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2 2 -1 1 }{CSTYLE "_cstyle3" -1 206 "Courier" 1 14 255 0 0 1 2 1 2 2 1 2 0 0 0 1 }{PSTYLE "_pstyle4" -1 205 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 2 2 2 0 0 0 1 }0 0 0 -1 -1 -1 1 0 1 0 2 2 -1 1 }} {SECT 0 {EXCHG {PARA 202 "" 0 "" {TEXT 204 15 "NEWTON'S METHOD" } {TEXT 204 0 "" }}{PARA 203 "" 0 "" {TEXT 205 77 "Newton's Method is a \+ procedure for numerically approximating solutions of an " }{TEXT 205 0 "" }}{PARA 203 "" 0 "" {TEXT 205 85 "equation f(x)=0. Suppose that f is a differentiable function and r is a root of " }{TEXT 205 0 " " }}{PARA 203 "" 0 "" {TEXT 205 83 "the equation f(x)=0. The proble m is to obtain an approximation of r. Suppose, " }{TEXT 205 0 "" }} {PARA 203 "" 0 "" {TEXT 205 82 "that x_0 is a known approximation of r. The idea of Newton's Method is to use" }{TEXT 205 0 "" }}{PARA 203 "" 0 "" {TEXT 205 80 "the linear approximation of f(x) near x_0 to obtain the next estimate x_1. " }{TEXT 205 0 "" }}{PARA 203 "" 0 "" {TEXT 205 28 "The linear approximation is " }{TEXT 205 0 "" }} {PARA 203 "" 0 "" {TEXT 205 54 " y=f'(x_0)* (x-x_0)+f(x_0). " }{TEXT 205 0 "" }}{PARA 203 "" 0 "" {TEXT 205 76 "Th e next approximation x_1 is taken to be the x -intercept of the lin ear " }{TEXT 205 0 "" }}{PARA 203 "" 0 "" {TEXT 205 13 "approximation" }{TEXT 205 0 "" }}{PARA 203 "" 0 "" {TEXT 205 52 " \+ x_1 = x_0 - f(x_0)/f'(x_0)." }{TEXT 205 0 "" }}{PARA 203 "" 0 "" {TEXT 205 82 "If we wish, we can now use x_1 as an estimate, take t he linear approximation of " }{TEXT 205 0 "" }}{PARA 203 "" 0 "" {TEXT 205 46 "f near x_0 and find the next approximation " }{TEXT 205 0 "" }}{PARA 203 "" 0 "" {TEXT 205 54 " x_ 2 = x_1 - f(x_1)/f'(x_1). " }{TEXT 205 0 "" }}{PARA 203 "" 0 "" {TEXT 205 75 "This process can be carried on until it succeeds by maki ng each successive " }{TEXT 205 0 "" }}{PARA 203 "" 0 "" {TEXT 205 82 "approximation differ by an insignificant amount from the previous ap proximation, " }{TEXT 205 0 "" }}{PARA 203 "" 0 "" {TEXT 205 66 "or fa ils by diverging. What happens is if after n steps we have" }{TEXT 205 0 "" }}{PARA 203 "" 0 "" {TEXT 205 54 " x_(n+ 1) is approximately x_n, " }{TEXT 205 0 "" }}{PARA 203 "" 0 "" {TEXT 205 20 "then we would have " }{TEXT 205 0 "" }}{PARA 203 "" 0 " " {TEXT 205 63 " x_n- f(x_n)/f'(x_n) is approximat ely x_n." }{TEXT 205 0 "" }}{PARA 203 "" 0 "" {TEXT 205 41 "Subtracti ng x_ n from both sides gives " }{TEXT 205 0 "" }}{PARA 203 "" 0 "" {TEXT 205 57 " f(x_n)/f'(x_n) is approximately 0, \+ " }{TEXT 205 0 "" }}{PARA 203 "" 0 "" {TEXT 205 2 "or" }{TEXT 205 0 " " }}{PARA 203 "" 0 "" {TEXT 205 49 " f(x_n) is a appr oximately 0. " }{TEXT 205 0 "" }}{PARA 203 "" 0 "" {TEXT 205 68 "Thu s we might hope that x_n approximates a root of the equation. " } {TEXT 205 0 "" }}{PARA 203 "" 0 "" {TEXT 205 57 "In this section we wi ll look an some examples of Newton's" }{TEXT 205 0 "" }}{PARA 203 "" 0 "" {TEXT 205 71 "Method. Notice the procedure can actually be regard ed as applying the" }{TEXT 205 0 "" }}{PARA 203 "" 0 "" {TEXT 205 9 " function " }{TEXT 205 0 "" }}{PARA 203 "" 0 "" {TEXT 205 42 " \+ x -> x - f(x)/ f'(x) " }{TEXT 205 0 "" }}{PARA 203 "" 0 "" {TEXT 205 56 "to each approximation to obtain the next approximation. " }{TEXT 205 0 "" }}{PARA 204 "> " 0 "" {MPLTEXT 1 206 14 "with(stude nt):" }{MPLTEXT 1 206 0 "" }}{PARA 204 "> " 0 "" {MPLTEXT 1 206 0 "" } }}{EXCHG {PARA 203 "" 0 "" {TEXT 205 64 "As a first illustration lets \+ approximate the value of sqrt\{2\}. " }{TEXT 205 0 "" }}{PARA 203 "" 0 "" {TEXT 205 64 "In this case we take f(x)=x^2-2, since one of the \+ roots of this" }{TEXT 205 0 "" }}{PARA 203 "" 0 "" {TEXT 205 22 "equat ion is sqrt\{2\}. " }{TEXT 205 0 "" }}{PARA 203 "" 0 "" {TEXT 205 0 " " }}{PARA 204 "> " 0 "" {MPLTEXT 1 206 14 "f := x->x^2-2;" }{MPLTEXT 1 206 0 "" }}}{EXCHG {PARA 204 "> " 0 "" {MPLTEXT 1 206 30 "plot(f(x), x=-5..5, y=-10..10);" }{MPLTEXT 1 206 0 "" }}}{EXCHG {PARA 204 "> " 0 "" {MPLTEXT 1 206 19 "showtangent(f,x=1);" }{MPLTEXT 1 206 0 "" }}} {EXCHG {PARA 203 "" 0 "" {TEXT 205 56 "We now illustrate Newton's meth od with two examples. In" }{TEXT 205 0 "" }}{PARA 203 "" 0 "" {TEXT 205 68 "the first example Newton's method works, but one can see that there" }{TEXT 205 0 "" }}{PARA 203 "" 0 "" {TEXT 205 68 "are possible pitfalls. In the second example the method will always" }{TEXT 205 0 "" }}{PARA 203 "" 0 "" {TEXT 205 5 "fail." }{TEXT 205 0 "" }}{PARA 203 "" 0 "" {TEXT 205 64 "EXAMPLE 1.: Use Newton's Method to approxim ate the zeros of the" }{TEXT 205 0 "" }}{PARA 203 "" 0 "" {TEXT 205 8 "function" }{TEXT 205 0 "" }}{PARA 203 "" 0 "" {TEXT 205 54 " \+ f(x) = (x-2)^2- 1/x ." }{TEXT 205 0 "" }} {PARA 203 "" 0 "" {TEXT 205 74 "SOLUTION: The first thing that we wil l do is to define f(x) in Maple V " }{TEXT 205 0 "" }}{PARA 203 "" 0 "" {TEXT 205 19 "and plot its graph." }{TEXT 205 0 "" }}{PARA 203 "" 0 "" {TEXT 205 0 "" }}{PARA 204 "> " 0 "" {MPLTEXT 1 206 18 "f:=x->(x- 2)^2-1/x;" }{MPLTEXT 1 206 0 "" }}}{EXCHG {PARA 204 "> " 0 "" {MPLTEXT 1 206 29 "plot(f(x),x=-4..4,y=-10..10);" }{MPLTEXT 1 206 0 "" }}}{EXCHG {PARA 203 "" 0 "" {TEXT 205 79 "Examination of this plot su ggests that f(x) has three zeros near the points " }{TEXT 205 0 "" }}{PARA 203 "" 0 "" {TEXT 205 84 "x=0.4, x= 1.0, and x=2.6, respec tively. Lets use Newton's Method to approximate " }{TEXT 205 0 "" }} {PARA 203 "" 0 "" {TEXT 205 167 "the zero near x=2.6. We will start \+ with a not so good guess of x_0=4 so that we can illustrate the proc edure graphically for a few steps. Thus the first step is to" }{TEXT 205 0 "" }}{PARA 203 "" 0 "" {TEXT 205 64 "obtain the linear approxima tion of f(x) at x=4. This is the" }{TEXT 205 0 "" }}{PARA 203 "" 0 "" {TEXT 205 9 "function " }{TEXT 205 0 "" }}{PARA 203 "" 0 "" {TEXT 205 50 " L(f)(x)= f'(4)*(x-4)+f(4). " } {TEXT 205 0 "" }}{PARA 203 "" 0 "" {TEXT 205 80 "The next plot is a gr aph of f(x) and this linear approximation on the interval" }{TEXT 205 0 "" }}{PARA 203 "" 0 "" {TEXT 205 7 " [2,4]." }{TEXT 205 0 "" }} {PARA 203 "" 0 "" {TEXT 205 0 "" }}{PARA 204 "> " 0 "" {MPLTEXT 1 206 39 "plot(\{f(x),D(f)(4)*(x-4)+f(4)\},x=2..4);" }{MPLTEXT 1 206 0 "" }} }{EXCHG {PARA 203 "" 0 "" {TEXT 205 66 "The procedure now calls for us to obtain the x- intercept of the" }{TEXT 205 0 "" }}{PARA 203 "" 0 "" {TEXT 205 47 "linear approximation as the next approximation." } {TEXT 205 0 "" }}{PARA 203 "" 0 "" {TEXT 205 0 "" }}{PARA 204 "> " 0 " " {MPLTEXT 1 206 35 "x1 := fsolve(D(f)(4)*(x-4)+f(4),x);" }{MPLTEXT 1 206 0 "" }}}{EXCHG {PARA 203 "" 0 "" {TEXT 205 67 "Now we use the line ar approximation of f(x) at x=3.076923077 to" }{TEXT 205 0 "" }} {PARA 203 "" 0 "" {TEXT 205 65 "obtain the next approximation. See th e next figure for a plot of" }{TEXT 205 0 "" }}{PARA 203 "" 0 "" {TEXT 205 69 " f(x) with the linear approximations through x=4 and x =3.076923077." }{TEXT 205 0 "" }}{PARA 203 "" 0 "" {TEXT 205 0 "" }} {PARA 204 "> " 0 "" {MPLTEXT 1 206 73 "plot(\{f(x),D(f)(x1)*(x-x1)+f(x 1),D(f)(4)*(x-4)+f(4)\},x=2..4,thickness=2);" }{MPLTEXT 1 206 0 "" }}} {EXCHG {PARA 203 "" 0 "" {TEXT 205 69 "As before, the procedure now ca lls for us to obtain the x- intercept" }{TEXT 205 0 "" }}{PARA 203 "" 0 "" {TEXT 205 66 "of the linear approximation as the next approximat ion. Rather than" }{TEXT 205 0 "" }}{PARA 203 "" 0 "" {TEXT 205 65 "ex plain each step we will simply apply the Maple V commands to do" } {TEXT 205 0 "" }}{PARA 203 "" 0 "" {TEXT 205 67 "this until two succes sive approximations differ by no more than ten" }{TEXT 205 0 "" }} {PARA 203 "" 0 "" {TEXT 205 7 "digits." }{TEXT 205 0 "" }}{PARA 203 "" 0 "" {TEXT 205 0 "" }}{PARA 204 "> " 0 "" {MPLTEXT 1 206 38 "x2 := fs olve(D(f)(x1)*(x-x1)+f(x1),x);" }{MPLTEXT 1 206 0 "" }}}{EXCHG {PARA 204 "> " 0 "" {MPLTEXT 1 206 38 "x3 := fsolve(D(f)(x2)*(x-x2)+f(x2),x) ;" }{MPLTEXT 1 206 0 "" }}}{EXCHG {PARA 204 "> " 0 "" {MPLTEXT 1 206 38 "x4 := fsolve(D(f)(x3)*(x-x3)+f(x3),x);" }{MPLTEXT 1 206 0 "" }}} {EXCHG {PARA 204 "> " 0 "" {MPLTEXT 1 206 38 "x5 := fsolve(D(f)(x4)*(x -x4)+f(x4),x);" }{MPLTEXT 1 206 0 "" }}}{EXCHG {PARA 204 "> " 0 "" {MPLTEXT 1 206 38 "x6 := fsolve(D(f)(x5)*(x-x5)+f(x5),x);" }{MPLTEXT 1 206 0 "" }}}{EXCHG {PARA 203 "" 0 "" {TEXT 205 66 "Thus we conclude \+ that there is a zero of f(x) that agrees with " }{TEXT 205 0 "" }} {PARA 203 "" 0 "" {TEXT 205 61 " \+ x = 2.618033989 " }{TEXT 205 0 "" }}{PARA 203 "" 0 "" {TEXT 205 62 "to ten digits. As a check lets evaluate f(x) at this point. \+ " }{TEXT 205 0 "" }}{PARA 203 "" 0 "" {TEXT 205 0 "" }}{PARA 204 "> " 0 "" {MPLTEXT 1 206 30 "f(x2);f(x3);f(x4);f(x5);f(x6);" }{MPLTEXT 1 206 0 "" }}}{EXCHG {PARA 203 "" 0 "" {TEXT 205 75 "Thus f(x6) is ve ry near zero. Using \"Digits\" we can illustrate the fact " }{TEXT 205 0 "" }}{PARA 203 "" 0 "" {TEXT 205 74 "that Newton's Method conver ges at a rate termed `\"quadratic convergence.\" " }{TEXT 205 0 "" }} {PARA 203 "" 0 "" {TEXT 205 55 "We set \"Digits\" equal to 50 and cont inue the process. " }{TEXT 205 0 "" }}}{EXCHG {PARA 203 "" 0 "" {TEXT 205 83 "Study the preceding Maple V segment. Notice that the num ber of digits of agreement " }{TEXT 205 0 "" }}{PARA 203 "" 0 "" {TEXT 205 87 "that each approximation has with its predecessor almost \+ doubles after each iteration. " }{TEXT 205 0 "" }}{PARA 203 "" 0 "" {TEXT 205 82 "This means, for example, that if two succesive approxim ations differ by less than" }{TEXT 205 0 "" }}{PARA 203 "" 0 "" {TEXT 205 66 " 10^(-10) then the next approximation differs from it predece ssor" }{TEXT 205 0 "" }}{PARA 203 "" 0 "" {TEXT 205 91 "by around 10^ (-20)=(10^(-10))^2. [Compare f(x5) with f(x6)]. This gives rise to the term " }{TEXT 205 0 "" }}{PARA 203 "" 0 "" {TEXT 205 24 "\"quadratic convergence\"." }{TEXT 205 0 "" }}{PARA 203 "" 0 "" {TEXT 205 83 "Par t II. Now suppose we want to estimate the root x = 2.6 by starting wit h x0 =1.5." }{TEXT 205 0 "" }}}{EXCHG {PARA 204 "> " 0 "" {MPLTEXT 1 206 41 "x1 := fsolve(D(f)(1.5)*(x-1.5)+f(1.5),x);" }{MPLTEXT 1 206 0 " " }}}{EXCHG {PARA 204 "> " 0 "" {MPLTEXT 1 206 36 "x2:=fsolve(D(f)(x1) *(x-x1)+f(x1),x);" }{MPLTEXT 1 206 0 "" }}}{EXCHG {PARA 204 "> " 0 "" {MPLTEXT 1 206 36 "x3:=fsolve(D(f)(x2)*(x-x2)+f(x2),x);" }{MPLTEXT 1 206 0 "" }}}{EXCHG {PARA 204 "> " 0 "" {MPLTEXT 1 206 36 "x4:=fsolve(D (f)(x3)*(x-x3)+f(x3),x);" }{MPLTEXT 1 206 0 "" }}}{EXCHG {PARA 204 "> \+ " 0 "" {MPLTEXT 1 206 36 "x5:=fsolve(D(f)(x4)*(x-x4)+f(x4),x);" } {MPLTEXT 1 206 0 "" }}}{EXCHG {PARA 204 "> " 0 "" {MPLTEXT 1 206 36 "x 6:=fsolve(D(f)(x5)*(x-x5)+f(x5),x);" }{MPLTEXT 1 206 0 "" }}}{EXCHG {PARA 203 "" 0 "" {TEXT 205 168 "We can see that the approximation is \+ actually getting close to the other root which is 1. Therefore, the in itial value will determine the outcomes of the approximation." }{TEXT 205 0 "" }}}{EXCHG {PARA 203 "" 0 "" {TEXT 205 0 "" }}{PARA 203 "" 0 " " {TEXT 205 30 "Example: page 328, number 12. " }{TEXT 205 0 "" }}} {EXCHG {PARA 204 "> " 0 "" {MPLTEXT 1 206 28 "f:=x->x^2*(4-x^2)-4/(x^2 +1);" }{MPLTEXT 1 206 0 "" }}}{EXCHG {PARA 204 "> " 0 "" {MPLTEXT 1 206 29 "plot(f(x),x=-5..5,y=-10..10);" }{MPLTEXT 1 206 0 "" }}}{EXCHG {PARA 203 "" 0 "" {TEXT 205 103 "We predict there are four roots. We w ill estimate the one near x=2, the rest will be left as exercises." } {TEXT 205 0 "" }}}{EXCHG {PARA 204 "> " 0 "" {MPLTEXT 1 206 35 "x1 := \+ fsolve(D(f)(2)*(x-2)+f(2),x);" }{MPLTEXT 1 206 0 "" }}}{EXCHG {PARA 204 "> " 0 "" {MPLTEXT 1 206 36 "x2:=fsolve(D(f)(x1)*(x-x1)+f(x1),x);" }{MPLTEXT 1 206 0 "" }}}{EXCHG {PARA 204 "> " 0 "" {MPLTEXT 1 206 36 "x3:=fsolve(D(f)(x2)*(x-x2)+f(x2),x);" }{MPLTEXT 1 206 0 "" }}}{EXCHG {PARA 204 "> " 0 "" {MPLTEXT 1 206 36 "x4:=fsolve(D(f)(x3)*(x-x3)+f(x3 ),x);" }{MPLTEXT 1 206 0 "" }}}{EXCHG {PARA 204 "> " 0 "" {MPLTEXT 1 206 36 "x5:=fsolve(D(f)(x4)*(x-x4)+f(x4),x);" }{MPLTEXT 1 206 0 "" }}} {EXCHG {PARA 204 "> " 0 "" {MPLTEXT 1 206 36 "x6:=fsolve(D(f)(x5)*(x-x 5)+f(x5),x);" }{MPLTEXT 1 206 0 "" }}}{EXCHG {PARA 204 "> " 0 "" {MPLTEXT 1 206 36 "f(x1);f(x2);f(x3);f(x4);f(x5);f(x6);" }{MPLTEXT 1 206 0 "" }}}{EXCHG {PARA 204 "> " 0 "" {MPLTEXT 1 206 0 "" }}}{PARA 205 "" 0 "" {TEXT -1 0 "" }}}{MARK "0 0 0" 0 }{VIEWOPTS 1 1 0 15 10 1804 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }