1. Sets A set is any collection of elements. Examples: - the set of even numbers between zero and the set of colors on the national flag.

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1 San Francisco State University Math Review Notes Michael Bar Sets A set is any collection of elements Eamples: a A {,,4,6,8,} - the set of even numbers between zero and b B { red, white, bule} - the set of colors on the national flag c C { SFSU students feamale, GPA 3} - the set of SFSU students that satisfy the conditions listed after the vertical bar, ie, female and GPA at least 3 d D {(, R y} - the set of vectors in the two dimensional Euclidean space, such that the -coordinate is equal to the y-coordinate e E {(, R, y, y} - the set of vectors in the two dimensional Euclidean space such that both coordinates are between and and the -coordinate is greater or equal to the y-coordinate It is useful to illustrate this set graphically 8 y6 4 E The set E is colored blue Cartesian product of A and B is the set of all ordered pairs such that the first element belongs to A and the second belongs to B We denote the Cartesian product by A B Eample: A {,,3 }, B {7,8}, then A B {(,7,(,8,(,7,(,8,(3,7,(3,8} Conve sets: A set B is conve if, y B ( y B [,] In words, a linear combination of any two elements in the set also belongs to the set

2 Functions of one variable A function f : A B consists of the domain set (A the range set (B and a rule that assigns to every element in the domain, a unique element in the range We can say that the function f maps from A into B Eamples: a Let A { R } Let B { A, A, B, B, B, C, C, C, D, D, F} The grading scale at the end of the syllabus is a rule that assigns a unique element in B (a letter grade to every numerical grade between and We can give this grading function a name - G, and writeg : A B, where A is the domain and B is the range b Let f : R R be a function from the set of real numbers into the set of real numbers, such that f ( The domain of this function is R, the range is also R and the function assigns to every number its square Graphs of Functions The graph of a function f : A B is a set that all ordered pairs of the form (, f ( such that A and f ( B Formally Gr( f {(, A B y f ( } Eample: let f : R R, and f ( Then Gr( f {(, R R y } is the graph of f Linear functions The general form of a linear function is f ( a b Here a is the intercept with the vertical ais and b is the slope of the function It is important to be able to plot linear functions given their equation The following formula gives the equation of a linear function based on two points in the plane Let (,, (, y be two points in R The equation for the line through those points is given by y y y ( y The quotient on the right hand side is the slope pf the line Hence, we can use this formula to find a linear function when given only one point and the slope Let (, y be a point in R and given that the slope of a line that passes through this point is b, the equation of the line is given by y b( y 3 Inverse functions Let f be a function Then f has an inverse if there is a function g such that the domain of g is the range of f and such that f ( y if and only if g( for all in the domain of f and all y in the range of f In this case, g is the inverse of f, and is designated by f A mapping that assigns possibly more than one value to every element in the domain is called correspondence

3 Thus f ( y if and only if for all in the domain of f and all y in the range of f 3 / 3 Eample: let f ( Then f ( f ( 4 Eponential and logarithmic functions Rules of eponentiation: a b ab a b ( ab a a a ( a a y Rules of logarithms: ln( ln( ln( ln( / ln( ln( ln( a aln( 5 Implicit and eplicit functions Some times when a function is defined, it is clear which variable is the function of the other, for eample: y : R R, y( Here it is clear that y is a function of and we call this form eplicit form or eplicit function On the other hand, in some cases it is not clear which variable is the function of the other variable For eample, when we see the equation y, it is not clear that y is a function of (in which case we can write y( or is a function of y (in which case we write ( 5y We call the form y implicit form or implicit function In general, an implicit function of two variables is of the form g(, c, where c is a constant 6 Derivatives of functions f ( h h The derivative of a function at the point is defined to be lim If such limit h h df ( eists, we say that f is differentiable at the point and denote this limit by or d f '( If f is differentiable at all point in the domain, we say that f is differentiable function The derivative of a function at is equal to the slope of the function at If we let vary over the domain, then the derivative itself is a function of and we call it the derivative function Important rules of differentiation d For constants, : ( d d For sums: [ f ( g( ] f '( g'( d 3

4 Power rule: d d ( n n n d Product rule: [ f ( g( ] f '( g( f ( g'( d d f ( f '( g( f ( g'( Quotient rule: d( g( [ g( ] d Chain rule: [ f ( g( ] f '( g( g'( d Logarithmic function: d ln( d d Eponential function: e e d d Eponential function, general base: a ln( a a d 7 Second derivatives The second derivative of a function is the derivative of the derivative function Suppose y f ( and the derivative function is f '( Then the second derivative of f ( with d f ( respect to is or f ''( d 3 Eample: f (, the (first derivative function of f is f '( 3 and the second derivative is f ''( 6 8 Integrals d Informally, integral is an inverse operation of derivative Suppose f ( F(, that d is f ( is the derivative function of F ( Then, F ( f ( d, that is, F ( is the indefinite integral of f ( For eample, consider the function derivative f ( F'( Then we have: d, that is, integral of Let F ( f ( d, where ( closed interval [ a, b] Then, the definite integral of ( b a F(, and its is the indefinite f is a continuous function defined on a f over that interval is f ( d F( b F( a Geometrically, the definite integral is the signed area between the -ais and the graph of the function f ( within the interval [ a, b] Important rules of integration For constants, : d For sums: [ f ( g( ] d f ( d g( d 4

5 n n Power rule: d n Logarithmic function: ln( d (ln( Eponential function: e d e a Eponential function, general base: a d ln(a Integration by parts: udv uv vdu You should verify the above rules, by differentiating the right hand side 3 Functions of several variables Our definition of a function in section was general and we did not restrict the domain to be one-dimensional Now we focus attention on functions of several variables Eamples: 3 a Let M : R G R, M f ( w, n, r, g be the function that describes the size of a muscle as a function of (w workout, (n nutrition, (r rest time, and (g genetic characteristics of the athlete This is a function of four variables where the first three are assumed to be epressed in terms of real variables and the last one belongs to a set of all possible genetic characteristics for human b Let : R R, f ( p, a be the profit function of a firm that depends on the price of the firm s product and the amount of advertising 3 Partial derivatives Let y f (,, n The partial derivative of f with respect to i is defined as f ( f (,, i h,, n f (,, n lim h i h Other notations: y / i or f i ( or f i ( Eamples a Let f (, 3 Since this is a function of two variables, the two f (, f (, partial derivatives are: 3 and 3 b In the following eample you need to use the chain rule Suppose that g (, and ( t, ( t That is g is a function of, and those variables are themselves functions of t Find the derivative g '( t dg [ ( t, ( t] g[ ( t, ( t] d ( t g[ ( t, ( t] d ( t Solution: dt dt dt The bold faced letters denote vectors That is,,, ( n 5

6 Sometimes you will see the derivative with simpler notations (for lazy people like dg( t myself: g[ ] '( t g [ ] '( t dt 3 Differential Let f (,, n be a function The full differential of this function is defined as follows: f ( f ( df ( d dn The differential gives the change in the value of the function f when we change by d, by d,, and n by d n The changes n d,,dn are assumed to be small Eample: consider the following profit function f (,, where is the firm s output and is the amount of advertising Suppose that initially the firm produces 4 units of output and units of advertising Question: What will be the change in the firm s profit if it increases the output by units and reduces advertising by 5 units? Solution: One way to answer this question is simply calculate the initial profit f ( 4, 3 and the profit after the change f ( 4, The change in profit is +875 Another way to calculate the approimate change in profit is using the differential f ( f ( df (, d d ( ( ( 5 ( 4 ( ( 5 3 The answer we get with the differential is approimation, and the smaller the changes in the variables, the closer we get to the true change in the value of the function 33 Concave and Conve functions f : A R is a concave function if for all, A, f ( ( f ( ( a f ( [,] A function is strictly concave if f ( ( f ( ( a f ( (, Intuitively, a function is concave if for every pair of points on its graph, the cord joining them lies on or below the graph Concave function: 6

7 For conve functions, reverse the inequalities on or above Intuitively, a function is conve if the cord connecting any two points lies on or above the graph of the function Conve function: 34 Theorem (characterization of concavity with second derivatives 34 Let f be twice continuously differentiable function of one variable Then f is concave function if and only if f ''( for all in the domain of f 34 Let f be twice continuously differentiable function of one variable Then f is strictly concave function if and only if f ''( for all in the domain of f For conve functions, reverse the inequality The analogous theorem for multivariate functions is omitted 35 Implicit function theorem (for functions of two variables, sloppy version Consider an epression f (, c This is an implicit form Now, suppose that we wish dy to find the derivative The implicit function theorem says that if f is continuously d f (, differentiable, and, then y a We can solve for y y(, ie, find the eplicit form for y dy( f (, / b d f (, / y f (, It should be obvious why the assumption is necessary for the second part y The implicit function theorem helps us to find the derivative of y with respect to even if we don t have the eplicit form for y as a function of dy Eample: Let f (, y r Find the derivative d 7

8 Solution: dy d f (, / f (, / y y y 4 Homogeneous functions A function f R n k : R is homogeneous of degree k if f ( f ( Eamples: 5 5 a Let f (, This function is homogeneous of degree since f (, ( ( f (, b Let f (, This function is homogeneous of degree since f (, ( ( f (, c Let f (, / This function is homogeneous of degree since f (, / / f (, f (, d Let f (, This function is not homogeneous 4 Euler s theorem for homogeneous functions Theorem: Let f : R n R be homogeneous of degree k Then k f( f ( f n ( k f ( Proof: By definition of homogeneous function of degree k k ( f (,,, n f (,,, n Take the derivative of both sides with respect to : k ( f( f ( f n ( n k f (,,, n Since ( is true for all it must be true for, which gives us the required result 5 Optimization 3 5 Unconstrained optimization with one variable A function f has a local maimum at the point if there is some interval that contains and f ( f ( for all in that interval A function f has a global maimum at the point if f ( f ( for all the domain of f If a function f has maimum at, then is called a maimizer of f We write arg ma f (, which means that is an argument that maimizes the function f The value of f at is called the maimum of f or the maimum value of f 5 Theorem (necessary conditions for maimum Let f ( be twice continuously differentiable function of one variable If f has a local maimum at then 3 In this section we restrict attention to differentiable function only 8

9 a f '( (First Order Necessary Condition, FONC b f ''( (Second Order Necessary Condition For local minimum, reverse the inequality in b What do we mean by necessary conditions? We mean that the above conditions must be satisfied at the point of maimum, but they cannot guarantee maimum If we check those conditions for some function and we find that they are satisfied for some point, we CANNOT conclude that we have a maimum at that point 3 Eample: let f ( At the above conditions are satisfied, but this function is monotone increasing and has neither maimum nor minimum 5 Theorem (sufficient conditions for maimum Let f ( be twice continuously differentiable function of one variable If f '( and f ''(, then f has a local maimum at For local minimum, reverse the last inequality The above two conditions ( f '( and f ''( are together sufficient for maimum This means that if they are satisfied at some point of the domain, then the function has local maimum at that point Eample: Let f ( Verify that f has a local maimum at 5 Solution: f '(, f ''( f '(5, f ''(5 5 Unconstrained optimization with two variables 5 Theorem (sufficient conditions for maimum Let f (, be twice continuously differentiable If *, * f i for i, and * * f (, and f f f f * *, then f has local maimum at (, 5 Theorem (sufficient conditions for minimum Let f (, be twice continuously differentiable If *, * f i for i, and * * * * f (, and f f f f, then f has local minimum at (, Eample: Find a critical point of the function f (, 4 3 and check whether it is maimum or minimum Solution: We remember form calculus that the point at which all partial derivatives are equal to zero is a critical point f 8 3 f 3 3 Solving this system gives us and 7 Now find the second partial derivatives: 8 (critical point 7 4 By Young s theorem, for any twice continuously differentiable function f f, therefore I could have written the last inequality as f f f 9

10 f 8, f 3, f 3, f f f f f 8 ( Hence, the sufficient conditions for maimum are satisfied 53 Constrained optimization This section is the most important for this course In fact, most of the problems in this course and in economics in general, are problems of constrained optimization It is crucial to understand that constrained optimization usually gives entirely different results than unconstrained optimization Suppose that in one case you can chose any car in the world and in another case you can choose any car whose price is below $6 It is likely that your choice would be different in both cases I demonstrate two methods for solving constrained optimization problems, through a particular eample Suppose you have feet of rope and you need to construct a rectangular frame with maimal area This is a problem of constrained maimization Solution: First, we write this problem in a precise mathematical language: ma f (, y, y s t y f is called the objective function In our case f represents the area of the rectangular, with the sides and y Under the ma we write the choice variables In our case these are the lengths of the sides st means such that or subject to After st follows the constraint, which in our case means that the sum of lengths of all sides is equal to fit It is very important in this course to be able write the problems in a clear way, such that a mathematician, who doesn t know economics, will understand what is written 53 Substitution method The idea is to substitute the constraint into the objective function First solve for y in the constraint: y, and then plug this in the objective function The resulting optimization problem is ma ( Notice that this is maimization problem with one variable First Order Necessary Condition: Second Order Necessary Condition: f ''( The necessary and sufficient conditions for maimum are satisfied The first order condition yields * 5 Substituting into the constraint we find the optimal value for y: y * 5 Hence, the maimizer of the objective function is: ( *, y* (5,5 This means that we should use the rope to construct a squared frame The maimum value of the objective function is squared fit

11 53 Lagrange method The following theorem is a special case of Lagrange theorem for maimization of functions of two variables with single equality constraint Obviously, there are generalizations to arbitrary number of variables and inequality constraints Theorem: Let f and g be continuously differentiable functions of two variables Suppose * * that (, is the solution of the problem ma f (,, s t g(, c * * Suppose further that (, is not a critical point 5 of g Then, there is a real number * * * such that (,, is a critical point of the Lagrange function L(,, f (, [ g(, c] * * * In other words, at (,, L L L,, The theorem allows us to convert constrained optimization problem with two unknowns, into an unconstrained optimization, but with three unknowns:,, Notice that the theorem applies to constrained maimization problems, as well as to constrained minimization problems The theorem provides the first order necessary conditions for optimum The number is called Lagrange multiplier or shadow price Now, lets apply the theorem to our problem with the rope The Lagrange function is L(, y, y [ y ] The first order necessary conditions are: L ( y, L ( y L (3 y Notice that the last condition just repeats the constraint From ( and ( it follows that y Plug this in the constraint to get * y* 5, which means that in order to maimize the area of the rectangle, we should make it a square The value of the Lagrange multiplier is 5 * Remark: there is an economic meaning to the Lagrange multiplier * gives the change of the optimal value of the objective function that results from small unit change in the 5 Critical point is a point at which all partial derivatives are equal to zero

12 constraint To understand this, suppose that we have more fit of rope to construct the rectangular What would be the change in the maimal area? According to the Lagrange multiplier, it should increase by 5 Indeed, our calculations say that we should form a square from fit The side of the square is 55 and its area is 7565 The Lagrange multiplier gave us a pretty good approimation for the answer, and the smaller the change in the constraint, the better the approimation Second Order Sufficient Condition: Form the Hessian matri for the Lagrange function: L L L3 H ( L L L L3 L3 L3 L33 The Second order sufficient condition for maimum says that the determinant of this Hessian should be positive (negative for minimum