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hw-due-10-24

# hw-due-10-24 - Calculus I Fall 2005 MW 8:55-10:45 Silver...

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Calculus I Fall 2005 MW 8:55-10:45, Silver 710 Section 3.9 6. Approximation using the derivative of f ( x ) := x 2 / 3 at the point x = 27: (26) 2 3 (27) 2 3 - 2 3(27) 1 3 = 79 9 = 8 . 7777 ... On the other hand, the calculator gives (26) 2 / 3 = 8 . 7763 ... . 12. Approximation using the derivative: sin 43 π 180 sin 45 π 180 - 2 π 180 cos 45 π 180 = 2 2 - 2 π 180 = 0 . 682424 ... The calculator gives sin(43 π/ 180) = 0 . 681998 ... . 26. (a) Approximating using the derivative we obtain ( x + h ) n - x n nhx n - 1 . Hence, if the error of the left hand side is required to be less than 1% of x n we get the approximate inequality n | hx n - 1 | | x n | 100 . From this we conclude that the error h when estimating x has to be (approximately) at most | x | 100 n , ie. approximately within 1 n % of the correct value x . (b) Proceeding in the same way as above we obtain for the approximation of x 1 /n that the error in estimating x should be (approximately) less than n % of the correct value x . 30. The Newton-Raphson method reads as x n +1 = x n - x 5 n - 30 5 x 4 n . Starting with x 1 = 2 we therefore obtain: x 2 = 79 40 = 1 . 975, x 3 = 1 . 9743509 ... , x 4 1 . 97435 . 32. The Newton-Raphson method reads as x n +1 = x n - sin x n - x 2 n cos x n - 2 x n . Thus, starting with x 1 = 1 with obtain x 2 = 0 . 891395995 ... , x 3 = 0 . 876984844 ... , x 4 0 . 876726. 34. (a) Since f is continuous everywhere and since f (1) = - 2 and f (2) = 3, by the intermediate- value theorem, f attains all values between - 2 and 3 on the interval [1 , 2]. In particular, it attains 0.

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(b) The Newton-Raphson method gives x n +1 = x n - 2 x 3 n - 3 x 2 n - 1 6 x 2 n - 6 x n . Since f (1) = 0 we cannot start the Newton-Raphson method with x 1 = 1. (c) However, starting with x 1 = 2 we obtain x 2 = 1 . 75, x 3 = 1 . 682539 ... , x 4 1 . 67768. Evaluating f ( x 4 ) = 0 . 000199874 we see that x 4 is a fairly good approximation for the root of f .
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