Explanation Definition used: Gini index of Income Concentration: If y = f ( x ) is the equation of a Lorentz curve, then Gini Index = 2 ∫ 0 1 [ x − f ( x ) ] d x . The Gini index is always a number between 0 and 1. Result used: The Taylor series at 0 for f ( x ) = 1 1 + x is 1 1 + x = 1 − x + x 2 − ⋅ ⋅ ⋅ + ( − 1 ) n x n + ⋅ ⋅ ⋅ . Theorem used: Error Estimation for Alternating Series If x 0 is a number in the interval of convergence for f ( x ) = a 0 + a 1 x + a 2 x 2 + ⋅ ⋅ ⋅ + a k x k + ⋅ ⋅ ⋅ and the terms in the series f ( x 0 ) = a 0 + a 1 x 0 + a 2 x 0 2 + ⋅ ⋅ ⋅ + a k x 0 k + ⋅ ⋅ ⋅ are alternating in sign and decreasing in absolute value, then the error in the approximation f ( x 0 ) ≈ a 0 + a 1 x 0 + a 2 x 0 2 + ⋅ ⋅ ⋅ + a n x 0 n is strictly less than the absolute value of the next term. That is, | R n ( x 0 ) | < | a n + 1 x 0 n + 1 | . Calculation: The given Lorenz curve is, f ( x ) = 11 x 5 10 + x 2 . Use the above mentioned definition to obtain the Gini Index of the Lorentz curve. Gini Index = 2 ∫ 0 1 [ x − f ( x ) ] d x = 2 ∫ 0 1 x d x − 2 ∫ 0 1 f ( x ) d x = 2 ∫ 0 1 x d x − 2 ∫ 0 1 11 x 5 10 + x 2 d x (1) Obtain the value of the first integral in (1) as follows. 2 ∫ 0 1 x d x = 2 [ x 2 2 ] 0 1 = 2 [ 1 2 − 0 2 ] = 2 ( 1 2 ) = 1 Use the Taylor series to approximate the integral ∫ 0 1 11 x 5 10 + x 2 d x as follows. Consider the expression 11 x 5 10 + x 2 = 11 x 5 ( 1 10 + x 2 ) . Substitute z = x 2 in 1 10 + x 2 . On simplification, 1 10 + z = 1 10 ( 1 1 + z 10 ) = 1 10 ( 1 1 + t ) , where t = z 10 The Taylor series at 0 for 1 1 + t is 1 1 + t = 1 − t + t 2 − ⋅ ⋅ ⋅ + ( − 1 ) n t n + ⋅ ⋅ ⋅ . Replace t by x 2 10 in 1 1 + t = 1 − t + t 2 − ⋅ ⋅ ⋅ + ( − 1 ) n t n + ⋅ ⋅ ⋅ and simplify it. 1 1 + x 2 10 = 1 − x 2 10 + ( x 2 10 ) 2 − ⋅ ⋅ ⋅ + ( − 1 ) n ( x 2 10 ) n + ⋅ ⋅ ⋅ = 1 − x 2 10 + x 4 100 − ⋅ ⋅ ⋅ + ( − 1 ) n ( x 2 10 ) n + ⋅ ⋅ ⋅ Substitute the above obtained value in 1 10 + x 2 = 1 10 ( 1 1 + x 2 10 ) . 1 10 ( 1 1 + x 2 10 ) = 1 10 ( 1 − x 2 10 + x 4 100 − ⋅ ⋅ ⋅ + ( − 1 ) n ( x 2 10 ) n + ⋅ ⋅ ⋅ ) = 1 10 − x 2 100 + x 4 1000 − ⋅ ⋅ ⋅ + ( − 1 ) n x 2 n 10 n + 1 + ⋅ ⋅ ⋅ Substitute the value 1 10 + x 2 = 1 10 − x 2 100 + x 4 1000 − ⋅ ⋅ ⋅ + ( − 1 ) n x 2 n 10 n + 1 + ⋅ ⋅ ⋅ in 11 x 5 ( 1 10 + x 2 ) and simplify it as follows. 11 x 5 ( 1 10 + x 2 ) = 11 x 5 ( 1 10 − x 2 100 + x 4 1000 − ⋅ ⋅ ⋅ + ( − 1 ) n x 2 n 10 n + 1 + ⋅ ⋅ ⋅ ) = 11 x 5 10 − 11 x 5 + 2 100 + 11 x 5 + 4 1000 − ⋅ ⋅ ⋅ + ( − 1 ) n 11 x 5 + 2 n 10 n + 1 + ⋅ ⋅ ⋅ = 11 x 5 10 − 11 x 7 100 + 11 x 9 1000 − ⋅ ⋅ ⋅ + ( − 1 ) n 11 x 5 + 2 n 10 n + 1 + ⋅ ⋅ ⋅ Integrate on both sides of the equation 11 x 5 ( 1 10 + x 2 ) = 11 x 5 10 − 11 x 7 100 + 11 x 9 1000 − ⋅ ⋅ ⋅ + ( − 1 ) n 11 x 5 + 2 n 10 n + 1 + ⋅ ⋅ ⋅

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ISBN: 9780135961438

Author: Barnett

Publisher: PEARSON CO

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Chapter 10.4, Problem 4MP

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To approximate: The index of income concentration for the given Lorenz curve f(x)=11x510+x2 with an error of no more than 0.005.

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Chapter 10.4, Problem 3MP

Chapter 10.4, Problem 1ED

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(d) At any time, find the mean age of insects in the population. (e) Sketch g(x), the p.d.f. of the age distribution of insects in the population. (f) No calculations are required for this part of the question.Now suppose that a population with the life table function Q(x) thatyou obtained in part (b) is stable but not stationary. Describe brieflyhow the age distribution of the population would differ from that of astationary population with the same life table function if the populationis (i) growing, (ii) declining. Draw rough sketches to show what thep.d.f. of the age distribution might look like in each case.

In a study conducted by a certain country's Economic Development Board, it was found that the Lorenz Curve for the distribution of income of stockbrokers was described by the function f(x)=11/12 x^2+1/12 x And that of high school teachers by the function g(x)=5/6 x^2+1/6 xa) Sketch the Lorentz curve for stock-broker income.

Chapter 10 Solutions

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Ch. 10.1 - Find the nth derivative of f(x)=lnx.Ch. 10.1 - Prob. 2MP Ch. 10.1 - Prob. 3MP Ch. 10.1 - Find the second-degree Taylor polynomial at a = 8...Ch. 10.1 - Prob. 5MP Ch. 10.1 - Prob. 1ED Ch. 10.1 - (A)Let p(x) be a polynomial of degree n 1....Ch. 10.1 - Prob. 1E Ch. 10.1 - Prob. 2E Ch. 10.1 - Prob. 3E

Ch. 10.1 - Prob. 4E Ch. 10.1 - Prob. 5E Ch. 10.1 - Prob. 6E Ch. 10.1 - Prob. 7E Ch. 10.1 - Prob. 8E Ch. 10.1 - Prob. 9E Ch. 10.1 - Prob. 10E Ch. 10.1 - Prob. 11E Ch. 10.1 - Prob. 12E Ch. 10.1 - Prob. 13E Ch. 10.1 - Prob. 14E Ch. 10.1 - In Problems 1316, find f(3)(x). 15.f(x)=ex Ch. 10.1 - In Problems 1316, find f(3)(x). 16.f(x)=x Ch. 10.1 - Prob. 17E Ch. 10.1 - In Problems 1720, find f4(x). 18.f(x)=e5x Ch. 10.1 - Prob. 19E Ch. 10.1 - In Problems 1720, find f4(x). 20.f(x)=12+x Ch. 10.1 - Prob. 21E Ch. 10.1 - Prob. 22E Ch. 10.1 - Prob. 23E Ch. 10.1 - In Problems 2128, find the indicated Taylor...Ch. 10.1 - Prob. 25E Ch. 10.1 - Prob. 26E Ch. 10.1 - Prob. 27E Ch. 10.1 - Prob. 28E Ch. 10.1 - Prob. 29E Ch. 10.1 - Prob. 30E Ch. 10.1 - Prob. 31E Ch. 10.1 - Prob. 32E Ch. 10.1 - Prob. 33E Ch. 10.1 - Prob. 34E Ch. 10.1 - Prob. 35E Ch. 10.1 - Prob. 36E Ch. 10.1 - Prob. 37E Ch. 10.1 - Prob. 38E Ch. 10.1 - Prob. 39E Ch. 10.1 - Use the third-degree Taylor polynomial at 0 for...Ch. 10.1 - Prob. 41E Ch. 10.1 - Use the third-degree Taylor polynomial at 4 for...Ch. 10.1 - Prob. 43E Ch. 10.1 - Prob. 44E Ch. 10.1 - Prob. 45E Ch. 10.1 - Prob. 46E Ch. 10.1 - Prob. 47E Ch. 10.1 - Prob. 48E Ch. 10.1 - Prob. 49E Ch. 10.1 - Prob. 50E Ch. 10.1 - Prob. 51E Ch. 10.1 - Prob. 52E Ch. 10.1 - Prob. 53E Ch. 10.1 - Prob. 54E Ch. 10.1 - Prob. 55E Ch. 10.1 - Prob. 56E Ch. 10.1 - Prob. 57E Ch. 10.1 - Prob. 58E Ch. 10.1 - Prob. 59E Ch. 10.1 - Prob. 60E Ch. 10.1 - Prob. 61E Ch. 10.1 - Prob. 62E Ch. 10.1 - Prob. 63E Ch. 10.1 - Prob. 64E Ch. 10.1 - Prob. 65E Ch. 10.1 - Prob. 66E Ch. 10.1 - Prob. 67E Ch. 10.1 - Prob. 68E Ch. 10.1 - Prob. 69E Ch. 10.1 - Prob. 70E Ch. 10.1 - Prob. 71E Ch. 10.1 - Consider f(x) = ln (1 + x) and its third-degree...Ch. 10.1 - Prob. 73E Ch. 10.1 - Prob. 74E Ch. 10.1 - Prob. 75E Ch. 10.1 - Prob. 76E Ch. 10.1 - Prob. 77E Ch. 10.1 - Prob. 78E Ch. 10.1 - Prob. 79E Ch. 10.1 - Prob. 80E Ch. 10.1 - Prob. 81E Ch. 10.1 - Average price. 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Prob. 21E Ch. 10.4 - Prob. 22E Ch. 10.4 - Prob. 23E Ch. 10.4 - Prob. 24E Ch. 10.4 - Prob. 25E Ch. 10.4 - Prob. 26E Ch. 10.4 - Prob. 27E Ch. 10.4 - Prob. 28E Ch. 10.4 - Prob. 29E Ch. 10.4 - Prob. 30E Ch. 10.4 - Prob. 31E Ch. 10.4 - Prob. 32E Ch. 10.4 - In Problems 938, use Theorem 1 to perform the...Ch. 10.4 - Prob. 34E Ch. 10.4 - Prob. 35E Ch. 10.4 - Prob. 36E Ch. 10.4 - Prob. 37E Ch. 10.4 - Prob. 38E Ch. 10.4 - Prob. 39E Ch. 10.4 - Prob. 40E Ch. 10.4 - Prob. 41E Ch. 10.4 - Prob. 42E Ch. 10.4 - Prob. 43E Ch. 10.4 - Prob. 44E Ch. 10.4 - In Problems 4548, use the second-degree Taylor...Ch. 10.4 - Prob. 46E Ch. 10.4 - In Problems 4548, use the second-degree Taylor...Ch. 10.4 - Prob. 48E Ch. 10.4 - Prob. 49E Ch. 10.4 - Prob. 50E Ch. 10.4 - Prob. 51E Ch. 10.4 - To estimate 01.511+x2dx a student takes the first...Ch. 10.4 - There are different ways to approximate a function...Ch. 10.4 - There are different ways to approximate a function...Ch. 10.4 - In Problems 5566, use Theorem 1 to perform the...Ch. 10.4 - Prob. 56E Ch. 10.4 - Prob. 57E Ch. 10.4 - Prob. 58E Ch. 10.4 - Useful life. 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To approximate: The index of income concentration for the given Lorenz curve f ( x ) = 11 x 5 10 + x 2 with an error of no more than 0.005.

Solution Summary: The author calculates the Gini index of income concentration for the given Lorenz curve with an error of no more than 0.005.

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To approximate: The index of income concentration for the given Lorenz curve f(x)=11x510+x2 with an error of no more than 0.005.

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Chapter 10 Solutions