Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
§3.6 Newton’s Method. The student will learn about Newton’s method of approximating roots and tangent line approximations. 1 Introduction to Newton’s Method Sometimes we are presented with a problem which cannot be solved by simple algebraic means. For instance, if we needed to find the roots of the polynomial , x 3 x 1 0 we would find that the tried and true techniques just wouldn't work. However, we will see that calculus through Newton’s Method gives us a way of finding approximate solutions. 2 An Easier Example to Start Let’s start by computing the √5. This is of course easy with your calculator but stay with me for this. First we rewrite the problem as an equation f (x) = x 2 – 5 = 0 Newton’s method is an iterative method. That means that you must first pick an initial value for the solution and then the method will yield a better value. The method may be repeated as often as necessary to get the accuracy needed. What would be a good initial value for √5? OK we will use 2. 3 An Easier Example to Start Before we continue let’s look at the method. Consider the drawing. If x Is the root and x n is an approximation then x n + 1 is a better approximation. Using the tangent line slope yn dy y y n y n 1 f '(xn ) dx x xn xn 1 xn xn 1 And solving for x n + 1 yields x n1 x n f (x n ) f '(x n ) 4 Back to our Example x n1 x n f (x n ) f '(x n ) We were trying to find √5 using f (x) = x 2 – 5 = 0 and x n = x 0 = 2 and f ′ (x) = 2x. x n1 1 2.25 or x 1 x 0 and x 1 2 xn 4 f '(x 0 ) f '(x n ) f (x n ) f (x 0 ) Probably not too impressed! Let’s find x 2. x 2 x1 f (x 1 ) f '(x 1 ) and 0.0625 x 2 2.25 2.23611111... 4.5 With just two iterations we have accuracy to 0.000043134. FACT: Newton described this method in a book he wrote in 1669! 5 Example 2 x n1 x n f (x n ) f '(x n ) Approximate the solution to cos x = x in the interval [0, 2]. First we rewrite the problem as f (x) = cos x – x = 0 We will let x 0 = 1 (midpoint of the interval) and we know f ′(x) = - sin x - 1 cos(1) 1 0.5403023059 1 x1 x0 1 1 0.7503668679 f '(x 0 ) sin(1) 1 0.8414709848 1 f (x 0 ) If we were to repeat the process we would get x 2 = 0.7391128909 x 3 = 0.7390851334 - accuracy to 9 places! A bit tedious BUT if you know a little programming your calculator or computer can do this easily. 6 Example 3 Approximate f (x) 3 x n1 x n f (x n ) f '(x n ) x 0 using x 0 = 1 OK it’s a silly example (Do you know the solution?) but stay with me while I make a point. xn 1 xn x 1 3 n 2 3 n 1 x 3 xn 3 xn 2 xn The computation is easy with x 0 = 1, x 1 = - 2, x 2 = 4, x 3 = - 8, x 4 = 16, etc. So the method fails. But, it fails spectacularly! 7 Failure Newton's method makes no guarantee on convergence. Indeed, convergence depends on the starting point and on the shape of the function. 8 Your Calculator x n1 x n f (x n ) f '(x n ) Calculators basically only know how to add and multiply. So, how does it find a ? Let’s use Newton. f (x) x2 a 0 and then f '(x) 2x xn a 2 xn 1 xn 2x n 1 a xn 2 x n Notice that the operations involved in the iteration are addition and multiplication and you computer can do that! Use x 0 = 2 and approximate √5 with two iterations. 9 Your Calculator xn 1 xn x n2 a 2x n x n1 x n f (x n ) f '(x n ) 1 a xn 2 xn Use x 0 = 2 and approximate √5 with two iterations. 1 5 9 x 1 2 2.25 2 2 4 1 5 x 2 2.25 2.23611... 2.236067977. 2 2.25 10 Tangent Line Approximations From our definition of derivative we know that f(x h) f (x) f(x h) f (x) f '(x) lim h0 h h When h is small. If we multiply both sides of the above by h, we get dy h f '(x) f(x h) f (x) y Δy is the exact change in y dy = h · f ′ (x) is an approximation of Δy and is called the differential. 11 Tangent Line Approximations Summary dy h f '(x) f(x h) f (x) y Approximate change exact change Another useful form: f(x h) f (x) h f '(x) 12 Tangent Line Approximations f(x h) f (x) h f '(x) Let’s use this form for a practical problem. Approximate √5 using the differential above. 5 41 With x = 4, h = 1, f (x) x f(x h) f (x) h f '(x) 1 2 1 21 and f '(x) x 2 f(4 1) f (4) 1 f '(4) 1 1 4 1 4 1 2 2.25 4 2 4 Does this look familiar? 13 Summary. • We learned how to use Newton’s Method to solve equations. • We developed the approximation formula using the differential dy, f (x + h) – f (x) ≈ h · f ′ (x) = dy 14 ASSIGNMENT §3.6; Page 66; 1 - 9, odd. 15