Consider the graphs of f(x) = x – 1 and f(x) = lnx. x y y = x – 1 y = lnx (1.0) Sections 2.6, 2.7, 2.8 We find that for any x > 0, x – 1 > lnx Now, suppose a 1, a 2, …, a n are all positive. Let T = a 1 a 2 … a n, and let y i = a i / T 1/n for each i = 1, 2, …, n. Then, y 1 y 2 … y n =, and ln(y 1 y 2 … y n ) =. Also, we see that10 (y 1 – 1) + (y 2 – 1) + … + (y n – 1) > ln(y 1 ) + ln(y 2 ) + … + ln(y n ) = ln(y 1 y 2 … y n ) = 0.

In Appendix III (page 394), find the definition of the arithmetic mean and the geometric mean. Then observe that we have proven that that the arithmetic mean must always be greater than or equal to the geometric mean (assuming of course only positive values). We now have that (y 1 – 1) + (y 2 – 1) + … + (y n – 1) > 0  y 1 + y 2 + … + y n > n  y 1 + y 2 + … + y n ——————— > 1  n a 1 + a 2 + … + a n ——————— > 1  T 1/n n a 1 + a 2 + … + a n ——————— > (a 1 a 2 …a n ) 1/n n

Amounts $500, $800, and $1000 are to be paid at respective times 3 years from today, 5 years from today, and 11 years from today. Suppose we would like to find the number of years t from today when one single payment of $500 + $800 + $1000 = $2300 would be equivalent to the individual payments made separately. If the rate of interest per annum is known to be i, then v = 1 / (1 + i), and the equation of value is 2300v t = 500v v v 11 In general, suppose amounts s 1, s 2, …, s n are to be paid at respective times t 1, t 2, …, t n. Also suppose we would like to find the time t when one single payment of s 1 + s 2 + … + s n would be equivalent to the individual payments made separately. If v = 1 / (1 + i) is known, then the equation of value is (s 1 + s 2 + … + s n )v t = s 1 v + s 2 v + … + s n v t 1 t 2 t n (One of the homework assignments will have you find an exact formula for the unknown value of t.)

(s 1 + s 2 + … + s n )v t = s 1 v + s 2 v + … + s n v t 1 t 2 t n With the exact t, this is the true present value for the equivalent single payment. An approximate value of t can be obtained from the method of equated time. This method estimates t with s 1 t 1 + s 2 t 2 + … + s n t n t =————————— s 1 + s 2 + … + s n With the approximation t, the estimated present value for the equivalent single payment is t (s 1 + s 2 + … + s n )v In order to discover an interesting property about how the exact value of t and the estimate t compare, let us imagine that we have a list of numbers where s 1 of the numbers are equal to v, s 2 of the numbers are equal to v, etc. t1t1 t2t2

t1t1 t2t2 t Then v can be written as s 1 t 1 + s 2 t 2 + … + s n t n ————————— s 1 + s 2 + … + s n v v … v s1t1s1t1 s2t2s2t2 sntnsntn 1 —————— s 1 + s 2 + … + s n This is the product of the list of numbers. and v can be written as t t 1 t 2 t n s 1 v + s 2 v + … + s n v —————————— s 1 + s 2 + … + s n This is the average of the list of numbers. From Appendix III, we see that this is the geometric mean of the list of numbers. From Appendix III, we see that this is the arithmetic mean of the list of numbers. v =

Since the arithmetic mean of positive numbers must always be larger than the geometric mean, we have that t v t v> t 1 t 2 t n s 1 v + s 2 v + … + s n v —————————— s 1 + s 2 + … + s n > t v t 1 t 2 t n s 1 v + s 2 v + … + s n v> t (s 1 + s 2 + … + s n )v We now see that the true present value (on the left) will always be larger than the estimated present value (on the right). Consequently, the estimated time t from the method of equated time must be an overestimation.

Amounts $500, $800, and $1000 are to be paid at respective times 3 years from today, 5 years from today, and 11 years from today. The effective rate of interest is 4.5% per annum. (a) (b) Use the method of equated time to find the number of years t from today when one single payment of $500 + $800 + $1000 = $2300 would be equivalent to the individual payments made separately. s 1 t 1 + s 2 t 2 + … + s n t n t =————————— = s 1 + s 2 + … + s n (500)(3) + (800)(5) + (1000)(11) ————————————— = 2300 Find the exact number of years t from today when one single payment of $500 + $800 + $1000 = $2300 would be equivalent to the individual payments made separately years With v = 1 / ( ), the equation of value is 2300v t = 500v v v v v v 11 v t =—————————— = t =6.917 years

Given a particular rate of interest, how long will it take an investment to double? By solving (1 + i) n = 2, we find the exact solution n = ln(2) ——— ln(1 + i) Observe that n  ——— = ln(1 + i) i ——— ———, i ln(1 + i) and when i = 0.08 in the second factor, we have n  0.72 —— = i This is called the rule of 72, and it gives amazingly accurate results for a wide range of interest rates! (See Table 2.1 on page 47 of the text.) 72 ——. 100i

Suppose $2000 is invested at a rate of 7% compounded semiannually. (a) (b) (c) Find the length of time for the investment to double by using the exact formula. Find the length of time for the investment to double by using the rule of 72. Find the length of time for the investment to double by using the interest tables. t = years  2000(1.035) 2t = 4000  t = years 72/3.5 = half years  t = years Interpolating from the tables, we obtain (2 – ) 20 + (21 – 20)————————— = half-years = ( – ) years

Suppose $2000 is invested at a rate of 7% compounded semiannually. (d) (e) Find the length of time for the investment to be worth $3000 by using the exact formula. Find the length of time for the investment to be worth $3000 by using the interest tables. t = years  2000(1.035) 2t = 3000  t = years Interpolating from the tables, we obtain (2 – ) 11 + (12 – 11)————————— = half-years = ( – ) years

Suppose it is the rate of interest i that is unknown in an equation of value. There are four different methods to be considered in solving for i. Depending on the type of situation, one method may be better suited than another. At what interest rate convertible semiannually would $500 accumulate to $800 in 4 years? When a single payment is involved, the method that works best is to solve for i directly in the equation of value (using exponential and logarithmic functions). 500(1 + j) 8 = j = (8/5) 1/8 j = (8/5) 1/8 – 1 = i (2) = 2j = (2)(0.0605) = = 12.1%

At what effective interest rate would the present value of $1000 at the end of 3 years plus $2000 at the end of 6 years be equal to $2700? When multiple payments are involved, the equation of value is generally a polynomial, and the method to solve for i is one for obtaining the roots of a polynomial. Sometimes this can be done using the quadratic formula. 1000v v 6 = 2700 We are only interested in v > 0, and the only positive root from the quadratic formula is v 3 =, from which we obtain v = v v 3 – 2700 = 0 20v v 3 – 27 = 0 v = 1 / (1 + i) = i =

Look at Example 2.9 (pages 49-50) in the textbook. When multiple payments are involved, the equation of value is generally a polynomial, and the method to solve for i is one for obtaining the roots of a polynomial. If this cannot be done using the quadratic formula, then some trial an error together with interpolation in the interest tables can be used. Look at Example 2.10 (pages 50-51) in the textbook. When multiple payments are involved, the equation of value is generally a polynomial, and the method to solve for i is one for obtaining the roots of a polynomial. For a more accurate answer than one obtained from interpolation in the interest tables, an iterative procedure can be used. The use of a computer generally makes an iterative procedure much easier to use. Use the Excel file Interest_Solver to find the solution in Example 2.10 as follows:

Section 2.8 of the textbook discusses some of the ambiguous interpretations of words used in the “real world” versus the precise meanings used in mathematics. Type the formula =1000*(1+j)^ *(1+j)^ in cell A5. Select options Tools > Solver to solve the equation of value as indicated below (and if necessary, first use options Tools > Add-Ins)