# (5) Chapter 5 Differential Calculus(1).ppt - Differential...

• 25

This preview shows page 1 - 11 out of 25 pages.

Chapter 5: Differential Calculus 1.1 Differential Calculus: Review of Calculus Chapter 5
Chapter 4 Review Questions Chapter 5: Differential Calculus 1.2
Inverting A Matrix |1 2 | 1 0| |3 4 | 0 1| Multiply 1 st row by –3 and add to 2 nd row: |1 2 1 0 | |0 -2 -3 1 | Divide 2 nd row by –2: |1 2 1 0 | |0 1 1.5 -.5 | Multiply 2 nd row by –2 and add to the 1 st row: |1 0 -2 1 | |0 1 1.5 -.5 | Answer: | -2 1 | | 1.5 -.5 | Chapter 5: Differential Calculus 1.3
| -2 1 1 0 | | 1.5 -.5 0 1 | Divide 1 st row by –2: | 1 -.5 -.5 0 | | 1.5 -.5 0 1 | Multiply 1 st row by –1.5 and add to 2 nd row: | 1 -.5 -.5 0 | | 0 0.25 0.75 1 | Multiply 2nd row by 4: | 1 -.5 -.5 0 | | 0 1 3 4 | Multiply 2nd row by 0.5 and add to the 1 st row: | 1 0 1 2 | | 0 1 3 4 | Answer: | 1 2 | | 3 4 | Chapter 5: Differential Calculus 1.4
Stock, Bond and Option Arbitrage Future cash flows states 1 & 2: Stock q = |\$10 | Bond b = |100| Call = | \$2 | State 1 |\$16 | |100| | \$8 | State 2 Prices = | \$12 | Stock | \$0.8889| Bond | ? | Option Solving for the pure security values: State: 1 2 | \$12 | = | 10 16 | | p 1 | | \$0.8889 | | 1 1 | | p 2 | Any 2 securities vs. 2 unknowns Chapter 5: Differential Calculus 1.5 Say the riskless payoff in one year is \$1 With present value = \$1/1.125 = \$0.8889 |10 16 | 1 0 | | 1 1 | 0 1| Multiply by -16 & add up = | -6 0 | 1 -16| Divide by -6 | 1 1 | 0 1 | = | 1 0 | -1/6 8/3| Mult by -1, add down | 1 1 | 0 1 | = | 1 0 | -1/6 8/3 | | 0 1 | 1/6 -5/3 |
Inner Products and Orthogonality Examine the inner products: | 1 1 1 | | 1 | = 1+9+-6 = 4 | 9 | | -6 | | 1 1 1 | | -6 | = -6+10+-4 = 0 Yes, orthogonal. | 10 | | -4 | | 1 1 1 | | -1 | = -1+-4+5 = 0Yes, orthogonal. | -4 | | 5 | Chapter 5: Differential Calculus 1.6
Chapter 5 Differential Calculus Chapter 5: Differential Calculus 1.7
Chapter 5: Differential Calculus 1.8 Functions and Limits Y = f(x) “y is a function of x” Associates with each element of a set X one and only one element of a set Y Inverse function : if f is a one-to-one function with domain X and range Y, then a function g with domain Y and range X is called the inverse function of f if f(g(x)) = x g(f(x)) = x Continuous functions : a function f is continuous at a number a if the following 3 conditions are satisfied: ( i) a is in the domain of f ( ii) lim(x a) f(x) exists (iii) lim(x a) f(x) = f(a) There are no discontinuities (gaps or jumps) in the function.
Chapter 5: Differential Calculus 1.9 Limits Lim(x a) f(x) = L In particular, Lim(m  ) (1+1/m) m = e = 2.718 Lim(m  ) (1+i/m) m*n = e i*n
Chapter 5: Differential Calculus 1.10 Slopes, Derivatives Defined
• • • 