Transcription
Hello and welcome to another lecture of Mathematics for Economics, part 1. So, we have been doing some tutorials in this series of lectures now. The tutorials have actually two purposes for the students. Number one, they help the students to solve the actual problems that the students will encounter in the tests. So, that is the practical part of the tutorial. The second purpose of doing these tutorials is that it also helps the students to review the entire syllabus, the different topics that have been covered in this course.
So, the topics that we have so far covered in our tutorials are the following: we have covered the real number system; after that, we talked about mathematical logic; and then we talked about the idea of proofs; and then we went to differentiation, the idea of limits, differentiability; and then we also talked about the idea of series and sequences, the idea of convergence; and then the last topic that we covered is about optimization. Optimization is a very important application of mathematics in economics, so we have covered the topic of optimization in our tutorials. We talked about the first-order condition and the second-order conditions, and we also talked about the practical problems of maximizing profit, for example, for a producer, for a farmer. So, those are the things that we have done.
Now, today what I propose to do is to take a further topic which we have covered in this course. So, after optimization, what we shall do today, as you can see on your screen, is this: this is the second series of tutorials, and here optimization has been done; we have done this before, and let me take you to the last problem that we have done. This was the last problem; this was about maximizing profits in two simultaneous markets; the producer is selling his product. So, in this case, how the producer decides how much good he will sell in these two markets and what will be the prices that he will charge from these two markets. This is called price discrimination that we have seen in the previous lecture. So, today we are going to start with this new topic of tutorial, which is integration.
So, let us straight away start with a problem. So, this is the problem: compute the area bounded by the function, x-axis, and the interval mentioned. So, a function is given here; this is the function in the first problem, f(x) = 2x². So, we have to compute the area bounded by this particular function, the x-axis, and the interval mentioned here in the first problem; the interval is 0 to 2. Just visualize this in terms of our diagram. So, we have this first quadrant here; along the horizontal axis, you have x, and suppose this is 0 to 2, that is the interval that we have been given. Now, here is the x-axis; here is the interval; so, just imagine two vertical lines at points x = 0 and x = 2. So, the three sides are known to us: the x-axis, this vertical line, and this vertical line; but what is bounding the area from above? So, what is bounding this required area from above is this function y = f(x) = 2x². Now, how does this function look like, 2x²? It is going to be something like this. It is going to be increasing because as x is rising, this f(x) is rising, and you can see we have x², so it is not a linear function; it is a quadratic function, in a particular quadratic form, 2x². So, it is rising and rising at an increasing rate; it is a convex function. So, we have to find out the area here; that is the problem.
Now, as we know, this can be easily found out by the application of integration. So, the main problem that we have to solve is that we have to find a function F(x) such that d/dx of F(x), that is the first derivative of F(x), will give me the small f(x), and what is small f(x)? Small f(x), as we know, it is 2x². So, we have to find out the anti-derivative of 2x². What is that function? If we take the derivative of that, it gives me 2x²; we have to find that function, so we are going in the opposite direction of derivative here; that is why we are calling it anti-derivative. So, anti-derivative, as we know, can be rephrased as indefinite integral of 2x². So, here you have the integral sign, and then the function that we have to integrate, that is called the integrand. Integrand is 2x², and you have dx; x is the variable of integration; and so, this will give me this expression here; we are simply using the formula that if you take the integration of xⁿ dx, then what you get is xⁿ⁺¹/(n+1) + c; c is the constant of integration. So, that is the formula that we have used here, and we have also noted that this 2 is constant, so if you have 2 here, so it will just get multiplied by 2. Whatever the constant is with x² or xⁿ, the result that we are getting here is going to get multiplied with that 2. So, here n = 2 itself, so the power of x is 2 here. In general, it can be n, so it will be xⁿ⁺¹, so you are getting 3, and, in the denominator, you are getting again n+1, that is 2+1 = 3. So, this is the result: 2x³/3 + c.
Now, this is the indefinite integral, but we have not yet found out the area of this shaded region; that we have to find out. Well, basically the idea is that we have to evaluate this function, that is 2x³/3, at these two numbers 0 and 2, and we have to take the difference. So, here we are doing that; remember here we are ignoring that constant part, that is c; that c actually will again get cancelled if I take the difference, so it is not making any appearance here. So, if I take the value of this function at 2, I will get 2/3 multiplied by 2³ minus, I evaluate this function at 0, that is going to give us 2/3 multiplied by 0³. So, here you are getting a 0, and ultimately, we are going to get 16/3. So, this is the answer: the area of this region, this shaded region, is 16/3.
And here is another function: f(x) = eˣ; so, it is an exponential function here, eˣ. And the question is similar, but here the interval is a little bit different; it is starting from -1. So, actually we are going to the left of the origin; so, suppose here you have 1, and here you have -1, and how this function looks like? This is a different function: 3eˣ; this function is not starting from the origin because if you put x = 0 here, then e⁰ = 1. So, at 0, the function is taking the value 3. So, it has a vertical intercept. So, the function will roughly look like this; so, what you have is this area. This is the area that we are required to estimate; the interval is from -1 to +1; the downward limit, that is the vertical limit in the bottom, is the x-axis itself; and the upward limit, that is the vertical limit on the top, is this function. So, this is the area. The method is going to be just the same as before; first, we find that F(x), which if differentiated gives me 3eˣ; and after I find f(x), then I will evaluate that F(x) at 1 and -1 and take the difference. So, this is the strategy. Now, what is the indefinite integral of this? As we can see, it is written as 3eˣ + c; so, this can be easily checked. So, what you have is 3eˣ dx. This is the indefinite integral, and again 3 is a constant, will come out; integration of eˣ dx, and what is the integration of ex? It is ex, plus there is this constant of integration. So, that is what I have written here: 3eˣ + c. The next stage is that we have to take the value of eˣ at -1 and +1 and take the difference. So, that is what we have done here: 3 multiplied by e¹ that is e, minus e⁻¹; what is e⁻¹? It is 1/e. So, 3 is taken as common; therefore, 3 multiplied by e - 1/e. Now, e, as we know, it is an irrational number, but it can be approximated by this number 2.718; we are taking up to the third decimal place, and I divide 1 by this number 2.718, and that will give me 0.368. So, I simplify this, and I get this answer: 7.05. So, that is my answer: that the area of this shaded region is a little bit over 7, 7.05. But again, this is an approximate value because at this stage we have taken the approximate value. So, this is how we apply the idea of integration to evaluate the area under a curve within a particular interval.
Here is another practical instance of application of integration. In the manufacturing of a product, the marginal cost of production is given by this: -q² + 2q - 3; the fixed cost is given as 12. Find the total cost function. So, marginal cost is given; fixed cost is given; we have to find out the total cost function. So, we start by assuming that the total cost function is given by this: c(q); so, this is the total cost function; so, if the total cost function is given to us, we know that the marginal cost can be found out by taking the derivative, first derivative; and we are also given the information. In this case, the marginal cost is -q² + 2q - 3. So, we combine these two things, and I will get c'(q) = -q² + 2q - 3. Now, our purpose is to find out what is c(q)? So, c(q) is the anti-derivative of this marginal cost function. So, therefore, we take the indefinite integral of this marginal cost; and so, that is what we are doing here: indefinite integral of -q² + 2q - 3; this should not be x; this should be q because q is the variable of integration. So, the next stage is just to find out the indefinite integral; at this stage, we are using that formula, that power rule kind of formula that we have encountered in the previous problem. So, that will give me -q³/3 + q². Why q²? Because there will be a 2 coming here, and that will get cancelled. So, we will be left with only q², and -3q + the constant of integration, which is c. So, this is the cost function: cost function c(q) = -q³/3 + q² - 3q + c. But here there is an arbitrary constant c, and we cannot leave it like that. In order to get rid of the arbitrary constant c, we used the fact that the fixed cost is 12. Now, fixed cost is 12 means what? That if the producer is not producing, he has to bear some cost, and that cost is 12. So, basically it means that at output 0, the producer's cost is 12. That is what I have written here in mathematics: c(0) = 12; and if you put q = 0 in this function, this part will become 0, and c(0) = 12; and therefore, c = 12. Therefore, c has been found out; and therefore, the cost function is c(q) = -q³/3 + q² - 3q + 12. So, this is another application of integration.
Here is another sort of application. The rate of extraction from an oil well is given by this: u(t) = at, where a > 0 is a constant; x₀ is the initial reserve of oil. What is the reserve of oil after time t? After what time will the oil reserve be completely exhausted? So, there are two parts to this question. Now, after time t, the oil that is left in the well is given by this formula, if you recall. So, let us assume that the oil that is left in the well after time t is given by x(t). So, x(t) is what? x(t) is the initial reserve; initial reserve is x₀, minus the oil that has been extracted already. How much time that has elapsed? It is t; so, from 0, 0 is the initial time, the point of origin, to the time t, how much oil has been extracted. If I subtract that from the initial reserve, I will get the oil that is left. Now, how much oil has been extracted? That is given by this. So, here comes the application of integration, and in this case, it is definite integral. So, from 0 to t, I am taking the integral, and what is the integrand? It is av, and I am integrating with v, so dv. Why av? Because this is the rate at which the extraction is taking place at. So, then the problem now becomes a simple problem of integration; the first term is x₀ - a is a constant, so it is coming out, and integration of v dv, that is simply v²/2, and there are these limits, the upper limits and the lower limits t and 0. I am writing that, and then I am evaluating this v²/2 at t and 0 and taking the difference; so, I will get x₀ - a multiplied by t²/2. So, that is what the answer is: x(t), that is the oil that is left after time t, is equal to x₀ - a(t²/2). As you can see, the first and the second derivatives of x(t) are negative, implying that the result goes on depleting at an increasing rate, at an increasing manner. So, as you can see, as t rises, this part is rising. So, x(t) is declining, and not only it is declining, it is declining in an increasing manner. So, just again imagine this in terms of a diagram: at t = 0, x(t) = x₀. So, you are starting from this point, vertical intercept, and it is declining like this. It is a concave function, but declining concave function. The second derivative is negative.
The second part was this: let the time it takes to completely exhaust the well be T. So, this was the question: how much time does it take to completely exhaust the oil well? So, that time we have to find out. In terms of this diagram, actually, we have to find out this time. So, this is the T that we are trying to find through our mathematics. So, what we do? So, in this equation, this is the equation of the oil that is left, we put the oil that is left equals to 0; that is left-hand side, suppose, is equal to 0; and at what time it is getting equal to 0? That time is T. That we have assumed; so, I put t = T, and I put the LHS to be equal to 0. So, that will give me this equation: 0 = x₀ - aT²/2. And then I just have to solve this for T; I take the positive root, and this gives me √(2x₀/a). So, as you can see, what is the intuition of this T? T rises if x₀ rises. And that is intuitive: if you have a large reserve to begin with, it will take more time to extract all the oil. So, that is why, as x₀ rises, the time it takes to exhaust the well also rises. As a rises, which is coming in the denominator, T actually declines. What is the intuition for that? The a is appearing here; that is where a is appearing. So, a is an indicator of how fast the extraction is taking place per unit of time. If a is high, that means the rate of extraction is high, and that will also mean that the exhaustion of the well will take place at a shorter point of time; that is the intuition. So, if you are extracting at a faster rate per unit of time, then it will take shorter time to completely take out the oil from the oil well. So, in that case, T will decline.
Here is a problem from macro-economics. Let K(t) be the capital stock of a country at point e. I(t) is the net investment at time t, so K̇(t) = I(t); K̇ means the derivative of the capital stock with respect to time, and this is also given to us: K(0) = K₀; that means at point 0, the stock of capital is given as K₀. So, there are two parts of this question: one, if I(t) is given, I(t) = αt + β, what is the capital stock? What is the rise in capital stock between time 0 and time T? And secondly, what is the capital stock at time t? Let us start with the first question; we have been given this particular function of the net investment: I(t) = αt + β. Now, how much will be the rise of capital stock, accumulation of capital stock between two points as t = 0 and t = T? That is what we have to find out, and that is easily found out by applying the idea of definite integral. So, we are integrating this function, the I(t) function, because we know this, that if I take the derivative of the capital stock, I get the net investment, I(t); so, here we are going the opposite direction, and the limits of integration are 0 and T. So, these are the upper and the lower limits; well, after I do the integration, what I get? I get αt²/2 + βt, but I have to take the upper and the lower limits, so those are written here, and this is coming out to be αT²/2 + βT. So, that is the answer. So, this is the rise in the capital stock between 0 and T. The second part is: what is the capital stock at time t? Well, that is found out by taking the definite indefinite integral of the investment function. So, what we are doing? We are taking the indefinite integral of αt + β. Here, remember, we are not interested in the rise of capital stock. We are just finding out what is the capital stock at a particular point of time? So, that is why we are doing the indefinite integral, and we get this result: αt²/2 + βt + c. But here there is an arbitrariness, which is c is there, and we have to get rid of that. That can be got rid of by using this information, that at t = 0, the capital stock is K₀. So, that information we use here; we put t = 0 here. So, here I will get K₀; K₀ is this, and that is equal to c because this part will just become 0. So, c, which was sort of arbitrary, is now given a value; it is equal to K₀; so, therefore, the stock of capital at any point of time t is given by this function: αt²/2 + βt + K₀.
In an economy, the marginal propensity to consume is given by this: MPC, marginal propensity to consume is given by c'(y) = 3/2y⁻½. The consumption level is 150 if y = 0. Find the consumption function? So, this is going to be a similar application of the idea that we have been using; the c'(y), that is marginal propensity to consume, is 3/2y⁻½. Now, as we know, the marginal propensity to consume is found out by taking the derivative of the consumption function, first derivative; so, the consumption function, therefore, is given by the integration of this MPC. We are taking the indefinite integral, and we just use the power rule, and I will get this form: 3√y + c; c is the constant of integration. We know that the consumption level is 150 if y = 0. So, I put y = 0 here in this form. So, on the left-hand side, I will get C₀; C₀ is 150. On the right-hand side, this part will drop out, and so I will get only c; therefore, c = 150. Thus, the consumption function is this: c(y) = 3√y + 150.
Here is a little bit, little bit difficult question compared to what we have been doing. A fund yields a return of a rupees per year in perpetuity. The rate of interest is r compounded continuously. What is the present value of this income stream? Now, the formula of present value is, if you have compounded continuously, then what is the formula? It is given by this: So, integration of 0 to infinity; why infinity? Because this return is coming in perpetuity, forever; and what is the return in each year? This is given by a; that is a, but this is in the future. So, therefore, I have to discount that. So, we are getting e⁻ʳᵗ and dt. So, here r is the rate of interest which is given to me. So, this is the present value, but this is an improper integral because the upper limit is infinity. So, I write this as this: limit t → ∞ integration 0 to t, a e⁻ʳu du. I have taken u to be the variable of integration, so as to not confuse with t. So, this is easy; so, I have to integrate a e⁻ʳu du. a can be taken out; a is a constant; and so, integration of e⁻ʳu. So, that is e⁻ʳu / -r, and t and 0 are the limits, and of course, there is a limit in the beginning; a/r can be taken out because this is a constant; so, within the limit, what you are getting is e⁰ - e⁻ʳᵗ. Remember, there is a minus sign here; that is why it is coming out as the lower limit minus the upper limit. And this part is just 1, and this part goes to 0. That is what I have written here: limit t → ∞ e⁻ʳᵗ; it is actually equal to 0. So, the second part is 0; the first part is 1; therefore, the answer is a/r. So, this is the present value of a fund which gives you a amount of money per year in perpetuity, and the interest rate is compounded continuously.
Find the present value and future value of a constant income stream of 100 rupees per year for 20 years from next year with the interest rate of 5 percent continuous annually. We have to find out two things: the present value and the future value. So, here the amount of income that we are getting annually is 100 rupees. The rate of interest is 5 percent, so it is 0.05. The time period is 20 years. It is compounded continuously, so the PDV will be this. This is the standard formula that we have just used in the previous problem also. So, here I just have to insert the values that are given to us: a = 100, r = 0.05, t = 20. So, we have put those values here. So, this is the integration that we have to do, and this again the same formula: e⁻ʳᵗ / -r. And the limits are 20 and 0. And this simplifies to, here I have a -0.05. So, that will come out to be -20, and within the brackets these are the...
Values if I take the limits and take the difference. So, I can just multiply this thing with the minus sign so it becomes 1 minus 1 divided by e, and this is approximately equal to 2000 multiplied by 1 minus 0.368. I have taken the approximate value here, and this is simplifying to be 1264 rupees; so this is the present discounted value.
Now we also have to find out the future value, not only the present value but also the future value. What do we do to get the future value? I multiply the PDV, that is the Present Discounted Value, with e to the power rt; r is known to me, 0.05, t is 20, so I just have to put those values here, and then I will simply get e multiplied by 1264. And that is approximately equal to this. So, this is the future discounted value. That actually completes our discussion of integration.
Now this was the last topic that we covered in our course; this is difference equations. So, let me take some problems from this last topic, and then I think we can call it a day. So, difference equations, here is the problem. Given the first-order difference equation, this, yt plus one minus 1/4 of yt is equal to 4, and y0 is equal to 5. Find the particular integral, complementary function, and the definite solution. Is the time path dynamically stable?
So, we start with this difference equation; this is a first-order difference equation; it is a non-homogeneous difference equation, and the degree is 1. So, first let us find out the particular integral. I will take the trial solution yt is equal to k, k is a constant, and that will give me this, k minus one-fourth of k is equal to 4, because yt is equal to k, which is a constant, so yt plus 1 is equal to yt, which is equal to k. So, that fact has been used here, and if I simplify this, k is equal to 16 divided by 3.
Second part is the complementary function. We again take this as the trial solution yt is equal to a * b to the power t and put that into this equation, only the homogeneous part. And therefore, I will get a * b to the power t plus 1 minus 1/4 of a * b to the power t is equal to 0. And this we have to solve for b, because a will get cancelled, and actually, we are going to get b minus 1/4 is equal to 0, that means b is equal to 1/4. Therefore, the complementary function, we had assumed this to be a * b to the power t, it is actually a * (1/4) to the power t because b is equal to one-fourth. Therefore, the general solution is this, yt is equal to a * (1/4) to the power t plus the particular integral, which is 16 divided by 3.
Now we use the fact that at t is equal to 0, y is equal to 5; y0 is equal to 5 means if you put t is equal to 0, y is equal to 5. So, that is what we have done here; we have put t is equal to 0, so that will give me 5 is equal to a plus 16 divided by 3, or a is equal to 5 minus 16 divided by 3, that is minus one-third. So, I put this in this form, and I will get this, yt is equal to minus one-third multiplied by (1/4) to the power t plus 16 divided by 3.
Last part was whether the time path is dynamically stable. Now in this case, b is 1/4, and 1/4 is less than 1, in absolute terms, and that basically means that the path is dynamically stable; it will converge to the particular integral 16 divided by 3 over time. So, that is the answer.
Here is another application of difference equations in the Cobweb Model. In a Cobweb Model, the demand and supply functions are as follows: this is the demand function, ydt is equal to 60, 18 minus 2pt; p is the price, q is the quantity, d stands for demand. And the second is the supply function, qst is equal to minus 3 plus pt minus 1. As you can see that here I have pt minus 1; there is a lag of one period. What is the inter-temporal equilibrium price, and is the equilibrium stable? So, these are the two questions.
Now we compared these functions, demand and supply functions, with the standard model; the standard model was this: this was the demand function, qdt is equal to alpha minus beta pt, and the supply function was qst is equal to minus gamma plus delta pt minus 1. So, we can immediately see what are the alpha, beta, gamma, delta in this particular set of equations. Now we know that for the standard model, the general solution or the time path of price is given by this, pt is equal to p0 minus (alpha plus gamma)/(beta plus delta) multiplied by minus delta/beta to the power t plus (alpha plus gamma)/(beta plus delta), where this is the particular integral or the inter-temporal equilibrium price. So, that we can find out immediately by putting the alpha and gamma, beta and delta here. Alpha here is what? Alpha is 18, 18, then gamma, gamma in this case is 3, so 3 divided by beta plus delta; beta is 2, 2 and delta is 1, so 1. So, this will simplify as 21 divided by 3, which is 7. So, the particular integral has been found out.
What about the complementary function? In this case, this b is very important; this is the b term, this b here is minus delta/beta; minus delta/beta as we have just seen, minus delta/beta will be minus of 1 divided by 2 because delta is this 1, 1 and beta is 2, 2. So, 2 is coming here, so it is minus 1/2; the b term here is minus 1/2. So, the time path is in short given by this, pt is equal to p0 minus 7 multiplied by (-1/2) to the power t plus 7. P0 remember is not known to us; it is a constant; it is the initial price, so unless we have some information of what the initial price was at time is equal to t is equal to 0, we cannot say anything about P0. So, but that does not hinder us from saying something about the stability, because the stability depends on this. The absolute value of minus 1/2 is actually 1/2, which is less than 1, therefore, the equilibrium is stable. So, the stability is there, and as we know in all Cobweb Models, this b term is always negative as long as the demand function is downward sloping, supply function is upward rising; the b will always be negative, that means there will be a price oscillation over time. So, oscillation is going to be there, but as we can see the value of b is less than 1 in absolute terms, so therefore, the equilibrium is stable.
In a market model with inventory, the demand, supply, and price adjustment functions are given as follows: find the time path of pt and comment on its stability. So, this is the other application of first-order difference equation that we discussed in this course; this is called the market model with inventory. Demand function is given, qdt is equal to 18 minus 2pt; the supply function is qst is equal to minus 6 plus pt; mind you, unlike the Cobweb Model, there is no lag here as far as the supply function is concerned. And finally, there is a price adjustment function, pt plus 1 is equal to pt minus 0.2 multiplied by qst minus qdt.
So, just to recall, to understand the logic of this model, let us refresh our memory. In the inventory model, there is no automatic clearance in the market. So, that was there in the Cobweb Model, that what happens to the price depends on the market demand and market supply, and the price adjusts automatically according to whether there is excess demand or excess supply in the market. That was determined by the market mechanism; here the market is not competitive in this model, the inventory model. Here the producers actually have some discretion over the price; they control the price, and the price adjustment they do is given by this. So, price in the next period, that is pt plus 1, will depend on the price of this period, but it also depends on the access supply in this period. So, this is the excess supply, qst minus qdt, and this is getting multiplied with minus 0.2. So, if you have excess supply in this period, then in the next period the price is going to be adjusted downwards compared to this period's price, and vice versa; if you have excess demand in this period, this will be negative, and negative minus negative is positive, so in the next period the price will be higher than this period's price. So, there is an element of time involved here; that is why we have difference equations to take care of that.
Again, what we are going to do in this problem is to compare this set of equations with the standard model. Well, what was the standard model? Standard model was qdt is equal to alpha minus beta pt, qst is equal to minus gamma plus delta pt, and Pt plus 1 is equal to pt minus sigma * (qst minus qdt). These were the notations, and we can just compare these equations or this functions with these equations and functions, and we can figure out what are the alpha, beta, gamma, delta, sigma in this particular given set of functions. The time path in the standard model was given by this equation, pt is equal to p0 minus (alpha plus gamma)/(beta plus delta) multiplied by (1 minus sigma * (beta plus delta)) to the power t plus (alpha plus gamma)/(beta plus delta). So, this was the time path. So, I just have to figure out this value here and this value here, and I can substitute those things in this function on this time path. That is going to be our strategy.
So, in this case, (alpha plus gamma)/(beta plus delta); alpha is 18, gamma is 6. So, this 18, 6; beta is 2 and delta is 1. So, I put everything here; 6 plus 18 is 24. 24 divided by 3 is giving me 8, and think about this term; sigma is 0.2 and beta plus delta that we have just seen, 2 plus 1. So, 0.2 multiplied by 3 that is 0.6, and remember we have to figure out what is 1 minus sigma; I am sorry, this is sigma, this is not true, this is sigma because that sigma is the standard notation that has been used in the model. So, this is sigma, so we have to figure out what is 1 minus sigma multiplied by beta plus delta, that is 1 minus 0.6, so that is 0.4. So, I will have to substitute 0.4 here. And this also we have found out; this is 8. So, therefore, the time path is given by this, pt is equal to p0 minus 8 multiplied by 0.4 to the power t plus 8. 8 here is the particular integral or the inter-temporal equilibrium price. And this 0.4 is very critical; this is that b term in the complementary function. Once again, I do not know the p0 here, that is the price at point t is equal to 0. So, that is the arbitrariness, but as far as the stability is concerned, I can say something about that. Here it is going to be a stable convergence to equilibrium. There is going to be stability, and there is convergence to the equilibrium, and the convergence is not kind of oscillating convergence because 0.4 is greater than 0; there is going to be a monotony convergence to the equilibrium price; the equilibrium price is 8. So, that is the answer.
So, I think with that I will conclude this series of tutorials that we have been discussing in the last few lectures. And I thank you once again for joining me, and all the best. Thank you.