Annuity is defined as a sequence of n periodical payments, where n can be either finite or infinite.

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1 1 Annuity Annuity is defined as a sequence of n periodical payments, where n can be either finite or infinite. In order to evaluate an annuity or compare annuities, we need to calculate their present value at a common reference time point. To find the present value of an annuity is to sum up the present value at the reference time of each payment. 1.1 Discrete Annuity Dot Product Strategy is a generalized algebraic tool to evaluate present value of annuity. With this tool, we can coherently generalize all formulas presented in chapter 2. Outlines of Dot Product Strategy for Discrete Annuity: 1. Construct the dated cash flow diagram, and define the unit of time period based on the payment frequency. 2. Find the effective interest rate per period for periods under consideration. For example, if ( ) n the assumed accumulation function a(t) is of the form (1 + j) t, then we can use 1 + i(n) n = ( ) m, 1 + i(m) m where n and m are the payment frequency and interest crediting frequency respectively. 3. Choose the appropriate reference time point, denoted by t Based on the cash flow diagram, define an n component vector, payment schedule, representing the periodical payment at each payment period. P = P 0 P 1 P 2... P n, where P k is the payment at time k. 5. Given t 0, define another n-component vector, called discount schedule, representing the 1

2 discount factor of corresponding payment at each payment period. ν t0 = a(t 0 ) a(0) a(t 0 ) a(1) a(t 0 ) a(2)... a(t 0 ) a(n) 6. P V t0 { P } = P ν t0 (1) a(t 0 ) = P 0 a(0) + P a(t 0 ) 1 a(1) + P a(t 0 ) 2 a(2) + + P a(t 0 ) n a(n) Note: a(t 0) a(k) is the accumulation factor for the payment P k if t 0 > k. Otherwise a(t 0) a(k) factor if t 0 < k. is present-value If a(t) = (1 + j) t, then we can rewrite P V t0 { P } as follow P V t0 { P } = P 0 (1 + j) t 0 + P 1 (1 + j) t P 2 (1 + j) t P n (1 + j) t 0 n Dot-product strategy separates the cashflow analysis and discounting mechanism Examples Example (Textbook Exercise ) A loan of 10,000 is being repaid by 10 semiannual payments, with the first payment made one half year after the loan. The first 5 payments are K each, and the final 5 are K each. What is K if i (2) =.06? Soln: 2

3 Let t 0 = 0 and j = i(2) 2 = P = = 0 K K K K K K K K = P 1 + P 2 ν t0 =0 = (1+j) (1+j) 10 P V 0 { P } = P ν t0 =0 = ( P 1 + P 2 ) ν t0 =0 = P 1 ν t0 =0 + P 2 ν t0 = = Ka 10 j a 5 j Example (Textbook Exercise 2.1.3) Smith makes deposits of 1000 on the last day of each month in an account earning interest at rate i (12) =.12. The first deposit is January 31, 2000 and the final deposit is December 31, The accumulated account is used ot make monthly payments of Y starting January 31, 2030 with the final one on December 31, Find Y. Soln: Note that there are deposits and withdrawals. Denote P in the payment schedule for deposits (cash-inflow) and P out the payment schedule for withdrawals (cash-outflow). Since the deposits or withdrawals are made monthly, the unit of time period is month. 3

4 Let t 0 = 360 and effective rate per period j = i(12) 12 = P in = P out = ν t0 =360 = = Y Y Y a(360) a(0) a(360) a(1) a(360) a(360) a(360) a(361) a(360) a(660) (1 + j) 360 (1 + j) (1+j) 1 (1+j) 300 P V t0 { P in } = P V t0 { P out } P in ν t0 =360 = P out ν t0 = s 360 j = Y a 300 j Y = 1000s 360 j a 300 j Remarks: 1. The annuity-immediate formula a n j is used to evaluate the present value of the uniform payment schedule at one period before the first non-zero payment. For example, the formula is the dot-product of the following payment and discount schedule. P = and ν = ν 0 ν 1 ν 2... ν n where P and ν are n + 1 component vectors. Note that there are n non-zero payments in the payment schedule above. 2. The perpetuity-immediate formula a j = lim n a n j 3. The annuity-due formula ä n j is used to evaluate the present value of the uniform payment schedule at the period of the first non-zero payment. For example, the formula is the dotproduct of the following payment and discount schedule. P = and ν = ν 0 ν 1 ν 2... ν n 4

5 where P and ν are n + 1 component vectors. Note that there are n non-zero payments in the payment schedule above. 4. The perpetuity-due formula ä j = lim n ä n j 5. The k -deferred annuity-immediate k a n j is used to evaluate the present value of the uniform payment schedule at k + 1 periods before the first non-zero payment. In other words, the first non-zero payment is deferred k periods. 6. The k -deferred annuity-due k ä n j is used to evaluate the present value of the uniform payment schedule at k periods before the first non-zero payment. In other words, the first non-zero payment is deferred k periods. 7. The annuity-immediate formula s n j is used to evaluate the present value of the uniform payment schedule at the period of the last non-zero payment. For example, the formula is the dot-product of the following payment and discount schedule. P = and ν = ν n ν 1 n ν 2 n... ν n n where P and ν are n + 1 component vectors. Note that there are n non-zero payments. 8. The annuity-due formula s n j is used to evaluate the present value of the uniform payment schedule at one period after the last non-zero payment. For example, the formula is the dot-product of the following payment and discount schedule. P = and ν = ν n ν 1 n ν 2 n... ν n n where P and ν are n + 1 component vectors. Note that there are n non-zero payments. 5

6 1.1.2 More Examples Example (Textbook Example 2.9) A perpetuity immediate pay X per year. Brain receives the first n payments, Colleen receives the next n payments, and Jeff receives the remaining payments. Brian s share of the present value of the original perpetuity is 40%, and Jeff s share is K. Calculate K. Let P denote the payment schedule for the perpetuity-immediate, PB denote the payment schedule for the portion that Brian receives, and P J denote the portion that Jeff receives. Let t 0 = 0. Then P = ν t0 =0 = 0 X X X 1 ν ν 2 ν 3 PV 0 { P } = P ν t0 =0 = Xa j As for Brian, PB = 0 X X X PV 0 { P B } = P B ν t0 =0 = Xa n j Xa n j = 0.4Xa j X 1 νn j = 0.4 X j ν n = 0.6 Similarly, PJ = X X PV 0 { P J } = P J ν t0 =0 = X 2n a n j = ν 2n X j = (ν n ) 2 X j = 0.36 PV 0{ P } 6

7 Example (Textbook Exercise 2.3.6) Smith has 100,000 with which he buys a perpetuity on January 1, Suppose that i = and the perpetuity has annual payments beginning January 1, The first three payments are 2000 each, the next three payments are 2000(1+r) each,..., increasing forever by a factor of 1+r every three years. What is r? Let t 0 = 0. Then P = ν = (1 + r) 2000(1 + r) 2000(1 + r) 2000(1 + r) (1 + r) 2 1 ν ν 2 ν 3, where ν = PV 0 { P } = P ν = = 2000ν ν ν (1 + r)ν (1 + r)ν (1 + r)ν 6 + = ( 2000ν ν ν 3) + ( 2000(1 + r)ν (1 + r)ν (1 + r)ν 6) + = 2000ν ( 1 + ν + ν 2) (1 + r)ν 4 ( 1 + ν + ν 2) (1 + r) 2 ν 7 ( 1 + ν + ν 2) + = ( 1 + ν + ν 2) 2000ν (1 + r)ν (1 + r) 2 ν 7 + = ( 1 + ν + ν 2) 2000ν(1 + (1 + r)ν 3 + (1 + r) 2 ν 6 + ) ( ) ( 1 ν 3 ) = 2000ν ω k, where ω = (1 + r)ν 3. 1 ν k=0 ( ) ( ) 1 ν = 2000ν 1 ν 1 ω 7

8 Example (Textbook Exercise ) A loan of 10,000 is made on January 1, 2005 at an interest rate i (12) =.12. The loan calls for payments of 500 on the first day of April, July, and October each year, with an additional fractional payment on the next scheduled payment date after the final regular payment of 500. Find the date and amount of the additional fractional payment. Since the payments are made on the first day of April, July, and October each year, the unit of year should be a quarter of year. j = i(4) 4 = = P = ν = ν ν 2 ν n, where ν = 1 1+j. PV 0 { P } = P ν = = 500ν + 500ν ν ν ν ν ν n ν n = (500ν + 500ν ν 3 ) + (500ν ν ν 7 ) + = 500(ν + ν 2 + ν 3 ) + 500ν 4 (ν + ν 2 + ν 3 ) ν 4n (ν + ν 2 + ν 3 ), where n is the number of years required to repay the loan = 500a 3 j + 500ν 4 a 3 j ν 4n a 3 j = 500a 3 j 1 + ν (ν 4 ) n 1 (ν 4 ) n = 500a 3 j 1 ν 4 n = Assignment: Complete the example 8

9 1.2 Continuous Annuity When the annuity is continuous, few modifications are required to evaluate the present value of a continuous annuity. First, as the counterpart of payment schedule, we define a payment function h(t). The payment function h(t) represents the instantaneous rate of payment at time t. Think of h(t) as a density function. The amount of payment made between t, t + t) is approximately equal to h(t) t. Of course, h(t) is not a density function since it does not integrate to 1. Next, we define a discount function as ν(t; t 0 ) = a(t 0) a(t), where t 0 is a chosen reference time point. The discount function ν(t; t 0 ) is the counter part of the discount schedule. Furthermore, given the force of interest δ(t), we know that the accumulation function a(t) = e t 0 δ(s)ds is differentiable, and thus continuous. The continuity of a(t) implies that ν(t; t 0 ) is also continuous, except at some time t such that a(t) = 0. The dot-product of two continuous functions on 0, ) is defined as f g = 0 f(x)g(x) dx. Therefore, P V t 0 {h} = h ν = 0 h(t)ν(t; t 0) dt. Next, I would like to derive an expression for the present value evaluated at the reference time t 0, where 0 t 0 <, given the payment function h(t) and the force of interest δ(t). Assuming that the payment function h(t) is defined on 0, ), I can calculate the present value at time t 0 of the payment function h(t), i.e. PV t0 {h} = 0 h(t)ν(t; t 0 ) dt, where ( ) ν(t; t 0 ) = a(t t0 exp 0) a(t) = 0 ( δ(s) ds ) t exp 0 δ(s) ds ( t0 t ) ( t0 ) = exp δ(s) ds δ(s) ds = exp δ(s) ds 0 0 t 9

10 1.2.1 Examples Example Define the payment function h(t) and the force of interest δ(t) as follow: h(t) = δ(t) = 1000 for t 0 0 otherwise 0.02t where 0 t < where t 3 0 otherwise Calculate the present value at time 0. ( 0 ) ( ν(t; 0) = exp δ(s) ds = exp t t 0 ) δ(s) ds = exp ( 0.01t 2) where 0 t < 3 exp ( (t 3)) where t 3 1 otherwise Then PV 0 {h} = = h(t)ν(t; 0) dt 1000 exp ( 0.01t 2) dt exp ( (t 3)) dt Example (SOA Exam FM Sample #21) Payments are made to an account at a continuous rate of (8k+tk), where 0 t 10. Interest is credited at a force of interest δ(t) = 1 8+t. After 10 years, the account is worth Calculate k. 10

11 h(t) = 8k + tk = k(8 + t) δ(t) = t t 0 = ν(t; t 0 = 10) = a(10) a(t) = exp{ 0 δ(s)ds} exp{ t 0 δ(s)ds} { 10 } = exp δ(s)ds = exp t = exp { ln(8 + s) 10 t = t P V 10 {h} = h ν = = 10 0 k = { 10 t } = exp { ln h(t)ν(t; t 0 = 10)dt k(8 + t) 18 dt = 180k 8 + t } s ds )} ( t Remarks: 1. When the payment function is of the following functional form, the annuity is said to be an unit n-year continuous annuity. 1 if 0 t n h 1 (t) = 0 otherwise The present value at time 0 is denoted by ā n ; the present value at time n is denoted by s n. 2. When the payment function is of the following functional form, the annuity is said to be an unit n-year continuous perpetuity. 1 if t 0 h 2 (t) = 0 otherwise The present value at time 0 is denoted by ā. 11

12 3. When the payment function is of the following functional form, the annuity is said to be an unit k-deferred n-year continuous annuity. 1 if k t n + k h 3 (t) = 0 otherwise The present value at time 0 is denoted by k ā n ; the present value at time n + k is denoted by k s n. 4. When the payment function is of the following functional form, the annuity is said to be an unit n-year increasing continuous annuity. t if 0 t n h 4 (t) = 0 otherwise The present value at time 0 is denoted by Īā n ; the present value at time n + k is denoted by Ī s n. 5. When the payment function is of the following functional form, the annuity is said to be an unit n-year increasing continuous perpetuity. t if t 0 h 5 (t) = 0 otherwise The present value at time 0 is denoted by Īā. 6. When the payment function is of the following functional form, the annuity is said to be an unit n-year decreasing continuous annuity. n t if 0 t n h 6 (t) = 0 otherwise The present value at time 0 is denoted by Dā n ; the present value at time n + k is denoted by D s n. 12

13 7. When the payment function is of the following functional form, the annuity is said to be an unit n-year step-increasing continuous annuity. t if 0 t n h 7 (t) = 0 otherwise The present value at time 0 is denoted by Iā n ; the present value at time n is denoted by I s n. 8. When the payment function is of the following functional form, the annuity is said to be an unit n-year step-decreasing continuous annuity n t if 0 t n h 8 (t) = 0 otherwise The present value at time 0 is denoted by Dā n ; the present value at time n is denoted by D s n. 2 Applications We are going two important applications regarding annuity. One is reinvestment rate; the other is loan repayment schemes. In the application of reinvestment rate, we are interested in the accumulated value of $1 invested at time 0 if the interest is reinvested at another rate. As for the loan repayment schemes, the aims are to understand different ways to pay off a loan and to calculate the amount of payment as well as the yield rate with respect to each schemes In Chapter 3, we are going to further study the loan/mortgage repayment system by decomposing each payment into two parts principal reduction and interest. 2.1 Reinvestment Rate Consider $1 deposited at time 0 with the compound-interest accumulation function a(t) = (1 + j) t, where j is the effective rate per period. One of the implicit assumptions associated with such an 13

14 accumulation function is that the interest is automatically reinvested at the same rate from which it was generated. Let relax this particular assumption. Suppose that the interest is reinvested at different rate j, instead of j. Image that at the time you open an bank account, which I will call the principal account, and so is another invisible account, called interest account. The interest account collects the amount of interest generated by the principal account. While the per-period interest rate of the principal account is j, the per-period of interest rate of the interest account is j. In other words, the interest, which generated by the rate j, is reinvested at the rate j. Note that j is call the investment rate, whereas j is called reinvestment rate. We are interested in the accumulated value at time n of $1 deposited at time 0. Remember the interest account exists but being invisible. Therefore, the accumulated value is the sum of the balances of the principal and interest account. Let A(t) denote the accumulated value at time t of $1 deposited at time 0, A P (t) denote the balance of the principal account at time t and A I (t) denote the balance of the interest account at time t. At time 0, $1 is deposited in the principal account. A P (0) = 1 and A I (0) = 0. At time 1, the amount of interest generated by $1 in the principal account is j, but $j is deposited (reinvested) into the interest account instead of being credited in the principal account. A P (1) = 1 and A I (1) = j. At time 2, the amount of interest generated by $1 in the principal account is also j (why?). Again $j is deposited (reinvested) into the interest account. A P (2) = 1 and A I (2) = j(1 + j ) + j. At time 3, the amount interest generated by A(2) is again $j, which is deposited (reinvested) into the interest account. A P (3) = 1 and A I (3) = j(1 + j ) 2 + j(1 + j ) + j. 14

15 . At time n, A P (n) = 1 and A I (n) = j(1 + j ) n 1 + j(1 + j ) n j(1 + j ) + j. Therefore, at time n, the accumulated value is A(n) = A P (n) + A I (n) = 1 + js n j. The average yield rate or rate of return per period j is then defined as A(n) = A(0)(1 + j) n For example, if j = j, then we know that A(n) = (1 + j) n and j = j. On the other hand, when j j, A(n) = A P (n) + A I (n) = A(0)(1 + j) n 1 + js n j = (1 + j) n. Remarks: 1. The account balance of principal account is always $1. 2. the cash-flow diagram for the interest account can be represented by P I = 0 j j... j. 3. j is always between j and j. (Q # ) 2.2 Loan Repayment Scheme and the Yield Rate Consider the following three schemes which a loan of L at the interest (investment) rate of j can be repaid: Present Value Scheme A sequence of n periodical payments is used to pay off the loan, which is the most common repayment scheme in the financial industry. It is understood implicitly that payments and interest are calculated based on the same investment rate. For instance, the periodical level mortgage payment P is calculated L = P a n j, where n is called the amortization period, and j is called the mortgage/loan rate. Note that each payment usually contains a portion to reduce the principal and a portion to pay off the interest. In Chapter 2, we study how to calculate the size of each payment under some specifications. In Chapter 3, we study how to decompose of these payments into principal reduction and interest portion. Sink Fund Scheme A borrower pays creditors the interest, $Lj, on the loan at the end of each period, and a lump sum equal to the original loan amount L at time n. In addition, the 15

16 borrow also makes a sequence of n periodical payments into another account, called sinking fund, so that the account balance of sinking fund equals exactly the original loan amount at time n. For example, L = P s s n j, where P s is the deposit (payment) to the sinking fund, and j is the interest rate of sinking fund. A borrower chooses the sinking fund scheme if there is a more favorable investment option, i.e. j > j. Lump Sum Scheme A borrower pays off the loan with a lump sum of L(1 + j) n at time n. 2.3 Borrower s Payments Dedicated to Loan Someday, you are going to be a borrower. It is then useful to choose the loan repayment scheme based on your ability to pay. In this section, I would like to illustrate the total amount of payment(s) that is dedicated to the loan at the end of each period under each repayment scheme. Set L = 10000, n = 10, and j = 5% = j. Present Value Scheme L = P a n j = P a P = a = Therefore, $1, is dedicated toward the loan at end of each period. Sinking Fund Scheme P = Lj P = = 500 L = P s s n j = P s s P s = s = Therefore, the amount of payments dedicated toward the loan is $1, (= P + P s ), which is the same amount under the PV scheme. 16

17 Now suppose the interest rate of sinking fund is slightly lower than the loan rate. Let s set L = 10000, n = 10, j = 5%, and j = 3%. Present Value Scheme L = P a n j = P a P = a = Therefore, $1, is dedicated toward the loan at end of each period. Sinking Fund Scheme P = Lj P = = 500 L = P s s n j = P s s P s = s = Therefore, the amount of payments dedicated toward the loan is $1, , which is higher than the payment under the PV scheme. 2.4 Lender s Yield Rate Keep in mind that the repayment scheme determine how the loan is repaid, and what amount the per-period payment is. How well the lend reinvests those payments plays a role on the yield rate as well. Let s consider the payment specified in each scheme, and the goal is to calculate the lender s average yield rate per period j. Remember that the initial investment for the lender is A(0) = L. In order to calculate the average yield rate j, we need the accumulated value at time n of the lender s initial investment. All calculations essentially involves the following 3 steps. 1. Calculate the size of payments which the borrower is supposed to pay under the specifications, i.e. 10 level payments per period for 10 periods at a loan rate of 5%. 2. Calculate the account balance of lender s account at the end of loan periods, i.e. A lender (n) 17

18 3. Calculate the average yield rate per period over n period based on A lender (n) = A lender (0)(1+ j) n. Denote the loan rate j, and the reinvestment rate j. Consider the case of j = j. For illustration purpose, let s set L = 10000, n = 10, and j =.05 = j. Present Value Scheme L = P a n j = P a P = a = A lender (n) = P s n j A lender (10) = s = A lender (n) = A lender (0)(1 + j) n = 10000(1 + j) 10 j =.05 Sinking Fund Scheme P = L j P = = 500 A lender (n) = L + P s n j A lender (10) = s = A lender (n) = A lender (0)(1 + j) 10 j =.05 Lump Sum Scheme A lender (n) = A lender (0)(1 + j) n A lender (10) = 10000(1.05) 10 A lender (n) = A lender (0)(1 + j) n j =.05 = j 18

19 Now suppose that L = 10000, n = 10, j =.05 and j = 0.3. Present Value Scheme L = P a n j = P a P = a = A lender (n) = P s n j A lender (10) = s = A lender (n) = A lender (0)(1 + j) n = 10000(1 + j) 10 j = Sinking Fund Scheme P = L j P = = 500 A lender (n) = L + P s n j A lender (10) = s = A lender (n) = A lender (0)(1 + j) = 10000(1 + j) 10 j = Lump Sum Scheme A lender (n) = A lender (0)(1 + j) n A lender (10) = 10000(1.05) 10 A lender (n) = A lender (0)(1 + j) n j =.05 = j Remark: 19

20 1. i is always between j and j. 2. When j = j, the average yield rate per period i is always equal to j, independent of repayment schemes. However, it is also shown that when j j, the average yield rate per period depends on the type of loan repayment scheme Examples Example (Textbook Exercise ) $10,000 can be invested under two options: Option 1 Deposit the 10,000 into a fund earning an effective annual rate of i; or Option 2 Purchase an annuity immediate with 24 level annual payments at an effective annual rate of 10%. These payments are then deposited into a fund earning an effective annual rate of 5%. Both options produce the same accumulated value at the end of 24 years. Calculate i. Let PV (2) 24 denote the present value at the end of 24 years under option 1, and PV(2) 24 denote the present value at the end of 24 years under option 2. Under Option 1, PV (2) 24 t = 0} = 10000(1 + i)24. Under Option 2, = P a 24 10% P = Therefore, at the end of each year, $1, is then being deposited into an account, earning at the rate of 5%. Immediately after the deposit of the 24th payment, the account balance is PV (2) 24 t = 0} = P s 24 5% = Hence i = %. 20

21 2.5 Borrower s Yield Rate Let s focus on the sinking fund scheme from the borrower s point of view. A borrower receives $L from the lender at the interest rate of j at time 0 and agrees to make a sequence of n periodical interest payment of Lj as well as a lump sum of L at the end of loan period (time n). Of course, a person does not borrow money so that he can just simply put the money under his pillow. Suppose that the borrower uses this $L to purchase an n year annuity, which makes $P 1 at time 1, $P 2 at time 2,..., P n at time n. At the end of period k, the borrows receives $P k and pays $Lj. The net amount is P k Lj. If the borrow deposits the net amount into a sinking fund, earning interest at the rate of j at the end of each periods, then account balance of sinking fund at time n will be just enough for the borrower to complete pay off the loan. In other words, A SF (n) = L. The sinking fund method is usually used to determine how much to borrow. In other words, determine L. The payment and the discount schedule of sinking fund are P SF = 0 P 1 Lj P 2 Lj... P n Lj = 0 P 1 P 2... P n 0 Lj Lj... Lj ν t0 =n = (1 + j ) n (1 + j ) n 1 (1 + j ) 1 1 If we assume that all payments are the same i.e. P k = P for k = 1, 2,..., n, then P SF = 0 P P... P 0 Lj Lj... Lj P V n { P SF } = P s n j Ljs n j = L L = P s n j 1 + js n j (**) The above equation involves 5 quantities, namely, L, P, n, j, and j. If any 4 of these are given, then the last unknown can be solved easily. Cost-conscious practitioners are often interested in determining the maximum amount of loan (the cost) relative to the potential returns that the 21

22 investment can bring. Therefore, n can usually be pre determined, j and j can be projected within a reasonable time frame in a politically stable economy, P or, more generally, {P k } n k=1 can be projected with experience and knowledge Examples Example (Textbook Example 2.30) A manufacturer is considering the purchase of some equipment to increase the production that will generate an extra income of $15,000 per year, payable at the end of each year for 8 years. After 8 years, the residual value or salvage value of the equipment is $0. What the price of equipment should be if the manufacturer wants to realize an average annual rate of return of 10% while recovering the principal in a sinking fund earning at 7%? Think of the manufacturer is both the lender and borrower. If the manufacturer wants to realize an average annual rate of return of 10%, then he must borrow $L from himself to purchase the equipment. Let L denote the price of equipment. j = 0.1, j = 0.07 n = 8, P k = for k = 1, 2,, 8 L = 15000s s 8.07 Example (Textbook Exercise ) A purchaser pays 245,000 for a mine which will be exhausted at the end of 18 years. What level annual revenue is required in order for the purchaser to receive a 5% annual return on his investment if he can recover his principal in a sinking fund earning 3.5% per year? 22

23 j = 0.05, j = n = 18, L = P s = s P = ( s ) s

n(n + 1) 2 1 + 2 + + n = 1 r (iii) infinite geometric series: if r < 1 then 1 + 2r + 3r 2 1 e x = 1 + x + x2 3! + for x < 1 ln(1 + x) = x x2 2 + x3 3

n(n + 1) 2 1 + 2 + + n = 1 r (iii) infinite geometric series: if r < 1 then 1 + 2r + 3r 2 1 e x = 1 + x + x2 3! + for x < 1 ln(1 + x) = x x2 2 + x3 3 ACTS 4308 FORMULA SUMMARY Section 1: Calculus review and effective rates of interest and discount 1 Some useful finite and infinite series: (i) sum of the first n positive integers: (ii) finite geometric

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