Use the approximation principle to estimate the value of in .NET framework

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21.2 Use the approximation principle to estimate the value of
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5 33.
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5 32 = 2, and
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1 1 4/5 1 1 1 1 1 x = . = 4/5 = 4 = = = 5 5 4 4 5 5(16) 80 5x 5( x) 5(2) 5( 32)
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21.3 Measurement of the side of a square room yields the result of 18.5 feet. Hence, the area is A = (18.5)2 = 342.25 square feet. If the measuring device has a possible error of at most 0.05 foot, estimate the maximum possible error in the area.
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The formula for the area is A = x 2 , where x is the side of the room. Hence, dA/dx = 2x. Let x = 18.5, and let 18.5 + x be the true length of the side of the room. By assumption, | x| 0.05. The approximation principle yields A(x + A(18.5 + |A(18.5 + dA x dx x) 342.25 2(18.5) x x) A(x) 37(0.05) = 1.85
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x) 342.25| |37 x|
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Hence, the error in the area should be at most 1.85 square feet, putting the actual area in the range of (342.25 1.85) square feet. See Problem 21.13.
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21.4 Use Newton s method to nd the positive solutions of x4 + x 3 = 0
Let f (x) = x 4 + x 3. Then f (x) = 4x 3 + 1. Since f (1) = 1 and f (2) = 15, the intermediate-value theorem tells us that there is a solution between 1 and 2. [The interval (1, 2) is suggested by drawing the graph of f with a graphing calculator.] Since f (x) > 0 for x 0, f is increasing for x > 0, and, therefore, there is exactly one positive real solution. Start with x0 = 1. Equation (21.5) becomes
4 x 4 + xn 3 4x 4 + xn (xn + xn 3) 3x 4 + 3 xn+1 = xn n 3 = n = n 3 3 4xn + 1 4xn + 1 4xn + 1
Successive calculations yield x1 = 1.2, x2 = 1.165 419 616, x3 = 1.164 037 269, x4 = 1.164 035 141, and x5 = 1.164 035 141. Thus, the approximate solution is x = 1.164 035 141.
21.5 Show that if Newton s method is applied to the equation x 1/3 = 0, with x0 = 1, the result is a divergent sequence of values (which certainly does not converge to the root x = 0).
Let f (x) = x 1/3 . So, f (x) = 1 and (21.5) becomes 3x 2/3 xn+1 = xn xn
1/3 2/3
1/(3xn )
= xn 3xn = 2xn
Hence, x1 = 2, x2 = 4, x3 = 8, and, in general, xn = ( 2)n .
CHAP. 21]
APPROXIMATION BY DIFFERENTIALS; NEWTON S METHOD
Note: If we are seeking a solution r of an equation f (x) = 0, then it can be shown that a suf cient condition that Newton s method yields a sequence of values that converges to r is that f (x) f (x) <1 [f (x)]2 for all x in an interval around r that includes x0 . However, this is not a necessary condition.
Supplementary Problems
21.6 Use the approximation principle to estimate the following quantities: (a) 51 (b) 78 (c) 3 123 (d) (8.35)2/3 (e) (33) 1/5 17 (g) 3 0.065 (f ) 4 (h) 80.5 (i) 3 215 81 21.7 The measurement of the side of a cubical container yields 8.14 centimeters, with a possible error of at most 0.005 centimeter. Give an estimate of the maximum possible error in the value of V = (8.14)3 = 539.353 14 cubic centimeters for the volume of the container. 21.8 It is desired to give a spherical tank 20 feet (240 inches) in diameter a coat of paint 0.1 inch thick. Use the approximation principle to estimate how many gallons of paint will be required. (V = 4 r 3 and 1 gallon is about 231 cubic inches.) 3 21.9 A solid steel cylinder has a radius of 2.5 centimeters and a height of 10 centimeters. A tight- tting sleeve is to be made that will extend the radius to 2.6 centimeters. Find the amount of steel needed for the sleeve: (a) by the approximation principle; (b) by an exact calculation. 21.10 If the side of a cube is measured with a percentage error of at most 3%, estimate the maximum percentage error in the volume of the cube. (If Q is the error in measurement of a quantity Q, then | Q/Q| 100% is the percentage error.) 21.11 Assume, contrary to fact, that the Earth is a perfect sphere, with a radius of 4000 miles. The volume of ice at the North and South Poles is estimated to be about 8 000 000 cubic miles. If this ice were melted and if the resulting water were distributed uniformly over the globe, approximately what would be the depth of the added water at any point of the Earth 21.12 (a) Let y = x 3/2 . When x = 4 and dx = 2, nd the value of dy. (b) Let y = 2x 1 + x 2 . When x = 0 and dx = 3, nd the value of dy. 21.13 For Problem 21.3, calulate exactly the largest possible error in the area. 21.14 Establish the very useful approximation formula (1 + u)r 1 + ru, where r is any rational exponent and |u| is small compared to 1. [Hint: Apply the approximation principle to f (x) = x r , letting x = 1 and x = u.] 21.15 GC Use Newton s method to approximate the following quantities: (a) 4 2 (b) 3 4 (c) 5 23 (d) 3 6 21.16 (a) Show that Newton s method for nding c yields the equation xn+1 = 1 c xn + 2 xn
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