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f (x) = It is clear that, for RT RT = x PR or RT = f (x) x df in .NET framework
f (x) = It is clear that, for RT RT = x PR or RT = f (x) x df QRCode Reader In Visual Studio .NET Using Barcode Control SDK for .NET Control to generate, create, read, scan barcode image in VS .NET applications. Paint QR Code ISO/IEC18004 In .NET Framework Using Barcode drawer for .NET Control to generate, create QR Code image in .NET framework applications. x very small, RT RQ, that is, df f (x + x) f (x) QR Code ISO/IEC18004 Reader In .NET Using Barcode decoder for .NET framework Control to read, scan read, scan image in VS .NET applications. Barcode Maker In Visual Studio .NET Using Barcode maker for .NET Control to generate, create barcode image in VS .NET applications. If the value of a function f is given by a formula, say, f (x) = x 2 + 2x 3 , let us agree that the differential df may also be written d(x 2 + 2x 3 ) = df = f (x) x = (2x 6x 4 ) x Recognizing Bar Code In Visual Studio .NET Using Barcode reader for Visual Studio .NET Control to read, scan read, scan image in .NET applications. QR Code 2d Barcode Drawer In C# Using Barcode encoder for Visual Studio .NET Control to generate, create Denso QR Bar Code image in .NET applications. Fig. 211 Denso QR Bar Code Maker In .NET Using Barcode encoder for ASP.NET Control to generate, create QR Code JIS X 0510 image in ASP.NET applications. QR Generation In VB.NET Using Barcode maker for VS .NET Control to generate, create Quick Response Code image in VS .NET applications. APPROXIMATION BY DIFFERENTIALS; NEWTON S METHOD
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If we repeat the same procedure, but now starting with x1 instead of x0 , we obtain a value x2 that should be still closer to the solution of (21.4), x2 = x1 f (x1 ) f (x1 ) If we keep applying this procedure, the resulting sequence of numbers x0 , x1 , x2 , , xn , is determined by the formula xn+1 = xn f (xn ) f (xn ) (21.5) This process for nding better and better approximations to a solution of the equation f (x) = 0 is known as Newton s method. It is not always guaranteed that the numbers generated by Newton s method will approach a solution of f (x) = 0. Some dif culties that may arise will be discussed in the problems below. EXAMPLE If we wish to approximate Then f (x) = 2x, and (21.5) becomes 2, we should try to nd an approximate solution of x 2 2 = 0. Here f (x) = x 2 2. 2 x2 2 2x 2 (xn 2) x2 + 2 xn+1 = xn n = n = n 2xn 2xn 2xn
(21.6) If we take the rst approximation x0 to be 1 (since we know that n = 0, 1, 2, . . . in (21.6),1 x1 =
2 is between 1 and 2), then we obtain by successively substituting
1+2 = 1.5 2 2.25 + 2 4.25 (1.5)2 + 2 = = 1.416 666 667 x2 = 2(1.5) 3 3 x3 x4 x5 (1.416 666 667)2 + 2 1.414 215 686 2(1.416 666 667) (1.414 215 686)2 + 2 1.414 213 562 2(1.414 215 686) (1.414 213 562)2 + 2 1.414 213 562 2(1.414 213 562) Since our approximations for x4 and x5 are equal, all future values will be the same, and we have obtained the best approximation to nine decimal places within the limits of our calculator. Thus, 2 1.414 213 562. Solved Problems
21.1 Using the approximation principle, estimate the value of
Letting f (x) = x and c = 62, choose x = 64 (the perfect square closest to 62). Then, x = c x = 62 64 = 2 f (x) = 64 = 8 1 1 1 1 = f (x) = = = 2(8) 16 2 x 2 64 1 The computations required by Newton s method are usually too tedious to be done by hand. A calculator, preferably a programmable calculator, should be used. APPROXIMATION BY DIFFERENTIALS; NEWTON S METHOD
[CHAP. 21
and (21.2) yields 1 7 1 62 8 + ( 2) = 8 = 7 = 7.875 16 8 8 62 = 7.8740 . . .. 5 x, c = 33, and x = 32. Then f (x) = Dx (x 1/5 ) = Hence, by (21.2), 1 (1) = 2.0125 80 Since the actual value is 2.0123 , the approximation is correct to three decimal places. 33 = f (c) 2 + 5 Actually,

