Transcript Document

Sections 2.1, 2.2, 2.3, 2.4, 2.5
Before the technological advances which made it easy to calculate
exponential and logarithmic functions, approximations were available.
Today, such approximations are useful primarily for “quick” calculations
when a calculator or computer is not available. One way to obtain such
approximations is to use a Maclaurin series expansion (which you should
recall from calculus):
(x – 0)2 f (0) (x – 0)3 f (0)
f(x – 0) = f(0) + (x – 0) f (0) + —————– + —————– + …
2!
3!
Suppose f(x) = (1 + x)k where k > 0. Then f (x) = k(1 + x)k–1
f (x) = k(k – 1)(1 + x)k–2
We now have (1 +
x)k
f (x) = k(k – 1)(k – 2)(1 + x)k–3
k(k – 1) 2 k(k – 1)(k – 2) 3
= 1 + kx + ——— x + —————— x + …
2!
3!
Suppose f(x) = ekx where k > 0. Then
f (x) = k2ekx
f (x) = kekx
f (x) = k3ekx
k2 2 k3 3
We now have ekx = 1 + kx + — x + — x + …
2!
3!
From the previously derived expansions, we may write the following:
k(k – 1)
k(k – 1)(k – 2)
(1 + i)k = 1 + ki + ——— i2 + —————— i3 + …
2!
3!
(if k is a positive integer, this is a binomial expansion)
(if k is not an integer, this is an infinite series expansion)
(k)2 (k)3
ek = 1 + k + —— + —— + …
2!
3!
Using only the first two terms in one of these expansions (i.e., terms up
to the first degree) would be considered a linear approximation; using
only the first three terms in one of these expansions (i.e., terms up to the
second degree) would be considered a quadratic approximation.
Find the accumulated value of $4000 invested for three years at 7.4%
convertible semiannually, by using
(a) an exact calculation,
4000(1 + 0.037)6 = $4974.31
(b) linear approximation,
4000(1 + 0.037)6  4000(1 + 6(0.037)) = $4888.00
(Note that this is the same as using simple interest!)
(c) quadratic approximation.
4000(1 +
0.037)6
(3)(5)
 4000 1 + 6(0.037) + ———(0.037)2 = $4970.14
2
Find the accumulated value of $4000 invested for three years with a
force of interest of 7.4%, by using
(a) an exact calculation,
4000e(3)(0.074) = $4994.29
(b) linear approximation,
4000e(3)(0.074) = 4000e0.222  4000(1 + 0.222)) = $4888.00
(Note that this is like using simple interest with i = .)
(c) quadratic approximation.
(0.222)2
4000e0.222  4000 1 + 0.222 + ——— = $4986.57
2
Suppose $4000 is invested at 7.4% per annum, and we would like the to
find the value of the investment at 3.75 years.
An exact answer could be found by obtaining 4000(1 + 0.074)3.75.
An approximation could be obtained by using linear interpolation.
In general, suppose k (0 < k < 1) represents the fraction of the period
desired between periods n and n + 1. Linear interpolation yields the
following approximation:
(1 + i)n+k  (1 – k)(1 + i)n + k(1 + i)n + 1 = (1 + i)n[(1 – k) + k(1 + i)] =
(1 + i)n(1 + ki)
This is the same as using simple interest over the final fractional period.
Find the accumulated value of $4000 invested for 3.75 years at 7.4% per
annum, by using
(a) an exact calculation, 4000(1 + 0.074)3.75 = $5227.89
(b) linear interpolation. 4000(1 + 0.074)3(1 + (0.75)(0.74)) = $5230.35
Three commonly encountered (but not the only) methods for counting
days in a period of investment:
The “actual/actual” method is to use the exact number of days for the
period of investment and to use 365 days in a year. (The table in
Appendix II (page 393) is useful with this method.) Simple interest
computed with this method is called exact simple interest.
The “30/360” method is to assume that each calendar month has 30 days
and that the calendar year has 360 days. Simple interest computed with
this method is called ordinary simple interest. The number of days
between two given dates can be found by using
360(Y2–Y1) + 30(M2–M1) + (D2–D1) where
Y1 = year of first date
Y2 = year of second date,
M1 = month of first date
M2 = month of second date
D1 = day of first date
D2 = day of second date
The “actual/360” method is to use the exact number of days for the
period of investment but to use only 360 days in a year. Simple interest
computed with this method is called the Banker’s Rule.
(A “30/actual” method or a “30/365” method could be defined, but rarely
is either one of these used in practice.)
Suppose that $2500 is deposited on March 8 and withdrawn on October
3 of the same year, and that the interest rate is 5%. Find the amount
of interest earned, if it is computed using
(a) exact simple interest,
With the help of Appendix II,
209
we obtain 2500 (0.05) —— = $71.58 .
365
(b) ordinary simple interest,
First we obtain the number of days from 30(10–3) + (3–8) = 205.
205
Then, we obtain 2500 (0.05) —— = $71.18 .
360
in part (a),
(c) the Banker’s Rule. With the counting done209
we obtain 2500 (0.05) —— = $72.60 .
360
Interest problems generally involve four quantities: principal(s),
investment period length(s), interest rate(s), accumulates value(s).
Two or more amounts of money payable at different points in time
cannot be compared until a common date, called the comparison date, is
established.
An equation of value accumulates or discounts each payment to the
comparison date. (A time diagram can be helpful in setting up an
equation of value.)
With compound interest, an equation of value will produce the same
answer for an unknown value regardless of what comparison date is
selected; however, this is not necessarily true for other patterns of
interest.
In return for a payment of $1200 at the end of 10 years, a lender agrees
to pay $200 immediately, $400 at the end of 6 years, and a final amount
at the end of 15 years. Find the amount of the final payment at the end
of 15 years if the nominal rate of interest is 9% converted semiannually.
Time Diagram:
$200
$400
$X
Lender
Borrower
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
$1200
Equation of Value:
v = 1 / (1 + 0.045)
(Time periods are counted in half-years,
since interest is converted semiannually.)
200 + 400v12 +Xv30 = 1200v20
1200v20 – 200 – 400v12
X = —————————
v30
=
$231.11
In return for a payments of $5000 at the end of 3 years and $4000 at the
end of 9 years, an investor agrees to pay $1500 immediately and to make
an additional payment at the end of 2 years. Find the amount of the
additional payment if i(4) = 0.08.
$1500 $X
Time Diagram:
Investor
Borrower
0 1 2 3 4 5 6 7 8 9 10
$5000
Equation of Value:
(Time periods are counted in quarter-years,
since interest is converted quarterly.)
v = 1 / (1 + 0.02) 1500 + Xv8 = 5000v12 + 4000v36
$4000
X = $5159.24