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

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Numerical Analysis

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