x x 2 < 5}. This exercise asks you to implement Newton s method on this problem, first without a line- search, and then with a line-search. Run your algorithm for θ = and for θ =, using the following starting points. x =(8, 9) T. x =(, 4) T. x =(5, 68.69) T. x =(, 2) T. (a) When you run Newton s method without a line-search for this problem and with these starting points, what behavior do you observe? (b) When you run Newton s method with a line-search for this problem, what behavior do you observe? 4. (Projected Newton s Method) Prove Proposition 6. of the notes on Projected Steepest Descent. 5. (Newton s Method) In class we described Newton s method as a method for finding a point x for which f(x ) =. Now consider the following setting, where we have n nonlinear equations in n unknowns x = (x,...,x n ):

12 which we conveniently write as g (x) =... g n (x) =, g(x) =. Let J (x) denote the Jacobian matrix (of partial derivatives) of g(x). Then at x = x we have g( x + d) g( x)+ J ( x)d, this being the linear approximation of g(x) at x = x. Construct a version of Newton s method for solving the equation system g(x) =. 6. (Newton s Method) Suppose that f (x) is a strictly convex twice-continuously differentiable function, and consider Newton s method with a linex, we compute the Newton direction d = [H( x)] f ( x) search. Given and the next iterate x is chosen to satisfy: x := arg min f ( x + αd). α Prove that the iterates of this method converge to the unique global minimum of f (x). 7. Prove Proposition 2. of the notes on Newton s method. 8. Bertsekas, Exercise.4., page 99. 2