PRELIMINARY MATHEMATICS LECTURE 7 Revision

On the pre-session exam 10-13.00 Friday 25 September 2015 Room V111 You will be given two answer books – one for math and one for statistics. The math part of the exam will be 1 1/2 hours, followed by another 90mins of statistics exam. For the math part, you will be given 5 questions, out of which you will be required to answer 3 questions. You will be allowed to use your electronic calculator in this examination provided that it cannot store text. The make and type of calculator must be stated clearly on the front of your answer book.

Equations confirmable for matrix representation

1. Simultaneous linear equations

1. Simultaneous linear equations

Quadratic forms For example, given For ease of matrix representation, we can rewrite this polynomial as

Quadratic forms The symmetric matrix representation can be obtained as: Note that in contrast to the previous example in which a system of linear equations is expressed in matrix which would allow us to test for the existence of a solution as well as to apply Cramer’s rule to obtain the solution, here we are expressing the quadratic form not in order to obtain solution to the variables (since there are three unknown variables in one equation, we cannot find unique solutions) but instead in order to find a parametrical condition for the sign of the quadratic form.

Other functional forms Consider

Other type of functions Consider We can rewrite this as

Other functional forms And express as

Other functional forms On the other hand, would not be suitable for matrix representation because we would end up with 4 variables and 3 equations.

Other functional forms In a similar spirit can be transformed to a system of linear equations

Other functional forms and hence be expressed as

Derivatives, differentials, total differentials, and second order total differentials

In a function y = f (x) the derivative Derivatives In a function y = f (x) the derivative denotes the limit of as Δ x approaches zero.

The differential of y can be expressed as Differentials The differential of y can be expressed as which measures the approximate change in y resulting from a given change in x.

Partial derivative In a function, such as z = f (x, y), z is the dependent variable; and x and y are the independent variables, partial derivative is used to measure the effect of changes in a single independent variable (x or y) on the dependent variable (z) in a multivariate function. The partial derivatives of z with respect to x measures the instantaneous rate of change of z with respect to x while y is held constant:

Partial derivative In a function, such as z = f (x, y), z is the dependent variable; and x and y are the independent variables, partial derivative is used to measure the effect of changes in a single independent variable (x or y) on the dependent variable (z) in a multivariate function. The partial derivatives of z with respect to y measures the instantaneous rate of change of z with respect to y while x is held constant:

Total differentials For a function of two or more independent variables, the total differential measures the change in the dependent variable brought about by a small change in each of the independent variables. If z = f (x, y), the total differential dz is expressed as

We can also derive the second-order total differential as Determinantal test We can also derive the second-order total differential as

Positive/ negative definiteness of q = d 2 z and the second order sufficient conditions for relative extrema

Positive and negative definiteness A quadratic form is said to be • positive definite if • positive semi-definite if • negative semi-definite if • negative definite if regardless of the values of the variables in the quadratic form, not all zero.

Positive and negative definiteness A quadratic form is said to be • positive definite if • positive semi-definite if • negative semi-definite if • negative definite if If q changes signs when the variables assume different values, q is said to be indefinite

Second-order sufficient conditions for minimum and maximum If d 2 z > 0 (positive definite), local minima If d 2 z < 0 (negative definite), local maxima

Second-order necessary conditions for minimum and maximum If d 2 z ≥ 0 (positive semi-definite), local minima If d 2 z ≤ 0 (negative semi-definite), local maxima

Indefinite q = d 2 z and saddle point When q = d 2 z is indefinite, we have a saddle point

Conversion and inversion of log base

Conversion of log base This rule can be generalised as where b ≠ c

Inversion of log base Also,

Example Find the derivative of We know that given,

Example Using the rule and

On the sign of lambda in the Lagrangian

The Lagrange multiplier method Max/Min z = f (x, y) (1) s.t. g (x, y) = c (2) Step 1. Rearrange the constraint in such a way that the right hand side of the equation equals a zero. Setting the constraint equal to zero: c – g (x, y) = 0

The Lagrange multiplier method Max/Min z = f (x, y) (1) s.t. g (x, y) = c (2) Step 2. Multiply the left hand side of the new constraint equation by λ (Greek letter lambda) and add it to the objective function to form the Lagrangian function Z. Z = f (x, y) + λ [ c – g (x, y) ]

The Lagrange multiplier method Z = f (x, y) + λ [ c – g (x, y) ] Step 3. The necessary condition for a stationary value of Z is obtained by taking the first order partial derivatives, set them equal to zero, and solve simultaneously: Zx = fx – λ gx = 0 Zy = fy – λ gy = 0 Zλ = c – g (x, y) = 0

The Lagrange multiplier method Max/Min z = f (x, y) (1) s.t. g (x, y) = c (2) Alternatively if we express the Lagrangian function Z as. Z = f (x, y) – λ [ c – g (x, y) ]

The Lagrange multiplier method Z = f (x, y) – λ [ c – g (x, y) ] In Step 3. The first order necessary conditions are now: Zx = fx + λ gx = 0 Zy = fy + λ gy = 0 Zλ = c – g (x, y) = 0 Is this a problem?

Example of cost minimizing firm Min c = 8 x2 – xy + 12y2 s.t. x + y = 42 Form the Lagrangian function C. C = 8 x2 – xy + 12y2 – λ (42 – x – y)

Example of cost minimizing firm C = 8 x2 – xy + 12y2 – λ (42 – x – y) Obtain the first order partial derivatives, C x = 16 x – y + λ C y = – x + 24y + λ C λ = 42 – x – y

Solution using matrix 16 x – y + λ = 0 – x + 24y + λ = 0 x + y = 42 Since the three first order conditions are linear equations, we can use matrix to obtain solutions:

Using the Cramer’s rule Solution using matrix Using the Cramer’s rule

Using the Cramer’s rule Solution using matrix Using the Cramer’s rule

Using the Cramer’s rule Solution using matrix Using the Cramer’s rule

Using the Cramer’s rule Solution using matrix Using the Cramer’s rule

Note that the sign of the determinants all changed except for Solution using matrix Note that the sign of the determinants all changed except for Recall from lecture 2… Property of the determinant (3) The multiplication of any one row (or one column) by a scalar k will change the value of the determinant k-fold.

Using the Cramer’s rule Solution using matrix Using the Cramer’s rule Solution for x and y are unchanged. λ is the same numerical value with a negative sign.