Explanation Formula used: Newton’s method formula: p n + 1 = p n − f ( p n ) f ′ ( p n ) Calculation: Consider the amount of money required to pay off a mortgage over a fixed period of time involve the formula, A = P i [ 1 − ( 1 + i ) − n ] P i [ 1 − ( 1 + i ) − n ] − A = 0 P [ 1 − ( 1 + i ) − n ] − A × i = 0 Here A is the amount mortgage, P is the amount of each payment, and I is the interest rate per period for n payments. First write the function f ( i ) . f ( i ) = P [ 1 − ( 1 + i ) − n ] − A i Now, differentiate the above function with respect to i : f ′ ( i ) = P d d i [ 1 − ( 1 + i ) − n ] − d d i ( A i ) = P [ n ( 1 + i ) − ( n + 1 ) ] − A Use the Newton’s method, calculate the value of i k + 1 . i k + 1 = i k − f ( i k ) f ′ ( i k ) Substitute f ( i ) = P [ 1 − ( 1 + i ) − n ] − A i and f ′ ( i ) = P [ n ( 1 + i ) − ( n + 1 ) ] − A in i k + 1 = i k − f ( i k ) f ′ ( i k ) . i k + 1 = i k − P [ 1 − ( 1 + i k ) − n ] − A i k P [ n ( 1 + i k ) − ( n + 1 ) ] − A Since the payment is done monthly, the number of periods n should be 12 × 30 = 360 . Substitute A = $ 135 , 000 , P = $ 1 , 000 , and n = 360 in i k + 1 = i k − P [ 1 − ( 1 + i k ) − n ] − A i k P [ n ( 1 + i k ) − ( n + 1 ) ] − A . i k + 1 = i k − 1 , 000 [ 1 − ( 1 + i k ) − 360 ] − 135 , 000 i k 1 , 000 [ 360 ( 1 + i k ) − 361 ] − 135 , 000 The solution of the above equation can be calculated using Newton’s method formula. The value of I lies between the interval [ 0 , 1 ] and start with the initial approximation i 0 = 0.5 . Now, calculate the first iteration by substituting k = 0 and i 0 = 0.5 in i k + 1 = i k − 1 , 000 [ 1 − ( 1 + i k ) − 360 ] − 135 , 000 i k 1 , 000 [ 360 ( 1 + i k ) − 361 ] − 135 , 000 . i 1 = i 0 − 1 , 000 [ 1 − ( 1 + i 0 ) − 360 ] − 135 , 000 i 0 1 , 000 [ 360 ( 1 + i 0 ) − 361 ] − 135 , 000 = 0.5 − 1 , 000 [ 1 − ( 1 + 0.5 ) − 360 ] − 135 , 000 ( 0.5 ) 1 , 000 [ 360 ( 1 + 0.5 ) − 361 ] − 135 , 000 = 0.00740740 Now, calculate the second iteration by substituting k = 1 and i 1 = 0.00740740 in i k + 1 = i k − 1 , 000 [ 1 − ( 1 + i k ) − 360 ] − 135 , 000 i k 1 , 000 [ 360 ( 1 + i k ) − 361 ] − 135 , 000

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ISBN: 9781305465350

Author: BURDEN

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Chapter 2.3, Problem 23ES

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The maximum interest rate, the borrower can afford to pay by using the given data.

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Chapter 2.3, Problem 22ES

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Ch. 2.1 - Use the Bisection method to find p3 for f(x)=xcosx...Ch. 2.1 - Let f(x) = 3(x +1)(x 12)(x 1) = 0. Use the...Ch. 2.1 - Use the Bisection method to find solutions...Ch. 2.1 - Use the Bisection method to find solutions...Ch. 2.1 - Use the Bisection method to find solutions...Ch. 2.1 - Prob. 7ES Ch. 2.1 - Prob. 8ES Ch. 2.1 - Prob. 9ES Ch. 2.1 - Prob. 10ES Ch. 2.1 - Prob. 11ES

Ch. 2.1 - Let f(x) = (x + 2)(x + 1)x(x 1)3(x 2). To which...Ch. 2.1 - Find an approximation to 253 correct to within 104...Ch. 2.1 - Find an approximation to 3 correct to within 104...Ch. 2.1 - A trough of length L has a cross section in the...Ch. 2.1 - Use Theorem 2.1 to find a bound for the number of...Ch. 2.1 - Prob. 18ES Ch. 2.1 - Prob. 19ES Ch. 2.1 - Let f(x) = (x 1)10, p = 1, and pn = 1 + 1/n. Show...Ch. 2.1 - The function defined by f(x) = sin x has zeros at...Ch. 2.1 - Prob. 1DQ Ch. 2.1 - Prob. 2DQ Ch. 2.1 - Is the Bisection method sensitive to the starting...Ch. 2.2 - Use algebraic manipulation to show that each of...Ch. 2.2 - a. Perform four iterations, if possible, on each...Ch. 2.2 - Let f(x) = x3 2x + 1. To solve f(x) = 0, the...Ch. 2.2 - Let f(x) = x4 + 3x2 2. To solve f(x) = 0, the...Ch. 2.2 - The following four methods are proposed to compute...Ch. 2.2 - Prob. 6ES Ch. 2.2 - Prob. 7ES Ch. 2.2 - Prob. 8ES Ch. 2.2 - Use Theorem 2.3 to show that g(x) = + 0.5...Ch. 2.2 - Use Theorem 2.3 to show that g(x) = 2x has a...Ch. 2.2 - Use a fixed-point iteration method to find an...Ch. 2.2 - Use a fixed-point iteration method to determine a...Ch. 2.2 - Use a fixed-point iteration method to determine a...Ch. 2.2 - Prob. 20ES Ch. 2.2 - Prob. 21ES Ch. 2.2 - a. Show that Theorem 2.3 is true if the inequality...Ch. 2.2 - a. Use Theorem 2.4 to show that the sequence...Ch. 2.2 - Prob. 24ES Ch. 2.2 - Prob. 25ES Ch. 2.2 - Suppose that g is continuously differentiable on...Ch. 2.3 - Let f(x) = x2 6 and p0 = 1. Use Newtons method to...Ch. 2.3 - Let f(x) = x3 cos x and p0 = 1. Use Newtons...Ch. 2.3 - Let f(x) = x2 6. With p0 = 3 and p1 = 2, find p3....Ch. 2.3 - Let f(x) = x3 cos x. With p0 = 1 and p1 = 0, find...Ch. 2.3 - Prob. 11ES Ch. 2.3 - Prob. 12ES Ch. 2.3 - The fourth-degree polynomial...Ch. 2.3 - Prob. 14ES Ch. 2.3 - Prob. 15ES Ch. 2.3 - Prob. 16ES Ch. 2.3 - Prob. 22ES Ch. 2.3 - Prob. 23ES Ch. 2.3 - Prob. 24ES Ch. 2.3 - Prob. 25ES Ch. 2.3 - Prob. 27ES Ch. 2.3 - A drug administered to a patient produces a...Ch. 2.3 - Prob. 30ES Ch. 2.3 - Prob. 32ES Ch. 2.3 - Prob. 1DQ Ch. 2.3 - Prob. 2DQ Ch. 2.3 - Prob. 3DQ Ch. 2.3 - Prob. 4DQ Ch. 2.4 - Prob. 6ES Ch. 2.4 - a. Show that for any positive integer k, the...Ch. 2.4 - Prob. 8ES Ch. 2.4 - a. Construct a sequence that converges to 0 of...Ch. 2.4 - Prob. 10ES Ch. 2.4 - Prob. 11ES Ch. 2.4 - Prob. 12ES Ch. 2.4 - Prob. 13ES Ch. 2.4 - Prob. 14ES Ch. 2.4 - Prob. 1DQ Ch. 2.4 - Prob. 2DQ Ch. 2.4 - Prob. 4DQ Ch. 2.5 - Let g(x) = cos(x 1) and p0(0) = 2. Use...Ch. 2.5 - Prob. 4ES Ch. 2.5 - Prob. 5ES Ch. 2.5 - Prob. 6ES Ch. 2.5 - Use Steffensens method to find, to an accuracy of...Ch. 2.5 - Prob. 8ES Ch. 2.5 - Prob. 9ES Ch. 2.5 - Use Steffensens method with p0 = 3 to compute an...Ch. 2.5 - Use Steffensens method to approximate the...Ch. 2.5 - Prob. 12ES Ch. 2.5 - Prob. 13ES Ch. 2.5 - Prob. 14ES

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Chapter 2 Solutions