f (x) = It is clear that, for RT RT = x PR or RT = f (x) x df in .NET framework

Creation QR Code JIS X 0510 in .NET framework f (x) = It is clear that, for RT RT = x PR or RT = f (x) x df

f (x) = It is clear that, for RT RT = x PR or RT = f (x) x df
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x very small, RT RQ, that is, df f (x + x) f (x)
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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
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Fig. 21-1
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APPROXIMATION BY DIFFERENTIALS; NEWTON S METHOD
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[CHAP. 21
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In particular, if f (x) = x, we shall write dx = df = f (x) x = 1 Since dx = x= x
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x, the de nition of the differential df can be rewritten as df = f (x)dx
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Assuming that dx =
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x = 0, we may divide both sides by dx, obtaining the result f (x) = df dx
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If we let y = f (x), this may explain the traditional notation dy/dx for the derivative. 21.3 NEWTON S METHOD Let us assume that we are trying to nd a solution of the equation f (x) = 0 (21.4)
and let us also assume that we know that x0 is close to a solution. If we draw the tangent line T to the graph of f at the point with abscissa x0 , then T will usually intersect the x-axis at a point whose abscissa x1 is closer to the solution of (21.4) than x0 (see Fig. 21-2).
Fig. 21-2 A point-slope equation of the tangent line T is y f (x0 ) = f (x0 )(x x0 ) If T intersects the x-axis at the point (x1 , 0), then 0 f (x0 ) = f (x0 )(x1 x0 ) If f (x0 ) = 0, x 1 x0 = (x0 ) f (x0 ) f (x0 ) x1 = x0 f (x0 )
CHAP. 21]
APPROXIMATION BY DIFFERENTIALS; NEWTON S METHOD
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,
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