1. PRELIMINARY MATHEMATICS
LECTURE 7
Revision

2. On the pre-session exam
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.

3. Equations confirmable for matrix representation

4. 1. Simultaneous linear equations

5. 1. Simultaneous linear equations

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

7. 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.

8. Other functional forms
Consider

9. Other type of functions
Consider
We can rewrite this as

10. Other functional forms
And express as

11. 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.

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

13. Other functional forms
and hence be expressed as

14. Derivatives, differentials, total differentials, and second order total differentials

15. 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.

16. 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.

17. 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:

18. 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:

19. 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

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

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

22. 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.

23. 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

24. 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

25. 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

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

27. Conversion and inversion of log base

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

29. Inversion of log base
Also,

30. Example
Find the derivative of
We know that given,

31. Example
Using the rule
and

32. On the sign of lambda in the Lagrangian

33. 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

34. 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) ]

35. 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

36. 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) ]

37. 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?

38. 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)

39. 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

40. 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:

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

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

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

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

45. 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.

46. 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.