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: 9780137554805

Author: Raymond Barnett, Michael Ziegler

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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

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>tt 1:32 > trend.1m 1m (sales > summary(trend.1m) - tt) #3###23 (i) #### Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 2107.220 57.997 36.332e-16 *** tt -43.500 3.067 -14.18 7.72e-15 *** > trend = ts (fitted (trend.1m), start-start (sales), freq-frequency (sales)) sales trend ###23%23 (ii) #### as.numeric((1:32 %% 4) > X > q1 > q2 > q3 > 94 = = = = - as.numeric((1:32 %% 4) as.numeric((1:32 %% 4) as.numeric((1:32 %% 4) == 1) 2) == == 3) == 0) > season.lm = 1m (resid (trend.1m) 0+q1 + q2 + q3 + q4) #3##23%23 (iii) #### > summary(season.1m) Coefficients: Estimate Std. Error t value Pr(>|t|) q1 -38.41 43.27 -0.888 0.38232 92 18.80 43.27 0.435 0.66719 q3 -134.78 43.27 -3.115 0.00422 ** 94 154.38 43.27 3.568 0.00132 ** > season = ts (fitted (season.lm), start=start (sales), freq=frequency (sales)) > Y X season %23%23%23%23 (iv) #### >ar (Y, aic=FALSE, order.max=1) #23%23%23%23 (v) #### Coefficients: 1 0.5704 Order selected 1 sigma 2 estimated as 9431 > ar(Y, aic=FALSE,…

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Chapter 10 Solutions

Pearson eText for Calculus for Business, Economics, Life Sciences, and Social Sciences -- Instant Access (Pearson+)

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. 4E Ch. 10.3 - Prob. 5E Ch. 10.3 - Prob. 6E Ch. 10.3 - Prob. 7E Ch. 10.3 - Prob. 8E Ch. 10.3 - Prob. 9E Ch. 10.3 - Prob. 10E Ch. 10.3 - Prob. 11E Ch. 10.3 - Prob. 12E Ch. 10.3 - Prob. 13E Ch. 10.3 - Prob. 14E Ch. 10.3 - Prob. 15E Ch. 10.3 - Prob. 16E Ch. 10.3 - Solve the problems by performing operations on the...Ch. 10.3 - Prob. 18E Ch. 10.3 - Prob. 19E Ch. 10.3 - Prob. 20E Ch. 10.3 - Prob. 21E Ch. 10.3 - Prob. 22E Ch. 10.3 - Prob. 23E Ch. 10.3 - Prob. 24E Ch. 10.3 - Prob. 25E Ch. 10.3 - Prob. 26E Ch. 10.3 - Prob. 27E Ch. 10.3 - Prob. 28E Ch. 10.3 - Prob. 29E Ch. 10.3 - Prob. 30E Ch. 10.3 - Prob. 31E Ch. 10.3 - Prob. 32E Ch. 10.3 - Prob. 33E Ch. 10.3 - Find the Taylor series at 0 for (A) f(x)=x1x2 (B)...Ch. 10.3 - Prob. 35E Ch. 10.3 - If f(x) satisfies f(x) = ln (1 + x2) and f(0) = 1,...Ch. 10.3 - Prob. 37E Ch. 10.3 - Prob. 38E Ch. 10.3 - Prob. 39E Ch. 10.3 - Prob. 40E Ch. 10.3 - Prob. 41E Ch. 10.3 - Prob. 42E Ch. 10.3 - Prob. 43E Ch. 10.3 - Prob. 44E Ch. 10.3 - Prob. 45E Ch. 10.3 - 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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. A computer store rents time on...Ch. 10.4 - Average price. Given the demand equation...Ch. 10.4 - Temperature. The temperature (in degrees Celsius)...Ch. 10.4 - Temperature. Repeat Problem 61 for...Ch. 10.4 - Prob. 63E Ch. 10.4 - Prob. 64E Ch. 10.4 - Prob. 65E Ch. 10.4 - Prob. 66E Ch. 10 - Prob. 1RE Ch. 10 - Prob. 2RE Ch. 10 - Prob. 3RE Ch. 10 - Prob. 4RE Ch. 10 - Prob. 5RE Ch. 10 - Prob. 6RE Ch. 10 - Prob. 7RE Ch. 10 - Use Theorem 1 of Section 10.2 to find the interval...Ch. 10 - Prob. 9RE Ch. 10 - Prob. 10RE Ch. 10 - In Problems 10 and 11, use the formula an =...Ch. 10 - Prob. 12RE Ch. 10 - Prob. 13RE Ch. 10 - Prob. 14RE Ch. 10 - Prob. 15RE Ch. 10 - Prob. 16RE Ch. 10 - Prob. 17RE Ch. 10 - Prob. 18RE Ch. 10 - Prob. 19RE Ch. 10 - Prob. 20RE Ch. 10 - Prob. 21RE Ch. 10 - Prob. 22RE Ch. 10 - Prob. 23RE Ch. 10 - Prob. 24RE Ch. 10 - In Problems 25 and 26, use the second-degree...Ch. 10 - Prob. 26RE Ch. 10 - Prob. 27RE Ch. 10 - In Problems 27 and 28, use a Taylor polynomial at...Ch. 10 - Prob. 29RE Ch. 10 - Prob. 30RE Ch. 10 - Prob. 31RE Ch. 10 - Prob. 32RE Ch. 10 - Prob. 33RE Ch. 10 - Prob. 34RE Ch. 10 - Prob. 35RE Ch. 10 - Prob. 36RE Ch. 10 - Prob. 37RE Ch. 10 - Prob. 38RE Ch. 10 - Prob. 39RE Ch. 10 - Prob. 40RE Ch. 10 - Medicine. The rate of healing for a skin wound (in...Ch. 10 - Prob. 42RE Ch. 10 - Prob. 43RE

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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