September Math Course: First Order Derivative

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1 September Math Course: First Order Derivative Arina Nikandrova Functions Function y = f (x), where x is either be a scalar or a vector of several variables (x,..., x n ), can be thought of as a rule which converts an input (denoted typically by x) into an output (denoted typically by y). Output y is a function of x if you can draw it from left to right without doubling back, i.e., only one value of y should correspond to a value of x. We will focus on functions where the input consists of many variables. Such functions are common in economics and finance. Example. A consumer s utility is a function of all the goods he consumes. So if there are n goods, then his utility is a function of the quantities (x, x,..., x n ) he consumes. We represent this by writing u (x, x,..., x n ). First Order Derivative of Univariate Functions Consider function of one variable, f (x). If this function is differentiable at a given point, x 0, it has both a value (its height ), y 0 = f (x 0 ), and a slope. The slope tells us the rate of change: how much y changes when x changes by a given amount. Example. (Linear Function) The simplest function to consider is a linear function of the form y = ax + b. Start at any point (x 0, y 0 ) on the line and move along the line so that the x-coordinate increases by one unit. The corresponding change in the y-coordinate is called the slope of the line. The slope tells us the rate of change: how much y changes when x changes by a given amount. The defining characteristic of a line is that this rate of change is constant: y x = a (x 0 + ) + b (ax 0 + b) = a. x 0 + x 0 a.nikandrova@bbk.ac.uk

2 (a) The change in y = x when x increases by is the same at any x. (b) The change in y = x when x increases by starting from x 0 = is smaller than that starting from x 0 =. Figure : The change in y as x changes by one unit. For non-linear functions the same change in x leads to different changes in y, depending on the starting point x 0. Example 3. Consider a quadratic function y = x. If we start at x 0 = and increase x by, then y changes by 3 (i.e., 4 ). If we start at x 0 =, however, then increasing x by changes y by 5 (i.e., 9 4). Thus the same change in x leads to different changes in y. Consequently, for non-linear functions we cannot define a global notion of the slope. However, it is possible to define a notion of the slope which is valid when the change in x is small. Example 4. Consider the quadratic function y = x. The line y = 4x 4 just touches the curve y = x at the point (x, y) = (, 4). This follows as 4 = y = x = and 4 = y = x 4 = 4 4. Such a line is called a tangent line. The tangent line has the property that it looks the same as the function around the point at which it just touches the function. The tangent line shows the rate of change in y at a point for small changes in x. The slope of the tangent line at point x 0 is called the derivative of function f (x) at the point x 0 and is denoted by f (x 0 ). Formally, if it exists, a slope or a derivative at x is given by: f f (x + h) f (x) (x) lim. h 0 h The definition implies that to find the slope of the tangent line at x, we find the limit of the slopes of the secant lines as they pass through points x and x + h where x + h gets ever

3 Figure : The curved line is the function f (x) = x. The straight solid line is the tangent of f (x) at the point (x, y) = (, 4). The slope of this line is the derivative of f (x) at x =. The total differential of f (x) at x 0, dy = f (x 0 ) dx, represents the main part of the change in f (x) with respect to any not necessarily small changes in x Figure 3: The slope of the tangent line at x is the limit of the slopes of the secant lines passing through x and some other point that gets ever closer to x 3

4 closer to x. That is, we find the limit of the slopes of the secant lines as h goes to zero (see Figure3). Given a function y = f (x), the derivative simply associates to every x the slope of the tangent line at x. So the derivative is a separate function of x. When we want to talk about the value of the derivative at a point x 0, we shall mention it by saying the derivative at x 0 is.... The derivative indicates whether a function is increasing or decreasing. A function f (x) is weakly decreasing at point x if f (x) 0; a function f (x) is weakly increasing at point x if f (x) 0. If the inequalities are strict, then the function is strictly decreasing or strictly increasing. The total differential of f (x) at x 0 represents the principal part of the change in a function y = f (x) with respect to changes in x and is defined by the following: dy = f (x 0 ) dx. The total differential is a way of understanding the local rate of change of the function f (x) around the point x 0. That is, it is an algebraic way of denoting the slope of a function (hence the alternative notation for a derivative: y dy, dx, d f dx ). Since the rate of change along a curve is changing constantly, the derivative has to be computed separately at each possible value of x. The derivative is thus a local phenomenon: it tells us something about the rate of change in the neighborhood of a point, but it gives no information about the rate of change globally. Example 5. The derivative of y = x (i.e., dy = xdx) at x = tells us that the rate of change in y is 4 when x is close to. It does not give any information about the rate of change at x = 0, and so on. The derivative discussed in this subsection is more precisely described as the first derivative. The second derivative is the derivative of the first derivative. The second derivative viewed as a function of x is typically denoted by f (x) or by d f or d y. The second dx dx derivative of a function f evaluated at x 0 is typically denoted by f (x 0 ) or by d f (x 0 ) or d y dx x=x0. If the second derivative of a function exists at x 0 then the function is said to be twice differentiable at x 0. If a second derivative exists at every point in some open interval A, then the function is said to be twice differentiable on A. Likewise, we can define third, fourth and nth derivatives and their notation is analogous to that of the second derivative. dx. A Derivative Might Not Exist A function f (x) defined on some open interval A is said to be differentiable on A if it differentiable at every point x A. However, a derivative fails to exist if function is not continuous or if it is continuous, but has kinks or becomes infinitely steep at some point. 4

5 (a) A Function that Explodes. (b) A Function that Becomes Infinitely Wiggly. Figure 4: Functions such that one or both of the limits lim x x0 f (x) and lim x x0 f (x) do not exist. Discontinuous Functions Informally, a function y = f (x) is a continuous function of x if you can draw it without removing your pencil from the page. Formally, a function f is said to be continuous at a point x 0 if f (x) f (x 0 ) as x x 0. To see why a function might not be continuous at x 0, it is useful to define lim x x0 f (x) as the limit of the function when x decreases to x 0. This is called the right-hand limit because graphically it is the limit approaching from the right-hand. Similarly, define lim x x0 f (x) as the limit of the function when x increases to x 0. This is called the left hand limit because graphically it is the limit approaching from the left-hand side. If f (x) f (x 0 ) as x x 0, it must be that both limits lim x x0 f (x) and lim x x0 f (x) exist and that they are both equal to f (x 0 ). By exist, we mean that there is some unique value that the function converges to and that this value is finite. There are two types of reasons that a function might not be continuous at x 0. The first reason is that one or both of the limits lim x x0 f (x) and lim x x0 f (x) might not exist. There are two reasons that this could happen. The usual reason is that f (x) goes to plus or minus infinity as x increases or decreases to x 0. An example of this, shown in Figure 4a, is f (x) = x. This function explodes at x 0 =, because of the division by zero. For this function, neitherlim x f (x) nor lim x f (x) exists. Functions that explode in this manner are not continuous. The other reason is that that the limits might not exist is that the function does not converge to a unique value: instead it becomes infinitely wiggly. An example of 5

6 (a) A Function that Jumps at x = 0: At x = 0 the value of the function is ; the red dot is the exclusion point. (b) A Function with a Hole at x = 0: At x = 0 the value of the function is 0; the red dot is the exclusion point. Figure 5: Functions such that both of the limits lim x x0 f (x) and lim x x0 f (x) exist, but one or both of them does not equal f (x 0 ). such a function is f (x) = sin x. This function, seen in Figure 4b, is not continuous at zero. You are not likely to encounter such a function in practice! The second reason that a function might not be continuous at x 0 is that, while both limits lim x x0 f (x) and lim x x0 f (x) exist, one or both of them does not equal f (x 0 ). There are two reasons that this can happen. The first is that the function can jump. Consider the function { if x 0 f (x) = if x > 0, shown in Figure 5a. The right hand limit at zero is equal to one and the left-hand limit is equal to minus one. This jump function is not continuous at zero. The second reason is that the function can have a hole. An example of this, shown in Figure 5b, is the function { if x = 0 f (x) = 0 if x = 0. This function has a hole at zero. Both the right-hand and the left-hand limits at zero are equal to minus one but the function evaluated at zero is one. A function is differentiable at x 0 if the slope of the secant lines converges to the same finite slope, irrespective of whether x increases or decreases to x 0. If a function is not 6

7 continuous at x 0, this does not happen. In Figure 4a the secant lines become vertical as x goes to one. A vertical line has an infinite slope. Thus the function drawn in Figure 4a is not differentiable at one. The functions with a jump and a hole, shown in Figures 5a and 5b, respectively, also suffer from this problem. As x decreases to zero the secants for the jump function become steeper and converge to a vertical line at zero. Approaching from either side for the function with a hole the secants become steeper and converge to a vertical line at zero. The pathological case in Figure 4b is too wiggly to be differentiable: the slope of the secants does not converge to a unique value. This suggests the intuition for the following important result: If a function is differentiable at x 0 then it is continuous at x 0. Continuous but Non-Differentiable Function While all functions that are differentiable at x 0 are continuous at x 0, not all functions that are continuous at x 0 are differentiable at x 0. There are two types of continuous functions that are not differentiable. The first type is a kinked function. Figure 6a depicts the function f (x) = x. The secants have a slope of plus one on the right-hand side of zero and a slope of minus one on the left-hand of zero. Thus, the right-hand side limit of their slopes is one and the left-hand side limit of their slopes is minus one. As the right-hand side limit does not equal the left-hand side limit, the function is not differentiable at zero. The second type is a function that has a point where the secants converge to a vertical line even though the function is continuous at that point. An example is the function graphed in Figure 6b. This f (x) = x /3,. Computing Derivatives: the Rules of Differentiation Differentiation is linear: For any functions f and g and any real numbers a and b the derivative of the function h (x) = a f (x) + bg (x) with respect to x is h (x) = a f (x) + bg (x). Power function rule: The derivative of power function h (x) = x n is Special cases include: h (x) = nx n. Constant rule: if f is the constant function f (x) = c, for any number c, then for all x, f (x) = 0. 7

8 (a) A Function with a Kink..0 (b) A Function that is too Steep. Figure 6: Function that is not Differentiable at Zero. If f (x) = x, then f (x) =. These special cases imply that the derivative of an affine function is constant, i.e., if f (x) = ax + b, then f (x) = a. This makes sense as shifting a function doesn t change its slope and so additive constants disappear. The product rule: For any functions f and g the derivative of the function h (x) = f (x) g (x) with respect to x is h (x) = f (x) g (x) + f (x) g (x). Quotient rule: The derivative of function h (x) = f (x), where g g(x) (x) = 0, is: h (x) = f (x) g (x) f (x) g (x) g. (x) The chain rule: The derivative of the function of a composite function h (x) = f (g (x)) with respect to x is h (x) = f (g (x)) g (x). The basic rules for differentiating exponential and logarithmic functions: The derivative of f (x) = e x is f (x) = e x, where e =.788 is the Euler s number. The derivative of f (x) = ln x is f (x) = /x, where ln is the natural logarithm with the base e =

9 The inverse function rule: If the function f has an inverse function g, meaning that g ( f (x)) = x and f (g (y)) = y, then g (y) = f (g (y)). Intuition for the inverse function rule: If y = f (x) is a strictly monotonic (or :) function, its inverse, x = f (y), is also a function. Formally: f (y) = {x : y = f (x)}. Thus inverse is a function if to each value of y corresponds only one value of x, e.g., parabola is ruled out (why?). Example 6. Inverse of y = f (x) = ax + b is function g (y) = y b a. Inverse of y = f (x) = x, where x > 0, is function g (y) = y. We can think of these two functions as inverses. If we take x as the input, apply f to it and then pass this output through the function g, we get back x. Computationally, we just express x from equation y = f (x) to obtain x = f (y) g (y). The derivatives of inverse functions are related to each other. If we apply chain rule to both sides of x = g ( f (x)), when g ( ) f ( ): g (y) = f (x). However, for the above display to make sense we need to express x in terms of y on the RHS. Example 7. If y = f (x) = x, where x > 0, then the derivative of its inverse is g (y) = f (x) = x = y, where the last equality follows as by the definition of inverse x = f (y) = y. 3 First Order Derivative of Multivariate Function We have considered functions of a single variable until now. Most economic problems involve more than one variable, so consider a function y = f (x, x,..., x n ). 9

10 Partial Derivatives The partial derivative of f with respect to x i is the derivative of f with respect to x i treating all other variables as constants and is denoted by f / x i or f i : f (x f (x, x,..., x n ) lim,..., x i, x i + h, x i+,..., x n ) f (x, x,..., x n ). x i h 0 h In order to calculate partial derivatives, we can apply the usual rules of differentiation. Example 8. Consider a function f (x, x ) = x x. Then, f x = x x f x = x. Mathematically, the partial derivative of f with respect to x i tells us the rate of change when only the variable x i is allowed to change. Economically, the partial derivatives give us useful information. For example, with a utility function, the partial derivative with respect to good x i tells us the rate at which the consumer s well being increases when she consumes additional amounts of x i holding constant her consumption of other goods, i.e., the marginal utility of that good. Total Differentials Partial derivatives are multivariate extensions of derivatives; total differentials are multivariate extensions of differentials. For functions of more than one independent variable, y = f (x, x,..., x n ), the partial differential of y with respect to any one of the variables x i is the principal part of the change in y resulting from a change dx i in that one variable. The partial differential is therefore y x i dx i involving the partial derivative of y with respect to x i. The sum of the partial differentials with respect to all of the independent variables is the total differential dy = y x dx + + y x n dx n, which is the principal part of the change in y resulting from changes in all independent variables. To gain some intuition about total differentials, suppose there are two variables and consider the plane y = a 0 + a x + a x. How does the function behave when we change x Recall that we motivated the notion of a derivative by saying that it was the slope of the line which looked like the function around the point x 0. When we have n variables, the natural notion of a line is given by the following linear function: y = a 0 + a x + a x a n x n. () In general, the function () is referred to the equation of a plane (it certainly is the equation of a plane when there are two variables, x and x ). 0

11 and x? Clearly, if dx and dx are the amounts by which we change x and x, we have, y dy = a dx + a dx. Note furthermore that the partials are, x = a and y x = a. We can then write total change in y as: dy = y x dx + y x dx. Rewriting this in matrix notation: dy = [ y x y x ] [ dx dx In the case of the plane, the vector of all partial derivatives is given by [ ] a a. This vector tells us the rates of change in the directions x and x. ]. [ y x ] y x = Now consider a more general two variable function, y = f (x, x ). With a general function, the idea is to find a plane which looks locally like the function around the point (x, x ). Since the partial derivatives give the rates of change in x and x, it makes sense to pick the appropriate plane which passes through the point (x, x ) and has slopes y/ x and y/ x in the two directions. The derivative of the function f (x, x ) at (x, x ) is simply the vector [ y x ] y x, where the partial derivatives are evaluated at the point (x, x ). We can interpret the derivative as the slopes in the two directions of the plane which looks like the function around the point (x, x ). For a general function of n variables, y = f (x, x,..., x n ), the derivative of f at point (x, x,..., x n ) is the vector of partial derivatives x.... This vector defines a linear map, which is the best linear approximation of the function f near the point (x, x,..., x n ). This linear map is thus the generalization the usual notion of derivative. [ y ] y x n Example 9. For function f (K, L) = K α L β, the vector of partial derivatives is ] = [ αk α L β βk α L β ]. [ f K f L Then total differential of f is: d f = [ αk α L β βk α L β ] [ dk dl ]. Total Derivatives While the partial derivative of f with respect to x i treats all other arguments of f as constants, the total derivative of f acknowledges that other arguments of f may also vary with x i due to some postulated relationship. Finding the total derivative relies on the chain rule.

12 0 5 y f x, x 00 0 f x,y x 5 (a) Function f (x, y) = x y (blue surface) and the tangent plane at point (4, 5) (red surface). The tangent plane, given by z = 8 (x 4) 0 (y 5) 4, looks like f (x, y) = x y around (4, 5). The derivative of f (x, y) = x y is the slopes in the two directions of the tangent plane (b) Cross-section when y = 5 : The slope of the red line represents the partial derivative of f (x, y) = x y with respect x at point (4, 5). Figure 7: Function f (x, y) = x y and its derivative. Definition. Consider function f (x, y, z, t), where x, y, and z depend on t. Then, the chain rule is given by: d f dt = f dx x dt + f dy y dt + f dz z dt + f t. In particular notice that d f dt = f t, as t has a direct effect on f, given by f t and an indirect effect through its effect on x, y and z. Example 0. Consider a function where y = 3x w, x = w + w + 4. Here w has a direct effect on y, given by y w and an indirect effect through its effect on x. Hence, the total derivative of y with respect to w is dy dw = y dx x dw + y w = 3 (4w + ) w.

13 Note that unless w = /4, A more complicated example. dy dw = y w. Example. Consider the function z = x y 0x t 3, where x = e y and t = 3y.. Find the partial derivative of z with respect to y.. Find the total derivative of z with respect to y, dz/dy. 3

14 September Math Course Unconstrained Optimization Arina Nikandrova Univariate Case We will consider the following maximization problem max x f (x) or minimization problem min x f (x). First Order Conditions: Necessary Conditions for Local Extrema If a differentiable function f (x) reaches its maximum or minimum at point x then f (x ) = 0. To see this consider the total differential: dy = f (x ) dx. If the function reaches a maximum or minimum at x then it must be impossible to increase or decrease the value of the function by small changes in x. However, if f (x ) = 0, then it is always possible to make by larger or smaller by making (small) appropriate changes in x. Therefore, we must have f (x ) = 0 at a maximum or a minimum. Any point satisfying the condition f (x ) = 0 may be referred to as a stationary point; when a point satisfying f (x ) = 0 is a minimum or a maximum, it is referred to as a critical value or extremum. We need to distinguish between local (or relative) extrema and global extrema. Figure a illustrates the difference, which is also explained in Definition. Definition. A point x is called a global maximum of the function f (x) if f (x ) f (x) for all x in the domain of f. A point x is called a local maximum of the function f (x) if there is a small interval centered at x such that f (x ) f (x) for all x in this small interval. A point x is called a global minimum of the function f (x) if f (x ) f (x) for all x in the domain of f. A point x is called a local minimum of the function f (x) if there is a small interval centered at x such that f (x ) f (x) for all x in this small interval. a.nikandrova@bbk.ac.uk

15 (a) Function f (x) defined for x [0, 6] : Each point where f (x) = 0 corresponds to either local minimum or local maximum, but condition f (x) = 0 does not identify global minimum or maximum. Moreover, condition f (x) = 0 on its own does not distinguish local maximum from local minimum. (b) The point where f (x) = 0 is the point of inflection. Figure : The first order condition f (x) = 0 is a necessary, but not sufficient condition for local minima and maxima The condition f (x ) = 0 at a maximum or minimum is valid only if x is in the interior of the domain of the function. This is because the argument for showing that f (x ) = 0 is a necessary condition for x to be a maximum or a minimum relies on the ability to make small changes in x around x. However, at a boundary point we cannot make certain changes. For instance, if the function is defined for all x in the interval [a, b], then at a, we can only increase x, while at b, we can only decrease x. Hence, it is possible that the maximum (or minimum) occurs at a or b and yet this boundary point does not satisfy the necessary condition for maximization (or minimization). For example, in Figure a the global minimum of a function defined for x [0, 6] occurs at point x = 0 and the global maximum occurs at point x = 6, neither of which satisfies the first order condition f (x ) = 0. The concept of a local maximum or minimum might seem restrictive. However, if a function only has one or a few local maxima then one only needs to compare these local maxima and pick out the one that is best. Then, compare this with what happens as x goes to the endpoints of the interval under consideration. For the function in Figure a, the function is defined on a closed interval [a, b]. We know that the unique local maximum is not a global maximum because the function s value is higher at b, the right endpoint of the interval on which the function is defined. This function s global maximum is at b. In general, a function defined on an entire real line may not have global extrema, but a continuous function defined on a closed interval always has global extrema. Proposition. If f is a continuous function defined a closed interval [a, b], then f has both a maximizing

16 (a) f (x) = (x 3) + 4: Point x = 3 is a maximum as f (x) is decreasing (changes sign from positive to negative) in the neighborhood of x = 3. (b) f (x) = (x 3) + 4: Point x = 3 is a minimum as f (x) is increasing (changes sign from negative to positive) in the neighborhood of x = 3. Figure : The second order conditions, i.e, the conditions on the sign of f (x), are sufficient for determining local minima and maxima and a minimizing value in [a, b]. Condition f (x ) = 0 is called a necessary condition because it cannot guarantee that x is indeed a maximum or minimum. It is entirely possible that f (x ) = 0 but x is neither a maximum nor a minimum. Example. Consider function f (x) = (x + ) Note that f ( ) = 0, but point x = is neither maximum, nor minimum (see Figure b). Second Order Conditions: Sufficient Conditions for Local Extrema Condition f (x ) = 0 on its own does not distinguish local maxima from local minima. To tell whether point x is a local maximum or a local minimum, we need to look at the sign of function f (x) in the immediate neighborhood of x, where neighborhood is defined as points immediately to the left and immediately to the right of x : Point x = x is a local maximum if in the neighborhood of x, f (x) is positive for x < x and is negative for x > x ; Point x = x is a local minimum if in the neighborhood of x, f (x) is negative for x < x and is positive for x > x ; Point x = x is neither a local maximum nor a local minimum if in the neighborhood of x, f (x) does not change sign. An equivalent way to express the above conditions is to say that Point x = x is a local maximum if in the neighborhood of x, f (x) is a decreasing function; 3

17 (a) f (x) = x is convex for all x (, ) (b) g (x) = ln x is concave for all x (0, ) Figure 3: An example of (a) a strictly convex and (b) a strictly concave function. Point x = x is a local minimum if in the neighborhood of x, f (x) is an increasing function; Point x = x is neither a local maximum nor a local minimum if in the neighborhood of x, f (x) is neither increasing nor decreasing. This last set of conditions can be expressed more succinctly in terms of second order derivatives, but it requires a few new definitions. Recall that f (x), the first derivative of the function f, indicates whether a function is increasing or decreasing. In particular, a function f (x) is weakly decreasing at point x if f (x) 0; a function f (x) is weakly increasing at point x if f (x) 0. If the inequalities are strict, then the function is strictly decreasing or strictly increasing. Since the derivative itself is a function, we can take its derivative. This is called the second derivative and denoted d f /dx or f (x). Formally, d f dx = d ( ) d f. dx dx The second derivative indicates whether the first derivative of a function is increasing or decreasing, thereby describing the curvature of the function. Definition. A differentiable function f (x) is called concave if f (x) 0 at all points of its domain; a differentiable function f (x) is called convex if f (x) 0 at all points of its domain. If the inequalities are strict, then the function is called strictly concave or strictly convex. A differentiable function f (x) is called concave at x if f (x ) 0; a differentiable function f (x) is called convex at x if f (x ) 0 Example. The function f (x) = x is convex on its domain; function g (x) = ln x is concave on the domain x > 0. A function may be neither concave nor convex on its entire domain. Example 3. Consider f (x) = x 3 /3 + 0x + 5 defined for x 0. In this case, f (x) = 4x + 0 and thus: 4

18 Figure 4: Function f (x) = x 3 /3 + 0x + 5: point x = 5 is an inflection point where the function changes its curvature from convex (for 0 < x < 5) to concave (for x > 5). for 0 < x 5, f (x) 0 and function is convex; for x > 5, f (x) < 0 and function is concave. Recall Figure b where f ( ) = 0, but point x = is neither maximum, nor minimum. This point is called the inflection point. Definition 3. The point where a function changes its curvature is called an inflection point. As an aside, note that since the second derivative is also a function, we can also take its derivative. This is called the third derivative and denoted f (x) to indicate that this function is found by three successive operations of differentiation, starting with the function f. One can continue this process, but we will typically not go beyond the second derivative. Example 4. Suppose that f (x) = x 5. Then, f (x) = 5x 4, f (x) = 0x 3 and f (x) = 60x. The observation that the second derivative indicates whether the first derivative of a function is increasing or decreasing leads to the following set of necessary and sufficient conditions for identifying maxima and minima: If f (x ) = 0 and f (x ) < 0, then x is a local maximum of f (x) ; If f (x ) = 0 and f (x ) > 0, then x is a local minimum of f (x). The necessary condition only identifies a local maximum or minimum, but not a global maximum or minimum. However, by Proposition a continuous function defined on a closed interval always has global extrema, which suggests the following algorithm: Algorithm for finding extrema: To find, say, the global maximum of a differentiable function f defined on a closed interval [a, b], we need to. find the critical points of f and check which ones are local maxima;. compute the value of the function at these local maxima and at the end points of the interval to determine the global maximum. 5

19 Furthermore, the local maxima of a function that is concave on its entire domain are also global maxima. Similarly, the local minima of globally convex functions are also global minima. That is: If f (x ) = 0 and f (x) < 0 for all x in the domain of f, then x is a global maximum of f (x) ; If f (x ) = 0 and f (x) > 0 for all x in the domain of f, then x is a global minimum of f (x). Function depicted in Figure a strictly concave on its entire domain and thus point x = 3 is a global maximum; function depicted in Figure b strictly convex on its entire domain and thus point x = 3 is a global minimum. Example 5. Consider following maximization problem: max q Π (q) = max (00 q) q q. q From the necessary first order conditions it follows that Π (q) = 00 4q = 0. So q = 5 is a candidate for a maximum. To check that this indeed is the maximum, we need to check the second order conditions for optimization. The second derivative Π (q) = 4 < 0 for all q and, in particular, for q = 5. Hence q = 5 is a global maximum. Another economic example. Example 6. Consider the following minimization problem: Then, first-order conditions imply: min C (q) = min 00/q + q. q q C (q) = 00/q + = 0 Therefore, q = 0 and q = 0. Since, C (q) = 00/q 3, C (q ) > 0 and q is a minimum, while C (q ) < 0 and q is a maximum. Multivariate Case Consider the general maximization problem: max f (x, x,..., x n ). x,...x n The first order conditions for maximization require the first order differential to be zero at the optimal point. That is, a vector of small changes (dx, dx,..., dx n ) should not change the value of the function. We thus have d f = f dx + + f dx n = 0. x x n 6

20 This can be satisfied if f x = 0, f x = 0,..., f x n = 0. These conditions are necessary conditions and they must also hold for minimization problems. As in the single variable case, we are really after maxima and minima. The first order conditions alone cannot distinguish between local maxima and local minima. Likewise, the first order conditions cannot identify whether a candidate solution is a local or global maxima. We thus need second order conditions to help us. For a point to be a (local) maximum, f needs to be a (locally) strictly concave function. Similarly, for a point to be a (local) minimum, f needs to be a (locally) strictly convex function. Definition 4. Point (x, x,..., x n) is a local maximum if for all i f x i x =x,...,x n=x n = 0 and function f (x, x,..., x n ) is concave at (x, x,..., x n). Point (x, x,..., x n) is a local minimum if for all i f x i x =x,...,x n=x n = 0 and function f (x, x,..., x n ) is convex at (x, x,..., x n). Point (x, x,..., x n) is a global maximum if for all i f x i x =x,...,x n=x n = 0 and function f (x, x,..., x n ) is concave for all (x, x,..., x n ) (see Figure 5a). Point (x, x,..., x n) is a global minimum if for all i f x i x =x,...,x n=x n = 0 and function f (x, x,..., x n ) is convex for all (x, x,..., x n ) (see Figure 5b). If at the point (x, x,..., x n) where for all i f x i x =x,...,x n=x n = 0, function f is neither convex, nor concave, then point (x, x,..., x n) is a saddle point (see Figure 6). Now we need tools for identifying whether a multivariate function is concave or convex. In the definition below, notation f x i x =x,...,x n=x n should be understood as a partial derivative of f with respect to x i evaluated at a point x = ( x, x,..., x n). 7

21 0 5 y 0 5 f x,y x (a) f (x, x ) = x x : the function is concave for all (x, x ), hence point (x, x ) = (0, 0) is a global maximum x 0 5 f x,x (b) f (x, x ) = x + x : the function is convex for all (x, x ), hence point (x, x ) = (0, 0) is a global minimum. Figure 5: An example of (a) a strictly concave and (b) a strictly convex function of two variables. 0 x 5 x f x,x x Figure 6: f (x, x ) = x x : the function is convex in the direction of x and concave in the direction of x, hence point (x, x ) = (0, 0) is a saddle point. 5 8

22 Higher-Order Derivatives of Multivariate Functions Recall that a single variable function f (x) is strictly concave is f (x) < 0 and is strictly convex if f (x) > 0. Notice that in the single-variable case, the second-order total differential is: d y = f (x) (dx). Hence, we can (equivalently) define a function of one variable to be strictly concave if d y < 0 and strictly convex if d y > 0. The advantage of writing it in this way is that we can extend this definition to functions of many variables: A multivariate function f (x, x,..., x n ) is strictly concave if d y < 0 and strictly convex if d y > 0. This imposes certain restrictions on its second-order partial derivatives. Second-order partial derivatives Given a function f (x, x,..., x n ), the second-order derivative f / x i x j is the partial derivative of f / x i with respect to x j. The above may suggest that the order in which the derivatives are taken matters and that the partial derivative of f / x i with respect to x j is different from the partial derivative of f / x j with respect to x i. While this can happen, it turns out that if the function f (x, x,..., x n ) is well-behaved then the order of differentiation does not matter. This result is called Young s Theorem. We will be dealing with well-behaved functions for which Young s Theorem holds. Example 7. Consider a function f (x, x ) = x x x + x 3. For this function we can evaluate the second-order partial derivative, f / x x, in two different ways. First, since f x = 4x x, taking the partial derivative of this with respect to x, we get, Alternatively, since f x x = ( ) f = 4x. x x y = x x + 3x, f = ( ) f = 4x. x x x x This illustrates Young s Theorem: no matter in which order we differentiate, we get the same answer. Concavity and convexity of a multivariate function Definition 5. A differentiable function y = f (x, x,..., x n ) is concave if d y 0 for all x and is convex if d y 0 for all x. If the function satisfies a stronger condition, d y < 0 for all x, then it is strictly concave. Analogously, if d y > 0 for all x, it is strictly convex. 9

23 Consider a two variable function, y = f (x, x ). Its differential is: dy = y x dx + y x dx, which again can be viewed as a function of x and x. Taking a differential, we obtain [ d (dy) = y dx + x which after collecting terms yields: d y = y x ] [ y dx dx + y x x (dx ) + y x x dx dx + y x ] dx + y x x x dx dx, (dx ). Thus, the second-order total differential depends on the second-order partial derivatives of f (x, x ). For a general function y = f (x, x,..., x n ), one can use a similar procedure to get the formula for the second-order total differential. This is a little more complicated but it can be written compactly as follows: d y = n i= n j= y x i x j dx i dx j. As things stand, it is not clear how to go about verifying that the second-order total differential of a function of n variables is never positive or never negative. However, notice that we can write the second order differential of a function of two variables, d y = y x (dx ) + y x x dx dx + y x (dx ), in matrix form in the following way: d y = [ ] dx dx y x y x x y x x y x [ dx dx ]. The matrix of second-order partial derivatives H y x y x x y x x y x is called Hessian matrix, which is symmetric by Young s Theorem. For a general function y = f (x, x,..., x n ), d y = [ ] dx dx... dx n y x y x x. y x n x y x x... y x. y x n x... y x x n y x x n y x n dx dx. dx n 0

24 and thus H y x y x x. y x n x y x x... y x. y x n x... y x x n y x x n Now it is clear that to determine whether a multivariate function is concave or convex, we need to know the sign of d y, i.e., we are interested in the sign of the quadratic form: y x n d y scalar = dx ( n) H (n n) dx (n ) = [ ] dx dx... dx n y x y x x. y x n x y x x... y x. y x n x... y x x n y x x n y x n dx dx. dx n. For a given symmetric matrix H and for any x R n five situations may arise: Definition 6. An (n n) matrix H is: positive definite if x Hx > 0 for any (n ) vector x R N, x = 0 n (note that x = 0 n means that at least one element of x is not equal 0 n ). positive semidefinite if x Hx 0 for any (n ) vector x R N, x = 0 n negative definite if x Hx < 0 for any (n ) vector x R N, x = 0 n negative semidefinite if x Hx 0 for any (n ) vector x R N, x = 0 n indefinite if x Hx > 0 for at least one vector x = 0 n and x Hx < 0 for at least one vector x = 0 n. From the discussion above, if the Hessian is negative definite for all (x,..., x n ), the function is strictly concave. If the Hessian is positive definite for all (x,..., x n ), the function is strictly convex. So to determine whether a function is concave or convex, we need to be able to determine whether the Hessian matrix is negative definite or positive definite. We can classify a symmetric matrix H in one of the above categories using either eigenvalue test or the principal minor test. Eigenvalue Test The quadratic form x Hx is: positive (semi)definite if and only if all the eigenvalues of H are strictly positive (nonnegative); negative (semi)definite if and only if all the eigenvalues of H are strictly negative (nonpositive).

25 Example 8. Consider matrix The characteristic equation is A = det (A λi) = ( λ) ( λ 8λ + ) 4 (8 4λ) + 6 (6λ 6) = 0. This equations of order three with no obvious factorization seems difficult to solve! Principal Minor Test Definition 7. Let H be an n n matrix. An i-th order principal minor of H is the determinant of a submatrix of H obtained by deleting n i rows and the n i columns with the same index. The i-th (order) leading principal minor of H is the determinant of the submatrix obtained from H by deleting the last n i rows and columns. Example 9. Let A be a 3 3 matrix Principal Minors A =. a a a 3 a a a 3 a 3 a 3 a 33 There is one third order principal minor of A, det (A). There are three second order principal minors: [ ] a a det, where the submatrix in the minor s calculation is obtained by deleting the a a third row and third column of A. [ ] a a det 3, where the submatrix in the minor s calculation is obtained by deleting the a 3 a 33 second row and second column of A. [ ] a a det 3, where the submatrix in the minor s calculation is obtained by deleting the a 3 a 33 first row and first column of A. There are also three first order principal minors: a formed by deleting the last two rows and columns; a formed by deleting the first and last rows and columns; and a 33 formed by deleting the first two rows and columns. Leading Principal Minors The ith leading principal minor of the determinant of the submatrix obtained from A by deleting all columns and all rows after the i-th. Thus. first l.p.m. = a [ ] a a second l.p.m. = det a a a a a 3 third l.p.m. = det a a a 3. a 3 a 3 a 33

26 Principal Minor Test: The quadratic form x Hx is positive definite if and only if all leading principal minors H are positive. The quadratic form x Hx is negative definite if and only if its leading principal minors of H alternate in sign, the first being negative (that is, every leading principal minor of H of odd order is < 0 and every leading principal minor of even order is > 0.) The quadratic form x Hx is positive semidefinite for every principal minor is 0. The quadratic form x Hx is negative semidefinite if every principal minor of H of odd order is 0 and every principal minor of even order is 0. Note that in the first two cases, it is enough to check the inequality for all the leading principal minors (i.e. for i n). In the last two ( cases, ) we must check for all principal minors (i.e for n each i with i n and for each of the principal minors of order i). i Example 0. Matrix is positive definite. [ 4 Matrix [ 4 is negative definite. Matrix [ 4 is neither positive definite nor negative definite. Matrix is indefinite. ] In the case of a function of two variables, y = f (x, x ) : d y is positive definite (and thus function is convex) if ] ] H = y x y x x y x y x x y x > 0 and ( = y x y x ) y > 0; x x 3

27 d y is negative definite (and thus function is concave) if H = y x y x x y x y x x y x < 0 and ( = y x y x ) y > 0. x x Note that the condition H > 0 implies that y x positive definite and negative definite H. Conditions for stationary point of y = f (x, x ) and y x should have the same sign for both FOC: Maximum Minimum Neither max nor min y y y x = 0, x = 0, x = 0, y y y x = 0 x = 0 x = 0 When SOC: y y x x y, y x x < 0, ( ) y x x > 0 y y x x y y x x y, y x x > 0, ( ) y x x > 0 ( ) y < 0, x x the point is either a saddle point or an inflection point. When y ( y ) y = 0, x x x x ( ) y y y x x x x < 0 the test fails and we need to check the other principal minors to determine whether the stationary point is a maximum, a minimum or neither. An Extended Example Find stationary points of the following function, and determine whether each is a maximum, minimum, or saddle point. Solution: FOC From () f (x, y) = 3x 6xy + y + y 4 f x f y = 6x 6y = 0 () = 6x + y + 4y 3 = 0. () x = y 4

28 and from () 4y ( y ) = 0. Thus, stationary points are (0, 0), (, ), (, ). SOC ( ) 6 6 H = 6 + y with det (H) = 6 ( + y ) 36 = 4 + 7y. At (0, 0), the first leading principal minor of H is positive, while the second is negative, hence H is indefinite and (0, 0) is a saddle point; At (, ) and (, ), the first and the second leading principal minors of H are positive, hence H is positive definite and both points are relative minima. 5

29 September Math Course: Constrained Optimization Arina Nikandrova Until now, we have considered unconstrained problems. Usually, economic agents face natural constraints. Example. Consumer s Problem: Suppose that a consumer has a utility function U (x, x ) = x / x /, the price of x is p, the price of x is p and the consumer has m in income. How much of the two goods should the consumer purchase to maximize her utility? In producer theory we are frequently interested in the following minimization problem: The above problem has the following mathematical structure: max x,...x n f (x, x,..., x n ) subject to g (x, x,..., x n ) = 0. We say that f (x, x,..., x n ) is the objective function, g (x, x,..., x n ) = 0 is the constraint and x, x,..., x n are the choice variables. We are interested in finding a solution to this problem x = x x. x n The value function for this problem is derived by substituting x into the objective function to obtain f (x, x,..., x n). It is also possible that instead of maximizing f (x, x,..., x n ) we could be minimizing f (x, x,..., x n ). Example. (Example continued) Utility maximization problem can be written as:. max x / x,x x / subject to p x + p x = m. The solution to the problem is a Marshallian demand as a function of prices and income, i.e., x = x (p, p, m) and x = x (p, p, m), while the objective function evaluated at the optimum is an indirect utility function: v (p, p, m) = (x ) / (x ) /. a.nikandrova@bbk.ac.uk

30 Direct Substitution When the constraint(s) are equalities, we can convert the problem from a constrained optimization to an unconstrained optimization problem by substituting for some of the variables. Example 3. (Example continued) In the consumer s utility maximization problem, suppose that the price of x is p =, the price of x is p = and the consumer has income m. Then consumer s budget constraint is x + x = m. Hence, Substituting this into the objective function, x = m x. ( max x m ) / x x /. This is a function of just x and we can now maximize this function with respect to x. By incorporating the constraint into the objective function, we transformed the constrained optimization problem into the unconstrained optimization problem, which we know how to solve. The first order conditions give: ( x / m ) / x ( 4 m ) / x x / = 0 Solving for x : m x = x Firm s problem can be solved similarly. = x = m = x = m x = 4 m. The Lagrangian Approach The substitution technique has serious limitations: In some cases, we cannot use substitution easily: for instance, suppose the constraint is x 4 + 5x 3 y + y x + x = 0. Here, it is not possible to solve this equation to get x as a function of y or vice versa. Moreover, in many cases, the economic constraints are written in the form g (x, x,..., x n ) 0 or g (x, x,..., x n ) 0. While the Lagrangian technique can be modified to take care of such cases, the substitution technique cannot be modified, or can be modified only with some difficulty.

31 Given a problem write down the Lagrangian function max f (x, x,..., x n ) subject to g (x, x,..., x n ) = 0 x,...x n L (x, x,..., x n, λ) = f (x, x,..., x n ) + λg (x, x,..., x n ). Note that the Lagrangian is a function of n + variables: (x, x,..., x n, λ). We then look for the stationary points of the Lagrangian, that is, points where all the partial derivatives of the Lagrangian are zero. Using a Lagrangian, we get n + first order conditions: L x i = 0, (i =,..., n) L λ = 0. Solving these equations will give us candidate solutions for the constrained optimization problem. Candidate solutions still need to be checked using the second-order conditions. Example 4. (Example continued) In the consumer s utility maximization problem: The first order conditions are given by: L (x, x, λ) = x / x / + λ (m x x ). L = x x / x / λ = 0 L = x x / x / λ = 0 L λ = x + x m = 0. Interpretation of FOC: If we divide the first two conditions, we get that MRS = U U =. This says that at the optimum point, the slope of the indifference curve must be equal to the slope of the budget line. To solve the problem, note that from the first two conditions it follows that or 4 x / x / = λ = x / x / Substituting this into the budget constraint yields: x = x. () x = 4 m. Substituting x back into () and solving for x yields: x = m. 3

32 Note that the technique is identical for both maximization and minimization problems. This means that the first order conditions identified so far are only necessary conditions and not sufficient conditions. We shall look at sufficient, or second order conditions later. The Lagrangian approach amounts to searching for points where: The constraint is satisfied. The constraint and the level curve of the objective function are tangent to one another. If we have more than two variables, then the same intuition can be extended. For instance, with three variables, the Lagrangian conditions will say: The rate of substitution between any two variables along the objective function must equal the rate of substitution along the constraint. The optimum point must be on the constraint. Intuition for the Lagrangian Method Consider the simplest case of the maximization of a function of two variables subject to one constraint: max x,x f (x, x ) subject to g (x, x ) = 0. Suppose that point x = ( x x is a constrained maximum. Therefore any small feasible change in x from this point, that is, a small movement along the constraint, should not be able to improve the value of the objective function. We represent small changes in x = (x, x ) T by differential notation ( dx dx = dx Then the first-order necessary conditions may be stated as follows: ) ). f x dx + f x dx = 0 () However, a feasible change in x does not change the value of the constraint. That is, the constraint g (x, x ) = 0 implies that g x dx + g x dx = 0 (3) and so dx and dx are no longer both arbitrary. We can take, e.g., dx as arbitrary, but then dx must be chosen to satisfy (3). Taking the ratio of () and (3), it is clear that at the optimum f x g x = f x g x λ. The Lagrange-multiplier method yields the same first-order necessary condition and the Lagrange multiplier λ makes sure that both () and (3) are simultaneously satisfied. 4

33 Economic Interpretation of the Lagrangian Multiplier Note that we did not compute λ in either consumer s problem or firm s problem. This is because our interest is in the values of x and x (or K and L). However, in some instances, it is useful to compute λ: this has an economic interpretation in terms of the shadow price of the constraint. Suppose we have the problem max f (x, x,..., x n ) subject to g (x, x,..., x n ) = 0 x,...x n Suppose we now relax this constraint: instead of requiring g (x, x,..., x n ) = 0, we require g(x, y) = δ where δ is a small positive number. Clearly, since the constraint has been changed, the value of the objective function must change. The question is: by how much? The answer to this question is given by λ. For this reason, λ is referred to as the shadow price of the constraint. It tells us the rate at which the objective function increases if the constraint is changed by a small amount. Example 5. (Example continued) In the consumer s utility maximization problem, we can compute λ = 4 x / x / = ( ) / ( 4 4 m m =. Thus, the shadow price of the constraint tells us that if we give a small amount of additional income to the consumer, then his utility will go up by a factor of Thus λ represents a marginal utility of income. λ =. 3 Second Order Conditions (Optional) As with the unconstrained case, we need to check the second-order conditions to ensure we have an optimum. As before, the second-order sufficient conditions for a maximum is d f < 0 and for a minimum is d f > 0. However, because of the constraint, it is no longer sufficient to look at the Hessian of f to verify these conditions. Suppose we have a two-variable constrained optimization problem max x,x ) / f (x, x ) or min x,x f (x, x ) subject to g (x, x ) = 0. The second order conditions for this problem differ slightly from the usual conditions because of the constraint g (x, x ) = 0 which implies that dx and dx must be chosen to satisfy (3). Thus the second-order sufficient conditions for a maximum is that d f < 0 subject to (3) and the secondorder sufficient conditions for a minimum is that d f > 0 subject to (3). In practice, to check the second-order sufficient conditions we need to compute the bordered Hessian matrix of the Lagrangian at the critical point that we want to check. The Lagrangian of the two-variable constrained optimization problem is L (x, x, λ) = f (x, x ) + λg (x, x ). 5

34 The bordered Hessian is the usual Hessian, bordered by the derivatives of the constraint with respect to the endogenous variables, here x and x. That is, The second order conditions state: H B = 0 g g g L L g L L If (x, x, λ ) corresponds to a constrained maximum, then H B evaluated at (x, x, λ ) must be positive. If (x, x, λ ) corresponds to a constrained minimum, then H B evaluated at (x, x, λ ) must be negative. Example 6. (Example continued) In the consumer s utility maximization problem,. 0 H B = 4 x 3/ x / 4 x / x / 4 x / x / 4 x/ x 3/ In a general n-variable problem with m (m < n) constraints, (x, λ ) that satisfies the first-order conditions is a local maximum if the last (n m) leading principle minors of H B alternate in sign beginning with that of ( ) m+ ; a local minimum if the last (n m) leading principle minors of H B are of the same sign as ( ) m. In both cases, H B must be evaluated at (x, λ ). There are also some global results for equality-constrained problems: If f (x,..., x n ) is concave and all constraints are linear in (x,..., x n ), then a solution to the constrained maximization problem is a global maximum. If f (x,..., x n ) is convex and all constraints are linear in (x,..., x n ), then a solution to the constrained minimization problem is a global minimum.. 6

35 September Math Course: The Envelope Theorem Arina Nikandrova We are interested in studying how the value function of an optimization problem changes when one of the parameters of the problem changes. A very powerful tool for such investigations if the envelope theorem. The Envelope Theorem for Unconstrained Optimization Suppose we have the unconstrained optimization problem max f (x, x ; α) x,x where α is some exogenous parameter. Suppose that (x (α), x (α)) solves this optimization problem. Note that the solution will depend upon α. The value function for this problem is derived by substituting (x (α), x (α)) into the objective function: V (α) = f (x (α), x (α) ; α). Notice that the value function is a function of the parameter α. Notice also that the value function depends on α in two different ways:. Direct dependence.. Indirect dependence through x (α) and x (α). We are interested in knowing how the value function changes when α changes. When we differentiate the value function, we get: dv dα = f x x α + f x x α + f α, where the partial derivatives of f are evaluated at the solution (x (α), x (α)). Now note that at the optimum (assuming we have an interior solution), it must be the case that f x x = x (α) = 0 x = x (α) and f x x = x (α) = 0. x = x (α) a.nikandrova@bbk.ac.uk

36 Hence, the first two terms drop out and we have dv dα = f α, where the partial derivative is evaluated at the point (x (α), x (α)). This result which is called the Envelope Theorem says in words: The total derivative of the value function with respect to the parameter α is the same as the partial derivative of the objective function evaluated at the optimal point. Example. Consider the unconstrained problem: The first order conditions: Solving: max 4x + αx x x,x x + x x. 4 x + x = 0 α x + x = 0. x = 8 + α 3 x = α (We also need to check the second order conditions.) Substituting x and x into the objective function, we obtain the value function: V (α) = α 3 By the Envelope Theorem: + α α (8 + α) 9 dv dα = x = α (α + 4) 9 + (8 + α) (α + 4). 9 The Envelope Theorem for Constrained Optimization (Optional) Now consider the constrained case. We can basically do the same as before. Consider the problem, The Lagrangian for this problem is, max f (x, x ; α) subject to g (x, x ; α) = 0. x,x L (x, x, λ; α) = f (x, x ; α) + λg (x, x ; α). Suppose that (x (α), x (α), λ (α)) solves the constrained optimization problem. The value function for this problem is defined as: Let us write the value function as: V (α) = f (x (α), x (α) ; α). V (α) = f (x (α), x (α) ; α) + λ (α) g (x (α), x (α) ; α).

37 Differentiating with respect to α : dv dα = f x x α + f x x α + f α + dλ dα g (x (α), x (α) ; α) +λ (α) [ g x x α + g x x α + g α ], where again all partial derivatives are evaluated at the solution (x (α), x (α), λ (α)). This can be written as [ dv f = + λ g ] x [ f dα x x α + + λ g ] x x x α + dλ dα g (x (α), x (α) ; α) + f α + λ (α) g α. Note that the first two terms on the right hand side drop out because (x, x, λ ) must satisfy the necessary conditions for constrained optimization. The third term drops out because g (x (α), x (α) ; α) = 0. We are left with the following: dv dα = f g + λ α α = L α (x, x, λ ). In words: The derivative of the value function with respect to the parameter α is the partial derivative of the Lagrangian function with respect to α evaluated at the solution (x, x, λ ). 3

38 September Math Course: Integration Arina Nikandrova Introduction The fundamental theorem of calculus is a theorem that links the concept of the derivative of a function with the concept of the integral. The first part of the theorem, sometimes called the first fundamental theorem of calculus, is that an indefinite integration can be reversed by a differentiation. This part of the theorem is also important because it guarantees the existence of anti-derivatives for continuous functions. The second part, sometimes called the second fundamental theorem of calculus, is that the definite integral of a function can be computed by using any one of its infinitely many anti-derivatives. This part of the theorem has key practical applications because it markedly simplifies the computation of definite integrals. Integration is useful in economics: In microeconomics, consumer surplus, i.e., the difference between what a consumer is willing to pay and what he actually pays, is an integral. In macroeconomics, stock variable (e.g., capital) is an integral of a flow variable (e.g., investment). In finance, stock price or net present value is an integral of a dividend flow. In probability and statistics, moments of random variables are integrals. There are two types of integrals: Indefinite integrals can be seen as anti-derivatives that recover the original function from the first derivative. Definite integrals calculate the area under a graph. In this form it is very similar to a sum, but of infinitely many, small parts. a.nikandrova@bbk.ac.uk

39 Indefinite Integrals We want to find a function F (x) that differentiates to f (x). Example. Consider f (x) = 3x In differentiation the Power Rule implies that if F (x) = x n, then F (x) = nx n. So guess: F (x) = x 3, then F (x) = 3x = f (x). Hence F (x) = x 3 is the anti-derivative primitive Integral of f (x) The first fundamental theorem of calculus: Let f be a continuous real-valued function defined on a closed interval [a, b]. Let F be the function defined, for all x in [a, b], by F (x) = x a f ( x) d x Then, F is continuous on [a, b], differentiable on the open interval (a, b), and for all x in (a, b). F (x) = f (x), In F (x) = f (x) dx, f (x) is known as the integrand. Is F (x) = x 3 the only anti-derivative of f (x) = 3x? No as d dx (F (x) + c) = f (x) for any constant c. This arbitrary constant is called the constant of integration.. Rules of Integration Integration is linear: ( f (x) + g (x)) dx = f (x) dx + g (x) dx. Power function rule: For n = f (x) = ax n F (x) = f (x) dx = a n + xn+ + c. Example. f (x) = 3x F(x) = f (x) = 5= 5x 0 F(x) = f (x)dx = 5x + c f (x)dx = x 3 + c

40 There is no general product rule, but k f (x) dx = k f (x) dx. Exponential Rule: Recall that d dx ekx = ke kx. Then, f (x) = ae kx F (x) = f (x) dx = a k ekx + c. Example 3. f (x) = 6e x F(x) = f (x)dx = 3e x + c Log Rule: Recall that d dx ln (x) = x. Then, f (x) = x F (x) = f (x) dx = ln (x) + c. Example 4. f (x) = 5 x+ F (x) = f (x) dx = 5 ln(x + ) + c The Substitution Rule: This technique operates through a change of a variable which converts an intractable integral into a form where it can be solved. f (u) du dx dx = f (u) du = F(u) + c. This is the inverse of the chain rule of differentiation. Example 5. Find 3x (x 3 + )dx. Let u = x 3, then du 3x (x 3 + )dx = (u + )dx dx = (u + ) du (by substitution rule) Integration by Parts: = u + u + c = x6 + x3 + c vdu = uv udv This is a direct consequence of the product rule of differentiation. Recall that (uv) = u v + uv. Integrating both sides of the above expression gives (uv) dx = u vdx + uv dx. Since by definition of an integral (uv) dx = uv, u vdx = uv uv dx. The first term on the RHS is the product of the integral of u and v and the second term is the integral of a product function which consists of the integral of u and the derivative of v. 3

41 Figure : A definite integral of f (x) over the interval [a, b] as an area under the curve. Example 6. Find ln (x) dx. Let v = ln (x), u = x dv = x dx, du = dx, then ln (x) dx = vdu = uv udv by integration by parts = x ln (x) dx = x ln (x) x + c. 3 Definite Integral Let f (x) be a continuous function on the interval [a, b], where a and b are real numbers with a < b. A definite integral of f (x) over the interval [a, b] gives the area underneath the graph of the function between a and b, where the parts below the x-axis are subtracted. What is the area bounded by the curve y = f (x), the vertical lines x = a and x = b and the x-axis? A first approximation of this area can be obtained by cutting the x-axis between a and b into intervals of equal length and thus creating rectangles of equal width, where the top right-hand corner touches the curve y = f (x) (see yellow rectangles in Figure ). Thus, if we split the interval [a, b] into 5 subintevals {[x 0, x ], [x, x ],..., [x 4, x 5 ]}, where x 0 = a and x 5 = b, the sum of the rectangle areas is (x x 0 ) f (x ) + (x x ) f (x ) (x 5 x 4 ) f (x 5 ) = 5 (x i x i ) f (x i ). i= However, this method of estimating the area leads to some errors thereby we either overestimate (like in Figure ) or underestimate the area (in Figure green rectangles underestimate the area as the height of the rectangle is the value of the function at the left-hand boundary of the subinterval). We can reduce these errors by creating many sub-intervals (see green rectangles in Figure ). This suggests that a definite integral of f (x) over the interval [a, b] can be viewed as the limit of the sum 4

42 Figure : A definite integral of f (x) over the interval [a, b] as a sum. of the areas of the rectangles as the size of each rectangle gets infinitesimally small and the number infinitely large. The intuition that the integral is the area under the graph is sufficient for (almost) all economics and finance. For example, consumer surplus is an area under the demand curve in price/quantity space. In macroeconomics and finance examples from Section flow variables are graphed against time (i.e., with time on x-axis). The second fundamental theorem of calculus: Let f and F be real-valued functions defined on a closed interval [a, b] such that the derivative of F is f, i.e., f and F are such that for all x in [a, b], Then, b a F (x) = f (x). f (x) dx = [F(x)] b a = F (b) F (a). As discussed earlier, if F (x) is an anti-derivative of f, then G (x) := F (x) + c is also an antiderivative of f for any constant c. However, the value of the definite integral does not depend on the choice of the anti-derivative, as G (b) G (a) = F (b) + c F (a) c = F (b) F (a). So in practical terms, we can then just ignore the constant term when evaluating definite integrals. Process of calculating a definite integral:. Determine indefinite integral. Evaluate at boundaries 3. Subtract F(a) from F(b) 5

43 Example 7. Find 0 xdx.. F(x) = xdx = x. F(0) = 0 3. [ 0 xdx = x = 0, F() = = ] 0 = 0 = Properties of definite integral a b a a f (x)dx = F(a) F(b) = (F(b) F(a)) = b a f (x)dx = F(a) F(a) = 0 c a f (x)dx = b a f (x)dx + c b b a k f (x)dx = k b a f (x)dx f (x)dx (a < b < c) b a [ f (x) + g(x)] dx = b a f (x)dx + b a g(x)dx Example 8. Calculate 4 9 Calculate 9 4 Another example: x dx: 4 x dx: 9 Example 9. Calculate e ln (x) dx: 9 4 x dx = [ x ] 4 9 = 4 9 f (x)dx = 3 =. x dx = [ x ] 9 4 = 9 4 = 3 = e ln (x) dx = [x ln (x) x] e = e ln (e) e ( ln () ) = e e 0 + = Sometimes we need to take integrals when the interval is not bounded. For example: Evaluating the present value of an infinite stream of benefits of a financial asset. Evaluating the consumer surplus of a constant elasticity demand function q = ap ɛ, as this demand curve it does not hit the y-axis. In this case, a f (x) dx = lim y F (y) F (a). 6

44 4 An Application: Continuous Compounding In finance, the present value of an asset can be approximated as a definite integral. Consider a continuous stream of income c for T years. Since a pound today is not the same as having it a year from now, we discount future income. If the discount rate is r, then the income c received t years into the future is worth c ( r) t in today s terms. Thus, the present value of an asset paying c every year into the future is PV = c ( r) 0 + c ( r) c ( r) T. When time becomes continuous it can be shown that the present value of an asset paying an amount c at time t into the future is ce rt. In this case, the present value of the asset is Note that for an infinitely lived asset PV = = c r T 0 T ce rt dt = c e rt dt 0 [ = c ] T r e rt 0 ( e rt). PV = ce rt dt 0 c = lim ( e rt) T r = c r. This follows because e rt goes to zero as T becomes very large. 7

45 September Math Course: Differential Equations Arina Nikandrova Introduction A differential equation specifies a relationship between a function and its derivatives. Differential Equations can be used to describe dynamic systems where time is an independent argument. Such systems occur frequently in economic models. For example, f (x (t), ẋ (t), t) = 0, where ẋ (t) or ẋ denotes dx dt, the total derivative of x with respect to time, is a differential equation and a solution to this equation is a function x (t) such that x (t) and ẋ (t) satisfy the preceding expression identically. There is no general way to solve differential equations, but some types of differential equations have known solutions. Example. The value V of an asset grows at a continuous rate of 0%. What is the value of the Asset as a function of time? a.nikandrova@bbk.ac.uk dv dt = 0.V dv V dt = 0. dv V dt dt = 0.dt V dv = 0.dt ln V + c = 0.t + c ln V = 0.t + c V = Ae 0.t,

46 where A is a constant. The solution of differential equations always involves some integration procedure, and as a result arbitrary integration constants appear. We know that e 0 =, so if the value at time 0 is V 0, then A = V 0 : V = V 0 e 0.t. More generally, a differential equation is separable if it can be written as f (x) ẋ = g (t). In this case, integrating both sides with respect to t yields: f (x) dx dt dt = f (x) dx = This expression can be used to obtain the solution x (t). g (t) dt. Example. Consider 3x ẋ = 4t 3. Then, 3x ẋdt = 3x dx = 4t 3 dt, or x 3 = t 4 + c so that x = ( t 4 + c) /3. Differential equations are classified along a number of dimensions: Order The order of a differential equation is given by the highest-order derivative entering the equation Degree The degree of a differential equation is given by the highest power to which the highest derivative is raised Homogeneity A homogenous equation is such that if x (t) is a solution of this equation, so is k x (t), where k is an arbitrary constant. Autonomous-ness Time does not appear explicitly in an autonomous equation Ordinary/Partial Ordinary differential equations contain only one independent variable and no partial derivatives We will learn only about first-order, first-degree, autonomous, ordinary equations.

47 Linear First Order Differential Equations with Constant Coefficients We consider equations of the form This equation is non-homogenous whenever a = 0. ẋ(t) = a + bx(t). () First, consider the case where x(t) = x is constant. But then ẋ = 0! 0 = a + bx x = a b This is called the Particular Solution. If x takes this value at any time, it will take the value at all later times also. Definition. A point x such that ẋ(t) = 0 is called stationary point. Now let x(t) = x + x(t). Then ẋ(t) = x(t). Substituting into the original differential equation: ẋ(t) = a + bx(t) x(t) = a + b(x + x(t)) x(t) = a + b( a b + x(t)) x(t) = a a + b x(t) x(t) = b x(t) We have ended up with a homogenous version of the original equation. x(t) = b x(t) x(t) x(t) = b d x(t) dt = b dt x(t) dt x(t) dx = b dt ln ( x(t)) + c = bt + c x(t) = Ae bt Solution to the homogenous equation is known as the complementary function. 3

48 But now x(t) = x + x(t). We can substitute back to get x(t) = a b + Aebt This is known as the general solution. It still contains an unknown, A. To find the unknown constant we need a boundary condition, a value of x at a particular time t. Assume, for instance, that we are given, x(0) = x 0. Then: This, finally, is the definite solution. x 0 = a b + Aeb0 x 0 = a b + A A = x 0 + a be bt x(t) = a b + ( x 0 + a b ) e bt Question Under what circumstances will x stay finite as time tends to infinity? Answer If b < 0, x will tend to the stationary point x = a b x will diverge. as t goes to infinity. If b > 0, Definition. A stationary point x is stable if for every x(0), lim t x (t) = x. Solution Method in Summary: To solve a linear, first order differential equation:. Assume x is constant to find the Particular Solution.. Solve the homogenous equation for the Complementary Function. 3. The sum of the two is the General Solution 4. Use boundary conditions to find the Definite Solution. Steps and can be combined by multiplying both sides of () by integrating factor e bt. Then the equation can be rewritten as: e bt ẋ(t) be bt x(t) = ae bt, where the LHS is the derivative of e bt x (t). Hence integrating both sides yields: or e bt x (t) = a b e bt + c x(t) = a b + Aebt. 4

49 3 Linear First Order Differential Equations with Variable Coefficients The coefficients of this equation depend on t so that the equation takes form: ẋ (t) = a (t) + b (t) x (t), where a (t) and b (t) are known functions of t. This equation can be solved by multiplying both sides by the integrating factor ( ) I (t) = exp b (t) dt. multiplying both sides by I (t) yields: I (t) ẋ (t) b (t) I (t) x (t) = a (t) I (t), where the LHS is the derivative of I (t) x (t).hence integrating both sides yields: I (t) x (t) = a (t) I (t) dt. Example 3. Consider the equation: Here a (t) = 5t and b (t) = t. Hence, Consequently, general solution is If x (t 0 ) = x 0, then I (t) = exp ẋ + tx = 5t. ( = exp ( tdt = e t. ) b (t) dt ) x (t) = a (t) I (t) dt I (t) = 5te t dt e t = ( ) 5 e t et + c = 5 + ce t. x 0 = 5 + ce t 0 5

50 so that and thus c = ( x 0 5 ) e t 0 x (t) = 5 ( + x 0 5 ) e (t t 0). 6

51 Professor Anne Sibert Autumn 03 V. Difference Equations This section is concerned with the properties of difference equations: the discrete- time analogue of differential equations. A linear first- order difference equation with a constant coefficient takes the form y t = ay t + x t, t Z, () where a 0 is a constant, { x t } t Z is an exogenous sequence of real numbers and { } t Z is the endogenous sequence to be determined. y t Upon completion of this section, students should know how to: Analyze autonomous linear first- order difference equations graphically. Solve linear first- order difference equations with a constant coefficient by backward and forward recursion. Solve linear first- order difference equations with a constant coefficient backward or forward using lag operators. Use a boundary condition to pick out a unique solution to a linear first- order difference equation with a constant coefficient. V.A. Graphical Analysis As was the case with differential equations, difference equations are said to be autonomous if they do not depend explicitly on time. The linear first- order difference equation in equation () is autonomous if it has the form where x is a constant. y t = ay t + x, t Z, () Z is the set of integers:, -, -, 0,,,.

52 The general form of an autonomous first- order difference equation is y t = F( y t ). Definition V.. A stationary point of an autonomous first- order differential equation y t = F( y t ) is a y * such that y * = F y * ( ). Definition V.. A stationary point is said to be stable if for every possible initial value, lim t y t y*. This subsection considers the graphical analysis of autonomous first- order difference equations. V.A.. Stable stationary points with monotonic convergence Equation () is graphed in Figure, below, for the case of 0 < a <. The horizontal axis represents y t and the vertical axis represents y t. The 45- degree line y t = y t is graphed in black and equation () is graphed in orange. The point where the black line and the orange line intersect is the stationary point y * = x / ( a). The slope of the orange line is strictly positive, as a > 0 and it is less steep than the black line, which has a slope of one, because a <. yt yt=yt- y=ayt- +x yt- y * Figure. Representing Equation () Graphically Suppose that y 0 > y *, as is shown in Figure, below. We can find y on the vertical axis by using the orange line, as shown in the figure. Given the position of y on the vertical axis, we can find y on the horizontal axis by using the black line, as is also

53 shown in the figure. This process can be repeated to find y, first on the vertical axis and then on the horizontal axis. yt yt=yt- y=ayt- +x y y y * y y y0 yt- Figure. Graphing the Solution Path If we continue the iteration process, it is seen that y t converges monotonically to y *, as is shown in Figure 3, below. Likewise if y 0 < y * it is seen that y t converges monotonically to y *. The stationary point is stable and convergence to it is monotonic. yt yt=yt- y=ayt- +x yt- y * Figure 3. Monotonic Convergence to the Stationary Point 3

54 V.A.. Stable stationary points with oscillating convergence Equation () is graphed as the orange line in Figure 4, below, for the case of < a < 0. The slope of this orange line is strictly negative, as a < 0 and it is flatter than the black line, which has a slope of one, because a <. As in the previous subsection, given y 0 we can find y on the vertical axis by using the orange line, as shown in the figure. Given the position of y on the vertical axis, we can find y on the horizontal axis by using the black line, as is also shown in the figure. This process can be repeated to find y, first on the vertical axis and then on the horizontal axis. yt yt=yt- y y=ayt- +x y yt- y y * y y0 Figure 4. Graphing the Solution Path yt yt=yt- y=ayt- +x y * yt- Figure 5. Oscillating Convergence to the Stationary Point 4