ε and ε > 0 we can find a δ > 0 such that

Transcription

1 John Riley June 5, 3 ANSWERS TO EXERCISES IN APPENDIX A SECTION A: MAPPINGS OF A SINGLE VARIABLE Eercise A-: Rules of limits (a) Limit of the sum = the sum of the limits We wish to estalish that for any ε > there eists δ > such that L + L ε < f ( ) + f ( ) < L + L + ε for all N(, δ ) () Since f i has a limit L i at, for any ˆ ε > there eists δ i > such that L ˆ ε < f ( ) < L + ˆ ε, for all N(, δ i ) i i i Choose ˆε = ε and δ = Min{ δ, δ} Then and L f L N ε < ( ) < + ε, (, δ) L ε < f < L + ε N δ ( ), (, ) Comining these inequalities it follows that () holds () Limit of the roduct = the roduct of the limits For simlicity assume oth L and L are strictly ositive Since oth functions have a limit at, it follows that for any ε and ε > we can find a δ > such that Then L ε < f < L + ε N δ i i i( ) i i, (, ) f f < ( L + ε)( L + ε ) = LL + Lε + L ε + εε () We need to show that for any ε > we can find a δ > such that LL ε < f( ) f ( ) < LL + ε for all ε Choose ε = Then L ε Lε + L ε + εε = ( L + ) ε + ε L N(, δ ) Answers to eercises in Aendi A age

2 John Riley June 5, 3 ε Finally choose ε to satisfy ( L+ ) ε = ε Then Lε+ L ε+ εε = ε and so, from L inequality (), f( ) f( ) < LL + ε for all N δ (, ) A similar argument estalishes the lower ound for the roduct SECTION A3: DERIVATIVES AND INTEGRALS Eercise A3-: Rules of Differentiation () If the functions f and f have limits L and L at LL Thus the limit of, then the roduct has a limit of g ( + h) g ( ) dg f( + h) is f( ) ( ) and the limit of h ( ) ( ) ( ) f + h g f df is g ( ) ( ) Since the limit of a sum is the sum of the limits it h d d df dg follows that y( ) = fg = g + f (c) To rove the Chain Rule note that as h aroaches zero so must k Then aeal to the roduct rule for limits Eercise A3-: Discontinuously Differentiale Function (a) Alying the Product Rule df d = (sin ) + sin The derivative of sin y is cos y and the derivative of is Alying the Chain d Rule, sin = cos Then df = (sin ) cos () Note that (sin ) 4 Thus the first term is continuous at = However the second term oscillates etween - and more and more raidly as aroaches zero An asterisk () indicates a somewhat harder question This articular question is designed for students with strong mathematical ackgrounds Answers to eercises in Aendi A age

3 John Riley June 5, 3 (c) Since the range of the sin function is [,], f( ) and f( ) Thus the function is continuous at = and f () = Then Since f( ) f() = (sin ) sin, it follows that f( ) f() Thus the function f has a derivative of zero at = From art () the derivative is less than - at oints aritrarily close to so the function is not continuously differentiale at = Eercise A3-3: Integrating y arts (a) d ln = (ln )() = (ln ) ( ln ) () = ln ln = ln ln ( ) γ + γ d + γ (ln )( ) (ln )( ) ( ln )( ) = + γ + γ + γ + γ γ ( ) ln ( ) ln = + γ + γ + γ + γ + γ + γ + γ = ( )( ln ln ) ( ) + γ ( + γ) (c) ( )( e ) = e () e = e e e + e Answers to eercises in Aendi A age 3

4 John Riley June 5, 3 SECTION A4 OPTIMIZATION Eercise A4-: First Order Conditions Consider Ma{ f ( ) = + } Differentiating f, df = If, then the sloe is negative for all >, thus f is maimized at = If > the sloe is ositive at = and reaches zero at = Thus if <, the solution is = and if > the solution is = Alternatively, if you are already comfortale with the Lagrange method, form the Lagrangian, FOC L= + + [ ] dl = with equality if > dl = with equality if d > Suose = Then from the first condition, For oth of these statements to hold, From the second condition = Suose net that < < From the second condition = Then from the first condition = Finally suose = From the first condition = Thus Which answer is more intuitive? Surely it is the first Eercise A4- : Consumer Choice I (a) The rolem to e analyzed is Ma{ f ( ) = B( ) + I } Answers to eercises in Aendi A age 4

5 John Riley June 5, 3 df db Differentiating the maimand, ( ) = ( ) Let e the maimizing value of (i) If df db = then ( ) () = I (ii) If < <, the solution is interior the sloe must e zero hence df db ( ) = ( ) = (iii) Arguing as in case (i), if the consumer urchases only commodity then db I ( ) () If B ( ) = ln, db = Note that case (i) is imossile since the derivative aroaches infinity as aroaches zero Case (iii) occurs if solution is interior and = so = If B ( ) = ln(5 + ), Case (i) occurs if db I ( ) =, that is, if I For all larger incomes the I db = 5 + db () = = Arguing as efore, case (iii) occurs if 5 db I ( ) =, that is, if I If these inequalities do not hold the solution I is interior Then db = = 5 + and so 5( ) = Answers to eercises in Aendi A age 5

6 John Riley June 5, 3 Eercise A4-3: Roinson Crusoe (a) Roinson s investment cannot eceed the endowment thus z ω Since = ω z and y = az, Roinson s utility is uz ( ) = + ln y= ln a+ ω z+ ln z du () Differentiating y z, = + This is ositive as long as z < Thus if dz z < ω the solution is interior and z = If ω Roinson invests his entire endowment in eriod A du (c) We have already seen that ( z ) > for sufficiently small z, thus investment is dz strictly ositive Eercise A4-4: Elasticities (a) ε ( a y) y d( a) y a y, = = = = ε(, y) a d( y) a dy dy () ε(, ) = d ln = d ln d ln y = ε(, ) ε( y, ) y d y d d y z dy z dy y dz dz (c) ε ( y, ) ε ( yz, ) = = = ε ( z, ) dln y dln y dln y = = = = d (d) ε y, ε ( y, ) (e) Using last result d ln y dln y y y ε, = ε, = = = ε ( y, ) Answers to eercises in Aendi A age 6

7 John Riley June 5, 3 SECTION A5 SUFFICIENT CONDITIONS FOR A MAXIMUM Eercise A5-: Profit maimizing firm (a) The rofit function is f ( ) = w hence df d f = 5 w and 5 3 = < It follows that the rofit function is concave () Note that the sloe of the rofit function increases without ound as aroaches df zero Thus the solution must e interior From the first order condition ( ) = it follows that 5 ( ) = wand hence laor demand, The firm s suly of outut is q = ( ) = w / 5 5 = ( ) w Eercise A5-: Cost minimizing inuts (a) Since q = 4K + L, the caital requirements are K q L = ( ) Since caital cannot 4 e negative () ( q L ) and hence L q Total cost is then 4 r C = rk + wl = ( q L ) + wl 4 dc r L w = + and dl 8 r 3 = L > dc Thus total cost is a conve function of L () If the solution is interior, the first order condition, dl 6 dc r r = L + w= L = ( ) dl 8 8w (c) We need to check that the oundary condition is satisfied We require that L q, that is r q 8w Answers to eercises in Aendi A age 7

8 John Riley June 5, 3 If this condition is violated K = and so q L = Hence L = q Eercise A5-3: Proerties of Concave Functions (a) Define h ( ) = f( ) + f( ) where the two function on the right hand side are concave Consider any, and any conve comination Then h ( ) = f( ) + f( ) Since f j is concave j j j f ( ) ( ) f ( ) + f ( ) Summing over j it follows that h( ) ( )( f ( ) + f ( )) + ( f ( ) + f ( )) = ( ) h ( ) + h ( ) () Define h ( ) = g( f( )) Suose that h ( ) h ( ) Since g is increasing it follows that f( ) f( ) Since f( ) is quasi-concave, for any conve comination Since g is increasing it then follows that f( ) f( ) g( f( )) g( f( )), that is, h ( ) h ( ) (c) Define h ( ) = f( g ( )) where f is an increasing concave function and g is concave Given the latter, g ( ) ( ) g ( ) + g ( ) Since f is increasing, it follows that h ( ) = f( g ( )) f(( ) g ( ) + g ( )) But f is also concave hence f(( ) g ( ) + g ( )) ( ) f( g ( )) + f( g ( )) = ( ) h ( ) + h ( ) Comining these inequalities it follows that h ( ) ( ) h ( ) + h ( ) (d) If f is concave then f( ) ( ) f( ) + f( ) = f( ) + ( f( ) f( )) It follows immediately that if f( ) f( ) then f( ) f( ) Eercise A5-4: Family of Concave Functions Answers to eercises in Aendi A age 8

9 John Riley June 5, 3 (i) d du ln = a ( ) Integrating, ln du = + k Then a k du + ( ) = e a Integrating again, + k a a U ( ) = ae + K = ak e + K where K = e k d du (ii) ln = a + du K Integrating, ln = ln( a+ ) + k = ln K ln( a+ ) = ln( ) a + du K Therefore = Integrating again, U( ) = Kln( a+ ) + K a + d du (iii) ln =, a + du Integrating, ln = ln( a + ) + ln K, hence ( a + ) Integrating again, U( ) = K + K ( ) du ( ) = K a + Eercise A5-5: Quasi-concavity (a) The grahs of the two functions elow y = and y = over the domain + are deicted Answers to eercises in Aendi A age 9

10 John Riley June 5, 3 The first is conve (increasing sloe) while the second is concave (decreasing sloe) Both are quasi-concave over + () The function f( ) is deicted elow It takes on the value 8 in the interval [,3] and is strictly increasing for all other Thus f( ) is a quasi-concave function 3 df (c) On the interval [,3] = and d f = Thus the first and second order necessary conditions hold for any oint in this interval The sufficient condition for a local d f maimum ( ( ) < ), is not satisfied in this case Answers to eercises in Aendi A age