Core: Finance and Recursion Notes PDF

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Detailed notes on the core financial coursework for VCE Further Maths...

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RECURSION AND FINANCIAL MODELLING

GENERAL TIPS/THINGS TO REMEMBER  ALWAYS look for the letter/value used to represent the rule or recurrence relation – e.g. An, Vn, Sn etc.  Use financial calculator WHEREVER POSSIBLE – it’s used for a reason  If annuity/perpetuity is not explicitly stated, DO NOT ASSUME – do what the question asks!!

SEQUENCES AND RECURRENCE RELATIONS Recurrence Relations Refers to a mathematical rule that can be used to generate a sequence. It has two parts: 1. A STARTING POINT: (V0) 2. A RULE FOR SUCCESSION EXAMPLE: 10, 15, 20, ... 1. V0 = 10 and 2. Vn+1 = Vn + 5

EXAMPLE: Write down the first five terms of the sequence defined by the recurrence relation V0 = 9, Vn+1 = Vn –4 showing the values of the first four iterations. V0 = 9 V1 = V0—4 =9–4=5 V2 = V1—4 =5–4=1 V3 = V2—4 = 1 – 4 = -3 V4 = V3—4 = 3 – 4 = -7 The sequence is 9, 5, 1, -3, -7, ... MODELLING LINEAR GROWTH AND DECAY Recurrence Model for Linear Growth and Decay As a general rule, if D is a constant, a recurrence relation rule of the form: Vn+1 = Vn + D can be used to model linear growth Vn+1 = Vn—D can be used to model linear decay where D = r/100 x V0 Simple Interest Loans and Investments If you deposit money into a bank account, the bank is effectively borrowing money from you. Interest = When you deposit money into a bank account, the bank pays you a fee for using your money. Simple interest investment = When a fixed amount of interest is paid into the account at regular time intervals. Simple interest loan = When you borrow money from the bank and are charged a fixed amount of interest after regular time periods. Simple interest is a special case of linear growth in which the starting value is the amount borrowed or invested. The amount borrowed or invested is called the principal. The amount added at each step is the interest and is usually a percentage of this principal. Recurrence Model for Simple Interest Let Vn be the value of the loan or investment after n years. Let r be the percentage interest rate. The recurrence relation for the value of the loan or investment after n years is V0 = principal, Vn+1 = Vn + D where D= r /100 × V0. EXAMPLE: Monica invests $5000 in an investment account that pays 4.8% per annum simple interest. Model this simple investment using a recurrence relation of the form:V0 = principal, Vn+1 = Vn + D where D = r/100 x V0. Vn = the value of the investment after n years. V0 = 5000 r = 4.8%. D = 4.8/100 x 5000 = 240.

RECURRENCE RELATION: V0 = 5000, Vn+1 = Vn + 240. (a.) Determine the value of the investment after 3 years. (b.) When will the investment first exceed $6000, state its value. (a. & b.) Use CAS steps on page 39 to determine that (a.) = $5720 and (b.) = After 5 years; $6200. Depreciation Depreciation = when an asset loses monetary value over time. Future Value = the estimated value of the asset after a certain period of time.  use flat-rate depreciation or unit-cost depreciation. Scrap Value = after a certain amount of time or when the value of the asset has depreciated to a certain amount where it is no longer considered a valued asset. Written off = When an asset is no longer considered of any substantial value and has reached the end of its usefulness. Flat-Rate Depreciation Flat-rate depreciation is very similar to simple interest, but instead of adding a constant amount of interest, a constant amount is subtracted to decay the value of the asset after every time period. This constant is called the depreciation amount and, like simple interest, it is often given as a percentage of the initial purchase price of the asset. Recurrence Model for Flat-Rate Depreciation Vn = the value of the asset after n years. r = the percentage depreciation rate. The recurrence relation for the value of the asset after n years is: V0 = initial value of the asset, Vn+1 = Vn - D where, D = r/100 x V0 EXAMPLE: A new car was purchased for $24000 in 2014. The car depreciates by 20% of its purchase price each year. Model the depreciating value of this car using a recurrence relation of the form: V0 = principal, Vn+1 = Vn - D where D = r/100 x V0. Answer: Vn = the value of the investment after n years. V0 = 24000 r = 20%. D = 20/100 x 24000 = 4800. RECURRENCE RELATION: V0 = 24000, Vn+1 = Vn - 4800. (a.) Determine the value of the car after 2 years. (b.) If purchased in 2014, in what year will the car’s value depreciate to zero? (a. & b.) Use CAS steps on page 39 to determine that (a.) = $14 000 and (b.) = In 2019, after 5 years depreciation. Unit Cost Depreciation When an item loses value because of how often its used rather than its age Recurrence Model for Unit Cost Depreciation Vn = the value of the asset after n years of use. D = the cost per use. The recurrence relation for the value of the asset a er n years is: V0 = initial value of the asset, Vn+1 = Vn - D EXAMPLE:A professional gardener purchased a lawn mower for $270. The mower depreciates in value by $3.50 each me it is used. Model the depreciating value of this mower using the recurrence relation: V0 = principal, Vn+1 = Vn - D where D = depreciation in value per use Vn is the value of the mower after being used to mow n lawns.

V0 =270, and D=3.50 RECURRENCE RELATION: V0 = 270, Vn+1 = Vn - 3.50 (a.) Determine the value of the mower after it has been used 3 times. (b.) How many times can the mower be used before its depreciated value is less than $250? (a. & b.) Use CAS steps on page 39 to determine that (a.) = $259.50 and (b.) = After six mows. RULES FOR THE NTH TERM IN A SEQUENCE MODELLING LINEAR GROWTH AND DECAY Rules for Nth Term for Linear G&D As a general rule, if D is a constant, a recurrence relation rule of the form: Vn = V0 + nD can be used to model linear growth Vn = V0 - nD can be used to model linear decay. D = r/100 x V0 This general rule can then be applied to simple interest investments and loans, flat-rate depreciation and unit-cost depreciation. Simple Interest Investments and Loans V0 = the initial value of the simple interest investment or loan r = the annual interest rate. The value of a simple interest investment or loan a er n years is: Vn = V0 + nD where D = r/100 x V0 or Vn = V0 + n x r/100 x V0 EXAMPLE: The following recurrence relation can be used to model a S.I.I: V0 = 3000, Vn+1 = Vn + 260 where Vn is the value of the investment after n years. (a.) What is the principal of the investment? How much interest is added to the investment each year? (b.) Write down the rule for the investment after n years. (c.) Use a rule to find the value of the investment after 15 years. (d.) Use a rule to find when the value of the investment first exceeds $10 000. (a.) Principal: $3000; Amount of Interest = $260 (b.) Vn = 3000 + n x 260 (c.) V15 = 3000 + 260 x 15 (d.) 10 000 = 3000 + 260n 50 7000 = 260n or n = 7000/260 =26.92 years. The value of the investment will first exceed $10 000 after 27 years. EXAMPLE: Susan invests $3000 in a S.I.I of paying interest at the rate of 6.5% per year. Use a rule to find the value of the investment after 10 years. V10 = 3000 + 10 x (6.5/100) x 3000 = $4950

Flat Rate of Depreciation V0 = the initial value of the asset.

r = the at rate of depreciation The value of asset after n years is: Vn = V0 – nD where D = r/100 x V0

or

Vn = V0 - n x (r/100 x V0)

EXAMPLE: The following recurrence relation can be used to model the F-R of depreciation of a set of office furniture: V0 = 12 000, Vn+1 = Vn - 1200 where Vn is the value of the furniture after n years. (a.) What is the inti al value of the furniture? By how much does the furniture decrease in value each year? (b.) Write down the rule for the value of the investment after n years. (c.) Use a rule to find the value of the investment after 6 years. (d.) How long does it take for the furniture’s value to decrease to zero? (e.) A photocopier in the office costs $6000 when new. It depreciates at the at rate of 17.5%. What is its value after 4 years? (a.) Initial value = $12 000, Depreciation = $1200 (b.) Vn = 12 000 - n x 1200 = 12 000—1200n (c.) V6 = 12 000 - 1200 x 6 = $4800 (d.) 0 = 12 000—n x 1200 so n = 10 years.The value of the furniture will depreciate to zero after 10years. (e.) V4 = 6000 - 4 x (17.5/100) x 6000 =$1800 Unit Cost V0 = the initial value of the asset D = the cost per unit of use The value of asset after n units of use is: Vn = V0 - nD EXAMPLE: A hairdryer in a salon was purchased for $850. The value of the hairdryer depreciates by 25 cents for every hour it is in use.Let Vn be the value of the hairdryer after n hours of use. (a.) Write down a rule to find the value of the hairdryer after n hours of use. (b.) What is the value of the hairdryer after 50 hours of use? (c.) On average, the salon will use the hairdryer for 17 hours each week. How many weeks will it take the value of the hairdryer to halve? (a.) 1 Identify the values of V0 and D. V0 = 850 and D = 0.25 2 Write down the rule for the value of the hairdryer after n hours of use Vn = 8500 - .25n (b.) 1 Decide on value of n and substitute into the rule; then write anwer After 50 hours of use, n = 50. V50 = 850 - 0.25 x 50 V50 = 837.50; After 50 hours of use, the hairdryer has a value of $837.50. (c.) 1 Find the value of Vn; halving the value of the hairdryer means it will be $425, so this is Vn Vn = 425 (as 850/2 = 425) 2 Write down the rule, with the value of the hairdryer, Vn = 425

425 = 850 - 0.25n 3 Solve the equation for n 0.25n = 850—425 0.25n = 425 n = 1700 4 Divide by 12 to get number of weeks and write down answer 1700/17 = 100 weeks; After 100 weeks, the hairdryer is expected to halve in value.

MODELLING GEOMETRIC GROWTH AND DECAY Recurrence Model for Geometric Growth and Decay As a general rule, if R is a constant, a recurrence relation rule of the form: Vn+1 = RVn for R > 1, can be used to model geometric growth. Vn+1 = RVn for R < 1, can be used to model geometric decay. EFFECTIVELY: Vn = principal, Vn+1 = 1 Vn REMINDER: > (left value is greater than right value) < (left value is less than right value) We read left to right, so we interpret the symbols from values left to right. Compound Interest Refers to geometric growth, where the any interest earned after one me period is added to the principal, which then contributes to the earning of interest in the next me period. Recurrence Model for Compound Interest Investments & Loans Vn = the value of the investment after n years r = the percentage of interest per compound period. The recurrence model for the value of the investment after n compounding periods: V0 = principal, Vn+1 = RVn where R = 1 + r/100 EXAMPLE: The following recurrence relation can be used to model a compound interest investment of $2000 paying interest at the rate of 7.5% per annum. V0 = 2000, Vn+1 = 1.075 x Vn where Vn is the value of the investment after n years. (a.) Use the recurrence relation to find the value of the investment after 1, 2 and 3 years (b.) Determine when the value of the investment will first exceed $2500 (c.) Write down the recurrence relation if $1500 was invested at a compound interest rate of 6.0% per annum. (a.) 1 Write down the principle of the investment, V0 V0 = 2000 2 Use the recurrence relation to calculate V1, V2 and V3 V1 = 1.075 x 2000 = 2150, V2 = 1.075 x 2150 = 2311.25, V3 = 1.075 x 2311.25 = 2484.59 (b.) After 4 years, the investment will first exceed $2500 (c.) 1 Identify the value of V0 and r

V0 = 1500 and r = 6 2 Calculate the value of R The value of the investment will grow over time so,R = 1 + (r/100) = 1 + (6/100), R = 1.06 3 Write answer V0 = 1500, Vn+1 = 1.06 x Vn EXAMPLE: Matthew borrows $5000 from a bank. He will pay interest at the rate of 4.5% per annum. Where Vn = the value of the loan after n compounding periods, write down a recurrence relation to model the value of Brian’s loan if interest is compounded:(a.) yearly (b.) quarterly (c.) monthly (a.) Vn = the value of Matthew’s loan after n years.The interest rate is 4.5% per annum. R = 1 + (4.5/100) = 1.045 RECURRENCE RELATION: V0 = 5000, Vn+1 = 1.045 x Vn (b.) Vn = the value of Matthew’s loan after n quarters.The interest rate is 4.5% per annum. The quarterly rate is 4.5/4 = 1.125R = 1 + (1.125/100) = 1.01125 RECURRENCE RELATION: V0 = 5000, Vn+1 = 1.01125 x Vn (c.) Vn = the value of Matthew’s loan after n months.The interest rate is 4.5% per annum. The monthly rate is (4.5/12) = 0.375.R = 1 + (0.375/100) = 1.00375 RECURRENCE RELATION: V0 = 5000, Vn+1 = 1.00375 x Vn Recurrence Model for Reducing Balance Depreciation Vn = the value of the asset after n years. r = the annual percentage depreciation. The recurrence model for the value of the asset after n years: V0 = initial value, Vn+1 = RVnwhere R=1- (r - 100) EXAMPLE: The following recurrence relation can be used to model the value of office furniture with a purchase price of $6900, depreciating at a reducing balance rate of 7% per annum. V0 = 6900, Vn+1 = 0.93 x Vn where Vn is the value of the office furniture after n years. (a.) Use the recurrence relation to find the value of the office furniture, correct to the nearest cent, after 1, 2 and 3 years. (b.) Determine when the value of the asset will first be less than $5000. (c.) Write down the recurrence relation if the furniture was initially valued at $7500 and is depreciating at a reducingbalance rate of 8.4% per annum. (a.) Generate the first 3 terms V1 = 0.93 x 6900 = 6417 V2 = 0.93 x 6417 = 5967.81 V3 = 0.93 x 5967.81 = 5550.06 (b.) The value of the furniture drops below $5000 after 5 years. (c.) Identify the value of V0, calculate R and write the answer V0 = 7500, Depreciation rate is 8.4% per annum R = 1 - (8.4/100) or R = 0.916 RECURRENCE RELATION: V0 = 7500, Vn+1 = 0.916 x Vn

RULES FOR THE NTH TERM IN A SEQUENCE MODELLING GEOMETRIC G&D For a geometric growth or decay recurrence relation: V0 = starting value, Vn+1 = RVn The value of the nth term of the sequences is generated by the rule: Vn = Rn x V0 This general rule can then be applied to compound interest investments and loans and reducing-balance depreciation. EXAMPLE: If $2000 is invested in a compound interest investment paying 5% interest per annum, compounding yearly, the investment will increase by the different amounts each year. Vn is the value of the investment after n years. If we let V0 = 2000, Vn+1 = 1.05Vn Using this recurrence relation, we can write out the sequence of terms generated as follows: V0 = 2000 V1 = 1.05V0 V2 = 1.05V1 = 1.05 (1.05V0) = 1.052 V0 V3 = 1.05V2 = 1.05 (1.052V0) = 1.053V0 V4 = 1.05V3 = 1.05 (1.053V0) = 1.054V0 Following this pattern, after n years’ interest has been added, we can write: Vn = 1.05nV0 So, using this rule, the value of the investment after 20 years would be: V 20 = 1.0520 x 2000 = $5306.60 (to nearest cent) EXAMPLE: A principal value of $10 000 is invested in an account earning compound interest at the rate of 9% per annum. The rule for the value of the investment after n years, Vn, is shown below.Vn = 1.09n x 10 000 (a.) Find the value of the investment after 4 years, correct to the nearest cent. (b.) Find the amount of interest earned after 4 years, correct to the nearest cent. (c.) Find the amount of interest earned in the fourth year, correct to the nearest cent. (d.) If the interest compounds monthly instead of yearly, write down a rule for the value of the investment made after n months. (e.) Use this rule to find the value of the investment after 4 years (48 months). (a.) V4 = 1.094 x 10 000, V4 = 14 115.816 After 4 years, the value of the investment is $14 115.82. (b.) Amount of interest = $14 115.82—$10 000 = $4115.82 After 4 years, the amount of interest earned is $4115.82. (c.) V3 = 1.093 x 10 000, V3 = 12 950.29 (nearest cent) 9% of V3 = 0.09 x 12950.29 = 1165.53 Interest of $1165.53 was earned in the fourth year. (d.) Let Vn be the value of the investment after n months.r = 9%/12 = 0.75%, R = 1 + (0.75/100) = 1.0075 Vn = 1.0075n x 10 000 (e.) V48 = 1.007548 x 10 000 = $14314.05 Compound Interest Loans and Investments V0 =the amount borrowed or invested (principal). r = the interest rate per compounding period. The value of a compound interest loan or investment after n compounding periods, Vn, is given by the rule:

Vn = (1+ r/100)n x V0 EXAMPLE: How many years will it take an investment of $2000, paying compound interest at 6% per annum, to grow above $3000? Write your answer to the nearest year. V0 = 2000 R = 1 + (6/100) = 1.06 Vn = 3000 Vn = Rn x V0 , 3000 = 1.06n x 2000 CAS solve function: solve (3000 = (1.06)n x 2000, n)n = 6.95851563317 After 6.95... years, the value of the investment is $3000. It will grow above $3000 after 7 years. Reducing Balance Depreciation V0 =the purchase price of the asset. r = the annual percentage rate of depreciation. The value of an asset after n years, Vn, is given by the rule: Vn = (1 – r/100)n x V0 EXAMPLE: A computer system costs $9500 to buy, and decreases in value with reducing-balance depreciation of 20% each year. A recurrence relation that can be used to model the value of the computer system a er n years, Vn, is shown below. V0 = 9500, Vn+1 = 0.8 x Vn (a.) Write down the rule for the value of the computer system after n years. (b.) Use a rule to find the value of the computer system after 8 years. Write your answer, correct to the nearest cent. (c.) Calculate the total depreciation of the computer after 8 years. (a.) V0 = 9500, R = 1 – (20/100) = 0.8 Vn =Rn x V0 Vn = 0.8n x 9500 (b.) V8 = 0.88 x 9500 After 8 years, the value of the computer system is $1593.84, correct to the nearest cent. (c.) Depreciation = $9500 - $1593.84 =$7906.16. After 8 years, the computer system has depreciated by $7906.16. EXAMPLE: An industrial weaving company purchased a new loom at a cost of $56 000. It has an estimated value of$15 000 after 10 years of operation. If the value of the loom is depreciated using a reducing-balanced method, what is the annual rate of depreciation? Write your answer correct to one decimal place. V0 = 56 000 Vn = 15 000 R = 1 - (r/100) n = 10 Vn = Rn x V0 15 000 = (1 - (r/100)10 x 56 000 CAS solve function: solve(15000 = (1 - r/100)10 x 56000, r) r = 12.3422491484 or r = 187.657750852 The annual rate of depreciation is 12.3%, correct to one decimal place. NOMINAL & EFFECTIVE INTEREST RATES Nominal Interest Rate Compounding interest rates are usually quoted as annual rates, or interest rate per annum. Nominal Interest Rate: Interest rate for the investment or loan. Sometimes an annual rate might be quoted, but the interest can be calculated and paid according to a different me period, e.g. monthly.

Compounding Period: the me period after which com- pound interest is calculated and paid. The terms of a compound interest loan or investment are usually quoted as a nominal interest rate followed by a compounding period. The interest rate for the compounding period is easily calculated using simple arithmetic. It may be assumed that there are: 12 equal months every year (despite varied numbers of days) 4 quarters in every year (1 quarter = 3 months) 26 fortnights in a year (even though there are slightly more) 52 weeks in a year (even though there are slightly more) 365 days in a year (ignore existence of leap years) A nominal interest rate is converted to a compounding interest rate by dividing by these numbers. EXAMPLE:An investment account will pay interest at the rate of 3.6% per annum. Convert this interest rate to a (a.) monthly rate (b.) fortnightly rate (c.) quarterly rate (a.) Monthly interest rate = 3.6/12 = 0.3% (b.) Fortnightly interest rate = 3.6/26 = 0.138% to 3 d.p. (c.) Quarterly interest rate = 3.6/4 = 0.9%

Effective Interest Rate As a general principle with compound interest, the more frequently interest is calculated and added to an investment or loan (the compounding period), the more rapidly the value of the investment or loan increases. Effective interest rate of a loan or investment: the interest earned after one year expressed as a percentage of the amount borrowed or invested. Can be used to compare the investment performance. r = the nominal interest rate per annum. reffective = the effective annual interest rate(actual interest rate applied). n = the number of times the interest compounds each year (compoundin...