Download GEOMETRIC SEQUENCES AND SERIES and more Exercises Mathematics in PDF only on Docsity!
GEOMETRIC SEQUENCES AND SERIES
Summary
- Geometric sequences .............................................................................................................. 2
- Exercise .................................................................................................................................... 6
- Geometric sequence applications to financial mathematics .................................................. 6
- Vocabulary ............................................................................................................................... 7
- Exercises : ................................................................................................................................ 8
- Geometric series...................................................................................................................... 8
- Exercises ................................................................................................................................ 11
- Geometric series applications in financial mathematics ....................................................... 12
During the duration of an investment, the value of an investment can vary in function of time. The study of an investment at different dates produces a sequence of values. The market index, for example, represents a random sequence in itself. At some point, you surely must have observed a curve of market tendencies like this one :
This curve is merely a visualization of the chronological sequence of values:
10 ‐juin 7542 11 ‐juin 7623 12 ‐juin 7743 13 ‐juin 7471 14 ‐juin 7443 15 ‐juin 7501
This section will cover the study of sequences and series. We will particularly study geometric sequences and series since these are the subject of most bank contracts (investments, loans, mortgages).
1. Geometric sequences
Definition: A sequence ሼa (^) ୬ ሽ୬ୀ^ ∞^ ൌ ሼa (^) , aଵ , a (^) ଶ , a (^) ଷ , … ሽ^ is an ordered set of numbers. The index of each term of the sequence indicates the position or order in which specific data is found. This order is very important. For example, the sequence ሼ1,3,5,7,9, … ሽ^ differs from the sequence ሼ9,7,5,3,1, …. ሽ, even if the terms are the same.
Definition: A sequence ሼa (^) ୬ ሽ^ ୬ୀ^ ∞^ ൌ ሼa (^) , aଵ , a (^) ଶ , a (^) ଷ , … ሽ^ is said to be geometric with common ratio ݎ if the terms satisfy the recurrent formula :
Example 1
The sequence ሼ1,2,4,8,16, … ሽ^ is a geometric sequence with common ratio 2, since each term is obtained from the preceding one by doubling. The sequence ሼ9,3,1,1/3, … ሽ is a geometric sequence with common ratio 1/3.
Standard form Generally, we prefer to express the term ܽ of a geometric sequence in function of ݎ and the initial term ܽ , as in the formula:
Example 2
Stocks of a company are initially issued at the price of 10 $. The value of the stock grows by 25 % every year. Show that the value of a stock follows a geometric sequence. Calculate the value of the stock ten years after the initial public offering. Plot a graph of the sequence over a period of 10 years after it was issued.
Example 3
Alberta’s crude oil reserves are diminishing by 10 % each year. Knowing that 100 000 Ml were the initial reserves, show that the crude oil reserves describe a decreasing geometric sequence and find the common ratio for it. Which volume will remain four years later? Plot a graph of the sequence for a period of 20 years.
Solution Each year, the volume decreases by 10 % compared to the previous year:
a (^) ୬ ൌ a (^) ୬ିଵ െ 0,10a (^) ୬ିଵ ൌ 0,90a (^) ୬ିଵ
This relation satisfies the recurrent form of a geometric sequence of common ratio 0,90. Moreover, the sequence is decreasing since 0 ൏ ݎ൏ 1. The initial volume of crude oil is ܽ ൌ 100 000. After 4 complete years, the crude oil reserves are
a (^) ସ ൌ a (^) r ସ^ ൌ 100 000ሺ0,90ሻସ^ ൌ 100 000 ൈ 0,6561 ൌ 65610
There are therefore 65 610 Ml of crude oil in the reserves after four years.
The recurrence formula also allows us to obtain the value of each element of a sequence without knowing ܽ but rather some element ܽ . Any term ܽ of a geometric sequence of common ratio ݎ is obtained from the term ܽ by the relation a (^) ୬ ൌ r ୬ି୩^ a (^) ୩.
Example 4
Gill Bate’s personal fortune doubles every year. If the value of his fortune was estimated at $ 32 000 000 in 2000, how much was it in 1995? At the end of which year will his fortune surpass one billion? ($ 1 000 000 000)?
Solution Each year, the amount of his fortune doubles with regards to the previous year a (^) ୬ ൌ 2a (^) ୬ିଵ. This is a geometric sequence of common ratio r ൌ 2. The initial value of the fortune is unknown, but this information is of no importance thanks to the relation a (^) ୬ ൌ r ୬ି୩^ a (^) ୩ :
aଵଽଽହ ൌ 2 ଵଽଽହିଶ^ a (^) ଶ ൌ 2ିହ^. 32 000 000 aଵଽଽହ ൌ
2 ହ^. 32 000 000 ൌ
aଵଽଽହ ൌ 1 000 000
To obtain the date when one billion will be surpassed, we need to find n such that ܽ ൌ 1 000 000 000.
a (^) ୬ ൌ 2 ୬ିଶ^ a (^) ଶ 1 000 000 000 ൌ 2୬ିଶ^ a (^) ଶ 2 ୬ିଶ^ ൌ
We have an equation in which the variable we want to solve for is in the exponent (see exponential equations). We must use a logarithmic transformation to solve this equation.
2 ୬ିଶ^ ൌ
32 → lnሺ^
୬ିଶ (^) ሻ ൌ ln ൬^1000 32 ൰ → ሺn െ 2000ሻ ln 2 ൌ ln ൬
→ ሺn െ 2000ሻ ൌ
ln ቀ^100032 ቁ ln 2
→ n ൌ 2000
ln ቀ^100032 ቁ ln 2 → n ൌ 2000 4,966 ൌ 2004,
The billion will be reached at the end of 2005.
In this case, it would be faster to create an iterative table in Excel allowing us to observe the temporal evolution of Gill Bate’s fortune. The recurrence formula a (^) ୬ ൌ 2a (^) ୬ିଵ is easily programmed:
will increase by an interest of 6 % at the end of each year. If the person leaves the interest in the account, the annual evolution of the investment is given in the following table:
Time passed deposit interest Balance ‐ 1000$ 0 ܽ ൌ 1000$ 1 year 0 0,06ሺ1000ሻ ൌ 60$ܽ (^) ଵ ൌ 1060$ 2 years 0 0,06ሺ1060ሻ ൌ 63,60$ܽ (^) ଶ ൌ 1123,60$ 3 years 0 0,06ሺ1123,60ሻ ൌ 67,42$ܽ (^) ଷ ൌ 1191,02$ 4 years 0 0,06ሺ1191,02ሻ ൌ 71,46$ܽ (^) ସ ൌ 1262,48$
The temporal evolution of the investment is a geometric sequence. Since a (^) ୬ ൌ a (^) ୬ିଵ 0,06a (^) ୬ିଵ ൌ 1,06a (^) ୬ିଵ , the sequence of accumulated values of the investment is geometric of common ratio 1,06.
4. Vocabulary
Interest dates : dates when the interests are deposited; Interest period : time interval between two interest dates; Capitalization : adding interests to the capital; Periodic interest rate(݅ ) : real interest rate per interest period; Nominal interest rate (݆ ) : This rate, calculated on an annual basis, is used to determine the periodic rate. IT is generally this rate that is posted. It should always be accompanied by a precision on the type of capitalization. Given
݉ൌ number of interest periods in the year ݀ ൌ duration of the period in the fraction of a year ݆ ൌ nominal rate
Then the periodic rate is given by ݅ ݆ൈ ݀ൌ ݉/ ݆ൌ. For example, a rate of "8 % biannually capitalized" signifies that the interest period is the half‐year (݉ ൌ 2 or ݀ ൌ 1/2 ) and that the periodic rate (biannually) is ݅ ൌ 8% 2⁄^ ൌ 4%. The nominal rate does not correspond to the real annual rate, unless the capitalization is annual; Effective rate : real annual interest rate;
In general, if ܸ is the initial amount invested at the periodic interest rate "݅ ", then the value of the investment after ݊ interest periods ܸ , is described by the relation
ܸ ܸൌ (^) ሺ1 ݅ ሻ
(if we let the interests capitalize). The sequence of the value of the investment ܸ ሼ (^) ܸ, (^) ଵ ܸ, (^) ଶ , … ሽ is geometric of common ratio 1 ݅.
Example
A student borrows 2 500 $. The bank loans this money at a rate of 9 %, capitalized monthly. What amount will the student have to reimburse two years later?
Solution When the interest rate is stated this way, it is the nominal rate. Since the capitalization is monthly, the interest period is one month and the number of periods in the year is m ൌ 12. The periodic rate is then i ൌ 0,09 / 12 ൌ 0,0075 per month. The student must reimburse the loan in two years, ݊ ൌ 24 interest periods later. He needs to reimburse
Vଶସ ൌ V ሺ1 iሻଶସ Vଶସ ൌ 2500ሺ1 0,0075ሻଶସ Vଶସ ൌ 2500 ∗ 1,1964 ൌ 2991,
5. Exercises
Problem 1 An investor deposits $ 15 000 in a bank account. The bank offers an interest rate of ݅ ൌ 4,1 % per year.
a) What is the value of the investment 4 years later? (answer : $ 17615,47) b) How much time is needed for the amount to double? (answer : 18 years)
Problem 2 A person wishes to buy a motorcycle worth $ 12 000. In order to collect this amount, he deposits an amount ܸ at the bank, and lets it flourish for 5 years at an interest rate of 5 %, capitalized biannually. Find ܸ . (answer : 9 374,38 $)
Problem 3 The price of a liter of milk in 1990 was $ 0,95. In 2000, the price of milk was fixed at $ 1,42. What was the annual inflation rate for this period? (answer : 4,1 %)
6. Geometric series
Given ܽሼ (^) , ሽ ൌ ሼܽ (^) ܽ, (^) ଵ ܽ,ݎ (^) ଶ ݎ ଶ^ , … ሽ, a geometric sequence of common ratio ݎ. A geometric series is the sum of the elements of a geometric sequence ܽ ܽ (^) ଵ ܽ ݎ (^) ଶ ݎ ଶ^ ..
A series can be finite (with a finite number of terms) or infinite. In order to reduce the writing of a series, we use the summation symbol (see section : The summation symbol) :
Formula to evaluate a finite geometric series :
Given ∑ ୬ୀ a (^) r ୬ൌ a (^) a (^) r a (^) r ଶ^ ⋯ a (^) r , a finite geometric series of common ratio ݎ and of initial a (^) . Then
ே
ୀ
Example 1
Evaluate the geometric series ∑^ ସ୬ୀ4ሺଵଶሻ୬
Solution You need to use the formula for the sum of a geometric series:
4ሺ
2 ሻ^
୬
ସ
୬ୀ
1 െ ሺ^12 ሻ ସାଵ
1 െ ሺ^12 ሻ
1 െ ሺ^12 ሻ
You can also verify that this answer is correct by adding the terms of the geometric series ∑^ ସ୬ୀ 4ሺଵଶ ሻ ୬ൌ 4 2 1 ଵଶ ଵସ ൌ 7,
Example 2
Calculate the sum of the series 9 ሺെ 3ሻ 1 ቀെ ଵଷቁ … ቀെ (^) ଶସଷଵቁ
Solution This is a geometric series of common ratio ݎൌ 1/3 with initial term ܽ ൌ 9. We must also identify the upper limit of summation (the exponent of ݎ of the last term. The last term, identified by ܽ ே is – 1/243 :
ܽ ே ܽൌ ݎ ே^ ⇒ െ
ே
ே
3 ଶ^. 3 ହ^ ൌ ൬െ
ே
ൌ ൬െ
ே
We can therefore represent the series in "sigma" notation and calculate the sum using the formula seen previously:
3 ሻ^
୬
୬ୀ
7. Exercises
Problem1 : Evaluate the following geometric series :
4ሺ
3 ሻ^
୧
୧ୀ 2,5ሺ
ଵ
୩ୀ
2 ሻ^
୬
ହ
୬ୀ
Problem 2 : Evaluate the following geometric series : aሻ 4 1 1/4 1/16 … 1/ bሻ ሺ 1/3ሻ 1 ሺ 3ሻ 9 … 729 cሻ 6 3/2 3/8 3/32 … 3/
ܸൌ ܯ ሺ 1 ሻ ݅^0 ܸ ሺ 1 ሻ ݅^1 ܸ ሺ 1 ሻ ݅^2 ⋯ ܸ ሺ 1 ሻ ݅^11 ܸ ൌ ሺ 1 ሻ ݅
11
ୀ 0
You will have noticed the form of a geometric series of common ratio 1 ݅. By applying the general formula
ܽ (^) ݎ
ே
ୀ
We obtain :
ܸ ൌ ܯ (^) ሺ 1 ሻ ݅ ^ ܸൌ (^) ቈ
1 െ ሺ 1 ሻ ݅^11 ା^1
11
ୀ 0
1 െ ሺ 1 ሻ^12
݅െ ܸൌ ^ ^ ቈ
ሺ 1 ሻ ݅^12 െ 1
Note : In financial mathematics, we generally denote the value of each deposit by "PMT = principal payment" and the number of deposits by "݊ ". Immediately after the last deposit, the acquired value (݁ݑ݈ܽݒ ݁ݎݑݐݑ ݂ൌ ܸܨ ) after a sequence of n equal deposits done at regular intervals at a periodic interest rate "݅ ", is given by the formula :
FV ൌ PMTsnՐi with snՐi ൌ ሺ^1 ାiሻ
nି (^1) i
The interest period considered for the rate "݅ " must correspond to the period between two consecutive deposits.
