Explanation Formula used: Newton’s method formula: p n + 1 = p n − f ( p n ) f ′ ( p n ) Calculation: Consider the accumulated value of a savings account based on regular periodic payments can be determined from the annuity due equation, A = P i [ ( 1 + i ) n − 1 ] A − P i [ ( 1 + i ) n − 1 ] = 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 ) = A − P i [ ( 1 + i ) n − 1 ] Now, differentiate the above function with respect to i : f ′ ( i ) = P d d i [ ( 1 + i ) n − 1 i ] + d d i ( A ) = P i 2 [ ( 1 + i ) n − 1 ] − P i [ n ( 1 + i ) ] Use the Newton’s method, calculate the value of P n + 1 . p n + 1 = p n − f ( p n ) f ′ ( p n ) Substitute f ( i ) = A − P i [ ( 1 + i ) n − 1 ] and f ′ ( i ) = P i 2 [ ( 1 + i ) n − 1 ] − P i [ n ( 1 + i ) ] in p n + 1 = p n − f ( p n ) f ′ ( p n ) . p n + 1 = p n − A − P p n [ ( 1 + p n ) n − 1 ] P ( p n ) 2 [ ( 1 + p n ) n − 1 ] − P p n [ n ( 1 + p n ) ] Substitute A = $ 750 , 000 , P = $ 1 , 500 , and n = 10 in p n + 1 = p n − A − P p n [ ( 1 + p n ) n − 1 ] P ( p n ) 2 [ ( 1 + p n ) n − 1 ] − P p n [ n ( 1 + p n ) ] . p n + 1 = p n − 750 , 000 − 1 , 500 p n [ ( 1 + p n ) 10 − 1 ] 1 , 500 ( p n ) 2 [ ( 1 + p n ) 10 − 1 ] − 1 , 500 p n [ 10 ( 1 + p n ) ] 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 n = 0 and p 0 = 0.5 in p n + 1 = p n − 750 , 000 − 1 , 500 p n [ ( 1 + p n ) 10 − 1 ] 1 , 500 ( p n ) 2 [ ( 1 + p n ) 10 − 1 ] − 1 , 500 p n [ 10 ( 1 + p n ) ] . p 1 = p 0 − 750 , 000 − 1 , 500 p 0 [ ( 1 + p 0 ) 10 − 1 ] 1 , 500 ( p 0 ) 2 [ ( 1 + p 0 ) 10 − 1 ] − 1 , 500 p 0 [ 10 ( 1 + p 0 ) ] = 0.5 − 750 , 000 − 1 , 500 0.5 [ ( 1 + 0.5 ) 10 − 1 ] 1 , 500 ( 0.5 ) 2 [ ( 1 + 0.5 ) 10 − 1 ] − 1 , 500 0.5 [ 10 ( 1 + 0.5 ) ] = 0.5 − 580 , 004.88 294 , 990.23 = − 1.46 Now, calculate the second iteration by substituting n = 1 and p 1 = − 1.46 in p n + 1 = p n − 750 , 000 − 1 , 500 p n [ ( 1 + p n ) 10 − 1 ] 1 , 500 ( p n ) 2 [ ( 1 + p n ) 10 − 1 ] − 1 , 500 p n [ 10 ( 1 + p n ) ] . p 2 = p 1 − 750 , 000 − 1 , 500 p 1 [ ( 1 + p 1 ) 10 − 1 ] 1 , 500 ( p 1 ) 2 [ ( 1 + p 1 ) 10 − 1 ] − 1 , 500 p 1 [ 10 ( 1 + p 1 ) ] = − 1

10th Edition

ISBN: 9781305253667

Author: Richard L. Burden, J. Douglas Faires, Annette M. Burden

Publisher: Cengage Learning

See similar textbooks

Concept explainers

Unitary Method

The word “unitary” comes from the word “unit”, which means a single and complete entity. In this method, we find the value of a unit product from the given number of products, and then we solve for the other number of products.

Speed, Time, and Distance

Imagine you and 3 of your friends are planning to go to the playground at 6 in the evening. Your house is one mile away from the playground and one of your friends named Jim must start at 5 pm to reach the playground by walk. The other two friends are 3 …

Profit and Loss

The amount earned or lost on the sale of one or more items is referred to as the profit or loss on that item.

Units and Measurements

Measurements and comparisons are the foundation of science and engineering. We, therefore, need rules that tell us how things are measured and compared. For these measurements and comparisons, we perform certain experiments, and we will need the experime…

Videos

Question

Chapter 2.3, Problem 24ES

To determine

The minimal interest rate, the amount can be invested by using the given data.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 2.3, Problem 23ES

Chapter 2.3, Problem 25ES

Students have asked these similar questions

Graph without using the calculator y-1 = | x+4 |

If h(x) = -2x-8 49x2-9 what is lim h(x)? x--00

Question Find the following limit. Select the correct answer below: ○ 0 ○ 3 ○ 6 ∞ 6x + 3e lim 00+2 x 2

Chapter 2 Solutions

Numerical Analysis

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

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, subject and related others by exploring similar questions and additional content below.

The minimal interest rate, the amount can be invested by using the given data.

Solution Summary: The author explains the minimum interest rate, the amount can be invested by using the given data.

Concept explainers

Videos

The minimal interest rate, the amount can be invested by using the given data.

Want to see the full answer?

Chapter 2 Solutions