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Doing the Mathematics of Economic

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Mathematics serves as foundation for modern economics. Almost every branch of mathematics finds its role in economic analysis. Arithmetic, Algebra; Analytical Geometry, and Calculus are liberally applied in economic theories. Study of economic concepts will be difficult without the understanding of mathematics that goes with it. Mathematics helps you learn economic functions, viz. demand, supply; production, cost; revenue, and consumption, logically and with clarity. It helps deal with all the basic and advanced equations used in economics.

Arithmetic, Analytical Geometry; Algebra, and Calculus

Arithmetic is used in economics to calculate percentage changes in prices and other functions. It finds use in comparing trends in key economic data and accounting ratios.

Analytical Geometry teaches you how to learn certain economic theories with the help of graphic representations. Complicated demand- supply and demand-price equations can be easily explained through graphs. Demand-supply and demand-price graphs give the rate of change of demand, the rate of change of price, the rate of change of supply etc. at any given point in time.

Algebra deals with equations. Equations mean whatever value implied on the left hand side is the same as what is on the right hand side. These are variable values. To maintain balance of the equation, whatever changes made to the right hand side has to be made to the left hand side and vice versa. Economics finds this theory’s application as it also deals in relationships between variable values. Algebra uses symbols to tell the variable values represented in economics.

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Calculus deals with limits and infinity. Calculus has greater application in economics. Students of economics have to do lots of practice to understand derivation in calculus as the derived value or derivative in calculus has its application in economics. Derivative is an important concept in economics that is concerned with decision making.  

2.36 Compound interest formula: Total amount = P (1 + (R/100))n where P is principal, R is rate, and n is number of years

$1,200 compounded annually for 5 years at 9% = $1,846.35

$1,000 compounded annually for 4 years at 9% = $1,411.58

$800 compounded annually for 3 years at 9% = $1,036.02

$600 compounded annually for 2 years at 9% =$712.86

Total amount in the fund immediately after the 5th deposit will be $1,846.35 + $1,141.58 + $1,036.02 + $712.86 + $400 (the 5th deposit) = $5,406.81

2.31 Loan payment formula: monthly payment = [rate + (rate/1 + rate)months – 1] x principal

Amortization schedule for $20,000 loan payable annually for 5 years at 10% compound interest

Year

Interest

Principal

Balance

2012

$1,729.28

$6,014.85

$13,985.15

2013

$1,099.44

$6,644.68

$7,340.47

2014

$403.66

$7,340.47

$0.00

Annual payment = $7,744.13

Interest in second year = $1,099.44

2.48 In the first diagram, the total cash flow in five years is $600. At 10% annual compound interest, the principal amount invested at the beginning of year one is $983. Since the second diagram is said to be equivalent to the first diagram, then the value of X is $120 ($600/5 years). This is because in the second diagram, the annual cash flows are equal for all the five years.

2.49 The total amount of cash flows in the first diagram is $220. The number of investment is 4 years. The value of C that makes the second investment to be economically equivalent at an interest rate of 10% is $36.667 [$220/(c+2c+2c+c)].

2.52 The total amount of inflow is $9,000 (for eight years). An equal amount of outflow is introduced at the beginning of ever year from year 2 to year 8. However, the initial outlay is as twice as the amount invested every year from year 2. This initial amount yield $800 at the end of year 1. Since is interest rate is given as 12% p.a, then the initial amount invested is $6,667 (refer to 2.36 for formula). If 2C = $6,667, then C = $3,333.5.

2.54 Formula for calculating present value of annuity:

Total = Amount [(1 + r)n+1 – 1/r] – amount

Present value of annual series of $5,000 for 10 years at 10% compound interest = $30,722.84

The amount $30,722.84 is equivalent to $8,104.5 invested for 5 years at 10 % compound interest.

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