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by: Mae Koelpin

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# Macro Econ Theory ECO 301

Marketplace > University of Miami > Economcs > ECO 301 > Macro Econ Theory
Mae Koelpin
UM
GPA 3.94

S. Akin

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COURSE
PROF.
S. Akin
TYPE
Class Notes
PAGES
12
WORDS
KARMA
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This 12 page Class Notes was uploaded by Mae Koelpin on Thursday September 17, 2015. The Class Notes belongs to ECO 301 at University of Miami taught by S. Akin in Fall. Since its upload, it has received 15 views. For similar materials see /class/205756/eco-301-university-of-miami in Economcs at University of Miami.

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Date Created: 09/17/15
Dr S Nuray Akin ECO 301 University of Miami Macroeconomic Theory Spring 2007 MATH REVIEW The notes I provide here are not meant to be a complete review of the algebra you have learned so far Rather my aim is to provide you examples of some very important algebraic de nitions and rules that we will make use of throughout our class For interested students the two references below are excellent sources of further information Chiang A Fundamental Methods of Mathematical Economics Third Edition McGraw Hill Publishers Takayama A Mathematical Economics Second Edition Cambridge University Press 1985 Percentage change in a variable Suppose that we would like to calculate how much GDP per capita grew in the US in the last 20 years If GDP per capita was 22000 in 1985 and 36000 in 2005 then the growth would be 14 000 36000 22000 X100 2 22 000 22 000 X 100 63 We can generalize this formula as follows let x7 be the beginning value and c1 be the ending value of variable x Then the percentage change in the value of x is ending value beginning value gtlt100 x1 x0 gtlt100 beginning value x7 Working with exponentials Suppose that we have four real numbers x y n and m Below are the rules that we follow when we work with powers 4 xquotyquot xyquot n 5 If xquot y39quot thenxym and yxquot39quot 6xmy1 xmn x2 x4 x24 x6 2 2y 2 5 x2 2x5 2 53 125 2 l 1 l 23y23y23 25 x23 x23 x6 4332 4332 492 Linear functions and their graphs The equation ofa linear function is given by y ax b Here a is called the slope and b is called the yaXis intercept To draw the graph of this function we determine the points at which the line crosses the X and yaxes and the connect those two points Let s see an example Example Suppose that the equation of our linear function is y 2x 5 To nd the point at which the line crosses the yaxis we set x0 This gives y2055 Hence the y intercept is 5 Then we nd the point at which the line crosses the xaxis To do this we set y0 Hence 02x5 which means that our xintercept is x52 We nish by connecting those two points with a straight line 52 7 Solving one equation in one unknown If in an in equation there is only one unknown variable there are a couple of strategies we can use to solve for the value of the unknown Let s see some examples 1 Solving Quadratic Equations Quadratic equations have the general form axz bx c 0 where a b and c are real numbers In order to solve quadratics there is a simple strategy we divide 0 into two factors c1 and c2 and a into two factors a1 and a such that ach a2 01 b Examples 1 x2 4x 4 0 1x 2 1x 2 In the above case al b4 and c4 We divided al into two factors 1 and 1 We divided c4 into two factors 2 and 2 Then the above equation can be written as x 2x 2 0 After factorization we equate each component to zero to nd the value of X x 2 0 3 x 2 2x2 4x 4 0 1x 2 1x 2 In the above case al b4 and c4 We divided al into two factors 1 and 1 We divided c4 into two factors 2 and 2 Then the above equation can be written as x 2x 2 0 After factorization we equate each component to zero to find the value of X x 2 0 3 x 2 3 2x2 8x60 2x 2 x 3 In the above case a2 b8 and c6 We divided a2 into two factors 2 and 1 We divided c6 into two factors 2 and 3 Then the above equation can be written as 2x 2x 3 0 After factorization we equate each component to zero to find the value of X 2x 2032x23x1 x 3 0 3 x 3 4 5x2 7x 6 0 x 2 5x 3 In the above case a5 b7 and c6 We divided a5 into two factors 5 and 1 We divided c6 into two factors 2 and 3 Then the above equation can be written as x 25x 3 0 After factorization we equate each component to zero to find the value of X x203 x 2 5x 303 x 35 II Cross Multiplication When the equality to be solved involves a fraction we use cross multiplication ie we multiply denominators and numerators of opposite sides of the equality Let s see eXamples 1 15 3132x5332x1033 102x3 72x3x 72 x x2 1 x5 1 1 2 523x223xj3x26x53 73 x33x 3x 6 x 6 3 2 l 41 1x2 4x34 x2 4x3x2 4x40 x 4x 4 x 2 x 2 3x2x20 3x 2 4 2x71 32x 12xx734x 2x2 7x3x27x 4x20 x 3x23x20 x 2 x 1 3x2x10 x20 or x1 3x 2 or X1 Derivatives of functions of one variable First Order Derivative Let f R gt R g R gt R and h R gt R be functions of one variable x where R d f 96 39 denotes the set of real numbers We denote the rst derivative of f by T or f x The following rules will be the useful throughout this class 1 The derivative of a constant is zero Example fx 5 Thenf39x 0 2 Iffx ax where a is a constant thenf39x a 3 Iffx xquot thenf39x rm 7 is a constant 4 Iffx lnx thenf39x1x 5 Iffx lngx thenf39x g39xgx Examples fx ln5x f39x 55x lx fx lnx2 3 f39x 2xx2 2x fx ln5x 1 2 f39x 55x 1 6 Summation rule If fx gx hx thenf39x g39x h39x Example fx x3 2x thenf39x 3x2 2 Similarly fx gx hx thenf39x g39x h39x 7 Iffx gx thenf x ngx 1 g39x Example fx 5 2x Then f39x 45 2x3 2 85 2x3 8 The Product Rule If fx gxhx thenf39x g39xhx gxh39x Example fx 2x lx 5 Then f39x 2x 5 2x l 5 2x 10 10x 5 8x15 9 The Quotient Rule fx thenf39x hx hx2 Example fx 2xl x 5 2x 5 2xl 5 x 52 2x 1010x5 x 52 12x5 x sr Then f 39x Second Order Derivative Second order derivative is the derivative of the first derivative So in order to find the second derivative of a function of one variable you must nd the first derivative then deOC quot 2 or f x take its derivative The second derivative is denoted by d x Examples 1 fx x3 2x Thenf39x 3x2 2 andfquotx 6x 2 fx In x Thenf39x l x and fquotx lx2 Using the quotient rule 3 fx 5 2904 Thenf39x 45 2x3 2 85 2x3 andquot96 835 2x2 2 485 2x2 Partial Derivatives of Functions of TwoVariables Let f R2 gt R be a function of two variables x and y Here I will review f1rst and secondorder partial derivatives off First order partial derivative The firstorder partial derivative of f with respect to x is denotes by afay or Similarly the partial derivative of f with respect to y is denoted by w y orfy When taking the partial derivative with respect to x we treat y like a constant We do the reverse if we want the partial derivative with respect to x Below are some examples Examples 1 fxy2x3y Then f 2andfy 3 2 fxyxy Then f yandfy x 3 fxyc 2gy2Then f 3x2 2y andfy 2x2y Optimiz ation I Unconstrained Optimization Suppose that we would like to nd the points at which a function reaches its maXima Such points are referred to as maximizers The way we nd maximizers of functions are slightly different depending on whether we have a function of one variable or two variables Maximizers of a Function of One Variable Suppose that f R gt R is a function of one variable x For a point 97 to be a maXimizer off the following suf cient conditions have to be satis ed 1 First order condition FOC f 39xt7 0 2 Second order condition SOC fquotx0 lt 0 Maximizers of a Function of Two Variables Suppose that f R2 gt R is a function oftwo variables x and y For a point 97 y7 to be a maXimizer off the following suf cient conditions have to be satis ed 1 x0y013x0y00 2 xy 13yx0ylt0 and xx0yfyyxgay yx0yjxxgayggt0 II Constrained Optimization Suppose that we would like to solve the problem of a utility maximizing individual who consumes two goods x and y That is we would like to nd the optimal consumption bundle X and y subject to the budget constraint As an example suppose that our consumer s utility function is given by U x y x y The price of good x is 2 and the price of goody is 4 The consumer has 10 dollars to spend Then our problem is maXU x y subject to the Budget Constraint Ivy max xy subject to 2x 4y 101 Ivy To solve such a problem we will use the Substitution Method The Substitution Method Step 1 Find y in terms of x from the budget constraint Step 2 Substitute the value of y in the utility function to write down the utility in terms of x only Step 3 Take the first derivative of the Utility function with respect to x and equate it to zero This will give you the value of x To nd y make use the relationship between x and y that you found in Step 1 Example Let s find the optimal consumption bundle for the consumer in our example above by following each step of the substitution method Step 1 The budget constraint is 2x 4y 10 So y 52 x2 Step 2 Substitute the value of y in the utility function U x x5 2 x 2 Step 3 The derivative with respect to x is U39x 52 x I used the product rule to find the derivative Equating the first derivative to zero will give us the optimal consumption of x for this consumer U39x 0 cgt 52 x 0 Hence 6 52 To f1ndywe use y 52 x2 from Step 1 Hence y 52 54 54 Step 4 Make sure that the solution is indeed a maximizer ie check the second order condition SOC SOC tells us to take the second derivative of U x and evaluate it at x52 Ifthe answer is less than zero x52 is a maximizer Uquotx 1 So Uquot52 l lt 0 This guarantees that our solution is a maximizer Example Find the optimal quantities of textbooks T and movies M for a student whose preferences for the two goods are represented by the utility function UMT1nM1nT 1 The budget constraint of a consumer who wants to allocate his total income I between the two goods x andy is with pricespX andpy is pix pyy Suppose a movie is 2 dollars and a textbook is 1 dollar The consumer has 4 dollars of income Solution The student s problem is maX lnMlnT subject to lT 2M 4 MT We follow each step of the substitution method Step 1 The budget constraint is T 2M 4 So T 4 2M Step 2 Substitute the value of T in the utility function U M lnM ln4 2M Step 3 The derivative with respect toM is U39M L 2 M 4 2M Equating the rst derivative to zero will give us the optimal consumption of x for this consumer U39M0cgti 2 0Hencei 2 24 2M2MM1 M 4 2M M 4 2M To nd T we use T 4 2M from Step 1 Hence T 4 21 2 Step 4 Check the SOC to guarantee a maximizer UquotM0cgt 12 222 12 4 2 M 4 2M M 4 2M 1 4 Hence Uquot1 2 2112lt0 1 4 21 Our solution is a maximizer Note When we logutility ie UXy lnxlny the SOC will always be satis ed Hence we may skip step 4 The Problem of a Consumer who Chooses between Consumption and Leisure One of the interesting questions in macroeconomics is unemployment and the labor supply of individuals Note that we as consumers make the decision on how much to work how much time to spend on leisurely activities and how many goods to buy with the wage we earn by working So how can we nd the optimal balance of work leisure and consumption for an individual Suppose that a consumer has to choose how many goods to consume C and how much time to spend on leisure each day We are interested in the allocation of time between work and leisure in one day Hence if the individual works L fraction of the time lL will be the fraction of the day spent on leisure Assume that this individual s preferences over consumption and leisure are given by UCL lnC 1nl L H J utility from consumption utility from leisure The price of one consumption good is p dollars and the wage rate per unit of time worked is w dollars Before formalizing and solving this problem let s first determine the budget constraint in this case Remember the budget constraint always tells us that total expenditure on consumption goods is equal to total income In this example when the consumer consumes C units of the good at per unit price p his total consumption expenditure would be p C Similarly if he works L fraction of the time earning wage w per unit of time worked his total income would be wL So we can write the budget constraint as pC wL Now we can formalize the problem 21310 lnC lnl L subject to pC wL This is a constrained optimization problem So we apply the steps of the substitution method Step 1 Use the budget constraint to eliminate C C 2L P Step 2 Substitute for C in the utility function max In 2L lnl L Lgt0 p Step 3 Write down the first order condition Take the first order derivative with respect to L and equate it to zero here w and p are treated as constants since we take the derivative with respect to variable L Hm LLmsi L i LcbLLEU2 l L L l L L l L El hlrslg Hence this person spends half of the day working ie he works for 12 hours a day To find how many goods he consumes use C 2L P E L p 2

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