Economics 307: Intermediate Macroeconomic Theory A Brief Mathematical Primer

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1 Economics 07: Intermediate Macroeconomic Theory A Brief Mathematical Primer Calculus: Much of economics is based upon mathematical models that attempt to describe various economic relationships. You have already seen a few in previous classes like the Keynesian cross and cost functions in the theory of the firm. Many of these models can be easily and more completely analyzed with a few tools from calculus than the cumbersome mathematics of introductory courses. This primer is designed to develop these tools and provide some eamples of their use. The fundamental tool that we will use is known as the derivative. The derivative allows one to easily calculate the slope of a mathematical function. There are many applications of this technique in economics but before discussing these applications it is important to first understand how to calculate the derivative. Consider the mathematical function y f(). This is a fancy way of saying that the variable y depends upon through some mathematical form. It is not to be read "y equals f times." We could be more eplicit and state eactly what the mathematical form is, such as: y + or y.5 45 or gross domestic product is a function of capital and labor [GDP f(capital, labor)]. Let us pick a simple form and use this to introduce the derivative. Consider: y f() In algebra, you learned how to graph this function: y We may be interested in finding the slope at any point on the plot of Remember the slope is given by Rise Run y. For instance, we could find the slope of our function by picking two points and plugging into the formula. This is a cumbersome way to find the slope and not entirely accurate. Instead, we will use what is known as the derivative. To find the derivative It is not accurate because there is a different slope at every point along this curve. When one finds a slope based upon two points, one is actually taking the average slope between those points rather than the slope at either one.

2 we take our function and multiply the coefficient in front of the by the eponent on that and subtract one from the eponent. Here goes: The derivative of - +4 is equal to -+4. How did I get this? Look at - first. In this case the coefficient is equal to negative one and the eponent is equal to. Negative one times gives me negative two (or the new coefficient on my derivative). To find the new eponent, I subtract one from the original eponent of two and get one. The derivative of - is then equal to - -. What about the 4 term? The same rules apply but this time the eponent is equal to one and so the coefficient remains four (one times four). Where did the go? Remember, we always subtract one from the eponent so the derivative of 4 is equal to Since any number to the zero power is simply equal to, Mathematicians will write what we just did in a succinct manner: ( + 4) + 4 d. The funny looking fraction with the d s in it just signifies that one is d going to take a derivative. Mathematicians read this as the derivative with respect to of negative squared plus four equals negative two plus four. So, how is this used? Well, we now have a simple method of calculating the slope of our original function What is the slope when? We just plug into our derivative, - +4, and find the answer: -*+4. The slope at is two (an increase in by one unit leads to an increase in y by two units at ). Note that this is upward sloping (since the slope is positive). What is the slope when 4? We can do the same thing: -*(4)+4-4. Note in this case the slope is downward sloping: eactly what we epect given our graph. Practice: d d +. ( ). d 0 ( ) d d. d (answers follow at the end of this primer). Finally, the derivative allows one to find the maimum (or minimum) of a function. Look at our graph of It seems like the highest point on the graph (the maimum) occurs when. Note, that at the maimum, the slope of the curve must be equal to zero. Think about the very top of a mountain: it is neither sloping down or up for if it was, it wouldn't be the top! So, if the maimum occurs when the slope is equal to zero, the derivative allows us a way to always locate the maimum: set the derivative equal to zero and solve for. For instance, we know the derivative of - +4 is -+4 so the maimum must occur at the that solves: With a little algebra, you may find that indeed the maimum occurs at. Practice finding the maimum (or minimum) on the following functions.

3 4. f() f() f() + So, what does this have to do with economics? Let's consider a pair of models you encountered in Economics 0 and 0: The Keynesian Cross Remember the equations that describe the relationship between consumption and income? Keynes thought they looked something like: C Y where C stands for consumption, Y represents income, and 00 is the autonomous consumption. A mathematician would say this is a special case of C f(y) (or Consumption is a function of income). You were probably introduced to the idea of the marginal propensity to consume or the MPC. If you remember, the marginal propensity to consume describes the relationship between how much etra consumption occurs given an etra unit of income. Mathematically, you probably learned MPC C. Notice, this is the same equation for the slope of a line in (,y) Y coordinates. Thus, since the marginal propensity to consume is simply the slope of the consumption function, the marginal propensity to consume is also simply the derivative of the consumption function! In the above eample, the MPC is simply the derivative of the consumption function with respect to Y: dc ( Y). 5 dy The slope of the consumption function is.5 indicating that an increase of income of $ leads to an increase in consumption of $.50! It turns out that any time you hear the word marginal, you should begin to think about the derivative. For instance, whenever you talked about marginal cost or marginal revenue in Economics 0, you were talking about the derivative of the cost and revenue curves. Just like when we talk of the marginal propensity to consume, we are talking about the derivative of the consumption curve. The Profit Function Economists believe that a firm always wants to maimize its profit. In Economics 0, you discussed profit functions and found they were determined by the price of the good sold and the price of the inputs the firm needs to produce its product. Perhaps you saw a profit function represented as: Profit P*Q - w*l where P is the price of the good sold, Q is the total amount of goods sold, w is the wage the firm needs to pay its workers, and L is the number of workers the firm hires. You may have also learned that the price of the firm's good depends upon the number of goods sold (remember the Law of Demand states that as the price increases the number of goods sold decrease). Let's simplify the problem a little and assume that the demand curve is given by P 0 - Q, the wage that is needed to be paid is w, and for each good produced, the firm needs to hire one workers so L Q. We can combine these equations and get the profit for a firm as being: Profit (0-Q)*Q-*Q or Profit -Q + 8Q. The firm would like

4 to know how many goods to produce in order to maimize its profits. In other words, we need to find the maimum of the profit function. Recall, we do that by finding the derivative and setting it equal to zero. The derivative of our profit function is: -Q+8 and setting this equal to zero, we find Q 4. The firm makes the most profit by producing four units of output. How much would it profit? Simply plug Q 4 into the profit function and find: Profit (0-4)*4 - *4. Answers: d 4. ( 4 + 8) 8 setting the derivative 8 equal to zero gives 8 0. This occurs only when d 0. In fact, you ve found the minimum of this function. It occurs when 0; something that is easy to see upon graphing d ( 4 8) d Setting this equal to zero gives Subtracting 8 from both sides and dividing by 8 gives -. Once again you ve found a minimum that occurs when -.. You should recognize this as an equation for a line. As all lines have a constant slope, one hopes the derivative of this function is a constant. In fact, the derivative of this IS a constant. d ( + ). This is similar to using derivatives to find the MPC. d Algebra: A number of students have had trouble in class before because they have forgotten some common algebra rules. Most of these rules involve the use of eponents; this section is intended to remind you of the correct use of eponents. Typical 07 problems that use eponents include multiplying and dividing numbers with eponents. For instance: 4 or 4 /. The rules dealing with eponents are that eponents, when pertaining to the same base number, are either added or subtracted when depending on if the base numbers are multiplied or divided. When the base numbers are multiplied, as in the case of 4, the eponents are added without a change of the base numbers When the base numbers are divided, as in the case of 4 /, the eponents are subtracted. 4 /. Another frequently forgotten procedure includes how to deal with negative eponents. A negative eponent indicates nothing more than an inverse. In other words, - is nothing more than (/) and - is equal to (/ /9). Practice with a few eponent problems:

5 y y Answers: Because (/) + (/4) 5/,. 8. This problem has no way of easily simplifying because the base numbers differ and the eponents differ. The only way to simplify is to actually take (9) and (8) and multiply them together (7) y. y.. We will use these procedures frequently when attempting to solve algebraic formulae that contain eponents. The most frequent type of problem we will encounter are of the form:. To solve this problem for, one must follow the common rules of algebra, the first being to get all of the common unknowns (in this case ) to one side. I do this by dividing from both sides and getting:. Since we have just seen a way to simplify the left-hand side of this equation, it is straightforward to find that. Here are some more for you to practice: Answers:.. Since we are solving for (and not to the one-half), we need to get rid of the one-half power eponent. To do this, we will square both sides of the equation: 5. What is? This is simply another

6 multiplication problem as above: +. The answer to this problem therefore is... Like above, we are solving for, and not to the three-halfs. How do we get rid of this eponent? Well, in the answer to, we squared both sides of the equation and found that we ended up with. Why did we square? Because we wanted to eliminate the one-half power and squaring did the job. It turns out, that we always want to take the inverse of the power we want to eliminate. In this case the power we want to eliminate is three-halfs so I will take the two-thirds power of both sides:. Now it turns out 0. but what is? It turns out that we multiply eponents when we have an eponent to an eponent. In this case. In this problem,