1.4 Numerical Solutions 35 1.4 Numerical Solutions The last section emphasized that arithmetic (especially division) on machines entails approximation errors, and (sometimes) it simply cannot be done...

Use Newton’s method to approximate the square root of 63. Set A=63 and x₀=9. Compute the values x₁, x₂, x₃. Show at least 5 decimal places

Extracted text: 1.4 Numerical Solutions 35 1.4 Numerical Solutions The last section emphasized that arithmetic (especially division) on machines entails approximation errors, and (sometimes) it simply cannot be done (absolutely) accurately. The best we can do is to calculate approximate answers. In this section, we will try to make that a virtue. 1.4.1 Newton's Method for Square Roots To find a “good" approximation of VA, when A is a given positive number: 1. Make a guess at the square root, xo. 2. Revise the last guess, X;, to obtain a new guess: // some positive number X; +A/x; Xi+1 // until ? Let's see what happens when A = 144 (so we know the answer) and xo 10. 10+ 144/10 10+14.4 24.4 = Ix 2 = 12.2 12.2 + 144/12.2 12.2 +11.80327869. .. 24.00327869... X2 = 2 = 12.001 639 344 2... // my calculator truncates to 12 digits Хз — 12.000 000 111 9... X4 = 12 then x5 will also be 12, and all other x;'s will be 12. Newton's method found the square root exactly after only 4 revisions of our initial guess of 10. // Isn't that magical? And worth looking at a bit more closely? // Would you believe that we only got the right answer because of roundoff error? // (and that if we did exact arithmetic, we would never reach 12.) Let E; denote the error in x; as an approximation of VA; that is, E; = x; – VA and so x; = v. VĀ+E;. Then the next error X; +A/x; X; +A/x; – 2VA E¡+1= xj+1 - VA VĀ 2 2 x} - (2VA)x; +A_ (xị – VA)_ E 2x; Xi 2x; 2x;