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15700	https://departments.central.edu/actsci/files/2011/08/Exam_FM_Study_GuideFinan.pdf	A Basic Course in the Theory of Interest and Derivatives Markets: A Preparation for the Actuarial Exam FM/2 Marcel B. Finan Arkansas Tech University c ⃝All Rights Reserved Preliminary Draft Last updated February 27, 2011 2 In memory of my parents August 1, 2008 January 7, 2009 Preface This manuscripts is designed for an introductory course in the theory of interest and annuity. This manuscript is suitablefor a junior level course in the mathematics of ﬁnance. A calculator, such as TI BA II Plus, either the solar or battery version, will be useful in solving many of the problems in this book. A recommended resource link for the use of this calculator can be found at The recommended approach for using this book is to read each section, work on the embedded examples, and then try the problems. Answer keys are provided so that you check your numerical answers against the correct ones. Problems taken from previous exams will be indicated by the symbol ‡. This manuscript can be used for personal use or class use, but not for commercial purposes. If you ﬁnd any errors, I would appreciate hearing from you: mﬁnan@atu.edu This project has been supported by a research grant from Arkansas Tech University. Marcel B. Finan Russellville, Arkansas March 2009 3 4 PREFACE Contents Preface 3 The Basics of Interest Theory 9 1 The Meaning of Interest . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 2 Accumulation and Amount Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 3 Eﬀective Interest Rate (EIR) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 4 Linear Accumulation Functions: Simple Interest . . . . . . . . . . . . . . . . . . . . . . 28 5 Date Conventions Under Simple Interest . . . . . . . . . . . . . . . . . . . . . . . . . . 35 6 Exponential Accumulation Functions: Compound Interest . . . . . . . . . . . . . . . . 41 7 Present Value and Discount Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 8 Interest in Advance: Eﬀective Rate of Discount . . . . . . . . . . . . . . . . . . . . . . 56 9 Nominal Rates of Interest and Discount . . . . . . . . . . . . . . . . . . . . . . . . . . 66 10 Force of Interest: Continuous Compounding . . . . . . . . . . . . . . . . . . . . . . . 78 11 Time Varying Interest Rates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93 12 Equations of Value and Time Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . 100 13 Solving for the Unknown Interest Rate . . . . . . . . . . . . . . . . . . . . . . . . . . 107 14 Solving for Unknown Time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116 The Basics of Annuity Theory 143 15 Present and Accumulated Values of an Annuity-Immediate . . . . . . . . . . . . . . . 144 16 Annuity in Advance: Annuity Due . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157 17 Annuity Values on Any Date: Deferred Annuity . . . . . . . . . . . . . . . . . . . . . 168 18 Annuities with Inﬁnite Payments: Perpetuities . . . . . . . . . . . . . . . . . . . . . . 176 19 Solving for the Unknown Number of Payments of an Annuity . . . . . . . . . . . . . . 184 20 Solving for the Unknown Rate of Interest of an Annuity . . . . . . . . . . . . . . . . . 192 21 Varying Interest of an Annuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202 22 Annuities Payable at a Diﬀerent Frequency than Interest is Convertible . . . . . . . . 206 23 Analysis of Annuities Payable Less Frequently than Interest is Convertible . . . . . . . 210 5 6 CONTENTS 24 Analysis of Annuities Payable More Frequently than Interest is Convertible . . . . . . 218 25 Continuous Annuities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 228 26 Varying Annuity-Immediate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234 27 Varying Annuity-Due . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 249 28 Varying Annuities with Payments at a Diﬀerent Frequency than Interest is Convertible 257 29 Continuous Varying Annuities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 268 Rate of Return of an Investment 275 30 Discounted Cash Flow Technique . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 276 31 Uniqueness of IRR . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 285 32 Interest Reinvested at a Diﬀerent Rate . . . . . . . . . . . . . . . . . . . . . . . . . . 292 33 Interest Measurement of a Fund: Dollar-Weighted Interest Rate . . . . . . . . . . . . 302 34 Interest Measurement of a Fund: Time-Weighted Rate of Interest . . . . . . . . . . . . 311 35 Allocating Investment Income: Portfolio and Investment Year Methods . . . . . . . . 320 36 Yield Rates in Capital Budgeting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 328 Loan Repayment Methods 333 37 Finding the Loan Balance Using Prospective and Retrospective Methods. . . . . . . . 334 38 Amortization Schedules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 342 39 Sinking Fund Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 353 40 Loans Payable at a Diﬀerent Frequency than Interest is Convertible . . . . . . . . . . 365 41 Amortization with Varying Series of Payments . . . . . . . . . . . . . . . . . . . . . . 371 Bonds and Related Topics 379 42 Types of Bonds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 380 43 The Various Pricing Formulas of a Bond . . . . . . . . . . . . . . . . . . . . . . . . . 384 44 Amortization of Premium or Discount . . . . . . . . . . . . . . . . . . . . . . . . . . . 396 45 Valuation of Bonds Between Coupons Payment Dates . . . . . . . . . . . . . . . . . . 405 46 Approximation Methods of Bonds’ Yield Rates . . . . . . . . . . . . . . . . . . . . . . 413 47 Callable Bonds and Serial Bonds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 420 Stocks and Money Market Instruments 427 48 Preferred and Common Stocks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 428 49 Buying Stocks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 433 50 Short Sales . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 438 51 Money Market Instruments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 445 CONTENTS 7 Measures of Interest Rate Sensitivity 453 52 The Eﬀect of Inﬂation on Interest Rates . . . . . . . . . . . . . . . . . . . . . . . . . . 454 53 The Term Structure of Interest Rates and Yield Curves . . . . . . . . . . . . . . . . . 459 54 Macaulay and Modiﬁed Durations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 468 55 Redington Immunization and Convexity . . . . . . . . . . . . . . . . . . . . . . . . . . 479 56 Full Immunization and Dedication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 486 An Introduction to the Mathematics of Financial Derivatives 493 57 Financial Derivatives and Related Issues . . . . . . . . . . . . . . . . . . . . . . . . . 494 58 Derivatives Markets and Risk Sharing . . . . . . . . . . . . . . . . . . . . . . . . . . . 500 59 Forward and Futures Contracts: Payoﬀand Proﬁt Diagrams . . . . . . . . . . . . . . 504 60 Call Options: Payoﬀand Proﬁt Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . 514 61 Put Options: Payoﬀand Proﬁt Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . 523 62 Stock Options . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 533 63 Options Strategies: Floors and Caps . . . . . . . . . . . . . . . . . . . . . . . . . . . . 540 64 Covered Writings: Covered Calls and Covered Puts . . . . . . . . . . . . . . . . . . . 547 65 Synthetic Forward and Put-Call Parity . . . . . . . . . . . . . . . . . . . . . . . . . . 553 66 Spread Strategies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 560 67 Collars . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 568 68 Volatility Speculation: Straddles, Strangles, and Butterﬂy Spreads . . . . . . . . . . . 574 69 Equity Linked CDs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 585 70 Prepaid Forward Contracts On Stock . . . . . . . . . . . . . . . . . . . . . . . . . . . 591 71 Forward Contracts on Stock . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 598 72 Futures Contracts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 610 73 Understanding the Economy of Swaps: A Simple Commodity Swap . . . . . . . . . . 618 74 Interest Rate Swaps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 629 75 Risk Management . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 638 Answer Key 645 BIBLIOGRAPHY 646 8 CONTENTS The Basics of Interest Theory A component that is common to all ﬁnancial transactions is the investment of money at interest. When a bank lends money to you, it charges rent for the money. When you lend money to a bank (also known as making a deposit in a savings account), the bank pays rent to you for the money. In either case, the rent is called “interest”. In Sections 1 through 14, we present the basic theory concerning the study of interest. Our goal here is to give a mathematical background for this area, and to develop the basic formulas which will be needed in the rest of the book. 9 10 THE BASICS OF INTEREST THEORY 1 The Meaning of Interest To analyze ﬁnancial transactions, a clear understanding of the concept of interest is required. Inter-est can be deﬁned in a variety of contexts, such as the ones found in dictionaries and encyclopedias. In the most common context, interest is an amount charged to a borrower for the use of the lender’s money over a period of time. For example, if you have borrowed $100 and you promised to pay back $105 after one year then the lender in this case is making a proﬁt of $5, which is the fee for borrowing his money. Looking at this from the lender’s perspective, the money the lender is investing is changing value with time due to the interest being added. For that reason, interest is sometimes referred to as the time value of money. Interest problems generally involve four quantities: principal(s), investment period length(s), inter-est rate(s), amount value(s). The money invested in ﬁnancial transactions will be referred to as the principal, denoted by P. The amount it has grown to will be called the amount value and will be denoted by A. The diﬀerence I = A −P is the amount of interest earned during the period of investment. Interest expressed as a percent of the principal will be referred to as an interest rate. Interest takes into account the risk of default (risk that the borrower can’t pay back the loan). The risk of default can be reduced if the borrowers promise to release an asset of theirs in the event of their default (the asset is called collateral). The unit in which time of investment is measured is called the measurement period. The most measurement period is one year but may be longer or shorter (could be days, months, years, decades, etc.). Example 1.1 Which of the following may ﬁt the deﬁnition of interest? (a) The amount I owe on my credit card. (b) The amount of credit remaining on my credit card. (c) The cost of borrowing money for some period of time. (d) A fee charged on the money you’ve earned by the Federal government. Solution. The answer is (c) Example 1.2 Let A(t) denote the amount value of an investment at time t years. (a) Write an expression giving the amount of interest earned from time t to time t + s in terms of A only. (b) Use (a) to ﬁnd the annual interest rate, i.e., the interest rate from time t years to time t + 1 years. 1 THE MEANING OF INTEREST 11 Solution. (a) The interest earned during the time t years and t + s years is A(t + s) −A(t). (b) The annual interest rate is A(t + 1) −A(t) A(t) Example 1.3 You deposit $1,000 into a savings account. One year later, the account has accumulated to $1,050. (a) What is the principal in this investment? (b) What is the interest earned? (c) What is the annual interest rate? Solution. (a) The principal is $1,000. (b) The interest earned is $1,050 - $1,000 = $50. (c) The annual interest rate is 50 1000 = 5% Interest rates are most often computed on an annual basis, but they can be determined for non-annual time periods as well. For example, a bank oﬀers you for your deposits an annual interest rate of 10% “compounded” semi-annually. What this means is that if you deposit $1000 now, then after six months, the bank will pay you 5% × 1000 = $50 so that your account balance is $1050. Six months later, your balance will be 5% × 1050 + 1050 = $1102.50. So in a period of one year you have earned $102.50. The annual interest rate is then 10.25% which is higher than the quoted 10% that pays interest semi-annually. In the next several sections, various quantitative measures of interest are analyzed. Also, the most basic principles involved in the measurement of interest are discussed. 12 THE BASICS OF INTEREST THEORY Practice Problems Problem 1.1 You invest $3,200 in a savings account on January 1, 2004. On December 31, 2004, the account has accumulated to $3,294.08. What is the annual interest rate? Problem 1.2 You borrow $12,000 from a bank. The loan is to be repaid in full in one year’s time with a payment due of $12,780. (a) What is the interest amount paid on the loan? (b) What is the annual interest rate? Problem 1.3 The current interest rate quoted by a bank on its savings accounts is 9% per year. You open an account with a deposit of $1,000. Assuming there are no transactions on the account such as depositing or withdrawing during one full year, what will be the amount value in the account at the end of the year? Problem 1.4 The simplest example of interest is a loan agreement two children might make:“I will lend you a dollar, but every day you keep it, you owe me one more penny.” Write down a formula expressing the amount value after t days. Problem 1.5 When interest is calculated on the original principal only it is called simple interest. Accumulated interest from prior periods is not used in calculations for the following periods. In this case, the amount value A, the principal P, the period of investment t, and the annual interest rate i are related by the formula A = P(1 + it). At what rate will $500 accumulate to $615 in 2.5 years? Problem 1.6 Using the formula of the previous problem, in how many years will 500 accumulate to 630 if the annual interest rate is 7.8%? Problem 1.7 Compounding is the process of adding accumulated interest back to the principal, so that interest is earned on interest from that moment on. In this case, we have the formula A = P(1 + i)t and we call i a yearly compound interest. You can think of compound interest as a series of back-to-back simple interest contracts. The interest earned in each period is added to the principal of the previous period to become the principal for the next period. You borrow $10,000 for three years at 5% annual interest compounded annually. What is the amount value at the end of three years? 1 THE MEANING OF INTEREST 13 Problem 1.8 Using compound interest formula, what principal does Andrew need to invest at 15% compounding annually so that he ends up with $10,000 at the end of ﬁve years? Problem 1.9 Using compound interest formula, what annual interest rate would cause an investment of $5,000 to increase to $7,000 in 5 years? Problem 1.10 Using compound interest formula, how long would it take for an investment of $15,000 to increase to $45,000 if the annual compound interest rate is 2%? Problem 1.11 You have $10,000 to invest now and are being oﬀered $22,500 after ten years as the return from the investment. The market rate is 10% compound interest. Ignoring complications such as the eﬀect of taxation, the reliability of the company oﬀering the contract, etc., do you accept the investment? Problem 1.12 Suppose that annual interest rate changes from one year to the next. Let i1 be the interest rate for the ﬁrst year, i2 the interest rate for the second year,· · · , in the interest rate for the nth year. What will be the amount value of an investment of P at the end of the nth year? Problem 1.13 Discounting is the process of ﬁnding the present value of an amount of cash at some future date. By the present value we mean the principal that must be invested now in order to achieve a desired accumulated value over a speciﬁed period of time. Find the present value of $100 in ﬁve years time if the annual compound interest is 12%. Problem 1.14 Suppose you deposit $1000 into a savings account that pays annual interest rate of 0.4% compounded quarterly (see the discussion at the end of page 11.) (a) What is the balance in the account at the end of year. (b) What is the interest earned over the year period? (c) What is the eﬀective interest rate? Problem 1.15 The process of ﬁnding the present value P of an amount A, due at the end of t years, is called discounting A. The diﬀerence A −P is called the discount on A. Notice that the discount on A is also the interest on P. For example, if $1150 is the discounted value of $1250, due at the end of 7 months, the discount on the $1250 is $100. What is the interest on $1150 for the same period of time? 14 THE BASICS OF INTEREST THEORY 2 Accumulation and Amount Functions Imagine a fund growing at interest. It would be very convenient to have a function representing the accumulated value, i.e. principal plus interest, of an invested principal at any time. Unless stated otherwise, we will assume that the change in the fund is due to interest only, that is, no deposits or withdrawals occur during the period of investment. If t is the length of time, measured in years, for which the principal has been invested, then the amount of money at that time will be denoted by A(t). This is called the amount function. Note that A(0) is just the principal P. Now, in order to compare various amount functions, it is convenient to deﬁne the function a(t) = A(t) A(0). This is called the accumulation function. It represents the accumulated value of a principal of 1 invested at time t ≥0. Note that A(t) is just a constant multiple of a(t), namely A(t) = A(0)a(t). That is, A(t) is the accumulated value of an original investment of A(0). Example 2.1 Suppose that A(t) = αt2 + 10β. If X invested at time 0 accumulates to $500 at time 4, and to $1,000 at time 10, ﬁnd the amount of the original investment, X. Solution. We have A(0) = X = 10β; A(4) = 500 = 16α + 10β; and A(10) = 1000 = 100α + 10β. Using the ﬁrst equation in the second and third we obtain the following system of linear equations 16α + X =500 100α + X =1000. Multiply the ﬁrst equation by 100 and the second equation by 16 and subtract to obtain 1600α + 100X −1600α −16X = 50000 −16000 or 84X = 34000. Hence, X = 34000 84 = $404.76 What functions are possible accumulation functions? Ideally, we expect a(t) to represent the way in which money accumulates with the passage of time. Hence, accumulation functions are assumed to possess the following properties: (P1) a(0) = 1. (P2) a(t) is increasing,i.e., if t1 < t2 then a(t1) ≤a(t2). (A decreasing accumulation function implies a negative interest. For example, negative interest occurs when you start an investment with $100 and at the end of the year your investment value drops to $90. A constant accumulation function 2 ACCUMULATION AND AMOUNT FUNCTIONS 15 implies zero interest.) (P3) If interest accrues for non-integer values of t, i.e. for any fractional part of a year, then a(t) is a continuous function. If interest does not accrue between interest payment dates then a(t) possesses discontinuities. That is, the function a(t) stays constant for a period of time, but will take a jump whenever the interest is added to the account, usually at the end of the period. The graph of such an a(t) will be a step function. Example 2.2 Show that a(t) = t2 + 2t + 1, where t ≥0 is a real number, satisﬁes the three properties of an accumulation function. Solution. (a) a(0) = 02 + 2(0) + 1 = 1. (b) a′(t) = 2t + 2 > 0 for t ≥0. Thus, a(t) is increasing. (c) a(t) is continuous being a quadratic function Example 2.3 Figure 2.1 shows graphs of diﬀerent accumulation functions. Describe real-life situations where these functions can be encountered. Figure 2.1 Solution. (1) An investment that is not earning any interest. (2) The accumulation function is linear. As we shall see in Section 4, this is referred to as “simple interest”, where interest is calculated on the original principal only. Accumulated interest from prior periods is not used in calculations for the following periods. (3) The accumulation function is exponential. As we shall see in Section 6, this is referred to as “compound interest”, where the fund earns interest on the interest. (4) The graph is a step function, whose graph is horizontal line segments of unit length (the period). A situation like this can arise whenever interest is paid out at ﬁxed periods of time. If the amount of interest paid is constant per time period, the steps will all be of the same height. However, if the amount of interest increases as the accumulated value increases, then we would expect the steps to get larger and larger as time goes 16 THE BASICS OF INTEREST THEORY Remark 2.1 Properties (P2) and (P3) clearly hold for the amount function A(t). For example, since A(t) is a positive multiple of a(t) and a(t) is increasing, we conclude that A(t) is also increasing. The amount function gives the accumulated value of k invested/deposited at time 0. Then it is natural to ask what if k is not deposited at time 0, say time s > 0, then what will the accumulated value be at time t > s? For example, $100 is deposited into an account at time 2, how much does the $100 grow by time 4? Consider that a deposit of $k is made at time 0 such that the $k grows to $100 at time 2 (the same as a deposit of $100 made at time 2). Then A(2) = ka(2) = 100 so that k = 100 a(2). Hence, the accumulated value of $k at time 4 (which is the same as the accumulated value at time 4 of an investment of $100 at time 2) is given by A(4) = 100a(4) a(2). This says that $100 invested at time 2 grows to 100a(4) a(2) at time 4. In general, if $k is deposited at time s, then the accumulated value of $k at time t > s is k × a(t) a(s), and a(t) a(s) is called the accumulation factor or growth factor. In other words, the accumulation factor a(t) a(s) gives the dollar value at time t of $1 deposited at time s. Example 2.4 It is known that the accumulation function a(t) is of the form a(t) = b(1.1)t + ct2, where b and c are constants to be determined. (a) If $100 invested at time t = 0 accumulates to $170 at time t = 3, ﬁnd the accumulated value at time t = 12 of $100 invested at time t = 1. (b) Show that a(t) is increasing. Solution. (a) By (P1), we must have a(0) = 1. Thus, b(1.1)0 + c(0)2 = 1 and this implies that b = 1. On the other hand, we have A(3) = 100a(3) which implies 170 = 100a(3) = 100[(1.1)3 + c · 32] Solving for c we ﬁnd c = 0.041. Hence, a(t) = A(t) A(0) = (1.1)t + 0.041t2. It follows that a(1) = 1.141 and a(12) = 9.042428377. Now, 100 a(t) a(1) is the accumulated value of $100 investment from time t = 1 to t > 1. Hence, 100a(12) a(1) = 100 × 9.042428377 1.141 = 100(7.925002959) = 792.5002959 2 ACCUMULATION AND AMOUNT FUNCTIONS 17 so $100 at time t = 1 grows to $792.50 at time t = 12. (b) Since a(t) = (1.1)t +0.041t2, we have a′(t) = (1.1)t ln 1.1+0.082t > 0 for t ≥0. This shows that a(t) is increasing for t ≥0 Now, let n be a positive integer. The nthperiod of time is deﬁned to be the period of time between t = n −1 and t = n. More precisely, the period normally will consist of the time interval n −1 ≤t ≤n. We deﬁne the interest earned during the nth period of time by In = A(n) −A(n −1). This is illustrated in Figure 2.2. Figure 2.2 This says that interest earned during a period of time is the diﬀerence between the amount value at the end of the period and the amount value at the beginning of the period. It should be noted that In involves the eﬀect of interest over an interval of time, whereas A(n) is an amount at a speciﬁc point in time. In general, the amount of interest earned on an original investment of $k between time s and t is I[s,t] = A(t) −A(s) = k(a(t) −a(s)). Example 2.5 Consider the amount function A(t) = t2 + 2t + 1. Find In in terms of n. Solution. We have In = A(n) −A(n −1) = n2 + 2n + 1 −(n −1)2 −2(n −1) −1 = 2n + 1 Example 2.6 Show that A(n) −A(0) = I1 + I2 + · · · + In. Interpret this result verbally. Solution. We have A(n)−A(0) = [A(1)−A(0)]+[A(2)−A(1)]+· · ·+[A(n−1)−A(n−2)]+[A(n)−A(n−1)] = I1 + I2 + · · · + In. Hence, A(n) = A(0) + (I1 + I2 + · · · + In) so that I1 + I2 + · · · + In is the interest earned on the capital A(0). That is, the interest earned over the concatenation of n periods is the 18 THE BASICS OF INTEREST THEORY sum of the interest earned in each of the periods separately. Note that for any 0 ≤t < n we have A(n) −A(t) = [A(n) −A(0)] −[A(t) −A(0)] = Pn j=1 Ij − Pt j=1 Ij = Pn j=t+1 Ij. That is, the interest earned between time t and time n will be the total interest from time 0 to time n diminished by the total interest earned from time 0 to time t. Example 2.7 Find the amount of interest earned between time t and time n, where t < n, if Ir = r for some positive integer r. Solution. We have A(n) −A(t) = n X i=t+1 Ii = n X i=t+1 i = n X i=1 i − t X i=1 i =n(n + 1) 2 −t(t + 1) 2 = 1 2(n2 + n −t2 −t) where we apply the following sum from calculus 1 + 2 + · · · + n = n(n + 1) 2 2 ACCUMULATION AND AMOUNT FUNCTIONS 19 Practice Problems Problem 2.1 An investment of $1,000 grows by a constant amount of $250 each year for ﬁve years. (a) What does the graph of A(t) look like if interest is only paid at the end of each year? (b) What does the graph of A(t) look like if interest is paid continuously and the amount function grows linearly? Problem 2.2 It is known that a(t) is of the form at2 + b. If $100 invested at time 0 accumulates to $172 at time 3, ﬁnd the accumulated value at time 10 of $100 invested at time 5. Problem 2.3 Consider the amount function A(t) = t2 + 2t + 3. (a) Find the the corresponding accumulation function. (b) Find In in terms of n. Problem 2.4 Find the amount of interest earned between time t and time n, where t < n, if Ir = 2r for some positive integer r. Hint: Recall the following sum from Calculus: Pn i=0 ari = a 1−rn+1 1−r , r ̸= 1. Problem 2.5 $100 is deposited at time t = 0 into an account whose accumulation function is a(t) = 1 + 0.03 √ t. (a) Find the amount of interest generated at time 4, i.e., between t = 0 and t = 4. (b) Find the amount of interest generated between time 1 and time 4. Problem 2.6 Suppose that the accumulation function for an account is a(t) = (1 + 0.5it). You invest $500 in this account today. Find i if the account’s value 12 years from now is $1,250. Problem 2.7 Suppose that a(t) = 0.10t2 + 1. The only investment made is $300 at time 1. Find the accumulated value of the investment at time 10. Problem 2.8 Suppose a(t) = at2 + 10b. If $X invested at time 0 accumulates to $1,000 at time 10, and to $2,000 at time 20, ﬁnd the original amount of the investment X. Problem 2.9 Show that the function f(t) = 225 −(t −10)2 cannot be used as an amount function for t > 10. 20 THE BASICS OF INTEREST THEORY Problem 2.10 For the interval 0 ≤t ≤10, determine the accumulation function a(t) that corresponds to A(t) = 225 −(t −10)2. Problem 2.11 Suppose that you invest $4,000 at time 0 into an investment account with an accumulation function of a(t) = αt2 + 4β. At time 4, your investment has accumulated to $5,000. Find the accumulated value of your investment at time 10. Problem 2.12 Suppose that an accumulation function a(t) is diﬀerentiable and satisﬁes the property a(s + t) = a(s) + a(t) −a(0) for all non-negative real numbers s and t. (a) Using the deﬁnition of derivative as a limit of a diﬀerence quotient, show that a′(t) = a′(0). (b) Show that a(t) = 1 + it where i = a(1) −a(0) = a(1) −1. Problem 2.13 Suppose that an accumulation function a(t) is diﬀerentiable and satisﬁes the property a(s + t) = a(s) · a(t) for all non-negative real numbers s and t. (a) Using the deﬁnition of derivative as a limit of a diﬀerence quotient, show that a′(t) = a′(0)a(t). (b) Show that a(t) = (1 + i)t where i = a(1) −a(0) = a(1) −1. Problem 2.14 Consider the accumulation functions as(t) = 1 + it and ac(t) = (1 + i)t where i > 0. Show that for 0 < t < 1 we have ac(t) ≈a1(t). That is (1 + i)t ≈1 + it. Hint: Expand (1 + i)t as a power series. Problem 2.15 Consider the amount function A(t) = A(0)(1 + i)t. Suppose that a deposit 1 at time t = 0 will increase to 2 in a years, 2 at time 0 will increase to 3 in b years, and 3 at time 0 will increase to 15 in c years. If 6 will increase to 10 in n years, ﬁnd an expression for n in terms of a, b, and c. 2 ACCUMULATION AND AMOUNT FUNCTIONS 21 Problem 2.16 For non-negative integer n, deﬁne in = A(n) −A(n −1) A(n −1) . Show that (1 + in)−1 = A(n −1) A(n) . Problem 2.17 (a) For the accumulation function a(t) = (1 + i)t, show that a′(t) a(t) = ln (1 + i). (b) For the accumulation function a(t) = 1 + it, show that a′(t) a(t) = i 1+it. Problem 2.18 Deﬁne δt = a′(t) a(t) . Show that a(t) = e R t 0 δrdr. Hint: Notice that d dr(ln a(r)) = δr. Problem 2.19 Show that, for any amount function A(t), we have A(n) −A(0) = Z n 0 A(t)δtdt. Problem 2.20 You are given that A(t) = at2 + bt + c, for 0 ≤t ≤2, and that A(0) = 100, A(1) = 110, and A(2) = 136. Determine δ 1 2. Problem 2.21 Show that if δt = δ for all t then in = a(n)−a(n−1) a(n−1) = eδ−1. Letting i = eδ−1, show that a(t) = (1+i)t. Problem 2.22 Suppose that a(t) = 0.1t2 + 1. At time 0, $1,000 is invested. An additional investment of $X is made at time 6. If the total accumulated value of these two investments at time 8 is $18,000, ﬁnd X. 22 THE BASICS OF INTEREST THEORY 3 Eﬀective Interest Rate (EIR) Thus far, interest has been deﬁned by Interest = Accumulated value −Principal. This deﬁnition is not very helpful in practical situations, since we are generally interested in compar-ing diﬀerent ﬁnancial situations to ﬁgure out the most proﬁtable one. In this section we introduce the ﬁrst measure of interest which is developed using the accumulation function. Such a measure is referred to as the eﬀective rate of interest: The eﬀective rate of interest is the amount of money that one unit invested at the beginning of a period will earn during the period, with interest being paid at the end of the period. If i is the eﬀective rate of interest for the ﬁrst time period then we can write i = a(1) −a(0) = a(1) −1. where a(t) is the accumulation function. Remark 3.1 We assume that the principal remains constant during the period; that is, there is no contribution to the principal or no part of the principal is withdrawn during the period. Also, the eﬀective rate of interest is a measure in which interest is paid at the end of the period compared to discount interest rate (to be discussed in Section 8) where interest is paid at the beginning of the period. We can write i in terms of the amount function: i = a(1) −a(0) = a(1) −a(0) a(0) = A(1) −A(0) A(0) = I1 A(0). Thus, we have the following alternate deﬁnition: The eﬀective rate of interest for a period is the amount of interest earned in one period divided by the principal at the beginning of the period. One can deﬁne the eﬀective rate of interest for any period: The eﬀective rate of interest in the nth period is deﬁned by in = A(n) −A(n −1) A(n −1) = In A(n −1) where In = A(n) −A(n −1). Note that In represents the amount of growth of the function A(t) in the nth period whereas in is the rate of growth (based on the amount in the fund at the beginning of the period). Thus, the eﬀective rate of interest in is the ratio of the amount of interest earned during the period to the amount of principal invested at the beginning of the period. Note that i1 = i = a(1) −1 and for any accumulation function, it must be true that a(1) = 1 + i. 3 EFFECTIVE INTEREST RATE (EIR) 23 Example 3.1 Assume that A(t) = 100(1.1)t. Find i5. Solution. We have i5 = A(5) −A(4) A(4) = 100(1.1)5 −100(1.1)4 100(1.1)4 = 0.1 Now, using the deﬁnition of in and solving for A(n) we ﬁnd A(n) = A(n −1) + inA(n −1) = (1 + in)A(n −1). Thus, the fund at the end of the nth period is equal to the fund at the beginning of the period plus the interest earned during the period. Note that the last equation leads to A(n) = (1 + i1)(1 + i2) · · · (1 + in)A(0). Example 3.2 If A(4) = 1000 and in = 0.01n, ﬁnd A(7). Solution. We have A(7) =(1 + i7)A(6) =(1 + i7)(1 + i6)A(5) =(1 + i7)(1 + i6)(1 + i5)A(4) = (1.07)(1.06)(1.05)(1000) = 1, 190.91 Note that in can be expressed in terms of a(t) : in = A(n) −A(n −1) A(n −1) = A(0)a(n) −A(0)a(n −1) A(0)a(n −1) = a(n) −a(n −1) a(n −1) . Example 3.3 Suppose that a(n) = 1 + in, n ≥1. Show that in is decreasing as a function of n. Solution. We have in = a(n) −a(n −1) a(n −1) = [1 + in −(1 + i(n −1))] 1 + i(n −1) = i 1 + i(n −1). Since in+1 −in = i 1 + in − i 1 + i(n −1) = − i2 (1 + in)(1 + i(n −1)) < 0 we conclude that as n increases in decreases 24 THE BASICS OF INTEREST THEORY Example 3.4 Show that if in = i for all n ≥1 then a(n) = (1 + i)n. Solution. We have a(n) = A(n) A(0) = (1 + i1)(1 + i2) · · · (1 + in) = (1 + i)n Remark 3.2 In all of the above discussion the interest rate is associated with one complete period; this will be contrasted later with rates−called “nominal”−that are stated for one period, but need to be applied to fractional parts of the period. Most loans and ﬁnancial products are stated with nominal rates such as a nominal rate that is compounded daily, or monthly, or semiannually, etc. To compare these loans, one compare their equivalent eﬀective interest rates. Nominal rates will be discussed in more details in Section 9. We pointed out in the previous section that a decreasing accumulated function leads to negative interest rate. We illustrate this in the next example. Example 3.5 You buy a house for $100,000. A year later you sell it for $80,000. What is the eﬀective rate of return on your investment? Solution. The eﬀective rate of return is i = 80, 000 −100, 000 100, 000 = −20% which indicates a 20% loss of the original value of the house 3 EFFECTIVE INTEREST RATE (EIR) 25 Practice Problems Problem 3.1 Consider the accumulation function a(t) = t2 + t + 1. (a) Find the eﬀective interest rate i. (b) Find in. (c) Show that in is decreasing. Problem 3.2 If $100 is deposited into an account, whose accumulation function is a(t) = 1 + 0.03 √ t, at time 0, ﬁnd the eﬀective rate for the ﬁrst period(between time 0 and time 1) and second period (between time 1 and time 2). Problem 3.3 Assume that A(t) = 100 + 5t. (a) Find i5. (b) Find i10. Problem 3.4 Assume that A(t) = 225 −(t −10)2, 0 ≤t ≤10. Find i6. Problem 3.5 An initial deposit of 500 accumulates to 520 at the end of one year and 550 at the end of the second year. Find i1 and i2. Problem 3.6 A fund is earning 5% simple interest (See Problem 1.5). Calculate the eﬀective interest rate in the 6th year. Problem 3.7 Given A(5) = 2500 and i = 0.05. (a) What is A(7) assuming simple interest (See Problem 1.5)? (b) What is a(10)? Answer: (a) 2700 (b) 1.5 Problem 3.8 If A(4) = 1200, A(n) = 1800, and i = 0.06. (a) What is A(0) assuming simple interest? (b) What is n? 26 THE BASICS OF INTEREST THEORY Problem 3.9 John wants to have $800. He may obtain it by promising to pay $900 at the end of one year; or he may borrow $1,000 and repay $1,120 at the end of the year. If he can invest any balance over $800 at 10% for the year, which should he choose? Problem 3.10 Given A(0) = $1, 500 and A(15) = $2, 700. What is i assuming simple interest? Problem 3.11 You invest $1,000 now, at an annual simple interest rate of 6%. What is the eﬀective rate of interest in the ﬁfth year of your investment? Problem 3.12 An investor purchases 1000 worth of units in a mutual fund whose units are valued at 4.00. The investment dealer takes a 9% “front-end load” from the gross payment. One year later the units have a value of 5.00 and the fund managers claim that the “fund’s unit value has experienced a 25% growth in the past year.” When units of the fund are sold by an investor, there is a redemption fee of 1.5% of the value of the units redeemed. (a) If the investor sells all his units after one year, what is the eﬀective annual rate of interest of his investment? (b) Suppose instead that after one year the units are valued at 3.75. What is the return in this case? Problem 3.13 Suppose a(t) = 1.12t −0.05 √ t. (a) How much interest will be earned during the 5th year on an initial investment of $12? (b) What is the eﬀective annual interest rate during the 5th year? Problem 3.14 Assume that A(t) = 100 + 5t, where t is in years. (a) Find the principal. (b) How much is the investment worth after 5 years? (c) How much is earned on this investment during the 5th year? Problem 3.15 If $64 grows to $128 in four years at a constant eﬀective annual interest rate, how much will $10,000 grow to in three years at the same rate of interest? Problem 3.16 Suppose that in = 5% for all n ≥1. How long will it take an investment to triple in value? 3 EFFECTIVE INTEREST RATE (EIR) 27 Problem 3.17 You have $1000 that you want to deposit in a savings account. Bank A computes the amount value of your investment using the amount function A1(t) = 1 + 0.049t whereas Bank B uses the amount function A2(t) = (1.4)12t. Where should you put your money? Problem 3.18 Consider the amount function A(t) = 12(1.01)4t. (a) Find the principal. (b) Find the eﬀective annual interest rate. Problem 3.19 Given i5 = 0.1 and A(4) = 146.41. Find A(5). Problem 3.20 Given i5 = 0.1 and I5 = 14.641. Find A(4). Problem 3.21 Suppose that in = 0.01n for n ≥1. Show that In = 0.01n(1.01)(1.02) · · · [1 + 0.01(n −1)]A(0). Problem 3.22 If A(3) = 100 and in = 0.02n, ﬁnd A(6). 28 THE BASICS OF INTEREST THEORY 4 Linear Accumulation Functions: Simple Interest Accumulation functions of two common types of interest are discussed next. The accumulation func-tion of “simple” interest is covered in this section and the accumulation function of “compound” interest is discussed in Section 6. Consider an investment of 1 such that the interest earned in each period is constant and equals to i. Then, at the end of the ﬁrst period, the accumulated value is a(1) = 1 + i, at the end of the second period it is a(2) = 1 + 2i and at the end of the nth period it is a(n) = 1 + in, n ≥0. Thus, the accumulation function is a linear function. The accruing of interest according to this function is called simple interest. Note that the eﬀective rate of interest i = a(1)−1 is also called the simple interest rate. We next show that for a simple interest rate i, the eﬀective interest rate in is decreasing . Indeed, in = a(n) −a(n −1) a(n −1) = [1 + in −(1 + i(n −1))] 1 + i(n −1) = i 1 + i(n −1), n ≥1 and in+1 −in = i 1 + in − i 1 + i(n −1) = − i2 (1 + in)(1 + i(n −1)) < 0. Thus, even though the rate of simple interest is constant over each period of time, the eﬀective rate of interest per period is not constant−it is decreasing from each period to the next and converges to 0 in the long run. Because of this fact, simple interest is less favorable to the investor as the number of periods increases. Example 4.1 A fund is earning 5% simple interest. Calculate the eﬀective interest rate in the 6th year. Solution. The eﬀective interest rate in the 6th year is i6 which is given by i6 = i 1 + i(n −1) = 0.05 1 + 0.05(5) = 4% Remark 4.1 For simple interest, the absolute amount of interest earned in each time interval, i.e., In = a(n) − a(n −1) is constant whereas in is decreasing in value as n increases; in Section 6 we will see that under compound interest, it is the relative amount of interest that is constant, i.e. in = a(n)−a(n−1) a(n−1) . 4 LINEAR ACCUMULATION FUNCTIONS: SIMPLE INTEREST 29 The accumulation function for simple interest has been deﬁned for integral values of n ≥0. In order for this function to have the graph shown in Figure 2.1(2), we need to extend a(n) for nonintegral values of n. This is equivalent to crediting interest proportionally over any fraction of a period. If interest accrued only for completed periods with no credit for fractional periods, then the accumulation function becomes a step function as illustrated in Figure 2.1(4). Unless stated otherwise, it will be assumed that interest is allowed to accrue over fractional periods under simple interest. In order to deﬁne a(t) for real numbers t ≥0 we will redeﬁne the rate of simple interest in such a way that the previous deﬁnition is a consequence of this general assumption. The general assumption states the following: Under simple interest, the interest earned by an initial investment of $1 in all time periods of length t + s is equal to the sum of the interest earned for periods of lengths t and s. Symbolically, a(t + s) −a(0) = [a(t) −a(0)] + [a(s) −a(0)] or a(t + s) = a(t) + a(s) −a(0) (4.1) for all non-negative real numbers t and s. Note that the deﬁnition assumes the rule is to hold for periods of any non-negative length, not just of integer length. Are simple interest accumulation functions the only ones which preserve property (4.1)? Suppose that a(t) is a diﬀerentiable function satisfying property (4.1). Then a′(t) = lim s→0 a(t + s) −a(t) s = lim s→0 a(t) + a(s) −a(0) −a(t) s = lim s→0 a(s) −a(0) s =a′(0), a constant Thus the time derivative of a(t) is shown to be constant. We know from elementary calculus that a(t) must have the form a(t) = a′(0)t + C where C is a constant; and we can determine that constant by assigning to t the particular value 0, so that C = a(0) = 1. 30 THE BASICS OF INTEREST THEORY Thus, a(t) = 1 + a′(0)t. Letting t = 1 and deﬁning i1 = i = a(1) −a(0) we can write a(t) = 1 + it, t ≥0. Consequently, simple interest accumulation functions are the only ones which preserve property (4.1). It is important to notice that the above derivation does not depend on t being a nonnegative integer, and is valid for all nonnegative real numbers t. Example 4.2 You invest $100 at time 0, at an annual simple interest rate of 10%. Find the accumulated value after 6 months. Solution. The accumulated function for simple interest is a continuous function. Thus, A(0.5) = 100[1 + 0.1(0.5)] = $105 Remark 4.2 Simple interest is in general inconvenient for use by banks. For if such interest is paid by a bank, then at the end of each period, depositors will withdraw the interest earned and the original deposit and immediately redeposit the sum into a new account with a larger deposit. This leads to a higher interest earning for the next investment year. We illustrate this in the next example. Example 4.3 Consider the following investments by John and Peter. John deposits $100 into a savings account paying 6% simple interest for 2 years. Peter deposits $100 now with the same bank and at the same simple interest rate. At the end of the year, he withdraws his balance and closes his account. He then reinvests the total money in a new savings account oﬀering the same rate. Who has the greater accumulated value at the end of two years? Solution. John’s accumulated value at the end of two years is 100(1 + 0.06 × 2) = $112. Peter’s accumulated value at the end of two years is 100(1 + 0.06)2 = $112.36. 4 LINEAR ACCUMULATION FUNCTIONS: SIMPLE INTEREST 31 Thus, Peter has a greater accumulated value at the end of two years Simple interest is very useful for approximating compound interest, a concept to be discussed in Section 6, for a short time period such as a fraction of a year. To be more speciﬁc, we will see that the accumulation function for compound interest i is given by the formula a(t) = (1 + i)t. Using the binomial theorem we can write the series expansion of a(t) obtaining (1 + i)t = 1 + it + t(t −1) 2! i2 + t(t −1)(t −2) 3! i3 + · · · . Thus, for 0 < t < 1 we can write the approximation (1 + i)t ≈1 + it. Example 4.4 $10,000 is invested for four months at 12.6% compounded annually, that is A(t) = 10000(1 + 0.126)t, where interest is computed using a quadratic to approximate an exact calculation. Find the accumulated value. Solution. We want to estimate A(1/3) = 10000(1 + 0.126) 1 3 using the ﬁrst three terms of the series expansion of (1 + i)t discussed above. That is, A(1/3) =10000(1.126) 1 3 ≈10000 1 + 1 3 × 0.126 + 1 3 −2 3 2! (0.126)2 ! = $10, 402.36 Remark 4.3 Under simple interest, what is the accumulated value at time t > s of 1 deposited at time s? The SOA/CAS approach is diﬀerent than the approach discussed in Section 2 right after Remark 2.1. According to the SOA/CAS, “simple interest” is generally understood to mean that the linear function starts all over again from the date of each deposit or withdrawal. We illustrate this approach in the next example Example 4.5 Suppose you make a deposit of $100 at time t = 0. A year later, you make a withdrawal of $50. Assume annual simple interest rate of 10%, what is the accumulated value at time t = 2 years? Solution. With the SOA/CAS recommended approach for simple interest, the answer is 100(1 + 0.1 × 2) −50(1 + 0.1 × 1) = $65 32 THE BASICS OF INTEREST THEORY Practice Problems Problem 4.1 You invest $100 at time 0, at an annual simple interest rate of 9%. (a) Find the accumulated value at the end of the ﬁfth year. (b) How much interest do you earn in the ﬁfth year? Problem 4.2 At what annual rate of simple interest will $500 accumulate to $615 in 21 2 years? Problem 4.3 In how many years will $500 accumulate to $630 at 7.8% annual simple interest? Problem 4.4 What principal will earn interest of 100 in 7 years at a simple interest rate of 6%? Problem 4.5 What simple interest rate is necessary for $10,000 to earn $100 interest in 15 months? Problem 4.6 At a certain rate of simple interest $1,000 will accumulate to $1110 after a certain period of time. Find the accumulated value of $500 at a rate of simple interest three fourths as great over twice as long a period of time. Problem 4.7 At time 0, you invest some money into an account earning 5.75% simple interest. How many years will it take to double your money? Problem 4.8 You invest $1,000 now, at an annual simple interest rate of 6%. What is the eﬀective rate of interest in the ﬁfth year of your investment? Problem 4.9 Suppose that the accumulation function for an account is a(t) = 1 + 3it. At time 0, you invest $100 in this account. If the value in the account at time 10 is $420, what is i? Problem 4.10 You have $260 in a bank savings account that earns simple interest. You make no subsequent deposits in the account for the next four years, after which you plan to withdraw the entire account balance and buy the latest version of the iPod at a cost of $299. Find the minimum rate of simple interest that the bank must oﬀer so that you will be sure to have enough money to make the purchase in four years. 4 LINEAR ACCUMULATION FUNCTIONS: SIMPLE INTEREST 33 Problem 4.11 The total amount of a loan to which interest has been added is $20,000. The term of the loan was four and one-half years. If money accumulated at simple interest at a rate of 6%, what was the amount of the loan? Problem 4.12 If ik is the rate of simple interest for period k, where k = 1, 2, · · · , n, show that a(n) −a(0) = i1 + i2 + · · · + in. Be aware that in is not the eﬀective interest rate of the nth period as deﬁned in the section! Problem 4.13 A fund is earning 5% simple interest. The amount in the fund at the end of the 5th year is $10,000. Calculate the amount in the fund at the end of 7 years. Problem 4.14 Simple interest of i = 4% is being credited to a fund. The accumulated value at t = n −1 is a(n−1). The accumulated value at t = n is a(n) = 1+0.04n. Find n so that the accumulated value of investing a(n −1) for one period with an eﬀective interest rate of 2.5% is the same as a(n). Problem 4.15 A deposit is made on January 1, 2004. The investment earns 6% simple interest. Calculate the monthly eﬀective interest rate for the month of December 2004. Problem 4.16 Consider an investment with nonzero interest rate i. If i5 is equal to i10, show that interest is not computed using simple interest. Problem 4.17 Smith has just ﬁled his income tax return and is expecting to receive, in 60 days, a refund check of 1000. (a) The tax service that helped him ﬁll out his return oﬀers to buy Smith’s refund check for 850. What annual simple interest rate is implied? (b) Smith negotiates and sells his refund check for 900. What annual simple interest rate does this correspond to? (c) Smith deposits the 900 in an account which earns simple interest at annual rate 9%. How many days would it take from the time of his initial deposit of 900 for the account to reach 1000? Problem 4.18 Let i be a simple interest rate and suppose that i6 = 0.04. Calculate i. 34 THE BASICS OF INTEREST THEORY Problem 4.19 A fund is earning 5% simple interest. If in = 0.04, calculate n. Problem 4.20 Suppose that an account earns simple interest with annual interest rate of i. If an investment of k is made at time s years, what is the accumulated value at time t > s years? Note that the answer is diﬀerent from k a(t) a(s). Problem 4.21 Suppose A(5) = $2, 500 and i = 0.05. (a) What is A(7) assuming simple interest? (b) What is a(10) assuming simple interest? Problem 4.22 If A(4) = $1, 200 and A(n) = $1, 800, (a) what is A(0), assuming a simple interest of 6%? (b) what is n if i = 0.06 assuming simple interest? Problem 4.23 What is A(15), if A(0) = $1, 100, simple interest is assumed and in = 0.01n? 5 DATE CONVENTIONS UNDER SIMPLE INTEREST 35 5 Date Conventions Under Simple Interest In the simple interest problems encountered thus far, the length of the investment has been an integral number of years. What happens if the time is given in days. In this section we discuss three techniques for counting the number of days in a period of investment or between two dates. In all three methods, time = # of days between two dates # of days in a year . In what follows, it is assumed, unless stated otherwise, that in counting days interest is not credited for both the starting date and the ending date, but for only one of these dates. Exact Simple Interest: The “actual/actual” method is to use the exact number of days for the period of investment and to use 365 days in a nonleap year and 366 for a leap year (a year divisible by 4). Simple interest computed with this method is called exact simple interest. For this method, it is important to know the number of days in each month. In counting days between two dates, the last, but not the ﬁrst, date is included. Example 5.1 Suppose that $2,500 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 exact simple interest. Assume non leap year. Solution. From March 8 (not included) to October 3 (included) there are 23+30+31+30+31+31+30+3 = 209 days. Thus, the amount of interest earned using exact simple interest is 2500(0.05) · 209 365 = $71.58 Ordinary Simple Interest: This method is also known as “30/360”. The 30/360 day counting scheme was invented in the days before computers to make the computations easier. The premise is that for the purposes of computation, all months have 30 days, and all years have 12 × 30 = 360 days. Simple interest computed with this method is called ordinary simple interest. The Public Securities Association (PSA) publishes the following rules for calculating the number of days between any two dates from M1/D1/Y1 to M2/D2/Y2: • If D1 (resp. D2) is 31, change D1 (resp. D2)to 30. • If M1 (resp. M2) is 2, and D1 (resp. D2) is 28 (in a non-leap year) or 29, then change D1 (resp. D2) to 30. Then the number of days, N is: N = 360(Y2 −Y1) + 30(M2 −M1) + (D2 −D1). 36 THE BASICS OF INTEREST THEORY For exampe, the number of days from February 25 to March 5 of the same year is 10 days. Like the exact simple interest, the ending date is counted and not the starting date. Example 5.2 Jack borrows 1,000 from the bank on January 28, 1996 at a rate of 15% simple interest per year. How much does he owe on March 5, 1996? Use ordinary simple interest. Solution. The amount owed at time t is A(t) = 1000(1 + 0.15t). Using ordinary simple interest with Y1 = Y2 = 1996, M1 = 1, M2 = 3, and D1 = 28 and D2 = 5 we ﬁnd t = 37 360 and the amount owed on March 5, 1996 is 1, 000 1 + 0.15 × 37 360 = $1, 015.42 Banker’s Rule: This method is also known as “actual/360”. This method uses the exact number of days for the period of investment and that the calendar year has 360 days. Simple interest computed with this method is called Banker’s rule. The number of days between two dates is found in the same way as for exact simple interest. In this method, we also count the last day but not the ﬁrst day. Example 5.3 Jack borrows 1,000 from the bank on January 1, 1996 at a rate of 15% simple interest per year. How much does he owe on January 17, 1996?Use Banker’s rule. Solution. Jack owes 1000 1 + 0.15 × 16 360 = $1, 006.67 Example 5.4 If an investment was made on the date the United States entered World War II, i.e., December 7, 1941, and was terminated at the end of the war on August 8, 1945, for how many days was the money invested under: 1. the “actual/actual” basis? 2. the “30/360” basis? Solution. 1. From December 7, 1941(not included) to December 31, 1941 (included) there were 24 days. From January 1, 1942 to December 31, 1944 ( including a leap year) there were 3(365) + 1 = 1096 days. 5 DATE CONVENTIONS UNDER SIMPLE INTEREST 37 From January 1, 1945 to August 8, 1945(included) the numbers of days is 31 + 28 + 31 + 30 + 31 + 30 + 31 + 8 = 220 days. The total number of days is 24 + 1096 + 220 = 1340. 2. We have 360(1945 −1941) + 30(8 −12) + (8 −7) = 1321 . Remark 5.1 If the time is given in months, reduce it to a fraction of a year on the basis of 12 months to the year, without changing to days. Example 5.5 A merchant is oﬀered $50 discount for cash payment of a $1200 bill due in two months. If he pays cash, at what rate may he consider his money to be earning interest in the next two months? Solution. The merchant would pay now $1150 in place of $1200 in two months. To ﬁnd the interest rate under which $1150 is the present value of $1200, due in two months, use the formula I = Pit which by substitution becomes 50 = 1150i 6 . Solving for i we ﬁnd i = 26.087% We end this section by pointing out that the methods discussed above do not only apply for simple interest rate problems but also to compound interest rate problems. Compound interest rates are introduced in the next section. Unless otherwise stated, in later sections we will assume always the actual/actual method is in use. 38 THE BASICS OF INTEREST THEORY Practice Problems Problem 5.1 Find the amount of interest that $2,000 deposited on June 17 will earn, if the money is withdrawn on September 10 in the same year and if the simple rate of interest is 8% using (a) exact simple interest, (b) ordinary simple interest, and (c) Banker’s rule. Assume non-leap year. Problem 5.2 A sum of 10,000 is invested for the months of July and August at 6% simple interest. Find the amount of interest earned: 1. Assuming exact simple interest (Assume non-leap year). 2. Assuming ordinary simple interest. 3. Assuming the Banker’s Rule. Problem 5.3 Show that the Banker’s Rule is always more favorable to the lender than is exact simple interest. Problem 5.4 (a) Show that the Banker’s Rule is usually more favorable to the lender than is ordinary simple interest. (b) Give an example in (a) for which the opposite relationship holds. Problem 5.5 Suppose that $2,500 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 (Assume non-leap year), (b) ordinary simple interest, (c) the Banker’s Rule. Problem 5.6 The sum of $ 5,000 is invested for the months of April, May, and June at 7% simple interest. Find the amount of interest earned (a) assuming exact simple interest in a non-leap year; (b) assuming exact simple interest in a leap year (with 366 days); (c) assuming ordinary simple interest; (d) assuming the Banker’s Rule. Problem 5.7 Fund A calculates interest using exact simple interest (actual/actual). Fund B calculates interest 5 DATE CONVENTIONS UNDER SIMPLE INTEREST 39 using ordinary simple interest (30/360). Fund C calculates interest using the Banker’s Rule (ac-tual/360). All Funds earn 5% simple interest and have the same amount of money deposited on January 1, 2005. Order the Funds based on the amount in the funds on March 1, 2005 from smallest to largest. Problem 5.8 Suppose you lend $60 to your sister on Sept 14, at an annual rate of simple interest of 10%, to be repaid on Dec 25. How much does she have to pay you back? Use actual/360 time measurement. Problem 5.9 John borrows $60 from Eddie. If he repays Eddie $63 after 5 weeks, what simple interest has John paid? Use actual/actual for days counting. Assume non-leap year. Problem 5.10 Henry invests 1000 on January 15 in an account earning simple interest at an annual eﬀective rate of 10%. On November 25 of the same year, Henry withdraws all his money. How much money will Henry withdraw if the bank counts days: (a) Using exact simple interest (ignoring February 29th) (b) Using ordinary simple interest (c) Using Banker’s Rule Problem 5.11 Let Ie denote the interest earned using the exact simple interest method and Ib the interest earned using the Banker’s rule. Find the ratio Ie Ib . Problem 5.12 Suppose that the interest earned using the Banker’s rule is $40.58. What is the interest earned using the exact simple interest method? Problem 5.13 Show that Ib = Ie + Ie 72 and Ie = Ib −Ib 73. Thus, the Banker’s rule is more favorable to the investor than the actual/actual rule. Problem 5.14 On January 1, 2000, you invested $1,000. Your investment grows to $1,400 by December 31, 2007. What was the exact simple interest rate at which you invested? Problem 5.15 A 3% discount is oﬀered for cash payment of a $2500 bill, due at the end of 90 days. At what rate of simple interest earned over the 90 days is cash payment is made? Use the Banker’s rule. 40 THE BASICS OF INTEREST THEORY Problem 5.16 The terms of payment of a certain debt are: net cash in 90 days or 2% discount for cash in 30 days. At what rate of simple interest is earned if the discount is taken advantage of? 6 EXPONENTIAL ACCUMULATION FUNCTIONS: COMPOUND INTEREST 41 6 Exponential Accumulation Functions: Compound Interest Simple interest has the property that the interest earned is not invested to earn additional interest. In contrast, compound interest has the property that the interest earned at the end of one period is automatically invested in the next period to earn additional interest. We next ﬁnd the accumulation function for compound interest. Starting with an investment of 1 and with compound interest rate i per period. At the end of the ﬁrst period, the accumulated value is 1 + i. At the end of the second period, the accumulated value is (1 + i) + i(1 + i) = (1 + i)2. Continuing this way, we ﬁnd that the accumulated value after t periods is given by the exponential function a(t) = (1 + i)t, for integral t ≥0. Interest accruing according to this function is called compound interest. We call i the rate of compound interest. From the above accumulation function, we can write in = a(n) −a(n −1) a(n −1) = (1 + i)n −(1 + i)n−1 (1 + i)n−1 = i. Thus, the eﬀective rate of interest for compound interest is constant. The accumulation function for compound interest has been deﬁned for nonnegative integers. In order to extend the domain to fractional periods we note that the compound interest accumulation function a(t) = (1 + i)t satisﬁes the property (1 + i)t+s = (1 + i)t · (1 + i)s. Thus, under compound interest we will require the accumulation function to satisfy the property a(t + s) = a(t) · a(s), t, s ≥0. (6.1) This formula says that under compound interest the amount of interest earned by an initial invest-ment of 1 over t+s periods is equal to the amount of interest earned if the investment is terminated at the end of t periods and the accumulated value at that point is immediately reinvested for an additional s periods. Are compound interest accumulation functions the only ones which preserve property (6.1)? Assum-ing a(t) is diﬀerentiable and satisfying the above property, then from the deﬁnition of the derivative 42 THE BASICS OF INTEREST THEORY we have a′(t) = lim s→0 a(t + s) −a(t) s = lim s→0 a(t)a(s) −a(t) s =a(t) lim s→0 a(s) −a(0) s =a(t)a′(0) Hence, a′(t) a(t) = (ln a(t))′ = a′(0) a constant for all t ≥0. Hence, ln a(t) = a′(0)t + C Since a(0) = 1, we obtain C = 0 so ln a(t) = a′(0)t. Setting t = 1 and recalling that a(1) −a(0) = i1 = i yields a′(0) = ln (1 + i) so a(t) = (1 + i)t and this is valid for all t ≥0. We thus see the graphical distinction between simple and compound interest: the graph of an accumulation function under simple interest is a straight line −a linear function; the graph of an accumulation function under compound interest is an exponential function. We next present a comparison result between simple interest and compound interest. Theorem 6.1 Let 0 < i < 1. We have (a) (1 + i)t < 1 + it for 0 < t < 1, (b) (1 + i)t = 1 + it for t = 0 or t = 1, (c) (1 + i)t > 1 + it for t > 1. 6 EXPONENTIAL ACCUMULATION FUNCTIONS: COMPOUND INTEREST 43 Proof. (a) Suppose 0 < t < 1. Let f(i) = (1 + i)t −1 −it. Then f(0) = 0 and f ′(i) = t(1 + i)t−1 −t. Since i > 0, it follows that 1 + i > 1, and since t < 1 we have t −1 < 0. Hence, (1 + i)t−1 < 1 and therefore f ′(i) = t(1 + i)t−1 −t < 0 for 0 < t < 1. It follows that f(i) < 0 for 0 < t < 1. (b) Follows by substitution. (c) Suppose t > 1. Let g(i) = 1 + it −(1 + i)t. Then g(0) = 0 and g′(i) = t −t(1 + i)t−1. Since i > 0, we have 1 + i > 1. Since t > 1 we have t −1 > 0. Hence (1 + i)t−1 > 1 and therefore g′(i) = t −t(1 + i)t−1 < t −t = 0. We conclude that g(i) < 0. This establishes part (c) Thus, simple and compound interest produce the same result over one measurement period. Com-pound interest produces a larger return than simple interest for periods greater than 1 and smaller return for periods smaller than 1. See Figure 6.1. Figure 6.1 It is worth observing the following (1) With simple interest, the absolute amount of growth is constant, that is, for a ﬁxed s the diﬀerence a(t + s) −a(t) = a(s) −1 does not depend on t. (2) With compound interest, the relative rate of growth is constant, that is, for a ﬁxed s the ratio [a(t+s)−a(t)] a(t) = a(s) −1 does not depend on t. Example 6.1 It is known that $600 invested for two years will earn $264 in interest. Find the accumulated value of $2,000 invested at the same rate of compound interest for three years. Solution. We are told that 600(1 + i)2 = 600 + 264 = 864. Thus, (1 + i)2 = 1.44 and solving for i we ﬁnd i = 0.2. Thus, the accumulated value of investing $2,000 for three years at the rate i = 20% is 2, 000(1 + 0.2)3 = $3, 456 44 THE BASICS OF INTEREST THEORY Example 6.2 At an annual compound interest rate of 5%, how long will it take you to triple your money? (Provide an answer in years, to three decimal places.) Solution. We must solve the equation (1 + 0.05)t = 3. Thus, t = ln 3 ln 1.05 ≈22.517 Example 6.3 At a certain rate of compound interest, 1 will increase to 2 in a years, 2 will increase to 3 in b years, and 3 will increase to 15 in c years. If 6 will increase to 10 in n years, ﬁnd an expression for n in terms of a, b, and c. Solution. If the common rate is i, the hypotheses are that 1(1 + i)a =2 →ln 2 = a ln (1 + i) 2(1 + i)b =3 →ln 3 2 = b ln (1 + i) 3(1 + i)c =15 →ln 5 = c ln (1 + i) 6(1 + i)n =10 →ln 5 3 = n ln (1 + i) But ln 5 3 = ln 5 −ln 3 = ln 5 −(ln 2 + ln 1.5). Hence, n ln (1 + i) = c ln (1 + i) −a ln (1 + i) −b ln (1 + i) = (c −a −b) ln (1 + i) and this implies n = c −a −b We conclude this section by noting that the three counting days techniques discussed in Section 5 for simple interest applies as well for compound interest. Example 6.4 Christina invests 1000 on April 1 in an account earning compound interest at an annual eﬀective rate of 6%. On June 15 of the same year, Christina withdraws all her money. Assume non-leap year, how much money will Christina withdraw if the bank counts days: (a) Using actual/actual method. (b) Using 30/360 method. (c) Using actual/360 method. 6 EXPONENTIAL ACCUMULATION FUNCTIONS: COMPOUND INTEREST 45 Solution. (a) The number of days is 29 + 31 + 15 = 75. Thus, the amount of money withdrawn is 1000(1 + 0.06) 75 365 = $1, 012.05. (b) Using the 30/360, we ﬁnd 1000(1 + 0.06) 74 360 = $1, 012.05. (c) Using the actual/360 method, we ﬁnd 1000(1 + 0.06) 75 360 = $1, 012.21 46 THE BASICS OF INTEREST THEORY Practice Problems Problem 6.1 If $4,000 is invested at an annual rate of 6.0% compounded annually, what will be the ﬁnal value of the investment after 10 years? Problem 6.2 Jack has deposited $1,000 into a savings account. He wants to withdraw it when it has grown to $2,000. If the interest rate is 4% annual interest compounded annually, how long will he have to wait? Problem 6.3 At a certain rate of compound interest, $250 deposited on July 1, 2005 has to accumulate to $275 on January 1, 2006. Assuming the interest rate does not change and there are no subsequent deposits, ﬁnd the account balance on January 1, 2008. Problem 6.4 You want to triple your money in 25 years. What is the annual compound interest rate necessary to achieve this? Problem 6.5 You invest some money in an account earning 6% annual compound interest. How long will it take to quadruple your account balance? (Express your answer in years to two decimal places.) Problem 6.6 An amount of money is invested for one year at a rate of interest of 3% per quarter. Let D(k) be the diﬀerence between the amount of interest earned on a compound interest basis, and on a simple interest basis for quarter k, where k = 1, 2, 3, 4. Find the ratio of D(4) to D(3). Problem 6.7 Show that the ratio of the accumulated value of 1 invested at rate i for n periods, to the accumulated value of 1 invested at rate j for n periods, where i > j, is equal to the accumulated value of 1 invested for n periods at rate r. Find an expression for r as a function of i and j. Problem 6.8 At a certain rate of compound interest an investment of $1,000 will grow to $1,500 at the end of 12 years. Determine its value at the end of 5 years. Problem 6.9 At a certain rate of compound interest an investment of $1,000 will grow to $1,500 at the end of 12 years. Determine precisely when its value is exactly $1,200. 6 EXPONENTIAL ACCUMULATION FUNCTIONS: COMPOUND INTEREST 47 Problem 6.10 ‡ Bruce and Robbie each open up new bank accounts at time 0. Bruce deposits 100 into his bank account, and Robbie deposits 50 into his. Each account earns the same annual eﬀective interest rate. The amount of interest earned in Bruce’s account during the 11th year is equal to X. The amount of interest earned in Robbie’s account during the 17th year is also equal to X. Calculate X. Problem 6.11 Given A(5) = $7, 500 and A(11) = $9, 000. What is A(0) assuming compound interest? Problem 6.12 If $200 grows to $500 over n years, what will $700 grow to over 3n years? Assuming same compound annual interest rate. Problem 6.13 Suppose that your sister repays you $62 two months after she borrows $60 from you. What eﬀective annual rate of interest have you earned on the loan? Assume compound interest. Problem 6.14 An investor puts 100 into Fund X and 100 into Fund Y. Fund Y earns compound interest at the annual rate of j > 0, and Fund X earns simple interest at the annual rate of 1.05j. At the end of 2 years, the amount in Fund Y is equal to the amount in Fund X. Calculate the amount in Fund Y at the end of 5 years. Problem 6.15 Fund A is invested at an eﬀective annual interest rate of 3%. Fund B is invested at an eﬀective annual interest rate of 2.5%. At the end of 20 years, the total in the two funds is 10,000. At the end of 31 years, the amount in Fund A is twice the amount in Fund B. Calculate the total in the two funds at the end of 10 years. Problem 6.16 Carl puts 10,000 into a bank account that pays an annual eﬀective interest rate of 4% for ten years. If a withdrawal is made during the ﬁrst ﬁve and one-half years, a penalty of 5% of the withdrawal amount is made. Carl withdraws K at the end of each of years 4, 5, 6, 7. The balance in the account at the end of year 10 is 10,000. Calculate K. Problem 6.17 ‡ Joe deposits 10 today and another 30 in ﬁve years into a fund paying simple interest of 11% per 48 THE BASICS OF INTEREST THEORY year. Tina will make the same two deposits, but the 10 will be deposited n years from today and the 30 will be deposited 2n years from today. Tina’s deposits earn an annual eﬀective rate of 9.15% . At the end of 10 years, the accumulated amount of Tina’s deposits equals the accumulated amount of Joe’s deposits. Calculate n. Problem 6.18 Complete the following table. Simple interest Compound interest a(t) = a(0) = a(1) = Period during which a(t) is greater in = Problem 6.19 On January 1, 2000, you invested $1,000. Your investment grows to $1,400 by December 31, 2007. What was the compound interest rate at which you invested? Problem 6.20 A certiﬁcate of deposit or CD is a ﬁnancial product commonly oﬀered to consumers by banks and credit unions. A CD has a speciﬁc, ﬁxed term (often three months, six months, or one to ﬁve years), and, usually, a ﬁxed interest rate. It is intended that the CD be held until maturity, at which time the money may be withdrawn together with the accrued interest. However, a type a penalty is imposed for early withdrawal. A two-year certiﬁcate of deposit pays an annual eﬀective rate of 9%. The purchaser is oﬀered two options for prepayment penalties in the event of early withdrawal: A −a reduction in the rate of interest to 7% B −loss of three months interest. In order to assist the purchaser in deciding which option to select, compute the ratio of the proceeds under Option A to those under Option B if the certiﬁcate of deposit is surrendered: (a) At the end of six months. (b) At the end of 18 months. 6 EXPONENTIAL ACCUMULATION FUNCTIONS: COMPOUND INTEREST 49 Problem 6.21 You invest $10,000 at time 0 into each of two accounts. Account A earns interest at an annual simple interest rate of 8%; Account B earns interest at an eﬀective annual compound interest rate of 6%. What is the diﬀerence in the amount of interest earned during the 5th year in these two accounts? Problem 6.22 On April 1, 2013, you will need $10,000. Assuming a 6% annual compound eﬀective rate of interest, what would you have to invest on October 1, 2008, in order for you to fulﬁll that need? Problem 6.23 On January 1, 2009, you invest $500 into an account earning an 8% annual compound eﬀective interest rate. On January 1, 2011, you deposit an additional $1,000 into the account. What is the accumulated value of your account on January 1, 2014? Problem 6.24 Let i be a compound eﬀective interest rate. Show that In = iA(n −1). Problem 6.25 Suppose that A(1) = $110, A(2) = $121, and A(10) = $259.37. (a) Find the eﬀective rate of interest. (b) Is the interest a simple interest or a compound interest? (c) Find I5. Problem 6.26 If A(5) = $1, 500 and A(n) = $2, 010.14, what is n if i = 0.05 assuming compound interest? Problem 6.27 How much interest is earned in the 9th year by $200 invested today under compound interest with i = 0.07? 50 THE BASICS OF INTEREST THEORY 7 Present Value and Discount Functions In this section we discuss the question of determining the today’s or present value of some amount in the past or in the future. Unless stated otherwise, we assume in this section that we are in a compound interest situation, where a(t) = (1 + i)t. Why would you be interested in present values? Suppose that in your personal ﬁnancial life, you need to save or invest money that will grow to a speciﬁed amount in a speciﬁed number of years. For example, you may want to start saving now to buy a car for $20,000 in two years, or to make a down payment of $25,000 on a home in 3 years, or to pay a year’s college expenses of $30,000 in 8 years. All of these questions involve ﬁnding the present value (PV), i.e., the amount invested today that will grow to the desired amount in the desired time. To start with, suppose we want to know what $1 a period ago, invested at compound interest rate i per period, worths today. If X is the accumulated value then we must have 1(1 + i) = X. Thus, $1 a period ago worths 1 + i dollars now. We call 1 + i the accumulation factor. Similarly, $1 a period from now invested at the rate i worths ν = 1 1+i today. We call ν the discount factor since it discounts the value of an investment at the end of a period to its value at the beginning of the period. The above discussion can be generalized to more than one period. For example, $1 invested t periods ago worths (1 + i)t today and $1 invested t periods from now is worth 1 (1+i)t today. We call 1 (1+i)t a discount function. It represents the amount that needs to be invested today at interest rate i per period to yield an amount of $1 at the end of t time periods. This function can be expressed in terms of the accumulation function a(t). Indeed, since [a(t)]−1a(t) = 1, the discount function is [a(t)]−1 = 1 (1+i)t = νt. In a sense, accumulation and discounting are opposite processes. The term (1 + i)t is said to be the accumulated value of $1 at the end of t time periods. The term νt is said to be the present value or discounted(back) value of $1 to be paid at the end of t periods. Example 7.1 What is the present value of $8,000 to be paid at the end of three years if the interest rate is 11% compounded annually? Solution. Let FV stands for the future value and PV for the present value. We want to ﬁnd PV. We have FV = PV (1 + i)3 or PV = FV (1 + i)−3. Substituting into this equation we ﬁnd PV = 8000(1.11)−3 ≈$5, 849.53 Example 7.2 Show that the current value of a payment of 1 made n > 1 periods ago and a payment of 1 to be made n periods in the future is greater than 2, if i > 0. 7 PRESENT VALUE AND DISCOUNT FUNCTIONS 51 Solution. We have (1 + i)n + (1 + i)−n = (1 + i) n 2 −(1 + i)−n 2 2 + 2 ≥2. Equality holds if and only if (1 + i) n 2 = (1 + i)−n 2 which is equivalent to (1 + i)n = 1. This happens when either i = 0 or n = 0. Hence, (1 + i)n + (1 + i)−n > 2 since by assumption n > 1 and i > 0 Example 7.3 Find an expression for the discount factor during the nth period from the date of investment, i.e. (1 + in)−1, in terms of the amount function. Solution. Recall that in = A(n)−A(n−1) A(n−1) . Thus, 1 1 + in = 1 1 + A(n)−A(n−1) A(n−1) = A(n −1) A(n) Remark 7.1 One should observe that νt extend the deﬁnition of a(t) = (1 + i)t to negative values of t. Thus, a graph of a(t) is given in Figure 7.1. Figure 7.1 What happens to the calculation of present values if simple interest is assumed instead of compound interest? The accumulation function is now a(t) = 1 + it. Hence, the present value of 1, t years in the future, is (a(t))−1 = 1 1 + it, t ≥0. Example 7.4 Find the present value of $3,000 to be paid at the end of 5 years with a rate of simple interest of 7% per annum. 52 THE BASICS OF INTEREST THEORY Solution. The answer is 3000[a(5)]−1 = 3000 1+0.07(5) ≈$2, 222.22 Example 7.5 Rework Example 7.4 using compound interest instead of simple interest. Solution. The answer is 3000 1.075 = $2, 138.96 7 PRESENT VALUE AND DISCOUNT FUNCTIONS 53 Practice Problems Problem 7.1 At an eﬀective rate of interest of 8% per annum, the present value of $100,000 due in X years is $65,322. Determine X. Problem 7.2 What deposit made today will provide for a payment of $1,000 in 1 year and $2,000 in 3 years, if the eﬀective rate of interest is 7.5%? Problem 7.3 The total amount of a loan to which interest has been added is $20,000. The term of the loan was four and one-half years. (a) If money accumulated at simple interest at a rate of 6%, what was the amount of the loan? (b) If the annual rate of interest was 6% and interest was compounded annually, what was the amount of the loan? Problem 7.4 It is known that an investment of $500 will increase to $4,000 at the end of 30 years. Find the sum of the present values of three payments of $10,000 each which will occur at the end of 20, 40, and 60 years. Problem 7.5 The sum of the present value of 1 paid at the end of n periods and 1 paid at the end of 2n periods is 1. Find (1 + i)2n. Problem 7.6 ‡ Sally has two IRA’s. IRA #1 earns interest at 8% eﬀective annually and IRA #2 earns interest at 10% eﬀective annually. She has not made any contributions since January 1, 1985, when the amount in IRA #1 was twice the amount in IRA #2. The sum of the two accounts on January 1, 1993 was $75,000. Determine how much was in IRA #2 on January 1, 1985. Problem 7.7 ‡ At an annual eﬀective interest rate of i, i > 0, the following are all equal: (i) the present value of 10,000 at the end of 6 years; (ii) the sum of the present values of 6,000 at the end of year t and 56,000 at the end of year 2t; and (iii) 5,000 immediately. Calculate the present value of a payment of 8,000 at the end of year t + 3 using the same annual eﬀective interest rate. 54 THE BASICS OF INTEREST THEORY Problem 7.8 The Kelly family buys a new house for $93,500 on May 1, 1996. How much was this house worth on May 1, 1992 if real estate prices have risen at a compound rate of 8% per year during that period? Problem 7.9 Find the present (discounted) value of $3,000 to be paid at the end of 5 years if the accumulation function is a(t) = 1 + t2 25. Problem 7.10 Suppose a(t) = αt2 + 10β. If $X invested at time 0 accumulates to $1,000 at time 10, and to $2,000 at time 20, ﬁnd the original amount of the investment X. Problem 7.11 You want to buy a sailboat when you retire 5 years from now. The sailboat you wish to buy costs $50,000 today and inﬂation is at 4% per year. What do you need to put into an account today to have enough funds to be able to purchase the boat 5 years from now if the account earns 7% per year? Problem 7.12 ‡ A store is running a promotion during which customers have two options for payment. Option one is to pay 90% of the purchase price two months after the date of sale. Option two is to deduct X% oﬀthe purchase price and pay cash on the date of sale. A customer wishes to determine X such that he is indiﬀerent between the two options when valuing them using an eﬀective annual interest rate of 8%. Which of the following equations would the customer need to solve? (A) X 100 1 + 0.08 6 = 0.90 (B) 1 − X 100 1 + 0.08 6 = 0.90 (C) X 100 (1.08) 1 6 = 0.90 (D) X 100 1.08 1.06 = 0.90 (E) 1 − X 100 (1.08) 1 6 = 0.90 Problem 7.13 Derive a formula for i in terms of ν where ν = 1 1+i. Problem 7.14 In this section the present value is assumed to be the value of an investment at time 0. We can determine the present value of an investment at any time n not necessarily at time 0. A payment of $10 is to be made at time 7 years. (a) Determine the present value of this payment at time 4 years if the annual compound interest is 6%. (b) Determine the present value of this payment at time 4 years if the annual simple interest is 6%. 7 PRESENT VALUE AND DISCOUNT FUNCTIONS 55 Problem 7.15 A treasury bill or T−Bill with face value $100 is a security which is exchangeable for $100 on the maturity date. Suppose that a T−Bill with face value $100 is issued on 08/09/2005 and matures on 03/09/2006 (there are 182 days between the dates). Assuming simple interest, given that the annual rate of interest is 2.810%, ﬁnd the price of the security. Problem 7.16 If we are promised $1,500 5 years from now, what is its present value today at: (a) 7% simple interest? (b) 7% compound interest? Problem 7.17 Show that the discount function of a simple interest rate i is larger than the discount function for a compound interest rate i for t ≥1. Problem 7.18 If i 1+i = 0.06, ﬁnd i and ν. Problem 7.19 What is the value 8 years from now of $500 today if ν = 0.96? Problem 7.20 ‡ At an eﬀective annual interest rate of i, i > 0, each of the following two sets of payments has present value K : (i) A payment of 121 immediately and another payment of 121 at the end of one year. (ii) A payment of 144 at the end of two years and another payment of 144 at the end of three years. Calculate K. 56 THE BASICS OF INTEREST THEORY 8 Interest in Advance: Eﬀective Rate of Discount The eﬀective rate of interest is deﬁned as a measure of the interest paid at the end of the period. In this section, we introduce the eﬀective rate of discount, denoted by d, which is a measure of interest where the interest is paid at the beginning of the period. What do we mean by a statement such as “A loan of $1,200 is made for one year at an eﬀective rate of discount of 5%”? This means that the borrower will pay the interest of 1200 × 0.05 = $60 (called the amount of discount)at the beginning of the year and repays $1,200 at the end of the year. So basically, the lender is getting the interest in advance from the borrower. In general, when $k is borrowed at a discount rate of d, the borrower will have to pay $kd in order to receive the use of $k. Therefore, instead of the borrower having the use of $k at the beginning of a period he will only have the use of $(k −kd). In our example above, we note that the eﬀective rate of discount 5% is just the ratio 5% = 1200 −1140 1200 . From this we can formulate the deﬁnition of eﬀective rate of discount: The eﬀective rate of discount is the ratio of the amount of discount during the period to the amount at the end of the period. Example 8.1 What is the diﬀerence between the following two situations? (1) A loan of $100 is made for one year at an eﬀective rate of interest of 5%. (2) A loan of $100 is made for one year at an eﬀective rate of discount of 5%. Solution. In both cases the fee for the use of the money is the same which is $5. That is, the amount of discount is the same as the amount of interest. However, in the ﬁrst case the interest is paid at the end of the period so the borrower was able to use the full $100 for the year. He can for example invest this money at a higher rate of interest say 7% and make a proﬁt of $2 at the end of the transaction. In the second case, the interest is paid at the beginning of the period so the borrower had access to only $95 for the year. So, if this amount is invested at 7% like the previous case then the borrower will make a proﬁt of $1.65. Also, note that the eﬀective rate of interest is taken as a percentage of the balance at the beginning of the year whereas the eﬀective rate of discount is taken as a percentage of the balance at the end of the year Keep in mind the diﬀerences between the interest model and the discount model: • Under the interest model, the payment for the use of the money is made at the end of the period, 8 INTEREST IN ADVANCE: EFFECTIVE RATE OF DISCOUNT 57 based on the balance at the beginning of the period. • Under the discount model, the payment for the use of the money is deducted at the beginning of the period from the ﬁnal amount which will be present unchanged at the end of the period. Even though an eﬀective rate of interest is not the same as an eﬀective rate of discount, there is a relationship between the two. Assume that $1 is invested for one year at an eﬀective rate of discount d. Then the original principal is $(1 −d). The eﬀective rate of interest i for the year is deﬁned to be the ratio of the amount of interest divided by the balance at the beginning of the year. That is, i = d 1 −d (8.1) Solving this last equation for d we ﬁnd d = i 1 + i. (8.2) There is a verbal interpretation of this result: d is the ratio of the amount of interest that 1 will earn during the year to the balance at the end of the year. Example 8.2 The amount of interest earned for one year when X is invested is $108. The amount of discount earned when an investment grows to value X at the end of one year is $100. Find X, i, and d. Solution. We have iX = 108, i 1+iX = 100. Thus, 108 1+i = 100. Solving for i we ﬁnd i = 0.08 = 8%. Hence, X = 108 0.08 = 1, 350 and d = i 1+i = 2 27 ≈7.41% Observe that several identities can be derived from Equations (8.1)-(8.2). For example, since ν = 1 1+i we have d = iν (8.3) that is, discounting i from the end of the period to the beginning of the period with the discount factor ν, we obtain d. Next, we have d = i 1 + i = 1 + i 1 + i − 1 1 + i = 1 −ν. (8.4) This is equivalent to ν = 1 −d. Both sides of the equation represent the present value of 1 to be paid at the end of the period. Now, from i = d 1−d we have i(1 −d) = d or i −id = d. Adding id −d to both sides we ﬁnd i −d = id (8.5) 58 THE BASICS OF INTEREST THEORY that is, the diﬀerence of interest in the two schemes is the same as the interest earned on amount d invested at the rate i for one period. Eﬀective rates of discount can be calculated over any particular measurement period: “The eﬀective rate of discount dn in the nth period is deﬁned to be the ratio of the amount of discount and the accumulated value at the end of the period”. That is dn = a(n) −a(n −1) a(n) . Since A(n) = A(0)a(n), we can write dn = A(n) −A(n −1) A(n) = In A(n). Example 8.3 If a(t) = 1 + t2 25, ﬁnd d3. Solution. We have d3 = a(3) −a(2) a(3) = 1 + 9 25 −1 −4 25 1 + 9 25 = 5 34 ≈14.71% Analogous to the eﬀective rate of interest in, dn may vary from period to period. Recall that compound interest implies a constant rate of eﬀective interest. A concept parallel to compound interest is compound discount. We say that d is a compound discount if it discounts $1 according to the model shown in Figure 8.1, where t is the number of periods. Figure 8.1 Let ac(t) be the accumulation function for compound discount d > 0 per period. From Figure 8.1, the original principal which will produce an accumulated value of 1 at the end of t periods is given by [ac(t)]−1 = (1 −d)t, t ≥0. Thus, the accumulation function for a compound discount d is ac(t) = 1 (1 −d)t, t ≥0. 8 INTEREST IN ADVANCE: EFFECTIVE RATE OF DISCOUNT 59 Example 8.4 An investor would like to have $5,000 at the end of 20 years. The annual compound rate of discount is 5%. How much should the investor deposit today to reach that goal? Solution. The investor should set aside 5000(1 −0.05)20 ≈$1, 792.43 In the following theorem we prove that compound discount implies a constant rate of eﬀective discount. Theorem 8.1 Assuming compound discount d > 0 per period, we have dn = d for all n ≥1. Proof. Compound interest and compound discount have equal accumulation functions. Indeed, (1 + i)t = 1 + d 1 −d t = 1 (1 −d)t, for all t ≥0. Hence, we see that dn = a(n) −a(n −1) a(n) = 1 (1−d)n − 1 (1−d)n−1 1 (1−d)n = d As for compound interest and compound discount, it is possible to deﬁne simple discount in a manner analogous to the deﬁnition of simple interest. We say that d is a simple discount if it discounts $1 according to the model shown in Figure 8.2. Figure 8.2 If as(t) is the accumulation function for simple discount d, then from Figure 8.2 the original principal which will produce an accumulated value of 1 at the end of t periods is given by [as(t)]−1 = 1 −dt, 0 ≤t < 1 d. Note that, unlike the situation for simple interest, we must restrict the length of time over which we propose to apply simple discount. This is necessary to keep [as(t)]−1 > 0. 60 THE BASICS OF INTEREST THEORY Thus, the accumulation function for simple discount d is as(t) = 1 1 −dt, 0 ≤t < 1 d. Example 8.5 Calculate the present value of a payment of 10,000 to be made in 17 years assuming a simple rate of discount of 3% per annum. Solution. The answer is PV = 10000[1 −0.03(17)] = $4, 900 It should be noted that formulas (15.3) - (8.5) assume eﬀective rates of compound interest and discount (since a(t) = ac(t)) and are not valid for simple rates of interest and discount unless the period of investment happens to be exactly one period. For example, if i and d are equivalent simple rate of interest and simple discount then 1 + it = 1 1−dt for all t ≥0. If (15.3) is satisﬁed then (1 + it)(1 −dt) = 1 is only valid for t = 0. Example 8.6 If i and d are equivalent rates of simple interest and simple discount over t periods, show that i −d = idt. Solution. Since i and d are equivalent we must have 1+it = 1 1−dt or (1−dt)(1+it) = 1. Thus, 1+ti−td−t2id = 1. This can be rearranged to obtain ti −td = t2id. Dividing through by t, the result follows Example 8.7 (a) Find d5 if the rate of simple interest is 10%. (b) Find d5 if the rate of simple discount is 10%. Solution. (a) We are given that i = 0.10. Then d5 = a(5) −a(4) a(5) = 1 + 0.10(5) −(1 + 0.10(4)) 1 + 0.10(5) = 1 15. (b) We are given d = 0.10. Then d5 = as(5) −as(4) as(5) = 1 1−0.10(5) − 1 1−0.10(4) 1 1−0.10(5) = 1 6 In the case of simple interest, in is a decreasing function of n. This is reversed for dn with a simple discount rate. 8 INTEREST IN ADVANCE: EFFECTIVE RATE OF DISCOUNT 61 Theorem 8.2 Assuming simple discount at a rate of discount d > 0, dn is an increasing function of n for 0 < n −1 < 1 d. Proof. Under simple discount at a rate d of discount, we have as(n) = 1 1−dn. Thus, dn = as(n) −as(n −1) as(n) = 1 1−dn − 1 1−d(n−1) 1 1−dn = d 1 −dn + d = d 1 + d(1 −n) > 0 since 0 < n −1 < 1 d. As n increases the denominator decreases so dn increases We next introduce the following deﬁnition: Two rates of interest and/or discount are said to be equivalent if a given amount of principal invested over the same period of time at each of the rates produces the same accumulated value. We have seen in the process of proving Theorem 8.1 that an eﬀective rate of compound interest i is equivalent to an eﬀective rate of compound discount d since a(t) = ac(t). Example 8.8 With i and d satisfying (8.1) - (8.5), show that d3 (1 −d)2 = (i −d)2 1 −ν . Solution. We have d3 (1 −d)2 =d d 1 −d 2 =di2 = (id)2 d =(i −d)2 1 −ν Simple and compound discount produce the same result over one period. Over a longer period, simple discount produces a smaller present value than compound discount, while the opposite is true over a shorter period. 62 THE BASICS OF INTEREST THEORY Theorem 8.3 Assuming 0 < d < 1, we have (a) (1 −d)t < 1 −dt if 0 < t < 1 (b) (1 −d)t = 1 −dt if t = 0 or t = 1 (c) (1 −d)t > 1 −dt if t > 1 Proof. (a) Writing the power series expansion of (1 −d)t we ﬁnd (1 −d)t = 1 −dt + 1 2t(t −1)d2 −1 6t(t −1)(t −2)d3 + · · · If t < 1 then all terms after the second are negative. Hence, (1 −d)t < 1 −dt for 0 < t < 1. (b) Straight forward by substitution. (c) Suppose t > 1. Let f(d) = 1 −dt −(1 −d)t. Then f(0) = 0 and f ′(d) = −t + t(1 −d)t−1. Since t > 1 we have t −1 > 0. Since 0 < d < 1, we have 1 −d < 1. Thus, (1 −d)t−1 < 1 and consequently f ′(d) < −t + t = 0. Hence, f(d) < 0. This completes a proof of the theorem Figure 8.3 compares the discount function under simple discount and compound discount. Figure 8.3 8 INTEREST IN ADVANCE: EFFECTIVE RATE OF DISCOUNT 63 Practice Problems Problem 8.1 The amount of interest earned on A for one year is $336, while the equivalent amount of discount is $300. Find A. Problem 8.2 1000 is to be accumulated by January 1, 1995, at a compound discount of 9% per year. (a) Find the present value on January 1, 1992. (b) Find the value of i equivalent to d. Problem 8.3 The amount of interest earned on A for one year is $562.50, while the equivalent amount of discount is $500. Find A. Problem 8.4 Let i be a compound interest with equivalent compound discount d. Show the following (a) 1 d −1 i = 1. (b) d 1 + i 2 = i 1 −d 2 . (c) i √ 1 −d = d√1 + i. Hint: Use i −d = id for all three identities. Problem 8.5 Calculate the present value of $2000 payable in 10 years using an annual eﬀective discount rate of 8%. Problem 8.6 Calculate the accumulated value at the end of 3 years of 15,000 payable now assuming an interest rate equivalent to an annual discount rate of 8%. Problem 8.7 An investor deposits $1,000 today. The annual compound rate of discount is 6%. What is the accumulated value of the investment at the end of 10 years? Problem 8.8 An investor would like to have $10,000 at the end of 5 years. The annual simple rate of discount is 3%. How much should the investor deposit today to reach that goal? 64 THE BASICS OF INTEREST THEORY Problem 8.9 An investor deposits $5,000 today. The annual simple rate of discount is 5%. What is the accumu-lated value of the investment at the end of 7 months? Problem 8.10 You are given that ν = 0.80. Calculate d. Problem 8.11 A fund earns interest at a rate equivalent to the rate of discount of d. Megan invested 10,000 in the fund. Eleven years later, Megan has 30,042. Calculate d. Problem 8.12 In Account A, an investment of $1,000 grows to $1,700 in four years at an eﬀective annual interest rate of x. In Account B, $3,000 is invested for ﬁve years, at an eﬀective rate of discount d = x. What is the ending balance in Account B? Problem 8.13 ‡ Bruce and Robbie each open up new bank accounts at time 0 . Bruce deposits 100 into his bank account, and Robbie deposits 50 into his. Each account earns an annual eﬀective discount rate of d. The amount of interest earned in Bruce’s account during the 11th year is equal to X. The amount of interest earned in Robbie’s account during the 17th year is also equal to X. Calculate X. Problem 8.14 Suppose that a(t) = 1.12t. What is the eﬀective rate of discount in the 7th year? Problem 8.15 A fund accumulates based on the following formula: a(t) = 1 + t 10 · 1.2t. Find d3. Problem 8.16 The eﬀective rate of discount is 5%. Linda will receive $500 one year from today, $1,000 two years from today, and $2,000 ﬁve years from today. Find the combined present value of her three future cash ﬂows. Problem 8.17 A business permits its customers to pay with a credit card or to receive a percentage discount of r for paying cash. For credit card purchases, the business receives 97% of the purchase price one-half month later. At an annual eﬀective rate of discount of 22%, the two payments are equivalent. Find r. 8 INTEREST IN ADVANCE: EFFECTIVE RATE OF DISCOUNT 65 Problem 8.18 A deposit of X is made into a fund which pays an annual eﬀective interest rate of 6% for 10 years. At the same time , X 2 is deposited into another fund which pays an annual eﬀective rate of discount of d for 10 years. The amounts of interest earned over the 10 years are equal for both funds. Calculate d. Problem 8.19 A signs a one-year note for $1000 and receives $920 from the bank. At the end of six months, A makes a payment of $288. Assuming simple discount, to what amount does this reduce the face amount of the note? Problem 8.20 A bank oﬀers a 272−day discounted loan at a simple discount rate of 12%. (a) How much money would a borrower receive if she asked for a $5000 loan? (b) What size loan should the borrower ask for in order to actually receive $5000? (c) What is the equivalent simple interest rate that is being charged on the loan? Problem 8.21 A discounted loan of $3000 at a simple discount rate of 6.5% is oﬀered to Mr. Jones. If the actual amount of money that Mr. Jones receives is $2869.11, when is the $3000 due to be paid back? Problem 8.22 Show that dn < in. Problem 8.23 At time t = 0, Paul deposits $3500 into a fund crediting interest with an annual discount factor of 0.96. Find the fund value at time 2.5. Problem 8.24 Method A assumes simple interest over ﬁnal fractional periods, while Method B assumes simple discount over ﬁnal fractional periods. The annual eﬀective rate of interest is 20%. Find the ratio of the present value of a payment to be made in 1.5 years computed under method A to that computed under Method B. 66 THE BASICS OF INTEREST THEORY 9 Nominal Rates of Interest and Discount When we speak of either the “eﬀective” rate of interest/discount we mean interest is paid once per measurement period, either at the end of the period( in the case of interest rate) or at the beginning of the period(in the case of discount rate). In this section, we consider situations where interest is paid more than once per measurement period. Rates of interest and discount in these cases are called nominal. In this section we deﬁne nominal rates of interest and discount and determine relationships between nominal rates and eﬀective rates as well as relationships between nominal rates of interest and discount. Compound interest or discount will always be assumed, unless speciﬁed otherwise. When interest is paid (i.e., reinvested) more frequently than once per period, we say it is “payable” (“convertible”, “compounded”) each fraction of a period, and this fractional period is called the interest conversion period. A nominal rate of interest i(m) payable m times per period, where m is a positive integer, represents m times the eﬀective rate of compound interest used for each of the mth of a period. In this case, i(m) m is the eﬀective rate of interest for each mth of a period. Thus, for a nominal rate of 12% compounded monthly, the eﬀective rate of interest per month is 1% since there are twelve months in a year. Suppose that 1 is invested at a nominal rate i(m) compounded m times per measurement period. That is, the period is partitioned into m equal fractions of a period. At the end of the ﬁrst fraction of the period the accumulated value is 1 + i(m) m . At the end of the second fraction of the period the accumulated value is 1 + i(m) m 2 . Continuing, we ﬁnd that the accumulated value at the end of the mth fraction of a period, which is the same as the end of one period, is 1 + i(m) m m and at the end of t years the accumulated value is a(t) = 1 + i(m) m mt . Figure 9.1 illustrates accumulation at a nominal rate of interest for one measurement period. Figure 9.1 9 NOMINAL RATES OF INTEREST AND DISCOUNT 67 Example 9.1 Find the accumulated value of 1,000 after three years at a rate of interest of 24% per year convertible monthly. Solution. The accumulated value is 1, 000 1 + 0.24 12 12×3 = 2, 039.89 Example 9.2 Find the accumulated value of $3,000 to be paid at the end of 8 years with a rate of compound interest of 5% (a) per annum; (b) convertible quarterly; (c) convertible monthly. Solution. (a) The accumulated value is 3, 000 1 + 0.05 1 8 ≈$4, 432.37. (b) The accumulated value is 3, 000 1 + 0.05 4 8×4 ≈$4, 464.39. (c) The accumulated value is 3, 000 1 + 0.05 12 8×12 ≈$4, 471.76 Next we describe the relationship between eﬀective and nominal rates. If i denotes the eﬀective rate of interest per one measurement period equivalent to i(m) then we can write 1 + i = 1 + i(m) m m since each side represents the accumulated value of a principal of 1 invested for one year. Rearrang-ing we have i = 1 + i(m) m m −1 and i(m) = m[(1 + i) 1 m −1]. For any t ≥0 we have (1 + i)t = 1 + i(m) m mt . Example 9.3 Given the nominal interest rate of 12%, compounded monthly. Find the equivalent eﬀective annual interest rate. 68 THE BASICS OF INTEREST THEORY Solution. The answer is i = 1 + 0.12 12 12 −1 = 12.7% Example 9.4 (a) Find the annual eﬀective interest rate i which is equivalent to a rate of compound interest of 8% convertible quarterly. (b) Find the compound interest rate i(2) which is equivalent to an annual eﬀective interest rate of 8%. (c) Find the compound interest rate i(4) which is equivalent to a rate of compound interest of 8% payable semi-annually. Solution. (a) We have 1 + i = 1 + 0.08 4 4 ⇒i = 1 + 0.08 4 4 −1 ≈0.08243216 (b) We have 1 + 0.08 = 1 + i(2) 2 2 ⇒i(2) = 2[(1.08) 1 2 −1] ≈0.07846. (c) We have 1 + i(4) 4 4 = 1 + i(2) 2 2 ⇒i(4) = 4[(1.04) 1 2 −1] ≈0.0792 In the same way that we deﬁned a nominal rate of interest, we could also deﬁne a nominal rate of discount, d(m), as meaning an eﬀective rate of discount of d(m) m for each of the mth of a period with interest paid at the beginning of a mth of a period. Figure 9.2 illustrates discounting at a nominal rate of discount for one measurement period. Figure 9.2 The accumulation function with the nominal rate of discount d(m) is a(t) = 1 −d(m) m −mt , t ≥0. 9 NOMINAL RATES OF INTEREST AND DISCOUNT 69 Example 9.5 Find the present value of $8,000 to be paid at the end of 5 years at a an annual rate of compound interest of 7% (a) convertible semiannually. (b) payable in advance and convertible semiannually. Solution. (a) The answer is 8, 000 1 + 0.07 2 5×2 ≈$5, 671.35 (b) The answer is 8000 1 −0.07 2 5×2 ≈$5, 602.26 If d is the eﬀective discount rate equivalent to d(m) then 1 −d = 1 −d(m) m m since each side of the equation gives the present value of 1 to be paid at the end of the measurement period. Rearranging, we have d = 1 − 1 −d(m) m m and solving this last equation for d(m) we ﬁnd d(m) = m[1 −(1 −d) 1 m] = m(1 −ν 1 m). Example 9.6 Find the present value of $1,000 to be paid at the end of six years at 6% per year payable in advance and convertible semiannually. Solution. The answer is 1, 000 1 −0.06 2 12 = $693.84 There is a close relationship between nominal rate of interest and nominal rate of discount. Since 1 −d = 1 1+i, we conclude that 1 + i(m) m m = 1 + i = (1 −d)−1 = 1 −d(n) n −n . (9.1) 70 THE BASICS OF INTEREST THEORY If m = n then the previous formula reduces to 1 + i(n) n = 1 −d(n) n −1 . Example 9.7 Find the nominal rate of discount convertible semiannually which is equivalent to a nominal rate of interest of 12% per year convertible monthly. Solution. We have 1 −d(2) 2 −2 = 1 + 0.12 12 12 . Solving for d(2) we ﬁnd d(2) = 0.11591 Note that formula (9.1) can be used in general to ﬁnd equivalent rates of interest or discount, either eﬀective or nominal, converted with any desired frequency. Example 9.8 Express d(4) as a function of i(3). Solution. We have 1 −d(4) 4 −4 = 1 + i(3) 3 3 . Thus, 1 −d(4) 4 = 1 + i(3) 3 −3 4 So d(4) = 4 " 1 − 1 + i(3) 3 −3 4# An analogous formula to i −d = id holds for nominal rates of interest and discount as shown in the next example. Example 9.9 Prove that i(m) m −d(m) m = i(m) m · d(m) m 9 NOMINAL RATES OF INTEREST AND DISCOUNT 71 Solution. We have 1 + i(m) m = 1 −d(m) m −1 which is equivalent to 1 + i(m) m · 1 −d(m) m = 1. Expanding we obtain 1 −d(m) m + i(m) m −i(m) m · d(m) m = 1. Hence, i(m) m −d(m) m = i(m) m · d(m) m Example 9.10 Show that i(m) = d(m)(1 + i) 1 m. Solution. We know that i(m) m −d(m) m = i(m) m · d(m) m Multiplying through by m and rearranging we ﬁnd i(m) = d(m) 1 + i(m) m = d(m)(1 + i) 1 m Example 9.11 (a) On occasion, interest is convertible less frequently than once a year. Deﬁne i( 1 m ) and d( 1 m ) to be the nominal annual rates of interest and discount convertible once every m years. Find a formula analogous to formula (9.1) relating i( 1 m ) and d( 1 m ). (b) Find the accumulated value of $100 at the end of two years if the nominal annual rate of discount is 6%, convertible once every four years. Solution. (a) If i( 1 m ) is convertible once every m years, then m · i( 1 m ) is the eﬀective interest rate over m years. Thus, by deﬁnition of equivalent rates we can write 1 + m · i( 1 m ) = (1 + i)m. 72 THE BASICS OF INTEREST THEORY Likewise, we have 1 −p · d( 1 p ) = (1 + i)−p. It follows that (1 + m · i( 1 m )) 1 m = (1 −p · d( 1 p ))−1 p. Note that this formula can be obtained from formula (9.1) by replacing m and p by their reciprocals. (b) If i is the annual eﬀective interest rate than the accumulated value at the end of two years is 100(1 + i)2 = 100(1 −4 × 0.06)−2 4 = $114.71 Remark 9.1 Nominal rates of interest or discount are not relevant under simple interest and simple discount. For example, if we let i(m) be the nominal interest rate and i be the equivalent simple interest rate then at the end of one year we ﬁnd 1 + i = 1 + i(m) or i = i(m). Similarly, we have d = d(m). 9 NOMINAL RATES OF INTEREST AND DISCOUNT 73 Practice Problems Problem 9.1 A man borrows $1,000 at an interest rate of 24% per year compounded monthly. How much does he owe after 3 years? Problem 9.2 If i(6) = 0.15, ﬁnd the equivalent nominal rate of interest convertible semiannually. Problem 9.3 Suppose that $100 is deposited into a savings account, earning at a discount rate of 0.15% biweekly, at the beginning of year 2006. (a) Find the nominal annual discount rate. (b) Find the eﬀective annual discount rate. (c) Find the amount of interest generated during the year. (d) Find the equivalent eﬀective annual interest rate. (e) Find the equivalent nominal annual interest rate, convertible monthly. Problem 9.4 Express i(6) as a function of d(2). Problem 9.5 Given that i(m) = 0.1844144 and d(m) = 0.1802608. Find m. Problem 9.6 It is known that 1 + i(n) n = 1 + i(4) 4 1 + i(5) 5 Find n. Problem 9.7 If r = i(4) d(4), express ν in terms of r. Problem 9.8 You deposit $1,000 into Account A and $750 into Account B. Account A earns an eﬀective annual interest rate of 5%. Account B earns interest at i(4). Ten years later, the two accounts have the same accumulated value. Find i(4). (Express as a percentage, to two decimal places.) 74 THE BASICS OF INTEREST THEORY Problem 9.9 ‡ Eric deposits X into a savings account at time 0, which pays interest at a nominal rate of i, compounded semiannually. Mike deposits 2X into a diﬀerent savings account at time 0, which pays simple interest at an annual rate of i. Eric and Mike earn the same amount of interest during the last 6 months of the 8th year. Calculate i. Problem 9.10 ‡ Calculate the nominal rate of discount convertible monthly that is equivalent to a nominal rate of interest of 18.9% per year convertible monthly. Problem 9.11 A bank credits interest on deposits quarterly at rate 2% per quarter. What is the nominal interest rate (per annum) for interest compounded quarterly? Find the annual eﬀective rate of interest. Problem 9.12 Interest on certain deposits is compounded monthly. Find the nominal rate of interest (per annum) for these deposits if the APR is 15%. Obtain the accumulation of an investment of $1,000 over a period of 6 months. How much interest has accrued by that time? Problem 9.13 Suppose that the eﬀective interest rate for 1 5 year is 2%. Determine the equivalent nominal interest rate, compounded every 6 years. Problem 9.14 Fund A earns interest at a nominal rate of 6% compounded monthly. Fund B earns interest at a nominal rate of discount compounded three times per year. The annual eﬀective rates of interest earned by both funds are equivalent. Calculate the nominal rate of discount earned by Fund B. Problem 9.15 Treasury bills, or T−bills, are sold in terms ranging from a few days to 26 weeks. Bills are sold at a discount from their face value. For instance, you might pay $990 for a $1,000 bill. When the bill matures, you would be paid $1,000. The diﬀerence between the purchase price and face value is interest. A bill for 100 is purchased for 96 three months before it is due. Find: (a) The nominal rate of discount convertible quarterly earned by the purchaser. (b) The annual eﬀective rate of interest earned by the purchaser. Problem 9.16 If i(8) = 0.16, calculate d( 1 2 ). 9 NOMINAL RATES OF INTEREST AND DISCOUNT 75 Problem 9.17 A fund earns a nominal rate of interest of 6% compounded every two years. Calculate the amount that must be contributed now to have 1000 at the end of six years. Problem 9.18 A deposit is made on January 1, 2004. The investment earns 6% compounded semi-annually. Calculate the monthly eﬀective interest rate for the month of December 2004. Problem 9.19 A deposit is made on January 1, 2004. The investment earns interest at a rate equivalent to an annual rate of discount of 6%. Calculate the monthly eﬀective interest rate for the month of December 2004. Problem 9.20 A deposit is made on January 1, 2004. The investment earns interest at a rate equivalent to a rate of discount of 6% convertible quarterly. Calculate the monthly eﬀective interest rate for the month of December 2004. Problem 9.21 Calculate the present value of $1,000 payable in 10 years using a discount rate of 5% convertible quarterly. Problem 9.22 Calculate the accumulated value at the end of 3 years of 250 payable now assuming an interest rate equivalent to a discount rate of 12% convertible monthly. Problem 9.23 Thomas pays 94 into a fund. Six months later, the fund pays Thomas 100. Calculate the nominal rate of discount convertible semi-annually. Problem 9.24 Thomas pays 94 into a fund. Six months later, the fund pays Thomas 100. Calculate the annual eﬀective rate of interest earned. Problem 9.25 Find the present value of 5000, to be paid at the end of 25 months, at a rate of discount of 8% convertible quarterly: (a) assuming compound discount throughout; (b) assuming simple discount during the ﬁnal frac-tional period. 76 THE BASICS OF INTEREST THEORY Problem 9.26 ‡ A bank oﬀers the following certiﬁcates of deposit: Nominal annual interest rate Term in years (convertible quarterly) 1 4% 3 5% 5 5.65% The bank does not permit early withdrawal. The certiﬁcates mature at the end of the term. During the next six years the bank will continue to oﬀer these certiﬁcates of deposit with the same terms and interest rates. An investor initially deposits $10,000 in the bank and withdraws both principal and interest at the end of six years. Calculate the maximum annual eﬀective rate of interest the investor can earn over the 6-year period. Problem 9.27 A bank oﬀers the following certiﬁcates of deposit: Nominal annual interest rate Term in years (convertible semi-annually) 1 5% 2 6% 3 7% 4 8% The bank does not permit early withdrawal. The certiﬁcates mature at the end of the term. During the next six years the bank will continue to oﬀer these certiﬁcates of deposit. An investor plans to invest 1,000 in CDs. Calculate the maximum amount that can be withdrawn at the end of six years. Problem 9.28 You are given (1) Fund X accumulates at an interest rate of 8% compounded quarterly (2) Fund Y accumulates at an interest rate of 6% compounded semiannually (3) at the end of 10 years, the total amount in the two funds combined is 1000 (4) at the end of 5 years, the amount in Fund X is twice that in Fund Y Calculate the total amount in the two funds at the end of 2 years. 9 NOMINAL RATES OF INTEREST AND DISCOUNT 77 Problem 9.29 A collection agency pays a doctor $5,000 for invoices that the doctor has not been able to collect on. After two years, the collection agency has collected $6,000 on the invoices. At what nominal rate of discount compounded monthly did the collection agency receive on this transaction? Problem 9.30 A trust company oﬀers guaranteed investment certiﬁcates paying i(2) = 8.9% and i(1) = 9%. Which option yields the higher annual eﬀective rate of interest? 78 THE BASICS OF INTEREST THEORY 10 Force of Interest: Continuous Compounding Eﬀective and nominal rates of interest and discount each measures interest over some interval of time. Eﬀective rates of interest and discount measure interest over one full measurement period, while nominal rates of interest and discount measure interest over mths of a period. In this section we want to measure interest at any particular moment of time. This measure of interest is called the force of interest. To start with, consider the case of a nominal compound interest i(m) converted m times a period. We can think of the force of interest, denoted by δ, as the limit of i(m) as the number of times we credit the compounded interest goes to inﬁnity. That is, δ = lim m→∞i(m). Letting i be the eﬀective interest rate equivalent to i(m) we have i(m) = m[(1 + i) 1 m −1] = (1 + i) 1 m −1 1 m . Hence, δ = lim m→∞ (1 + i) 1 m −1 1 m . The above limit is of the form 0 0 so that we can apply L’Hopital’s rule to obtain δ = lim m→∞ d dm h (1 + i) 1 m −1 i d dm 1 m = lim m→∞[(1 + i) 1 m ln (1 + i)] = ln (1 + i) since limm→∞(1 + i) 1 m = 1. Remark 10.1 The eﬀective interest rate i can be written as a series expansion of δ : i = eδ −1 = δ + δ2 2! + · · · + δn n! + · · · . Similarly, δ = ln (1 + i) = i −i2 2 + · · · + (−1)nin n + · · · . 10 FORCE OF INTEREST: CONTINUOUS COMPOUNDING 79 Example 10.1 Given the nominal interest rate of 12%, compounded monthly. Find the equivalent force of interest δ. Solution. The eﬀective annual interest rate is i = (1 + 0.01)12 −1 ≈0.1268250. Hence, δ = ln (1 + i) = ln (1.1268250) ≈0.119404. Remark 10.2 Intuitively, δ represents a nominal interest rate which is converted continuously, a notion of more theoretical than practical importance. Indeed, in theory the most important measure of interest is the force of interest. In practice, however, eﬀective and nominal rates of interest tend to be used more frequently because most ﬁnancial transactions involve discrete, not continuous, processes. However, δ can be used in practice as an approximation to interest converted very frequently, such as daily. Having related δ and i immediately relates δ and the other measures of interested introduced in the previous sections. Indeed, we have the following important set of equalities 1 + i(m) m m = 1 + i = (1 −d)−1 = 1 −d(p) p −p = eδ. Example 10.2 Using a constant force of interest of 4.2%, calculate the present value of a payment of $1,000 to be made in 8 years’ time. Solution. The present value is 1, 000(1 + i)−8 = 1, 000e−8δ = 1, 000e−8×0.042 ≈$714.62 Example 10.3 A loan of $3,000 is taken out on June 23, 1997. If the force of interest is 14%, ﬁnd each of the following (a) The value of the loan on June 23, 2002. (b) The value of i. (c) The value of i(12). 80 THE BASICS OF INTEREST THEORY Solution. (a) 3, 000(1 + i)5 = 3, 000e5δ = 3, 000e0.7 ≈$6, 041.26. (b) i = eδ −1 = e0.14 −1 ≈0.15027. (c) We have 1 + i(12) 12 12 = 1 + i = e0.14. Solving for i(12) we ﬁnd i(12) ≈0.14082 Now, using the accumulation function of compound interest a(t) = (1 + i)t we notice that δ = ln (1 + i) = d dta(t) a(t) . The deﬁnition of force of interest in terms of a compound interest accumulation function can be extended to any accumulation function. That is, for an accumulation function a(t) we deﬁne the force of interest at time t by δt = a′(t) a(t) . Let’s look closely at the expression on the right. From the deﬁnition of the derivative we have d dta(t) a(t) = lim n→∞ a(t+ 1 n )−a(t) a(t) 1 n . Now the expression a(t + 1 n) −a(t) a(t) is just the eﬀective rate of interest over a very small time period 1 n so that a(t+ 1 n )−a(t) a(t) 1 n is the nominal annual interest rate converted n periods a year with each time period of length 1 n corresponding to that eﬀective rate, which agrees with Remark 10.2. Note that, in general, the force of interest can be a function of t. However, in the case of compound interest δt is a constant. Also, we notice that, since A(t) = A(0)a(t), we can write δt = A′(t) A(t) . 10 FORCE OF INTEREST: CONTINUOUS COMPOUNDING 81 Example 10.4 Show that, for any amount function A(t), we have Z n 0 A(t)δtdt = A(n) −A(0) = I1 + I2 + · · · + In Interpret this result verbally. Solution. We have Z n 0 A(t)δtdt = Z n 0 A′(t)dt = A(n) −A(0) = I1 + I2 + · · · + In where we used Example 2.6. The term A(n) −A(0) is the amount of interest earned over n measurement periods. The term δtdt represents the eﬀective rate of interest over the inﬁnitesimal “period of time” dt. Hence, A(t)δtdt is the amount of interest in this period and R n 0 A(t)δtdt represents the total amount of interest earned over the entire n periods which is A(n) −A(0) Example 10.5 You are given that A(t) = at2 + bt + c, for 0 ≤t ≤2, and that A(0) = 100, A(1) = 110, and A(2) = 136. Determine the force of interest at time t = 1 2. Solution. The condition A(0) = 100 implies c = 100. From A(1) = 110 and A(2) = 136 we ﬁnd the linear system of equations a + b = 10 and 4a + 2b = 36. Solving this system we ﬁnd a = 8 and b = 2. Hence, A(t) = 8t2 + 2t + 100. It follows that δt = A′(t) A(t) = 16t + 2 8t2 + 2t + 100. Hence, δ0.5 = 10 103 = 0.09709 Example 10.6 Find δt in the case of simple interest, that is, when a(t) = 1 + it. Solution. From the deﬁnition of δt we have δt = a′(t) a(t) = i 1 + it 82 THE BASICS OF INTEREST THEORY Note that δt is a decreasing function of t We have now a method for ﬁnding δt given a(t). What if we are given δt instead, and we wish to derive a(t) from it? From the deﬁnition of δt we can write d dr ln a(r) = δr. Integrating both sides from 0 to t we obtain Z t 0 d dr ln a(r)dr = Z t 0 δrdr. Hence, ln a(t) = Z t 0 δrdr. From this last equation we ﬁnd a(t) = e R t 0 δrdr. Also, we can derive an expression for A(t). Since A(t) = A(0)a(t), we can write A(t) = A(0)e R t 0 δrdr. Example 10.7 A deposit of $10 is invested at time 2 years. Using a force of interest of δt = 0.2 −0.02t, ﬁnd the accumulated value of this payment at the end of 5 years. Solution. The accumulated value is A(5) = 10a(5) a(2) = 10e R 5 2 (0.2−0.02t)dt = 10e[0.2t−0.01t2] 5 2 ≈$14.77 Note that a compound interest rate i implies a constant force of interest (equals to ln (1 + i)). The converse is also true as shown in the next example. Example 10.8 Show that if δt = δ for all t then a(t) = (1 + i)t for some i. 10 FORCE OF INTEREST: CONTINUOUS COMPOUNDING 83 Solution. Since δt = δ for all t, we obtain R t 0 δrdr = tδ. Now, for all integer values n ≥1 we have in = a(n) −a(n −1) a(n −1) = enδ −e(n−1)δ e(n−1)δ = eδ −1 = i. In this case, a(t) = eδt = (eδ)t = (1 + i)t The above example shows that a constant force of interest implies a constant eﬀective interest rate. The converse of this result is not always true. See Problem 10.32. Replacing the accumulation function by the discount function in the deﬁnition of δt we obtain the force of discount: δ′ t = − d dt[a(t)]−1 [a(t)]−1 . The negative sign is necessary in order to make the force of discount a positive quantity since the discount function is a decreasing function of t. It is possible to dispence with δ′ t and just use δt according to the following theorem. Theorem 10.1 For all t we have δ′ t = δt. Proof. we have δ′ t = −[a−1(t)]′ a−1(t) =a−2(t)[a(t)]′ a−1(t) =a−2(t)a(t)δt a−1(t) =δt Example 10.9 Find the force of discount under a simple discount rate d. 84 THE BASICS OF INTEREST THEORY Solution. Recall that in the case of a simple discount the discount function is given by [a(t)]−1 = 1 −dt for 0 ≤t < 1 d. Thus, δt =δ′ t = − d dt[a(t)]−1 [a(t)]−1 = − d dt(1 −dt) 1 −dt = d 1 −dt It follows that the force of interest is increasing with simple discount in contrast to simple interest where it is decreasing. See Example 10.6 Example 10.10 Find δ′ t in the case of compound discount. Solution. We have δ′ t = −[(1 −d)t]′ (1 −d)t = −ln (1 −d) Example 10.11 Show that limm→∞d(m) = δ Solution. We have 1 −d(m) m m = (1 + i)−1 = e−δ. Solving for d(m) we ﬁnd d(m) = m[1 −e−δ m]. Using power series expansion of e−δ m we obtain d(m) =m " 1 − 1 + −δ m + 1 2! −δ m 2 + 1 3! −δ m 3 + · · · !# =m δ m −1 2! δ2 m2 + 1 3! δ3 m3 −· · · =δ −1 2! δ2 m + 1 3! δ3 m2 −· · · →δ as m →∞ 10 FORCE OF INTEREST: CONTINUOUS COMPOUNDING 85 Example 10.12 An investor invests $1,000 into an investment account at time 1. The interest is computed based on the following force of interest: δt =    0.02t 0 ≤t < 3 0.025 t ≥3 0 otherwise (a) Find the amount of interest generated between time 2.5 and 3.5. (b) Calculate eﬀective discount rate during the third period, i.e. d3. Solution. (a) We know that A(t) = 1000 a(t) a(1) = 1000e R t 1 δrdr. Thus, the amount of interest generated between time 2.5 and 3.5 is A(3.5) −A(2.5) = 1000 e R 3.5 1 δtdt −e R 2.5 1 δtdt . But Z 3.5 1 δtdt = Z 3 1 0.02tdt + Z 3.5 3 0.025dt = 0.0925 and Z 2.5 1 δtdt = Z 2.5 1 0.02tdt = 0.0525 Hence, A(3.5) −A(2.5) = 1000(e0.0925 −e0.0525) ≈$43.01. (b) d3 = A(3)−A(2) A(3) = e0.08−e0.03 e0.08 ≈4.877% Example 10.13 Find an expression of t in terms of δ so that f(t) = (1 + it) −(1 + i)t is maximum. Solution. Since f ′(t) = i −(1 + i)t ln (1 + i) = i −δ(1 + i)t we see that f ′(t) = 0 when (1 + i)t = i δ. Solving for t we ﬁnd t = ln i−ln δ ln (1+i) = ln i−ln δ δ . Now, f ′′(t) = −δ2(1 + i)t < 0 so that the critical point of f is a maximum Example 10.14 On March 15, 2003, a student deposits X into a bank account. The account is credited with simple interest where i = 7.5% 86 THE BASICS OF INTEREST THEORY On the same date, the student’s professor deposits X into a diﬀerent bank account where interest is credited at a force of interest δt = 2t t2 + k, t ≥0 From the end of the fourth year until the end of the eighth year, both accounts earn the same dollar amount of interest. Calculate k. Solution. The interest earned by the student is X[1 + 0.075(8)] −X[1 + 0.075(4)] = 0.3X The interest earned by the professor is Xe R 8 0 δtdt −Xe R 4 0 δtdt = Xe R 8 0 2t t2+k dt −Xe R 4 0 2t t2+k dt = Xe[ln (t2+k)] 8 0 −Xe[ln (t2+k)] 4 0 = X 64 + k k −X 16 + k k = 48 k X. Thus, 0.3X = 48 k X ⇒k = 48 0.3 = 160 Example 10.15 Show that d < d(m) < δ < i(m) < i, m > 1. Solution. We have d dm(i(m)) = e δ m 1 −δ m −e−δ m . From the inequality 1 −x ≤e−x for x ≥0 we conclude that d dm(i(m)) = e δ m 1 −δ m −e−δ m < 0. That is, i(m) is a decreasing function of m. Since i(1) = i and i(∞) = δ we obtain δ < i(m) < i. Similarly, d dm(d(m)) = e−δ m(e δ m −1 −δ m) > 0 10 FORCE OF INTEREST: CONTINUOUS COMPOUNDING 87 so that d(m) is an increasing function of m. Since d(1) = d and d(∞) = δ we can write d < d(m) < δ. Combining these inequalities we ﬁnd d < d(m) < δ < i(m) < i, m > 1 88 THE BASICS OF INTEREST THEORY Practice Problems Problem 10.1 A deposit of $500 is invested at time 5 years. The constant force of interest is 6% per year. Determine the accumulated value of the investment at the end of 10 years. Problem 10.2 If the constant force of interest is 6%, what is the corresponding annual eﬀective rate of interest? Problem 10.3 Assume that the force of interest varies with time and is given by δt = a + b t. Find the formula for the accumulation of one unit money from time t1 to time t2. Problem 10.4 You need $500 on Jan 1, 2012. To save for this amount, you invest x on Jan 1, 2008 and 2x on July 1, 2008. The force of interest is δt = 0.02t where t is 0 on Jan 1, 2008. Find x. Problem 10.5 At a constant force of interest, $200 accumulates to $240 over the course of 8 years. Find the force of interest δ. Problem 10.6 ‡ Bruce deposits 100 into a bank account. His account is credited interest at a nominal rate of interest of 4% convertible semiannually. At the same time, Peter deposits 100 into a separate account. Peter’s account is credited interest at a force of interest of δ. After 7.25 years, the value of each account is the same. Calculate δ. Problem 10.7 ‡ Bruce deposits 100 into a bank account. His account is credited interest at a nominal rate of interest i convertible semiannually. At the same time, Peter deposits 100 into a separate account. Peter’s account is credited interest at a force of interest of δ. After 7.25 years, the value of each account is 200. Calculate i −δ. Problem 10.8 ‡ At time t = 0, 1 is deposited into each of Fund X and Fund Y. Fund X accumulates at a force of interest δt = t2 k . Fund Y accumulates at a nominal rate of discount of 8% per annum convertible semiannually. At time t = 5, the accumulated value of Fund X equals the accumulated value of Fund Y. Determine k. 10 FORCE OF INTEREST: CONTINUOUS COMPOUNDING 89 Problem 10.9 ‡ At time 0, K is deposited into Fund X, which accumulates at a force of interest δt = 0.006t2. At time m, 2K is deposited into Fund Y, which accumulates at an annual eﬀective interest rate of 10%. At time n, where n > m, the accumulated value of each fund is 4K. Determine m. Problem 10.10 Suppose $2,000 is invested for 3 years with a constant force of interest equal to 9%. (a) Find the accumulation function. (b) Find the eﬀective rate of interest. (c) Find the value of the investment after 3 years. Problem 10.11 ‡ Ernie makes deposits of 100 at time 0, and X at time 3 . The fund grows at a force of interest δt = t2 100, t > 0. The amount of interest earned from time 3 to time 6 is X. Calculate X. Problem 10.12 If the force of interest δ is 0.1, obtain the nominal rate of interest (per annum) on deposits com-pounded (a) monthly; (b) quarterly; (c) annually. Problem 10.13 The force of interest depends upon time, with δ(t) = 1 10(1+t), 0 ≤t ≤4. Find the accumulated amount after four years for an initial investment of $1,000 at time zero. Problem 10.14 The force of interest δ is constant. The corresponding nominal rate of discount when interest is compounded p-thly is d(p). Express d(p) in terms of p. Show that d(p) is an increasing function of p. Problem 10.15 ‡ Tawny makes a deposit into a bank account which credits interest at a nominal interest rate of 10% per annum, convertible semiannually. At the same time, Fabio deposits 1,000 into a diﬀerent bank account, which is credited with simple interest. At the end of 5 years, the forces of interest on the two accounts are equal, and Fabio’s account has accumulated to Z. Determine Z. Problem 10.16 Fund A earns interest at a nominal rate of interest of 12% compounded quarterly. Fund B earns interest at a force of interest δ. John invested $1,000 in each fund ﬁve years ago. Today, the amount in Fund A is equal to the amount in Fund B. Calculate δ. 90 THE BASICS OF INTEREST THEORY Problem 10.17 A fund earns interest at a force of interest of δ = 0.01t. Calculate the value at the end of 10 years of 8,000 invested at the end of 5 years. Problem 10.18 A fund earns interest at a force of interest of δ = 0.01t. Calculate the eﬀective rate of interest in the 10th year. Problem 10.19 Calculate the eﬀective annual interest rate equivalent to a nominal rate of discount of 6% com-pounded continuously. Problem 10.20 Fund A accumulates at a simple interest rate of 10%. Find B accumulates at a simple discount rate of 5%. Find the point in time at which the force of interest on the two funds are equal. Problem 10.21 An investment is made for one year in a fund with a(t) = a + bt + ct2. The nominal rate of interest earned during the ﬁrst half of the year is 5% convertible semi-annually. The eﬀective rate of interest earned for the entire year is 7%. Find δ0.5. Problem 10.22 If a(t) = 1 + .01(t2 + t), calculate δ5. Problem 10.23 You are given that δ = 0.05. Calculate the amount that must be invested at the end of 10 years to have an accumulated value at the end of 30 years of $1,000. Problem 10.24 Calculate k if a deposit of 1 will accumulate to 2.7183 in 10 years at a force of interest given by: δt = kt 0 < t ≤5 0.04kt2 t > 5 Problem 10.25 A deposit is made on January 1, 2004. The investment earns interest at a constant force of interest of 6%. Calculate the monthly eﬀective interest rate for the month of December 2004. 10 FORCE OF INTEREST: CONTINUOUS COMPOUNDING 91 Problem 10.26 A fund earns interest at a constant force of interest of δ. Kathy invested 10,000 in the fund. Twenty two years later, Kathy has 30,042. Calculate δ. Problem 10.27 A fund earns interest at a force of interest of δt = kt. Ryan invests 1,000 at time t = 0. After 8 years, Ryan has 1,896.50 . Calculate k. Problem 10.28 You are given that i(12) = 0.09. Calculate δ. Problem 10.29 You are given that d(2) = 0.04. Calculate δ. Problem 10.30 ‡ At time 0, deposits of 10,000 are made into each of Fund X and Fund Y. Fund X accumulates at an annual eﬀective interest rate of 5%. Fund Y accumulates at a simple interest rate of 8%. At time t, the forces of interest on the two funds are equal. At time t, the accumulated value of Fund Y is greater than the accumulated value of Fund X by Z. Determine Z. Problem 10.31 Find an expression for the fraction of a period at which the excess of present values computed at simple discount over compound discount is a maximum. Hint: See Example 10.13. Problem 10.32 Consider the following problem: Invest 1 at a compound interest i for integral periods and at a simple interest for fractional periods. (a) Show that ik = i for k = 1, 2, · · · , n. (b) Show that δt1 ̸= δt2 for k < t1 < t2 < k + 1, k = 1, 2, · · · n −1. Thus, a constant eﬀective rate of interest i does not necessarily imply a constant force of interest. Problem 10.33 Which of the following are true? (I) d dd(i) = ν−2 (II) d di(i(m)) = ν−( m−1 m ) (III) d dδ(i) = 1 + i. 92 THE BASICS OF INTEREST THEORY Problem 10.34 You are given that δt = 0.2t 1+0.1t2. Calculate i2. Problem 10.35 Simplify the following expression d dνδ d did . Problem 10.36 An investment is made in an account in which the amount function is given by A(t) = t2 + t + 2.25, 0 ≤t ≤4. Compute the ratio i3 δ2. Problem 10.37 Fund A accumulates at a force of interest 0.05 1 + 0.05t at time t ≥0. Fund B accumulates at a force of interest 0.05. You are given that the amount in Fund A at time zero is 1,000, the amount in Fund B at time zero is 500, and that the amount in Fund C at any time t is equal to the sum of the amount in Fund A and Fund B. Fund C accumulates at force of interest δt. Find δ2. Problem 10.38 Show that d dtδt = A′′(t) A(t) −δ2 t . Problem 10.39 Find δt if A(t) = K2t3t252t. Problem 10.40 Show that R n 0 δtdt = ln a(n). 11 TIME VARYING INTEREST RATES 93 11 Time Varying Interest Rates In this section we consider situations involving varying interest. The ﬁrst involves a continuously varying force of interest δt. In this case, the accumulated value at time t is given by A(t) = A(0)e R t 0 δrdr. Example 11.1 Find the accumulated value of $200 invested for 5 years if the force of interest is δt = 1 8+t. Solution. The accumulated value is A(5) = 200a(5) = 200e R 5 0 dt 8+t = 200e[ln (8+t)]5 0 = 200 × 13 8 = $325 The second situation involves changes in the eﬀective rate of interest over a period of time. Let-ting in denote the eﬀective rate of interest during the nth period from the date of investment, the accumulated value for integral t is given by a(t) = (1 + i1)(1 + i2) · · · (1 + it) (11.1) and the present value is given by (a(t))−1 = (1 + i1)−1(1 + i2)−1 · · · (1 + in)−1. (11.2) Example 11.2 Find the accumulated value of $500 invested for 9 years if the rate of interest is 5% for the ﬁrst 3 years, 5.5% for the second 3 years, and 6.25% for the third 3 years. Solution. The accumulated value is 500(1 + 0.05)3(1 + 0.055)3(1 + 0.0625)3 = $815.23 Formulas (11.1) and (11.2) can be extended to include nominal rates of interest or discount. We illustrate this in the next example. Example 11.3 A fund will earn a nominal rate of interest of 5% compounded quarterly during the ﬁrst two years, a nominal rate of discount of 4% compounded monthly during years 3 and 4, and a constant force of interest of 3% during the ﬁfth and sixth year. Calculate the amount that must be invested today in order to accumulate 5,000 after 6 years. 94 THE BASICS OF INTEREST THEORY Solution. The amount that must be invested is 5000 1 + 0.05 4 −8 1 −0.04 12 24 e−0.03×2 = $3, 935.00 A common question involving varying interest is the question of ﬁnding an equivalent level rate to the rate that vary. Moreover, the answer depends on the period of time chosen for the compari-son . We illustrate this point in the next example. Example 11.4 (a) Find the eﬀective interest rate over the ﬁrst four-year period from now if the eﬀective rate of interest is 5% for the ﬁrst three years from now, 4.5% for the next three years, and 4% for the last three years. (b) What about over the six-year period from now? Solution. (a) We have (1 + i)4 = (1.05)3(1.045). Solving this equation for i we ﬁnd i = [(1.05)3(1.045)] 1 4 −1 = 4.87% (b) We have (1 + i)6 = (1.05)3(1.045)3. Solving this equation for i we ﬁnd i = [(1.05)3(1.045)3] 1 6 −1 = 4.75% Example 11.5 Find the accumulated value of 1 at the end of n periods where the eﬀective rate of interest for the kth period, 1 ≤k ≤n, is deﬁned by ik = (1 + r)k(1 + i) −1. Solution. We know that a(n) = (1 + i1)(1 + i2) · · · (1 + in) = (1 + r)(1 + i)(1 + r)2(1 + i) · · · (1 + r)n(1 + i) = (1 + r)1+2+···+n(1 + i)n = (1 + r) n(n+1) 2 (1 + i)n 11 TIME VARYING INTEREST RATES 95 Practice Problems Problem 11.1 Find the eﬀective interest rate over a three-year period which is equivalent to an eﬀective rate of discount of 8% the ﬁrst year, 7% the second year, and 6% the third year. Problem 11.2 Find the accumulated value of 1 at the end of n years if the force of interest is δt = 1 1+t. Problem 11.3 Find the accumulated value of 1 at the end of 19 years if the force of interest is δt = 0.04(1 + t)−2. Problem 11.4 If δt = 0.01t, 0 ≤t ≤2, ﬁnd the equivalent annual eﬀective rate of interest over the interval 0 ≤t ≤2. Problem 11.5 An investment account earns 4% the ﬁrst year, 5% the second year, and X% the third year. Find X if the three year average interest rate is 10%. Problem 11.6 An investment of $500 was made three years ago. For the ﬁrst six months the annual eﬀective rate of interest paid on the account was 5%, but it then increased to 7% and did not change over the following two and a half years. Find the present accumulation of the investment. Problem 11.7 The annual eﬀective interest rate for a given year is determined by the function: in = 0.02n, where n = 1, 2, 3, · · · . Brad’s initial investment of $400 earns interest every year according to the preceding. Find the accumulated value of his investment after six years. Problem 11.8 ‡ At a force of interest δt = 2 (k+2t) (i) a deposit of 75 at time t = 0 will accumulate to X at time t = 3; and (ii) the present value at time t = 3 of a deposit of 150 at time t = 5 is also equal to X. Calculate X. Problem 11.9 ‡ Jennifer deposits 1000 into a bank account. The bank credits interest at a nominal annual rate of i compounded semi-annually for the ﬁrst 7 years and a nominal annual rate of 2i convertible quarterly for all years thereafter. The accumulated amount in the account after 5 years is X. The accumulated amount in the account at the end of 10.5 years is 1,980. Calculate X. 96 THE BASICS OF INTEREST THEORY Problem 11.10 Suppose δt = t3 100. Find 1 a(3). Problem 11.11 In Fund X money accumulates at a force of interest δt = 0.01t + 0.1, 0 ≤t ≤20. In Fund Y money accumulates at an annual eﬀective interest rate i. An amount of 1 is invested in each fund for 20 years. The value of Fund X at the end of 20 years is equal to the value of Fund Y at the end of 20 years. Calculate the value of Fund Y at the end of 1.5 years. Problem 11.12 If the eﬀective rate of discount in year k is equal to 0.01k + 0.06 for k = 1, 2, 3, ﬁnd the equivalent rate of simple interest over the three-year period. Problem 11.13 A savings and loan association pays 7% eﬀective on deposits at the end of each year. At the end of every 3 years a 2% bonus is paid on the balance at that time. Find the eﬀective rate of interest earned by an investor if the money is left on deposit (a) Two years. (b) Three years. (c) Four years. Problem 11.14 On July 1, 1999 a person invested 1,000 in a fund for which the force of interest at time t is given by δt = .02(3 + 2t) where t is the number of years since January 1, 1999. Determine the accumulated value of the investment on January 1, 2000. Problem 11.15 You are given that δt = t 100. Calculate the present value at the end of the 10 year of an accumulated value at the end of 15 years of $1,000. Problem 11.16 Calculate k if a deposit of 1 will accumulate to 2.7183 in 10 years at a force of interest given by: δt = kt 0 < t ≤5 0.04kt2 t > 5 11 TIME VARYING INTEREST RATES 97 Problem 11.17 The annual eﬀective interest rate for year t is it = 1 (10+t). Calculate the current value at the end of year 2 of a payment of 6,000 at the end of year 7. Problem 11.18 A fund pays a nominal rate of 12%. The nominal rate is compounded once in year 1, twice in year 2, 3 times in year 3, etc. Calculate the amount that must be invested today in order to accumulate 5,000 after 6 years. Problem 11.19 You are given δt = 2 t−1, for 2 ≤t ≤20. For any one year interval between n and n + 1, with 2 ≤n ≤9, calculate d(2) n+1. Problem 11.20 Investment X for 100,000 is invested at a nominal rate of interest j convertible semi-annually. After four years it accumulates to 214,358.88. Investment Y for 100,000 is invested at a nominal rate of discount k convertible quarterly. After two years, it accumulates to 232,305.73. Investment Z for 100,000 is invested at an annual eﬀective rate of interest equal to j in year one and an annual eﬀective rate of discount equal to k in year two. Calculate the value of investment Z at the end of two years. Problem 11.21 A bank agrees to lend John 10,000 now and X three years later in exchange for a single repayment of 75,000 at the end of 10 years. The bank charges interest at an annual eﬀective rate of 6% for the ﬁrst 5 years and at a force of interest δt = 1 t+1 for t ≥5. Determine X. Problem 11.22 John invests 1000 in a fund which earns interest during the ﬁrst year at a nominal rate of K convertible quarterly. During the 2nd year the fund earns interest at a nominal discount rate of K convertible quarterly. At the end of the 2nd year, the fund has accumulated to 1173.54. Calculuate K. Problem 11.23 At time 0, 100 is deposited into FundX and also into Fund Y. FundX accumulates at a force of interest δt = 0.5(1 + t)−2. Fund Y accumulates at an annual eﬀective interest rate of i. At the end of 9 years, the accumulated value of Fund X equals the accumulated value of Fund Y. Determine i. 98 THE BASICS OF INTEREST THEORY Problem 11.24 ‡ An investor deposits 1000 on January 1 of year x and deposits 1000 on January 1 of year x + 2 into a fund that matures on January 1 of year x + 4. The interest rate on the fund diﬀers every year and is equal to the annual eﬀective rate of growth of the gross domestic product (GDP) during the fourth quarter of the previous year. The following are the relevant GDP values for the past 4 years. Year x −1 x x + 1 x + 2 Quarter III 800.0 850.0 900.0 930.0 Quarter IV 808.0 858.5 918.0 948.6 What is the internal rate of return earned by the investor over the 4-year period? Problem 11.25 A deposit of 10,000 is made into a fund at time t = 0. The fund pays interest at a nominal rate of discount of d compounded quarterly for the ﬁrst two years. Beginning at time t = 2, interest is credited using a nominal rate of interest of 8% compounded quarterly. At time t = 5, the accumulated value of the fund is 14,910. Calculate d. Problem 11.26 Amin deposits 10,000 in a bank. During the ﬁrst year the bank credits an annual eﬀective rate of interest i. During the second year the bank credits an annual eﬀective rate of interest (i −5%). At the end of two years she has 12,093.75 in the bank. What would Amin have in the bank at the end of three years if the annual eﬀective rate of interest were (i + 9%) for each of the three years? Problem 11.27 Fund X starts with 1,000 and accumulates with a force of interest δt = 1 15 −t for 0 ≤t < 15. Fund Y starts with 1,000 and accumulates with an interest rate of 8% per annum compounded semi-annually for the ﬁrst three years and an eﬀective interest rate of i per annum thereafter. Fund X equals Fund Y at the end of four years. Calculate i. Problem 11.28 Amin puts 100 into a fund that pays an eﬀective annual rate of discount of 20% for the ﬁrst two years and a force of interest of rate δt = 2t t2 + 8 < 2 ≤t ≤4 11 TIME VARYING INTEREST RATES 99 for the next two years. At the end of four years, the amount in Amin’s account is the same as what it would have been if he had put 100 into an account paying interest at the nominal rate of i per annum compounded quarterly for four years. Calculate i. Problem 11.29 On January 1, 1980, Jack deposited 1,000 into Bank X to earn interest at the rate of j per annum compounded semi-annually. On January 1, 1985, he transferred his account to Bank Y to earn interest at the rate of k per annum compounded quarterly. On January 1, 1988, the balance at Bank Y is 1,990.76. If Jack could have earned interest at the rate of k per annum compounded quarterly from January 1, 1980 through January 1, 1988, his balance would have been 2,203.76. Calculate the ratio k j . 100 THE BASICS OF INTEREST THEORY 12 Equations of Value and Time Diagrams Interest problems generally involve four quantities: principal(s), investment period length(s), in-terest rate(s), accumulated value(s). If any three of these quantities are known, then the fourth quantity can be determined. In this section we introduce equations that involve all four quantities with three quantities are given and the fourth to be found. In calculations involving interest, the value of an amount of money at any given point in time depends upon the time elapsed since the money was paid in the past or upon time which will elapse in the future before it is paid. This principle is often characterized as the recognition of the time value of money. We assume that this principle reﬂects only the eﬀect of interest and does not include the eﬀect of inﬂation. Inﬂation reduces the the purchasing power of money over time so investors expect a higher rate of return to compensate for inﬂation. As pointed out, we will neglect the eﬀct of inﬂation when applying the above mentioned principle. As a consequence of the above principle, various amounts of money payable at diﬀerent points in time cannot be compared until all the amount are accumulated or discounted to a common date, called the comparison date, is established. The equation which accumulates or discounts each payment to the comparison date is called the equation of value. One device which is often helpful in the solution of equations of value is the time diagram. A time diagram is a one-dimensional diagram where the only variable is time, shown on a single coordinate axis. We may show above or below the coordinate of a point on the time-axis, values of money intended to be associated with diﬀerent funds. A time diagram is not a formal part of a solution, but may be very helpful in visualizing the solution. Usually, they are very helpful in the solution of complex probems. Example 12.1 In return for a payment of $1,200 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. Solution. The comparison date is chosen to be t = 0. The time diagram is given in Figure 12.1. Figure 12.1 12 EQUATIONS OF VALUE AND TIME DIAGRAMS 101 The equation of value is 200 + 400(1 + 0.045)−12 + X(1 + 0.045)−30 = 1200(1 + 0.045)−20. Solving this equation for X we ﬁnd X ≈$231.11 With compound interest, an equation of value will produce the same answer for an unknown value regardless of what comparison date is selected. This is due to the fact that multiplying an equation by a power of an expression yields an equivalent equation. We illustrate this in the next example. Example 12.2 In return for a promise to receive $600 at the end of 8 years, a person agrees to pay $100 at once, $200 at the end of 5 years, and to make further payment at the end of 10 years. Find the payment at the end of 10 years if the nominal interest rate is 8% convertible semi-annually. Solution. The equation of value at t = 0 is 100 + 200(1 + 0.04)−10 + X(1 + 0.04)−20 = 600(1 + 0.04)−16. Solving for X we ﬁnd X ≈186.75. The equation of value at t = 10 is 100(1 + 0.04)20 + 200(1 + 0.04)10 + X = 600(1 + 0.04)4. Solving for X we ﬁnd X ≈186.75. Note that by multiplying the last equation by (1 + 0.04)−20 one gets the equation of value at t = 0 Whereas the choice of a comparison date has no eﬀect on the answer obtained with compound interest, the same cannot be said of simple interest or simple discount. Example 12.3 Find the amount to be paid at the end of 10 years which is equivalent to two payments of 100 each, the ﬁrst to be paid immediately, and the second to be paid at the end of 5 years. Assume 5% simple interest is earned from the date each payment is made, and use a comparison date 1. The end of 10 years. 2. The end of 15 years. 102 THE BASICS OF INTEREST THEORY Solution. 1. With a comparison date at t = 10, the payment at the end of 10 years will be 100(1 + 10 × 0.05) + 100(1 + 5 × 0.05) = 275. 2. With a comparison date at t = 15, the amount P that must be paid at the end of 10 years satisﬁes the equation of value 100(1 + 15 × 0.05) + 100(1 + 10 × 0.05) = P(1 + 5 × 0.05) which implies that P = 260 Example 12.4 Investor A deposits 1,000 into an account paying 4% compounded quarterly. At the end of three years, he deposits an additional 1,000. Investor B deposits X into an account with force of interest δt = 1 6+t. After ﬁve years, investors A and B have the same amount of money. Find X. Solution. Consider investor A’s account ﬁrst. The initial 1,000 accumulates at 4% compounded quarterly for ﬁve years; the accumulated amount of this piece is 1, 000 1 + 0.04 4 4×5 = 1000(1.01)20. The second 1,000 accumulates at 4% compounded quarterly for two years, accumulating to 1, 000 1 + 0.04 4 4×2 = 1000(1.01)8. The value in investor A’s account after ﬁve years is A = 1000(1.01)20 + 1000(1.01)8. The accumulated amount of investor B’s account after ﬁve years is given by B = Xe R 5 0 dt 6+t = Xeln( 11 6 ) = 11 6 X. The equation of value at time t = 5 is 11 6 X = 1000(1.01)20 + 1000(1.01)8. Solving for X we ﬁnd X ≈$1, 256.21 12 EQUATIONS OF VALUE AND TIME DIAGRAMS 103 Practice Problems Problem 12.1 In return for payments of $5,000 at the end of 3 years and $4,000 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. Problem 12.2 At a certain interest rate the present values of the following two payment patterns are equal: (i) 200 at the end of 5 years plus 500 at the end of 10 years; (ii) 400.94 at the end of 5 years. At the same interest rate 100 invested now plus 120 invested at the end of 5 years will accumulate to P at the end of 10 years. Calculate P. Problem 12.3 An investor makes three deposits into a fund, at the end of 1, 3, and 5 years. The amount of the deposit at time t is 100(1.025)t. Find the size of the fund at the end of 7 years, if the nominal rate of discount convertible quarterly is 4 41. Problem 12.4 ‡ Brian and Jennifer each take out a loan of X. Jennifer will repay her loan by making one payment of 800 at the end of year 10. Brian will repay his loan by making one payment of 1,120 at the end of year 10. The nominal semi-annual rate being charged to Jennifer is exactly one-half the nominal semi-annual rate being charged to Brian. Calculate X. Problem 12.5 Fund A accumulates at 6% eﬀective, and Fund B accumulates at 8% eﬀective. At the end of 20 years the total of the two funds is 2,000. At the end of 10 years the amount in Fund A is half that in Fund B. What is the total of the two funds at the end of 5 years? Problem 12.6 Louis has an obligation to pay a sum of $3,000 in four years from now and a sum of $5,000 in six years from now. His creditor permits him to discharge these debts by paying $X in two years from now, $1000 in three years from now, and a ﬁnal payment of $2X in nine years from now. Assuming an annual eﬀective rate of interest of 6%, ﬁnd X. Problem 12.7 Every Friday in February (7, 14, 21,28) Vick places a 1,000 bet, on credit, with his oﬀ-track bookmaking service, which charges an eﬀective weekly interest rate of 8% on all credit extended. 104 THE BASICS OF INTEREST THEORY Vick looses each bet and agrees to repay his debt in four installments to be made on March 7, 14, 21, and 28. Vick pays 1,100 on March 7, 14, and 21. How much must Vick pay on March 28 to completely repay his debt? Problem 12.8 A borrower is repaying a loan by making payments of 1,000 at the end of each of the next 3 years. The interest rate on the loan is 5% compounded annually. What payment could the borrower make at the end of the ﬁrst year in order to extinguish the loan? Problem 12.9 An investor purchases an investment which will pay 2,000 at the end of one year and 5,000 at the end of four years. The investor pays 1,000 now and agrees to pay X at the end of the third year. If the investor uses an interest rate of 7% compounded annually, what is X? Problem 12.10 Today is New Year’s Day. In return for payments of 1,500 at the end of January, February, and March, and of 3,000 at the end of May, July, and September, an investor agrees to pay now the total value of the 6 payments, and to either make or receive an additional payment at the end of December. Find the amount of that additional payment if it is known that the nominal annual interest rate is 6%, compounded monthly. Problem 12.11 A debt of 7,000 is due at the end of 5 years. If 2,000 is paid at the end of 1 year, what single payment should be made at the end of the 2nd year to liquidate the debt, assuming interest at the rate of 6.5% per year, compounded quarterly. Problem 12.12 George agrees to buy his brother’s car for 7,000. He makes a down payment of 4,000 now, and agrees to pay two equal payments, one at the end of 6 months, and the other at the end of a year. If interest is being charged at 5% per annum eﬀective, how large should each of the equal payments be? Problem 12.13 ‡ Jeﬀdeposits 10 into a fund today and 20 ﬁfteen years later. Interest is credited at a nominal discount rate of d compounded quarterly for the ﬁrst 10 years, and at a nominal interest rate of 6% compounded semiannually thereafter. The accumulated balance in the fund at the end of 30 years is 100. Calculate d. 12 EQUATIONS OF VALUE AND TIME DIAGRAMS 105 Problem 12.14 ‡ The parents of three children, ages 1, 3, and 6, wish to set up a trust fund that will pay X to each child upon attainment of age 18, and Y to each child upon attainment of age 21. They will establish the trust fund with a single investment of Z. Find the equation of value for Z. Problem 12.15 An investment fund is established at time 0 with a deposit of 5000. 1000 is added at the end of 4 months, and an additional 4000 is added at the end of 8 months. No withdrawals are made. The fund value, including interest, is 10560 at the end of 1 year. The force of interest at time t is k 1+(1−t)k, for 0 ≤t ≤1. Determine k. Problem 12.16 A loan of 1000 is made at an interest rate of 12% compounded quarterly. The loan is to be repaid with three payments: 400 at the end of the ﬁrst year, 800 at the end of the ﬁfth year, and the balance at the end of the tenth year. Calculate the amount of the ﬁnal payment. Problem 12.17 You are given δt = 0.5 1+t for t > 0. A payment of 400 at the end of 3 years and 800 at the end of 15 years has the same present value as a payment of 300 at the end of 8 years and X at the end of 24 years. Calculate X. Problem 12.18 A loan of 12 is to be repaid with payments of 10 at the end of 3 years and 5 at the end of 6 years. Calculate the simple discount rate that is being charged on the loan. Problem 12.19 A deposit of K3 at time t = 0 accumulates to 8000 after 3n years using an annual eﬀective discount rate of d. Using the same annual eﬀective discount rate, a deposit of K2 at time t = 0 accumulates to X after 2n years. Calculate X. Problem 12.20 An investor puts 1000 into Fund X and 1000 into Fund Y. Fund Y earns compound interest at the annual rate of j > 0, and Fund X earns simple interest at the annual rate of 1.10j. At the end of 2 years, the amount in Fund Y is 30 more than the amount in Fund X. After 5 years, the amount in Fund Y exceeds the amount in Fund X by E. Determine E. Problem 12.21 You are given two loans, with each loan to be repaid by a single payment in the future. Each payment includes both principal and interest. The ﬁrst loan is repaid by a 3,000 payment at the 106 THE BASICS OF INTEREST THEORY end of four years. The interest is accrued at 10% per annum compounded semi-annually. The second loan is repaid by a 4,000 payment at the end of ﬁve years. The interest is accrued at 8% per annum compounded semi-annually. These two loans are to be consolidated. The consolidated loan is to be repaid by two equal installments of X, with interest at 12% per annum compounded semi-annually. The ﬁrst payment is due immediately and the second payment is due one year from now. Calculate X. Problem 12.22 An investment fund accrues interest with force of interest δt = K 1 + (1 −t)K for 0 ≤t ≤1. At time zero, there is 100,000 in the fund. At time one there is 110,000 in the fund. The only two transactions during the year are a deposit of 15,000 at time 0.25 and a withdrawal of 20,000 at time 0.75. Calculate K. Problem 12.23 Paul borrows 1000 at an annual interest rate of 12%. He repays the loan in full by making the following payments: (1) 500 at the end of 4 months (2) 200 at the end of 14 months (3) R at the end of 18 months Calculate R. 13 SOLVING FOR THE UNKNOWN INTEREST RATE 107 13 Solving for the Unknown Interest Rate In this section, it is the rate of interest i that is unknown in an equation of value. We discuss four ways for detrmining i. Method 1: Direct Method 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). In this situation, the equation of value takes either the form A = P(1 + i)n or A = Peδn. In the ﬁrst case, the interest is found by using the exponential function obtaining i = A P 1 n −1. In the second case, the interest is found using the logarithmic function obtaining δ = 1 n ln A P . Example 13.1 At what interest rate convertible semiannually would $500 accumulate to $800 in 4 years? Solution. Let j = i(2) 2 be the eﬀective rate per six months. We are given that 500(1 + j)8 = 800. Solving for j and using a calculator we ﬁnd j = 8 5 1 8 −1 ≈0.06051. Thus, i(2) = 2j = 0.121 = 12.1% Method 2: Analytical Method When multiple payments are involved, the equation of value takes the form f(i) = 0. In this case, the problem is the classical problem of determining the nonnegative solutions to the equation f(i) = 0 with f(i) being a diﬀerentiable function. If f(i) is a polynomial then the problem reduces to solving polynomial equations. Some known algebraic methods can be used such as the rational zero test, or the quadratic formula, etc. Example 13.2 At what eﬀective interest rate would the present value of $1000 at the end of 3 years plus $2,000 at the end of 6 years be equal to $2,700? Solution. The equation of value at time t = 0 is 2000ν6 + 1000ν3 = 2700. Dividing through by 100 we ﬁnd 108 THE BASICS OF INTEREST THEORY 20ν6 +10ν3 = 27. Using the quadratic formula and the fact that ν3 > 0 we obtain ν3 = 0.9385 from which we obtain ν ≈0.9791. Hence, i = 1 0.9791 −1 ≈0.0213 = 2.13% Method 3: Linear Interpolation If the equation f(i) = 0 can not be solved known algebraic methods, then approximation methods can be used such as linear interpolation. We next introduce this concept. Suppose that we know the values of a function f(x) at distinct points x1 and at x2, and that f(x1) ̸= f(x2). If \|x1 −x2\| is small, it may be reasonable to assume that the graph of f is approxi-mately linear between x1 and x2. That is equivalent to assuming that f(x) = f(x1) + f(x2) −f(x1) x2 −x1 (x −x1) for x1 < x < x2. If now we wish to determine an approximate value of x where f(x) has a speciﬁc value y0, we may use this equation for that purpose and ﬁnd x = x1 + (x2 −x1) y0 −f(x1) f(x2) −f(x1). In particular, if we wish to ﬁnd a zero of f near the given points (i.e. y0 = 0), and if f(x1) ̸= f(x2), a good approximation could be x ≈x1 −f(x1) x2 −x1 f(x2) −f(x1). When we apply these formulas for points between two points x1 and x2, we speak of linear inter-polation. The following example illustrates the use of linear interpolation in determing the unknown rate. Example 13.3 At what interest rate convertible semiannually would an investment of $1,000 immediately and $2,000 three years from now accumulate to $5,000 ten years from now? Solution. The equation of value at time t = 10 years is 1000(1 + j)20 + 2000(1 + j)14 = 5000 where j = i(2) 2 . We will use linear interpolation to solve for j. For that purpose, we deﬁne f(j) = 1000(1 + j)20 + 2000(1 + j)14 −5000. 13 SOLVING FOR THE UNKNOWN INTEREST RATE 109 We want to ﬁnd j such that f(j) = 0. By trial and error we see that f(0.03) = −168.71 and f(0.035) = 227.17. Since f is continuous, j is between these two values. Using linear interpolation we ﬁnd j ≈0.03 + 168.71 × 0.005 227.17 + 168.71 ≈0.0321 and f(0.0321) = −6.11. Hence, i(2) = 2j = 0.0642 = 6.42% Method 4: Successive Iterations Methods In order to achieve a higher level of accuracy than the one provided with linear interpolation, it-eration methods come to mind. We will discuss two iteration methods: the bisection method and the Newton-Raphson method. The bisection method is based on the fact that a diﬀerentiable function f that satisﬁes f(α)f(β) < 0 must satisfy the equation f(x0) = 0 for some x0 between α and β. The ﬁrst step in the methods consists of ﬁnding two starting values x0 < x1 such that f(x0)f(x1) < 0. Usually, these values are found by trial and error. Next, bisect the interval by means of the mid-point x2 = x0+x1 2 . If f(x0)f(x2) < 0 then apply the bisection process to the interval x0 ≤x ≤x2. If f(x1)f(x2) < 0 then apply the bisection process to the interval x2 ≤x ≤x1. We continue the bisection process as many times as necessary to achieve the desired level of accuracy. We illustrate this method in the next example. Example 13.4 At what interest rate convertible semiannually would an investment of $1,000 immediately and $2,000 three years from now accumulate to $5,000 ten years from now? Use the bisection method. Solution. The equation of value at time t = 10 years is 1000(1 + j)20 + 2000(1 + j)14 = 5000 where j = i(2) 2 . Deﬁne f(j) = 1000(1 + j)20 + 2000(1 + j)14 −5000. Descartes’ rule of signs says that the maximum number of positive roots to a polynomial equation is equal to the number of sign changes in the coeﬃcients of the polynomial. In our case, Descartes’ rule asserts the existence of one positive root to the equation f(j) = 0. Now, by trial and error we see that f(0) = −2000 and f(0.1) = 9322.50. Thus, j is between these two values. Let j2 = 0+0.1 2 = 0.05. Then f(0.05) = 1613.16 so that f(0)f(0.05) < 0. Now bisect the interval [0, 0.05] by means of the point j3 = 0.5(0 + 0.05) = 0.025 and note that f(j3) = −535.44. Continue this process to obtain the table 110 THE BASICS OF INTEREST THEORY n jn f(jn) 0 0 −2000 1 0.1 9322.50 2 0.05 1613.16 3 0.025 −535.44 4 0.0375 436.75 5 0.03125 −72.55 6 0.034375 176.02 7 0.0328125 50.25 8 0.03203125 −11.52 9 0.032421875 19.27 10 0.0322265625 3.852 11 0.0321289063 −3.84 12 0.0321777344 0.005 Thus, j ≈0.032178 accurate to six decimal places compared to 0.0321 found in Example 13.3. Hence, i2 = 2j = 0.06436 = 6.436% The problem with the bisection method is that the rate of convergence is slow. An iteration method with a faster rate of convergence is the Newton-Raphson method given by the iteration formula jn+1 = jn −f(jn) f ′(jn). This method is discussed in more details in Section 20. Example 13.5 Rework Example 13.3 using the Newton-Raphson method. Solution. We have f(j) = 1000(1 + j)20 + 2000(1 + j)14 −5000. and f ′(j) = 20000(1 + j)19 + 28000(1 + j)13. Thus, jn+1 = jn −(1 + jn)20 + 2(1 + jn)14 −5 20(1 + jn)19 + 28(1 + jn)13 . 13 SOLVING FOR THE UNKNOWN INTEREST RATE 111 Let j0 = 0. Then j1 =0 −1 + 2 −5 20 + 28 = 0.0416666667 j2 =0.0416666667 −1.041666666720 + 2(1.0416666667)14 −5 20(1.0416666667)19 + 28(1.0416666667)13 = 0.0328322051 j3 =0.0328322051 −1.032832205120 + 2(1.0328322051)14 −5 20(1.0328322051)19 + 28(1.0328322051)13 = 0.0321809345 j4 =0.0321809345 −1.032180934520 + 2(1.0321809345)14 −5 20(1.0321809345)19 + 28(0.0321809345)13 = 0.032177671 Thus, in four iterations we ﬁnd j = 0.032178 and hence i(2) = 6.436% 112 THE BASICS OF INTEREST THEORY Practice Problems Problem 13.1 Find the nominal rate of interest convertible semiannually at which the accumulated value of 1,000 at the end of 15 years is 3,000. Problem 13.2 Find the exact eﬀective rate of interest at which payments of 300 at the present, 200 at the end of one year, and 100 at the end of two years will accumulate to 700 at the end of two years. Problem 13.3 The sum of the accumulated value of 1 at the end of three years at a certain eﬀective rate of interest i, and the present value of 1 to be paid at the end of three years at an eﬀective rate of discount numerically equal to i is 2.0096. Find the rate i. Problem 13.4 ‡ David can receive one of the following two payment streams: (i) 100 at time 0, 200 at time n, and 300 at time 2n (ii) 600 at time 10 At an annual eﬀective interest rate of i, the present values of the two streams are equal. Given νn = 0.75941, determine i. Problem 13.5 ‡ Mike receives cash ﬂows of 100 today, 200 in one year, and 100 in two years. The present value of these cash ﬂows is 364.46 at an annual eﬀective rate of interest i. Calculate i. Problem 13.6 ‡ At a nominal interest rate of i convertible semi-annually, an investment of 1,000 immediately and 1,500 at the end of the ﬁrst year will accumulate to 2,600 at the end of the second year. Calculate i. Problem 13.7 ‡ Eric deposits 100 into a savings account at time 0, which pays interest at a nominal rate of i, compounded semiannually. Mike deposits 200 into a diﬀerent savings account at time 0, which pays simple interest at an annual rate of i. Eric and Mike earn the same amount of interest during the last 6 months of the 8th year. Calculate i. 13 SOLVING FOR THE UNKNOWN INTEREST RATE 113 Problem 13.8 ‡ A deposit of 100 is made into a fund at time t = 0. The fund pays interest at a nominal rate of discount of d compounded quarterly for the ﬁrst two years. Beginning at time t = 2, interest is credited at a force of interest δt = 1 t+1. At time t = 5, the accumulated value of the fund is 260. Calculate d. Problem 13.9 ‡ A customer is oﬀered an investment where interest is calculated according to the following force of interest: δt = 0.02t 0 ≤t ≤3 0.045 3 < t. The customer invests 1,000 at time t = 0. What nominal rate of interest, compounded quarterly, is earned over the ﬁrst four-year period? Problem 13.10 A manufacturer sells a product to a retailer who has the option of paying 30% below the retail price immediately, or 25% below the retail price in six months. Find the annual eﬀective rate of interest at which the retailer would be indiﬀerent between the two options. Problem 13.11 An investor deposits 10,000 in a bank. During the ﬁrst year, the bank credits an annual eﬀective rate of interest i. During the second year the bank credits an annual eﬀective interest i −0.05. At the end of two years the account balance is 12,093.75. What would the account balance have been at the end of three years if the annual eﬀective rate of interest were i + 0.09 for each of the three years? Problem 13.12 You are given δt = 2 t−1 for any t ∈[2, 10]. For any one year interval between n and n + 1; with 2 ≤n ≤9; calculate the equivalent nominal rate of discount d(2). Problem 13.13 It is known that an investment of $1,000 will accumulate to $1,825 at the end of 10 years. If it is assumed that the investment earns simple interest at rate i during the 1st year, 2i during the second year, · · · , 10i during the 10th year, ﬁnd i. Problem 13.14 It is known that an amount of money will double itself in 10 years at a varying force of interest δt = kt. Find k. 114 THE BASICS OF INTEREST THEORY Problem 13.15 A fund pays 1 at time t = 0, 2 at time t = 2n and 1 at time t = 4n. The present value of the payments is 3.61. Calculate (1 + i)n. Problem 13.16 The present value of 300 in 3 years plus 600 in 6 years is equal to 800. Calculate i. Problem 13.17 A savings and loan association pays 7% eﬀective on deposits at the end of each year. At the end of every three years a 2% bonus is paid on the balance at that time. Find the eﬀective rate of interest earned by an investor if the money is left on deposit: (a) Two years. (b) Three years. (c) Four years. Problem 13.18 Consider the following equation of value 25j = 1 −(1 + j)−40. Use linear interpolation to estimate j. Problem 13.19 Consider the following equation of value 0.0725[1 −(1 + 0.01j)−50] = 0.01j. Use linear interpolation to estimate j. Problem 13.20 Two funds, X and Y, start with the same amount. You are given (1) Fund X accumulates at a force of interest 5% (2) Fund Y accumulates at a rate of interest j compounded semi-annually (3) At the end of eight years, Fund X is 1.05 times as large as Fund Y. Calculate j. Problem 13.21 You are given a loan on which interest is charged over a 4 year period, as follows (1) an eﬀective rate of discount of 6% for the ﬁrst year (2) a nominal rate of discount of 5% compounded every 2 years for the second year (3) a nominal rate of interest of 5% compounded semiannually for the third year (4) a force of interest of 5% for the fourth year. Calculate the annual eﬀective rate of interest over the 4 year period. 13 SOLVING FOR THE UNKNOWN INTEREST RATE 115 Problem 13.22 An investor deposits 10 into a fund today and 20 ﬁfteen years later. Interest is credited at a nominal discount rate of d compounded quarterly for the ﬁrst 10 years, and at a nominal interest rate of 6% compounded semi-annually thereafter. The accumulated balance in the fund at the end of 30 years is 100. Calculate d. 116 THE BASICS OF INTEREST THEORY 14 Solving for Unknown Time As pointed out in Section 12, if any three of the four basic quantities: principal(s), investment period length(s), interest rate(s), accumulated value(s) are given, then the fourth can be determined. In this section, we consider the situation in which the length of the investment period is the unknown. Example 14.1 A single payment of $3,938.31 will pay oﬀa debt whose original repayment plan was $1,000 due on January 1 of each of the next four years, beginning January 1, 2006. If the eﬀective annual rate is 8%, on what date must the payment of $3,938.31 be paid? Solution. The equation of value with comparison date January 1, 2006 is 1, 000[1 + (1 + 0.08)−1 + (1 + 0.08)−2 + (1 + 0.08)−3] = 3938.31(1 + 0.08)−t. Solving this equation for t we ﬁnd t = ln {1000(3938.31)−1[1 + (1 + 0.08)−1 + (1 + 0.08)−2 + (1 + 0.08)−3]} ln 1.08 ≈1.25. Therefore, the single payment of 3938.31 must be paid 1.25 years after January 1, 2006, or on March 31, 2007 In the above example, the answer was obtained by using a pocket calculator with exponential and logarithmic functions. An alternative approach to using calculator is the use of linear interpolation in the interest tables as illustrated in the next example. Example 14.2 Find the length of time necessary for $1000 to accumulate to $1500 if invested at 6% per year compounded semiannually: (a) by a direct method, i.e. using a calculator; (b) by interpolating in the interest tables Solution. Let t be the comparison date. Then we must have 1000(1.03)2t = 1000 or (1.03)2t = 1.5. (a) Using a pocket calculator we ﬁnd t = 1 2 ln 1.5 ln 1.03 = 6.859 years. 14 SOLVING FOR UNKNOWN TIME 117 (b) From interest tables, (1.03)13 = 1.46853 and (1.03)14 = 1.51259. Thus, 13 < 2t < 14. Performing a linear interpolation we ﬁnd 2t = 13 + 1.5 −1.46853 1.51259 −1.46853 = 13.714. Thus, t = 6.857 years The Rule of 72 Next, we consider the following problem: Given a particular rate of compound interest, how long will it take an investment to double? Starting with 1 invested at an annually compounded interest i, let n be the time needed for the accumulated value to become 2. That is, (1 + i)n = 2. Taking the natural logarithm of both sides we ﬁnd n ln (1 + i) = ln 2 and solving for n we obtain n = ln 2 ln (1 + i). We would like to ﬁnd an approximation that is pretty good for an interest rate of i = 8%. For that purpose, we start by writing the Taylor series approximation of x ln (1+x) around x = 0.08 to obtain x ln (1 + x) =1.039486977 + 0.4874162023(x −0.08) −0.07425872903(x −0.08)2 + · · · ≈1.039486977 for x very “close” to 0.08. Thus, the time necessary for the original principal to double is given by t = ln 2 ln (1 + i) = ln 2 i · i ln (1 + i) ≈ln 2 i (1.039486977) ≈0.72 i . This is called the rule of 72, since t can be written in the form t = 72 100i. Here’s a table that shows the actual number of years required to double your money based on diﬀerent interest rates, along with the number that the rule of 72 gives you. 118 THE BASICS OF INTEREST THEORY % Rate Actual Rule 72 1 69.66 72 2 35.00 36 3 23.45 24 4 17.67 18 5 14.21 14.4 6 11.90 12 7 10.24 10.29 8 9.01 9 9 8.04 8 10 7.27 7.2 · · · · · · · · · 15 4.96 4.8 20 3.80 3.6 25 3.11 2.88 30 2.64 2.4 40 2.06 1.8 50 1.71 1.44 75 1.24 0.96 100 1.00 0.72 Example 14.3 Suppose $2,000 is invested at a rate of 7% compounded semiannually. (a) Find the length of time for the investment to double by using the exact formula. (b) Find the length of time for the investment to double by using the rule of 72. Solution. (a) By using the exact formula for compound interest we ﬁnd 2000(1 + 0.035)2t = 4000. Solving for t we ﬁnd t ≈10.074 years. (b) Using the Rule of 72 we ﬁnd t = 72 7 ≈10.286 The Method of Equated Time We consider the following problem: Suppose amounts s1, s2, · · · , sn are to be paid at respective times t1, t2, · · · , tn with annual interest rate i. What would be the time t when one single payment 14 SOLVING FOR UNKNOWN TIME 119 of s1 + s2 + · · · + sn would be equivalent to the individual payments made separately? A time diagram for this situation is shown in Figure 14.1 Figure 14.1 In this case, the equation of value at t = 0 is (s1 + s2 + · · · + sn)νt∗= s1νt1 + s2νt2 + · · · + snνtn. (14.1) Taking the natural logarithm of both sides we ﬁnd t∗ln ν + ln (s1 + s2 + · · · + sn) = ln (s1νt1 + s2νt2 + · · · + snνtn). Solving for t∗and recalling that ln ν = −δ we ﬁnd t∗= −1 δ ln s1νt1 + s2νt2 + · · · + snνtn s1 + s2 + · · · + sn =1 δ " ln n X k=1 sk ! −ln n X k=1 skνtk !# In order to provide an approximation to t∗we recall the binomial series expansion (1 + x)α = 1 + α 1!x + α(α −1) 2! x2 + · · · + α(α −1) · · · (α −n + 1) n! xn + · · · where α ∈R and −1 < x < 1. Using this series expansion we can write νt∗= (1 + i)−t∗= 1 + (−t∗) 1! i + (−t∗)(−t∗−1) 2! i2 + · · · ≈1 −it∗. Similarly, we have νtk = (1 + i)−tk = 1 + (−tk) 1! i + (−tk)(−tk −1) 2! i2 + · · · ≈1 −itk. 120 THE BASICS OF INTEREST THEORY Substituting these approximations in the equation νt∗= s1νt1 + s2νt2 + · · · + snνtn s1 + s2 + · · · + sn we ﬁnd 1 −it∗≈s1(1 −it1) + s2(1 −it2) + · · · + sn(1 −itn) s1 + s2 + · · · + sn which can be written in the form 1 −it∗≈1 −is1t1 + s2t2 + · · · + sntn s1 + s2 + · · · + sn . Solving for t∗we obtain the approximation t∗≈s1t1 + s2t2 + · · · + sntn s1 + s2 + · · · + sn = t. Estimating t∗with t is known as the method of equated time. Note that the formula for t is similar to the center of mass that is covered in a calculus course where the weights are replaced by the various amounts paid. Example 14.4 Payments of $100, $200, and $500 are due at the ends of years 2,3, and 8, respectively. Assuming an eﬀective rate of interest of 5% per year, ﬁnd the point in time at which a payment of $800 would be equivalent using the method of equated time. Solution. Using the method of equating time we ﬁnd t = 100 · 2 + 200 · 3 + 500 · 8 100 + 200 + 500 = 6 years With the approximation t, the estimated present value for the equivalent single payment is PV = (s1 + s2 + · · · + sn)νt. The following result shows that t is an overestimate of t. Also, the result shows that the true present value exceeds the present value given by the method of equated time. Theorem 14.1 With s1, s2, · · · , sn, t1, t2, · · · , tn, t, and t as deﬁned above we have s1νt1 + s2νt2 + · · · + snνtn > (s1 + s2 + · · · + sn)νt. Hence, t > t. 14 SOLVING FOR UNKNOWN TIME 121 Proof. Consider s1 payments each equal to νt1, s2 payments each equal to νt2, · · · , sn payments each equal to νtn. The arithmetic mean of these payments is s1νt1 + s2νt2 + · · · + snνtn s1 + s2 + · · · + sn and the geometric mean of these payments is νs1t1νs2t2 · · · νsntn 1 s1+s2+···+sn = ν s1t1+s2t2+···+sntn s1+s2+···+sn = νt. In Problem 14.16 we show that the arithmetic mean is always greater than the geometric mean. Hence, s1νt1 + s2νt2 + · · · + snνtn s1 + s2 + · · · + sn > νt or s1νt1 + s2νt2 + · · · + snνtn > (s1 + s2 + · · · + sn)νt. The left-hand side is the true present value and the right-hand side is the estimated present value. Note that this last equation implies that νt > νt by (14.1). Thus, νt−t > 1 and since ν < 1 we must have t > t Remark 14.1 As seen above, the solution by the method of equated time is relatively easy to compute, and is generally fairly accurate. Also, an interesting thing to notice is that the interest rate plays no role in the computations when the method of equated time is used. Example 14.5 A loan is negotiated with the lender agreeing to accept $1,000 after 10 years, $2,000 after 20 years, and $3,000 after 30 years in full repayments. The borrower wishes to liquidate the loan with a single $6,000 payment. Let T1 represent the time of the $6,000 payment calculated by an equation of value. Let T2 represent the time determined by the method of equated time. If i = 0.01, ﬁnd T2 −T1 to the nearest penny. Solution. The equation of value at time t = 0 is 1, 000ν10 + 2, 000ν20 + 3, 000ν30 = 6, 000νT1. Solving we ﬁnd T1 = ln (ν10 + 2ν20 + 3ν30) −ln 6 −ln 1.01 = 23.05. 122 THE BASICS OF INTEREST THEORY Now, using the method of equated time we ﬁnd T2 = 1, 000 × 10 + 2, 000 × 20 + 3, 000 × 30 6, 000 = 23.33. Thus, T2 −T1 = 23.33 −23.05 = 0.28 14 SOLVING FOR UNKNOWN TIME 123 Practice Problems Problem 14.1 The present value of a payment of $5,000 to be made in t years is equal to the present value of a payment of $7,100 to be made in 2t years. If i = 7.5% ﬁnd t. Problem 14.2 The present value of two payments of 100 each to be made at the end of n years and 2n years is 100. If i = 0.08, ﬁnd n. Problem 14.3 You are asked to develop a rule of n to approximate how long it takes money to triple. Find n. Problem 14.4 ‡ Joe deposits 10 today and another 30 in ﬁve years into a fund paying simple interest of 11% per year. Tina will make the same two deposits, but the 10 will be deposited n years from today and the 30 will be deposited 2n years from today. Tina’s deposits earn an annual eﬀective rate of 9.15% . At the end of 10 years, the accumulated amount of Tina’s deposits equals the accumulated amount of Joe’s deposits. Calculate n. Problem 14.5 Find how long $1,000 should be left to accumulate at 6% eﬀective in order that it will amount to twice the accumulated value of another 1,000 deposited at the same time at 4% eﬀective. Problem 14.6 If an investment will be doubled in 8 years at a force of interest δ, in how many years will an investment be tripled at a nominal rate of interest numerically equal to δ and convertible once every three years? Problem 14.7 Amounts $500, $800, and $1,000 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) Use the method of equated time to ﬁnd the number of years t from today when one single payment of $500 + $800 + $1000 = $2,300 would be equivalent to the individual payments made separately. (b) Find the exact number of years t from today when one single payment of $500 + $800 + $1000 = $2,300 would be equivalent to the individual payments made separately. 124 THE BASICS OF INTEREST THEORY Problem 14.8 A payment of n is made at the end of n years, 2n at the end of 2n years, · · · , n2 at the end of n2 years. Find the value of t by the method of equated time. Problem 14.9 Fund A accumulates at a rate of 12% convertible monthly. Fund B accumulates with a force of interest δt = t 6. At time t = 0 equal deposits are made in each fund. Find the next time that the two funds are equal. Problem 14.10 A loan requires the borrower to repay $1,000 after 1 year, 2000 after 2 years, $3,000 after 3 years, and $4,000 after 4 years. At what time could the borrower make a single payment of $10,000 according to the method of equated time? Problem 14.11 A series of payments of 100 are made with the ﬁrst payment made immediately. The second payment is made at the end of year 2. The third payment is made at the end of year 4, etc with the last payment being made at the end of year 12. Calculate t using the method of equated time. Problem 14.12 A series of payments are made at the end of each year for 50 years. The amount of the payment at the end of year n is n. Calculate t using the method of equated time. Problem 14.13 A fund earns interest at a nominal rate of interest of 12% compounded monthly. If $1,000 is invested in the fund, it will grow to be $5,320.97 after n years. Calculate n. Problem 14.14 A fund earns interest at a force of interest of δt = 0.05t. Lauren invests $2,000 at time t = 0. After n years, Lauren has $4,919.20 . Calculate n. Problem 14.15 A fund earns interest at a force of interest of δt = 0.01t2. Calculate the amount of time until the fund triples. Problem 14.16 Consider the positive numbers a1, a2, · · · , an where n is a positive integer. The arithmetic mean is deﬁned by a1 + a2 + · · · + an n 14 SOLVING FOR UNKNOWN TIME 125 whereas the geometric mean is deﬁned by [a1 · a2 · · · an] 1 n. Show that a1 + a2 + · · · + an n ≥[a1 · a2 · · · an] 1 n. Hint: Use the fact that ex ≥1 + x for all x. This implies e ai µ −1 ≥ai µ , 1 ≤i ≤n where µ is the arithmetic mean. Problem 14.17 A fund starts with a zero balance at time zero. The fund accumulates with a varying force of interest δt = 2t t2+1, for t ≥0. A deposit of 100,000 is made at time 2. Calculate the number of years from the time of deposit for the fund to double. Problem 14.18 At time t = 0 Billy puts 625 into an account paying 6% simple interest. At the end of year 2, George puts 400 into an account paying interest at a force of interest δt = 1 6+t, for t ≥2. If both accounts continue to earn interest indeﬁnitely at the levels given above, the amounts in both accounts will be equal at the end of year n. Calculate n. Problem 14.19 Payment of 300, 500, and 700 are made at the end of years ﬁve, six, and eight, respectively. Interest is accumulated at an annual eﬀective interest rate of 4%. You are to ﬁnd the point in time at which a single payment of 1500 is equivalent to the above series of payments. You are given: (i) X is the point in time calculated using the method of equated time. (ii) Y is the exact point in time. Calculate X + Y. Problem 14.20 The present value of a payment of 1004 at the end of T months is equal to the present value of 314 after 1 month, 271 after 18 months, and 419 after 24 months. The eﬀective annual interest rate is 5%. Calculate T to the nearest integer. Problem 14.21 Using the method of equated time, a payment of 400 at time t = 2 plus a payment of X at time t = 5 is equivalent to a payment of 400 + X at time t = 3.125. At an annual eﬀective interest rate of 10%, the above two payments are equivalent to a payment of 400 + X at time t = k using the exact method. Calculate k. 126 THE BASICS OF INTEREST THEORY Problem 14.22 Investor A deposits 100 into an account that earns simple interest at a rate of 10% per annum. Investor B makes two deposits of 50 each into an account that earns compound interest at an annual eﬀective rate of 10%. B′s ﬁrst deposit occurs n years after A′s deposit. B′s second deposit occurs 2n years after A′s deposit. The balances in the two accounts are equal 10 years after Investor A′s deposit. What is the value of n? 14 SOLVING FOR UNKNOWN TIME 127 128 THE BASICS OF INTEREST THEORY 14 SOLVING FOR UNKNOWN TIME 129 130 THE BASICS OF INTEREST THEORY 14 SOLVING FOR UNKNOWN TIME 131 132 THE BASICS OF INTEREST THEORY 14 SOLVING FOR UNKNOWN TIME 133 134 THE BASICS OF INTEREST THEORY 14 SOLVING FOR UNKNOWN TIME 135 136 THE BASICS OF INTEREST THEORY 14 SOLVING FOR UNKNOWN TIME 137 138 THE BASICS OF INTEREST THEORY 14 SOLVING FOR UNKNOWN TIME 139 140 THE BASICS OF INTEREST THEORY 14 SOLVING FOR UNKNOWN TIME 141 142 THE BASICS OF INTEREST THEORY The Basics of Annuity Theory A series of payments made at equal intervals of time is called an annuity. Common examples are house rents, mortgage payments on homes, and installments payments on automobiles. An annuity where payments are guaranteed to occur for a ﬁxed period of time is called an annuity−certain. For example, mortgage payments on a home. The ﬁxed period of time for which payments are made is called term of the annuity. For example, in the case of a home mortgage a term can be either a 15−year loan or a 30−year loan. Annuities that are not certain are called contingent annuities. For example, a pension is paid so long as the person survives. That is, regular payments are made as long as the person is alive. Pension is an example of contingent annuity also called life annuity. Unless otherwise indicated, the annuity−certain is the type of annuity we will assume in this book, and the “certain” will be dropped from the name. The interval between annuity payments is called a payment period, often just called a period. In Sections 15 through 21, we consider annuities for which the payment period and the interest conversion period are equal. Also, payments are of level amount, i.e. have the same ﬁxed monetary value for each period. In Sections 22−29, we will discuss annuities for which payments are made more or less frequently than interest is converted and annuities with varying payments. With level annuities, we will most of the time assume payments of 1 since any other level annuity can be obtained from this by a simple multiplication. 143 144 THE BASICS OF ANNUITY THEORY 15 Present and Accumulated Values of an Annuity-Immediate An annuity under which payments of 1 are made at the end of each period for n periods is called an annuity−immediate or ordinary annuity. The cash stream represented by the annuity can be visualized on a time diagram as shown in Figure 15.1 . Figure 15.1 The ﬁrst arrow shows the beginning of the ﬁrst period, at the end of which the ﬁrst payment is due under the annuity. The second arrow indicates the last payment date−just after the payment has been made. Let i denote the interest rate per period. The present value of the annuity at time 0 will be denoted by an i or simply an. See Figure 15.2. Figure 15.2 Using the equation of value with comparison date at time t = 0, we can write an = ν + ν2 + · · · + νn. That is, the present value of the annutiy is the sum of the present values of each of the n payments. We recognize the expression on the right−hand side as a geometric progression. Thus, multiplying both sides by ν to obtain νan = ν2 + ν3 + · · · + νn + νn+1. Subtracting this from the previous equation we ﬁnd (1 −ν)an = ν(1 −νn). Hence, an = ν · 1 −νn 1 −ν = ν · 1 −νn iν = 1 −(1 + i)−n i . (15.1) 15 PRESENT AND ACCUMULATED VALUES OF AN ANNUITY-IMMEDIATE 145 Example 15.1 Calculate the present value of an annuity−immediate of amount $100 paid annually for 5 years at the rate of interest of 9%. Solution. The answer is 100a5 = 1001−(1.09)−5 0.09 ≈388.97 Formula (15.1) can be rewritten as 1 = νn + ian. This last equation is the equation of value at time t = 0 of an investment of $1 for n periods during which an interest of i is received at the end of each period and is reinvested at the same rate i, and at the end of the n periods the original investment of $1 is returned. Figure 15.3 describes the time diagram of this transaction. Figure 15.3 Next, we will determine the accumulated value of an annuity−immediate right after the nth payment is made. It is denoted by sn. See Figure 15.4. Figure 15.4 Writing the equation of value at the comparison date t = n we ﬁnd sn = 1 + (1 + i) + (1 + i)2 + · · · + (1 + i)n−1. That is, sn is the sum of the accumulated value of each of the n payments. Using the deﬁnition of sn and the reasoning used to establish the formula for an we can write sn =1 + (1 + i) + (1 + i)2 + · · · + (1 + i)n−1 =(1 + i)n −1 (1 + i) −1 = (1 + i)n −1 i . 146 THE BASICS OF ANNUITY THEORY This last equation is equivalent to 1 + isn = (1 + i)n. This last equation is the equation of value at time t = n of an investment of $1 for n periods during which an interest of i is received at the end of each period and is reinvested at the same rate i, and at the end of the n periods the original investment of $1 is returned. Figure 15.5 describes the time diagram of this transaction. Figure 15.5 Example 15.2 Calculate the future value of an annuity−immediate of amount $100 paid annually for 5 years at the rate of interest of 9%. Solution. The answer is 100s5 = 100 × (1.09)5−1 0.09 ≈$598.47 Example 15.3 Show that am+n = am + νman = an + νnam. Interpret this result verbally. Solution. We have am + νman = 1−νm i + νm · 1−νn i = 1−νm+νm−νm+n i = 1−νm+n i = am+n. A verbal interpretation is as follows: The present value of the ﬁrst m payments of an (m + n)−year annuity−immediate of 1 is am. The remaining n payments have value an at time t = m; discounted to the present, these are worth νman at time t = 0 Example 15.4 At an eﬀective annual interest rate i, you are given (1) the present value of an annuity immediate with annual payments of 1 for n years is 40 (2) the present value of an annuity immediate with annual payments of 1 for 3n years is 70. Calculate the accumulated value of an annuity immediate with annual payments of 1 for 2n years. Solution. Using Example 15.3 we can write a3n = a2n + ν2nan = an + νnan + ν2nan = an(1 + νn + ν2n). Hence, we obtain the quadratic equation ν2n + νn + 1 = 7 4. Solving this equation we ﬁnd νn = 1 2. 15 PRESENT AND ACCUMULATED VALUES OF AN ANNUITY-IMMEDIATE 147 Again, using Example 15.3 we can write a3n = an +νna2n which implies that νna2n = 70−40 = 30 and therefore a2n = 60. Finally, s2n = ν−2na2n = 4(60) = 240 We next establish a couple of relationships between an and sn. Theorem 15.1 With an and sn as deﬁned above we have (i) sn = (1 + i)nan. That is, the accumulated value of a principal of an after n periods is just sn. (ii) 1 an = 1 sn + i. Proof. (i) We have sn = (1+i)n−1 i = (1 + i)n · 1−(1+i)−n i = (1 + i)nan. (ii) We have 1 sn + i = i (1 + i)n −1 + i = i + i(1 + i)n −i (1 + i)n −1 = i 1 −νn = 1 an A verbal interpretation of (ii) will be introduced when we discuss the concepts of amortization and sinking funds. Example 15.5 For a given interest rate i, an = 8.3064 and sn = 14.2068. (a) Calculate i. (b) Calculate n. Solution. (a) Using part (ii) of the previous theorem we ﬁnd i = 1 8.3064 − 1 14.2068 = 5%. (b) Using part (i) of the previous theorem we ﬁnd n = 1 ln (1 + i) ln sn an = 11 The type of annuity discussed in this section involves compound interest rate. It is possible to deﬁne annuity−immediate not involving compound interest such as simple interest rate, simple discount rate, and force of interest. For example, we wish to ﬁnd the present value of an n−period annuity immediate in which each payment is invested at simple interest rate i. The present value is equal to the sum of the present value of the individual payments. Thus, we obtain an = 1 1 + i + 1 1 + 2i + · · · + 1 1 + ni. 148 THE BASICS OF ANNUITY THEORY The accumulated value of such an annuity is equal to the accumulated value of the individual payments. That is, sn = 1 + (1 + i) + (1 + 2i) + · · · + [1 + (n −1)i]. Example 15.6 Find an expression for an assuming each payment of 1 is valued at simple discount rate d. Solution. The present value is the sum of the present value of individual payments. That is, an = (1 −d) + (1 −2d) + · · · + (1 −nd) = n −d(1 + 2 + · · · + n) = n −n(n + 1) 2 d 15 PRESENT AND ACCUMULATED VALUES OF AN ANNUITY-IMMEDIATE 149 Practice Problems Problem 15.1 Consider an investment of $5,000 at 6% convertible semiannually. How much can be withdrawn each half−year to use up the fund exactly at the end of 20 years? Problem 15.2 The annual payment on a house is $18,000. If payments are made for 40 years, how much is the house worth assuming annual interest rate of 6%? Problem 15.3 If d = 0.05, calculate a12. Problem 15.4 Calculate the present value of 300 paid at the end of each year for 20 years using an annual eﬀective interest rate of 8%. Problem 15.5 If an = x and a2n = y, express d as a function of x and y. Problem 15.6 (a) Given: a7 = 5.153, a11 = 7.036, a18 = 9.180. Find i. (b) You are given that an = 10.00 and a3n = 24.40. Determine a4n. Problem 15.7 Show that sm+n = sm + (1 + i)msn = sn + (1 + i)nsm. Interpret this result verbally. Problem 15.8 A grandmother has a granddaughter entering university next year. Her granddaughter expects to remain in school for ten years and receive a PhD. This grandmother wishes to provide $1,000 a year to her granddaughter for entertainment expenses. Assuming a 3.5% eﬀective annual interest rate, how much does the grandmother have to deposit today to provide ten annual payments starting one year from now and continuing for ten years? Problem 15.9 In the previous problem, if the PhD student saves the income from her grandmother in an account also paying 3.5% eﬀective annual interest, how much will she have when she receives the ﬁnal $1,000 payment? 150 THE BASICS OF ANNUITY THEORY Problem 15.10 Find the present value of an annuity which pays $200 at the end of each quarter−year for 12 years if the rate of interest is 6% convertible quarterly. Problem 15.11 Compare the total amount of interest that would be paid on a $3,000 loan over a 6−year period with an eﬀective rate of interest of 7.5% per annum, under each of the following repayment plans: (a) The entire loan plus accumulated interest is paid in one lump sum at the end of 6 years. (b) Interest is paid each year as accrued, and the principal is repaid at the end of 6 years. (c) The loan is repaid with level payments at the end of each year over the 6-year period. Problem 15.12 A loan of $20,000 to purchase a car at annual rate of interest of 6% will be paid back through monthly installments over 5 years, with 1st installment to be made 1 month after the release of the loan. What is the monthly installment? Problem 15.13 Over the next 20 years, you deposit money into a retirement account at the end of each year according to the following schedule: Time Amount invested each year 1 - 5 $2000 6 - 10 $3000 11 - 20 $5000 The eﬀective annual rate of interest is 9%. Find the accumulated value of your account at time 20. Problem 15.14 A family wishes to accumulate 50,000 in a college education fund by the end of 20 years. If they deposit 1,000 into the fund at the end of each of the ﬁrst 10 years, and 1, 000 + X at the end of each of the second 10 years, ﬁnd X to the nearest unit if the fund earns 7% eﬀective. Problem 15.15 An annuity provides a payment of n at the end of each year for n years. The annual eﬀective interest rate is 1 n. What is the present value of the annuity? Problem 15.16 The cash price of a new automobile is 10,000. The purchaser is willing to ﬁnance the car at 18% convertible monthly and to make payments of 250 at the end of each month for 4 years. Find the down payment which will be necessary. 15 PRESENT AND ACCUMULATED VALUES OF AN ANNUITY-IMMEDIATE 151 Problem 15.17 You have $10,000 down payment on a $20,000 car. The dealer oﬀers you the following two options: (a) paying the balance with end−of−month payments over the next three years at i(12) = 0.12; (b) a reduction of $500 in the price of the car, the same down payment of $10,000, and bank ﬁnancing of the balance after down payment, over 3 years with end−of−month payments at i(12) = 0.18. Which option is better? Problem 15.18 Nancy has 10,000 in a bank account earning 6% compounded monthly. Calculate the amount that she can withdraw at the end of each month from the account if she wants to have zero in the account after 12 months. Problem 15.19 Megan purchased a new car for 18,000. She ﬁnances the entire purchase over 60 months at a nominal rate of 12% compounded monthly. Calculate Megan’s monthly payment. Problem 15.20 ‡ Seth, Janice, and Lori each borrow 5,000 for ﬁve years at a nominal interest rate of 12%, compounded semiannually. Seth has interest accumulated over the ﬁve years and pays all the interest and principal in a lump sum at the end of ﬁve years. Janice pays interest at the end of every six−month period as it accrues and the principal at the end of ﬁve years. Lori repays her loan with 10 level payments at the end of every six−month period. Calculate the total amount of interest paid on all three loans. Problem 15.21 If d(12) = 12%, calculate the accumulated value of 100 paid at the end of each month for 12 months. Problem 15.22 The accumulated value of an n year annuity−immediate is four times the present value of the same annuity. Calculate the accumulated value of 100 in 2n years. Problem 15.23 You are given the following information: (i) The present value of a 6n−year annuity−immediate of 1 at the end of every year is 9.7578. (ii) The present value of a 6n−year annuity−immediate of 1 at the end of every second year is 4.760. (iii) The present value of a 6n−year annuity−immediate of 1 at the end of every third year is K. Determine K assuming an annual eﬀective interest rate of i. 152 THE BASICS OF ANNUITY THEORY Problem 15.24 ‡ Which of the following does not represent a deﬁnition of an? (a) νn h (1+i)n−1 i i (b) 1−νn i (c) ν + ν2 + · · · + νn (d) ν 1−νn 1−ν (e) sn (1+i)n−1 Problem 15.25 ‡ To accumulate 8000 at the end of 3n years, deposits of 98 are made at the end of each of the ﬁrst n years and 196 at the end of each of the next 2n years. The annual eﬀective rate of interest is i. You are given (1 + i)n = 2.0. Determine i. Problem 15.26 ‡ For 10,000, Kelly purchases an annuity−immediate that pays 400 quarterly for the next 10 years. Calculate the annual nominal interest rate convertible monthly earned by Kelly’s investment. Hint: Use linear interpolation. Problem 15.27 ‡ Susan and Jeﬀeach make deposits of 100 at the end of each year for 40 years. Starting at the end of the 41st year, Susan makes annual withdrawals of X for 15 years and Jeﬀmakes annual withdrawals of Y for 15 years. Both funds have a balance of 0 after the last withdrawal. Susan’s fund earns an annual eﬀective interest rate of 8%. Jeﬀ’s fund earns an annual eﬀective interest rate of 10%. Calculate Y −X. Problem 15.28 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 + 200 each. What is K if i(2) = 0.06? Problem 15.29 Smith makes deposits of 1,000 on the last day of each month in an account earning interest at rate i(12) = 0.12. The ﬁrst deposit is January 31, 2005 and the ﬁnal deposit is December 31, 2029. The accumulated account is used to make monthly payments of Y starting January 31, 2030 with the ﬁnal one on December 31, 2054. Find Y. Problem 15.30 A loan of $1,000 is to be repaid with annual payments at the end of each year for the next 20 years. For the ﬁrst ﬁve years the payments are k per year; the second 5 years, 2k per year; the third 5 years, 3k per year; and the fourth 5 years, 4k per year. Find an expression for k. 15 PRESENT AND ACCUMULATED VALUES OF AN ANNUITY-IMMEDIATE 153 Problem 15.31 ‡ Happy and ﬁnancially astute parents decide at the birth of their daughter that they will need to provide 50,000 at each of their daughter’s 18th, 19th, 20th and 21st birthdays to fund her college education. They plan to contribute X at each of their daughter’s 1st through 17th birthdays to fund the four 50,000 withdrawals. If they anticipate earning a constant 5% annual eﬀective rate on their contributions, which of the following equations of value can be used to determine X, assuming compound interest? (A) X(ν + ν2 + · · · + ν17) = 50, 000(ν + ν2 + ν3 + ν4) (B) X[(1.05)16 + (1.05)15 + · · · + (1.05)] = 50, 000(1 + ν + ν2 + ν3) (C) X[(1.05)17 + (1.05)16 + · · · + (1.05) + 1] = 50, 000(1 + ν + ν2 + ν3) (D) X[(1.05)17 + (1.05)16 + · · · + (1.05)] = 50, 000(1 + ν + ν2 + ν3) (E) X(1 + ν + ν2 + · · · + ν17) = 50, 000(ν18 + ν19 + ν20 + ν21 + ν22) Problem 15.32 For time t > 0, the discount function is deﬁned by [a(t)]−1 = 1 1 + 0.01t. Consider a ﬁve-year annuity with payments of 1 at times t = 1, 2, 3, 4, 5. Consider the following: A calculates a5 as the sum of the present value of the individual payments. However, B accumulates the payments according to the accumulation function a(t) = 1 + 0.01t and then multiplies the result by 1 a(5). By how much do the answers of A and B diﬀer? Remark 15.1 It can be shown that the above two processes do not produce the same answers in general for any pattern of interest other than compound interest. That’s why, it is always recommended to avoid dealing with annuities not involving compound interest, if possible. Problem 15.33 Simplify the sum P40 n=15 sn. Problem 15.34 A 20 year annuity pays 100 every other year beginning at the end of the second year, with additional payments of 300 each at the ends of years 3, 9, and 15. The eﬀective annual interest rate is 4%. Calculate the present value of the annuity. 154 THE BASICS OF ANNUITY THEORY Problem 15.35 An annuity pays 1 at the end of each 4−year period for 40 years. Given a8 i = k, ﬁnd the present value of the annuity. Problem 15.36 Smith is negotiating a price for a new car. He is willing to pay $250 at the end of each month for 60 months using the 4.9% compounded monthly interest rate that he qualiﬁes for. Smith estimates that tax, title, and license for the new car will increase the negotiated price by 10%, and he estimates that he will receive $500 trade−in value for his current car. Calculate the highest negotiated price that Smith is willing to pay for the car. Problem 15.37 An account is credited interest using 6% simple interest rate from the date of each deposit into the account. Annual payments of 100 are deposited into this account. Calculate the accumulated value of the account immediately after the 20th deposit. Problem 15.38 A homeowner signs a 30 year mortgage that requires payments of $971.27 at the end of each month. The interest rate on the mortgage is 6% compounded monthly. If the purchase price of the house is $180,000 then what percentage down payment was required? Problem 15.39 A 25−year−old worker begins saving for retirement, making level annual deposits at the end of each year. The savings are invested at an annual eﬀective interest rate of 8%, and are of an amount that is projected to equal 1,000,000 when the worker is 65 (after the 40th deposit is made on that date). After making ﬁve annual deposits, the worker becomes unemployed for a period of time and, as a result, skips the next three annual deposits. Assuming that the account earns an 8% annual eﬀective rate in all 40 years, what amount will the worker need to deposit in each of the remaining 32 years in order to achieve the original goal of a 1,000,000 balance at age 65? Problem 15.40 Paul lends 8000 to Peter. Peter agrees to pay it back in 10 annual installments at 7% with the ﬁrst payment due in one year. After making 4 payments, Peter renegotiates to pay oﬀthe debt with 4 additional annual payments. The new payments are calculated so that Paul will get a 6.5% annual yield over the entire 8−year period. Determine how much money Peter saved by renegotiating. Problem 15.41 Mario deposits 100 into a fund at the end of each 2 year period for 20 years. The fund pays interest at an annual eﬀective rate of i. The total amount of interest earned by the fund during the 19th and 20th years is 250. Calculate the accumulated amount in Marios account at the end of year 20. 15 PRESENT AND ACCUMULATED VALUES OF AN ANNUITY-IMMEDIATE 155 Problem 15.42 Show that 1 1−ν10 = 1 s10 s10 + 1 i . Problem 15.43 If a4 = 3.2397 and s4 = 4.5731 what is the value of a8? Problem 15.44 Smith borrows $5,000 on January 1, 2007. He repays the loan with 20 annual payments, starting January 1, 2008. The payments in even-number year are 2X each; the payments in odd-number years are X each. If d = 0.08, ﬁnd the total amount of all 20 payments. Problem 15.45 (a) Show that am−n = am −νmsn where 0 < n < m. (b) Show that sm−n = sm −(1 + i)man where 0 < n < m. Problem 15.46 Show that sn ≥n ≥an. Thus far, we have only considered annuities where the payments are made at the end of the period, but it is possible that the circumstances may be such that the annuity is to run for a given number of periods and a portion of a period. The purpose of the remaining problems is to deﬁne terms such as an+k and sn+k where n is a positive integer and 0 < k < 1. Problem 15.47 Let k be a positive real number. Find an expression for the error involved in approximating (1+i)k−1 i by the number k. Hint: Use the Taylor series expansion of (1 + i)k around i = 0. Problem 15.48 (a) Find the interest accrued of a principal of 1 at t = n to time t = n + k, assuming compound interest. (b) Show that i(ν + ν2 + ν3 + · · · + νn) + [(1 + i)k −1]νn+k + νn+k = 1. (c) Let an+ 1 m denote the present value of an annuity consisting of n payments of 1 at the end of each period and a ﬁnal payment of (1+i) 1 m −1 i at the end of the 1 mth fraction of the (n + 1) period. Deﬁne an+ 1 m = 1−νn+ 1 m i . Show that an+ 1 m = an + νn+ 1 m " (1 + i) 1 m −1 i # . 156 THE BASICS OF ANNUITY THEORY Thus, an+k is the sum of the present value of an n−period annuity−immediate of 1 per period, plus a ﬁnal payment at time n + k of (1+i)k−1 i . Problem 15.49 Show that an+k ≈an + kνn+ 1 m. Hint: Use Problems 15.47 and 15.48. Problem 15.50 Compute a5.25 if i = 5% using the following deﬁnitions: (a) The formula established in Problem 15.48. (b) A payment of 0.25 at time 5.25. (c) A payment of 0.25 at time 6. Problem 15.51 (a) Deﬁne sn+k = (1+i)n+k−1 k . Show that sn+k = (1+i)ksn + (1+i)k−1 i . Thus, sn+k can be interprested as the accumulated value of an n−period annuity−immediate at time t = n + k with an additional payment of (1+i)k−1 i at time t = n + k. (b) Show that sn+k ≈(1 + i)ksn + k. 16 ANNUITY IN ADVANCE: ANNUITY DUE 157 16 Annuity in Advance: Annuity Due An annuity−due is an annuity for which the payments are made at the beginning of the payment periods. The cash stream represented by the annuity can be visualized on a time diagram as shown in Figure 16.1 . Figure 16.1 The ﬁrst arrow shows the beginning of the ﬁrst period at which the ﬁrst payment is made under the annuity. The second arrow appears n periods after arrow 1, one period after the last payment is made. Let i denote the interest rate per period. The present value of the annuity at time 0 will be denoted by ¨ an. To determine ¨ an, we can proceed as in the case of determining an. In this case, we use the time diagram shown in Figure 16.2 Figure 16.2 Considering the equation of value at time t = 0 we can write ¨ an = 1 + ν + ν2 + · · · + νn−1. That ¨ an is is equal to the sum of the present values of each of the n payments. We recognize the expression on the right−hand side as a geometric progression. Thus, multiplying both sides by ν to obtain ν¨ an = ν + ν2 + ν3 + · · · + νn−1 + νn. Subtracting this from the previous equation we ﬁnd (1 −ν)¨ an = (1 −νn). 158 THE BASICS OF ANNUITY THEORY Hence, ¨ an = 1 −νn 1 −ν . Since 1 −ν = d, we have ¨ an = 1 −νn d = 1 −(1 + i)−n d = 1 −(1 −d)n d . Example 16.1 Find ¨ a8 if the eﬀective rate of discount is 10%. Solution. Since d = 0.10, we have ν = 1 −d = 0.9. Hence, ¨ a8 = 1−(0.9)8 0.1 = 5.6953279 Example 16.2 What amount must you invest today at 6% interest rate compounded annually so that you can withdraw $5,000 at the beginning of each year for the next 5 years? Solution. The answer is 5000¨ a5 = 5000 · 1−(1.06)−5 0.06(1.06)−1 = $22, 325.53 Now, let ¨ sn denote the accumulated value of an annuity−due at the end of n periods. To de-termine ¨ sn we proceed in a way analogous to determining sn. We consider the time diagram shown in Figure 16.3. Figure 16.3 Writing the equation of value at time t = n we ﬁnd ¨ sn =(1 + i) + (1 + i)2 + · · · + (1 + i)n−1 + (1 + i)n =(1 + i)(1 + i)n −1 (1 + i) −1 =(1 + i)n −1 iν = (1 + i)n −1 d 16 ANNUITY IN ADVANCE: ANNUITY DUE 159 Example 16.3 What amount will accumulate if we deposit $5,000 at the beginning of each year for the next 5 years? Assume an interest of 6% compounded annually. Solution. The answer is 5000¨ s5 = 5000 · (1.06)5−1 0.06(1.06)−1 = $29, 876.59 The following theorem provides a couple of relationships between ¨ an and ¨ sn Theorem 16.1 (a) ¨ sn = (1 + i)n¨ an (b) 1 ¨ an = 1 ¨ sn + d Proof. (a) We have ¨ sn = (1 + i)n −1 d = (1 + i)n · 1 −(1 + i)−n d = (1 + i)n¨ an (b) 1 ¨ sn + d = d (1 + i)n −1 + d(1 + i)n −1 (1 + i)n −1 =d + d[(1 + i)n −1] (1 + i)n −1 = d(1 + i)n (1 + i)n −1 = d 1 −(1 + i)−n = 1 ¨ an The result in (a) says that if the present value at time 0, ¨ an, is accumulated forward to time n, then you will have its future value, ¨ sn. Example 16.4 For a given interest rate i, ¨ an = 8.3064 and ¨ sn = 14.2068. (a) Calculate d. (b) Calculate n. Solution. (a) Using part (b) of the previous theorem we ﬁnd d = 1 8.3064 − 1 14.2068 = 5%. (b) Using part (a) of the previous theorem we ﬁnd n = 1 ln (1 + i) ln ¨ sn ¨ an = 10.463 It is possible to relate the annuity-immediate and the annuity-due as shown in the following theorem. 160 THE BASICS OF ANNUITY THEORY Theorem 16.2 (a) ¨ an = (1 + i)an . (b) ¨ sn = (1 + i)sn . Proof. The two results can be obtained directly by using the time diagram shown in Figure 16.4 Figure 16.4 An algebraic justiﬁcation is given next. (a) Since d = i i+1, we have ¨ an = 1−(1+i)−n d = (1 + i) · 1−(1+i)−n i = (1 + i)an . (b) Since d = i i+1, we have: ¨ sn = (1+i)n−1 d = (1 + i) · (1+i)n−1 i = (1 + i)sn An annuity−due starts one period earlier than an annuity−immediate and as a result, earns one more period of interest, hence more proﬁt is made. Example 16.5 Over the next 30 years, you deposit money into a retirement account at the beginning of each year. The ﬁrst 10 payments are 200 each. The remaining 20 payments are 300 each. The eﬀective annual rate of interest is 9%. Find the present value of these payments. Solution. The time diagram of this situation is shown in Figure 16.5 Figure 16.5 The answer is 300¨ a30 −100¨ a10 = 1.09(300a30 −100a10 ) = $2659.96 More relationships between annuity-immediate and annuity-due are given below. 16 ANNUITY IN ADVANCE: ANNUITY DUE 161 Theorem 16.3 (a) ¨ an = 1 + an−1 . (b) sn = ¨ sn−1 + 1. Proof. (a) We have ¨ an = 1−(1+i)−n d = i+1 i [1 −(1 + i)−n] = 1−(1+i)−n+1+i i = 1 + an−1 . The result has the following verbal interpretation: An additional payment of 1 at time 0 results in an−1 becoming n payments that now commence at the beginning of each year whose present value is ¨ an. (b) We have ¨ sn−1 = (1+i)n−1−1 d = i+1 i [(1 + i)n−1 −1] = (1+i)n−1−i i = sn −1 An interpretation of (b) is as follows: A withdrawal of 1 at time n results in sn becoming n −1 payments that commence at the beginning of each year (starting at t = 1) whose accumulated value at time t = n is ¨ sn−1 Remark 16.1 Most compound interest tables do not include values of annuities−due. Thus, the formulas of Theorems 16.2 - 16.3 must be used in ﬁnding numerical values for annuities−due. Example 16.6 An investor wishes to accumulate $3000 at the end of 15 years in a fund which earns 8% eﬀective. To accomplish this, the investor plans to make deposits at the beginning of each year, with the ﬁnal payment to be made one year prior to the end of the investment period. How large should each deposit be? Solution. Let R be the payment at the beginning of each year. The accumulated value of the investment at the end of the investment period is to be $3000, so the equation of value at time t = 15 is 3000 = R¨ s14 . Solving for R we ﬁnd R = 3000 ¨ s14 = 3000 s15 −1 = $114.71 Example 16.7 Show that ¨ an = an + 1 −νn. Interpret the result verbally. Solution. We have ¨ an = (1 + i)an = an + ian = an + i · 1−νn i = an + 1 −νn. An additional payment of 1 at time 0 results in an becoming n + 1 payments that now commence at the beginning of the year and whose present value is ¨ an + νn 162 THE BASICS OF ANNUITY THEORY Example 16.8 Show that ¨ sn = sn −1 + (1 + i)n. Interpret the result verbally. Solution. We have ¨ sn = (1 + i)sn = sn + i · (1+i)n−1 i = sn −1 + (1 + i)n. An additional payment of 1 at time n results in ¨ sn becoming n + 1 payments that now commence at the beginning of the year and whose accumulated value at time t = n is sn + (1 + i)n Remark 16.2 The names annuity−immediate and annuity-due are used traditionally, although they do not seem to be logical. The ﬁrst payment of an annuity−immediate is not made immediately at the beginning of the ﬁrst payment period, it is due at the end of it. 16 ANNUITY IN ADVANCE: ANNUITY DUE 163 Practice Problems Problem 16.1 An 8-year annuity due has a present value of $1,000. If the eﬀective annual interest rate is 5%, then what is the value of the periodic payment? Problem 16.2 An 8-year annuity due has a future value of $1,000. Find the periodic payment of this annuity if the eﬀective annual interest rate is 5%. Problem 16.3 A 5-year annuity due has periodic cash ﬂows of $100 each year. Find the accumulated value of this annuity if the eﬀective annual interest rate is 8%. Problem 16.4 A 5-year annuity due has periodic cash ﬂows of $100 each year. Find the present value of this annuity if the eﬀective annual interest rate is 8%. Problem 16.5 Calculate the accumulated value immediately after the last payment of a 20 year annuity due of annual payments of 500 per year. The annual eﬀective interest rate is 7%. Problem 16.6 Megan wants to buy a car in 4 years for 18,000. She deposits X at the beginning of each month for four years into an account earning 6% compounded monthly. Calculate X. Problem 16.7 Kathy wants to accumulate a sum of money at the end of 10 years to buy a house. In order to accomplish this goal, she can deposit 80 per month at the beginning of the month for the next ten years or 81 per month at the end of the month for the next ten years. Calculate the annual eﬀective rate of interest earned by Kathy. Problem 16.8 An annuity pays $500 every six months for 5 years. The annual rate of interest is 8% convertible semiannually. Find each of the following: (a) The PV of the annuity six months (one period) before the ﬁrst payment, (b) the PV of the annuity on the day of the ﬁrst payment, (c) the FV of the annuity on the day of the last payment, (d) and the FV of the annuity six months after the last payment. 164 THE BASICS OF ANNUITY THEORY Problem 16.9 You will retire in 30 years. At the beginning of each month until you retire, you will invest X earning interest at 9% convertible monthly. Starting at year 30, you will withdraw $4,000 at the beginning of each month for the next 15 years. Also, starting at year 30, your fund will only earn interest at 6% convertible monthly. Find X such that your account will be empty after the last withdrawal. Problem 16.10 ‡ Chuck needs to purchase an item in 10 years. The item costs 200 today, but its price increases by 4% per year. To ﬁnance the purchase, Chuck deposits 20 into an account at the beginning of each year for 6 years. He deposits an additional X at the beginning of years 4, 5, and 6 to meet his goal. The annual eﬀective interest rate is 10%. Calculate X. Problem 16.11 Which of the following are true: (i) ¨ s9 + 1 = s10 (ii) a10 −a3 = v2a7 (iii) a5 (1 + i) −1 = a4 ? Problem 16.12 Find the present value of payments of $200 every six months starting immediately and continuing through four years from the present, and $100 every six months thereafter through ten years from the present, if i(2) = 0.06. Problem 16.13 A worker aged 40 wishes to accumulate a fund for retirement by depositing $1,000 at the beginning of each year for 25 years. Starting at age 65 the worker plans to make 15 annual withdrawals at the beginning of each year. Assuming all payments are certain to be made, ﬁnd the amount of each withdrawal starting at age 65 to the nearest dollar, if the eﬀective rate of interest is 8% during the ﬁrst 25 years but only 7% thereafter. Problem 16.14 Which one is greater sn −an or ¨ sn −¨ an ? Problem 16.15 A (2n −1)−payment annuity-immediate has payments 1, 2, · · · , n −1, n, n −1, · · · , 2, 1. Show that the present value of the annuity one payment period before the ﬁrst payment is an · ¨ an . 16 ANNUITY IN ADVANCE: ANNUITY DUE 165 Problem 16.16 Jeﬀmakes payments at the end of each year into an account for 10 years. The present value of Jeﬀ’s payments is 5,000. Ryan makes a payment equal to Jeﬀ’s payment at the beginning of each year for 11 years into the same account. The present value of Ryan’s payments is 5,900. Calculate the amount of Jeﬀ’s payment. Problem 16.17 A person deposits 100 at the beginning of each year for 20 years. Simple interest at an annual rate of i is credited to each deposit from the date of deposit to the end of the twenty year period. The total amount thus accumulated is 2,840. If instead, compound interest had been credited at an eﬀective annual rate of i, what would the accumulated value of these deposits have been at the end of twenty years? Problem 16.18 You plan to accumulate 100,000 at the end of 42 years by making the following deposits: X at the beginning of years 1-14 No deposits at the beginning of years 15-32; and Y at the beginning of years 33-42. The annual eﬀective interest rate is 7%. Suppose X −Y = 100. Calculate Y. Problem 16.19 ‡ Jim began saving money for his retirement by making monthly deposits of 200 into a fund earning 6% interest compounded monthly. The ﬁrst deposit occurred on January 1, 1985. Jim became unemployed and missed making deposits 60 through 72. He then continued making monthly deposits of 200. How much did Jim accumulate in his fund on December 31, 1999 ? Problem 16.20 ‡ An investor accumulates a fund by making payments at the beginning of each month for 6 years. Her monthly payment is 50 for the ﬁrst 2 years, 100 for the next 2 years, and 150 for the last 2 years. At the end of the 7th year the fund is worth 10,000. The annual eﬀective interest rate is i, and the monthly eﬀective interest rate is j. Which of the following formulas represents the equation of value for this fund accumulation? (a) ¨ s24 i(1 + i)[(1 + i)4 + 2(1 + i)2 + 3] = 200 (b) ¨ s24 i(1 + j)[(1 + j)4 + 2(1 + j)2 + 3] = 200 (c) ¨ s24 j(1 + i)[(1 + i)4 + 2(1 + i)2 + 3] = 200 (d) s24 j(1 + i)[(1 + i)4 + 2(1 + i)2 + 3] = 200 (e) s24 i(1 + j)[(1 + j)4 + 2(1 + j)2 + 3] = 200 166 THE BASICS OF ANNUITY THEORY Problem 16.21 ‡ Carol and John shared equally in an inheritance. Using his inheritance, John immediately bought a 10-year annuity-due with an annual payment of 2,500 each. Carol put her inheritance in an investment fund earning an annual eﬀective interest rate of 9%. Two years later, Carol bought a 15-year annuity-immediate with annual payment of Z. The present value of both annuities was determined using an annual eﬀective interest rate of 8%. Calculate Z. Problem 16.22 ‡ Jerry will make deposits of 450 at the end of each quarter for 10 years. At the end of 15 years, Jerry will use the fund to make annual payments of Y at the beginning of each year for 4 years, after which the fund is exhausted. The annual eﬀective rate of interest is 7% . Determine Y. Problem 16.23 If a(t) = 1 log2 (t+2)−log2 (t+1), ﬁnd an expression for ¨ an by directly taking the present value of the payments. Problem 16.24 The accumulated value of an n year annuity-due is four times the present value of the same annuity. Calculate the accumulated value of 100 in 2n years. Problem 16.25 Show that sn · ¨ an > n2 for i > 0 and n > 1. Problem 16.26 ‡ A man turns 40 today and wishes to provide supplemental retirement income of 3000 at the begin-ning of each month starting on his 65th birthday. Starting today, he makes monthly contributions of X to a fund for 25 years. The fund earns a nominal rate of 8% compounded monthly. Each 1000 will provide for 9.65 of income at the beginning of each month starting on his 65th birthday until the end of his life. Calculate X. Problem 16.27 An annuity pays 3 at the beginning of each 3 year period for 30 years. Find the accumulated value of the annuity just after the ﬁnal payment, using i(2) = 0.06. Problem 16.28 A worker aged 30 wishes to accumulate a fund for retirement by depositing $3,000 at the beginning 16 ANNUITY IN ADVANCE: ANNUITY DUE 167 of each year for 35 years. Starting at age 65 the worker plans to make 20 equal annual withdrawals at the beginning of each year. Assuming that all payments are certain to be made, ﬁnd the amount of each withdrawal starting at age 65, if the annual eﬀective rate of interest is 9% during the ﬁrst 35 years but only 6% thereafter. Problem 16.29 Irene deposits 100 at the beginning of each year for 20 years into an account in which each deposit earns simple interest at a rate of 10% from the time of the deposit. Other than these deposits Irene makes no other deposits or withdrawals from the account until exactly 25 years after the ﬁrst deposit was made, at which time she withdraws the full amount in the account. Determine the amount of Irene’s withdrawal. Problem 16.30 Show that ¨ s2n ¨ sn + ¨ sn ¨ s2n −¨ s3n ¨ s2n = 1. Problem 16.31 If ¨ ap = x and sq = y, show that ap+q = νx+y 1+iy . 168 THE BASICS OF ANNUITY THEORY 17 Annuity Values on Any Date: Deferred Annuity Evaluating annuities thus far has always been done at the beginning of the term( either on the date of, or one period before the ﬁrst payment) or at the end of the term (either on the date of, or one period after the last payment). In this section, we shall now consider evaluating the (1) present value of an annuity more than one period before the ﬁrst payment date, (2) accumulated value of an annuity more than one period after the last payment date, (3) current value of an annuity between the ﬁrst and last payment dates. We will assume that the evaluation date is always an integral number of periods from each payment date. (1) Present values more than one period before the ﬁrst payment date Consider the question of ﬁnding the present value of an annuity−immediate with periodic interest rate i and m + 1 periods before the ﬁrst payment date. Figure 17.1 shows the time diagram for this case where “?” indicates the present value to be found. Figure 17.1 The present value of an n−period annuity−immediate m + 1 periods before the ﬁrst payment date (called a deferred annuity since payments do not begin until some later period) is the present value at time m discounted for m time periods, that is, vman . It is possible to express this answer strictly in terms of annuity values. Indeed, am+n −am = 1 −νm+n i −1 −νm i = νm −νm+n i = νm1 −νn i = νman . Such an expression is convenient for calculation, if interest tables are being used. Example 17.1 Exactly 3 years from now is the ﬁrst of four $200 yearly payments for an annuity-immediate, with an eﬀective 8% rate of interest. Find the present value of the annuity. Solution. The answer is 200ν2a4 = 200(a6 −a2 ) = 200(4.6229 −1.7833) = $567.92 17 ANNUITY VALUES ON ANY DATE: DEFERRED ANNUITY 169 The deferred−annuity introduced above uses annuity−immediate. It is possible to work with a deferred annuity−due. In this case, one can easily see that the present value is given by νm¨ an = ¨ am+n −¨ am . Example 17.2 Calculate the present value of an annuity−due paying annual payments of 1200 for 12 years with the ﬁrst payment two years from now. The annual eﬀective interest rate is 6%. Solution. The answer is 1200(1.06)−2¨ a12 = 1200(¨ a14 −¨ a2) = 1200(9.8527 −1.9434) ≈9, 491.16 (2) Accumulated values more than 1 period after the last payment date Consider the question of ﬁnding the accumulated value of an annuity−immediate with periodic interest rate i and m periods after the last payment date. Figure 17.2 shows the time diagram for this case where “?” indicates the sought accumulated value. Figure 17.2 The accumulated value of an n−period annuity−immediate m periods after the last payment date is the accumulated value at time n accumulated for m time periods, that is, (1+i)msn . Notice that sm+n −sm =(1 + i)m+n −1 i −(1 + i)m −1 i =(1 + i)m+n −(1 + i)m i = (1 + i)m(1 + i)n −1 i = (1 + i)msn Example 17.3 For four years, an annuity pays $200 at the end of each year with an eﬀective 8% rate of interest. Find the accumulated value of the annuity 3 years after the last payment. Solution. The answer is 200(1 + 0.08)3s4 = 200(s7 −s3 ) = 200(8.9228 −3.2464) = $1135.28 It is also possible to work with annuities−due instead of annuities−immediate. The reader should verify that (1 + i)m¨ sn = ¨ sm+n −¨ sm . 170 THE BASICS OF ANNUITY THEORY Example 17.4 A monthly annuity−due pays 100 per month for 12 months. Calculate the accumulated value 24 months after the ﬁrst payment using a nominal rate of 4% compounded monthly. Solution. The answer is 100 1 + 0.04 12 12¨ s12 0.04 12 = 1, 276.28 (3) Current value between the ﬁrst and last payment date Next, we consider the question of ﬁnding the present value of an n−period annuity−immediate after the payment at the end of mth period where 1 ≤m ≤n. Figure 17.3 shows the time diagram for this case. Figure 17.3 The current value of an n−period annuity−immediate immediately upon the mth payment date is the present value at time 0 accumulated for m time periods which is equal to the accumulated value at time n discounted for n −m time periods, that is, (1 + i)man = νn−msn . One has the following formula, (1 + i)man = νn−msn = sm + an−m . To see this, (1 + i)man =(1 + i)m · 1 −(1 + i)−n i =(1 + i)m −(1 + i)m−n i =(1 + i)m −1 i + 1 −(1 + i)m−n i =sm + an−m . 17 ANNUITY VALUES ON ANY DATE: DEFERRED ANNUITY 171 Example 17.5 For four years, an annuity pays $200 at the end of each half−year with an 8% rate of interest convertible semiannually. Find the current value of the annuity immediately upon the 5th payment (i.e., middle of year 3). Solution. The answer is 200(1.04)5a8 0.04 = 200(s5 0.04 + a3 0.04) = 200(5.4163 + 2.7751) = $1, 638.28 For annuity−due we have a similar formula for the current value (1 + i)m¨ an = νn−m¨ sn = ¨ sm + ¨ an−m . Example 17.6 Calculate the current value at the end of 2 years of a 10 year annuity due of $100 per year using a discount rate of 6%. Solution. We have (1 + i)−1 = 1 −d = 1 −0.06 = 0.94 and i = 0.06 0.94. Thus, 100(.94)−2¨ a10 = $870.27 Up to this point, we have assumed that the date is an integral number of periods. In the case the date is not an integral number of periods from each payment date, the value of an annuity is found by ﬁnding the value on a date which is an integral number of periods from each payment date and then the value on this date is either accumulated or discounted for the fractional period to the actual evaluation date. We illustrate this situation in the next example. Example 17.7 An annuity−immediate pays $1000 every six months for three years. Calculate the present value of this annuity two months before the ﬁrst payment using a nominal interest rate of 12% compounded semiannually. Solution. The present value at time t = 0 is 1000a6 0.06 = 10001 −(1.06)−6 0.06 = $4917.32. Let j be the interest rate per 2-month. Then 1 + j = (1 + 0.06) 1 3. The present value two months before the ﬁrst payment is made is 4917.32(1.06) 2 3 = $5112.10 172 THE BASICS OF ANNUITY THEORY Practice Problems Problem 17.1 For four years, an annuity pays $200 at the end of each half-year with an 8% rate of interest convertible semiannually. Find the current value of the annuity three months after the 5th payment (i.e., nine months into year 3). Problem 17.2 Which of the following is equal to the current value shown on the time diagram? (i) s3 + a4 (ii) ¨ s3 + a4 (iii) ¨ a6 (1 + i)3 (iv) v4s6 (v) ¨ s2 + (1 + i)a4 Problem 17.3 Calculate the present value of an annuity immediate with 20 annual payments of 500 if the ﬁrst payment of the annuity immediate starts at the end of the ﬁfth year. The annual eﬀective interest rate is 8%. Problem 17.4 Calculate the current value at the end of 5 years of an annuity due paying annual payments of 1200 for 12 years. The annual eﬀective interest rate is 6%. Problem 17.5 A monthly annuity due pays 100 per month for 12 months. Calculate the accumulated value 12 months after the last payment using a nominal rate of 4% compounded monthly. Problem 17.6 A monthly annuity immediate pays 100 per month for 12 months. Calculate the accumulated value 12 months after the last payment using a nominal rate of 4% compounded monthly. 17 ANNUITY VALUES ON ANY DATE: DEFERRED ANNUITY 173 Problem 17.7 Annuities X and Y provide the following payments: End of Year Annuity X Annuity Y 1 - 10 1 K 11 - 20 2 0 21 - 30 1 K Annuities X and Y have equal present values and an annual eﬀective interest rate of i such that v10 = 0.5. Determine K. Problem 17.8 Payments of $100 per quarter are made from June 7, Z through December 7, Z + 11, inclusive. If the nominal interest convertible quarterly is 6%: (a) Find the present value on September 7, Z −1. (b) Find the current value on March 7, Z + 8. (c) Find the accumulated value on June 7, Z + 12. Problem 17.9 Find the current value to the nearest dollar on January 1 of an annuity which pays $2,000 every six months for ﬁve years. The ﬁrst payment is due on the next April 1 and the rate of interest is 9% convertible semiannually. Problem 17.10 John buys a series of payments. The ﬁrst payment of 50 is in six years. Annual payments of 50 are made thereafter until 14 total payments have been made. Calculate the price John should pay now to realize an annual eﬀective return of 7%. Problem 17.11 Which of the following are true? (i) ¨ a10 −¨ a3 = a9 −a2 (ii) ν3¨ a3 = ν2a3 (iii) ν8s12 = ¨ a3 + ¨ s9 Problem 17.12 A loan of 1,000 is to be repaid by annual payments of 100 to commence at the end of the ﬁfth year and to continue thereafter for as long as necessary. The eﬀective rate of discount is 5%. Find the amount of the ﬁnal payment if it is to be larger than the regular payments. 174 THE BASICS OF ANNUITY THEORY Problem 17.13 Using an annual eﬀective interest rate i > 0, you are given: (i) The present value of 2 at the end of each year for 2n years, plus an additional 1 at the end of each of the ﬁrst n years, is 36. (ii) The present value of an n−year deferred annuity-immediate paying 2 per year for n years is 6. Calculate i. Problem 17.14 Show that P15 t=10(¨ st −st) = s16 −s10 −6. Problem 17.15 It is known that a7 a11 = a3 +sx ay +sz . Find x, y, and z. Problem 17.16 Simplify a15(1 + ν15 + ν30) to one symbol. Problem 17.17 The present value of an annuity-immediate which pays $200 every 6 months during the next 10 years and $100 every 6 months during the following 10 years is $4,000. The present value of a 10-year deferred annuity-immediate which pays $250 every 6 months for 10 years is $2,500. Find the present value of an annuity-immediate which pays $200 every 6 months during the next 10 years and $300 every 6 months during the following 10 years. Problem 17.18 ‡ At an annual eﬀective interest rate of i, i > 0, both of the following annuities have a present value of X : (i) a 20-year annuity-immediate with annual payments of 55; (ii) a 30-year annuity-immediate with annual payments that pays 30 per year for the ﬁrst 10 years, 60 per year for the second 10 years, and 90 per year for the ﬁnal 10 years. Calculate X. Problem 17.19 Tom borrows 100 at an annual eﬀective interest rate of 4% and agrees to repay it with 30 annual installments. The amount of each payment in the last 20 years is set at twice that in the ﬁrst 10 years. At the end of 10 years, Tom has the option to repay the entire loan with a ﬁnal payment X, in addition to the regular payment. This will yield the lender an annual eﬀective rate of 4.5% over the 10 year period. Calculate X. 17 ANNUITY VALUES ON ANY DATE: DEFERRED ANNUITY 175 Problem 17.20 You are given an annuity-immediate with 11 annual payments of 100 and a ﬁnal larger payment at the end of 12 years. At an annual eﬀective interest rate of 3.5%, the present value at time 0 of all payments is 1000. Using an annual eﬀective interest rate of 1%, calculate the present value at the beginning of the ninth year of all remaining payments. Round your answer to the nearest integer. Problem 17.21 The proceeds of a 10,000 death beneﬁt are left on deposit with an insurance company for seven years at an eﬀective annual interest rate of 5%. The balance at the end of seven years is paid to the beneﬁciary in 120 equal monthly payments of X, with the ﬁrst payment made immediately. During the payout period, interest is credited at an annual eﬀective rate of 3%. Problem 17.22 The present value today of a 20 year annuity-immediate paying 500 every 6 months but with the ﬁrst payment deferred t years has a present value of 5,805.74. If i(12) = .09 is the interest rate used to calculate the present value, ﬁnd t. Problem 17.23 John buys an enterprise that sells dental equipment in the amount of 60 million dollars. He decides to ﬁnance the purchase by making 20 semiannual payments with the ﬁrst payment in two years. Find the value of each payment if the nominal interest rate is 12% compounded semiannually. Problem 17.24 Peter wants to accumulate $20,000 in ﬁve years by making monthly payment at the end of each month for the ﬁrst three years into a savings account that pays annual interest rate of 9% com-pounded monthly and then leave the accumulated amount in the account for the remaining two years. What is the value of his regular monthly payment? 176 THE BASICS OF ANNUITY THEORY 18 Annuities with Inﬁnite Payments: Perpetuities A perpetuity is an annuity whose term is inﬁnite, i.e., an annuity whose payments continue forever with the ﬁrst payment occurs either immediately (perpetuity−due) or one period from now (perpetuity−immediate). Thus, accumulated values of perpetuities do not exist. Let us determine the present value of a perpetuity−immediate at the time one period before the ﬁrst payment, where a payment of 1 is made at the end of each period. The present value will be denoted by a∞. A time diagram of this case is given in Figure 18.1 Figure 18.1 Using the equation of value at time t = 0 we ﬁnd a∞=ν + ν2 + · · · =inﬁnite geometric progression with ν < 1 = ν 1 −ν = ν iν = 1 i . The verbal interpretation of this formula is as follows: If the periodic eﬀective rate of interest is i then one can invest a principal of 1 i for one period and obtain a balance of 1 + 1 i at the end of the ﬁrst period $1. A payment of $1 is made and the remaining balance of 1 i is reinvested for the next period. This process continues forever. Now, since an = 1−νn i and limn→∞νn = 0 for 0 < ν < 1 we have a∞= lim n→∞an = 1 i . Example 18.1 Suppose a company issues a stock that pays a dividend at the end of each year of $10 indeﬁnitely, and the the companies cost of capital is 6%. What is the the value of the stock at the beginning of the year? Solution. The answer is 10 · a∞= 10 · 1 0.06 = $166.67 18 ANNUITIES WITH INFINITE PAYMENTS: PERPETUITIES 177 Analogously to perpetuity−immediate, we may deﬁne a perpetuity−due to be an inﬁnite sequence of equal payments where each payment is made at the beginning of the period. Let ¨ a∞denote the present value of a perpetuity−due at the time of ﬁrst payment is made. A time diagram describing this case is given in Figure 18.2. Figure 18.2 The equation of value at time t = 0 is ¨ a∞= 1 + ν + ν2 + · · · = 1 1 −ν = 1 d = lim n→∞¨ an . This formula can be obtained from ﬁnding the present value of a payment of $1 at the beginning of the ﬁrst period and an annutiy−immediate. Indeed, 1 + a∞= 1 + 1 i = 1 + i i = 1 d = ¨ a∞. Example 18.2 What would you be willing to pay for an inﬁnite stream of $37 annual payments (cash inﬂows) beginning now if the interest rate is 8% per annum? Solution. The answer is 37¨ a∞= 37 0.08(1.08)−1 = $499.50 Remark 18.1 We pointed out at the beginning of the section that accumulated values for perpetuities do not exist since payments continue forever. We can argue mathematically as follows: If a perpetuity−immediate has an accumulated value, denoted by s∞, then we expect to have s∞= limn→∞sn. But lim n→∞sn = lim n→∞ (1 + i)n −1 i does not exist since 1 + i > 1 and limn→∞(1 + i)n = ∞. 178 THE BASICS OF ANNUITY THEORY Perpetuities are useful in providing verbal explanations of identities. For example, the formula an = 1 −νn i = 1 i −νn i = a∞−νna∞. can be interpreted as the diﬀerence between payments for two perpetuities each paying 1 at the end of each period; the ﬁrst payment of the ﬁrst perpetuity is one period from now and the ﬁrst payment of the second perpetuity is n + 1 periods from now. The present value of the ﬁrst perpetuity is 1 i and that of the second perpetuity is νn i . An identical relationship holds for perpetuity due. That is ¨ an = 1 −νn d = 1 d −νn d = ¨ a∞−νn¨ a∞. Example 18.3 You can receive one of the following two sets of cash ﬂows. Under Option A, you will receive $5,000 at the end of each of the next 10 years. Under Option B, you will receive X at the beginning of each year, forever. The annual eﬀective rate of interest is 10%. Find the value of X such that you are indiﬀerent between these two options. Solution. The equation of value at time t = 0 is 5000a10 =X d = X 1 i + 1 30722.84 =11X X =$2, 792.99 Similar to deferred annuities, one can discuss deferred perpetuities. The present value P0 of a deferred perpetuity−immediate with periodic payment of 1 that starts in n periods time, with a ﬁrst cash ﬂow at the beginning of period n + 1, is given by the equation of value at time t = n (1 + i)nP0 = a∞. Example 18.4 Fifty dollars is paid at the end of each year forever starting six years from now. Assume the annual eﬀective rate of interest is 10%, ﬁnd the present value of the investment. Solution. The deferred period is t = 5. The answer is 50(1 + i)−5a∞= 50(1.1)−5 · 1 0.1 = $310.46 18 ANNUITIES WITH INFINITE PAYMENTS: PERPETUITIES 179 Example 18.5 A deferred perpetuity-immediate paying $2000 monthly is bought for a sum of $229,433.67. Find the deferred period if the interest rate compounded monthly is 6%. Solution. The monthly interest rate is 0.06 12 = 0.005. Let n be the deferred period, i.e. the time it takes $229,433.67 to accumulate to 2000a∞. Thus, n satisﬁes the equation 2000a∞= 229, 433.67(1.005)n ⇒2000 · 1 0.005 = 229, 433.67(1.005)n. Solving this equation for n we ﬁnd n = 111.45 months 180 THE BASICS OF ANNUITY THEORY Practice Problems Problem 18.1 What would you be willing to pay for an inﬁnite stream of $37 annual payments (cash inﬂows) beginning one year from today if the interest rate is 8%? Problem 18.2 Which of the following are equal to a3. (i) ν + ν2 + ν3 (ii) ¨ a4 −1 (iii) a∞(1 −ν3) Problem 18.3 An annuity−due pays 100 at the beginning of each year forever. The eﬀective annual rate of interest is 25%. What is the present value of the annuity? Problem 18.4 A special perpetuity pays $200 at the end of each year for the ﬁrst 10 years and $100 at the end of each year thereafter. If the eﬀective annual rate of interest is 10%, ﬁnd the present value of the perpetuity. Problem 18.5 ‡ A perpetuity−immediate pays X per year. Brian receives the ﬁrst n payments, Colleen receives the next n payments, and Jeﬀreceives the remaining payments. Brian’s share of the present value of the original perpetuity is 40%, and Jeﬀ’s share is K. Calculate K. Problem 18.6 A sum P is used to buy a deferred perpetuity−due of 1 payable annually. The annual eﬀective interest rate is i > 0. Find an expression for the deferred period. Problem 18.7 Deposits of $1,000 are placed in a fund at the beginning of each year for the next 20 years. After 30 years annual payments commence and continue forever, with the ﬁrst payment at the end of the 30th year. Find an expression for the amount of each payment. Problem 18.8 A benefactor leaves an inheritance to four charities, A, B, C, and D. The total inheritance is a series of level payments at the end of each year forever. During the ﬁrst n years A, B, and C share each payment equally. All payments after n years revert to D. If the present values of the shares of A, B, C, and D are all equal, ﬁnd (1 + i)n. 18 ANNUITIES WITH INFINITE PAYMENTS: PERPETUITIES 181 Problem 18.9 A level perpetuity−immediate is to be shared by A, B, C, and D. A receives the ﬁrst n payments. B the second n payments, C the third n payments, and D all payments thereafter. It is known that the ratio of the present value of C′s share to A′s share is 0.49. Find the ratio of the present value of B′s share to D′s share. Problem 18.10 Adam buys a perpetuity due of 1,000 per month for 100,000. Calculate the annual eﬀective rate of interest used to calculate the price of this perpetuity. Problem 18.11 The present value of a perpetuity immediate where the payment is P is 1,000 less than the present value of a perpetuity due where the payment is P. Calculate P. Problem 18.12 A trust has been established such that RJ will receive a perpetuity of 1000 a year with the ﬁrst payment at the end of 5 years. Calculate the present value of the perpetuity at a discount rate of d = 8%. Problem 18.13 Julie, Chris, and Allen will share an annual perpetuity immediate of 1200. Julie will receive the ﬁrst 9 payments. Chris will receive the next 16 payments. Allen will receive all remaining payments. At an annual eﬀective interest rate of 5%, order the value of each person’s share of the perpetuity. Problem 18.14 John is receiving annual payments from a perpetuity immediate of 12,000. Krista is receiving annual payments from a perpetuity due of 10,000. If the present value of each perpetuity is equal, calculate i. Problem 18.15 The present value of an annual perpetuity immediate of 150 is equal to the present value of an annual perpetuity immediate that pays 100 at the end of the ﬁrst 20 years and 200 at the end of year 21 and each year thereafter. Calculate i. Problem 18.16 ‡ An estate provides a perpetuity with payments of X at the end of each year. Seth, Susan, and Lori share the perpetuity such that Seth receives the payments of X for the ﬁrst n years and Susan receives the payments of X for the next m years, after which Lori receives all the remaining payments of X. Which of the following represents the diﬀerence between the present value of Seth’s 182 THE BASICS OF ANNUITY THEORY and Susan’s payments using a constant rate of interest? (a) X[an −νnam] (b) X[¨ an −νn¨ am] (c) X[an −νn+1am] (d) X[an −νn−1am] (e) X[van −νn+1am] Problem 18.17 ‡ Which of the following are characteristics of all perpetuities?(including non−level payments) I. The present value is equal to the ﬁrst payment divided by the annual eﬀective interest rate. II. Payments continue forever. III. Each payment is equal to the interest earned on the principal. Problem 18.18 ‡ A perpetuity paying 1 at the beginning of each 6-month period has a present value of 20 . A second perpetuity pays X at the beginning of every 2 years. Assuming the same annual eﬀective interest rate, the two present values are equal. Determine X. Problem 18.19 What is the present value of receiving $30 at the end of each year forever, starting 9 years from now? Assume an annual rate of interest of 7% Problem 18.20 Jim buys a perpetuity of 100 per year, with the ﬁrst payment 1 year from now. The price for the perpetuity is 975.61, based on a nominal yield of i compounded semiannually. Immediately after the second payment is received, the perpetuity is sold for 1642.04, to earn for the buyer a nominal yield of j compounded semiannually. Calculate i −j. Problem 18.21 Victor wants to purchase a perpetuity paying 100 per year with the ﬁrst payment due at the end of year 11. He can purchase it by either (1) paying 90 per year at the end of each year for 10 years, or (2) paying K per year at the end of each year for the ﬁrst 5 years and nothing for the next 5 years. Calculate K. Problem 18.22 A perpetuity pays 1 at the end of every year plus an additional 1 at the end of every second year. The present value of the perpetuity is K for i ≥0. Determine K. 18 ANNUITIES WITH INFINITE PAYMENTS: PERPETUITIES 183 Problem 18.23 The accumulated value just after the last payment under a 12-year annuity-immediate of 1000 per year, paying interest at the rate of 5% per annum eﬀective, is to be used to purchase a perpetuity at an interest rate of 6%, with ﬁrst payment to be made 1 year after the last payment under the annuity. Determine the size of the payments under the perpetuity. 184 THE BASICS OF ANNUITY THEORY 19 Solving for the Unknown Number of Payments of an An-nuity In this section we consider the question of ﬁnding the number of payments n given the regular payment R, the interest per period i and either the present value or the accumulated value of an annuity. We will assume an annuity−immediate. A similar calculation applies for annuity−due. Let P be the present value of an annuity−immediate. A time diagram is given in Figure 19.1 Figure 19.1 The equation of value at time t = 0 is P = Ran i, where P, R, and i are known quantities and n is the unknown. Solving this equation for n we ﬁnd P = Ran i = R 1 −νn i ⇒νn = 1 −i P R ⇒n = ln 1 −i P R ln ν . This last expression is not necessary a positive integer. Thus, the equation P = Ran is replaced by P = Ran+k where n is a positive integer and 0 < k < 1. In this case, using the results of Problems 15.47-15.48 we can write P = Ran+k i = R 1 −νn+k i ⇒νn+k = 1 −i P R which implies n + k = ln 1 −i P R ln ν . It follows that, for the annuity to have the present value P, n regular payments of R must be made and an additional one payment in the amount of R (1 + i)k −1 i to be made at time t = n + k, that is, at the fractional period k of the (n + 1)th period. In practice, the last smaller payment is made either at the same time as the last regular payment making the last payment larger than the regular payment (such a payment is called a balloon 19 SOLVING FOR THE UNKNOWN NUMBER OF PAYMENTS OF AN ANNUITY 185 payment) or at the end of the period following the last regular payment. In this case the smaller payment is called drop payment. The following examples illustrate how to deal with the question of ﬁnding the unknown number of payments. Example 19.1 An investment of $80,000 is to be used to make payments of $5000at the end of every six months for as long as possible. If the fund earns a nominal interest rate of 12% compounded semiannually, ﬁnd how many regular payments can be made and ﬁnd the amount of the smaller payment: (a) to be paid along the last regular payment, (b) to be paid six months after the last regular payment, (c) to be paid during the six months following the last regular payment. Solution. We ﬁrst solve the equation 8000 = 5000an+k 0.06 as follows 80000 = 5000an+k 0.06 = 5000 1 −(1.06)−(n+k) i ⇒1 −(1.06)−(n+k) = 80000 5000 (0.06) = 0.96 from which we ﬁnd 1 −0.96 = 0.04 = (1.06)−(n+k) ⇒n + k = −ln 0.04 ln 1.06 = 55.242. Thus, n = 55 and k = 0.242. (a) Let X be the amount of the smaller payment to be be made at the end of the 55th period. A time diagram of this situation is given in Figure 19.2. Figure 19.2 Note that every period on the time diagram consists of six months. The equation of value at time t = 0 is 80000 = 5000a55 0.06 + X(1.06)−55 ⇒80000 = 79952.715 + (0.0405674)X ⇒X = $1165.59. Thus, in this case, we have 54 payments of $5000 each, and a last payment of 5000 + 1165.59 = 6165.59 186 THE BASICS OF ANNUITY THEORY (b) This situation is illustrated in Figure 19.3. Figure 19.3 The equation of value at time t = 0 is 80000 = 5000a55 0.06 + Y (1.06)−56 ⇒80000 = 79952.715 + (0.0382712)Y ⇒Y = $1235.53. In this case, we have 55 payments of $5000 each and a last payment of $1235.53 six months after the last regular payment. (c) In this case we have 55 payments of $5000 each and one last payment of 5000 (1 + 0.06)0.242 −1 0.06 = $1182.13 to be made 44 days (0.242 × 365 2 = 44.165) after the last regular payment A similar type of calculation can be done with known accumulated value instead of present value as illustrated in the next example. Example 19.2 In order to accumulate $5,000, you will deposit $100 at the end of each month for as long as necessary. Interest is 15% compounded monthly. (a) Find the number of regular payments and the fractional period that are required to accumulate the $5,000? (b) If a ﬁnal fractional payment will be added to the last regular payment, what must this fractional payment be? (c) If a ﬁnal fractional payment will be made one month after the last regular payment, what must this fractional payment be? (d) If the fractional payment to be made during the month following the last regular payment. Solution. (a) The monthly interest rate is 0.15 12 = 0.0125. The equation of value at time t = n + k is 5000 = 100sn+k 0.0125 = 100 · (1 + i)n+k −1 i = 100 · (1.0125)n+k −1 0.0125 19 SOLVING FOR THE UNKNOWN NUMBER OF PAYMENTS OF AN ANNUITY 187 Solving for n + k we ﬁnd n + k = ln 1 + 5000(0.0125) 100 ln 1.0125 = 39.08 Thus, there will be 39 regular payments and a payment at the fractional time 0.08 = 0.08 × 365 12 = 2.433, i.e., two days after the last regular payment. (b) Let X be the fractional payment added to the last regular payment. The time diagram of this case is shown in Figure 19.4. Figure 19.4 The equation of value at time t = 39 is 5000 = 100s39 0.0125 + X. Solving for x we ﬁnd X = $13.38. (c) Let Y be such a payment. The time diagram of this case is shown in Figure 19.5. Figure 19.5 The equation of value at time t = 40 is 5000 = 100 s39(1.0125) + Y. Solving for Y we ﬁnd Y = −48.96. What does this negative answer mean? If 39 regular payments are made at the end of each month, then at time 39 (measured in months) the account will have accumulated 100s39 0.0125 = 4986.62. This is extremely close to the 5000 that we wanted in the ﬁrst place. If no additional payment is made from time 39 to time 40, then interest alone will make the account have 4986.62(1.0125) = 5048.96. This means that there will be no additional payment required at time 40. In fact, $48.96 would have to be withdrawn from the account to make the account have only $5000 at time 40. (d) In tis case, 39 regular payments of $100 each to be made and one additional payment in the amount of 100 (1 + 0.0125)0.08 −1 0.0125 = $7.95 to be made 2 days after the last regular payment 188 THE BASICS OF ANNUITY THEORY Example 19.3 An investment of $10,000 is to be used to make payments of $700 at the end of every year for as long as possible. If the eﬀective annual rate if interest is 6%, ﬁnd: (a) the number of regular payments the fund will make, (b) the size of the smaller ﬁnal payment if the payment will be made in addition to the last regular payment, (c) the size of the smaller ﬁnal payment if the payment will be made one year after the last regular payment. (d) the size of the smaller ﬁnal payment if the payment is to be made during the year following the last regular payment. Solution. (a) We have 10000 = 700an or 10000 = 700 · 1−(1+0.06)−n 0.06 . Solving for n we ﬁnd n = ln 1 −10000(0.06) 700 −ln 1.06 = 33.40 Thus, there will be 33 regular payments. (b) Let x be the size of the payment. The equation of value at t = 0 is 10000 = 700a33 + v33x or 10000(1.06)33 = 700s33 + x. Solving for x we ﬁnd x = 265.68. (c) Let x be the size of the payment. The equation of value at time t = 0 is 10000 = 700a33 + v34x. Solving for x we ﬁnd x = 281.62. (d) The equation of value at time t = 0 is 700a33+k = 10000 or 0.07a33+k = 1 where 0 < k < 1. (See Problem 15.29). Thus, we have 0.07 1−v33+k i = 1 ⇒v33+k = 1 −6 7 = 1 7 ⇒ 33 + k = ln 7 ln 1.06 ≈33.395 ⇒k = 0.395. In this case, the equation of value at time t = 0 is 700a33 + (1.06)−33.395x = 10000. Solving for x we ﬁnd x = $271.87. That is, the ﬁnal irregular payment of $271.87 is paid at time 33.395 19 SOLVING FOR THE UNKNOWN NUMBER OF PAYMENTS OF AN ANNUITY 189 Practice Problems Problem 19.1 Allison pays $27,506.28 for an annuity due today. This annuity will pay her $1,750 at the beginning of each month for n months with the ﬁrst payment coming today. If i(12) = 12%, ﬁnd n. Round your answer to the nearest integer. Problem 19.2 In order to accumulate $2,000, you will deposit $100 at the end of each month for as long as necessary. Interest is i(12) = 4%. (a) How many regular payments are required to accumulate the $2,000? (b) If a ﬁnal fractional payment will be added to the last regular payment, what must this fractional payment be? Problem 19.3 ‡ The present values of the following three annuities are equal: 1. perpetuity-immediate paying 1 each year, calculated at an annual eﬀective interest rate of 7.25%; 2. 50-year annuity-immediate paying 1 each year, calculated at an annual eﬀective interest rate of j%; 3. n−year annuity-immediate paying 1 each year, calculated at an annual eﬀective interest rate of (j −1)%. Calculate n. Problem 19.4 Jenna is the beneﬁciary of a fund of 20,000 that pays her 1,000 at the end of each month. The fund earns 6% compounded monthly. The ﬁnal payment to exhaust the fund will be a balloon payment. Calculate the amount of the balloon payment. Problem 19.5 Jordan inherits 50,000. This inheritance is invested in a fund earning an annual rate of interest of 6%. He withdraws 5,000 per year beginning immediately. Once Jordan can no longer withdraw a full 5,000, he will withdraw a ﬁnal payment one year after the prior regular payment. Calculate the ﬁnal payment. Problem 19.6 A loan of 1,000 is to be repaid by annual payments of 100 to commence at the end of the 5th year, and to continue thereafter for as long as necessary. Find the time and amount of the ﬁnal payment if the ﬁnal payment is to be larger than the regular payments. Assume i = 4.5%. 190 THE BASICS OF ANNUITY THEORY Problem 19.7 A fund of 2,000 is to be accumulated by n annual payments of 50 by the end of each year, followed by n annual payments of 100 by the end of each year, plus a smaller ﬁnal payment made 1 year after the last regular payment. If the eﬀective rate of interest is 4.5%, ﬁnd n and the amount of the ﬁnal irregular payment. Problem 19.8 One annuity pays 4 at the end of each year for 36 years. Another annuity pays 5 at the end of each year for 18 years. The present values of both annuities are equal at eﬀective rate of interest i. If an amount of money invested at the same rate i will double in n years, ﬁnd n. Problem 19.9 A fund earning 8% eﬀective is being accumulated with payments of 500 at the beginning of each year for 20 years. Find the maximum number of withdrawals of 1,000 which can be made at the end of each year under the condition that once withdrawals start they must continue through the end of the 20-year period. Problem 19.10 A borrower has the following two options for repaying a loan: (i) Sixty monthly payments of 100 at the end of each month. (ii) A single payment of 6,000 at the end of K months. Interest is at the nominal annual rate of 12% convertible monthly. The two options have the same present value. Find K to the nearest integer. Problem 19.11 ‡ An annuity pays 1 at the end of each year for n years. Using an annual eﬀective interest rate of i, the accumulated value of the annuity at time (n + 1) is 13.776 . It is also known that (1 + i)n = 2.476. Calculate n. Problem 19.12 Ten annual deposits of 1,000 each are made to account A, starting on January 1, 1986. Annual deposits of 500 each are made to account B indeﬁnitely, also starting on January 1, 1986. Interest on both accounts is i = 0.05, with interest credited on December 31. On what date will the balance in account B be larger than the balance in account A? Assume the only transactions are deposits and interest credited to the accounts on December 31. Problem 19.13 ‡ You are given an annuity-immediate with 11 annual payments of 100 and a ﬁnal payment at the end of 12 years. At an annual eﬀective interest rate of 3.5%, the present value at time 0 of all payments is 1,000. Calculate the ﬁnal payment. 19 SOLVING FOR THE UNKNOWN NUMBER OF PAYMENTS OF AN ANNUITY 191 Problem 19.14 A loan of 10,000 is being repaid with payment of 300 at the end of each month for as long as necessary plus an additional payment at the time of the last regular payment. What is the amount of the additional payment using an interest rate of 9% compounded monthly. Problem 19.15 A loan of 10,000 is being repaid with payments of 500 starting one month after the loan is made and lasting as long as necessary. A ﬁnal smaller payment is made one month after the last regular payment of 500. What is the amount of the additional smaller payment using an interest rate of 12% compounded monthly? Problem 19.16 You are given an annuity immediate with 11 annual payments of 100 and a ﬁnal balloon payment at the end of 12 years. At an annual eﬀective interest rate of 3.5%, the present value at time 0 of all the payments is 1,000. Using an annual eﬀective interest rate of 1%, calculate the present value at the beginning of the ninth year of all remaining payments. Problem 19.17 On the ﬁrst day of every January, April, July and October Smith deposits 100 in an account earning i(4) = 0.16. He continues the deposits until he accumulates a suﬃcient balance to begin withdrawals of 200 every 3 months, starting three months after the ﬁnal deposit such that he can make twice as many withdrawals as he made deposits. How many deposits are needed? Problem 19.18 On the ﬁrst day of the month, starting January 1, 1995, Smith deposits 100 in an account earning i(12) = 0.09, with interest credited the last day of each month. In addition, Smith deposits 1,000 in the account every December 31. Around what date does the account ﬁrst exceed 100,000? 192 THE BASICS OF ANNUITY THEORY 20 Solving for the Unknown Rate of Interest of an Annuity In this section we consider the problem of ﬁnding the interest rate given the number of payments and either the present value or the accumulated value of the payments. That is, we want to solve an equation like an = k or sn = k for i. We consider three methods in determining an unknown rate of interest. The ﬁrst is to solve for i by algebraic techniques. For example, the expression k = an = ν + ν2 + · · · + νn is a polynomial in ν of degree n. If the roots of this polynomial can be determined algebraically, then i is immediately determined. That is, i = ν−1 −1. This method is generally practical only for small values of n. Example 20.1 If a2 = 1.75, ﬁnd an exact expression for i > 0. Solution. We have a2 = 1.75 ⇔ν + ν2 = 1.75 ⇔ν2 + ν −1.75 = 0 ⇔ν = −1 + √ 8 2 . Since (1 + i)−1 = −1+ √ 8 2 , solving this for i we ﬁnd i = 2 √ 8−1 −1 ≈0.09384 Example 20.2 Solve s3 i = 3.31 for i. Solution. We have (1 + i)3 −1 i = 3.31. This reduces to i3 + 3i2 −0.31i = 0 and can be factored to i(i + 3.1)(i −0.1) = 0. i = 0 is clearly an extraneous solution and is not the correct answer. Hence, since i > 0 the correct answer is i = 0.1 = 10% 20 SOLVING FOR THE UNKNOWN RATE OF INTEREST OF AN ANNUITY 193 Example 20.3 The Honors Society decides to set up a scholarship for university students. They deposit $1,000 in an account for 11 years with the ﬁrst payment a year from now, and then transfer the accumulated value to a perpetuity-immediate paying $500 each year to a deserving student. Assuming the account and the perpetuity are at the same eﬀective annual interest rate i, ﬁnd i. Solution. The accumulated value in the account after the last deposit is 100s11 . The present value of the perpetuity at that time is 500a∞; the accumulated value and the value of the perpetuity must be equal. 1000s11 =500a∞ 2 · (1 + i)11 −1 i =1 i (1 + i)11 =1.5 i =1.5 1 11 −1 ≈3.75% The second method for determining the unknown rate is to use linear interpolation (see Section 13). Example 20.4 An annuity−immediate pays you $100 in each of the next eight years. Its present value is $680. Using linear interpolation of the interest table, ﬁnd an estimate of the annual eﬀective rate i. i s8 a8 2.5% 8.7361 7.1701 3% 8.8923 7.0197 3.5% 9.0517 6.8740 4% 9.2142 6.7327 4.5% 9.3800 6.5959 5% 9.5491 6.4632 6% 9.8975 6.2098 7% 10.2598 5.9713 Solution. From the given table we see that a8 0.035 = 6.8740 and a8 0.04 = 6.7327. Using linear interpolation (See Section 13), we ﬁnd i ≈0.035 + (0.04 −0.035) · 6.80 −6.8740 6.7327 −6.8740 = 3.762% 194 THE BASICS OF ANNUITY THEORY Example 20.5 At what interest rate, convertible quarterly, is $16,000 the present value of $1,000 paid at the end of every quarter for ﬁve years? Solution. Let j = i(4) 4 so that the equation of value at time t = 0 becomes 1000a20 j = 16, 000 or a20 j = 16. From the tables of interest we see that a20 0.02 = 16.3514 and a20 0.0250 = 15.5892. Thus, using linear interpolation we ﬁnd j = 0.02 + (0.0250 −0.0200) · 16 −16.3514 15.5892 −16.3514 = 0.0223 which gives i(4) = 4(0.0223) = 0.0892 = 8.92% i a20 1.0% 18.0456 1.5% 17.1686 2.0% 16.3514 2.5% 15.5892 The third method is to use the Newton-Raphson iteration method. This method is used to approxi-mate the zeros of the equation f(x) = 0 where f is a diﬀerentiable function. The idea of the method is to start with an initial guess x0 and then ﬁnd the equation of the tangent line at x0. We next ﬁnd the x−intercept of this line say x1, which is closer to the real solution than x0. See Figure 20.1. One can easily see that x1 = x0 −f(x0) f ′(x0). Next, we ﬁnd the equation of the tangent line at x = x1 and ﬁnd the x−intercept of this line. Say, x2 = x1 −f(x1) f ′(x1). The iteration continues so that we generate a sequence x0, x1, · · · , xn, · · · with xn+1 = xn −f(xn) f ′(xn) 20 SOLVING FOR THE UNKNOWN RATE OF INTEREST OF AN ANNUITY 195 xn converging to the solution of f(x) = 0. Figure 20.1 In our context, we want to solve an = k for i using the above mentioned method. Equivalently, we want to solve the equation f(i) = 1 −(1 + i)−n i −k = 0. The iteration formula in this case is given by is+1 =is − [1 −(1 + is)−n]i−1 s −k n(1 + is)−n−1i−1 s + [1 −(1 + is)−n]i−2 s =is 1 + 1 −(1 + is)−n −kis 1 −(1 + is)−n−1{1 + is(n + 1)} The next question regarding this third method is the selection of the initial guess i0. From the power series of (1 + i)−n we ﬁnd an =1 i (1 −(1 + i)−n) =1 i 1 − 1 −ni + (−n)(−n −1) 2! i2 + · · · =n 1 −n + 1 2! i + (n + 1)(n + 2) 3! i2 −· · · 196 THE BASICS OF ANNUITY THEORY Thus, 1 an =1 n 1 − n + 1 2! i −(n + 1)(n + 2) 3! i2 + · · · −1 =1 n " 1 + n + 1 2! i −(n + 1)(n + 2) 3! i2 + · · · + n + 1 2! i −(n + 1)(n + 2) 3! i2 + · · · 2 + · · · # =1 n 1 + n + 1 2! i + n2 −1 12 i2 −n(n2 −1) 24 i3 + · · · . The rate of convergence of this series is much faster than the one for the series expansion of an. Now, for the initial guess we use the approximation 1 k ≈1 n 1 + n + 1 2 i0 . Solving for i0 we ﬁnd i0 = 2(n −k) k(n + 1). In practice, the iterations will be carried out until is+1 = is to the required degree of accuracy. Example 20.6 Rework Example 20.5 using Newton-Raphson iterations Solution. Our starting value is i0 = 2(n−k) k(n+1) = 2(20−16) 16(20+1) = 0.0238. Finding several iterations we obtain i1 =0.0238 1 + 1 −(1.0238)−20 −16(0.0238) 1 −(1.0238)−21{1 + 0.0238 × 21} = 0.022246 i2 =0.022246 1 + 1 −(1.022246)−20 −16(0.022246) 1 −(1.022246)−21{1 + 0.022246 × 21} = 0.0222623 i3 =0.0222623 1 + 1 −(1.0222623)−20 −16(0.0222623) 1 −(1.0222623)−21{1 + 0.0222623 × 21} = 0.0222623 Thus, a more accurate interest rate to Example 20.5 is i(4) = 4(0.0222623) = 8.9049% Next, consider the problem of ﬁnding i solution to the equation sn i = k. 20 SOLVING FOR THE UNKNOWN RATE OF INTEREST OF AN ANNUITY 197 For this, we let f(i) = (1 + i)n −1 i −k. The Newton-Raphson iterations are given by is+1 = is 1 + (1 + is)n −1 −kis (1 + is)n−1{1 −is(n −1)} −1 and the initial guess is i0 ≈2(k −n) k(n −1). See Problem 20.20 198 THE BASICS OF ANNUITY THEORY Practice Problems Problem 20.1 (a) The accumulated value of a ﬁve-year annuity-immediate with semiannual payments of $2,000 is $25,000. Use linear interpolation of the interest table to ﬁnd an estimate for i. (b) The present value of a 11-year annuity-due with annual payments of $2,000 is $18,000. Use linear interpolation of the interest table to ﬁnd an estimate for i. Problem 20.2 ‡ To accumulate 8,000 at the end of 3n years, deposits of 98 are made at the end of each of the ﬁrst n years and 196 at the end of each of the next 2n years. The annual eﬀective rate of interest is i. You are given (1 + i)n = 2.0. Determine i. Problem 20.3 ‡ For 10,000, Kelly purchases an annuity-immediate that pays 400 quarterly for the next 10 years. Calculate the annual nominal interest rate convertible monthly earned by Kelly’s investment. Problem 20.4 ‡ A 10-year loan of $20,000 is to be repaid with payments at the end of each year. It can be repaid under the following two options: (X) Equal annual payments at the annual eﬀective rate of interest of 8%; (Y) Installments of $2,000 each year plus interest on the unpaid balance at an annual eﬀective rate of i. The sum of the payments under option (X) equals the sum of the payments under option (Y). Determine i. Problem 20.5 Given a∞= X and ¨ a∞= 1.25X. Calculate the interest rate used. Problem 20.6 A perpetuity pays 1 at the beginning of every year. The present value is 10. Calculate the annual eﬀective rate of interest earned by the perpetuity. Problem 20.7 If ¨ a2 · s2 = 4.05, calculate i. Problem 20.8 A deferred perpetuity pays 500 annually beginning at the end of year 5. The present value of the deferred perpetuity is 4,992. Calculate the annual eﬀective interest rate used to calculate the present value. 20 SOLVING FOR THE UNKNOWN RATE OF INTEREST OF AN ANNUITY 199 Problem 20.9 An annuity due which pays 100 per month for 12 years has a present value of 7,908. Calculate the annual eﬀective interest rate used to determine the present value. Problem 20.10 An annuity which pays 200 at the end of each quarter for 5 years has a present value of 3,600. Calculate the nominal rate of interest compounded quarterly. Problem 20.11 An annuity immediate pays 750 per year for 15 years. The accumulated value of the annuity after 15 years is 15,000. Calculate the annual eﬀective rate of interest used to calculate the accumulated value. Problem 20.12 Adam buys a perpetuity due of 1000 per month for 100,000. Calculate the annual eﬀective rate of interest used to calculate the price of this perpetuity. Problem 20.13 A beneﬁciary receives a 10,000 life insurance beneﬁt. If the beneﬁciary uses the proceeds to buy a 10-year annuity-immediate, the annual payout will be 1,538. If a 20-year annuity-immediate is purchased, the annual payout will be 1,072. Both calculations are based on an annual eﬀective interest rate of i. Find i. Problem 20.14 If ¨ sn = 11 and sn+1 = 2an+1 , ﬁnd i. Problem 20.15 If sn = 10 and s2n = 24, ﬁnd νn. Problem 20.16 ‡ Kathryn deposits 100 into an account at the beginning of each 4-year period for 40 years. The account credits interest at an annual eﬀective interest rate of i. The accumulated amount in the account at the end of 40 years is X, which is 5 times the accumulated amount in the account at the end of 20 years. Calculate X. Problem 20.17 The following two investment plans result in the same accumulated amount at the end of 10 years. 1. You lend $10,000, to be repaid in 10 equal installments at the end of each year, at the annual rate of interest of 6%. The installments are deposited to a savings account paying 3% eﬀective annual rate of interest. 2. You deposit $10,000 to an account paying the annual eﬀective rate of interest i. Find i. 200 THE BASICS OF ANNUITY THEORY Problem 20.18 ‡ At an annual eﬀective interest rate of i, i > 0%, the present value of a perpetuity paying 10 at the end of each 3-year period, with the ﬁrst payment at the end of year 3, is 32. At the same annual eﬀective rate of i, the present value of a perpetuity paying 1 at the end of each 4-month period, with ﬁrst payment at the end of 4 months, is X. Calculate X. Problem 20.19 A fund of $17,000 is to be accumulated at the end of ﬁve years with payments at the end of each half-year. The ﬁrst ﬁve payments are $1000 each, while the second ﬁve are $2000 each. Find the nominal rate of interest convertible semiannually earned on the fund. Problem 20.20 Consider solving the equation sn = k. (a) Show that sn = n 1 + n−1 2! i + (n−1)(n−2) 3! i2 + · · · . (b) Show that [sn]−1 = 1 n 1 −n−1 2 i + · · · . (c) Show that the Newton-Raphson iterations are given by is+1 = is 1 + (1 + is)n −1 −kis (1 + is)n−1{1 −(n −1)is} −1 . (d) Show that an initial guess for Newton-Raphson method is given by i0 ≈2(n −k) k(n −1). Problem 20.21 Use the Newton-Raphson iteration method with three iterations to estimate the rate of interest if s20 i = 25. Problem 20.22 Eric receives 12000 from a life insurance policy. He uses the fund to purchase two diﬀerent annuities, each costing 6000. The ﬁrst annuity is a 24 year annuity immediate paying K per year to himself. The second annuity is an 8 year annuity immediate paying 2K per year to his son. Both annuities are based on an annual eﬀective interest rate of i > 0. Determine i. Problem 20.23 Jeﬀdeposits 100 at the end of each year for 13 years into Fund X. Antoinette deposits 100 at the end of each year for 13 years into Fund Y. Fund X earns an annual eﬀective rate of 15% for the ﬁrst 5 years and an annual eﬀective rate of 6% thereafter. Fund Y earns an annual eﬀective rate of i. At the end of 13 years, the accumulated value of Fund X equals the accumulated value of Fund Y. Calculate i. 20 SOLVING FOR THE UNKNOWN RATE OF INTEREST OF AN ANNUITY 201 Problem 20.24 Dottie receives payments of X at the end of each year for n years. The present value of her annuity is 493. Sam receives payments of 3X at the end of each year for 2n years. The present value of his annuity is 2748. Both present values are calculated at the same annual eﬀective interest rate. Determine νn. Problem 20.25 ‡ A discount electronics store advertises the following ﬁnancing arrangement: “We don’t oﬀer you confusing interest rates. We’ll just divide your total cost by 10 and you can pay us that amount each month for a year.” The ﬁrst payment is due on the date of sale and the remaining eleven payments at monthly intervals thereafter. Calculate the eﬀective annual interest rate the store’s customers are paying on their loans. 202 THE BASICS OF ANNUITY THEORY 21 Varying Interest of an Annuity In this section we consider situations in which interest can vary each period, but compound interest is still in eﬀect. Let ik denote the rate of interest applicable from time k −1 to time k. We consider ﬁrst the present value of an n−period annuity-immediate. We consider the following two variations. The ﬁrst is when ik is applicable only for period k regardless of when the payment is made. That is, the rate ik is used only in period k for discounting all payments. In this case the present value is given by an = (1 + i1)−1 + (1 + i1)−1(1 + i2)−1 + · · · + (1 + i1)−1(1 + i2)−1 · · · (1 + in)−1. The second variation is when the rate ik is used as the eﬀective rate for each period i ≤k when a payment is made at time k. In this case, the present value is an = (1 + i1)−1 + (1 + i2)−2 + · · · + (1 + in)−n. Present value of annuity-due can be obtained from present value of annuity-immediate by using the formula ¨ an = 1 + an−1 . We now turn to accumulated values. We will consider an annuity-due. Again we consider two diﬀerent situations. If ik is applicable only for period k regardless of when the payment is made, then the accumulated value is given by ¨ sn = (1 + i1)(1 + i2) · · · (1 + in) + · · · + (1 + in−1)(1 + in) + (1 + in). For ik applicable for all periods i ≥k, the accumulated value is ¨ sn = (1 + i1)n + (1 + i2)n−1 + · · · + (1 + in). Accumulated values of annuity-immediate can be obtained from the accumulated values of annuity-due using the formula sn+1 = ¨ sn + 1. Example 21.1 Find the accumulated value of a 12-year annuity-immediate of $500 per year, if the eﬀective rate of interest (for all money) is 8% for the ﬁrst 3 years, 6% for the following 5 years, and 4% for the last 4 years. 21 VARYING INTEREST OF AN ANNUITY 203 Solution. The accumulated value of the ﬁrst 3 payments to the end of year 3 is 500s3 0.08 = 500(3.2464) = $1623.20. The accumulated value of the ﬁrst 3 payments to the end of year 8 at 6% and then to the end of year 12 at 4% is 1623.20(1.06)5(1.04)4 = $2541.18. The accumulated value of payments 4, 5, 6, 7, and 8 at 6% to the end of year 8 is 500s5 0.06 = 500(5.6371) = $2818.55. The accumulated value of payments 4, 5, 6, 7, and 8 to the end of year 12 at 4% is 2818.55(1.04)4 = $3297.30. The accumulated value of payments 9, 10, 11, and 12 to the end of year 12 at 4% is 500s4 0.04 = 500(4.2465) = $2123.23. The accumulated value of the 12-year annuity immediate is 2541.18 + 3297.30 + 2123.23 = $7961.71 Example 21.2 How much must a person deposit now into a special account in order to withdraw $1,000 at the end of each year for the next ﬁfteen years, if the eﬀective rate of interest is equal to 7% for the ﬁrst ﬁve years, and equal to 9% for the last ten years? Solution. The answer is PV =1000(a5 0.07 + a10 0.09(1.07)−5) =1000(4.1002 + 4.5757) = $8675.90 Example 21.3 Find s5, if δt = 0.02t for 0 ≤t ≤5. Solution. s5 is equal to the sum of the accumulated value of the individual payments. That is, s5 = e R 5 1 δtdt + e R 5 2 δtdt + · · · + 1 Since e R 5 r δtdt = e R 5 r 0.02tdt = e.25−0.01r2 we ﬁnd s5 = e0.24 + e0.21 + e0.16 + e0.09 + 1 ≈5.7726 204 THE BASICS OF ANNUITY THEORY Practice Problem Problem 21.1 James deposits 1,000 into an account at the end of each year for the next 6 years. The account earns 5% interest. James also deposits 1,000 at the end of each of years 7 through 10 into another account earning 4%. Calculate the total amount James will have in both accounts at the end of ten years. Problem 21.2 A fund earns 5% during the next six years and 4% during years 7 through 10. James deposits 1,000 into the account now. Calculate his accumulated value after 10 years. Problem 21.3 A fund earns 5% during the next six years and 4% during years 7 through 10. James deposits 1,000 into the account at the end of each year for the next ten years. Calculate his accumulated value after 10 years. Problem 21.4 A perpetuity pays $1,200 at the beginning of each year with the ﬁrst payment being made imme-diately. The trust funding the perpetuity will earn an annual eﬀective interest rate of 10% for the ﬁrst 10 years, 8% for the second 10 years and 5% thereafter. Calculate the amount needed to fund the perpetuity immediately before the ﬁrst payment. Problem 21.5 You are given: (i) X is the current value at time 2 of a 20-year annuity-due of 1 per annum. (ii) The annual eﬀective interest rate for year t is 1 8+t. Find X. Problem 21.6 Fund A pays interest of 4% on all money deposited in the ﬁrst 5 years and 5% on all money deposited thereafter. Fund B pays interest of 4% during the ﬁrst 5 years and 5% thereafter without regard to the date the deposit was made. (a) If Kevin deposits 500 into each fund now, how much will he have after 10 years in each fund? (b) Heather deposits 100 at the start of each year for 10 years into Fund B. Lisa deposits 100 at the end of each year into Fund A. Who will have more after 10 years and how much more will that person have? 21 VARYING INTEREST OF AN ANNUITY 205 Problem 21.7 A loan P is to be repaid by 10 annual payments beginning 6 months from the date of the loan. The ﬁrst payment is to be half as large as the others. For the ﬁrst 41 2 years interest is i eﬀective; for the remainder of the term interest is j eﬀective. Find an expression for the ﬁrst payment. Problem 21.8 Find the present value of an annuity-immediate which pays 1 at the end of each half-year for ﬁve years, if the rate of interest is 8% convertible semiannually for the ﬁrst three years and 7% convertible semiannually for the last two years. Problem 21.9 Find the present value of an annuity-immediate which pays 1 at the end of each half-year for ﬁve years, if the payments for the ﬁrst three years are discounted at 8% convertible semiannually and the payments for the last two years are discounted at 7% convertible semiannually. Problem 21.10 Given that δt = 1 20−t, t ≥0, ﬁnd s10. Problem 21.11 What is the cost of an annuity of $300 per year for 15 years where the future interest rate for the ﬁrst 5 years is 4%, the rate for the next 5 years is 5%, and the rate for the last 5 years is 6%. Problem 21.12 A perpetuity pays $200 each year, with the ﬁrst payment scheduled one year from now. Assume that the eﬀective annual interest rate for the next ﬁve years is 8%, and thereafter it is 5%. Find the present value of the perpetuity. 206 THE BASICS OF ANNUITY THEORY 22 Annuities Payable at a Diﬀerent Frequency than Interest is Convertible In this section we address annuities for which payment period and the interest conversion period diﬀer and for which the payments are level amount. For example, an annuity-due with monthly payment of $2,000 and with interest rate say 12% convertible quarterly. The approach we use in this section for solving this type of problem consists of the following two steps: (i) Find the rate of interest convertible at the same frequency as payments are made, which is equivalent to the given rate of interest. In the example above, we need to ﬁnd the interest rate convertible monthly which is equivalent to the rate of 12% convertible quarterly. In this case, we ﬁnd i(12) = 12 h1 + 0.12 4 4 12 −1 i = 11.9%. (ii) Using this new interest rate, ﬁnd the value of the annuity using the techniques discussed in the previous sections. We illustrate this approach in the following examples. Example 22.1 You want to accumulate $250,000. You intend to do this by making deposits of $2,000 into an investment account at the beginning of each month. The account earns 12% interest, convertible quarterly. How many months will it take to reach your goal? (Round to the nearest month.) Solution. We are given an interest rate of 12% per quarter. Let j be the equivalent rate of interest per month, which is the payment period. We have (1 + j)12 = 1 + 0.12 4 4 . Solving for j we ﬁnd j = (1.03) 1 3 −1 = 0.9901634%. Thus, 250,000 = 2000¨ sn j →(1+j)n−1 j j+1 = 125 →(1+j)n = 2.22557 → n = ln 2.22557 ln 1.009901634 = 81.195 ≈81 months Example 22.2 A loan of $3,000 is to be repaid with quarterly installments at the end of each quarter for 5 years. If the rate of interest charged on the loan is 10% convertible semi-annually, ﬁnd the amount of each quarterly payment. Solution. We are given an interest rate of 5% per half-year. Let j be the equivalent rate of interest per quarter, which is the payment period. We have (1 + j)4 = (1 + 0.05)2 →j = (1.05) 1 2 −1 = 0.024695. If R denote the quarterly payment then the equation of value at time t = 0 is Ra20 j = 3000. Solving for R we ﬁnd R = 3000 15.6342 = $191.89 22 ANNUITIES PAYABLE AT A DIFFERENT FREQUENCY THAN INTEREST IS CONVERTIBLE207 Example 22.3 The nominal rate of interest, convertible semiannually, is 6%. An annuity-immediate pays $50 each month for ﬁve years. Find the accumulated value of this annuity at the time of the last payment. Solution. We are given an interest rate of 3% per half-year. Let j be the equivalent rate of interest per month, which is the payment period. We have (1 + j)12 = (1 + 0.03)2 →j = (1.03) 1 6 −1 = 0.49386%. The accumulated value is 50s60 j = 50 · (1+j)60−1 j = $3, 481.91 Example 22.4 At what annual eﬀective rate of interest will payments of $100 at the end of every quarter accumulate to $2500 at the end of ﬁve years? Solution. Let j be the eﬀective rate per quarter. We are given that 100s20 j = 2500 or (1 +j)20 −1 = 25j. Let f(j) = (1+j)20 −25j −1. By trial and error we ﬁnd f(0.02) = −0.014053 and f(0.024) = 0.006938. Using linear interpolation we ﬁnd j ≈0.02 + 0.014053 × 0.024 −0.02 0.006938 + 0.014053 = 0.02268. Let i be the annual eﬀective rate of interest. Then i = (1 + 0.02268)4 −1 = 9.39%. Alternatively, we can use the Newton-Raphson iterations to estimate j. Letting j0 = 2(k−n) k(n−1) = 2(25−20) 25(20−1) ≈0.021053 be the starting value of the interations, the ﬁrst three iterations of Newton-Raphson method are j1 =0.021053 1 + 1.02105320 −1 −25(0.021053) (1.021053)19(1 −0.021053 × 19) −1 = 0.02288 j2 =0.02288 1 + 1.0228820 −1 −25(0.02288) (1.02288)19(1 −0.02288 × 19) −1 = 0.02277 j3 =0.02277 1 + 1.0227720 −1 −25(0.02277) (1.02277)19(1 −0.02277 × 19) −1 = 0.02285 j4 =0.02285 1 + 1.0228520 −1 −25(0.02285) (1.02285)19(1 −0.02285 × 19) −1 = 0.02285 Thus, j = 2.285% and consequently i = (1.02285)4 −1 = 0.0946 = 9.46% 208 THE BASICS OF ANNUITY THEORY Practice Problems Problem 22.1 Calculate the present value of an annuity due that pays 500 per month for 10 years. The annual eﬀective interest rate is 6%. Problem 22.2 Calculate the present value of an annuity immediate of 100 per quarter for 6 years using a nominal interest rate of 9% compounded monthly. Problem 22.3 Calculate the accumulated value of an annuity which pays 1,000 at the beginning of each year for 10 years. Use an interest rate of i(12) = 0.08. Problem 22.4 A perpetuity pays 1,000 at the end of each quarter. Calculate the present value using an annual eﬀective interest rate of 10%. Problem 22.5 A perpetuity pays 100 at the beginning of each quarter. Calculate the present value using a force of interest of 0.06. Problem 22.6 A perpetuity due pays 6,000 at the start of each year. Calculate the present value using i(4) = 0.06. Problem 22.7 Find the accumulated value at the end of four years of an investment fund in which $100 is deposited at the beginning of each quarter for the ﬁrst two years and $200 is deposited at the beginning of each quarter for the second two years, if the fund earns 12% convertible monthly. Problem 22.8 A 20-year annuity-due pays $1,000 each year. If i(4) = 8%, ﬁnd the present value of this annuity three years before the ﬁrst payment. Problem 22.9 Find the accumulated value 18 years after the ﬁrst payment is made of an annuity-immediate on which there are 8 payments of $2,000 each made at two-year intervals. The nominal interest rate is 7% convertible semiannually. Round your answer to the nearest dollar. 22 ANNUITIES PAYABLE AT A DIFFERENT FREQUENCY THAN INTEREST IS CONVERTIBLE209 Problem 22.10 Find the present value of a ten-year annuity which pays $400 at the beginning of each quarter for the ﬁrst 5 years, increasing to $600 per quarter thereafter. The annual eﬀective rate of interest is 12%. Round to the nearest dollar. Problem 22.11 A sum of $100 is placed into a fund at the beginning of every other year for eight years. If the fund balance at the end of eight years is $520, ﬁnd the rate of simple interest earned by the fund. Problem 22.12 An annuity pays 100 at the end of each quarter for 10 years. Calculate the present value of the annuity using an annual eﬀective interest rate of 12%. Problem 22.13 An annuity pays 100 at the end of each quarter for 10 years. Calculate the present value of the annuity using a nominal interest rate of 12% compounded monthly. Problem 22.14 An annuity pays 100 at the end of each quarter for 10 years. Calculate the present value of the annuity using a nominal interest rate of 12% compounded six times per year. Problem 22.15 A 30−year annuity pays $1,000 every quarter. If i(12) = 6%, what is the present value of this annuity six months prior to the ﬁrst payment? Problem 22.16 You deposit $2,500 into an account at the beginning of every month for 15 years. The interest rate on the account is i(2) = 8%. Find the accumulated value in the account nine months after the last deposit. Problem 22.17 A perpetuity will make annual payments of $3,000, with the ﬁrst payment occurring 5 months from now. The interest rate is 12% convertible monthly. Find the present value of this annuity. Problem 22.18 You want to accumulate $2,000,000 over the next 30 years. You intend to do this by making deposits of X into an investment account at the end of each month, for 30 years. The account earns 12% convertible semi-annually. Find X. 210 THE BASICS OF ANNUITY THEORY 23 Analysis of Annuities Payable Less Frequently than In-terest is Convertible In this section we analyze annuities where the compounding frequency is larger than the payment frequency. We start with the case of annuity−immediate. Let k be the number of interest conversion periods in one payment period. Consider an annuity−immediate that pays 1 at the end of each payment period. Let i be the rate per conversion period and n the total number of conversion periods for the term of the annuity. We will assume that each payment period contains an integral number of interest conversion periods so that n and k are positive in-tegers and also we assume that n is divisible by k. The total number of annuity payments made is then n k. Let L denote the present value of an annuity−immediate which pays 1 at the end of each k interest conversion periods for a total of n interest conversion periods. A time diagram of this sitatuation is shown in Figure 32.1. Figure 23.1 Using the equation of value at time t = 0 we ﬁnd L =νk + ν2k + · + ν n k ·k =νk(1 + νk + ν2k + · + ν( n k −1)k) =νk × 1 −(νk) n k 1 −νk = νk 1 −νn 1 −νk = 1 −νn (1 + i)k −1 = 1−νn i (1+i)k−1 i = an sk . The accumulated value of this annuity immediately after the last payment is made is (1 + i)nan sk = sn sk . 23 ANALYSIS OF ANNUITIES PAYABLE LESS FREQUENTLY THAN INTEREST IS CONVERTIBLE211 Remark 23.1 The annuity of making a payment of $1 at the end of each k interest conversion periods for a total of n k payments is equivalent to the annuity that consists of a periodic payment of 1 sk for n periods. A time diagram illustrating this annuity is given in Figure 23.2. Figure 23.2 Example 23.1 Find the present value and the accumulated value of an annuity−immediate with payments of $1,500 every 4 years from years 4 to 40. The interest earned is 8% converted annually. Solution. With i = 0.08, the present value is 1500 a40 s4 = 1500 · 11.9246 4.5061 = $3, 969.48 and the accumulated value is (1.08)40(3969.48) = $86, 235.05 Next, we consider the case of an annuity−due. Let ¨ L be the present value of an annuity which pays 1 at the beginning of each k interest conversion periods for a total of n conversion interest periods. Then we have the following time diagram. Figure 23.3 Using the equation of value at time t = 0 we ﬁnd 212 THE BASICS OF ANNUITY THEORY ¨ L =1 + νk + ν2k + · + νn−k =1 + (νk) + (νk)2 + · + (νk) n−k k =1 −(νk) n−k k +1 1 −νk = 1 −νn 1 −νk = 1−νn i 1−νk i = an ak . The accumulated value of this annuity k interest conversion periods after the last payment is (1 + i)nan ak = sn ak . Remark 23.2 It is easy to see that the annuity described above is equivalent to the annuity that consists of n payments of 1 ak at the end of each interest conversion period. Example 23.2 An annuity pays 1,000 at time 0, time 3, time 6, etc until 20 payments have been made. Calculate the present value and the accumulated value using an annual eﬀective interest rate of 6%. Solution. With i = 0.06, the present value is 1000 a60 a3 = 1000· 16.16142771 2.67301194 = 6046.15 and the accumulated value is (1.06)60(6046.15) = 199, 448.48 In the case of a perpetuity−immediate, the present value is νk + ν2k + · · · = νk 1 −νk = 1 (1 + i)k −1 = 1 i · sk = lim n→∞ an sk . Example 23.3 The present value of a perpetuity paying 1 at the end of every 3 years is 125 91 . Find i. 23 ANALYSIS OF ANNUITIES PAYABLE LESS FREQUENTLY THAN INTEREST IS CONVERTIBLE213 Solution. We have 1 i·s3 = 125 91 . Thus, 1 (1+i)3−1 = 125 91 →(1 + i)3 = 6 5 3 →1 + i = 6 5 →i = 6 5 −1 = 0.2 = 20% In the case of a perpetuity-due, the present value is 1 + νk + ν2k + · · · = 1 1 −νk =1 i · i 1 −(1 + i)−k = 1 i · ak = lim n→∞ an ak . Example 23.4 A perpetuity paying 1 at the beginning of each year has a present value of 20. If this perpetuity is exchanged for another perpetuity paying R at the beginning of every 2 years with the same eﬀective annual rate as the ﬁrst perpetuity, ﬁnd R so that the present values of the two perpetuities are equal. Solution. An equation of value now for the ﬁrst perpetuity is 1 d = 20 implying i = 1 19. Also, given that 20 = R ia2 = R 1−ν2 = R (1−ν)(1+ν) = R (0.05)(1.95) →R = (20)(0.0975) = 1.95 Another problem that comes under the category of annuities payable less frequently than inter-est is convertible is the problem of ﬁnding the value of a series of payments at a given force of interest δ. We illustrate this in the next example. Example 23.5 Find an expression for the present value of an annuity on which payments are 100 per quarter for 5 years, just before the ﬁrst payment is made, if δ = 0.08. Solution. Let j be the rate per quarter equivalent to δ. Then (1 + j)4 = e0.08 or 1 + j = e0.02. Thus, PV = 100¨ a20 j = 100 h 1−(1+j)−20 1−(1+j)−1 i = 100 h 1−e−0.40 1−e−0.02 i In some cases, the number of conversion periods in a payment period is not an integral, i.e. k > 1 but k is not an integer. In this case, we handle this problem using the basic principles, i.e. to ﬁnd the present value or the accumulated value we write a sum of present values or accumulated values of the individual payments. An illustration of this type is shown next. 214 THE BASICS OF ANNUITY THEORY Example 23.6 Find an expression for the present value of an annuity on which payments are 1 at the beginning of each 4−month period for 12 years, assuming a rate of interest per 3−month period i. Solution. Let i be the eﬀective rate of interest per 3-month period and j the eﬀective interest rate per 4−month. Then (1 + j)3 = (1 + i)4 ⇒1 + j = (1 + i) 4 3. The present value is given by ¨ a36 j = 1−(1+j)−36 1−(1+j)−1 = 1−ν48 1−ν 4 3 We conclude this section by pointing out that the approach discussed in this section can be general-ized to ﬁnding annuities payable less frequently than interest is convertible on any date, as discussed in Section 17. Example 23.7 Find the present value of an annuity−immediate in which there are a total of 5 payments of 1, the ﬁrst to be made at the end of seven years, and the remaining four payments at three-year intervals, at an annual eﬀective rate of 6%. Solution. The time diagram for this example is given in Figure 23.4. Figure 23.4 The present value is given by ν4a15 s3 = (1.06)−41 −(1.06)−15 (1.06)3 −1 = $2.416 23 ANALYSIS OF ANNUITIES PAYABLE LESS FREQUENTLY THAN INTEREST IS CONVERTIBLE215 Practice Problems Problem 23.1 An annuity pays 1,000 at time 0, time 3, time 6, etc until 20 payments have been made. Calculate the accumulated value immediately after the last payment using an annual eﬀective interest rate of 6%. Problem 23.2 Give an expression of the the present value, 3 years before the ﬁrst payment is made, of an annuity on which there are payments of 200 every 4 months for 12 years: (a) expressed as an annuity-immediate; (b) expressed as an annuity-due. Assume monthly rate of interest. Problem 23.3 Find an expression for the present value of an annuity-due of 600 per annum payable semiannually for 10 years, if d(12) = 0.09. Problem 23.4 An annuity pays 500 at the start of each year for 15 years. Calculate the accumulated value of the annuity after 15 years assuming a constant force of interest of 5%. Problem 23.5 A perpetuity pays 100 at the end of every third year. The ﬁrst payment is in three years. Calculate the present value of the perpetuity using a constant force of interest of 10%. Problem 23.6 Find an expression for the present value of an annuity in which there are a total of r payments of 1, the ﬁrst to be made at the end of seven years, and the remaining payments at three-year intervals, at an annual eﬀective rate i, expressed as (a) annuity-immediate (b) annuity-due. Problem 23.7 A 30-year annuity-immediate with 2,000 payable at the end of every 6 months and a 30-year annuity-immediate with 10,000 payable at the end of every 6 years are to be replaced by a perpetuity paying R every 3 months. You are given that i(4) = .08. Find R. 216 THE BASICS OF ANNUITY THEORY Problem 23.8 The payments you received from a 20 year annuity−immediate paying 500 every 6 months have been left to accumulate in a fund and are now worth 40,000. If i(12) = 0.06 is the rate earned by your fund, calculate how long it is since the last annuity payment was made. Problem 23.9 A perpetuity of $1,000 payable at the end of every 6 months and a perpetuity of $10,000 payable at the end of every 6 years are to be replaced by a 30 year annuity paying R every 3 months. You are given that i(4) = 0.08. Find R. Problem 23.10 An investment of $1,000 is used to make payments of $100 at the end of each year for as along as possible with a smaller payment to be made at the time of the last regular payment. If interest is 7% convertible semiannually, ﬁnd the number of payments and the amount of the total ﬁnal payment. Problem 23.11 Show that the present value at time 0 of 1 payable at times 7, 11, 15, 19, 23, and 27, where the eﬀective rate per annum is i, is given by a28 −a4 s3 + a1 . Problem 23.12 ‡ Fence posts set in soil last 9 years and cost $Y each while fence posts set in concrete last 15 years and cost $(Y + X). The posts will be needed for 35 years. Find an expression of X such that a fence builder would be indiﬀerent between the two types of posts? Problem 23.13 A perpetuity of $750 payable at the end of every year and a perpetuity of $750 payable at the end of every 20 years are to be replaced by an annuity of R payable at the end of every year for 30 years. If i(2) = 0.04, ﬁnd an expression for R. Problem 23.14 Given that δt = 2 10+t, t ≥0, ﬁnd a4. Problem 23.15 At a nominal rate of interest i, convertible semiannually, the present value of a series of payments of 1 at the end of every 2 years forever, is 5.89. Calculate i. 23 ANALYSIS OF ANNUITIES PAYABLE LESS FREQUENTLY THAN INTEREST IS CONVERTIBLE217 Problem 23.16 Gus deposits 25 into a bank account at the beginning of each 3-year period for 18 years (i.e. there is no deposit in year 18). The account credits interest at an annual eﬀective rate of interest of i. The accumulated amount in the account after 18 years is X, which is four times as large as the accumulated amount in the account after 9 years (excluding the deposit made at time t = 9). Calculate X. Problem 23.17 Find the present value of an annuity−due in which there are a total of 5 payments of 1, the ﬁrst to be made at the end of seven years, and the remaining four payments at three-year intervals, at an annual eﬀective rate of 6%. Problem 23.18 The present value today of a 20 year annuity−immediate paying 500 every 6 months but with the ﬁrst payment deferred t years has a present value of 5,805.74. If i(12) = 0.09 is the interest rate used to calculate the present value, ﬁnd t. Problem 23.19 A 20−year annuity−due pays $1,000 each year. If i(4) = 8%, ﬁnd the present value of this annuity three years before the ﬁrst payment. 218 THE BASICS OF ANNUITY THEORY 24 Analysis of Annuities Payable More Frequently than In-terest is Convertible In a discussion parallel to the one in section 23, we discuss annuities payable more frequently than interest is convertible. We ﬁrst consider the case of an annuity-immediate. Let m denote the number of payments per one interest conversion period. Let n be the total number of conversion periods in the term of the annuity. Then the total number of payments for the term of the annuity is mn. Let i be the interest rate per conversion period. We will assume that the number of payments per conversion period is an integral number. Payments of 1 are being made per interest conversion period with 1 m being made at the end of each mth of an interest conversion period. The present value of such an annuity will be denoted by a(m) n . We have the following time diagram. Figure 24.1 The formula for the present value is derived as follows a(m) n = 1 m h ν 1 m + ν 2 m + · · · + νn−1 m + νni = ν 1 m m " 1 −(ν 1 m)mn 1 −ν 1 m # = 1 −νn m h (1 + i) 1 m −1 i = 1 −νn i(m) . The accumulated value of this annuity immediately after the last payment is made is given by s(m) n = (1 + i)na(m) n = (1 + i)n −1 i(m) . Formulas of a(m) n and s(m) n in terms of an and sn are a(m) n = i i(m)an = s(m) 1 an and s(m) n = i i(m)sn = s(m) 1 sn . 24 ANALYSIS OF ANNUITIES PAYABLE MORE FREQUENTLY THAN INTEREST IS CONVERTIBLE219 One consideration which is important in practice involves the proper coeﬃcients for annuity payable mthly. Each payment made is of amount 1 m, while the coeﬃcient of the symbol a(m) n is 1. In general, the proper coeﬃcient is the total amount paid during one interest conversion period, and not the amount of each actual payment. The total amount paid during one interest conversion period is called periodic rent of the annuity. Example 24.1 A loan of $3,000 is to be repaid with quarterly installments at the end of each quarter for ﬁve years. If the rate of interest charged is 10% converted semiannually, ﬁnd the amount of each quarterly payment. Solution. Let R be the amount of quarterly payment. In one interest conversion period there are two payments. Then 2Ra10 (2) 0.05 = 3000 →R = 1500 1−(1.05)−10 2[(1.05)0.5−1] = 191.89 Example 24.2 What is the accumulated value following the last payment of $100 deposited at the end of each month for 30 years into an account that earns an eﬀective annual rate of 10%? Solution. Each conversion period contains 12 payment periods. The term of the loan is 30 interest conversion periods. Thus, 12 · 100 · (1.1)30−1 12[(1.1) 1 12 −1] = 206, 284.33 Example 24.3 The accumulated amount of an annuity-immediate of $R per year payable quarterly for seven years is $3317.25. Find R if i(1) = 0.05. Solution. Each R is paid quarterly at R 4 . The equation of value at time t = 7 years is Rs(4) 7 = 3317.25. But s(4) 7 = i i(4)s7. On the other hand, 1 + i(4) 4 4 = 1.05 ⇒i(4) = 4[(1.05) 1 4 −1] = 0.0491. 220 THE BASICS OF ANNUITY THEORY Hence, we obtain 3317.25 = R 0.05 0.0491(8.1420). Solving for R; we ﬁnd R = $400 In the case of annuity-due, the amount 1 m is payable at the beginning of the mth period of an interest conversion period for a total of n interest conversion period. The present value is denoted by ¨ a(m) n . A time diagram of this case is shown in Figure 24.2. Figure 24.2 The present value is determined as follows: ¨ a(m) n = 1 m[1 + ν 1 m + ν 2 m + · · · + νn−1 m] = 1 m " 1 −(ν 1 m)mn 1 −ν 1 m # = 1 −νn m[1 −(1 + i)−1 m] = 1 −νn d(m) . The accumulated value one mth of an interest conversion period after the last payment is made is given by ¨ s(m) n = (1 + i)n¨ a(m) n = (1 + i)n −1 d(m) . It follows from the formulas of an , sn , ¨ a(m) n , and ¨ s(m) n that ¨ a(m) n = i d(m)an = ¨ s(m) 1 an and ¨ s(m) n = i d(m)sn = ¨ s(m) 1 sn . Example 24.4 What is the present value of the payment of $150 per quarter for 10 years if the eﬀective annual interest rate is 6% and the ﬁrst payment is due today. 24 ANALYSIS OF ANNUITIES PAYABLE MORE FREQUENTLY THAN INTEREST IS CONVERTIBLE22 Solution. The answer is 600· ¨ a10 (4) 0.06 = 600·a10 0.06 · i d(4) = 600 · 7.3601 · 1.037227 = $4, 580.46 Example 24.5 Express ¨ a(12) n in terms of a(2) n with an adjustment factor. Solution. We have ¨ a(12) n = i d(12)an and a(2) n = i i(2)an. Thus, ¨ a(12) n = i(2) d(12)a(2) n The following identities involving an , sn , ¨ a(m) n , and ¨ s(m) n are analogous to the identity involving an , sn , ¨ an , and ¨ sn discussed in earlier sections. Theorem 24.1 (a) 1 a(m) n = 1 s(m) n + i(m) (b) 1 ¨ a(m) n = 1 ¨ s(m) n + d(m) (c) ¨ a(m) n = (1 + i) 1 ma(m) n = i i(m) + i m an (d) ¨ s(m) n = (1 + i) 1 ms(m) n = i i(m) + i m sn (e) ¨ a(m) n = 1 m + a(m) n−1 m (f) ¨ s(m) n = s(m) n+ 1 m −1 m Proof. See Problem 24.10 Next, consider an inﬁnite payment of 1 m at the end of mth of an interest conversion period. Let a(m) ∞ denote the present value. A time diagram describing this case is shown in Figure 24.3. 222 THE BASICS OF ANNUITY THEORY The present value is determined as follows. a(m) ∞= 1 m[ν 1 m + ν 2 m + ν 3 m + · · · ] = 1 m ∞ X p=1 ν p m = 1 m ν 1 m 1 −ν 1 m = 1 m(1 + i) 1 m(1 −ν 1 m) = 1 m((1 + i) 1 m −1) = 1 i(m) Figure 24.3 Example 24.6 An annuity pays 100 at the end of each month forever. Calculate the present value of the annuity using an annual eﬀective interest rate of 8%. Solution. The present value is 12 · 100a(12) ∞ = 1200 12[(1.08) 1 12 −1] = 15, 542.36 Now, consider an inﬁnite payment of 1 m at the beginning of mth of an interest conversion pe-riod. Let the present value be denoted by ¨ a(m) ∞. Figure 24.4 describes this situation. Figure 24.4 The present value is determined as follows. ¨ a(m) ∞= 1 m[1 + ν 1 m + ν 2 m + ν 3 m + · · · ] = 1 m ∞ X p=0 ν p m = 1 m[1 −ν 1 m] = 1 d(m), ν 1 m < 1 24 ANALYSIS OF ANNUITIES PAYABLE MORE FREQUENTLY THAN INTEREST IS CONVERTIBLE223 Example 24.7 At what annual eﬀective rate of interest is the present value of a series of payments of $1 for every six months forever, with the ﬁrst payment made now, equal to $10? Solution. We have 10 = 2¨ a(2) ∞= 2 d(2) →d(2) = 0.2 →i = (1 −0.1)−2 −1 = 23.46% In some cases, each conversion period does not contain an integral number of payments, i.e. m > 1 but m is not an integer. In this case, we handle this problem using the basic principles, i.e. to ﬁnd the present value or the accumulated value we write a sum of present values or accumulated values of the individual payments. An illustration of this type is shown next. Example 24.8 Find an expression for the present value of an annuity which pays 1 at the beginning of each 3-month period for 12 years, assuming a rate of interest per 4-month period. Solution. Working in years the annual payments are 4 × 1 = $4. Thus, the present value is 4¨ a(4) 12 = 4 · 1 −(1 + i)−12 d(4) . But 1 + i = 1 −d(4) 4 −4 ⇒d(4) = 4[1 −(1 + i)−1 4]. Thus 4¨ a(4) 10 =4 1 −(1 + i)−12 4[1 −(1 + i)−1 4] =1 −ν12 1 −ν 1 4 =1 −ν48 j 1 −νj =1 −ν36 k 1 −ν 3 4 k where j is the eﬀective interest rate per 3-month and k is the eﬀective interest rate per 4-month Finally, it is possible to generalize the approach to ﬁnding annuity values on any date, as discussed in Section 17, to annuities payable more frequently than interest is convertible. 224 THE BASICS OF ANNUITY THEORY Example 24.9 Payments of $400 at the end of each month are made over a ten-year period. Find expressions for (a) the present value of these payments two years prior to the ﬁrst payment; (b) the accumulated value three years after the ﬁnal payment. Use symbols based on an eﬀective rate of interest i. Solution. (a) The present value is given by 12 · 400νa(12) 10 . (b) The accumulated value is given by 12 · 400s(12) 10 (1 + i)3 24 ANALYSIS OF ANNUITIES PAYABLE MORE FREQUENTLY THAN INTEREST IS CONVERTIBLE225 Practice Problems Problem 24.1 The present value of an annuity is denoted by 120a(12) n . What is the amount of the monthly payment of this annuity? Problem 24.2 The present value of an annuity that pays 100 at the end of each year for n years using an annual eﬀective interest rate of 10.25% is 1,000. Calculate the present value of an annuity that pays 100 at the end of every six months for n years using the same interest rate. (Note: You should work this problem without ﬁnding n.) Problem 24.3 Suppose you deposit $200 per month, at the end of each month, in a savings account paying the eﬀective annual rate of interest 6%. How much will be in the account after ten years? (i.e. just after your 120th deposit?) Problem 24.4 A loan of $150,000 is repaid by equal installments at the end of each month, for 25 years. Given that the nominal annual rate of interest, convertible semiannually, is 8%, compute the amount of monthly installment. Problem 24.5 Suppose you deposit $300 at the end of every three months for 10 years into a bank account paying the annual eﬀective rate of interest 4%. What is the accumulated amount after 10 years? Problem 24.6 An annuity pays 100 at the end of each month for 20 years. Using a nominal rate of interest of 4% compounded quarterly, calculate the current value of the annuity at the end of the 3rd year. Problem 24.7 Which of the following are true? (i) an = i(m) i a(m) n (ii) s(m) n = (1 + i)na(m) n (iii) a(∞) ∞ = ¨ a(∞) ∞ Problem 24.8 If 3a(2) n = 2a(2) 2n = 45s(2) 1 , ﬁnd i. 226 THE BASICS OF ANNUITY THEORY Problem 24.9 A sum of 10,000 is used to buy a deferred perpetuity-due paying $500 every six months forever. Find an expression for the deferred period expressed as a function of d, the annual discount rate. Problem 24.10 Prove the following identities: (a) 1 a(m) n = 1 s(m) n + i(m) (b) 1 ¨ a(m) n = 1 ¨ s(m) n + d(m) (c) ¨ a(m) n = (1 + i) 1 ma(m) n = i i(m) + i m an (d) ¨ s(m) n = (1 + i) 1 ms(m) n = i i(m) + i m sn (e) ¨ a(m) n = 1 m + a(m) n−1 m (f) ¨ s(m) n = s(m) n+ 1 m −1 m Problem 24.11 Find the present value of a ten-year annuity which pays $400 at the beginning of each quarter for the ﬁrst 5 years, increasing to $600 per quarter thereafter. The annual eﬀective rate of interest is 12%. Round to the nearest dollar. Use the approach developed in this section. Problem 24.12 A family wishes to provide an annuity of $100 at the end of each month to their daughter now entering college. The annuity will be paid for only nine months each year for four years. Prove that the present value one month before the ﬁrst payment is 1200¨ a4a(12) 9/12. Problem 24.13 The nominal rate of interest, convertible semiannually, is 6%. An annuity−immediate pays $50 each month for ﬁve years. Find the accumulated value of this annuity at the time of the last payment. Problem 24.14 You deposit $2,500 into an account at the beginning of every month for 15 years. The interest rate on the account is i(2) = 8%. Find the accumulated value in the account nine months after the last deposit. 24 ANALYSIS OF ANNUITIES PAYABLE MORE FREQUENTLY THAN INTEREST IS CONVERTIBLE22 Problem 24.15 You want to accumulate $1,000,000. You intend to do this by making deposits of $5,000 into an investment account at the end of each month, until your account balance equals $1,000,000. The account earns 8% convertible quarterly. Determine the number of monthly deposits you will need to make to achieve your goal. 228 THE BASICS OF ANNUITY THEORY 25 Continuous Annuities In this section we consider annuities with a ﬁnite term and an inﬁnte frequency of payments . Formulas corresponding to such annuities are useful as approximations corresponding to annuities payable with great frequency such as daily. Consider an annuity in which a very small payment dt is made at time t and these small payments are payable continuously for n interest conversion periods. Let i denote the periodic interest rate. Then the total amount paid during each period is Z k k−1 dt = [t]k k−1 = $1. Let an denote the present value of an annuity payable continuously for n interest conversion periods so that 1 is the total amount paid during each interest conversion period. Then the present value can be found as follows: an = Z n 0 νtdt = νt ln ν n 0 =νn −1 ln ν = νn −1 −ln (1 + i) =1 −νn δ . It is easy to see the following an = lim m→∞a(m) n = lim m→∞ 1 −νn i(m) = 1 −νn δ since lim m→∞i(m) = δ. Similarly, an = lim m→∞¨ a(m) n = lim m→∞ 1 −νn d(m) = 1 −νn δ since lim m→∞d(m) = δ. Moreover, an = i δan = d δ ¨ an = 1 −e−nδ δ . 25 CONTINUOUS ANNUITIES 229 Example 25.1 Starting four years from today, you will receive payment at the rate of $1,000 per annum, payable continuously, with the payment terminating twelve years from today. Find the present value of this continuous annuity if δ = 5%. Solution. The present value is PV = 1000v4 · a8 = 1000e−0.20 · 1−e−0.40 0.05 = $5, 398.38 The next example exhibit the situation when the constant force above is replaced by a variable force of interest. Example 25.2 Find an expression for an if δt = 1 1+t. Solution. The present value of the payment dt at the exact time t with variable force of interest is e− R t 0 δrdrdt = e−ln (1+t)dt = (1 + t)−1dt. Hence, an = Z n 0 dt 1 + t = ln (1 + t)\|n 0 = ln (1 + n) Next, let sn denote the accumulated value at the end of the term of an annuity payable continuously for n interest conversion periods so that 1 is the total amound paid during each interest conversion period. Then sn =(1 + i)nan = Z n 0 (1 + i)n−tdt = −(1 + i)n−t ln (1 + i) n 0 =(1 + i)n −1 δ . It is easy to see that sn = lim m→∞s(m) n = lim m→∞¨ s(m) n = enδ −1 δ = i δsn = d δ ¨ sn. Example 25.3 Find the force of interest at which the accumulated value of a continuous payment of 1 every year for 8 years will be equal to four times the accumulated value of a continuous payment of 1 every year for four years. 230 THE BASICS OF ANNUITY THEORY Solution. We have s8 =4s4 e8δ −1 δ =4 · e4δ −1 δ e8δ −4e4δ + 3 =0 (e4δ −3)(e4δ −1) =0 If e4δ = 3 then δ = ln 3 4 ≈0.275 = 27.5%. If e4δ = 1 then δ = 0, an extraneous solution Example 25.4 The annual eﬀective rate is i = 5%. Find the accumulated value after six years of an annuity that oﬀers payments of $1,000 per annum, convertible continuously for ﬁve years. After ﬁve years, payments terminate, but the balance still earns interest during the sixth year. Solution. The accumulated value is AV = 1000s5 (1.05). But δ = ln (1 + i) = ln 1.05 = 0.0487902. Thus, AV = 1000 · (1.05)5−1 0.0487902 · (1.05) = $5, 945.78 The present value of a perpetuity payable continuously with total of 1 per period is given by a∞= lim n→∞an = 1 δ. Example 25.5 A perpetuity paid continuously at a rate of 100 per year has a present value of 800. Calculate the annual eﬀective interest rate used to calculate the present value. Solution. The equation of value at time t = 0 is 800 = 100 δ = 100 ln (1 + i). Thus, i = e 1 8 −1 = 13.3% 25 CONTINUOUS ANNUITIES 231 Practice Problems Problem 25.1 Calculate the present value of a continuous annuity of 1,000 per annum for 8 years at: (a) An annual eﬀective interest rate of 4%; (b) A constant force of interest of 4%. Problem 25.2 Find the force of interest at which s20 = 3s10 . Problem 25.3 If an = 4 and sn = 12, ﬁnd δ. Problem 25.4 There is $40,000 in a fund which is accumulating at 4% per annum convertible continuously. If money is withdrawn continuously at the rate of $2,400 per annum, how long will the fund last? Problem 25.5 Annuity A oﬀers to pay you $100 per annum, convertible continuously, for the next ﬁve years. Annuity B oﬀers you $ X at the end of each year for ten years. The annual eﬀective interest rate i is 8%. Find X such that you are indiﬀerent between the two annuities. Problem 25.6 Given δ = 0.1. Evaluate a(12) 10 a10 . Problem 25.7 You are given d dtst = (1.02)2t. Calculate δ. Problem 25.8 Given an = n −4 and δ = 10%, ﬁnd R n 0 at dt. Problem 25.9 Payments will be made to you at a continuous rate of $100 per year for the next ﬁve years. The eﬀective rate of interest over this period is i = 0.06. Calculate the present value of this payment stream. Problem 25.10 Show that d dtat = νt s1 . 232 THE BASICS OF ANNUITY THEORY Problem 25.11 Show that an < a(m) n < an < ¨ a(m) n < ¨ an. Hint: See Example 10.15. Problem 25.12 Find an expression for t, 0 < t < 1, such that 1 paid at time t is equivalent to 1 paid continuously between 0 and 1. Problem 25.13 A bank makes payments continuously at a rate of $400 a year. The payments are made between 5 and 7 years. Find the current value of these payments at time 2 years using an annual rate of discount of 4%. Problem 25.14 A company makes payments continuously at a rate of $200 per year. The payments are made between 2 and 7 years. Find the accumulated value of these payments at time 10 years using an annual rate of interest of 6.5%. Problem 25.15 Lauren is being paid a continuous perpetuity payable at a rate of 1000 per year. Calculate the present value of the perpetuity assuming d(12) = 0.12. Problem 25.16 If i = 0.04, calculate the accumulated value of a continuous annuity payable at a rate of 100 per year for 10 years. Problem 25.17 If δ = 0.06, calculate the present value of a continuous annuity of 1 payable for 20 years. Problem 25.18 You are given R n 0 atdt = 100. Calculate an. Problem 25.19 An n−year continuous annuity that pays at a rate of $748 per year has a present value of $10,000 when using an interest rate of 6%, compounded continuously. Determine n. Problem 25.20 Which of the following are true? (I) an −d δ (1 + i) = an−1. 25 CONTINUOUS ANNUITIES 233 (II) The present value of a 10 year annuity immediate paying 10 per month for the ﬁrst eight months of each year is 120a10a(12) 8/12. (III) The present value of a perpetuity paying one at the end of each year, except paying nothing every fourth year, is s3 is4 . Problem 25.21 Payments of $3650 per year are made continuously over a ﬁve-year period. Find the present value of this continuous annuity two years prior to the ﬁrst payment. Given the nominal rate i(365) = 8%. 234 THE BASICS OF ANNUITY THEORY 26 Varying Annuity-Immediate Thus far in this book, all annuities that we have considered had level series of payments, that is, payments are all equal in values. In this and the next three sections we consider annuities with a varying series of payments. In this section, we assume that the payment period and interest conversion period coincide. Annuities with varying payments will be called varying annuities. Any type of annuities can be evaluated by taking the present value or the accumulated value of each payment seperately and adding the results. There are, however, several types of varying annuities for which relatively simple compact expressions are possible. The only general types we will study vary in either arithmetic progression or geometric progression. Payments Varying in an Arithmetic Progression First, let us assume that payments vary in arithmetic progression. That is, the ﬁrst payment is P and then the payments increase by Q thereafter, continuing for n years as shown in the time diagram of Figure 26.1. Figure 26.1 If PV is the present value for this annuity-immediate, then PV = Pν + (P + Q)ν2 + (P + 2Q)ν3 + · · · + [P + (n −1)Q]νn. (26.1) Multiplying Equation (26.1) by (1 + i), we obtain (1 + i)PV = P + (P + Q)ν + (P + 2Q)ν2 + · · · + [P + (n −1)Q]νn−1. (26.2) Subtracting Equation(26.1) from Equation (26.2) we obtain iPV = P(1 −νn) + (ν + ν2 + · · · + νn)Q −nνnQ or PV = Pan + Q[an −nνn] i . (26.3) The accumulated value of these payments at time n is AV = (1 + i)nPV = Psn + Q[sn −n] i . (26.4) 26 VARYING ANNUITY-IMMEDIATE 235 Two special cases of the above varying annuity often occur in practice. The ﬁrst of these is the increasing annuity where P = Q = 1 as shown in Figure 26.2. Figure 26.2 The present value of such an annuity is (Ia)n = an + an −nνn i = (1 + i)an −nνn i = ¨ an −nνn i . (26.5) The accumulated value at time n is given by (Is)n = (1 + i)n(Ia)n = ¨ sn −n i = sn+1 −(n + 1) i . Example 26.1 The following payments are to be received: $500 at the end of the ﬁrst year, $520 at the end of the second year, $540 at the end of the third year and so on, until the ﬁnal payment is $800. Using an annual eﬀective interest rate of 2% (a) determine the present value of these payments at time 0; (b) determine the accumulated value of these payments at the time of the last payment. Solution. In n years the payment is 500 + 20(n −1). So the total number of payments is 16. The given payments can be regarded as the sum of a level annuity immediate of $480 and an increasing annuity-immediate $20, $40, · · · , $320. (a) The present value at time t = 0 is 480a16 + 20(Ia)16 = 480(13.5777) + 20(109.7065) = $8, 711.43. (b) The accumulated value at time t = 16 is 480s16 + 20(Is)16 = 480(18.6393) + 20(150.6035) = $11, 958.93 Example 26.2 Show that (Ia)n = Pn−1 t=0 νtan−t. 236 THE BASICS OF ANNUITY THEORY Solution. We have (Ia)n = n−1 X t=0 νtan−t = n−1 X t=0 νt1 −νn−t i =1 i n−1 X t=0 νt −νn i n−1 X t=0 1 =¨ an i −nνn i = ¨ an −nνn i . The second special case is the decreasing annuity-immediate where P = n and Q = −1 as shown in Figure 26.3. Figure 26.3 In this case, the present value one year before the ﬁrst payment (i.e., at time t = 0) is given by (Da)n = nan −an −nνn i = n −nνn −an + nνn i = n −an i . (26.6) The accumulated value at time n is given by (Ds)n = (1 + i)n(Da)n = n(1 + i)n −sn i = (n + 1)an −(Ia)n . Example 26.3 John receives $400 at the end of the ﬁrst year, $350 at the end of the second year, $300 at the end of the third year and so on, until the ﬁnal payment of $50. Using an annual eﬀective rate of 3.5%, calculate the present value of these payments at time 0. Solution. In year n the payment is 400−50(n−1). Since the ﬁnal payment is 50 we must have 400−50(n−1) = 50. Solving for n we ﬁnd n = 8. Thus, the present value is 50(Da)8 = 50 · 8 −a8 0.035 = $1, 608.63 26 VARYING ANNUITY-IMMEDIATE 237 Example 26.4 Calculate the accumulated value in Example 26.3. Solution. The answer is 50(Ds)8 = 50 · 8(1.035)8−s8 0.035 = $2, 118.27 Example 26.5 Show that (Ia)n + (Da)n = (n + 1)an. Solution. We have (Ia)n + (Da)n =an + an −nνn i + n −an i =an + nan = (n + 1)an Besides varying annuities immediate, it is also possible to have varying perpetuity−immediate. Consider a perpetuity−immediate with payments that form an arithmetic progression (and of course P > 0 and Q > 0). The present value for such a perpetuity with the ﬁrst payment at the end of the ﬁrst period is PV = lim n→∞ Pan + Q[an −nνn] i == P lim n→∞an + Q lim n→∞ [an −nνn] i =P lim n→∞an + Q[limn→∞an −limn→∞nνn] i =Pa∞+ Qa∞ i = P i + Q i2 since a∞= 1 i and limn→∞nνn = 0 (by L’Hopital’s rule). For the special case P = Q = 1 we ﬁnd (Ia)∞= 1 i + 1 i2. Example 26.6 Find the present value of a perpetuity-immediate whose successive payments are 1, 2, 3, 4,· · · at an eﬀective rate of 6%. Solution. We have (Ia)∞= 1 i + 1 i2 = 1 0.06 + 1 0.062 = $294.44 238 THE BASICS OF ANNUITY THEORY Payments Varying in a Geometric Progression Next, we consider payments varying in a geometric progression. Consider an annuity-immediate with a term of n periods where the interest rate is i per period, and where the ﬁrst payment is 1 and successive payments increase in geometric progression with common ratio 1 + k. The present value of this annuity is ν + ν2(1 + k) + ν3(1 + k)2 + · · · + νn(1 + k)n−1 =ν · 1 −[(1 + k)ν]n 1 −(1 + k)ν = 1 1 + i · 1 − 1+k 1+i n 1 − 1+k 1+i = 1 − 1+k 1+i n i −k provided that k ̸= i. If k = i then the original sum consists of a sum of n terms of ν which equals to nν. Example 26.7 The ﬁrst of 30 payments of an annuity occurs in exactly one year and is equal to $500. The payments increase so that each payment is 5% greater than the preceding payment. Find the present value of this annuity with an annual eﬀective rate of interest of 8%. Solution. The present value is given by PV = 500 · 1 − 1.05 1.08 30 0.08 −0.05 = $9, 508.28 For an annuity-immediate with a term of n periods where the interest rate is i per period, and where the ﬁrst payment is 1 and successive payments decrease in geometric progression with common ratio 1 −k. The present value of this annuity is ν + ν2(1 −k) + ν3(1 −k)2 + · · · + νn(1 −k)n−1 =ν · 1 −[(1 −k)ν]n 1 −(1 −k)ν = 1 1 + i · 1 − 1−k 1+i n 1 − 1−k 1+i = 1 − 1−k 1+i n i + k 26 VARYING ANNUITY-IMMEDIATE 239 provided that k ̸= i. If k = i then the original sum becomes ν + ν2(1 −i) + ν3(1 −i)2 + · · · + νn(1 −i)n−1 = 1 2i 1 − 1 −i 1 + i n Finally, we consider a perpetuity with payments that form a geometric progression where 0 < 1 + k < 1 + i. The present value for such a perpetuity with the ﬁrst payment at the end of the ﬁrst period is ν + ν2(1 + k) + ν3(1 + k)2 + · · · = ν 1 −(1 + k)ν = 1 i −k. Observe that the value for these perpetuities cannot exist if 1 + k ≥1 + i. Example 26.8 What is the present value of a stream of annual dividends, which starts at 1 at the end of the ﬁrst year, and grows at the annual rate of 2%, given that the rate of interest is 6% ? Solution. The present value is 1 i−k = 1 0.06−0.02 = 25 240 THE BASICS OF ANNUITY THEORY Practice Problems Problem 26.1 Smith receives $400 in 1 year, $800 in 2 years, $1,200 in 3 years and so on until the ﬁnal payment of $4,000. Using an eﬀective interest rate of 6%, determine the present value of these payments at time 0. Problem 26.2 Find an expression for the present value at time 0 of payments of $55 at time 1 year, $60 at time 2 years, $65 at time 3 years and so on, up to the last payment at time 20 years. Problem 26.3 Find an expression for the accumulated value at time 20 years of payments of $5 at time 1 year, $10 at time 2 years, $15 at time 3 years, and so on, up to $100 at time 20 years. Problem 26.4 Find the present value of a perpetuity-immediate whose successive payments are $5, $10, $15, · · · assuming an annual eﬀective rate of interest of 5%. Problem 26.5 An annuity pays $100 at the end of one month. It pays $110 at the end of the second month. It pays $120 at the end of the third month. The payments continue to increase by $10 each month until the last payment is made at the end of the 36th month. Find the present value of the annuity at 9% compounded monthly. Problem 26.6 An annual annuity-immediate pays $100 at the end of the ﬁrst year. Each subsequent payment is 5% greater than the preceding payment. The last payment is at the end of the 20th year. Calculate the accumulated value at: (a) an annual eﬀective interest rate of 4%; (b) an annual eﬀective interest rate of 5%. Problem 26.7 A perpetuity pays $100 at the end of the ﬁrst year. Each subsequent annual payment increases by $50. Calculate the present value at an annual eﬀective interest rate of 10%. Problem 26.8 A small business pays you an annual proﬁt at the end of each year for 20 years. The payments grow at an annual rate of 2.5%, the ﬁrst payment is $10,000. What is the present value of this stream of payments at the annual eﬀective rate of interest 6%? 26 VARYING ANNUITY-IMMEDIATE 241 Problem 26.9 What is the present value of a 30 year immediate annuity at 10% interest where the ﬁrst payment is $100 and each payment thereafter is increased by 5%? Problem 26.10 A perpetuity-immediate has annual payments of 1, 3, 5, 7, · · · . If the present values of the 6th and 7th payments are equal, ﬁnd the present value of the perpetuity. Problem 26.11 If X is the present value of a perpetuity of 1 per year with the ﬁrst payment at the end of the 2nd year and 20X is the present value of a series of annual payments 1, 2, 3, · · · with the ﬁrst payment at the end of the 3rd year, ﬁnd d. Problem 26.12 There are two perpetuities-immediate. The ﬁrst has level payments of p at the end of each year. The second is increasing such that the payments are q, 2q, 3q, · · · . Find the rate of interest that will make the diﬀerence in present value between these perpetuities a) zero; b) a maximum. Problem 26.13 An annuity pays 10 at the end of the ﬁrst year, 20 at the end of the second year, 30 at the end of the third year, etc. The last payment is made at the end of the 12th year. Calculate the accumulated value of this annuity immediately after the last payment using an annual eﬀective rate of 4%. Problem 26.14 A 20 year annuity-immediate pays 500 + 50t at the end of year t. Calculate the present value of this annuity using an annual eﬀective rate of 6%. Problem 26.15 An annuity pays 10 at the end of year 2, and 9 at the end of year 4. The payments continue decreasing by 1 each two year period until 1 is paid at the end of year 20. Calculate the present value of the annuity at an annual eﬀective interest rate of 5%. Problem 26.16 Find an expression for the present value of a perpetuity under which a payment of 1 is made at the end of the 1st year, 2 at the end of the 2nd year, increasing until a payment of n is made at the end of the nth year, and thereafter payments are level at n per year forever. 242 THE BASICS OF ANNUITY THEORY Problem 26.17 Find an expression for the present value of an annuity-immediate where payments start at 1, increase by 1 each period up to a payment of n, and then decrease by 1 each period up to a ﬁnal payment of 1. Problem 26.18 An 11-year annuity has a series of payments 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1, with the ﬁrst payment made at the end of the second year. The present value of this annuity is 25 at interest rate i. A 12-year annuity has a series of payments 1, 2, 3, 4, 5, 6, 6, 5, 4, 3, 2, 1, with the ﬁrst payment made at the end of the ﬁrst year. Calculate the present value of the 12-year annuity at interest rate i. Problem 26.19 Olga buys a 5-year increasing annuity for X. Olga will receive 2 at the end of the ﬁrst month, 4 at the end of the second month, and for each month thereafter the payment increases by 2 . The nominal interest rate is 9% convertible quarterly. Calculate X. Problem 26.20 An annuity-immediate has semiannual payments of 800, 750, 700, · · · , 350, at i(2) = 0.16. If a10 0.08 = A, ﬁnd the present value of the annuity in terms of A. Problem 26.21 ‡ A perpetuity costs 77.1 and makes annual payments at the end of the year. The perpetuity pays 1 at the end of year 2, 2 at the end of year 3,· · · , n at the end of year (n + 1). After year (n + 1), the payments remain constant at n. The annual eﬀective interest rate is 10.5%. Calculate n. Problem 26.22 ‡ A perpetuity-immediate pays 100 per year. Immediately after the ﬁfth payment, the perpetuity is exchanged for a 25-year annuity-immediate that will pay X at the end of the ﬁrst year. Each subsequent annual payment will be 8% greater than the preceding payment. The annual eﬀective rate of interest is 8%. Calculate X. Problem 26.23 ‡ A perpetuity-immediate pays 100 per year. Immediately after the ﬁfth payment, the perpetuity is exchanged for a 25-year annuity-immediate that will pay X at the end of the ﬁrst year. Each subsequent annual payment will be 8% greater than the preceding payment. Immediately after the 10th payment of the 25-year annuity, the annuity will be exchanged for a perpetuity-immediate paying Y per year. The annual eﬀective rate of interest is 8%. Calculate Y. 26 VARYING ANNUITY-IMMEDIATE 243 Problem 26.24 ‡ Mike buys a perpetuity-immediate with varying annual payments. During the ﬁrst 5 years, the payment is constant and equal to 10. Beginning in year 6, the payments start to increase. For year 6 and all future years, the current year’s payment is K% larger than the previous year’s payment. At an annual eﬀective interest rate of 9.2%, the perpetuity has a present value of 167.50. Calculate K, given K < 9.2. Problem 26.25 ‡ A company deposits 1000 at the beginning of the ﬁrst year and 150 at the beginning of each subsequent year into perpetuity. In return the company receives payments at the end of each year forever. The ﬁrst payment is 100. Each subsequent payment increases by 5%. Calculate the company’s yield rate for this transaction. Problem 26.26 ‡ Megan purchases a perpetuity-immediate for 3250 with annual payments of 130. At the same price and interest rate, Chris purchases an annuity-immediate with 20 annual payments that begin at amount P and increase by 15 each year thereafter. Calculate P. Problem 26.27 ‡ The present value of a 25-year annuity-immediate with a ﬁrst payment of 2500 and decreasing by 100 each year thereafter is X. Assuming an annual eﬀective interest rate of 10%, calculate X. Problem 26.28 ‡ The present value of a series of 50 payments starting at 100 at the end of the ﬁrst year and increasing by 1 each year thereafter is equal to X. The annual eﬀective rate of interest is 9%. Calculate X. Problem 26.29 ‡ An annuity-immediate pays 20 per year for 10 years, then decreases by 1 per year for 19 years. At an annual eﬀective interest rate of 6%, the present value is equal to X. Calculate X. Problem 26.30 ‡ At an annual eﬀective interest rate of i, the present value of a perpetuity-immediate starting with a payment of 200 in the ﬁrst year and increasing by 50 each year thereafter is 46,530. Calculate i. Problem 26.31 ‡ An insurance company has an obligation to pay the medical costs for a claimant. Average annual claims costs today are $5,000, and medical inﬂation is expected to be 7% per year. The claimant is expected to live an additional 20 years. Claim payments are made at yearly intervals, with the ﬁrst claim payment to be made one year from today. Find the present value of the obligation if the annual interest rate is 5%. 244 THE BASICS OF ANNUITY THEORY Problem 26.32 ‡ Sandy purchases a perpetuity-immediate that makes annual payments. The ﬁrst payment is 100, and each payment thereafter increases by 10. Danny purchases a perpetuity-due which makes annual payments of 180. Using the same annual eﬀective interest rate, i > 0, the present value of both perpetuities are equal. Calculate i. Problem 26.33 ‡ Joe can purchase one of two annuities: Annuity 1: A 10-year decreasing annuity-immediate, with annual payments of 10, 9, 8, · · · , 1. Annuity 2: A perpetuity-immediate with annual payments. The perpetuity pays 1 in year 1, 2 in year 2, 3 in year 3, · · · , and 11 in year 11 . After year 11, the payments remain constant at 11. At an annual eﬀective interest rate of i, the present value of Annuity 2 is twice the present value of Annuity 1 . Calculate the value of Annuity 1. Problem 26.34 ‡ Mary purchases an increasing annuity-immediate for 50,000 that makes twenty annual payments as follows: i) P, 2P, · · · , 10P in years 1 through 10; and ii) 10(1.05)P, 10(1.05)2P, · · · , 10(1.05)10P in years 11 through 20. The annual eﬀective interest rate is 7% for the ﬁrst 10 years and 5% thereafter. Calculate P. Problem 26.35 Find the present value of an annuity-immediate such that payments start at 1, each payment thereafter increases by 1 until reaching 10, and then remain at that level until 25 payments in total are made. Problem 26.36 Find an expression of the present value of an annuity that pays 10 at the end of the ﬁfth year and decreases by 1 thereafter until reaching 0. Problem 26.37 Annual deposits are made into a fund at the beginning of each year for 10 years. The ﬁrst ﬁve deposits are $1000 each and deposits increase by 5% per year thereafter. If the fund earns 8% eﬀective, ﬁnd the accumulated value at the end of 10 years. Problem 26.38 Find the present value of a 20-year annuity-immediate with ﬁrst payment of $600 and each subse-quent payment is 5% greater than the preceding payment. The annual eﬀective rate of interest is 10.25%. 26 VARYING ANNUITY-IMMEDIATE 245 Problem 26.39 A perpetuity pays 200 at the end of each of the ﬁrst two years, 300 at the end of years 3 and 4, 400 at the end of years 5 and 6, etc. Calculate the present value of this perpetuity if the annual eﬀective rate of interest is 10%. Problem 26.40 Janice is receiving a perpetuity of 100 at the end of each year. Megan is receiving a perpetuity that pays 10 at the end of the ﬁrst year, 20 at the end of the second year, 30 at the end of the third year, etc. The present values of the two perpetuities are equal. Calculate the annual eﬀective interest rate used to determine the present values. Problem 26.41 Gloria borrows 100,000 to be repaid over 30 years. You are given (1) Her ﬁrst payment is X at the end of year 1 (2) Her payments increase at the rate of 100 per year for the next 19 years and remain level for the following 10 years (3) The eﬀective rate of interest is 5% per annum Calculate X. Problem 26.42 You are given a perpetual annuity immediate with annual payments increasing in geometric pro-gression, with a common ratio of 1.07. The annual eﬀective interest rate is 12%. The ﬁrst payment is 1. Calculate the present value of this annuity. Problem 26.43 An annuity immediate pays 10 at the ends of years 1 and 2, 9 at the ends of years 3 and 4, etc., with payments decreasing by 1 every second year until nothing is paid. The eﬀective annual rate of interest is 5%. Calculate the present value of this annuity immediate. Problem 26.44 Barbara purchases an increasing perpetuity with payments occurring at the end of every 2 years. The ﬁrst payment is 1, the second one is 2, the third one is 3, etc. The price of the perpetuity is 110. Calculate the annual eﬀective interest rate. Problem 26.45 You are given (Ia)n−1 = K d , where d is the annual eﬀective discount rate. Calculate K. Problem 26.46 Francois purchases a 10 year annuity immediate with annual payments of 10X. Jacques purchases 246 THE BASICS OF ANNUITY THEORY a 10 year decreasing annuity immediate which also makes annual payments. The payment at the end of year 1 is equal to 50. At the end of year 2, and at the end of each year through year 10, each subsequent payment is reduced over what was paid in the previous year by an amount equal to X. At an annual eﬀective interest rate of 7.072%, both annuities have the same present value. Calculate X, where X < 5. Problem 26.47 ‡ 1000 is deposited into Fund X, which earns an annual eﬀective rate of 6%. At the end of each year, the interest earned plus an additional 100 is withdrawn from the fund. At the end of the tenth year the fund is depleted. The annual withdrawals of interest and principal are deposited into Fund Y, which earns an annual eﬀective rate of 9%. Determine the accumulated value of Fund Y at the end of year 10. Problem 26.48 You are given two series of payments. Series A is a perpetuity with payments of 1 at the end of each of the ﬁrst 2 years, 2 at the end of each of the next 2 years, 3 at the end of each of the next 2 years, and so on. Series B is a perpetuity with payments of K at the end of each of the ﬁrst 3 years, 2K at the end of the next 3 years, 3K at the end of each of the next 3 years, and so on. The present values of the two series of payments are equal. Calculate K. Problem 26.49 John deposits 100 at the end of each year for 20 years into a fund earning an annual eﬀective interest rate of 7%. Mary makes 20 deposits into a fund at the end of each year for 20 years. The ﬁrst 10 deposits are 100 each, while the last 10 deposits are 100 + X each. The fund earns an annual eﬀective interest rate of 8% during the ﬁrst 10 years and 6% annual eﬀective interest thereafter. At the end of 20 years, the amount in John’s fund equals the amount in Mary’s fund. Calculate X. Problem 26.50 An annuity-immediate pays an initial beneﬁt of 1 per year, increasing by 10.25% every four years. The annuity is payable for 40 years. Using an annual eﬀective interest rate of 5%, determine an expression for the present value of this annuity. Problem 26.51 An increasing perpetuity with annual payments has a present value of 860 at an annual eﬀective discount rate, d. The initial payment at the end of the ﬁrst year is 3 and each subsequent payment is 2 more than its preceding payment. Calculate d. 26 VARYING ANNUITY-IMMEDIATE 247 Problem 26.52 A 10-year increasing annuity-immediate paying 5 in the ﬁrst year and increasing by 5 each year thereafter has a present value of G. A 10-year decreasing annuity-immediate paying y in the ﬁrst year and decreasing by y 10 each year thereafter has a present value of G. Both present values are calculated using an annual eﬀective interest rate of 4%. Calculate y to the nearest integer. Problem 26.53 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 + 200 each. Given i(2) = 0.6, determine K. Problem 26.54 An annuity immediate has the following payment pattern: 1, 2, · · · , n −1, n, n −1, · · · , 2, 1. The present value of this annuity using an annual eﬀective interest rate of 9% is 129.51. Determine n. Problem 26.55 An increasing 25-year annuity immediate has an initial payment in year one of 4 and each subsequent annual payment is 3 more than the preceding one. Find the present value of this annuity using an annual eﬀective interest rate of 7%. Problem 26.56 A 10-year annuity-immediate has a ﬁrst payment of 2,000. Each subsequent annual payment is 100 less than the preceding payment. (The payments are 2,000; 1,900; 1,800; etc.) At an annual eﬀective interest rate of 5%, what is the accumulated value of this stream of payments on the date of the ﬁnal payment? Problem 26.57 A payment of 100 purchases an increasing perpetuity with a ﬁrst payment of 10 at the end of the ﬁrst year. Each annual payment thereafter is 5% greater than the prior year’s payment. What is the annual interest rate for this stream of payments? Problem 26.58 You are given a perpetuity, with annual payments as follows: (1) Payments of 1 at the end of the ﬁrst year and every three years thereafter. (2) Payments of 2 at the end of the second year and every three years thereafter. (3) Payments of 3 at the end of the third year and every three years thereafter. (4) The interest rate is 5% convertible semi-annually. Calculate the present value of this perpetuity. 248 THE BASICS OF ANNUITY THEORY Problem 26.59 (a) You just paid $50 for a share of stock which pays annual dividends. You expect to receive the ﬁrst dividend payment of $5 one year from now. After that, you expect dividends to increase by $X each year, forever. The price you paid implies that you would be happy with an annual return of 12%. What is X? (b) What value $Y could you sell your share for if the new buyer has the same desired annual return but expects the annual dividend increase to be X + 1? Problem 26.60 A 30-year annuity-immediate pays 20,000 in the ﬁrst year. Thereafter, the payments decrease 500 each year. If d(4) is 6%, what is the present value of this annuity? Problem 26.61 You are about to buy shares in a company. The ﬁrst annual dividend, of $3 per share, is expected to be paid 3 years from now. Thereafter, you expect that the dividend will grow at the rate of 10% per year for ever. Assuming your desired return is 12%, what are you willing to pay per share for this stock? Problem 26.62 A perpetuity with annual payments is payable beginning 10 years from now. The ﬁrst payment is 50. Each annual payment thereafter is increased by 10 until a payment of 150 is reached. Subsequent payments remain level at 150. This perpetuity is purchased by means of 10 annual premiums, with the ﬁrst premium of P due immediately. Each premium after the ﬁrst is 105% of the preceding one. The annual eﬀective interest rates are 5% during the ﬁrst 9 years and 3% thereafter. Calculate P. Problem 26.63 Show that (Da)n = Pn t=1 at. Problem 26.64 Simplify P10 t=1(t + 1)νt. 27 VARYING ANNUITY-DUE 249 27 Varying Annuity-Due In this section, we examine the case of an increasing annuity-due. Consider an annuity with the ﬁrst payment is P at the beginning of year 1 and then the payments increase by Q thereafter, continuing for n years. A time diagram of this situation is given in Figure 27.1 Figure 27.1 The present value for this annuity-due is PV = P + (P + Q)ν + (P + 2Q)ν2 + · · · + [P + (n −1)Q]νn−1. (27.1) Multiplying Equation (28.1) by ν, we obtain νPV = Pν + (P + Q)ν2 + (P + 2Q)ν3 + · · · + [P + (n −1)Q]νn. (27.2) Subtracting Equation(28.2) from Equation (28.1) we obtain (1 −ν)PV = P(1 −νn) + (ν + ν2 + · · · + νn)Q −nνnQ or PV = P¨ an + Q[an −nνn] d . The accumulated value of these payments at time n is AV = (1 + i)nPV = P ¨ sn + Q[sn −n] d . In the special case when P = Q = 1 we ﬁnd (I¨ a)n = ¨ an −nνn d and (I¨ s)n = ¨ sn −n d = sn+1 −(n + 1) d . 250 THE BASICS OF ANNUITY THEORY Example 27.1 Determine the present value and future value of payments of $75 at time 0, $80 at time 1 year, $85 at time 2 years, and so on up to $175 at time 20 years. The annual eﬀective rate is 4%. Solution. The present value is 70¨ a21 + 5(I¨ a)21 = $1, 720.05 and the future value is (1.04)21(1, 720.05) = $3919.60 In the case of a decreasing annuity-due where P = n and Q = −1, the present value at time 0 is (D¨ a)n = n −an d and the accumulated value at time n is (D¨ s)n = (1 + i)n(D¨ a)n = n(1 + i)n −sn d . Example 27.2 Calculate the present value and the accumulated value of a series of payments of $100 now, $90 in 1 year, $80 in 2 years, and so on, down to $10 at time 9 years using an annual eﬀective interest rate of 3%. Solution. The present value is 10(D¨ a)10 = 10· 10−8.530203 0.03/1.03 = $504.63 and the accumulated value is (1.03)10(504.63) = $678.18 Next, we consider a perpetuity-due with payments that form an arithmetic progression (and of course P > 0 and Q > 0). The present value for such a perpetuity with the ﬁrst payment at time 0 is PV = lim n→∞ P¨ an + Q[an −nνn] d =P lim n→∞¨ an + Q lim n→∞ [an −nνn] d =P lim n→∞¨ an + Q[limn→∞an −limn→∞nνn] d =P¨ a∞+ Qa∞ d = P d + Q(1 + i) i2 27 VARYING ANNUITY-DUE 251 since a∞= 1 i , ¨ a∞= 1 d and limn→∞nνn = 0 (by L’Hopital’s rule). In the special case P = Q = 1 we ﬁnd (I¨ a)∞= 1 d2. Example 27.3 Determine the present value at time 0 of payments of $10 paid at time 0, $20 paid at time 1 year, $30 paid at time 2 years, and so on, assuming an annual eﬀective rate of 5%. Solution. The answer is 10(I¨ a)∞= 10 d2 = 10 1.05 0.05 2 = $4, 410.00 Next, we consider payments varying in a geometric progression. Consider an annuity-due with a term of n periods where the interest rate is i per period, and where the ﬁrst payment is 1 at time 0 and successive payments increase in geometric progression with common ratio 1 + k. The present value of this annuity is 1 + ν(1 + k) + ν2(1 + k)2 + · · · + νn−1(1 + k)n−1 =1 −[(1 + k)ν]n 1 −(1 + k)ν = 1 − 1+k 1+i n 1 − 1+k 1+i =(1 + i) 1 − 1+k 1+i n i −k provided that k ̸= i. If k = i then the original sum consists of a sum of n terms of 1 which equals to n. Example 27.4 An annual annuity due pays $1 at the beginning of the ﬁrst year. Each subsequent payment is 5% greater than the preceding payment. The last payment is at the beginning of the 10th year. Calculate the present value at: (a) an annual eﬀective interest rate of 4%; (b) an annual eﬀective interest rate of 5%. Solution. (a) PV = (1.04) h 1−( 1.05 1.04) 10i (0.04−0.05) = $10.44. (b) Since i = k, PV = n = $10.00 252 THE BASICS OF ANNUITY THEORY For an annuity-due with n payments where the ﬁrst payment is 1 at time 0 and successive payments decrease in geometric progression with common ratio 1 −k. The present value of this annuity is 1 + ν(1 −k) + ν2(1 −k)2 + · · · + νn−1(1 −k)n−1 =1 −[(1 −k)ν]n 1 −(1 −k)ν = 1 − 1−k 1+i n 1 − 1−k 1+i =(1 + i) 1 − 1−k 1+i n i + k provided that k ̸= i. If k = i then the original sum is 1 + ν(1 −i) + ν2(1 −i)2 + · · · + νn−1(1 −i)n−1 = 1 2d 1 − 1 −i 1 + i n . Example 27.5 ‡ Matthew makes a series of payments at the beginning of each year for 20 years. The ﬁrst payment is 100. Each subsequent payment through the tenth year increases by 5% from the previous payment. After the tenth payment, each payment decreases by 5% from the previous payment. Calculate the present value of these payments at the time the ﬁrst payment is made using an annual eﬀective rate of 7%. Solution. The present value at time 0 of the ﬁrst 10 payments is 100 " 1 − 1.05 1.07 10 0.07 −0.05 # · (1.07) = 919.95. The value of the 11th payment is 100(1.05)9(0.95) = 147.38. The present value of the last ten payments is 147.38 " 1 − 0.95 1.07 10 0.07 + 0.05 # · (1.07)(1.07)−10 = 464.71. The total present value of the 20 payments is 919.95 + 464.71 = 1384.66 Finally, the present value of a perpetuity with ﬁrst payment of 1 at time 0 and successive pay-ments increase in geometric progression with common ration 1 + k is 1 + ν(1 + k) + ν2(1 + k)2 + · · · = 1 1 −(1 + k)ν = 1 + i i −k. Observe that the value for these perpetuities cannot exist if 1 + k ≥1 + i. 27 VARYING ANNUITY-DUE 253 Example 27.6 Perpetuity A has the following sequence of annual payments beginning on January 1, 2005: 1, 3, 5, 7, · · · Perpetuity B is a level perpetuity of 1 per year, also beginning on January 1, 2005. Perpetuity C has the following sequence of annual payments beginning on January 1, 2005: 1, 1 + r, (1 + r)2, · · · On January 1, 2005, the present value of Perpetuity A is 25 times as large as the present value of Perpetuity B, and the present value of Perpetuity A is equal to the present value of Perpetuity C. Based on this information, ﬁnd r. Solution. The present value of Perpetuity A is 1 d + 2(1+i) i2 . The present value of Perpetuity B is 1 d. The present value of Perpetuity C is 1+i i−r. We are told that 1 + i i + 2(1 + i) i2 = 25(1 + i) i . This is equivalent to 12i2 + 11i −1 = 0. Solving for i we ﬁnd i = 1 12. Also, we are told that 1 + i i + 2(1 + i) i2 = 1 + i i −r or 25(12)(1 + 1 12) = 1 + 1 12 1 12 −r. Solving for r we ﬁnd r = 0.08 = 8% 254 THE BASICS OF ANNUITY THEORY Practice Problems Problem 27.1 A 20 year increasing annuity due pays 100 at the start of year 1, 105 at the start of year 2, 110 at the start of year 3, etc. In other words, each payment is 5% greater than the prior payment. Calculate the present value of this annuity at an annual eﬀective rate of 5%. Problem 27.2 A perpetuity pays 1000 at the beginning of the ﬁrst year. Each subsequent payment is increased by inﬂation. If inﬂation is assumed to be 4% per year, calculate the present value of this perpetuity using an annual eﬀective interest rate of 10%. Problem 27.3 A perpetuity pays 1000 at the beginning of the ﬁrst year. Each subsequent payment is increased by inﬂation. If inﬂation is assumed to be 4% per year, calculate the present value of this perpetuity using an annual eﬀective interest rate of 4%. Problem 27.4 A 10-year decreasing annuity-due makes a payment of 100 at the beginning of the ﬁrst year, and each subsequent payment is 5 less then the previous. What is the accumulated value of this annuity at time 10 (one year after the ﬁnal payment)? The eﬀective annual rate of interest is 5%. Problem 27.5 A 10 year annuity due pays 1,000 as the ﬁrst payment. Each subsequent payment is 50 less than the prior payment. Calculate the current value of the annuity at the end of 5 years using an annual eﬀective rate of 8%. Problem 27.6 You buy an increasing perpetuity-due with annual payments starting at 5 and increasing by 5 each year until the payment reaches 100. The payments remain at 100 thereafter. The annual eﬀective interest rate is 7.5%. Determine the present value of this perpetuity. Problem 27.7 Debbie receives her ﬁrst annual payment of 5 today. Each subsequent payment decreases by 1 per year until time 4 years. After year 4, each payment increases by 1 until time 8 years. The annual interest rate is 6%. Determine the present value. 27 VARYING ANNUITY-DUE 255 Problem 27.8 Determine the accumulated value at time 20 years of payments of $10 at time 0, $20 at time 1 year, $30 at time 2 years, and so on, up to $200 at time 19 years. The annual eﬀective rate of interest is 4%. Problem 27.9 Perpetuity X has payments of 1, 2, 3, · · · at the beginning of each year. Perpetuity Y has payments of q, q, 2q, 2q, 3q, 3q, · · · at the beginning of each year. The present value of X is equal to the present value of Y at an eﬀective annual interest rate of 10%. Calculate q. Problem 27.10 Kendra receives $900 now, $970 in 1 year, $1040 in 2 years, $1,110 in 3 years, and so on, until the ﬁnal payment of $1600. Using an annual eﬀective rate of interest of 9%, ﬁnd (a) the present value of these payments at time 0; (b) the accumulated value at time 11 years. Problem 27.11 Mary is saving money for her retirement. She needs $750,000 in 10 years to purchase a retirement apartment in Florida. She invests X, now, X −5, 000 in 1 year, X −10, 000 in 2 years, and so on, down to X −45, 000 in 9 years. Using an annual eﬀective rate of interest of 5%, ﬁnd X. Problem 27.12 Chris makes annual deposits into a bank account at the beginning of each year for 20 years. Chris’ initial deposit is equal to 100, with each subsequent deposit k% greater than the previous year’s deposit. The bank credits interest at an annual eﬀective rate of 5%. At the end of 20 years, the accumulated amount in Chris’ account is equal to 7276.35. Given k > 5, calculate k. Problem 27.13 You are given: (i) The present value of an annuity-due that pays 300 every 6 months during the ﬁrst 15 years and 200 every 6 months during the second 15 years is 6000. (ii) The present value of a 15-year deferred annuity-due that pays 350 every 6 months for 15 years is 1580. (iii) The present value of an annuity-due that pays 100 every 6 months during the ﬁrst 15 years and 200 every 6 months during the next 15 years is X. The same interest rate is used in all calculations. Determine X. Problem 27.14 A 20 year annuity-due with annual payments has a ﬁrst payment of 100 and each subsequent annual payment is 5% more than its preceding payment. Calculate the present value of this annuity at a nominal interest rate of 10% compounded semiannually. 256 THE BASICS OF ANNUITY THEORY Problem 27.15 The payments of a perpetuity-due are three years apart. The initial payment is 7 and each sub-sequent payment is 5 more than its previous payment. Find the price (present value at time 0) of this perpetuity-due, using an annual eﬀective discount rate of 6%. Problem 27.16 You are considering the purchase of a share of XY Z stock. It pays a quarterly dividend. You expect to receive the ﬁrst dividend, of $2.50, 3 months from now. You expect the dividend to increase by 6% per year, forever, starting with the 5th dividend. If the desired return on your investment is 15%, what is the amount your are willing to pay for a share of XY Z stock? Problem 27.17 You plan to make annual deposits at the start of each year for 25 years into your retirement fund. The ﬁrst 5 deposits are to be 2,000 each. Thereafter, you plan to increase the deposits by X per year. Assuming the fund can earn an eﬀective annual interest rate of 6%, what must X be for you to have accumulated 250,000 at the end of the 25 years? 28 VARYING ANNUITIES WITH PAYMENTS AT A DIFFERENT FREQUENCY THAN INTEREST IS CO 28 Varying Annuities with Payments at a Diﬀerent Fre-quency than Interest is Convertible In Sections 26 and 27, we discussed varying annuities where the payment period coincide with the interest conversion period. In this section we consider varying annuities with payments made more or less frequently than interest is convertible. We will limit our discussion to increasing annuities. Decreasing annuities can be handled in a similar fashion. Varying Annuities Payable Less Frequently Than Interest is Convertible Consider annuities payable less frequently than interest is convertible. We let k be the number of interest conversion periods in one payment period, n the term of the annuity measured in interest conversion periods, and i the rate of interest per conversion period. It follows that the number of payments over the term of the annuity is given by n k which we assume is a positive integer. Let PV be the present value of a generalized increasing annuity-immediate with payments 1, 2, 3, etc. occurring at the end of each interval of k interest conversion periods. Then, PV = νk + 2ν2k + 3ν3k + · · · + n k −1 νn−k + n k νn. (28.1) Multiply Equation (28.1) by (1 + i)k to obtain (1 + i)kPV = 1 + 2νk + 3ν2k + · · · + n k −1 νn−2k + n k νn−k. (28.2) Now subtracting Equation (28.1) from Equation (28.2) we ﬁnd PV [(1 + i)k −1] = 1 + νk + ν2k + · · · + νn−k −n k nun = an ak −n k νn. Hence, PV = an ak −n kνn isk . The accumulated value at time t = n is AV = (1 + i)nPV = (1 + i)n an ak −n kνn isk . Example 28.1 A 10-year annuity-immediate pays 1 in two years, 2 in four years, 3 in six years, 4 in eight years, and 5 in ten years. Using an eﬀective annual interest rate 5%, ﬁnd the present value and the accumulated value of this annuity. 258 THE BASICS OF ANNUITY THEORY Solution. The present value of this annuity at time 0 year is PV = a10 a2 −5(1.05)−10 0.05s2 = 7.721735 1.85941 −5(1.05)−10 (1.05)2 −1 = 10.568. The accumulated value is 10.568(1.05)10 = $17.21 Now, consider an increasing annuity-due with payments 1, 2, 3, etc. occurring at the start of each interval of k interest conversion periods. Then PV = 1 + 2νk + 3ν2k + · · · + n k −1 νn−2k + n k νn−k. (28.3) Multiply Equation (28.3) by νk to obtain νkPV = νk + 2ν2k + 3ν3k + · · · + n k −1 νn−k + n k νn. (28.4) Now subtracting Equation (28.4) from Equation (28.3) we ﬁnd PV [1 −νk] = 1 + νk + ν2k + · · · + νn−k −n k νn = an ak −n k νn. Hence, PV = an ak −n kνn iak . The accumulated value at time t = n is AV = (1 + i)n an ak −n kνn iak . Example 28.2 Find the present value and the accumulated value of a 5-year annuity in which payments are made at the beginning of each half-year, with the ﬁrst payment of $50, the second payment of $100, the third payment of $150, and so on. Interest is 10% convertible quarterly. Solution. The present value is PV = 50 a20 a2 −10ν20 0.025a2 = 15.58916 1.92742 −10(1.025)−20 1 −(1.025)−2 = $2060.15. 28 VARYING ANNUITIES WITH PAYMENTS AT A DIFFERENT FREQUENCY THAN INTEREST IS CO The accumulated value at time t = 20 is (2060.15)(1.025)20 = $3375.80 Varying Annuities Payable More Frequently Than Interest is Convertible We next consider increasing annuities where payments are made more frequently than interest is convertible. We examine two diﬀerent types. The ﬁrst is the increasing mthly annuity where the increase occurs once per conversion period. The second is the mthly increasing mthly annuity where the increase takes place with each mthly payment. Annuities Payable mthly Consider ﬁrst an increasing annuity-immediate with constant payments during each interest con-version period with increases occurring only once per interest conversion period. Let the annuity be payable mthly with each payment in the ﬁrst period equal to 1 m, each payment in the second period equal to 2 m, · · · , each payment in the nth period equal to n m as shown in Figure 28.1. Figure 28.1 Let i be the rate per one conversion period and i(m) be the nominal interest rate payable m times per interest conversion period. Making the mthly payments of k m in period k is the same as making one payment equal to the accumulated value of the mthly payments at the end of the period, that is, one payment of k msm j = k m · (1 + j)m −1 j = k i i(m) where j = i(m) m . The time diagram of this is given in Figure 28.2. Figure 28.2 260 THE BASICS OF ANNUITY THEORY Consequently, the increasing annuity payable mthly is the same as an increasing annuity with P = Q = i i(m) which implies the present value (Ia)(m) n = i i(m)(Ia)n = ¨ an −nνn i(m) . The accumulated value is (Is)(m) n = (1 + i)n(Ia)(m) n = ¨ sn −n i(m) . Example 28.3 Determine the present value and the accumulated value of 1 at the end of each quarter in the ﬁrst year, 2 at the end of each quarter in the second year, 3 at the end of each quarter in the third year, 4 at the end of each quarter in the fourth year, and 5 at the end of each quarter in the ﬁfth year. The annual eﬀective interest rate is 4% Solution. The factor (Ia)(4) 5 values a payment of 1/4 at the end of each quarter during the ﬁrst year, 2/4 at the end of each quarter during the second year, and so on. So, the present value is 4(Ia)(4) 5 = 4 · ¨ a5 −5ν5 i(4) . But i(4) =4[(1.04) 1 4 −1] = 3.9414% ¨ a5 =1 −(1.04)−5 0.04/1.04 = 4.629895 (Ia)(4) 5 =4.629895 −5(1.04)−5 0.039414 = 13.199996 4(Ia)(4) 5 =52.80. The accumulated value is (Is)(4) 5 = (1.04)5(Ia)(4) 5 = 1.045 × 52.80 = 64.24 In the case of an annuity-due, one can easily show that (I¨ a)(m) n = ¨ an −nνn d(m) and (I¨ s)(m) n = (1 + i)n(I¨ a)(m) n = ¨ sn −n d(m) . 28 VARYING ANNUITIES WITH PAYMENTS AT A DIFFERENT FREQUENCY THAN INTEREST IS CO Example 28.4 Determine the present value and accumulated value of 10 at the start of each quarter in the ﬁrst year, 20 at the start of each quarter in the second year, and so on for 5 years. The annual eﬀective interest rate is 4%. Solution. The present value is 40(I¨ a)(4) 5 =40 × ¨ a5 −5ν5 d(4) =40 × 4.629895 −5(1.04)−5 0.039029 =533.20 where d(4) = 4[1 −(1.04)−1 4] = 3.9029%. The accumulated value is (1.04)5 × 533.20 = 648.72 mthly Increasing mthly Payable Annuities Consider next a situation in which payments vary within each interest conversion period as shown in Figure 28.3. Such an annuity is referred to as mthly increasing mthly annuity. Figure 28.3 Denoting the present value of such an annuity by (I(m)a)(m) n we ﬁnd (I(m)a)(m) n = 1 m2[ν 1 m + 2ν 2 m + · · · + mnν mn m ]. (28.5) Multiply both sided by (1 + i) 1 m to obtain (1 + i) 1 m(I(m)a)(m) n = 1 m2[1 + 2ν 1 m + · · · + mnνn−1 m]. (28.6) 262 THE BASICS OF ANNUITY THEORY Subtract Equation (28.5) from Equation (28.6) to obtain (I(m)a)(m) n [(1 + i) 1 m −1] = 1 m2[1 + ν 1 m + · · · + νn−1 m −nmνn] = 1 m[¨ a(m) n −nνn] (I(m)a)(m) n = ¨ a(m) n −nνn m[(1 + i) 1 m −1] =¨ a(m) n −nνn i(m) where ¨ a(m) n = 1 −νn d(m) = i d(m)an. The accumulated value is (I(m)s)(m) n = (1 + i)n(I(m)a)(m) n = ¨ s(m) n −n i(m) where ¨ s(m) n = (1 + i)m −1 d(m) . Example 28.5 Determine the present value and the accumulated value of 1 at the end of the ﬁrst quarter, 2 at the end of the second quarter, 3 at the end of the third quarter, and so on for 5 years. The annual eﬀective interest rate is 4%. Solution. The factor (I(4)a)(4) 5 values a payment of 1 16 at the end of the ﬁrst quarter, 2 16 at the end of the second quarter, and so on. So the present value is 16(I(4)a)(4) 5 = 16 × ¨ a(4) 5 −5ν5 i(4) = 16 × 4.56257 −5(1.04)−5 0.039414 = 183.87 and the accumulated value is (1.04)5 × 183.87 = 223.71 In the case of an annuity-due, it can be easily shown that (I(m)¨ a)(m) n = ¨ a(m) n −nνn d(m) 28 VARYING ANNUITIES WITH PAYMENTS AT A DIFFERENT FREQUENCY THAN INTEREST IS CO and (I(m)¨ s)(m) n = (1 + i)n(I(m)¨ a)(m) n = ¨ s(m) n −n d(m) . Example 28.6 Determine the present value and the accumulated value of 1 at the start of the ﬁrst quarter, 2 at the start of the second quarter, 3 at the start of the third quarter, and so on for 5 years. The annual eﬀective interest rate is 4%. Solution. The present value is 16(I(5)¨ a)(4) 5 = 16 × ¨ a(4) 5 −5ν5 d(4) = 4.56257 −5(1.04)−5 0.03903 = 185.68 and the accumulated value is (1.04)5 × 185.68 = 225.91 Other Considerations Types of annuities like the one discussed above but that vary in geometric progression can be handled by expressing the annuity value as a summation of the present value or accumulated value of each payment. This summation is a geometric progression which can be directly evaluated. We illutrate this in the next example. Example 28.7 Find the accumulated value at the end of eight years of an annuity in which payments are made at the beginning of each quarter for four years. The ﬁrst payment is $3,000, and each of the other payments is 95% of the previous payment. Interest is credited at 6% convertible semiannually. Solution. We count periods in quarters of a year. If j is the interest rate per quarter then 1 + j = (1.03) 1 2. The accumulated value at t = 8 × 4 = 32 is 3000[(1.03)16 + (0.95)(1.03)15.5 + (0.95)2(1.03)15 + · · · + (0.95)15(1.03)8.5] =3000(1.03)16 " 1 + 0.95 √ 1.03 + 0.95 √ 1.03 2 + · · · + 0.95 √ 1.03 15# =3000(1.03)16 · 1 − 0.95 √ 1.03 16 1 − 0.95 √ 1.03 = $49, 134.18 264 THE BASICS OF ANNUITY THEORY Example 28.8 Find an expression for the present value of a perpetuity which pays 1 at the end of the third year, 2 at the end of the sixth year, 3 at the end of the ninth year and so on. Solution. The present value satisﬁes the equation PV = ν3 + 2ν6 + 3ν9 + · · · Thus, PV (1 −ν3) = ν3 + ν6 + ν9 + · · · = ν3 1 −ν3 or PV = ν3 (1 −ν3)2 Some of the problems can be solved using the basic principles. Example 28.9 Scott deposits 1 at the beginning of each quarter in year 1, 2 at the beginning of each quarter in year 2, · · · , 8 at the beginning of each quarter in year 8. One quarter after the last deposit Scott withdraws the accumulated value of the fund and uses it to buy a perpetuity immediate with level payments of X at the end of each year. All calculations assume a nominal interest rate of 10% per annum compounded quarterly. Calculate X. Solution. Regard the annuity as an annuity-due with 8 annual payments: k¨ a4 0.025 where 1 ≤k ≤8. The annual eﬀective rate of interest is 1 + i = (1.025)4. The accumulated value at time t = 32 is ¨ a4 0.025(I¨ s)8 i = ¨ a4 0.025 s9 i−9 i = 196.77. On the other hand, we have X (1.025)4 −1 = 196.77. Solving this equation for X we ﬁnd X = 20.43 28 VARYING ANNUITIES WITH PAYMENTS AT A DIFFERENT FREQUENCY THAN INTEREST IS CO Practice Problems Problem 28.1 (a) Find the total sum of all the payments in (Ia)(12) 2 . (b) Find the total sum of all the payments in (I(12)a)(12) 2 . Problem 28.2 Find an expression for (I(m)a)(m) ∞. Problem 28.3 An annuity pays $100 at the end of each month in the ﬁrst year, $200 at the end of each month in the second year, and continues to increase until it pays 1000 at the end of each month during the 10th year. Calculate the present value of the annuity at an annual eﬀective interest rate of 6%. Problem 28.4 A monthly annuity due pays $1 at the beginning of the ﬁrst month. Each subsequent payment increases by $1. The last payment is made at the beginning of the 240th month. Calculate the accumulated value of the annuity at the end of the 240th month using: (a) an interest rate of 6% compounded monthly; (b) an annual eﬀective interest rate of 6%. Problem 28.5 A 20 year annuity pays 10 at the beginning of each quarter during the ﬁrst year, 20 at the beginning of each quarter during the second year, etc with 200 being paid at the beginning of each quarter during the last year. Calculate the present value of the annuity assuming an annual eﬀective interest rate of 12%. Problem 28.6 A 20 year annuity pays 10 at the beginning of each quarter during the ﬁrst year, 20 at the beginning of each quarter during the second year, etc with 200 being paid at the beginning of each quarter during the last year. Calculate the accumulated value of the annuity assuming a nominal interest rate of 6% compounded monthly. Problem 28.7 Show that the present value of a perpetuity on which payments are 1 at the end of the 5th and 6th years, 2 at the end of the 7th and 8th years, 3 at the end of the 9th and 10th years, etc, is ν4 i −νd. 266 THE BASICS OF ANNUITY THEORY Problem 28.8 A perpetuity has payments at the end of each four-year period. The ﬁrst payment at the end of four years is 1. Each subsequent paymet is 5 more than the previous payment. It is known that v4 = 0.75. Calculate the present value of this perpetuity. Problem 28.9 A perpetuity provides payments every six months starting today. The ﬁrst payment is 1 and each payment is 3% greater than the immediately preceding payment. Find the present value of the perpetuity if the eﬀective rate of interest is 8% per annum. Problem 28.10 Find the accumulated value at the end of ten years of an annuity in which payments are made at the beginning of each half-year for ﬁve years. The ﬁrst payment is $2,000, and each of the other payments is 98% of the previous payment. Interest is credited at 10% convertible quarterly. Problem 28.11 Find the present value of payments of 5 now, 10 in 6 months, 15 in one year, 20 in 18 months, and so on for 6 years. The nominal discount rate is 12% convertible semiannually. Problem 28.12 Marlen invests $ X now in order to receive $5 in 2 months, $10 in 4 months, $15 in 6 months, and so on. The payments continue for 10 years. The annual eﬀective rate of interest is 8%. Determine X. Problem 28.13 A 10-year annuity has the following schedule of payments: $100 on each January 1, $200 on each April 1, $300 on each July 1, and $400 on each October 1. Show that the present value of this annuity on January 1 just before the ﬁrst payment is made is 1600¨ a10(I(4)¨ a)(4) 1 . Problem 28.14 Calculate the accumulated value at time 15 years of payments of 35 at the start of every quarter during the ﬁrst year, 70 at the start of every quarter during the second year, 105 at the start of every quarter during the third year, and so on for 15 years. The nominal interest is 12% convertible quarterly. Problem 28.15 A perpetuity pays 1000 immediately. The second payment is 97% of the ﬁrst payment and is made at the end of the fourth year. Each subsequent payment is 97% of the previous payment and is paid four years after the previous payment. Calculate the present value of this annuity at an annual eﬀective rate of 8%. 28 VARYING ANNUITIES WITH PAYMENTS AT A DIFFERENT FREQUENCY THAN INTEREST IS CO Problem 28.16 A 20 year annuity pays 10 at the end of the ﬁrst quarter, 20 at the end of the second quarter, etc with each payment increasing by 10 until the last quarterly payment is made at the end of the 20th year. Calculate the present value of the annuity assuming a nominal interest rate of 6% compounded monthly. Problem 28.17 A 20 year annuity pays 10 at the end of the ﬁrst quarter, 20 at the end of the second quarter, etc with each payment increasing by 10 until the last quarterly payment is made at the end of the 20th year. Calculate the present value of the annuity assuming an annual eﬀective interest rate of 12%. Problem 28.18 Joan has won a lottery that pays 1000 per month in the ﬁrst year, 1100 per month in the second year, 1200 per month in the third year, and so on. Payments are made at the end of each month for 10 years. Using an eﬀective interest rate of 3% per annum, calculate the present value of this prize. Round your answer to the nearest integer. Problem 28.19 On his 65th birthday Smith would like to purchase a ten-year annuity-immediate that pays 6000 per month for the ﬁrst year, 5500 per month for the second year, 5000 per month for the third year, and so on. Using an annual eﬀective interest rate of 5%, how much will Smith pay for such an annuity? Problem 28.20 On Susan’s 65th birthday, she elects to receive her retirement beneﬁt over 20 years at the rate of 2,000 at the end of each month. The monthly beneﬁt increases by 4% each year. Assuming an annual eﬀective interest rate of 8.16%, and letting j denote the equivalent monthly eﬀective interest rate, ﬁnd the present value on Susan’s 65th birthday of her retirement beneﬁt. 268 THE BASICS OF ANNUITY THEORY 29 Continuous Varying Annuities In this section we look at annuities in which payments are being made continuously at a varying rate. Consider an annuity for n interest conversion periods in which payments are being made contin-uously at the rate f(t) at exact moment t and the interest rate is variable with variable force of interest δt. Then f(t)e− R t 0 δrdrdt is the present value of the payment f(t)dt made at exact moment t. Hence, the present value of this n−period continuous varying annuity is PV = Z n 0 f(t)e− R t 0 δrdrdt. (29.1) Example 29.1 Find an expression for the present value of a continuously increasing annuity with a term of n years if the force of interest is δ and if the rate of payment at time t is t2 per annum. Solution. Using integration by parts process, we ﬁnd Z n 0 t2e−δtdt = −t2 δ e−δt n 0 + 2 δ Z n 0 te−δtdt = −n2 δ e−δn − 2t δ2e−δt n 0 + 2 δ2 Z n 0 e−δtdt = −n2 δ e−δn −2n δ2 e−δn − 2 δ3e−δt n 0 = −n2 δ e−δn −2n δ2 e−δn −2 δ3e−δn + 2 δ3 = 2 δ3 −e−δn n2 δ + 2n δ2 + 2 δ3 Under compound interest, i.e., δt = ln (1 + i), formula (29.1) becomes PV = Z n 0 f(t)νtdt. 29 CONTINUOUS VARYING ANNUITIES 269 Under compound interest and with f(t) = t (an increasing annuity), the present value is (Ia)n = Z n 0 tνtdt = tνt ln ν n 0 − Z n 0 νt ln ν dt = tνt ln ν n 0 − νt (ln ν)2 n 0 = −nνn δ −νn δ2 + 1 δ2 =1 −νn δ2 −nνn δ =an −nνn δ and the accumulated value at time n years is (Is)n = (1 + i)n(Ia)n = sn −n δ . The above two formulas can be derived from the formulas of (I(m)a)(m) n and (I(m)s)(m) n . Indeed, (Ia)n = lim m→∞(I(m)a)(m) n = lim m→∞ ¨ a(m) n −nνn i(m) = an −nνn δ and (Is)n = lim m→∞(I(m)s)(m) n = lim m→∞ ¨ s(m) n −n i(m) = sn −n δ . Example 29.2 Sam receives continuous payments at an annual rate of 8t + 5 from time 0 to 10 years. The continuously compounded interest rate is 9%. (a) Determine the present value at time 0. (b) Determine the accumulated value at time 10 years. Solution. (a) The payment stream can be split into two parts so that the present value is 8(Ia)10 + 5a10 . 270 THE BASICS OF ANNUITY THEORY Since i =e0.09 −1 = 9.4174% a10 =1 −(1.094174)−10 0.09 = 6.59370 (Ia)10 =6.59370 −10(1.094174)−10 0.09 = 28.088592 we obtain 8(Ia)10 + 5a10 = 8 × 28.088592 + 5 × 6.59370 = 257.68. (b) The accumulated value at time 10 years is 257.68 × (1.094174)10 = 633.78 Example 29.3 If δ = 0.06, calculate the present value of 10 year continuous annuity payable at a rate of t at time t. Solution. The present value is (Ia)10 = 1 −ν10 δ2 −10ν10 δ = 1 −e−10×0.06 0.062 −10e−10×0.06 0.06 = 33.86 For a continuous payable continuously increasing perpetuity (where f(t) = t), the present value at time 0 is (Ia)∞= lim n→∞ an −nνn δ = lim n→∞ 1−(1+i)−n δ −n(1 + i)−n δ = 1 δ2. Example 29.4 Determine the present value of a payment stream that pays a rate of 5t at time t. The payments start at time 0 and they continue indeﬁnitely. The annual eﬀective interest rate is 7%. Solution. The present value is 5(Ia)∞= 5 [ln (1.07)]2 = 1, 092.25 We conclude this section by considering the case of a continuously decreasing continuously payable stream in which a continuous payment is received from time 0 to time n years. The rate of payment 29 CONTINUOUS VARYING ANNUITIES 271 at time t is f(t) = n −t, and the force of interest is constant at δ. The present value is (Da)n =nan −(Ia)n =n1 −νn δ −an −nνn δ =n −an δ Example 29.5 Otto receives a payment at an annual rate of 10 −t from time 0 to time 10 years. The force of interest is 6%. Determine the present value of these payments at time 0. Solution. Since i = e0.06 −1 = 6.184% a10 = 1 −(1.06184)−10 0.06 = 7.5201 the present value is then (Da)10 = 10 −7.5201 0.06 = 41.33 Example 29.6 Using the information from the previous example, determine the accumulated value at time 10 years of the payments received by Otto. Solution. The accumulated value at time 10 years is (1.06184)10 × (Da)10 = (1.06184)10 × 41.33 = 75.31 272 THE BASICS OF ANNUITY THEORY Practice Problems Problem 29.1 Calculate the present value at an annual eﬀective interest rate of 6% of a 10 year continuous annuity which pays at the rate of t2 per period at exact moment t. Problem 29.2 Calculate the accumulated value at a constant force of interest of 5% of a 20 year continuous annuity which pays at the rate of t+1 2 per period at exact moment t. Problem 29.3 Evaluate (Ia)∞if δ = 0.08. Problem 29.4 Payments under a continuous perpetuity are made at the periodic rate of (1 + k)t at time t. The annual eﬀective rate of interest is i, where 0 < k < i. Find an expression for the present value of the perpetuity. Problem 29.5 A perpetuity is payable continuously at the annual rate of 1 + t2 at time t. If δ = 0.05, ﬁnd the present value of the perpetuity. Problem 29.6 A one-year deferred continuous varying annuity is payable for 13 years. The rate of payment at time t is t2 −1 per annum, and the force of interest at time t is (1 + t)−1. Find the present value of the annuity. Problem 29.7 If an = a and a2n = b, express (Ia)n in terms of a and b. Problem 29.8 A 10 year continuous annuity pays at a rate of t2 + 2t + 1 at time t. Calculate the present value of this annuity if δt = (1 + t)−1. Problem 29.9 A 30 year continuous annuity pays at a rate of t at time t. If δ = 0.10, calculate the current value of the annuity after ten years. Problem 29.10 A 10 year continuous annuity pays at a rate of t at time t. If νt = .94t, calculate the accumulated value of the annuity after ten years. 29 CONTINUOUS VARYING ANNUITIES 273 Problem 29.11 Nancy receives a payment stream from time 2 to time 7 years that pays an annual rate of 2t −3 at time t. The force of interest is constant at 6% over the period. Calculate the accumulated value at time 7 years. Problem 29.12 Find the ratio of the total payments made under (Ia)10 during the second half of the term of the annuity to those made during the ﬁrst half. Problem 29.13 ‡ 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 δt = 1 8+t. After 10 years, the account is worth 20,000. Calculate k. Problem 29.14 Show that (Ia)n < (I(m)a)(m) n < (Ia)(m) n whenever nδ < 1. Hint: Use Problem 25.11. Problem 29.15 An investment fund is started with an initial deposit of 1 at time 0. New deposits are made continuously at the annual rate 1 + t at time t over the next n years. The force of interest at time t is given by δt = (1 + t)−1. Find the accumulated value in the fund at the end of n years. Problem 29.16 Money was deposited continuously at the rate of 5,000 per annum into a fund for 7 years. If the force of interest is 0.05 from t = 0 to t = 10, what will the fund balance be after 10 years? Problem 29.17 A 5−year annuity makes payments continuously at a rate f(t) = e t2 24 +t . The force of interest varies continuously at a rate of δt = t 12. Find the present value of this annuity. 274 THE BASICS OF ANNUITY THEORY Rate of Return of an Investment In this chapter, we present a number of important results for using the theory of interest in more complex “real-world” contexts than considered in the earlier chapters. In this chapter, the concept of yield rates is introduced: A yield rate (also known as the internal rate of return) is that interest rate at which the present value of all returns from the investment is equal to the present value of all contributions into the investment. Thus, yield rates are solutions to the equation NPV (i) = 0 where NPV stands for the net present value and is the diﬀerence between the present value of all returns and that of all contributions. Techniques for measuring the internal rate of return will be discussed. Yield rates are used as an index to see how favorable or unfavorable a particular transaction is. When lower, yield rates favor the borrower and when higher they favor the lender. We point out here, as always the case thus far, that the eﬀect of taxes will be ignored throughout the discussions. 275 276 RATE OF RETURN OF AN INVESTMENT 30 Discounted Cash Flow Technique In this section we consider the question of proﬁtability of investment projects that involve cash ﬂows. More speciﬁcally, we look at the present values of projects consisting of cash ﬂows. Such projects can be considered as a generalization of annuities with any pattern of payments and/or withdrawals compared to annuities previously discussed where the annuities consisted of regular series of payments. The process of ﬁnding the present value of the type of projects consisting of a stream of cash ﬂows will be referred to as the discounted cash ﬂow technique, abbreviated by DCF. We examine two DCF measures for assessing a project: net present value and internal rate of return. Consider an investment project with contributions (outﬂow) C0, C1, · · · , Cn at time t0 < t1 < t2 < · · · < tn and returns (inﬂow) R0, R1, · · · , Rn at the same point of time. Let ct = Rt −Ct be the net change ( or net cash ﬂow) in the account at time t which can be positive or negative. From the vantage point of the lender, if ct > 0, then there is a net cash deposit into the investment at time t. If ct < 0 then there is a net cash withdrawal from the investment at time t. Thus, if at time 1 a deposit of $1,000 is made and at the same time a withdrawal of $2,000 is made then c1 = 1000 −2000 = −1, 000 so there is a net cash withdrawal of $1,000 from the investment at that time. We have adopted the vantage point of the lender. From the vantage point of the borrower the signs in the above discussion are switched. The following example illustrates these deﬁnitions. Example 30.1 In order to develop a new product and place it in the market for sale, a company has to invest $80,000 at the beginning of the year and then $10,000 for each of the next three years. The product is made available for sale in the fourth year. For that, a contribution of $20,000 must be made in the fourth year. The company incurs maintenance expenses of $2000 in each of the next ﬁve years. The project is expected to provide an investment return of $12,000 at the end of the fourth year, $30,000 at the end of the ﬁfth year, $40,000 at the end of the sixth year, $35,000 at the end of the seventh year, $25,000 at the end of the eigth year, $15,000 at the end of the ninth year and $8,000 at the end of the tenth year. After the tenth year, the product is withdrawn from the market. Create a chart to describe the cash ﬂows of this project. Solution. Table 30.1 describes the cash ﬂows of this project 30 DISCOUNTED CASH FLOW TECHNIQUE 277 Year Contributions Returns ct 0 80,000 0 −80, 000 1 10,000 0 −10, 000 2 10,000 0 −10, 000 3 10,000 0 −10, 000 4 20,000 12,000 −8, 000 5 2,000 30,000 28,000 6 2,000 40,000 38,000 7 2,000 35,000 33,000 8 2,000 25,000 23,000 9 2,000 15,000 13,000 10 0 8,000 8,000 Table 30.1 Now, suppose that the rate of interest per period is i. This interest rate is sometimes referred to as the required return of the investment or the cost of capital. Using the discounted cash ﬂow technique, the net present value at rate i of the investment is deﬁned by NPV (i) = n X k=0 νtkctk, ν = (1 + i)−1. Thus, the net present value of a series of cash ﬂows is the present value of cash inﬂows minus the present value of cash outﬂows. Expressed in another way, NPV is the sum of the present values of the net cash ﬂows over n periods. The value of NPV (i) can be positive, negative, or zero depending on how large or small the value of i is. Example 30.2 Find the net present value of the investment discussed in the previous example. Solution. The net present value is NPV (i) = −80, 000 −10, 000ν −10, 000ν2 −10, 000ν3 −8, 000ν4 + 28, 000ν5 + 38, 000ν6 + 33, 000ν7 + 23, 000ν8 + 13, 000ν9 + 8, 000ν10 Notice that NPV (0.03) = 1488.04 > 0, NPV (0.032180) = 0, and NPV (0.04) = −5122.13 < 0 The value of i for which NPV (i) = 0 is called the yield rate (or internal rate of return); 278 RATE OF RETURN OF AN INVESTMENT that is, the yield rate is the rate of interest at which the present value of contributions from the investment is equal to the present value of returns into the investment. Stated diﬀerently, the yield rate is the rate that causes investment to break even. According to the eqution NPV (i) = 0, yield rates are the same from either the borrower’s or lender’s perspective. They are totally determined by the cash ﬂows deﬁned in the transaction and their timing. Yield rates are often used to measure how favorable or unfavorable a transaction might be. From a lender’s perspective, a higher yield rate makes a transaction more favorable. From a borrower’s perspective, a lower yield rate makes a transaction more favorable. Based on ﬁnancial factors alone, projects with positive NPV are considered acceptable. Projects with NPV (i) < 0 should be rejected. When NPV (i) = 0, the investment would neither gain nor lose value; We should be indiﬀerent in the decision whether to accept or reject the project. Example 30.3 Determine the net present value of the investment project with the following cash ﬂow, at a cost of capital of 4.8%. Time 0 1 2 3 4 5 Flow −1000 70 70 70 70 1070 Solution. The answer is NPV (i) = −1000 + 70a5 + 1000ν5 = 95.78 Example 30.4 Assume that cash ﬂows from the construction and sale of an oﬃce building is as follows. Year 0 1 2 Amount −150000 −100000 300000 Given a 7% cost of capital, create a present value worksheet and show the net present value, NPV. Solution. t vt ct PV 0 1.0 −150000 −150000 1 1 1.07 = 0.935 −100000 −93500 2 1 1.072 = 0.873 300000 261900 NPV= 18400 30 DISCOUNTED CASH FLOW TECHNIQUE 279 Finding the yield rate may require the use of various approximation methods, since the equations that have to be solved may be polynomials of high degree. Example 30.5 An investment project has the following cash ﬂows: Year Contributions Returns 0 100 0 1 200 0 2 10 60 3 10 80 4 10 100 5 5 120 6 0 60 (a) Calculate the net present value at 15%. (b) Calculate the internal rate of return (i.e. the yield rate) on this investment. Solution. (a) NPV (0.15) = −100 −200(1.15)−1 + 50(1.15)−2 + 70(1.15)−3 + 90(1.15)−4 + 115(1.15)−5 + 60(1.15)−6 = −$55.51. (b) We must solve the equation −100 −200ν + 50ν2 + 70ν3 + 90ν4 + 115ν5 + 60ν6 = 0. Using linear interpolation with the points (7%, 4.53) and (8%, −4.52) one ﬁnds i ≈0.07 −4.53 × 0.08 −0.07 −4.52 −4.53 = 7.50% Yield rate needs not be unique as illustrated in the following example. Example 30.6 In exchange for receiving $230 at the end of one year, an investor pays $100 immediately and pays $132 at the end of two years. Find the yield rate. Solution. The net present value is NPV (i) = −100 −132ν2 + 230ν. To ﬁnd the IRR we must solve 132ν2 − 230ν + 100 = 0. Solving this equation by the quadratic formula, we ﬁnd i = 10% or i = 20% Consider the situation where a bank makes a loan to an individual. In this case the bank acts as a lender or investor, and the individual acts as the borrower. Payments that are received by the investor (cash inﬂows) are taken to be positive cash ﬂows. Payments that are made by the investor (cash outﬂows) are considered to be negative. For the borrower, the cash ﬂows will be of the opposite sign. 280 RATE OF RETURN OF AN INVESTMENT Example 30.7 A bank lends Diane $8,000 now. She repays $600 at the end of each quarter for 5 years. Find the annual eﬀective rate of return. Solution. Let j be the eﬀective IRR per quarter. Then j satisﬁes the equation −8000 + 600a20 j = 0 or 3(1 −(1 + j)−20) −40j = 0. Using linear interpolation with the points (4%, 0.030839) and (4.5%, −0.04392) we ﬁnd j ≈0.04 −0.030839 × 0.045 −0.04 −0.04392 −0.030839 = 0.042. Thus, the annual rate of return is i = 1.0424 −1 = 17.89% Yield rates need not be positive. If a yield rate is 0, then the investor(lender) received no return on investment. If a yield rate is negative, then the investor(lender) lost money on the investment. We will assume that such a negative yield rate satisﬁes −1 < i < 0. A yield rate i < −1 implies full loss of the investment. Example 30.8 Find the yield rates of Example 30.1 Solution. Using a scientiﬁc calculator one ﬁnds the two real solutions i = −1.714964 and i = 0.032180. Since, i > −1, the only yield rate is i = 3.218% 30 DISCOUNTED CASH FLOW TECHNIQUE 281 Practice Problems Problem 30.1 Assuming a nominal rate of interest, convertible quarterly, of 12%, ﬁnd the net present value of a project that requires an investment of $100,000 now, and returns $16,000 at the end of each of years 4 through 10. Problem 30.2 The internal rate of return for an investment with contributions of $3,000 at time 0 and $1,000 at time 1 and returns of $2,000 at time 1 and $4,000 at time 2 can be expressed as 1 n. Find n. Problem 30.3 An investor enters into an agreement to contribute $7,000 immediately and $1,000 at the end of two years in exchange for the receipt of $4,000 at the end of one year and $5,500 at the end of three years. Find NPV (0.10). Problem 30.4 An investment project has the following cash ﬂows: Year Contributions Returns 0 100,000 0 1 5,000 0 2 4,000 10,000 3 2,000 10,000 4 0 20,000 5 0 40,000 6 0 60,000 7 0 80,000 (a) Calculate the net present value at 15%. (b) Calculate the internal rate of return on this investment. Problem 30.5 You are the President of XYZ Manufacturing Company. You are considering building a new factory. The factory will require an investment of 100,000 immediately. It will also require an additional investment of 15,000 at the beginning of year 2 to initiate production. Finally, maintenance costs for the factory will be 5,000 per year at the beginning of years 3 through 6. The factory is expected to generate proﬁts of 10,000 at the end of year one, 15,000 at the end of year two, 20,000 at the end of year 3, and 30,000 at the end of years 4 through 6. Calculate the internal rate of return on the potential factory. 282 RATE OF RETURN OF AN INVESTMENT Problem 30.6 An investment project has the following cash ﬂows: Year Contributions Returns 0 100,000 0 1 0 10,000 2 0 20,000 3 0 30,000 4 0 20,000 5 0 10,000 Calculate the internal rate of return on this investment. Problem 30.7 What is the internal rate of return of a project that requires an investment of $1,000,000 now, and returns $150,000 at the end of each of years 1 through 15? Problem 30.8 Assuming an annual eﬀective discount rate of 6%, what is the net present value of a project that requires an investment of $75,000 now, and returns 2, 000(13 −t) at times t = 1, 2, · · · , 12 (i.e, $24,000 at time 1, $22,000 at time 2, $20,000 at time 3, etc.)? Problem 30.9 What is the internal rate of return on a project that requires a $10,000 investment now, and returns $4,000 four years from now and $12,000 eight years from now? Problem 30.10 A project has the following out-ﬂows and in-ﬂows: out-ﬂows: 12,000 at t = 0, 6,000 at t = 3, 9,000 at t = 6 and 12,000 at t = 9 in-ﬂows: 3,000 at t = 2 through t = 14 plus 12,000 at t = 15. If your desired internal rate of return was 8%, would you accept this project? Problem 30.11 A company pays $120,000 to purchase a property. The company pays $3,000 at the end of each of the next 6 months to renovate the property. At the end of the eigth month, the company sells the property for $150,000. Find the net present value of this project for the company at an annual eﬀective rate of 8%. Problem 30.12 What is the project’s monthly internal rate of return of the previous problem? 30 DISCOUNTED CASH FLOW TECHNIQUE 283 Problem 30.13 A project requires an initial investment of $50,000. The project will generate net cash ﬂows of $15,000 at the end of the ﬁrst year, $40,000 at the end of the second year, and $10,000 at the end of the third year. The project’s cost of capital is 13%. According to the DCF technique, should you invest in the project? Problem 30.14 A ten-year investment project requires an initial investment of $100,000 at inception and mainte-nance expenses at the beginning of each year. The maintenance expense for the ﬁrst year is $3,000, and is anticipated to increase 6% each year thereafter. Projected annual returns from the project are $30,000 at the end of the ﬁrst year decreasing 4% per year thereafter. Find c6 = R6 −C6. Problem 30.15 Determine the net present value for a project that costs $104,000 and would yield after−tax cash ﬂows of $16,000 the ﬁrst year, $18,000 the second year, $21,000 the third year, $23,000 the fourth year, $27,000 the ﬁfth year, and $33,000 the sixth year. Your ﬁrm’s cost of capital is 12%. Problem 30.16 Determine the net present value for a project that costs $253,494.00 and is expected to yield after−tax cash ﬂows of $29,000 per year for the ﬁrst ten years, $37,000 per year for the next ten years, and $50,000 per year for the following ten years. Your ﬁrm’s cost of capital is 12%. Problem 30.17 An investment project has the following cash ﬂows: Year Contributions Returns 0 10,000 0 1 5,000 0 2 1,000 0 3 1,000 0 4 1,000 0 5 1,000 0 6 1,000 8,000 7 1,000 9,000 8 1,000 10,000 9 1,000 11,000 10 0 12,000 Find the yield rate of this investment project. 284 RATE OF RETURN OF AN INVESTMENT Problem 30.18 Suppose a sum of money could be invested in project A which pays a 10% eﬀective rate for six years, or be invested in project B which pays an 8% eﬀective rate for 12 years. (a) Which option we choose as an investor? (b) Find the eﬀective rate of interest necessary for the six years after project A ends so that investing in project B would be equivalent. Problem 30.19 ‡ Project P requires an investment of 4000 at time 0. The investment pays 2000 at time 1 and 4000 at time 2. Project Q requires an investment of X at time 2. The investment pays 2000 at time 0 and 4000 at time 1. The net present values of the two projects are equal at an interest rate of 10%. Calculate X. Problem 30.20 X corporation must decide whether to introduce a new product line. The new product will have startup costs, operational costs, and incoming cash ﬂows over six years. This project will have an immediate (t = 0) cash outﬂow of $100,000 (which might include machinery, and employee training costs). Other cash outﬂows for years 1-6 are expected to be $5,000 per year. Cash inﬂows are expected to be $30,000 per year for years 1-6. All cash ﬂows are after-tax, and there are no cash ﬂows expected after year 6. The required rate of return is 10%. Determine the net present value at this rate. Problem 30.21 An investment account is established on which it is estimated that 8% can be earned over the next 20 years. If the interest each year is subject to income tax at a 25% tax rate, ﬁnd the percentage reduction in the accumulated interest at the end of 20 years. 31 UNIQUENESS OF IRR 285 31 Uniqueness of IRR As pointed out in Example 30.4, an internal rate of return may not be unique. In this section, we look at conditions under which the equation Pn i=0 ctiνti = 0 has exactly one solution. For the ease of the analysis we choose time to be equally spaced so that NPV (i) = Pn t=0 ctνt = 0, that is, the net present value function is a polynomial of degree n in ν. By the Fundamental Theorem of Algebra, NPV (i) = 0 has n roots, counting repeated roots and complex roots. Example 31.1 Consider an investment with C0 = 1, C1 = 0, C2 = 1.36, R0 = 0, R1 = 2.3, R2 = 0. Find the internal rate of return. Solution. We have c0 = −1, c1 = 2.3, c2 = −1.36. The internal rate of return is a solution to the equation −1 + 2.3ν −1.36ν2 = 0. Solving this equation using the quadratic formula we ﬁnd ν = 2.3±i √ 0.0111 2.72 so no real root for i Example 31.2 Consider an investment with C0 = 1, C1 = 0, C2 = 1.32, R0 = 0, R1 = 2.3, R2 = 0. Find the internal rate of return. Solution. We have c0 = −1, c1 = 2.3, c2 = −1.32. Thus, the equation of value at time t = 0 is −1 + 2.3ν − 1.32ν2 = 0. Solving this equation using the quadratic formula we ﬁnd ν = −2.3±0.1 −2.64 so either ν = 0.833 or ν = 0.91. Hence, either i = 20% or i = 10% so two internal rates of return Example 31.3 Consider an investment with C0 = 1, C1 = 0, C2 = 1.2825, R0 = 0, R1 = 2.3, R2 = 0. Find the internal rate of return. Solution. We have c0 = −1, c1 = 2.3, c2 = −1.2825. Thus, the equation of value at time t = 0 is −1 + 2.3ν − 1.2825ν2 = 0. Solving this equation we ﬁnd ν = −2.3±0.4 −2.565 so either ν = 0.74 or ν = 1.053. Hence, either i = 35% or i = −5% What we would like to do next is to formulate conditions on the ct that guarantee a unique yield rate i > −1. A ﬁrst set of conditions is given in the following theorem. 286 RATE OF RETURN OF AN INVESTMENT Theorem 31.1 Let k be an integer such that 0 < k < n. Suppose that either (i) ct ≤0 for 0 ≤t ≤k and ct ≥0 for k + 1 ≤t ≤n; or (ii) ct ≥0 for 0 ≤t ≤k and ct ≤0 for k + 1 ≤t ≤n. Then there is a unique interest rate i > −1 such that NPV (i) = 0. This theorem states that a unique internal rate of return exists for a transaction in which the net payments are all of one sign for the ﬁrst portion of the transaction and then have the opposite sign for the remainder of the transaction. To prove the theorem, we require the concept of Descartes’ Rule of signs which is a technique for determining the number of positive or negative roots of a polynomial: If the terms of a single-variable polynomial with real coeﬃcients and a nonzero constant term are ordered by descending variable exponent, then the number of positive roots counting multiplicity of a polynomial f(x) is either equal to the number of sign changes between adjacent nonzero coeﬃcients, or less than it by a multiple of 2. The number of negative zeros of a polynomial f(x) counting multiplicity is equal to the number of positive zeros of f(−x), so it is the number of changes of signs of f(−x) or that number decreased by an even integer. Complex roots always come in pairs. That’s why the number of positive or number of negative roots must decrease by two. Example 31.4 Use Descartes’ rule of sign to ﬁnd the possible number of positive roots and negative roots of the polynomial f(x) = x2 + x + 1. Solution. The polynomial f(x) has no sign change between adjacent coeﬃcients. Thus, f(x) has no positive roots. The polynomial f(−x) = x2 −x + 1 has two sign changes between adjacent coeﬃcients so that f(x) has either two negative roots or zero negative roots. Thus, the two possibilities are either two negative roots or two complex roots Example 31.5 Use Descartes’ rule of sign to ﬁnd the possible number of positive roots and negative roots of the polynomial f(x) = x3 + x2 −x −1. Solution. The polynomial f(x) has one sign change between the second and third terms. Therefore it has at most one positive root. On the other hand, f(−x) = −x3 +x2 +x−1 has two sign changes, so f(x) has 2 or 0 negative roots. Since complex roots occur in conjugate, the two possibilities are either one positive root and two negative roots or one positive root and two complex roots. Thus, f(x) 31 UNIQUENESS OF IRR 287 has exactly one positive root Proof of Theorem 31.1. According to either (i) and (ii), the coeﬃcients of NPV (i) changes signs once. So by Descartes’ rule of signs, there will be at most one positive real root. If such a root exists, then ν > 0 implies i > −1. For such values of i we deﬁne the function f(i) = (1 + i)kNPV (i) = k X t=0 ct(1 + i)k−t + n X t=k+1 ct(1 + i)k−t. Taking the ﬁrst derivative we ﬁnd f ′(i) = k−1 X t=0 ct(k −t)(1 + i)k−t−1 + n X t=k+1 ct(k −t)(1 + i)k−t−1. If (i) is satisﬁed then ct(k−t) ≤0 for 0 ≤t ≤n. Hence, f ′(i) ≤0 and the function f(i) is decreasing. If (ii) is satisﬁed then f ′(i) ≥0 and f(i) is increasing. In case (i), the function f(i) is decreasing. Moreover, as i →−1+ the ﬁrst sum in f(i) approaches ck whereas the second sum approaches ∞. Hence, limi→−1+ f(i) = ∞. Similarly, as i approaches +∞ the ﬁrst sum in f(i) approaches −∞whereas the second sum approaches 0. Hence, limi→∞f(i) = −∞. Since f is decreasing, there is a unique i > −1 such that NPV (i) = 0. A similar argument holds for case (ii) Example 31.6 A project requires an initial investment of $10,000 and it produces net cash ﬂows of $10,000 one year from now and $2,000 two years from now. Show that there is a unique internal rate of return. Solution. The net present value is NPV (i) = −10, 000+10, 000ν +2, 000ν2. The result follows from Theorem 31.1(i) with k = 1 The following theorem establishes uniqueness of yield rates under a broader set of conditions than the one given in the previous theorem. Theorem 31.2 288 RATE OF RETURN OF AN INVESTMENT Let i > −1 be a solution to NPV (i) = 0. Suppose that B0 =c0 > 0 B1 =c0(1 + i) + c1 > 0 B2 =c0(1 + i)2 + c1(1 + i) + c2 > 0 . . . Bn−1 =c0(1 + i)n−1 + c1(1 + i)n−2 + · · · + cn−1 > 0 Then (i) Bn = c0(1 + i)n + c1(1 + i)n−1 + · · · + cn = 0 (ii) i is unique. Proof. (i) We are given that Pn t=0 ctνt = 0. This is an nth degree polynomial in ν and could be written as an nth degree polynomial in i by simply multiplying both sides by (1 + i)n to obtain c0(1 + i)n + c1(1 + i)n−1 + · · · + cn = 0. Thus, the assumption of (i) is established. (ii) Suppose that j satisﬁes NPV (j) = 0 and j > i. Then we have B′ 0 =c0 = B0 > 0 B′ 1 =B′ 0(1 + j) + c1 = c0(1 + j) + c1 > B1 > 0 B′ 2 =B′ 1(1 + j) + c2 = c0(1 + j)2 + c1(1 + j) + c2 > B2 > 0 . . . B′ n−1 =c0(1 + j)n−1 + c1(1 + j)n−2 + · · · + cn−1 > Bn−1 > 0 B′ n =B′ n−1(1 + j) + cn = c0(1 + j)n + c1(1 + j)n−1 + · · · + cn > Bn = 0 The last inequality shows that j is not a yield rate, a contradiction. A similar argument holds for −1 < j < i. Hence, we conclude that j = i Remark 31.1 Note that Bt is the outstanding balance at time t. According to the theorem above, if the outstanding balance is positive throughout the period of investment, then the yield rate will be unique. Also, note that c0 > 0 and cn = −Bn−1(1 + i) < 0, but that ct for t = 1, 2, · · · , n −1 may be either positive, negative, or zero. 31 UNIQUENESS OF IRR 289 Example 31.7 Show that we cannot guarantee uniqueness of the yield rate if c0 and cn have the same sign. Solution. The conclusion of uniqueness depends on the outstanding balance being positive at all time during the investment. If c0 > 0 and cn > 0 then the outstanding balance must be negative prior to t = n. The negative outstanding balance implies no guarantee of uniqueness. Similarly, if c0 < 0 and cn < 0, the outstanding balance at the start of the investment is negative so that no guarantee of uniqueness of yield rate The discussion of this section focused on the existence of yield rates. However, it is possible that no yield rate exists or all yield rates are imaginary (See Example 31.1). Example 31.8 An investor is able to borrow $1,000 at 8% eﬀective for one year and immediately invest the $1,000 at 12% eﬀective for the same year. Find the investor’s yield rate on this transaction. Solution. The proﬁt at the end of the year is $40, but there is no yield rate, since the net investment is zero 290 RATE OF RETURN OF AN INVESTMENT Practice Problems Problem 31.1 Use Descartes’ rule of signs to determine the maximum number of real zeroes of the polynomial f(x) = x5 −x4 + 3x3 + 9x2 −x + 5. Problem 31.2 Consider the following transaction: Payments of $100 now and $108.15 two years from now are equivalent to a payment of $208 one year from now. Find the absolute value of the diﬀerence of the two rates that result from the equation of value at time t = 0. Problem 31.3 An investor pays $100 immediately and $X at the end of two years in exchange for $200 at the end of one year. Find the range of X such that two yield rates exist which are equal in absolute value but opposite in sign. Problem 31.4 What is the yield rate on a transaction in which an investor makes payments of $100 immediately and $101 at the end of two years, in exchange for a payment of $200 at the end of one year? Problem 31.5 ‡ A company deposits 1,000 at the beginning of the ﬁrst year and 150 at the beginning of each subsequent year into perpetuity. In return the company receives payments at the end of each year forever. The ﬁrst payment is 100. Each subsequent payment increases by 5%. Calculate the company’s yield rate for this transaction. Problem 31.6 An investor enters into an agreement to contribute $7,000 immediately and $1,000 at the end of two years in exchange for the receipt of $4,000 at the end of one year and $5,500 at the end of three years. What is the maximum number of possible yield rates using Descartes’ rule of signs? Problem 31.7 An investment costs $85 and it will pay $100 in 5 years. Determine the internal rate of return for this investment. Problem 31.8 A project requires an initial investment of $50,000. The project will generate net cash ﬂows of $15,000 at the end of the ﬁrst year, $40,000 at the end of the second year, and $10,000 at the end of the third year. Show that an internal rate of return exists and is unique. 31 UNIQUENESS OF IRR 291 Problem 31.9 Suppose payments of 100 now and 109.20 two years from now are to be made in return for receiving 209 one year from now. Determine the internal rate of return. Problem 31.10 What is the internal rate of return (in terms of an eﬀective annual interest rate) on a project that requires a $1,000 investment now, and provides returns to the investor of $500 one year from now and $800 two years from now? Problem 31.11 Suppose a project requires you to invest $10,000 now and $20,000 one year from now. The project returns $29,000 six months from now. Find all the yield rates (internal rates of return) of this project. Express these yield rates as annual eﬀective rates. Problem 31.12 For the previous problem, ﬁnd the range of annual interest rates which will produce a net present value greater than zero. Problem 31.13 Two growing perpetuities, each with annual payments, have the same yield rate (internal rate of return). The ﬁrst perpetuity has an initial payment of $50 one year from now, and each subsequent annual payment increases by $5. The present value of this ﬁrst perpetuity is $2,000. The second perpetuity (a perpetuity-due) has an initial payment of $100 now, and each subsequent annual payment increases by 3%. Find the present value, now, of the second perpetuity. 292 RATE OF RETURN OF AN INVESTMENT 32 Interest Reinvested at a Diﬀerent Rate In this section we consider transactions where interest may be reinvested at a rate which may or may not be equal to the original investment rate. We analyze two situations in which reinvestment rates are directly taken into consideration. First, consider an investment of 1 for n periods at rate i where the interest is reinvested at rate j. The time diagram of this situation is shown in Figure 32.1. Figure 32.1 The accumulated value at the end of the n periods is equal to the principal plus the accumulated value of the annuity-immediate with periodic payments of i at the end of each period and periodic rate j. That is, AV = 1 + isn j. Note that in the case i = j, the formula reduces to the familiar formula AV = (1 + i)n. That is, for compound interest, the reinvestment rate is equal to the original rate. Example 32.1 100 is invested in a Fund A now. Fund A will pay interest at 10% each year. When the interest is paid in Fund A, it will be immediately removed and invested in Fund B paying 8 %. Calculate the sum of the amounts in Fund A and Fund B after 10 years. Solution. At the end of 10 years, Fund A will just have the original principal: $100. Fund B will have the accumulated value of the interest payments at the rate 8%: 10s10 0.08 = 10(1.0810 −1) 0.08 = 144.87. So the total amount of money in both funds is about $244.87 Example 32.2 Brown deposits 5000 in a 5-year investment that pays interest quarterly at 8%, compounded quar-terly. Upon receipt of each interest payment, Brown reinvests the interest in an account that earns 6%, compounded quarterly. Determine Brown’s yield rate over the 5 year period, as a nominal interest rate compounded quar-terly. 32 INTEREST REINVESTED AT A DIFFERENT RATE 293 Solution. The equation of value after 20 quarters is 5000 1 + i(4) 4 20 = 5000 + 5000(0.08)s20 0.015. Solving this equation we ﬁnd i(4) = 4 [1 + 0.08s20 0.015] 1 20 −1 = 7.68% Second, consider an investment of 1 at the end of each period for n periods at rate i where the interest is reinvested at rate j. The time diagram of this situation is shown in Figure 32.2. Figure 32.2 The accumulated value at the end of the n periods is the sum of the annuity payments and the accumulated value of the interest,i.e. AV = n + i(Is)n−1 j = n + i sn j −n j . Note that the above formula simpliﬁes to the familiar result AV = sn when j = i. Example 32.3 Payments of $1,000 are invested at the end of each year for 10 years. The payments earn interest at 7% eﬀective, and the interest is reinvested at 5% eﬀective. Find the (a) amount in the fund at the end of 10 years, (b) purchase price an investor(a buyer) must pay for a yield rate (to the seller) of 8% eﬀective. Solution. (a) The amount in the fund at the end of 10 years is 1000 10 + 0.07 s10 0.05 −10 0.05 =1000 10 + 0.07 12.5779 −10 0.05 =$13, 609.06. (b) The purchase price is 13, 609.06(1.08)−10 = $6, 303.63 294 RATE OF RETURN OF AN INVESTMENT Example 32.4 Esther invests 100 at the end of each year for 12 years at an annual eﬀective interest rate of i. The interest payments are reinvested at an annual eﬀective rate of 5%. The accumulated value at the end of 12 years is 1748.40. Calculate i. Solution. The equation of value at time t = 12 is 1748.40 = 1200 + 100i s12 0.05 −12 0.05 . Solving this equation for i we ﬁnd i = 1748.40 −1200 100[s12 0.05 −12] = 7% If the payments of 1 are made at the beginning of each period (instead of at the end), the accumu-lated value at the end of the n periods is AV = n + i(Is)n j = n + i sn+1 j −(n + 1) j . Example 32.5 David pays 1,000 at the beginning of each year into a fund which earns 6%. Any interest earned is reinvested at 8%. Calculate the total that David will have at the end of 7 years. Solution. Note that payment are made at the beginning of the year (annuity-due). The answer is AV = 7, 000 + 1000(0.06) " 1.088−1 0.08 −8 0.08 # = 8, 977.47 An important consideration to a lender (investor) is the rate of repayment by the borrower. A faster rate of repayment by the borrower, results in a higher yield rate for the investor. This is illustrated in the following example. Example 32.6 Three loan repayment plans are described for a $3,000 loan over a 6-year period with an eﬀective rate of interest of 7.5%. If the repayments to the lender can be reinvested at an eﬀective rate of 32 INTEREST REINVESTED AT A DIFFERENT RATE 295 6%, ﬁnd the yield rates (for the lender) (a) if the entire loan plus accumulated interest is paid in one lump sum at the end of 6 years; (b) if interest is paid each year as accrued and the principal is repaid at the end of 6 years; (c) if the loan is repaid by level payments over the 6-year period. Solution. (a) The lump sum is 3000(1.075)6 = $4629.90. Since there is no repayment to reinvest during the 6-year period, the yield rate is obviously 7.5%. This answer can be also found by solving the equation 3000(1 + i)6 = 4629.90. (b) At the end of each year during the 5-year period, the payment is 3000(0.075) = $225. The accumulated value of all payments at the end of the 6-year period is 3000 + 225s6 0.06 = 3000+225(6.9753) = $4569.44. To ﬁnd the yield rate, we solve the equation 3000(1+i)6 = 4569.44 → i = 0.07265 = 7.265% (c) Each year during the 6-year period, the payment is R where 3000 = Ra6 0.075 →R = 639.13. The accumulated value of all payments at the end of the 6-year period is 639.13s6 0.06 = 639.13(6.9753) = $4458.12. To ﬁnd the yield rate, we solve the equation 3000(1 + i)6 = 4458.12 →i = 0.06825 = 6.825% 296 RATE OF RETURN OF AN INVESTMENT Practice Problems Problem 32.1 Vanessa invests $5,000 at the end of each of the next 15 years and her investment earns interest at an annual eﬀective rate of 10%. The interest that she receives at the end of each year is reinvested and earns interest at an annual eﬀective rate of 5%. Calculate the accumulated value at time 15 years. Problem 32.2 Lauren has 100,000 invested in a fund earning 5%. Each year the interest is paid to Lauren who is able to invest the interest at 8%. Calculate the amount that Lauren will have at the end of 10 years. Problem 32.3 Julie has a sum of money, S, invested in a fund which earns 10%. Each year the fund pays the interest earned to Julie. Julie can only reinvest the interest at an annual eﬀective interest rate of 6%. After 20 years, Julie has 100,000 total including the amount in the fund plus the reinvested interest. Calculate S. Problem 32.4 Megan invests 500 at the end of each year. The investment earns 8% per year which is paid to Megan who reinvests the interest at 6%. Calculate how much Megan will have after 5 years. Problem 32.5 Thomas invests X into Fund 1 at the beginning of each year for 10 years. Fund 1 pays interest annually into Fund 2. Fund 1 earns 7% annually while Fund 2 earns 6% annually. After 10 years, Thomas has a total of 50,000. Calculate X. Problem 32.6 Chris is investing 1,000 at the beginning of each year into Fund A. Fund A earns interest at a nominal interest rate of 12% compounded monthly. Fund A pays Chris interest monthly. Chris reinvests that interest in Fund B earning an annual eﬀective rate of 8%. Calculate the total amount in Fund A and Fund B after 10 years. Problem 32.7 John invests 100 at the end of year one, 200 at the end of year two, etc until he invests 1,000 at the end of year ten. The investment goes into a bank account earning 4%. At the end of each year, the interest is paid into a second bank account earning 3%. Calculate the total amount John will have after 10 years. 32 INTEREST REINVESTED AT A DIFFERENT RATE 297 Problem 32.8 Kathy pays 1,000 at the end of each year into Fund A which earns interest at an annual eﬀective interest rate of i. At the end of each year, the interest earned is transferred to Fund B earning 10% interest. After 10 years. Kathy has 15,947.52. Calculate i. Problem 32.9 It is desired to accumulate a fund of 1,000 at the end of 10 years by equal deposits at the beginning of each year. If the deposits earn interest at 8% eﬀective, but the interest can only be reinvested at only 4% eﬀective, show that the deposit necessary is 1000 2s11 0.04 −12. Problem 32.10 A loan of 10,000 is being repaid with payments of 1,000 at the end of each year for 20 years. If each payment is immediately reinvested at 5% eﬀective, ﬁnd the yield rate of this investment. Problem 32.11 An investor purchases a 5-year ﬁnancial instrument having the following features: (i) The investor receives payments of 1,000 at the end of each year for 5 years. (ii) These payments earn interest at an eﬀective rate of 4% per annum. At the end of the year, this interest is reinvested at the eﬀective rate of 3% per annum. Find the purchase price to the investor to produce a yield rate of 4%. Problem 32.12 An investor deposits 1,000 at the beginning of each year for ﬁve years in a fund earning 5% eﬀective. The interest from this fund can be reinvested at only 4% eﬀective. Show that the total accumulated value at the end of ten years is 1250(s11 0.04 −s6 0.04 −1) Problem 32.13 A invests 2,000 at an eﬀective interest rate of 17% for 10 years. Interest is payable annually and is reinvested at an eﬀective rate of 11%. At the end of 10 years the accumulated interest is 5,685.48. B invests 150 at the end of each year for 20 years at an eﬀective interest rate of 14%. Interest is payable annually and is reinvested at an eﬀective rate of 11%. Find B’s accumulated interest at the end of 20 years. Problem 32.14 ‡ Victor invests 300 into a bank account at the beginning of each year for 20 years. The account pays out interest at the end of every year at an annual eﬀective interest rate of i%. The interest is reinvested at an annual eﬀective rate of i 2 %. The yield rate on the entire investment over the 20 year period is 8% annual eﬀective. Determine i. 298 RATE OF RETURN OF AN INVESTMENT Problem 32.15 ‡ Sally lends 10,000 to Tim. Tim agrees to pay back the loan over 5 years with monthly payments payable at the end of each month. Sally can reinvest the monthly payments from Tim in a savings account paying interest at 6%, compounded monthly. The yield rate earned on Sally’s investment over the ﬁve-year period turned out to be 7.45%, compounded semi-annually. What nominal rate of interest, compounded monthly, did Sally charge Tim on the loan? Problem 32.16 Paul invests 1,000 at the beginning of each year for 10 years into a Fund A earning Y% each year. At the end of the year, any interest earned is moved to Fund B which earns 5%. Combining both Funds, Paul has a total of 11,924.08 at the end of 10 years. Determine Y. Problem 32.17 Lisa invests 1200 at the beginning of each year for 8 years into an account earning 8%. The interest earned each year is transferred to an account earning 6%. At the end of 8 years, the total amount is paid out. Calculate the amount an investor would pay now for the ﬁnal payout if the investor wanted a return of 10%. Problem 32.18 Amy invests 1000 at an eﬀective annual rate of 14% for 10 years. Interest is payable annually and is reinvested at an annual eﬀective rate of i. At the end of 10 years the accumulated interest is 2341.08. Bob invests 150 at the end of each year for 20 years at an annual eﬀective rate of 15%. Interest is payable annually and is reinvested at an annual eﬀective rate of i. Find Bob’s accumulated interest at the end of 20 years. Problem 32.19 ‡ An investor pays $100,000 today for a 4-year investment that returns cash ﬂows of $60,000 at the end of each of years 3 and 4. The cash ﬂows can be reinvested at 4.0% per annum eﬀective. If the rate of interest at which the investment is to be valued is 5.0%, what is the net present value of this investment today? Problem 32.20 ‡ An investor wishes to accumulate 10,000 at the end of 10 years by making level deposits at the beginning of each year. The deposits earn a 12% annual eﬀective rate of interest paid at the end of each year. The interest is immediately reinvested at an annual eﬀective interest rate of 8%. Calculate the level deposit. 32 INTEREST REINVESTED AT A DIFFERENT RATE 299 Problem 32.21 ‡ Payments of X are made at the beginning of each year for 20 years. These payments earn interest at the end of each year at an annual eﬀective rate of 8%. The interest is immediately reinvested at an annual eﬀective rate of 6%. At the end of 20 years, the accumulated value of the 20 payments and the reinvested interest is 5600. Calculate X. Problem 32.22 A deposit of 1 is made at the end of each year for 30 years into a bank account that pays interest at the end of each year at j per annum. Each interest payment is reinvested to earn an annual eﬀective interest rate of j 2. The accumulated value of these interest payments at the end of 30 years is 72.88. Determine j. Problem 32.23 ‡ Susan invests Z at the end of each year for seven years at an annual eﬀective interest rate of 5%. The interest credited at the end of each year is reinvested at an annual eﬀective rate of 6%. The accumulated value at the end of seven years is X. Lori invests Z at the end of each year for 14 years at an annual eﬀective interest rate of 2.5%. The interest credited at the end of each year is reinvested at an annual eﬀective rate of 3%. The accumulated value at the end of 14 years is Y. Calculate Y/X. Problem 32.24 ‡ Jason deposits 3960 into a bank account at t = 0. The bank credits interest at the end of each year at a force of interest δt = 1 8 + t. Interest can be reinvested at an annual eﬀective rate of 7%. The total accumulated amount at time t = 3 is equal to X. Calculate X. Problem 32.25 ‡ Eric deposits 12 into a fund at time 0 and an additional 12 into the same fund at time 10. The fund credits interest at an annual eﬀective rate of i. Interest is payable annually and reinvested at an annual eﬀective rate of 0.75i. At time 20 the accumulated amount of the reinvested interest payments is equal to 64. Calculate i, i > 0. Problem 32.26 ‡ At time t = 0, Sebastian invests 2000 in a fund earning 8% convertible quarterly, but payable an-nually. He reinvests each interest payment in individual separate funds each earning 9% convertible quarterly, but payable annually. The interest payments from the separate funds are accumulated in 300 RATE OF RETURN OF AN INVESTMENT a side fund that guarantees an annual eﬀective rate of 7%. Determine the total value of all funds at t = 10. Problem 32.27 A loan of 1000 is being repaid in 10 years by semiannual installments of 50, plus interest on the unpaid balance at 4% per annum compounded semiannually. The installments and interest payments are reinvested at 5% per annum compounded semiannually. Calculate the annual eﬀective yield rate of the loan. Problem 32.28 John invests a total of 10,000. He purchases an annuity with payments of 1000 at the beginning of each year for 10 years at an eﬀective annual interest rate of 8%. As annuity payments are received, they are reinvested at an eﬀective annual interest rate of 7%. The balance of the 10,000 is invested in a 10-year certiﬁcate of deposit with a nominal annual interest rate of 9%, compounded quarterly. Calculate the annual eﬀective yield rate on the entire 10,000 investment over the 10 year period. Problem 32.29 Brown deposits $1000 at the beginning of each year for 5 years, but makes no additional deposits in years 6 through 10. The fund pays interest annually at an annual eﬀective interest rate of 5%. Interest is reinvested at only a 4% annual eﬀective interest rate. What is Brown’s total accumulated value at the end of 10 years? Problem 32.30 You have $20,000 to invest. Alternative A allows you to invest your money at 8% compounded annually for 12 years. Alternative B pays you simple interest at the end of each year of 6%. At what rate must you reinvest the interest in alternative B to have the same accumulation after 12 years? Problem 32.31 Henry invests 2000 at the beginning of the year in a fund which credits interest at an annual eﬀective rate of 9%. Henry reinvests each interest payment in a separate fund, accumulating at an annual eﬀective rate of 8%. The interest payments from this fund accumulate in a bank account that guarantees an annual eﬀective rate of 7%. Determine the sum of the principal and interest at the end of 10 years. Problem 32.32 Jim borrowed 10,000 from Bank X at an annual eﬀective rate of 8%. He agreed to repay the bank with ﬁve level annual installments at the end of each year. At the same time, he also borrowed 32 INTEREST REINVESTED AT A DIFFERENT RATE 301 15,000 from Bank Y at an annual eﬀective rate of 7.5%. He agreed to repay this loan with ﬁve level annual installments at the end of each year. He lent the 25,000 to Wayne immediately in exchange for four annual level repayments at the end of each year, at an annual eﬀective rate of 8.5%. Jim can only reinvest the proceeds at an annual eﬀective rate of 6%. Immediately after repaying the loans to the banks in full, determine how much Jim has left. 302 RATE OF RETURN OF AN INVESTMENT 33 Interest Measurement of a Fund: Dollar-Weighted Inter-est Rate An investment fund usually is a ﬁrm that invests the pooled funds of investors for a fee. In this section we examine a method of determining the interest rate earned by an investment fund. In practice, it is common for a fund to be incremented with new principal deposits, decremented with principal withdrawals, and incremented with interest earnings many times throughout a pe-riod. Since these occurrences are often at irregular intervals, we devise notation for the purpose of obtaining the eﬀective rate of interest i over one measurement period: A = the amount in the fund at the beginning of the period,i.e. t = 0. B = the amount in the fund at the end of the period, i.e. t = 1. I = the amount of interest earned during the period. ct = the net amount of principal contributed at time t (inﬂow less outﬂow at time t) where 0 ≤t ≤1. C = P t ct = total net amount of principal contributed during the period (if negative, this is an indication of net withdrawal). The amount at the end of a period is equal to the amount at the beginning of the period plus the net principal contributions plus the interest earned: B = A + C + I. (33.1) To be consistent with the deﬁnition of the eﬀective rate we will assume that all the interest earned I is received at the end of the period. Assuming compound interest i throughout the period, the exact equation of value for the interest earned over the period 0 ≤t ≤1 is I = iA + X 0≤t≤1 ct[(1 + i)1−t −1]. (33.2) That is, the amount of interest earned is the sum of amounts of interest earned by each individual contribution, plus the amount of interest earned on the balance at the beginning. Note that (1 + i)1−t −1 is the eﬀective rate for period from t to 1,i.e., A(1) −A(t) A(t) = A(1) A(t) −1 = (1 + i)1−t −1. Substituting equation(33.2) into equation (33.1) we ﬁnd the equation of value B = A(1 + i) + X 0≤t≤1 ct(1 + i)1−t. (33.3) 33 INTEREST MEASUREMENT OF A FUND: DOLLAR-WEIGHTED INTEREST RATE 303 The interest i satisfying equation (33.3) is called the dollar-weighted rate of interest. Note that the required return of the investment, cost of capital and dollar weighted rate of return stand for the same quantity. Example 33.1 At the beginning of a year, an investment fund was established with an initial deposit of $3,000. At the end of six months, a new deposit of $1,500 was made. Withdrawals of $500 and $800 were made at the end of four months and eight months respectively. The amount in the fund at the end of the year is $3,876. Set up the equation of value to calculate the dollar-weighted rate of interest. Solution. The equation of value is 3, 000(1 + i) + 1, 500(1 + i)0.5 −500(1 + i)(1−4 12) −800(1 + i)(1−8 12) = 3, 876 Finding i from the equation (33.2) requires approximation methods and this is not an easy problem. However, in practice one uses a simple interest approximation (1 + i)1−t −1 = 1 + (1 −t)i −1 = (1 −t)i obtaining I ≈iA + X 0≤t≤1 ct(1 −t)i which leads to the approximation i ≈ I A + P 0≤t≤1 ct(1 −t). (33.4) It is shown that this approximation is very good as long as the ct’s are small in relation to A which is often the case in practice. However, if the c′ ts are not small in relation to A, then the error can become signiﬁcant. The denominator in the above approximation is commonly called the exposure associated with i. Alternatively, by using equation (33.3) one ﬁnds i by solving the equation B = A(1 + i) + X 0≤t≤1 ct[1 + (1 −t)i]. Example 33.2 Using Example 33.1, ﬁnd the approximate eﬀective rate of interest earned by the fund during the year using the dollar-weighted rate of return formula. 304 RATE OF RETURN OF AN INVESTMENT Solution. The interest earned I is computed to be I = 3876 + 500 + 800 −(3000 + 1500) = 676. The exposure associated with i is 3000 + (−500) 1 −1 3 + (1500) 1 −1 2 + (−800) 1 −2 3 = 3150. Thus, the approximate eﬀective rate is 676 3150 = 0.2146 = 21.46%. Alternatively, i can be found by solving the equation 3, 000(1 + i) + 1, 500(1 + 0.5i) −500(1 + 2 3i) −800(1 + i 3) = 3, 876 giving i = 21.46% Formula (33.4) is in a form which can be directly calculated. However, it is tedious to use it because of the summation term in the denominator. Therefore, estimating the summation is fa-vored. One way for doing that is to assume that the net principal contributions C occur at time k that can be approximated by the arithmetic weighted average k = 1 C X 0≤t≤1 t · ct obtaining a simpler approximation i ≈ I A + (1 −k)C = I A + (1 −k)(B −A −I) = I kA + (1 −k)B −(1 −k)I . Example 33.3 A fund has 100,000 on January 1 and 125,000 on December 31. Interest earned by the fund during the year totaled 10,000. Calculate the net yield earned by the fund assuming that the net contributions occurred on April 1. 33 INTEREST MEASUREMENT OF A FUND: DOLLAR-WEIGHTED INTEREST RATE 305 Solution. Using the formula, i ≈ I [kA+(1−k)B−(1−k)I], we have i ≈ 10, 000 [(1/4)100, 000 + (3/4)125, 000 −(3/4)10, 000] = 10, 000 111, 250 = 8.98876404% One useful special case is when the principal deposits or withdrawals occur uniformly throughout the period. Thus, on average, we might assume that net principal contributions occur at time k = 1 2, in which case we have i ≈ I 0.5A + 0.5B −0.5I = 2I A + B −I . (33.5) This formula has been used for many years by insurance company regulators in both Canada and the US to allow them to get a feel for the investment returns a particular company is obtaining. Example 33.4 You have a mutual fund. Its balance was $10,000 on 31/12/1998. You made monthly contributions of $100 at the start of each month and the ﬁnal balance was $12,000 at 31/12/1999. What was you approximate return during the year? Solution. We are given that A = $10, 000, B = $12, 000, and C = $1, 200. Thus, I = B −(A + C) = $800 and i = 2(800) 10, 000 + 12, 000 −800 = 7.54717% Example 33.5 Find the eﬀective rate of interest earned during a calendar year by an insurance company with the following data: Assets, beginning of year 10,000,000 Premium income 1,000,000 Gross investment income 530,000 Policy beneﬁts 420,000 Investment expenses 20,000 Other expenses 180,000 306 RATE OF RETURN OF AN INVESTMENT Solution. We are given A =10, 000, 000 B =10, 000, 000 + 1, 000, 000 + 530, 000 −(420, 000 + 20, 000 + 180, 000) =10, 910, 000 I =530, 000 −20, 000 = 510, 000 Thus, i = 2I A + B −I = 2(510, 000) 10, 000, 000 + 10, 910, 000 −510, 000 = 5% 33 INTEREST MEASUREMENT OF A FUND: DOLLAR-WEIGHTED INTEREST RATE 307 Practice Problems Problem 33.1 A fund has 10,000 at the start of the year. During the year $5,000 is added to the fund and $2,000 is removed. The interest earned during the year is $1,000. Which of the following are true? (I) The amount in the fund at the end of the year is $14,000. (II) If we assume that any deposits and withdrawals occur uniformly throughout the year, i is approximately 8.33%. (III) If the deposit was made on April 1 and the withdrawal was made on August 1, then i is approximately 7.74%. Problem 33.2 The funds of an insurance company at the beginning of the year were $500,000 and at the end of the year $680,000. Gross interest earned was $60,000, against which there were investment expenses of $5,000. Find the net eﬀective rate of interest yielded by the fund. Problem 33.3 A fund earning 4% eﬀective has a balance of $1,000 at the beginning of the year. If $200 is added to the fund at the end of three months and if $300 is withdrawn from the fund at the end of nine months, ﬁnd the ending balance under the assumption that simple interest approximation is used. Problem 33.4 At the beginning of the year an investment fund was established with an initial deposit of $1,000. A new deposit of $500 was made at the end of four months. Withdrawals of $200 and $100 were made at the end of six months and eight months, respectively. The amount in the fund at the end of the year is $1,272. Find the approximate eﬀective rate of interest earned by the fund during the year using the dollar-weighted rate of return formula. Problem 33.5 ‡ An association had a fund balance of 75 on January 1 and 60 on December 31. At the end of every month during the year, the association deposited 10 from membership fees. There were withdrawals of 5 on February 28, 25 on June 30, 80 on October 15, and 35 on October 31. Calculate the dollar-weighted rate of return for the year. Problem 33.6 ‡ An insurance company earned a simple rate of interest of 8% over the last calendar year based on the following information: 308 RATE OF RETURN OF AN INVESTMENT Assets, beginning of year 25,000,000 Sales revenue X Net investment income 2,000,000 Salaries paid 2,200,000 Other expenses paid 750,000 All cash ﬂows occur at the middle of the year. Calculate the eﬀective yield rate. Problem 33.7 ‡ At the beginning of the year, an investment fund was established with an initial deposit of 1,000. A new deposit of 1,000 was made at the end of 4 months. Withdrawals of 200 and 500 were made at the end of 6 months and 8 months, respectively. The amount in the fund at the end of the year is 1,560. Calculate the dollar-weighted yield rate earned by the fund during the year. Problem 33.8 On January 1, an investment account is worth $500,000. On April 1, the account value has increased to $530,000, and $120,000 of new principal is deposited. On August 1, the account value has decreased to $575,000, and $250,000 is withdrawn. On January 1 of the following year, the account value is $400,000. Compute the yield rate using the dollar-weighted method. Problem 33.9 A fund has 10,000 at the beginning of the year and 12,000 at the end of the year. Net contributions of 1,000 were made into the fund during year. Calculate the net yield earned by the fund assuming that the net contributions were contributed uniformly throughout the year. Problem 33.10 A fund has 100,000 on January 1 and 125,000 on December 31. Interest earned by the fund during the year totaled 10,000. The net yield earned by the fund during the year was 9.6385%. Two contributions were made to the fund during the year and there were no withdrawals. The contributions were for equal amounts made two months apart. Determine the date of the ﬁrst contribution. Problem 33.11 An investor fund has a balance on January 1 of $273,000 and a balance on December 31 of $372,000. The amount of interest earned during the year was $18,000 and the computed yield rate on the fund was 6%. What was the average date for contributions to and withdrawals from the fund? 33 INTEREST MEASUREMENT OF A FUND: DOLLAR-WEIGHTED INTEREST RATE 309 Problem 33.12 You deposited 5,000 into a fund on January 1, 2002. On March 1,2002, you withdrew 1,000 from the fund. On August 1, 2002, you deposited 700 into the fund. On October 1, 2002, you deposited another 750 into the fund. If your annual dollar-weighted rate of return on your investment was 12.00%, what was your fund worth on December 31, 2002? Problem 33.13 At the beginning of the year a fund has deposits of 10,000. A deposit of 1000 is made after 3 months and a withdrawal of 2000 is made after 9 months. The amount in the fund at the end of the year is 9500. Find the dollar weighted rate of interest (a) using formula (33.3) (b) using formula (33.4) (c) using formula (33.5). Problem 33.14 Deposits of 10000 are made into an investment fund at time zero and 1. The fund balance at time 2 is 21000. Compute the eﬀective yield using the dollar weighted method. Problem 33.15 An investor puts $100 into a mutual fund in the ﬁrst year and $50 in the second year. At the end of the ﬁrst year the mutual pays a dividend of $10. The investor sells the holdings in the mutual fund at the end of the second year for $180. Find the dollar weighted return. Problem 33.16 A fund has 100,000 at the start of the year. Six months later, the fund has a value of 75,000 at which time, Stuart adds an additional 75,000 to the fund. At the end of the year, the fund has a value of 200,000. Calculate the exact dollar weighted rate of return( i.e. using (33.3)). Problem 33.17 A fund has $2,000,000 at the beginning of the year. On June 1 and September 1, two withdrawals of $100,000 each were made. On November 1, a contribution of $20,000 was made. The value in the fund at the end of the year is $1,900,000. Find (a) the exact dollar weighted rate of return; (b) the dollar weigthed rate of return using formula (33.4). Problem 33.18 The following table shows the history of account balances, deposits, and withdrawals for an account 310 RATE OF RETURN OF AN INVESTMENT during 2008: Date Value before dep/withd Dep/Withd January 1,2008 10,000 0 March 1,2008 9,500 3,000 June 1,2008 14,000 −2, 000 September 1,2008 12,000 −2, 000 January 1,2009 12,000 What is this account’s dollar-weighted rate of return?(Assume 30-day months.) Problem 33.19 On January 1, 2000, the balance in account is $25200. On April 1, 2000, $500 are deposited in this account and on July 1, 2001, a withdraw of $1000 is made. The balance in the account on October 1, 2001 is $25900. What is the annual rate of interest in this account according with the dollarweighted method? Problem 33.20 You invest $10,000 in a fund on January 1, 2008. On May 1, 2008, you withdraw $3,000 from the fund. On November 1, 2008, you deposit $4,000 into the fund. On December 31, 2008, your fund is worth $13,500. What was the annual dollar-weighted rate of return on your investment? Problem 33.21 You invest $5,000 in a fund on 1/1/07. On 3/1/07, you withdraw X from the fund. On 7/1/07, you deposit $2,500 into the fund. On 9/1/07, you withdraw $1,000 from the fund. On 12/31/07, your fund is worth $5,700. The annual dollar−weighted rate of return on your investment was 12.0%. Find X. 34 INTEREST MEASUREMENT OF A FUND: TIME-WEIGHTED RATE OF INTEREST 311 34 Interest Measurement of a Fund: Time-Weighted Rate of Interest The dollar-weighted rate of interest depends on the precise timing and amount of the cash ﬂows. In practice, professional fund managers who direct investment funds have no control over the timing or amounts of the external cash ﬂows. Therefore, if we are comparing the performance of diﬀerent fund managers, the dollar weighted rate of interest doesn’t always provide a fair comparison. In this section, we consider an alternative measure of performance that does not depend on the size or the timing of the cash ﬂows, namely the time-weighted rate of interest also known as the time-weighted rate of return. Consider the following one-year investment. An amount X is invested in a fund at the beginning of the year. In six months, the fund is worth X 2 at which time the investor can decide to add to the fund, to withdraw from the fund, or to do nothing. At the end of the year, the fund balance is double the balance at six months. We next examine three situations related to the above investment: Consider three investors A, B, and C. Investor A initially invests X = $1, 000. At the end of six months his investment is worth $500. He decideds not to withdraw from the fund or deposits into the fund. His fund at the end of the year worths $1,000. Since the interest earned is I = 0, the dollar-weighted rate of interest is i = 0. Next, investor B invests initially X = $1, 000. Again, after six months the balance is $500. At the end of six months the investor withdraws half the fund balance (i.e. $250). His balance in the fund at the end of the year is $500. The equation of value at t = 1 for this transaction is 1000(1 + i) −250(1 + i) 1 2 = 500. Solving this quadratic equation we ﬁnd (1 + i) 1 2 = 0.84307 and this implies a yield rate of i = −0.2892 or −28.92%. Finally, investor C invests initially X = $1, 000 and at six months deposits an amount equal to the fund balance(i.e. $500). Hence, his balance at the end of the year is $2,000. The equation of value at time t = 1 is 1000(1 + i) + 500(1 + i) 1 2 = 2000. Solving this quadratic equation we ﬁnd (1 + i) 1 2 = 1.18614. Hence, the yield rate in this case is i = 0.4069 or 40.69%. The yield rate for investor C is so much better than the yield rate for either investor A or investor B, because of the decisions made by each investor. The dollar-weighted rate of interest measures both the behavior of the fund and the skills of the investor. If the investor did not perform any transactions, then the dollar-weighted rate of interest would be close to 0%. Now how can we evaluate the decisions made by the fund manager? In the illustration above, we ﬁnd the yield rate for the ﬁrst six months to be j1 = −50% and for the second six months to be j2 = 100%. If i is the yield rate for the entire year then 1 + i = (1 + j1)(1 + j2) = 1 312 RATE OF RETURN OF AN INVESTMENT which implies that i = 0 and this is regardless of when cash is deposited or withdrawn. This indicates that the manager did a poor job of maintaining the fund. Yield rates computed by considering only changes in interest rate over time (which is what was previously done to evaluate the fund manager’s performance) are called time-weighted rates of interest. We can generalize the above approach as follows. Suppose m −1 deposits or withdrawals are made during a year at times t1, t2, · · · , tm−1 (so that no contributions at t0 = 0 and tm = 1). Thus, the year is divided into m subintervals. For k = 1, 2, · · · , m we let jk be the yield rate over the kth subinterval. For k = 1, · · · , m −1, let Ctk be the net contribution at exact time tk and Btk the value of the fund before the contribution at time tk. Note that C0 = Cm = 0 and B0 is the initial investment and B1 is the value of the fund at the end of the year. The yield rate of the fund from time tk−1 to time tk satisﬁes the equation of value Btk = (1 + jk)(Btk−1 + Ctk−1) or 1 + jk = Btk Btk−1 + Ctk−1 , k = 1, 2, · · · , m. The overall yield rate i for the entire year is given by 1 + i = (1 + j1)(1 + j2) · · · (1 + jm) or i = (1 + j1)(1 + j2) · · · (1 + jm) −1. We call i the time-weighted rate of return. Example 34.1 Suppose that an investor makes a series of payments and withdrawals, as follows: Date Flow Balance before Balance after 01/01/2003 0 100,000 100,000 30/06/2003 +1,000,000 74,681 1,074,681 31/12/2003 0 1,474,081 1,474,081 (a) Compute the dollar-weighted rate of interest that the investor has realized. (b) Compute the time-weigted rate of interest. Solution. (a) We assume that the contribution of $1,000,000 occurred exactly in the middle of the year. We 34 INTEREST MEASUREMENT OF A FUND: TIME-WEIGHTED RATE OF INTEREST 313 have A = $100, 000, B = $1, 474, 081, C = $1, 000, 000. Hence, I = 1, 474, 081 −1, 000, 000 − 100, 000 = 374, 081 and i = 2(374, 081) 100, 000 + 1, 474, 081 −374, 081 = 62.35%. (b) We have i = 74, 681 100, 000 1, 474, 081 1, 074, 681 −1 = 2.44% It follows from the previous example that the time-weighted method does not provide a valid measure of yield rate as does the dollar weighted method. However, the time-weighted method does provide a better indicator of underlying investment performance than the dollar-weighted method. Example 34.2 Your balance at time 0 is $2,000. At time t = 1 3, the balance has grown to $2,500 and a contribution of $1,000 is made. At time t = 2 3, the balance has dropped to $3,000 and a withdrawal of $1,500 is made. At the end of the year, the fund balance is $2,000. What is the time weighted rate of return? Solution. We create the following chart k tk Btk Ctk jk 0 0.0000 $2,000 $0 0% 1 1/3 $2,500 $1,000 25.00% 2 2/3 $3,000 -$1,500 -14.29% 3 1 $2,000 $0 33.33% The time weighted rate of return is i = (1.25)(.8571)(1.3333) −1 = .428464 Example 34.3 You are given the following information about an investment account: Date June 30,01 Sep. 30,01 Dec. 31,01 March 31,02 June 30,02 Bk 12,000 10,000 14,000 13,000 X Deposit 2,000 2,000 Withdrawal 2,000 2,000 If the eﬀective annual dollar-weighted rate of return from June 30, 2001 to June 30, 2002 was exactly 10%, what was the eﬀective annual time-weighted rate of return during that same period? 314 RATE OF RETURN OF AN INVESTMENT Solution. First, we ﬁnd X : X = 10, 000(1.1) + 2, 000(1.1)0.75 −2, 000(1.1)0.5 + 2, 000(1.1)0.25 = 13, 098.81 The time-weighted rate of return is then i = 10, 000 12, 000 −2, 000 · 14, 000 10, 000 + 2, 000 · 13, 000 14, 000 −2, 000 · 13, 098.81 13, 000 −2, 000 −1 = 10.37% 34 INTEREST MEASUREMENT OF A FUND: TIME-WEIGHTED RATE OF INTEREST 315 Practice Problems Problem 34.1 Which of the following are true? (I) Time-weighted method provides better indicator for the fund underlying performance than the dollar-weighted method. (II) Time weighted rates of interest will always be higher than dollar weighted rates of interest. (III) Dollar weighted rate of interest provide better indicators of underlying investment performance than do time weighted rates of interest. (IV) Dollar weighted rates of interest provide a valid measure of the actual investment results. Problem 34.2 A fund has 1,000 at beginning of the year. Half way through the year, the fund value has increased to 1,200 and an additional 1,000 is invested. At the end of the year, the fund has a value of 2,100. (a) Calculate the exact dollar weighted rate of return using compound interest. (b) Calculate the estimated dollar weighted rate of return using the simple interest assumptions. (c) Calculate the time weighted rate of return. Problem 34.3 A fund has 100,000 at the start of the year. Six months later, the fund has a value of 75,000 at which time, Stuart adds an additional 75,000 to the fund. At the end of the year, the fund has a value of 200,000. Calculate the time weighted rate of return. Problem 34.4 A fund has 100,000 at the start of the year. Six months later, the fund has a value of 75,000 at which time, Stuart adds an additional 75,000 to the fund. At the end of the year, the fund has a value of 200,000. Calculate the exact dollar weighted rate of return. Problem 34.5 A fund has 10,000 at the start of the year. Six months later, the fund has a value of 15,000 at which time, Stuart removes 5,000 from the fund. At the end of the year, the fund has a value of 10,000. Calculate the exact dollar weighted rate of return less the time weighted rate of return. Problem 34.6 ‡ You are given the following information about an investment account: 316 RATE OF RETURN OF AN INVESTMENT Date Value Immediately Deposit Before Deposit January 1 10 July 1 12 X December 31 X Over the year, the time-weighted return is 0%, and the dollar-weighted (money-weighted) return is Y. Determine Y. Problem 34.7 ‡ An investor deposits 50 in an investment account on January 1 . The following summarizes the activity in the account during the year: Date Value Immediately Deposit Before Deposit March 15 40 20 June 1 80 80 October 1 175 75 On June 30, the value of the account is 157.50 . On December 31, the value of the account is X. Using the time-weighted method, the equivalent annual eﬀective yield during the ﬁrst 6 months is equal to the (time-weighted) annual eﬀective yield during the entire 1-year period. Calculate X. Problem 34.8 On January 1, an investment account is worth $100,000. On May 1, the account value has increased to $112,000, and $30,000 of new principal is deposited. On November 1, the account value has decreased to $125,000, and $42,000 is withdrawn. On January 1 of the following year, the account value is $100,000. Compute the yield rate using (a) the dollar-weighted method and (b) the time-weighted method. Problem 34.9 In Problem 34.8, change May 1 to June 1 and November 1 to October 1. (a) Would the yield rate change when computed by the dollar-weighted method? (b) Would the yield rate change when computed by the time-weighted method? Problem 34.10 In Problem 34.8, assume that everything is unchanged except that an additional $5,000 is withdrawn on July 1. (a) Would the yield rate change when computed by the dollar-weighted method? (b) Explain why the yield rate cannot be computed by the time-weighted method. 34 INTEREST MEASUREMENT OF A FUND: TIME-WEIGHTED RATE OF INTEREST 317 Problem 34.11 A mutual fund account has the balance $100 on January 1. On July 1 (exactly in the middle of the year) the investor deposits $250 into the account. The balance of the mutual fund at the end of the year is $400. (a) Calculate the exact dollar-weighted annual rate of interest for the year. (b) Suppose that the balance of the fund immediately before the deposit on July 1 is $120. Calculate the time-weighted annual rate of interest for the year. Problem 34.12 You are given the following information about two funds: Fund X: January 1, 2005: Balance 50,000 May 1, 2005: Deposit 24,000; Pre-deposit Balance 50,000 November 1, 2005: Withdrawal 36,000; Pre-withdrawal Balance 77,310 December 31, 2005: Balance 43,100 Fund Y: January 1, 2005: Balance 100,000 July 1, 2005: Withdrawal 15,000; Pre-withdrawal Balance 105,000 December 31, 2005: Balance F. Fund Y’s time-weighted rate of return in 2005 is equal to Fund X’s dollar-weighted rate of return in 2005. Calculate F. Problem 34.13 You invested 10,000 in a fund on January 1, 2000. On April 1, 2000, your fund was worth 11,000, and you added 2,000 to it. On June 1, 2000, your fund was worth 12,000 and you withdrew 2,000 from the fund. On December 31, 2000, your fund was worth 9,500. What was the time-weighted rate of return on your investment from January 1, 2000 to December 31, 2000? Problem 34.14 Deposits of $1000 are made into an investment fund at time 0 and 1. The fund balance is $1200 at time 1 and $2200 at time 2. (a) Compute the annual eﬀective yield rate computed by a dollar-weighted method. (b) Compute the annual eﬀective yield rate which is equivalent to that produced by a time-weighted method. Problem 34.15 Let A be the fund balance on January 1, B the fund balance on June 30, and C the balance on December 31. 318 RATE OF RETURN OF AN INVESTMENT (a) If there are no deposits or withdrawals, show that yield rates computed by the dollar-weighted method and the time-weighted method are both equal. (b) If there was a single deposit of D immediately after the June 30 balance was calculated, ﬁnd expressions for the dollar-weighted and time-weighted yield rates. (c) If there was a single deposit of D immediately before the June 30 balance was calculated, ﬁnd expressions for the dollar-weighted and time-weighted yield rates. (d) Compare the dollar-weighted yield rates in (b) and (c). (e) Compare the time-weighted yield rates in (b) and (c). Problem 34.16 100 is deposited into an investment account on January 1, 1998. You are given the following information on investment activity that takes place during the year: April 19,1998 October 30,1998 Value immediately prior to deposit 95 105 deposit 2X X The amount in the account on January 1, 1999 is 115. During 1998, the dollar-weighted return is 0% and the time-weighted return is Y . Calculate Y. Problem 34.17 An investment manager’s portfolio begins the year with a value of 100,000. Eleven months through the year a withdrawal of 50,000 is made and the value of the portfolio after the withdrawal is 57,000. At the end of the year the value of the portfolio is 60,000. Find the time-weighted yield rate less the dollar-weighted yield rate. Problem 34.18 On January 1, 1997 an investment account is worth 100,000. On April 1, 1997 the value has increased to 103,000 and 8,000 is withdrawn. On January 1, 1999 the account is worth 103,992. Assuming a dollar weighted method for 1997 and a time weighted method for 1998 the annual eﬀective interest rate was equal to x for both 1997 and 1998. Calculate x. Problem 34.19 On January 1, 1999, Luciano deposits 90 into an investment account. On April 1, 1999, when the amount in Luciano’s account is equal to X, a withdrawal of W is made. No further deposits or withdrawals are made to Luciano’s account for the remainder of the year. On December 31, 1999, the amount in Luciano’s account is 85. The dollar weighted return over the 1 year period is 20%. The time weighted return over the 1 year period is 16%. Calculate X. 34 INTEREST MEASUREMENT OF A FUND: TIME-WEIGHTED RATE OF INTEREST 319 Problem 34.20 You are given the following information about the activity in two diﬀerent investment accounts. Account K Activity Date Fund value before activity Deposit Withdrawal January 1,1999 100 July 1,1999 125 X October 1,1999 110 2X December 31,1999 125 Account L Actyivity Date Fund value before activity Deposit Withdrawal January 1,1999 100 July 1,1999 125 X December 31,1999 105.8 During 1999 the dollar weighted return for investment account K equals the time weighted return for investment account L, which equals i. Calculate i. Problem 34.21 The following table shows the history of account balances, deposits, and withdrawals for an account during a particular year: Date Value before dep/withd Dep/Withd January 1,X 1,000 0 May 1,X 1,050 1,100 September 1,X 2,250 −900 January 1,X+1 16,00 What is the relationship among this account’s dollar-weighted rate of return (DW), its time-weighted rate of return (TW), and its internal rate of return (IRR)?Assume 30-day months. 320 RATE OF RETURN OF AN INVESTMENT 35 Allocating Investment Income: Portfolio and Investment Year Methods Consider an investment fund that involves many investors with seperate accounts. An example would be a pension fund for retirement in which each plan participant has an account but with commingled pool of assets. Each investor owns shares in the fund. In this section we consider the issue of allocating investment income to the various accounts. Two common methods are in use: the portfolio method and the investment year method also known as the new money method. The Portfolio Method Under the portfolio method an average rate based on the earnings of the entire fund is computed and credited to each account. All new deposits will earn this same portfolio rate. There are disadvantages of this method during time of ﬂuctuating interest rates. For example, sup-pose that the market rates have been rising signiﬁcantly during the recent years while the average portfolio rate is being lowered. In this case, there is a higher tendency for less new deposits to the fund and more withdrawals from the fund. The investment year method is an attempt to address this problem and to allocate investment income in a more equitable manner. Investment Year Method (IYM) We will describe this method on an annual basis, although typically smaller time intervals would be used. Under the investment year method, an investors’s contribution will be credited during the year with the interest rate that was in eﬀect at the time of the contribution. This interest rate is often referred to as the new money rate. For example, a portion of the IYM chart for a fund might look something like the one shown in Table 35.1. Suppose we invested $1,000 on January 1, 1994 and $500 on January 1, 1995. Then our total accumulation on Jan. 1, 2003 would be (1.064)(1.068)(1.071)(1.069)(1.073)(1.051)(1.049)(1.048)(1.047)1000 = 1688.75 from our 1994 contribution, plus (1.069)(1.070)(1.070)(1.074)(1.050)(1.049)(1.048)(1.047)500 = 794.31 from our 1995 contribution, for a total of 1688.75 + 794.31 = 2483.06. 35 ALLOCATING INVESTMENT INCOME: PORTFOLIO AND INVESTMENT YEAR METHODS321 Purchase Year 1994 1995 1996 1997 1998 1999 2000 2001 2002 1994 6.4% 6.8% 7.1% 6.9% 7.3% 5.1% 4.9% 4.8% 4.7% 1995 6.9% 7.0% 7.0% 7.4% 5.0% 4.9% 4.8% 4.7% 1996 7.1% 7.3% 7.3% 5.5% 5.4% 4.8% 4.7% 1997 7.0% 7.4% 5.4% 5.2% 4.6% 4.7% 1998 7.2% 5.7% 5.5% 4.5% 4.4% 1999 5.8% 5.1% 4.3% 4.7% 2000 5.0% 4.1% 4.6% 2001 4.0% 4.5% 2002 4.1% Table 35.1 Notice that from the chart above, investments made between 1994-1996 all earned the same rate of return both in 2001 (4.8%) and 2002 (4.7%). This illustrates the principle that under the investment year method, funds on investment longer than a certain ﬁxed period (5 years in our example) are assumed to grow at the overall yield rate for the fund (the portfolio rate), regardless of when the funds were invested. To better indicate the switch from the investment year method to the portfolio method, the data from the chart above would typically be displayed by a two-dimensional table such as the one in Table 35.2 If y is the calendar year of deposit, and m is the number of years for which the investment year method is applicable, then the rate of interest credited for the tth year of investment is denoted as iy t for t = 1, 2, · · · , m. For t > m, the portfolio method is applicable, and the portfolio rate of interest credited for calendar year y is denoted as iy. The investment year method is more complicated than the portfolio method, but it was deemed necessary to attract new deposits and to discourage withdrawals during periods of rising interest rates. We illustrate the use of the investment year method in the following example. Example 35.1 You are given the following table of interest rates (in percentages). A person deposits 1,000 on January 1, 1997. Find the accumulated amount on January 1, 2000 (a) under the investment year method; (b) under the portfolio method; (c) when the balance is withdrawn at the end of every year and is reinvested at the new money rate. 322 RATE OF RETURN OF AN INVESTMENT y iy 1 iy 2 iy 3 iy 4 iy 5 iy+5 Portfolio Year 1992 8.25 8.25 8.4 8.5 8.5 8.35 1997 1993 8.5 8.7 8.75 8.9 9.0 8.6 1998 1994 9.0 9.0 9.1 9.1 9.2 8.85 1999 1995 9.0 9.1 9.2 9.3 9.4 9.1 2000 1996 9.25 9.35 9.5 9.55 9.6 9.35 2001 1997 9.5 9.5 9.6 9.7 9.7 1998 10.0 10.0 9.9 9.8 1999 10.0 9.8 9.7 2000 9.5 9.5 2001 9.0 Table 35.2 Solution. (a) The sequence of interest rates beginning from a given year of investment runs horizontally through the row for that year and then down the last column of rates. An investment made at the beginning of 1992 would earn the investment year rate of 8.25% in 1992, the money rate of 8.25% in 1993, 8.4% in 1994, 8.5% in 1995 and 1996. Starting in 1997, an investment made at the beginning of 1992 would earn the portfolio average rates. The 1992 investment would earn the portfolio rate of 8.35% in 1997, the portfolio rate of 8.6% in 1998, and so on. An investment made at the beginning of 2001 would earn the investment year rate of 9% in 2001-2005. Starting the year 2006, the investment made in 2001 would earn the portfolio rate of 8.35% in 2006, the portfolio rate of 8.6% in 2007, and so on. (a) The accumulated value is 1000 · (1.095)(1.095)(1.096) = 1, 314.13. (b) The accumulated value is Q = 1000 · (1.0835)(1.086)(1.0885) = 1, 280.82. (c) The accumulated value is 1000 · (1.095)(1.1)(1.1) = 1, 324.95 Example 35.2 An investment of 1000 is made on January 1996 in an investment fund crediting interest according to Table 35.2. How much interest is credited from January 1, 1999 to January 1, 2001? Solution. The accumulated value on January 1, 1999 is 1000(1.0925)(1.0935)(1.095) = 1308.14. The accumulated value on January 1, 2001 is 1000(1.0925)(1.0935)(1.095)(1.0955)(1.096) = 1570.64. 35 ALLOCATING INVESTMENT INCOME: PORTFOLIO AND INVESTMENT YEAR METHODS323 Thus, the total amount of interest credited is 1570.64 −1308.14 = 262.50 Example 35.3 Using Table 35.2, ﬁnd (a) the interest rates credited in calendar year 2000 for deposits made in 1994 - 2000; (b) the new money rates credited in the ﬁrst year of investment for deposits made in 1992 - 2001. Solution. (a) 9.5% credited for new deposits made in 2000, 9.8% for deposits made in 1999, 9.9% for deposits made in 1998, 9.7% for deposits made in 1997, 9.6% for deposits made in 1996, 9.1% for deposits made in 1994 - 1995. Thus, the interest credited in the calendar year 2000 appear on an upwardly diagonal within the table. (b) The new money rates credited in the ﬁrst year of investment appear in the ﬁrst column of the table 324 RATE OF RETURN OF AN INVESTMENT Practice Problems Problem 35.1 Using Table 35.2, what were the interest rates credited in calendar year 2001 for deposits made in 2001 and before 2001? Problem 35.2 If 100 is invested in the fund at the beginning of 2002, ﬁnd the accumulated amount at the end of 2003 using: (a) The investment year method (b) The portfolio method (c) The investment year method if the amount is withdrawn and then reinvested at the end of 2002. y iy 1 iy 2 iy 3 iy 4 iy+4 Portfolio Year 1998 7.0 6.5 6.0 5.8 5.9 2002 1999 6.4 6.1 5.8 5.9 6.0 2003 2000 6.2 6.0 5.9 6.0 6.2 2004 2001 6.1 5.9 6.1 6.4 6.6 2005 2002 6.0 6.1 6.3 6.6 2003 6.4 6.6 6.7 2004 6.8 7.0 2005 7.5 Problem 35.3 The following table lists the interest rate credited under an investment year method of crediting interest. y iy 1 iy 2 iy 3 iy 4 iy 5 iy+5 1999 7.0 6.75 6.5 6.25 6.0 5.5 2000 6.0 5.5 5.25 5.1 5.0 2001 5.0 4.8 4.6 4.3 2002 4.0 3.75 3.5 2003 3.0 3.2 2004 4.0 Becky invests $1,000 on January 1, 1999 and an additional $500 on January 1, 2003. How much money does Becky have on December 31, 2004? Use the following chart of interest rates to answer Problems 35.4 through 35.6. 35 ALLOCATING INVESTMENT INCOME: PORTFOLIO AND INVESTMENT YEAR METHODS325 y iy 1 iy 2 iy 3 iy 4 iy+4 Calendar year of portfolio rate y + 4 1997 0.0650 0.0625 0.0600 0.0575 0.0550 2001 1998 0.0600 0.0575 0.0550 0.0525 0.0500 2002 1999 0.0500 0.0475 0.0460 0.0450 0.0450 2003 2000 0.0450 0.0440 0.0430 0.0420 0.0410 2004 2001 0.0400 0.0390 0.0380 0.0370 2002 0.0300 0.0300 0.0325 2003 0.0300 0.0325 2004 0.0300 Problem 35.4 Jordan invests 1,000 in a fund on January 1, 1998. The fund uses the investment year method of determining interest rates. Calculate the amount that Jordan will have at the end of 2004. Problem 35.5 Jenna invests 1,000 in a fund on January 1, 2002. The fund uses the portfolio method of determining interest rates. Calculate the amount that Jenna will have at the end of 2004. Problem 35.6 James invests 1,000 into a fund at the beginning of each year from 2002 to 2004. The fund pays interest using the investment year interest rate. Calculate the amount of money that James will have at the end of 2004. Use the following chart of interest rates to answer Questions 35.7 through 35.10. y iy 1 iy 2 iy 3 iy+3 1995 3.7% 3.6% 3.5% 6.0% 1996 3.2% 3.1% 3.0% 5.5% 1997 2.7% 2.6% 2.5% 5.0% 1998 2.2% 2.1% 2.0% 4.5% 1999 1.7% 1.6% 1.5% 4.0% Problem 35.7 A deposit of $100 is made at the beginning of 1997. How much interest was credited during 1998? What is the accumulated value at the end of 2002? 326 RATE OF RETURN OF AN INVESTMENT Problem 35.8 How much interest is credited in the calendar years 1997 through 1999 inclusive to a deposit of $100 made at the beginning of 1995? Problem 35.9 What were the interest rates credited in calendar year 1999 for deposits made in 1999, 1998, 1997, 1996, and 1995? Problem 35.10 What were the new money rates credited in the ﬁrst year of investment for deposits made in 1995, 1996, 1997, and so on? Problem 35.11 The following table shows the annual eﬀective interest rates credited to an investment fund by calendar year of investment. The investment year method applies for the ﬁrst two years, after which a portfolio method is used. y iy 1 iy 2 iy+2 1995 t 5.5% 4.5% 1996 6.0% 6.1% 5.0% 1997 7.0% t + 2.5% 5.5% An investment of $100 is made at the beginning of 1995 and 1997. The total amount of interest credited by the fund during the year 1998 is $13.81. Find t. Problem 35.12 It is known that 1 + iy t = (1.08 + 0.005t)1+0.01y for integral t, 0 ≤t ≤5, and integral y, 0 ≤y ≤10. If $1,000 is invested for three years beginning in year y = 5, ﬁnd the equivalent eﬀective rate of interest. Problem 35.13 For a certain portfolio, iy t = (1.06 + 0.004t)1+0.01y for 1 ≤t ≤5 and 0 ≤y ≤10. (a) What is the accumulated value at the end of year 4 of a single investment of 1,000 at the beginning of year 2? (b) What is the accumulated value at the end of year 4 of a deposit of 1,000 at the beginning of each of years 2 through 4? Problem 35.14 ‡ The following table shows the annual eﬀective interest rates being credited by an investment account, by calendar year of investment. The investment year method is applicable for the ﬁrst 3 years, after which a portfolio rate is used. 35 ALLOCATING INVESTMENT INCOME: PORTFOLIO AND INVESTMENT YEAR METHODS327 y iy 1 iy 2 iy 3 iy+3 1990 10% 10% t% 8% 1991 12% 5% 10% (t −1)% 1992 8% (t −2)% 12% 6% 1993 9% 11% 6% 9% 1994 7% 7% 10% 10% An investment of 100 is made at the beginning of years 1990, 1991, and 1992. The total amount of interest credited by the fund during the year 1993 is equal to 28.40. Calculate t. Problem 35.15 Find ¨ s5 from the data in Table 35.2 assuming the ﬁrst payment is made in calendar year 1995. Problem 35.16 You are given the following table of interest rates: y iy 1 iy 2 iy+5 Portfolio Year 2004 9.00% 10.00% 11.00% 2006 2005 7.00% 8.00% 2006 5.00% 1000 is invested at the beginning of calendar years 2004, 2005, and 2006. What is the total amount of interest credited for calendar year 2006? Problem 35.17 You are given: y iy 1 iy 2 iy 3 iy+5 2001 5.0 5.5 6.0 3.0 2002 4.5 5.0 5.5 2.5 2003 3.0 3.5 4.0 2.0 2004 4.0 X 5.0 2005 5.0 5.5 6.0 2006 4.5 5.0 5.5 The accumulated value on 1/1/2007 of a deposit on 1/1/2001 is 9.24% greater than the accumulated value on 1/1/2007 of the same amount deposited on 1/1/2004. Determine X. 328 RATE OF RETURN OF AN INVESTMENT 36 Yield Rates in Capital Budgeting One of the use of yield rates is in capital budgeting. An investor always is faced with the need to allocate an amount of capital among various alternative investments in order to achieve the highest possible level of return on the investments. This process of making such ﬁnancial decisions is referred to as capital budgeting. In this section we discuss brieﬂy this concept. Moreover, we will assume that risk is non-existent in the alternative investments being compared. The two major approaches to capital budgeting that are encountered in practice are the yield rate method and the net present value method. In the yield rate method the investor computes the yield rate(s) for each alternative investment by solving the equation NPV (i) = 0. Then these rates are compared to an interest preference rate set by the investor. This is usually the minimum rate of return acceptable by the investor. All yield rates that are lower than the interest preference rate are rejected. The yield rates that are higher than the interest preference rateare the only one considered. They are ranked from highest to lowest and are selected in descending order until the amount of capital available for investment is exhausted. In the net present value method or the NPV method, the investor computes NPV (i) for each alternative investment, where i is the interest preference rate. Negative values of NPV (i) are rejected and only the positive values are considered since the present value of returns is larger than that of the contributions. Capital is then allocated among those investments with positive NPV (i) in such a manner that the total present value of returns from the investment (computed at the interest preference rate) minus the contributions to the investment is maximized. It has been shown in Finance Theory that the NPV method usually provides better decisions than other methods when making capital investments. Consequently, it is the more popular evaluation method of capital budgeting projects. Example 36.1 Consider the investment project given in the table below. (a) Find the yield rate of this project. (b) Assuming an interest preference rate of 3%, would you accept this project when using the yield rate method? the net present value method? Solution. (a) Solving the equation NPV (i) = −80, 000 −10, 000ν −10, 000ν2 −10, 000ν3 −8, 000ν4 + 28, 000ν5 + 38, 000ν6 + 33, 000ν7 + 23, 000ν8 + 13, 000ν9 + 8, 000ν10 = 0 we ﬁnd the two solutions i = −1.714964 < −1 and i = 0.032180. Thus, the yield rate is 3.218%. (b) Using the yield rate method, since 3.218% > 3%, the investor would accept this project 36 YIELD RATES IN CAPITAL BUDGETING 329 for consideration. Using the net present value method the investor would also accept it, since NPV (0.03) = 1488.04 > 0 Year Contributions Returns ct 0 80,000 0 −80, 000 1 10,000 0 −10, 000 2 10,000 0 −10, 000 3 10,000 0 −10, 000 4 20,000 12,000 −8, 000 5 2,000 30,000 28,000 6 2,000 40,000 38,000 7 2,000 35,000 33,000 8 2,000 25,000 23,000 9 2,000 15,000 13,000 10 0 8,000 8,000 Example 36.2 Repeat the same problem as above but with an interest preference rate of 4%. Solution. Using the yield rate method, since 4% > 3.218%, the investor would reject this project. Using the net present value method the investor would also reject it, since NPV (0.04) = −5122.13 < 0 330 RATE OF RETURN OF AN INVESTMENT Practice Problems Problem 36.1 An investment project has the following cash ﬂows: Year Contributions Returns 0 100 0 1 200 0 2 10 60 3 10 80 4 10 100 5 5 120 6 0 60 (a) Using the net present value method with an interest preference rate of 15%, should the invest-ment be accepted or rejected? (b) Answer the same question when using the yield rate method. Problem 36.2 An investor enters into an agreement to contribute $7,000 immediately and $1,000 at the end of two years in exchange for the receipt of $4,000 at the end of one year and $5,500 at the end of three years. (a) Calculate NPV (0.09) and NPV (0.10). (b) If an investor’s interest preference rate is 12%, should the investment be accepted or rejected? Problem 36.3 A used car can be purchased for $5000 cash or for $2400 down and $1500 at the end of each of the next two years. Should a purchaser with an interest preference rate of 10% pay cash or ﬁnance the car? Problem 36.4 Consider an investment in which a person makes payments of $100 immediately and $132 at the end of one year. What method would be better to use, the yield rate method or the net present value method if the preference interest rate is 15%? Problem 36.5 Consider an investment in which a person makes payments of $100 immediately and $101 at the end of two years in exchange for a payment of $200 at the end of one year. Explain why the yield rate method is not applicable in this case. 36 YIELD RATES IN CAPITAL BUDGETING 331 Problem 36.6 A borrower needs $800. The funds can be obtained in two ways: (i) By promising to pay $900 at the end of the period. (ii) By borrowing $1000 and repaying $1120 at the end of the period. If the interest preference rate for the period is 10%, which option should be chosen? 332 RATE OF RETURN OF AN INVESTMENT Loan Repayment Methods Loan is an arrangement in which a lender gives money or property to a borrower, and the borrower agrees to return the property or repay the money, usually along with interest, at some future point(s) in time. Various methods of repaying a loan are possible. We will consider two of them: The amortization method and the sinking fund method. The amortization method: In this method the borrower makes installment payments to the lender. Usually these payments are at a regularly spaced periodic intervals; the progressive reduction of the amount owed is described as the amortization of the loan. Examples include car loan, home mortgage repayment. The sinking fund method: In this method the loan will be repaid by a single lump sum payment at the end of the term of the loan. The borrower pays interest on the loan in installments over this period. However, the borrower may prepare himself for the repayment by making deposits to a fund called a sinking fund to accumulate the repayment amount. 333 334 LOAN REPAYMENT METHODS 37 Finding the Loan Balance Using Prospective and Retro-spective Methods. When using the amortization method, the payments form an annuity whose present value is equal to the original amount of the loan. In this section, we want to determine the unpaid balance, also referred to as the outstanding loan balance or unpaid principal at any time after the inception of the loan. There are two approaches used in ﬁnding the amount of the outstanding balance: the prospective and the retrospective method. With the prospective method, the outstanding loan balance at any point in time is equal to the present value at that date of the remaining payments. With the retrospective method, the outstanding loan balance at any point in time is equal to the original amount of the loan accumulated to that date less the accumulated value at that date of all payments previously made. In general, the two approaches are equivalent. At the time of inception of the loan we have the following equality Present Value of All Payments = Amount of Loan Accumulate each side of the equation to the date at which the outstanding loan balance is desired, obtaining Current Value of Payments = Accumulated Value of Loan Amount But payments can be divided into past and future payments giving Accumulated Value of Past Payments + Present Value of Future Payments = Accumulated Value of Loan Amount or Present Value of Future Payments = Accumulated Value of Loan Amount −Accumulated Value of Past Payments. But the left side of this equation represents the prospective approach and the right side represents the retrospective approach. We can prove that the two methods are equivalent algebraically as follows. Let Bp t and Br t denote the outstanding loan balances at time t using the prospective and retrospective methods respectively. We denote the initial loan by L. 37 FINDING THE LOAN BALANCE USING PROSPECTIVE AND RETROSPECTIVE METHODS.335 Suppose we want to repay the loan by level payments of P for n periods at a periodic interest of i.. Then P satiﬁes the equation P = L an . For 0 < t < n, the outstanding loan balance at time t computed after making the tth payment is Bp t = Pan−t by the prospective method and Br t = L(1 + i)t −Pst by the retrospective method. We will show that Br t = Bp t . Indeed, Br t =L(1 + i)t −Pst =Pan(1 + i)t −Pst =P (1 + i)t −(1 + i)−(n−t) −(1 + i)t + 1 i =P 1 −νn−t i = Pan−t = Bp t . Example 37.1 ‡ A loan is being repaid with 25 annual payments of 300 each. With the 10th payment, the borrower pays an extra 1,000, and then repays the balance over 10 years with a revised annual payment. The eﬀective rate of interest is 8%. Calculate the amount of the revised annual payment. Solution. The balance after 10 years, prospectively, is Bp 10 = 300a15 = $2, 567.84. If the borrower pays an additional $1,000, the balance becomes $1,567.84. An equation of value for the revised payment, denoted by R, is Ra10 = 1, 567.84 or R = 1, 567.84 a10 = $233.66 336 LOAN REPAYMENT METHODS Example 37.2 A loan is being repaid with 16 quarterly payments, where the ﬁrst 8 payments are each $200 and the last 8 payments are each $400. If the nominal rate of interest convertible quarterly is 10%, use both the prospective method and the retrospective method to ﬁnd the outstanding loan balance immediately after the ﬁrst six payments are made. Solution. With the prospective method, we have Bp 6 =200(ν + ν2) + 400ν2(ν + ν2 + · · · + ν8) =400(ν + ν2 + · · · + ν10) −200(ν + ν2) =400a10 −200a2 =400(8.7521) −200(1.9274) = $3, 115.36. With the retrospective method, we have that the original loan amount is L =200(ν + ν2 + · · · + ν8) + 400ν8(ν + ν2 + · · · + ν8) =400(ν + ν2 + · · · + ν16) −200(ν + ν2 + · · · + ν8) =400a16 −200a8 =400(13.0550) −200(7.1701) = $3, 787.98 The outstanding loan balance is Br 6 = 3787.98(1.025)6 −200s6 = 4392.90 −200(6.3877) = $3, 115.36 Example 37.3 Megan is buying a car for 30,000 using a 60-month car loan with an interest rate of 9% compounded monthly. For the ﬁrst two years, Megan makes the required payment. Beginning with the ﬁrst payment in the third year, Megan begins paying twice the required payment. Megan will completely pay oﬀher loan by making a smaller ﬁnal payment. Determine the total number of payments that Megan will make. Solution. The monthly eﬀective interest rate is 0.09 12 = 0.0075. The original required payment is P such that 30000 = Pa60 0.0075 = P (1−1.0075−60) 0.0075 →P = 30000×0.0075 (1−1.0075−60) = 622.7506568. At the end of the ﬁrst 2 years, Megan’s outstanding loan balance, prospectively, is 622.7506568a36 0.0075 = 622.7506568(1 −1.0075−36) 0.0075 = 19583.51862 37 FINDING THE LOAN BALANCE USING PROSPECTIVE AND RETROSPECTIVE METHODS.337 From this point on, Megan pays 2P = 1245.501314 every month. Then 19583.51862 = 1245.501314an 0.0075 = 1245.501314(1−1.0075−n) 0.0075 , where n is the number of months until the loan is repaid. Thus, 0.1179255197 = (1 −1.0075−n) →1.0075−n = 0.8820744803 →n = −ln 0.8820744803 ln 1.0075 = 16.79315649. So Megan will make 16 full payments and 1 smaller payment for a total of 17 additional payments and 17 + 24 = 41 total payments over the course of repaying the loan Example 37.4 A loan of 1,000 is being repaid with quarterly payments at the end of each quarter for 5 years, at 6% convertible quarterly. Find the outstanding loan balance at the end of the 2nd year. Solution. The quarterly eﬀective interest rate is 0.06 4 = 0.015. Let P be the quarterly payment. Then 1000 = Pa20 0.015 = P (1−1.015−20) 0.015 →P = 1000×0.015 (1−1.015−20) = 58.24573587. The outstanding loan balance at the end of the 2nd year, prospectively, is 58.24573587a12 0.015 = 58.24573587(1−1.015−12) 0.015 = $635.32 338 LOAN REPAYMENT METHODS Practice Problems Problem 37.1 A loan is being repaid with level annual payments of $1,000. Calculate the outstanding balance of the loan if there are 12 payments left. The next payment will be paid one year from now and the eﬀective annual interest rate is 5%. Problem 37.2 A loan of 10,000 is being repaid will 20 non-level annual payments. The interest rate on the loan is an annual eﬀective rate of 6%. The loan was originated 4 years ago. Payments of 500 at the end of the ﬁrst year, 750 at the end of the second year, 1,000 at the end of the third year and 1,250 at the end of the fourth year have been paid. Calculate the outstanding balance immediately after the fourth payment. Problem 37.3 Calculate the outstanding balance to the loan in the previous problem one year after the fourth payment immediately before the ﬁfth payment. Problem 37.4 Julie bought a house with a 100,000 mortgage for 30 years being repaid with payments at the end of each month at an interest rate of 8% compounded monthly. What is the outstanding balance at the end of 10 years immediately after the 120th payment? Problem 37.5 If Julie pays an extra 100 each month, what is the outstanding balance at the end of 10 years immediately after the 120th payment? Problem 37.6 A loan 20,000 is being repaid with annual payments of 2,000 at the end of each year. The interest rate charged on the loan is an annual eﬀective rate of 8%. Calculate the outstanding balance of the loan immediately after the 5th payment. Problem 37.7 The interest rate on a 30 year mortgage is 12% compounded monthly. The mortgage is being repaid by monthly payments of 700. Calculate the outstanding balance at the end of 10 years. Problem 37.8 The interest rate on a 30 year mortgage is 12% compounded monthly. Lauren is repaying the mortgage by paying monthly payments of 700. Additionally, to pay oﬀthe loan early, Lauren has made additional payments of 1,000 at the end of each year. Calculate the outstanding balance at the end of 10 years. 37 FINDING THE LOAN BALANCE USING PROSPECTIVE AND RETROSPECTIVE METHODS.339 Problem 37.9 A loan is being repaid with 20 payments of 1,000. The total interest paid during the life of the loan is 5,000. Calculate the amount of the loan. Problem 37.10 A loan of 10,000 is being repaid by installments of 2,000 at the end of each year, and a smaller ﬁnal payment made one year after the last regular payment. Interest is at the eﬀective rate of 12%. Find the amount of outstanding loan balance remaining when the borrower has made payments equal to the amount of the loan. Problem 37.11 A loan is being repaid by quarterly installments of 1,500 at the end of each quarter, at 10% con-vertible quarterly. If the loan balance at the end of the ﬁrst year is 12,000, ﬁnd the original loan balance. Problem 37.12 A loan is being repaid by 15 annual payments at the end of each year. The ﬁrst 5 installments are 4,000 each, the next 5 are 3,000 each, and the ﬁnal 5 are 2,000 each. Find expressions for the outstanding loan balance immediately after the second 3,000 installment. (a) prospectively; (b) retrospectively. Problem 37.13 A loan is to be repaid with level installments payable at the end of each half-year for 31 2 years, at a nominal rate of interest of 8% convertible semiannually. After the fourth payment the outstanding loan balance is 5,000. Find the initial amount of the loan. Problem 37.14 A 20,000 loan is to be repaid with annual payments at the end of each year for 12 years. If (1 + i)4 = 2, ﬁnd the outstanding balance immediately after the fourth payment. Problem 37.15 ‡ A 20-year loan of 1,000 is repaid with payments at the end of each year. Each of the ﬁrst ten payments equals 150% of the amount of interest due. Each of the last ten payments is X. The lender charges interest at an annual eﬀective rate of 10%. Calculate X. Problem 37.16 ‡ A loan is amortized over ﬁve years with monthly payments at a nominal interest rate of 9% com-pounded monthly. The ﬁrst payment is 1,000 and is to be paid one month from the date of the loan. Each succeeding monthly payment will be 2% lower than the prior payment. Calculate the outstanding loan balance immediately after the 40th payment is made. 340 LOAN REPAYMENT METHODS Problem 37.17 A mortgage loan is being repaid with level annual payments of 5,000 at the end of the year for 20 years. The interest rate on the mortgage is 10% per year. The borrower pays 10 payments and then is unable to make payments for two years. Calculate the outstanding balance at the end of the 12th year. Problem 37.18 ‡ An investor took out a loan of 150,000 at 8% compounded quarterly, to be repaid over 10 years with quarterly payments of 5483.36 at the end of each quarter. After 12 payments, the interest rate dropped to 6% compounded quarterly. The new quarterly payment dropped to 5134.62. After 20 payments in total, the interest rate on the loan increased to 7% compounded quarterly. The investor decided to make an additional payment of X at the time of his 20th payment. After the additional payment was made, the new quarterly payment was calculated to be 4265.73, payable for ﬁve more years. Determine X. Problem 37.19 ‡ A small business takes out a loan of 12,000 at a nominal rate of 12%, compounded quarterly, to help ﬁnance its start-up costs. Payments of 750 are made at the end of every 6 months for as long as is necessary to pay back the loan. Three months before the 9th payment is due, the company reﬁnances the loan at a nominal rate of 9%, compounded monthly. Under the reﬁnanced loan, payments of R are to be made monthly, with the ﬁrst monthly payment to be made at the same time that the 9th payment under the old loan was to be made. A total of 30 monthly payments will completely pay oﬀthe loan. Determine R. Problem 37.20 A loan of $1,000 is being repaid with quarterly payments at the end of each quarter for 5 years, at 6% convertible quarterly. Find the outstanding loan balance at the end of the second year. Problem 37.21 A 20,000 mortgage is being repaid with 20 annual installments at the end of each year. The borrower makes 5 payments, and then is temporarily unable to make payments for the next 2 years. Find an expression for the revised payment to start at the end of the 8th year if the loan is still to be repaid at the end of the original 20 years. Problem 37.22 A loan of 1 was originally scheduled to be repaid by 25 equal annual payments at the end of each 37 FINDING THE LOAN BALANCE USING PROSPECTIVE AND RETROSPECTIVE METHODS.341 year. An extra payment K with each of the 6th through the 10th scheduled payments will be suﬃcient to repay the loan 5 years earlier than under the original schedule. Show that K = a20 −a15 a25a5 . Problem 37.23 Bob takes out a loan of 1000 at an annual eﬀective interest rate of i. You are given: (a) The ﬁrst payment is made at the end of year 6. (b) Ten equal annual payments are made to repay the loan in full at the end of 15 years. (c) The outstanding principal after the payment made at the end of year 10 is 908.91. Calculate the outstanding principal at the end of year 5. Problem 37.24 The original amount of an inheritance was just suﬃcient at 3.5% eﬀective to pay $10,000 at the end of each year for 10 years. The payments were made as scheduled for the ﬁrst ﬁve years even though the fund actually earned 5% eﬀective. How much excess interest was in the fund at the end of the ﬁfth year? 342 LOAN REPAYMENT METHODS 38 Amortization Schedules When a loan is being repaid with the amortization method, each payment is partially a repayment of principal and partially a payment of interest. Determining the amount of each for a payment can be important (for income tax purposes, for example). An amortization schedule is a table which shows the division of each payment into principal and interest, together with the outstanding loan balance after each payment is made. Consider a loan of an at interest rate i per period being repaid with payments of 1 at the end of each period for n periods. At the end of period 1 (i.e. after the ﬁrst payment), the interest paid is ian = 1−νn so that the principal repaid is νn, and the outstanding loan balance is an −νn = an−1. Next, at the end of the second period, the interest paid is ian−1 = 1 −νn−1 so that the principal repaid is νn−1, and the outstanding loan balance is an−1 −νn−1 = an−2. Continuing this process, we see that at the end of period k, the interest paid is ian−k+1 = 1 −νn−k+1 and the principal repaid is νn−k+1. The outstanding loan balance is an−k+1 −νn−k+1 = an−k = Bp k. The amortization table is shown below. Payment Interest Principal Outstanding Period amount paid repaid loan balance 0 an 1 1 ian = 1 −νn νn an −νn = an−1 2 1 ian−1 = 1 −νn−1 νn−1 an−1 −νn−1 = an−2 . . . . . . . . . . . . . . . k 1 ian−k+1 = 1 −νn−k+1 νn−k+1 an−k+1 −νn−k+1 = an−k . . . . . . . . . . . . . . . n −1 1 ia2 = 1 −ν2 ν2 a2 −ν2 = a1 n 1 ia1 = 1 −ν ν a1 −ν = 0 Total n n −an an Observe each of the following from this table: First, it should be noted that the outstanding loan balance agrees with that obtained by the prospective method. Second, the sum of the principal repayments equals to the original amount of the loan. Third, the sum of interest payments is equal to the diﬀerence between the sum of the total payments and the sum of the principal repayments. Fourth, the sum of principal repayments is a geometric progression with common ration (1 + i). Example 38.1 Create an amortization schedule for a loan of $1,000 repaid over four years if the annual eﬀective rate of interest is 8%. 38 AMORTIZATION SCHEDULES 343 Solution. If R is the periodic payment then R = 1000 a4 = $301.92. Payment Interest Principal Outstanding Period amount paid repaid loan balance 0 1000.00 1 301.92 80.00 221.92 778.08 2 301.92 62.25 239.67 538.41 3 301.92 43.07 258.85 279.56 4 301.92 22.36 279.56 0 Table 38.1 In the previous example, the last line exactly balances. That is, after the last payment is made the loan is paid oﬀ. In practice, there will be rounding errors as the table is generated line by line, and the last line may not lead to a zero balance. Standard practice is to adjust the last payment so that it is exactly equal to the amount of interest for the ﬁnal period plus the outstanding loan balance at the beginning of the ﬁnal period, in order to bring the outstanding loan balance to 0. Example 38.2 Create an amortization schedule for a loan of $10,000 repaid over three years if the annual eﬀective rate of interest is 7%. Solution. If R is the periodic payment then R = 10, 000 a3 = $3, 810.52. Payment Interest Principal Outstanding Period amount paid repaid loan balance 0 10,000.00 1 3,810.52 700.00 3,110.52 6,889.48 2 3,810.52 482.26 3,328.26 3,561.22 3 3,810.50 249.28 3,561.22 0 It should be noted that if it is desired to ﬁnd the amount of principal and interest in one particular payment, it is not necessary to construct the entire amortization schedule. The outstanding loan balance at the beginning of the period in question can be determined by either the retrospective or the prospective methods, and then that one line of the schedule can be created. 344 LOAN REPAYMENT METHODS Example 38.3 Jones borrows $20,000 from Smith and agrees to repay the loan with equal quarterly installments of principal and interest at 10% convertible quarterly over eight years. At the end of three years, Smith sells the right to receive future payments to Collins at a price which produces a yield rate of 12% convertible quarterly for Smith. Find the total amount of interest received (a) by Collins, and (b) by Smith. Solution. (a) Each quarterly payment by Jones is 20000 a32 0.025 = $915.37. Total payments by Jones to Collins over the last ﬁve years are (20)915.37 = $18307.40. After three years, the price Collins pays to Smith is 915.37a20 0.03 = 915.37(14.8775) = $13618.42. Total amount of interest received by Collins is 18307.40 −13618.42 = $4688.98. (b) Total payments by Jones to Smith over the ﬁrst three years are (12)915.37 = $10984.44. After three years, the outstanding loan balance is 915.37a20 0.025 = 915.37(15.5892) = $14269.89. Recall from (a) that the price Collins pays to Smith after three years is $13618.42. Total amount of interest received by Smith is 13618.42 + 10984.44 −20000 = $4602.86 Let It denote the amount of interest paid in the tth installment; Pt be the amount of principal and Bt be the loan balance. 38 AMORTIZATION SCHEDULES 345 Example 38.4 Show that for n = 100 and with a periodic payment of R, we have P11 + P12 + · · · + P50 = B10 −B50 and I11 + I12 + · · · + I50 = 40R −(B10 −B50). Solution. Let P = P11 + P12 + · · · + P50. Then P =R 50 X k=11 ν100−k+1 = Rν101 50 X k=11 (1 + i)k =Rv101( 50 X k=0 (1 + i)k − 10 X k=0 (1 + i)k) =R(1 + i)−50 −(1 + i)−90 i =Ra90 −Ra50 = B10 −B50 Now, let I = I11 + I12 + · · · + I50 then I = P50 j=11 R −P50 j=11 Pj = 40R −(B10 −B50) Amortization schedules of perpetuities do not exist since the entire payment represents interest and therefore the loan balance remains unchanged. Example 38.5 A $5,000 loan is being repaid by payments of $X at the end of each half year for as long as necessary until a smaller ﬁnal payment is made. The nominal rate of interest convertible semiannually is 14%. (a) If X = $400 ﬁnd the amount of principal and the interest in the sixth payment. (b) If X = $350, ﬁnd the principal in the sixth payment, and interpret this. Solution. (a) The outstanding balance at the beginning of the sixth half-year is Br 5 = 5000(1.07)5 −400s5 = 7012.76 −400(5.7507) = $4712.48. The interest in the sixth payment is (0.07)(4712.48) = $329.87 and the principal is 400 −329.87 = $70.13. (b) The outstanding balance at the beginning of the sixth half-year is Br 5 = 5000(1.07)5 −350s5 = 7012.76 −350(5.7507) = $5000.01. 346 LOAN REPAYMENT METHODS If X ≤350, then the loan will never be paid oﬀ, because every payment will count only toward interest Thus far, in creating an amortization schedule we have assumed a constant rate of interest, the conversion period and payment period are the same, and the payments are leveled. Example 38.2 is an example where the payments are not all leveled. It is possible to create an amortization schedule with varying interest rate as illustrated in the next example. Example 38.6 An amount is invested at an annual eﬀective rate of interest i which is exactly suﬃcient to pay 1 at the end of each year for n years. In the ﬁrst year, the fund earns rate i and 1 is paid at the end of the year. However, in the second year, the fund earns rate j > i. If X is the revised payment which could be made at the end of years 2 to n, then ﬁnd X assuming that (a) the rate reverts back to i again after this one year, (b) the rate earned remains at j for the rest of the n−year period. Solution. (a) From the amortization table, the balance at the end of the ﬁrst year is an−1 i, and therefore the balance after two years must be (1 + j)an−1 i −X. However, after two years, the balance must be equal to the present value of all future payments. Consequently, we have that (1 + j)an−1 i −X =Xan−2 i X(1 + an−2 i) =(1 + j)an−1 i X(1 + i)an−1 i =(1 + j)an−1 i X =1 + j 1 + i . (b) As in (a), the balance at the end of two years is (1 + j)an−1 i −X. However, after two years, the balance must be equal to the present value of all future payments. Consequently, we have that (1 + j)an−1 i −X =Xan−2 j X(1 + an−2 j) =(1 + j)an−1 i X(1 + j)an−1 j =(1 + j)an−1 i X =an−1 i an−1 j Situations where payment period and interest conversion period do not coincide are best handled 38 AMORTIZATION SCHEDULES 347 from ﬁrst principles rather than from developing formulas. With an amortization schedule in which payments are made at a diﬀerent frequency than interest is convertible, the following two-step procedure can be followed: (1) Find the rate of interest which is equivalent to the given rate of interest and convertible at the same frequency as payments are made. (2) Construct the amortization schedule with the rate of interest in step 1. Example 38.7 A 25-year mortgage for $100,000 bears an interest rate of 5.5% compounded semi-annually. The payments are made at the end of each month. Find the size of the payment. Solution. The monthly interest rate is j = (1.0275) 1 6 −1 = .00453168. The size of each payment is R = 100, 000 a300 j = 610.39 Example 38.8 A loan of $3,000 is being amortized by 20 quarterly payments. Payments 11 and 12 are not made. At the designated time of the 12th payment, the loan is renegotiated so that the 13th payment is $N and payments 14, 16, 18, and 20 are each $40 more that the preceding payment. If the rate of interest is 8% convertible semiannually, ﬁnd the value of N which would provide that the loan be completely repaid after 20 quarters. Solution. The quarterly eﬀective rate of interest is j = (1.02) 1 2 −1. The original quareterly payment isR = 3000 a20 j . The outstanding balance at time t = 10, by the prospective method, is Ra10 j. The outstanding balance at time t = 12 is Ra10 j(1 + j)2 = 3000a10 j a20 j (1 + j)2 = 1712.46. Now, the new payments stream starting at time t = 13 are: N, N + 40, N + 40, N + 80, N + 80, N + 120, N + 120, N + 160. By the prospective method 1712.46 = Na8 j + 40(ν2 j + ν3 j + 2ν4 j + 2ν5 j + 3ν6 j + 3ν7 j + 4ν8 j ). Solving this equation for N we ﬁnd N = 155.73 348 LOAN REPAYMENT METHODS Practice Problems Problem 38.1 A loan of 10,000 is being repaid with annual payments of 1500 for 11 years. Calculate the amount of principal paid over the life of the loan. Problem 38.2 A loan of 10000 is being repaid with annual payments of 1500 for 11 years. Calculate the amount of interest paid over the life of the loan. Problem 38.3 A loan is being repaid with level annual payments based on an annual eﬀective interest rate of 8%. If the amount of principal in the 10th payment is 100, calculate the amount of principal in the 5th payment. Problem 38.4 A loan is being repaid with level annual payments based on an annual eﬀective interest rate of 8%. If the outstanding balance immediately after the 10th payment is 1000, calculate the amount of interest in the 11th payment. Problem 38.5 A loan of 10,000 is being repaid with annual payments of 1500 for n years. The total principal paid in the ﬁrst payment is 685.58. Calculate the interest rate on the loan. Problem 38.6 For a loan with level annual payments, the principal repaid by the 10th payment is 10,000 while the principle repaid by the 11th payment is 11,000. Calculate the principal repaid by the 15th payment. Problem 38.7 A 60-month loan is to be repaid with level payments of 1000 at the end of each month. The interest in the last payment is 7.44. Calculate the total interest paid over the life of the loan. Problem 38.8 A 60-month loan is to be repaid with level payments of 1000 at the end of each month. The principal in the ﬁrst payment is 671.21. Calculate the eﬀective annual interest rate. Problem 38.9 Jenna is repaying a 120-month loan with interest compounded monthly at 12%. Calculate the payment in which the absolute value of the diﬀerence between the interest paid and the principal repaid is minimized. 38 AMORTIZATION SCHEDULES 349 Problem 38.10 A loan is being repaid with level payments at the end of each year for 20 years. The principal repaid in the 10th payment is 1000 and the principal repaid in the 15th payment is 1200. Calculate the amount of the loan. Problem 38.11 A loan of 10,000 is being repaid with annual payments over 10 years at 8% compound monthly. Create the amortization table for this loan. Problem 38.12 A loan is being repaid with quarterly installments of 1,000 at the end of each quarter for 5 years at 12% convertible quarterly. Find the amount of principal in the 6th installment. Problem 38.13 Consider a loan which is being repaid with installments of 1 at the end of each period for n periods. Find an expression at issue for the present value of the interest which will be paid over the life of the loan. Problem 38.14 A loan of 10,000 is being repaid with 20 installments at the end of each year, at 10% eﬀective. Show that the amount of interest in the 11th installment is 1000 1 + ν10. Problem 38.15 A loan is being repaid with 20 installments at the end of each year at 9% eﬀective. In what installment are the principal and interest portions most nearly equal to each other? Problem 38.16 A loan is being repaid with a series of payments at the end of each quarter, for 5 years. If the amount of principal in the 3rd payment is 100, ﬁnd the amount of principal in the last 5 payments. Interest is at the rate of 10% convertible quarterly. Problem 38.17 A loan is being repaid with installments of 1 at the end of each year for 20 years. Interest is at eﬀective rate i for the ﬁrst 10 years, and eﬀective rate j for the second 10 years. Find expressions for (a) the amount of interest paid in the 5th installment; (b) the amount of principal repaid in the 15th installment. 350 LOAN REPAYMENT METHODS Problem 38.18 A loan of 25,000 is being repaid with annual payments of 2,243 at the end of the year. The interest rate on the loan 7.5%. Calculate the interest in the 5th payment. Problem 38.19 ‡ Seth borrows X for four years at an annual eﬀective interest rate of 8%, to be repaid with equal payments at the end of each year. The outstanding loan balance at the end of the second year is 1076.82 and at the end of the third year is 559.12 . Calculate the principal repaid in the ﬁrst payment. Problem 38.20 ‡ A bank customer takes out a loan of 500 with a 16% nominal interest rate convertible quarterly. The customer makes payments of 20 at the end of each quarter. Calculate the amount of principal in the fourth payment. Problem 38.21 ‡ A loan is repaid with level annual payments based on an annual eﬀective interest rate of 7%. The 8th payment consists of 789 of interest and 211 of principal. Calculate the amount of interest paid in the 18th payment. Problem 38.22 ‡ Kevin takes out a 10-year loan of L, which he repays by the amortization method at an annual eﬀective interest rate of i. Kevin makes payments of 1000 at the end of each year . The total amount of interest repaid during the life of the loan is also equal to L. Calculate the amount of interest repaid during the ﬁrst year of the loan. Problem 38.23 ‡ Ron is repaying a loan with payments of 1 at the end of each year for n years. The amount of interest paid in period t plus the amount of principal repaid in period t + 1 equals X. Calculate X. Problem 38.24 A 35-year loan is to be repaid with equal installments at the end of each year. The amount of interest paid in the 8th installment is 135. The amount of interest paid in the 22nd installment is 108. Calculate the amount of interest paid in the 29th installment. Problem 38.25 A 1000 loan is repaid with equal payments at the end of each year for 20 years. The principal portion of the 13th payment is 1.5 times the principal portion of the 5th payment. Calculate the total amount of interest paid on the loan. 38 AMORTIZATION SCHEDULES 351 Problem 38.26 A loan is being amortized with payments at the end of each quarter for 25 years. If the amount of principal repaid in the third payment is $100, ﬁnd the total amount of principal repaid in the forty payments consisting of payments eleven through ﬁfty. Interest is at the rate of 8% convertible quarterly. Problem 38.27 A loan is being amortized using an interest rate of 8% convertible quarterly with level payments at the end of each quarter. If the amount of principal repaid in the third payment is $100, ﬁnd the total amount of principal repaid in the ﬁve payments consisting of payments eleven through ﬁfteen. Problem 38.28 A 100,000 loan is to be repaid by 30 equal payments at the end of each year. The outstanding balance is amortized at 4%. In addition to the annual payments, the borrower must pay an origination fee at the time the loan is made. The fee is 2% of the loan but does not reduce the loan balance. When the second payment is due, the borrower pays the remaining loan balance. Determine the yield to the lender considering the origination fee and the early pay-oﬀof the loan. Problem 38.29 A loan of 10,000 is amortized by equal annual payments for 30 years at an annual eﬀective interest rate of 5%. Determine the year in which the interest portion of the payment is most nearly equal to one-third of the payment. Problem 38.30 A loan of 1000 at a nominal rate of 12% convertible monthly is to be repaid by six monthly payments with the ﬁrst payment due at the end of 1 month. The ﬁrst three payments are X each, and the ﬁnal three payments are 3X each. Determine the sum of the principal repaid in the third payment and the interest paid in the ﬁfth payment. Problem 38.31 Carl borrows $10,000 at 12% compounded monthly, and Carl will repay the loan with 60 monthly amortization payments beginning at the end of the ﬁrst month. Find: (a) B20 (b) B30 (c) P30 j=21 Pj (d) P30 j=21 Ij Problem 38.32 ‡ A bank customer borrows X at an annual eﬀective rate of 12.5% and makes level payments at the end of each year for n years. The interest portion of the ﬁnal payment is 153.86. The total principal repaid as of time (n −1) is 6009.12. The principal repaid in the ﬁrst payment is Y. Calculate Y. 352 LOAN REPAYMENT METHODS Problem 38.33 A loan is to be amortized by n level annual payments of X where n > 5. You are given (1) The amount of interest in the ﬁrst payment is 604.00 (2) The amount of interest in the third payment is 593.75 (3) The amount of interest in the ﬁfth payment is 582.45 Calculate X. Problem 38.34 A loan is to be repaid by annual installments of X at the end of each year for 10 years. You are given (1) the total principal repaid in the ﬁrst 3 years is 290.35 (2) the total principal repaid in the last 3 years is 408.55 Calculate the total amount of interest paid during the life of the loan. Problem 38.35 Iggy borrows X for 10 years at an annual eﬀective rate of 6%. If he pays the principal and accumulated interest in one lump sum at the end of 10 years, he would pay 356.54 more in interest than if he repaid the loan with 10 level payments at the end of each year. Calculate X. Problem 38.36 A loan is being amortized by means of level monthly payments at an annual eﬀective interest rate of 8%. The amount of principal repaid in the 12th payment is 1000 and the amount of principal repaid in the tth payment is 3700. Calculate t. Problem 38.37 John is repaying a loan with payments of $3,000 at the end of every year over an unknown period of time. if the amount on interest in the third installment is $2,000, ﬁnd the amount of principal in the sixth installment. Assume the interest is 10% convertible quarterly. 39 SINKING FUND METHOD 353 39 Sinking Fund Method An alternative for repaying a loan in installments by the amortization method, a borrower can accumulate a fund which will exactly repay the loan in one lump sum at the end of a speciﬁed period of time. This fund is called a sinking fund. It is generally required that the borrower periodically pay interest on the loan, sometimes referred to as a service. In some cases payments into a sinking fund may vary irregularly at the discretion of the borrower. However, we will be dealing with sinking funds with regular contributions. Because the balance in the sinking fund could be applied against the loan at any point, the net amount of the loan is equal to the original amount of the loan minus the sinking fund balance. We next show that if the rate of interest paid on the loan equals the rate of interest earned on the sinking fund, then the amortization method and the sinking fund method are equivalent. To see this, Suppose that the eﬀective annual rate on the loan is i, and the sinking fund earns the same rate. Suppose that the amount of the loan is 1 , and the loan term is n periods. With the amortization method, the payment at the end of each period is 1 an . With the sinking fund method, to accumulate the amount of 1 in the sinking fund, the borrower deposits of 1 sn at the end of each year for n years. At the same moment the borrower also pays i per period to the lender. That is, payments of size i (interest) plus payments of size 1 sn are required. But we know from Section 15 that 1 sn + i = 1 an . Thus, the two methods are equivalent. As was seen in Section 38, a way to visualize the activity of an amortization method is the creation of an amortization schedule. The same idea can be used with sinking funds. The following example illustrates the creation of a sinking fund schedule. Example 39.1 Create a sinking fund schedule for a loan of $1000 repaid over four years if the annual eﬀective rate of interest is 8% Solution. If R is the sinking fund deposit then R = 1000 s4 = $221.92. 354 LOAN REPAYMENT METHODS Interest earned Interest Sinking fund on sinking Amount in Net amount Period paid deposit fund sinking fund of loan 0 1000.00 1 80.00 221.92 0 221.92 778.08 2 80.00 221.92 17.75 461.59 538.41 3 80.00 221.92 36.93 720.44 279.56 4 80.00 221.92 57.64 1000.00 0 Table 39.1 Notice from this table that the amount in the sinking fund after the tth payment is found by multiplying the sinking fund deposit by st j where j is the sinking fund rate of interest. Comparing Table 38.1 and Table 39.1 we notice the following: (1) For each period, Interest paid + Sinking Fund deposit (in Table 39.1) = Payment amount (in Table 38.1). (2) For each period, Interest paid −Interest earned on sinking fund (in Table 39.1) = Interest paid (in Table 38.1). (3) For each period, Sinking fund deposit + Interest earned on sinking fund (in Table 39.1) = principal repaid (in Table 38.1). (4) For each period, the Net amount of loan (in Table 39.1) = Outstanding loan balance in (Table 38.1) Notice that the net amount of loan concept plays the same role for the sinking fund method that the outsanding loan balance does for the amortization method. Next, we consider the situation in which the interest rate on the loan and the interest rate earned on the sinking fund diﬀers. The rate on the loan is denoted by i, and the rate on the sinking fund is denoted by j. Usually, j is less than i because the sinking fund wouldn’t normally be riskier than the loan. In this case, an amount of i will be deducted from the sinking fund deposit and the remaining amount will be invested in the sinking fund at a rate of j. Let an i&j be the present value of an annuity which pays 1 at the end of each period for n periods with i and j as previously deﬁned. If a loan of 1 is made, by the amortization method the amount of the loan will be repaid by interest payments of ian i&j at the end of each year for n years together with yearly deposits of 1 −ian i&j into a sinking fund which earns interest at the eﬀective annual rate of interest j. The sinking ﬁnd should accumulate an i&j at the end of n years. That is, (1 −ian i&j)sn j = an i&j or an i&j = sn j 1 + isn j 39 SINKING FUND METHOD 355 which is equivalent to 1 an i&j = 1 sn j + i. In other words, loan of 1 can be repaid by interest payments of i to the lender and depositing 1 sn j into the sinking fund at the end of each year for n years. Now since 1 an j = 1 sn j + j it follows that 1 an i&j = 1 an j + (i −j) or an i&j = an j 1 + (i −j)an j . It should be noted that if i = j, then an i&j = an i. Example 39.2 ‡ John borrows 10,000 for 10 years at an annual eﬀective interest rate of 10%. He can repay this loan using the amortization method with payments of 1,627.45 at the end of each year. Instead, John repays the 10,000 using a sinking fund that pays an annual eﬀective interest rate of 14%. The deposits to the sinking fund are equal to 1,627.45 minus the interest on the loan and are made at the end of each year for 10 years. Determine the balance in the sinking fund immediately after repayment of the loan. Solution. Under the amortization method, the periodic installment is 1627.45. The periodic interest payment on the loan is 0.1(10000) = 1000. Thus, deposits into the sinking fund are 1627.45 −1000 = 627.45. Thus, the amount in sinking fund immediately after repayment of the loan is 627.45s10 0.14−10, 000 = 2130 In general, the sinking fund schedule at two rates of interest is identical to the sinking fund schedule at one rate of interest equal to the rate of interest earned on the sinking fund, except that a constant addition of (i −j) times the amount of the original loan is added to the interest paid column. We illustrate this in the next example. Example 39.3 Create a sinking fund schedule for a loan of $1,000 repaid over four years if the annual eﬀective rate of interest is 10% and the sinking fund interest of 8% 356 LOAN REPAYMENT METHODS Solution. If R is the sinking fund deposit then R = 1, 000 s4 0.08 = $221.92. Interest earned Interest Sinking fund on sinking Amount in Net amount Period paid deposit fund sinking fund of loan 0 1000.00 1 100.00 221.92 0 221.92 778.08 2 100.00 221.92 17.75 461.59 538.41 3 100.00 221.92 36.93 720.44 279.56 4 100.00 221.92 57.64 1000 0 Note that this table is identical to Table 39.1 except each entry in the interest paid column in the table above is equal to each entry in the interest paid column in Table 39.1 increased by a constant equals to 20 which is (0.10 −0.08) × 1000 The following is an important example whose results are to keep in mind. Example 39.4 A loan of 1 yields the lender rate i per period for n periods, while the borrower replaces the capital in a sinking fund earning rate j per period. Find expressions for the following if 1 ≤t ≤n : (a) Periodic interest paid to the lender. (b) Periodic sinking fund deposit. (c) Interest earned on sinking fund during the tth period. (d) Amount in sinking fund at end of the tth period. (e) Net amount of loan at the end of the tth period. (f) Net interest paid in period t. (g) Principal repaid in period t. Solution. (a) The yield rate is i, so the lender must be receiving an amount of i each period. (b) The capital is 1, so the borrower is depositing (sn j)−1 regularly into the sinking fund. (c) At the beginning of the tth period the balance in the sinking fund is (st−1 j/sn j); during the period it earns interest in the amount of j · st−1 j sn j payable at the end of the period, i.e., at time t. (d) The amount in sinking fund at end of the tth period is st j sn j . 39 SINKING FUND METHOD 357 (e) The net amount of loan at the end of the tth period is the excess of 1 over the balance in the sinking fund, i.e., 1 − st j sn j . (f) The net interest paid in the tth period is the excess of interest paid over interest earned, i.e. i −j st−1 j sn j . (g) By (e) above, the change in the amount of the loan between the (t −1)th and the tth payment is 1 −st j sn j − 1 − st−1 j sn j = (1 + i)t−1 sn j In creating the sinking fund schedule the payment period and the conversion interest period can be diﬀerent. These cases can be handled from basic principles as illustrated in the next two examples. Remember to use the rate of interest which is equivalent to the given rate of interest and convertible at the same frequency as payments are made. Example 39.5 John borrows $5,000 for 10 years at 10% convertible quarterly. John does not pay interest currently and will pay all accrued interest at the end of 10 years together with the principal. Find the annual sinking fund deposit necessary to liquidate the loan at the end of 10 years if the sinking fund earns 7% convertible semi-annually. Solution. Let j be the annual interest rate. Then j = (1.025)4 −1. The loan balance at the end of 10 years is 5000(1 + j)10 = 5000(1.025)40 = 13425.32. Hence, the annual sinking fund deposit is R = 13425.32 s10 (1.035)2−1 = 966.08 Example 39.6 Create the sinking fund table for the following: a 3 year loan of $10,000, with interest payable semiannually at the nominal interest rate of 8.00% is to be retired by a sinking fund funded by quarterly deposits earning an eﬀective nominal interest rate of 6.00% compounded semiannually. Solution. Every six months the interest paid on the loan is 0.04(10, 000) = 400. The quarterly rate on the sinking fund is j = 1.030.5 −1. The quarterly sinking fund deposit is D = 10, 000 s12 j = 767.28. 358 LOAN REPAYMENT METHODS Interest earned Interest Sinking fund on sinking Amount in Net amount Period paid deposit fund sinking fund of loan 0 10,000.00 0.25 0 767.28 0 767.28 9,232.72 0.50 400.00 767.28 11.42 1,545.98 8,454.02 0.75 0 767.28 23.02 2,336.27 7,663.73 1.00 400.00 767.28 34.79 3,138.33 6,861.67 1.25 0 767.28 46.73 3,952.33 6,047.67 1.50 400.00 767.28 58.85 4,778.45 5,221.55 1.75 0 767.28 71.15 5,616.88 4,383.12 2.00 400.00 767.28 83.63 6,467.78 3,532.22 2.25 0 767.28 96.30 7,331.36 2,668.64 2.50 400.00 767.28 109.16 8,207.79 1,792.21 2.75 0 767.28 122.21 9,097.27 902.73 3.00 400.00 767.28 135.45 10,000.00 0.00 39 SINKING FUND METHOD 359 Practice Problems Problem 39.1 A loan of 10,000 is being repaid with annual payments for 10 years using the sinking fund method. The loan charges 10% interest and the sinking fund earns 8%. (a) Calculate the interest payment that is paid annually to service the loan. (b) Calculate the sinking fund payment made annually. (c) Calculate the amount in the sinking fund immediately after the deposit made at the end of 5 years. (d) Create the sinking fund schedule for the loan. Problem 39.2 If the loan in Problem 39.1 was repaid using the amortization method, but the annual payment was equal to the sum of the interest payment and the sinking fund deposit, calculate the interest rate under the amortization method. Problem 39.3 A loan of 20,000 is being repaid with annual payments for 5 years using the sinking fund method. The loan charges 10% interest compounded twice a year. The sinking fund earns 8% compounded monthly. Calculate the interest payment that is paid annually to service the loan and the sinking fund deposit. Problem 39.4 A loan of 20,000 is being repaid with monthly payments for 5 years using the sinking fund method. The loan charges 10% interest compounded twice a year. The sinking fund earns an annual eﬀective interest rate of 8%. (a) Calculate the interest payment that is paid monthly to service the loan and the sinking fund deposit paid monthly. (b) Calculate the amount in the sinking fund immediately after the 30th payment. Problem 39.5 Julie agrees to repay a loan of 10,000 using the sinking fund method over 10 years. The loan charges an annual eﬀective interest rate of 7% while the sinking fund earns 6%. Calculate the amount paid into the sinking fund each year less the amount of interest paid on the loan each year. Problem 39.6 Kathy can take out a loan of 50,000 with Bank A or Bank B. With Bank A, she must repay the loan with 60 monthly payments using the amortization method with interest at 7% compounded 360 LOAN REPAYMENT METHODS monthly. With Bank B, she can repay the loan with 60 monthly payments using the sinking fund method. The sinking fund will earn 6.5% compounded monthly. What interest rate can Bank B charge on the loan so that Kathy’s payment will be the same under either option? Problem 39.7 Lauren is repaying a loan of 100,000 using the sinking fund method. At the end of each year she pays 7,000 into a sinking fund earning 8%. At the end of 5 years, Lauren pays oﬀthe loan using the sinking fund plus an additional payment of X. Calculate X. Problem 39.8 Lauren is repaying a loan of 100,000 using the sinking fund method. At the end of each year she pays 7,000 into a sinking fund earning 8%. At the end of year Y, Lauren will have suﬃcient money in the sinking fund to repay the loan. Calculate Y. Problem 39.9 Ryan takes out a loan of 100,000 and agrees to repay it over 10 years using the sinking fund method. Ryan agrees to pay interest to the lender at the end of each year. The interest rate is .01(11 −t) in year t. The sinking fund will earn 5% per year. Determine the amount in the sinking fund after 10 years if the total payment made by Ryan at the end of each year is 12,000. Problem 39.10 On a loan of 10,000, interest at 9% eﬀective must be paid at the end of each year. The borrower also deposits X at the beginning of each year into a sinking fund earning 7% eﬀective. At the end of 10 years the sinking fund is exactly suﬃcient to pay oﬀthe loan. Calculate X. Problem 39.11 A borrower is repaying a loan with 10 annual payments of 1,000. Half of the loan is repaid by the amortization method at 5% eﬀective. The other half of the loan is repaid by the sinking fund method, in which the lender receives 5% eﬀective on the investment and the sinking fund accumulates at 4% eﬀective. Find the amount of the loan. Problem 39.12 A borrows 12,000 for 10 years, and agrees to make semiannual payments of 1,000. The lender receives 12% convertible semiannually on the investment each year for the ﬁrst 5 years and 10% convertible semiannually for the second 5 years. The balance of each payment is invested in a sinking fund earning 8% convertible semiannually. Find the amount by which the sinking fund is short of repaying the loan at the end of the 10 years. 39 SINKING FUND METHOD 361 Problem 39.13 A borrower takes out a loan of 3000 for 10 years at 8% convertible semiannually. The borrower replaces one-third of the principal in a sinking fund earning 5% convertible semiannually, and the other two-thirds in a sinking fund earning 7% convertible semiannually. Find the total semiannual payment. Problem 39.14 A payment of 36,000 is made at the end of each year for 31 years to repay a loan of 400,000. If the borrower replaces the capital by means of a sinking fund earning 3% eﬀective, ﬁnd the eﬀective rate paid to the lender on the loan. Problem 39.15 A loan of 30,000 is to be repaid using the sinking fund method over 6 years. The interest on the loan is paid at the end of each year and the interest rate is 10%. The sinking fund payment is made at the beginning of each year with the sinking fund earning 6%. Calculate the amount paid into the sinking fund each year. Problem 39.16 You have two equivalent ways to repay a loan of $100,000 over 20 years. (a) Using the sinking fund method, where the rate of interest on the loan is 8% and the rate of interest earned by the sinking fund is 6%. (b) Using the amortization method where the rate of interest is k. Find k. Problem 39.17 ‡ A 12-year loan of 8000 is to be repaid with payments to the lender of 800 at the end of each year and deposits of X at the end of each year into a sinking fund. Interest on the loan is charged at an 8% annual eﬀective rate. The sinking fund annual eﬀective interest rate is 4%. Calculate X. Problem 39.18 ‡ A 20-year loan of 20,000 may be repaid under the following two methods: i) amortization method with equal annual payments at an annual eﬀective rate of 6.5% ii) sinking fund method in which the lender receives an annual eﬀective rate of 8% and the sinking fund earns an annual eﬀective rate of j Both methods require a payment of X to be made at the end of each year for 20 years. Calculate j. Problem 39.19 ‡ Lori borrows 10,000 for 10 years at an annual eﬀective interest rate of 9%. At the end of each year, she pays the interest on the loan and deposits the level amount necessary to repay the principal to a sinking fund earning an annual eﬀective interest rate of 8%. The total payments made by Lori over the 10-year period is X. Calculate X. 362 LOAN REPAYMENT METHODS Problem 39.20 John borrows 10,000 for 10 years and uses a sinking fund to repay the principal. The sinking fund deposits earn an annual eﬀective interest rate of 5%. The total required payment for both the interest and the sinking fund deposit made at the end of each year is 1445.04. Calculate the annual eﬀective interest rate charged on the loan. Problem 39.21 Joe repays a loan of 10,000 by establishing a sinking fund and making 20 equal payments at the end of each year. The sinking fund earns 7% eﬀective annually. Immediately after the ﬁfth payment, the yield on the sinking fund increases to 8% eﬀective annually. At that time, Joe adjusts his sinking fund payment to X so that the sinking fund will accumulate to 10,000 20 years after the original loan date. Determine X. Problem 39.22 A corporation borrows 10,000 for 25 years, at an eﬀective annual interest rate of 5%. A sinking fund is used to accumulate the principal by means of 25 annual deposits earning an eﬀective annual interest rate of 4%. Calculate the sum of the net amount of interest paid in the 13th installment and the increment in the sinking fund for the ninth year. Problem 39.23 A loan of 1,000 is taken out at an annual eﬀective interest rate of 5%. Level annual interest payments are made at the end of each year for 10 years, and the principal amount is repaid at the end of 10 years. At the end of each year, the borrower makes level annual payments to a sinking fund that earns interest at an annual eﬀective rate of 4%. At the end of 10 years the sinking fund accumulates to the loan principal. Calculate the diﬀerence between the interest payment on the loan and the interest earned by the sinking fund in the ﬁfth year. Problem 39.24 John borrows 10,000 for 10 years at an annual eﬀective interest rate of i. He accumulates the amount necessary to repay the loan by using a sinking fund. He makes 10 payments of X at the end of each year, which includes interest on the loan and the payment into the sinking fund, which earns an annual eﬀective rate of 8%. If the annual eﬀective rate of the loan had been 2i, his total annual payment would have been 1.5X. Calculate i. Problem 39.25 Jason and Margaret each take out a 17 year loan of L. Jason repays his loan using the amortization method, at an annual eﬀective interest rate of i. He makes an annual payment of 500 at the end of each year. Margaret repays her loan using the sinking fund method. She pays interest annually, also at an annual eﬀective interest rate of i. In addition, Margaret makes level annual deposits at 39 SINKING FUND METHOD 363 the end of each year for 17 years into a sinking fund. The annual eﬀective rate on the sinking fund is 4.62% and she pays oﬀthe loan after 17 years. Margaret’s total payment each year is equal to 10% of the original loan amount. Calculate L. Problem 39.26 A 10 year loan of 10,000 is to be repaid with payments at the end of each year consisting of interest on the loan and a sinking fund deposit. Interest on the loan is charged at a 12% annual eﬀective rate. The sinking fund’s annual eﬀective interest rate is 8%. However, beginning in the sixth year the annual eﬀective interest rate on the sinking fund drops to 6%. As a result, the annual payment to the sinking fund is then increased by X. Calculate X. Problem 39.27 NTL Corp. has just taken out a 15 year, 200,000 loan on which it has to make semi-annual interest payments of 7,000. As part of the loan agreement, NTL Corp. needs to make a deposit at the end of every 6 months into a sinking fund in order to retire the loan at the end of the 15th year. NTL Corp. plans to make semi-annual deposits of 4,929.98 into the sinking fund. (a) What eﬀective annual interest rate is NTL Corp. assuming it can earn on the sinking fund? (b) What is the net interest cost in the 6 months following the 10th sinking fund deposit? Problem 39.28 E Corp has a $100,000 loan outstanding on which it has been paying semi-annual interest payments of $4,000. E Corp has also been accumulating deposits made at the end of every 6 months in a sinking fund earning 5% eﬀective annual interest so that it can retire the loan at the end of the 15th year. The lender has oﬀered to accept 103% of the sinking fund balance immediately after the 29th deposit in exchange for the outstanding loan balance and the remaining loan interest payment. At what eﬀective annual rate was the lender calculating the present value of the remaining amounts in order to make this oﬀer equivalent in value? Problem 39.29 A borrower takes out a loan of $2,000 for two years. Create a sinking fund schedule if the lender re-ceives 10% eﬀective on the loan and if the borrower replaces the amount of the loan with semiannual deposits in a sinking fund earning 8% convertible quarterly. Problem 39.30 John borrows $10,000 for ﬁve years at 12% convertible semi-annually. John replaces the principal by means of deposits at the end of every year for ﬁve years into a sinking fund which earns 8% eﬀective. Find the total dollar amount that John must pay over the ﬁve-year period to completely repay the loan. 364 LOAN REPAYMENT METHODS Problem 39.31 A borrower is repaying a loan of 300,000 by the sinking fund method. The sinking fund earns an annual eﬀective interest rate of 6.75%. Payments of $22,520 are made at the end of each year for 20 years to repay the loan. These payments consist of both the interest payment to the lender and also the sinking fund deposit. What is the annual eﬀective interest rate paid to the lender of the loan? 40 LOANS PAYABLE AT A DIFFERENT FREQUENCY THAN INTEREST IS CONVERTIBLE365 40 Loans Payable at a Diﬀerent Frequency than Interest is Convertible In Sections 37 - 39, we considered loans where the payment period coincides with the interest conversion period. In this section we examine loans where the payments are made at a diﬀerent frequency than interest is convertible. The same observations made about the amortization method and the sinking fund method apply to the loans considered in this section. We will restrict our discussion to payments made at the end of an interest conversion period. A similar argument holds for payments made at the beginning of an interest conversion period. Consider ﬁrst the amortization schedule of a loan with payments made less frequently than interest is convertible. Consider a loan of an sk at interest rate i per period being repaid with payments of 1 at the end of each k interest conversion periods for a total of n interest conversion periods. Thus, the total number of payments is n k which we assume is an integral number. At the end of k interest conversion periods a payment of 1 is made, the interest paid is [(1 + i)k − 1] an sk = 1−νn so that the principal repaid is νn, and the outstanding loan balance is an sk −νn = an−k sk . Next, at the end of 2k interest conversion periods, the interest paid is [(1 + i)k −1] an−k sk = 1 −νn−k so that the principal repaid is νn−k, and the outstanding loan balance is an−k sk −νn−k = an−2k sk . Continuing this process, we see that at the end of period mk, the interest paid is [(1 + i)k − 1] an−(m−1)k sk = 1 −νn−(m−1)k and the principal repaid is νn−(m−1)k. The outstanding loan balance is an−(m−1)k sk −νn−(m−1)k = an−mk sk . The amortization table is shown below. Payment Interest Principal Outstanding Period amount paid repaid loan balance 0 an sk k 1 [(1 + i)k −1] an sk = 1 −νn νn an sk −νn = an−k sk 2k 1 [(1 + i)k −1] an−k sk = 1 −νn−k νn−k an−k sk −νn−k = an−2k sk . . . . . . . . . . . . . . . mk 1 [(1 + i)k −1] an−(m−1)k sk = 1 −νn−(m−1)k νn−(m−1)k an−(m−1)k sk −νn−(m−1)k = an−mk sk . . . . . . . . . . . . . . . n −k 1 [(1 + i)k −1] a2k sk = 1 −ν2k ν2k a2k sk −ν2k = ak sk n 1 [(1 + i)k −1] ak sk = 1 −νk νk ak sk −νk = 0 Total n k n k −an sk an sk 366 LOAN REPAYMENT METHODS Note that the principal repaid column is a geometric progression with common ratio (1 + i)k. Example 40.1 A loan of $15,000 is to be repaid by means of six annual payments of $3719.18 each. The nominal interest rate is 12% compounded monthly. Create an amortization schedule for this transaction. Solution. The schedule is given below. Payment Interest Principal Outstanding Period amount paid repaid loan balance 0 15,000 12 3719.18 1902.37 1816.81 13,183.19 24 3719.18 1671.96 2047.22 11,135.97 36 3719.18 1412.32 2306.86 8829.11 48 3719.18 1119.75 2599.43 6229.68 60 3719.18 790.08 2929.10 3300.58 72 3719.18 418.60 3300.58 0 Next, consider a loan of a(m) n at interest rate i per period being repaid with payments of 1 m at the end of each mth of an interest conversion period for a total of n interest conversion periods. Thus, the total number of payments is mn which we assume is an integral number. At the end of the ﬁrst mth of an interest conversion period a payment of 1 m is made, the interest paid is i i(m)a(m) n = 1 m(1 −νn) so that the principal repaid is 1 mνn, and the outstanding loan balance is a(m) n −1 mνn = a(m) n−1 m . Next, at the end of second mth of an interest conversion period, the interest paid is i i(m)a(m) n−1 m = 1 m(1 −νn−1 m) so that the principal repaid is 1 mνn−1 m, and the outstanding loan balance is a(m) n−1 m −1 mνn−1 m = a(m) n−2 m . Continuing this process, we see that at the end of period t n, the interest paid is i i(m)a(m) n−(t−1) m = 1 m(1 −νn−t−1 m ) and the principal repaid is 1 mνn−(t−1) m . The outstanding loan balance is a(m) n−t−1 m −1 mνn−t−1 m = a(m) n−t m . The amortization table is shown below. 40 LOANS PAYABLE AT A DIFFERENT FREQUENCY THAN INTEREST IS CONVERTIBLE367 Payment Interest Principal Outstanding Period amount paid repaid loan balance 0 a(m) n 1 m 1 m i i(m)a(m) n = 1 m(1 −νn) 1 mνn a(m) n −1 mνn = a(m) n−1 m 2 m 1 m i i(m)a(m) n−1 m = 1 m(1 −νn−1 m) 1 mνn−1 m a(m) n−1 m −1 mνn−1 m = a(m) n−2 m . . . . . . . . . . . . . . . t m 1 m i i(m)a(m) n−(t−1) m = 1 m(1 −νn−t−1 m ) 1 mνn−(t−1) m a(m) n−t−1 m −1 mνn−t−1 m = a(m) n−t m . . . . . . . . . . . . . . . n −1 m 1 m i i(m)a(m) 2 m = 1 m(1 −ν 2 m) 1 mν 2 m a(m) n−2 m −1 mνn−2 m = a(m) n−1 m n 1 m i i(m)a(m) 1 m = 1 m(1 −ν 1 m) 1 mν 1 m a(m) n−1 m −1 mνn−1 m = a(m) n−1 m −1 mν 1 m = 0 Total n n −a(m) n a(m) n Note that the principal repaid column is a geometric progression with common ratio (1 + i) 1 m. Example 40.2 A debt is being amortized by means of monthly payments at an annual eﬀective rate of interest of 11%. If the amount of principal in the third payment is $1000, ﬁnd the amount of principal in the 33rd payment. Solution. Recall that the principal repaid column is a geometric progression with common ratio (1+i) 1 m. The interval of time from the 3rd payment to the 33rd payment is 33−3 12 = 2.5 years. Thus, the principal in the 33rd payment is 1000(1.11)2.5 = $1298.10 Example 40.3 A loan of $10,000 is to be repaid by means of twelve monthly payments. The nominal interest rate is 12% compounded quarterly. Create an amortization schedule for this transaction. Solution. Let R be the amount of monthly payment. Then R satisﬁes the equation 10000 = 3Ra4 (3) 0.03. Solving this equation for R we ﬁnd R = $887.94. Moreover, we have i(3) 3 a4 (3) 0.03 = 0.02970490211 3 = 99.02. The schedule is given below. 368 LOAN REPAYMENT METHODS Payment Interest Principal Outstanding Period amount paid repaid loan balance 0 10,000 1 3 887.94 99.02 788.92 9211.08 2 3 887.94 91.20 796.74 8414.34 1 887.94 83.32 804.62 7609.72 4 3 887.94 75.35 812.59 6797.13 5 3 887.94 67.30 820.64 5976.49 2 887.94 59.18 828.76 5147.73 7 3 887.94 50.97 836.97 4310.76 8 3 887.94 42.68 845.26 3465.50 3 887.94 34.31 853.63 2611.87 10 3 887.94 25.86 862.08 1749.79 11 3 887.94 17.33 870.61 879.18 4 887.94 8.71 879.23 0 It is not recommended that the reader depends upon the memorization of formulas in amortization and sinking fund schedules. It is rather more preferable to create these schedules using the basic principles. We illustrate the use of basic principles in creating a sinking fund schedule in the next example. Example 40.4 John borrows $2000 for two years at an annual eﬀective interest rate of 10%. He replaces the principal by means of semiannual deposits for two years in a sinking fund that earns 8% convertible quarterly. Create a sinking fund schedule for this transaction. Solution. The interest payment on the loan is 2000 × 10% = $200 at the end of each year. If R is the semiannual deposit in the sinking fund then Rs8 0.02 s2 0.02 = 2000. Solving this equation for R we ﬁnd R = 2000s2 0.02 s8 0.02 = $470.70. The sinking fund schedule is given below. 40 LOANS PAYABLE AT A DIFFERENT FREQUENCY THAN INTEREST IS CONVERTIBLE369 Interest earned Interest Sinking fund on sinking Amount in Net amount Period paid deposit fund sinking fund of loan 0 1/4 0 0 0 0 2000.00 1/2 0 470.70 0 470.70 1529.30 3/4 0 0 9.41 480.11 1519.89 1 200.00 470.70 9.60 960.41 1039.59 1 1/4 0 0 19.21 979.62 1020.38 1 1/2 0 470.70 19.59 1469.91 530.09 1 3/4 0 0 29.40 1499.31 500.69 2 200.00 470.70 29.99 2000.00 0 370 LOAN REPAYMENT METHODS Practice Problems Problem 40.1 A debt is being amortized by means of annual payments at a nominal interest rate of 12% com-pounded semiannually. If the amount of principal in the third payment is $1000, ﬁnd the amount of principal in the 33rd payment. Problem 40.2 A borrower is repaying a loan by making quarterly payments of $500 over 10 years. Portion of each payment is principal and the rest is interest. Assume an annual eﬀective rate of 8%, how much interest is paid by the borrower over the 10-year period? Problem 40.3 A lender receives payments of $3000 at the end of every year over an unknown number of years. If the interest portion of the third payment is $2000, ﬁnd the amount of principal in the sixth payment. Assume a nominal interest rate of 10% payable quarterly. Problem 40.4 A borrows $10,000 for ﬁve years at 12% convertible semiannually. A replaces the principal by means of deposits at the end of every year for ﬁve years into a sinking fund which earns 8% eﬀective. Find the total dollar amount which A must pay over the ﬁve-year period to completely repay the loan. Problem 40.5 A borrows $5000 for 10 years at 10% convertible quarterly. A does not pay interest currently and will pay all accrued interest at the end of 10 years together with the principal. Find the annual sinking fund deposit necessary to liquidate the loan at the end of 10 years if the sinking fund earns 7% convertible quarterly. Problem 40.6 A borrows $10,000 for ﬁve years at 12% convertible semiannually. A replaces the principal by means of deposits at the end of every year for ﬁve years into a sinking fund which earns 8% eﬀective. Create a sinking fund schedule for this loan. 41 AMORTIZATION WITH VARYING SERIES OF PAYMENTS 371 41 Amortization with Varying Series of Payments In this section, we consider amortization methods with more general patterns of payments, not necessarily leveled. We will keep assuming that the interest conversion period and the payment period are equal and coincide. Consider a loan L to be repaid with n periodic payments (that include principal and interest) R1, R2, · · · , Rn. The equation of value at time t = 0 is L = n X t=1 νtRt. In most cases the series of payments Rt follows some regular pattern encountered with annuties so that the results of Section 26 can be used. Example 41.1 A borrower is repaying a loan with payments at the end of each year for 10 years, such that the payment the ﬁrst year is $200, the second year is $190, and so forth, until the 10th year it is $110. Find an expression of the amount of the loan. Solution. The amount of the loan is L = 100a10 + 10(Da)10 For the type of installments considered in this section, amortization schedules can be constructed from ﬁrst principles as discussed in Section 38. Furthermore, the outstanding loan balance column can be found retrospectively or prospectively as in Section 37, from which the remaining columns of interest paid and principal repaid can be found. Example 41.2 A loan is repaid with payments which start at $200 the ﬁrst year and increase by $50 per year until a payment of $1000 is made, at which time payments cease. If the interest is 4% eﬀective, ﬁnd the amount of interest and principal paid in the fourth payment. Solution. The fourth payment is R4 = $350.00. The outstanding loan balance after making the third payment is Bp 3 = 300a14 + 50(Ia)14. 372 LOAN REPAYMENT METHODS The interest paid after the fourth payment is I4 =iBp 3 = 300(1 −ν14) + 50(¨ a14 −14ν14) =300(1 −0.57747) + 50(10.98565 −8.08465) =$271.81 The principal portion of the fourth payment is P4 = R4 −I4 = 350.00 −271.81 = $78.19 One common pattern of periodic installments is when the borrower makes level payments of princi-pal. Clearly, this leads to successive total payments (consisting of interest and principal) to decrease due to the decrease in successive outstanding loan balance (which in turn results in a decrease in successive interest paid). We illustrate this in the following example. Example 41.3 A borrower is repaying a $1000 loan with 10 equal payments of principal. Interest at 6% compounded semiannually is paid on the outstanding balance each year. Find the price to yield an investor 10% convertible semiannually. Solution. The period principal payment is $100 each. The periodic interest payments are :30, 27, 24, · · · , 9, 3. Thus, the price at time t = 0 is just the present value of the principal and interest paid. That is, Price = 100a10 0.05 + 3(Da)10 0.05 = 908.87 It is possible that when using varying payments, the loan payment for a period is less than the interest charged over that period. This leads to an increase in the outstanding balance which in turn means a negative principal. In ﬁnance, such a case is referred to as negative amortization or deferred interest. Credit card interest rates and payment plans are examples of negative amortization. Most credit cards carry high interest rates yet require a low minimum payment. Example 41.4 A loan of 1000 is being repaid with annual payments over 10 years. The payments in the last ﬁve years are 5 times the payments in the ﬁrst 5 years. If i = 0.08, calculate the principal amortized in the ﬁfth payment. Solution. Let K be the size of the payment in the ﬁrst 5 years. Then 1000 = K (1 −1.08−5) 0.08 + 5K(1.08−5)(1 −1.08−5) 0.08 41 AMORTIZATION WITH VARYING SERIES OF PAYMENTS 373 or 1000 = 17.57956685K →K = 56.88422296. The outstanding loan balance after making the 4th payment is 56.88422296 1.08 + 5(56.88422296)(1.08−1)(1 −1.08−5) 0.08 = 1104.16228. The outstanding loan balance after making the 5th payment is 5(56.88422296)(1 −1.08−5) 0.08 = 1135.61104. The principal amortized in the ﬁfth payment is 1104.16228 −1135.61104 = −$31.45 Next, consider a varying series of payments with the sinking fund method. We will assume that the interest paid to the lender is constant each period so that only the sinking fund deposits vary. Let the varying payments be denoted by R1, R2, · · · , Rn and assume that i ̸= j. Let L denote the amount of the loan. Then the sinking fund deposit for the tth period is Rt −iL. Since the accumulated value of the sinking fund at the end of n periods must be L, we have L =(R1 −iL)(1 + j)n−1 + (R2 −iL)(1 + j)n−2 + · · · + (Rn −iL) = n X t=1 Rt(1 + j)n−t −iLsn j where i is rate of interest paid on the loan and j rate of interest earned on the sinking fund. Solving for L we obtain L = Pn t=1 Rt(1 + j)n−t 1 + isn j = Pn t=1 νt jRt 1 + (i −j)an j . Note that if i = j then L = Pn t=1 Rtνt. Thus, the amortization method and the sinking fund method are equivalent. Example 41.5 A borrower is repaying a loan at 5% eﬀective with payments at the end of each year for 10 years, such that the payment the ﬁrst year is $200, the second year is $190, and so forth, until the 10th year it is $110. Find the amount of the loan if the borrower pays 6% eﬀective on the loan and accumulates a sinking fund to replace the amount fo the loan at 5% eﬀective. Solution. We have L = 100a10 0.05 + 10(Da)10 0.05 1 + (0.06 −0.05)a10 0.05 = 1227.83 1 + (0.01)(7.7217) = $1139.82 374 LOAN REPAYMENT METHODS Practice Problems Problem 41.1 A borrows $10,000 from B and agrees to repay it with a series of 10 installments at the end of each year such that each installment is 20% greater than the preceding installment. The rate of interest on the loan is 10% eﬀective. Find the amount of principal repaid in the ﬁrst three installments. Problem 41.2 John borrows 10,000 from Tony and agrees to repay the loan with 10 equal annual installments of principal plus interest on the unpaid balance at 5% eﬀective. Immediately after the loan is made, Tony sells the loan to Chris for a price which is equal to the present value of John’s future payments calculated at 4%. Calculate the price that Chris will pay for the loan. Problem 41.3 A loan is being repaid with 10 payments of 1000t at the end of year t. (In other words, the ﬁrst payment is 1000, the second payment is 2000, etc.) The interest rate on the loan is 6%. Calculate the outstanding principal immediately after the third payment. Problem 41.4 A loan of 100,000 is being repaid with annual payments at the end of each year for 10 years. The interest rate on the loan is 10.25%. Each annual payment increases by 5% over the previous annual payment. Calculate the principal in the ﬁfth payment. Problem 41.5 ‡ A 30-year loan of 1000 is repaid with payments at the end of each year. Each of the ﬁrst ten payments equals the amount of interest due. Each of the next ten payments equals 150% of the amount of interest due. Each of the last ten payments is X. The lender charges interest at an annual eﬀective rate of 10%. Calculate X. Problem 41.6 ‡ A loan is repaid with 20 increasing annual installments of 1,2,3,· · · ,20. The payments begin one year after the loan is made. Find the principal contained in the 10th payment, if the annual interest rate is 4%. Problem 41.7 In order to repay a school loan, a payment schedule of 200 at the end of the year for the ﬁrst 5 years, 1200 at the end of the year for the next 5 years, and 2200 at the end of the year for the ﬁnal 5 years is agreed upon. If interest is at the annual eﬀective rate of i = 6%, what is the loan value? 41 AMORTIZATION WITH VARYING SERIES OF PAYMENTS 375 Problem 41.8 Create an amortization schedule of the previous problem for the ﬁrst seven years. Problem 41.9 A borrows $20,000 from B and agrees to repay it with 20 equal annual installments of principal in addition to the interest on the unpaid balance at 3% eﬀective. After 10 years B sells the right to future payments to C, at a price which yields C 5% eﬀective over the next 5 years and 4% eﬀective over the ﬁnal 5 years. Find the price which C should pay to the nearest dollar. Problem 41.10 A loan of 3000 at an eﬀective quarterly interest rate of j = .02 is amortized by means of 12 quarterly payments , beginning one quarter after the loan is made. Each payment consists of a principal repayment of 250 plus interest due on the previous quarter’s outstanding balance. Construct the amortization schedule. Problem 41.11 A loan is being repaid with 10 annual payments. The ﬁrst payment is equal to the interest due only, the second payment is twice the ﬁrst, the third payment is three time the ﬁrst, and so forth. Prove that at the rate of interest on the loan is: (Ia)10 = a∞. Problem 41.12 A loan is being repaid with 10 payments. The ﬁrst payment is 10, the second 9, the third is 8, and so forth with the tenth payment being 1. Find an expression for the amount of interest in the sixth payment. Problem 41.13 A has money invested at eﬀective rate i. At the end of the ﬁrst year A withdraws 162.5% of the interest earned; at the end of the second year A withdraws 325% of the interest earned, and so forth with the withdrawal factor increasing in arithmetic progression. At the end of 16 years the fund is exhausted. Find i. Problem 41.14 A loan is amortized over ﬁve years with monthly payments at a nominal interest rate of 9% com-pounded monthly. The ﬁrst payment is 1000 and is to be paid one month from the date of the loan. Each succeeding monthly payment will be 2% lower than the prior payment. Calculate the outstanding loan balance immediately after the 40th payment is made. Problem 41.15 June borrows 20,000 from April and agrees to repay it with a series of ten installments at the end 376 LOAN REPAYMENT METHODS of each year, such that each installment is 15% greater than the preceding installment. The rate of interest on the loan is 10% annual eﬀective. Let Pj denote the amount of principal repaid in year j. Calculate P1 + P2. Problem 41.16 Smith borrows 30,000 at an annual eﬀective interest rate of 10%. He agrees to make annual payments at the end of each year for 9 years, and an additional balloon payment, X, at the end of the tenth year. Each of the ﬁrst 9 payments equals 25% more than what is owed in interest at the time of the payment. The balloon payment, X, equals the amount needed to pay oﬀthe loan. Determine X. Problem 41.17 A loan is being repaid by the amortization method using 10 semiannual payments with the ﬁrst payment due six months after the loan is made. The ﬁrst payment is 50 and each subsequent payment is 50 more than its previous payment. The interest rate on the loan is 10%, compounded semiannually. Determine the amount of principle repaid in the seventh payment. Problem 41.18 A 10 year loan with an eﬀective annual interest rate of 5% is to be repaid with the following payments (1) 100 at the end of the second year (2) 200 at the end of the fourth year (3) 300 at the end of the sixth year (4) 400 at the end of the eighth year (5) 500 at the end of the tenth year Calculate the amount of interest included in the second payment. Problem 41.19 John takes out a 10 year loan. The loan is repaid by making 10 annual repayments at the end of each year. The ﬁrst loan repayment is equal to X, with each subsequent repayment 10.16% greater than the previous year’s repayment. The annual eﬀective interest rate being charged on the loan is 8%. The amount of interest repaid during the ﬁrst year is equal to 892.20. Calculate X. Problem 41.20 Don takes out a 10 year loan of L which he repays with annual payments at the end of each year using the amortization method. Interest on the loan is charged at an annual eﬀective rate of i. Don repays the loan with a decreasing series of payments. He repays 1000 in year one, 900 in year two, 800 in year three, · · · , and 100 in year ten. The amount of principal repaid in year three is equal to 600. Calculate L. 41 AMORTIZATION WITH VARYING SERIES OF PAYMENTS 377 Problem 41.21 ‡ Betty borrows 19,800 from Bank X. Betty repays the loan by making 36 equal payments of principal at the end of each month. She also pays interest on the unpaid balance each month at a nominal rate of 12%, compounded monthly. Immediately after the 16th payment is made, Bank X sells the rights to future payments to Bank Y. Bank Y wishes to yield a nominal rate of 14%, compounded semi-annually, on its investment. What price does Bank X receive? Problem 41.22 Two loans for equal amounts are repaid at an eﬀective interest rate of 4%. Loan A is repaid with 30 equal annual payments. Loan B is to be repaid by 30 annual payments, each containing equal principal amounts and an interest amount based on the unpaid balance. Payments are made at the end of each year. The annual payment for Loan A ﬁrst exceeds the annual payment for Loan B with the kth payment. Find k. Problem 41.23 A borrows $2000 at an eﬀective rate of interest of 10% per annum and agrees to repay the loan with payments at the end of each year. The ﬁrst payment is to be $400 and each payment thereafter is to be 4% greater than the preceding payment, with a smaller ﬁnal payment made one year after the last regular payment. (a) Find the outstanding loan balance at the end of three years. (b) Find the principal repaid in the third payment. 378 LOAN REPAYMENT METHODS Bonds and Related Topics This chapter is about bonds and the valuation of bonds. We will learn how to determine the price of a bond that should be paid by an investor for a desired given yield rate, determine the yield rate given the price, how a bond is amortized, and ﬁnally determine the value of a bond on a given date after it has been purchased. 379 380 BONDS AND RELATED TOPICS 42 Types of Bonds Perhaps the simplest way for a company or a government agency to raise cash is for them to sell bonds to the public. A bond is an interest bearing security which promises to pay a stated amount (or amounts) of money at some future date (or dates). The company or government branch which is issuing the bond outlines how much money it would like to borrow and speciﬁes a length of time, along with an interest rate it is willing to pay. Investors who then lend the requested money to the issuer become the issuer’s creditors through the bonds that they hold. The term of the bond is the length of time from the date of issue until the date of ﬁnal payment. The date of the ﬁnal payment is called the maturity date. Bonds with an inﬁnite term are called perpetuals. The detailed conditions of the bond may permit the bond to be called at some date prior to the maturity date, at which time the issuer (i.e., the lender) will repay the commitment, possibly with some additional amounts. Such a bond is callable. Any date prior to, or including, the maturity date on which a bond may be redeemed is termed a redemption date. The par value or face value of a bond is the amount that the issuer agrees to repay the bondholder by the maturity date. There are several classiﬁcation of bonds: • Accumulation bond is one in which the redemption price includes the original loan plus accu-mulated interest. Examples of such bonds are the Series E Savings bonds. • Bonds with coupons are periodic payments of interest made by the issuer of the bond prior to redemption. It is called a “coupon” because some bonds literally have coupons attached to them. Holders receive interest by stripping oﬀthe coupons and redeeming them. This is less common today as more records are kept electronically. In what follows, we we use the term “bond” we mean bonds with coupons. Zero coupon bonds are bonds that pay no periodic interest payments. It just pays a lump sum at redemption date. • Registered and Unregistered bonds. A registered bond is a bond issued with the name of the owner printed on the face of the certiﬁcate. If the owner decides to sell the bond, the change must be reported to the borrower. The coupon payments are paid by the borrower to the owners of record on each coupon payment date. An unregistered bond or bearer bond is one in which the lender is not listed in the records of the borrower. In this case, the bond belongs to whomever has legal possession of it. Again, these bonds are occasionally called coupon bonds, due to the physically attached coupons. • Fixed-rate and ﬂoated-rate bonds. A ﬁxed-rate bond is a bond that has a ﬁxed rate over the term of the bond. On the other hand, a bond that has a ﬂuctuated interest rate over the term of the bond is called a ﬂoating-rate bond. • Mortgage and debenture bonds. A mortgage bond is a more secured bond backed by a 42 TYPES OF BONDS 381 collateral such as a mortgage on a property. Mortgage bonds are backed by real estate or physical equipment that can be liquidated. A debenture bond is an unsecured bond issued by a civil or governmental corporation or agency and backed only by the credit standing of the issuer. • Income or adjustment bonds. They are a type of high risk bonds in which coupons are paid only if the borrower has earned suﬃcient income to pay them. • Junk bonds. A junk bond is a high-risk bond of default in payments. The risk that a bond issuer does not pay the coupon or principal payments is called default risk. Because of this risk, these bonds typically pay higher yields than better quality bonds in order to make them attractive to investors. • Convertible bond. A convertible bond is a bond that can be converted into the common stock of the company at the option of the bond owner. The owner of the bond is compensated with the ability to convert the bond to common stock, usually at a substantial discount to the stock’s market value. • Serial bonds. A set of bonds issued at the same time but having diﬀerent maturity dates. These are used when a borrower is in need of a large amount of money. • Treasury bonds. Issued by the US Treasury. Terms of seven or more years. • Treasury bills. Short term debt with maturities of 13, 26, or 52 weeks. T-bills yields are com-puted as rates of discount. These yields are computed on a simple discount basis. The basis for counting time periods for T-bills is actual/360. • Municipal bonds. These are bonds issued by state and local governments to ﬁnance large, long−term capital projects (e.g., hospitals, highways, schools). Example 42.1 A 10 year bond matures for its par value of 5000. The price of the bond is 4320.48 at an 8% yield convertible semi-annually. Calculate the coupon rate convertible semi-annually. Coupon payment is deﬁned to be the product of the face value and the coupon periodic rate. Also, the price of a bond is the present value of all future payments. Solution. Let k be the semiannual eﬀective coupon rate. Then 4320.48 = 5000 1.0420 + 5000k(1 −1.04−20) 0.04 or 5000k(1 −1.04−20) 0.04 = 2038.545269 Thus, k = 2038.545269 0.04 [5000(1 −1.04−20)] = 0.029999 382 BONDS AND RELATED TOPICS the coupon rate convertible semi-annually is 2k = 0.059999 = 6.00% Example 42.2 A $1000 par value bond bearing a 6% coupon rate payable semi-annually will be redeemed at 105% at the end of 15 years. Find the price to yield an investor 5% eﬀective. Solution. The price is P = 30a30 j + 1050(1.05)−15 = $1135.54 where j = (1.05) 1 2 −1 Example 42.3 A 10-year accumulation bond with an initial par value(i.e. face value) of 1000 earns interest of 8% compounded semiannually. Find the price to yield an investor 10% eﬀective. Solution. The only return payment is at maturity. The price to yield 10% interest will therefore be (1.10)−10[1000(1.04)20] = $844.77 Example 42.4 Find the price which should be paid for a zero coupon bond which matures for 1000 in 10 years to yield: (a) 10% eﬀective (b) 9% eﬀective (c) Thus, a 10% reduction in the yield rate causes the price to increase by what percentage? Solution. (a) The bond is now worth 1000(1.10)−10 = $385.54 (b) The bond is now worth 1000(1.09)−10 = $422.41 (c) The 10% decrease in the interest rate thereby increases the price by 422.41 −385.54 385.54 = 9.6% 42 TYPES OF BONDS 383 Practice Problems Problem 42.1 A zero bond will pay $1000 at the end of 10 years, and is currently selling for $500. What is the annual return earned by the purchaser? Problem 42.2 Steve purchases a 10-year zero-coupon bond for $400 and has par value of $1,000. Find the yield rate convertible semiannually that would earned by Steve. Problem 42.3 John buys a 13-week T-bill with par value $10,000 at a simple discount rate of 7.5%. What price did he pay for the bond? Problem 42.4 An investor purchases a 26-week T-bill with face value of $10,000 and with a discount yield of d for $9,600. (a) Find d. (b) Even though T-bills are not usually quoted on an annual eﬀective yield basis, the annual eﬀective yield can still be determined. Find the annual eﬀective rate of interest of this investment, assuming the investment period is exactly half a year. Problem 42.5 An investor purchases a 26-week T-bill with face value $10,000 for $X and earns an annual eﬀective rate of 4.17%. Determine X. Problem 42.6 An investor buys a zero-coupon bond with par value $1,000 for $493.63. The maturity date is Y years and the yield rate convertible semi-annually is 8%. Calculate Y. Problem 42.7 A 20 year bond matures for its par value of 10,000. The coupon payable semi-annually is 400. Calculate the price of the bond at a 6% yield rate convertible semi-annually. Problem 42.8 A 20 year bond matures for its par value of 10,000. The coupon payable semi-annually is 400. Calculate the price of the bond at a 6% annual eﬀective yield rate. Problem 42.9 A $1000 par value bond bearing a 5% coupon rate payable semi-annually will be redeemed at 108% at the end of 7 years. Find the price to yield an investor 5% convertible semi-annually. 384 BONDS AND RELATED TOPICS 43 The Various Pricing Formulas of a Bond In this section we consider the question of determining the purchase price which will produce a given yield rate to an investor. We have already encountered this question in the previous section. When considering the price of a bond, we make the following assumptions: (1) All obligations will be paid by the bond issuer on the speciﬁed dates of payments. Any possi-bility of default will be ignored. (2) The bond has a ﬁxed maturity date. (3) The price of the bond is decided immediately after a coupon payment date, or alternatively, at issue date if the bond is brand new. The following symbols and notation will be used in connection with bonds valuation: P = the price paid for a bond. F = The par value or face value or face amount. This amount is is usually printed on the front of the bond and is often the amount payable at the maturity date. C = the redemption value of a bond, i.e. the amount of money paid at a redemption date to the holder of the bond. r = The coupon rate is the eﬀective rate per coupon payment period used in determining the amount of the coupon. The default payment period is a half-year. For example, r = 0.035 for a 7% nominal coupon paid semi-annually. Fr = The amount of a coupon payment. g = The modiﬁed coupon rate. It is deﬁned by g = Fr C . Thus, g is the coupon rate per unit of redemption value, rather than per unit of par value. i = The yield rate of a bond, or the yield to maturity. The interest rate actually earned by the investor, assuming the bond is held until redemption or maturity. n = The number of coupon payment periods from the date of calculation until maturity date or redemption date K = the present value, computed at the yield rate, of the redemption value at the maturity date, or a redemption date, i.e. K = Cνn at the yield rate i. G = The base amount of a bond. It is deﬁned by Gi = Fr or G = Fr i . Thus, G is the amount which, if invested at the yield rate i, would produce periodic interest payments equal to the coupons on the bond. The quantities F, C, r, g, and n are given by the terms of a bond and remain ﬁxed throughout the bonds life. On the other hand, P and i will vary throughout the life of the bond. Price and yield rate have a precise inverse relationship to each other,i.e. as the price rises the yield rate falls and 43 THE VARIOUS PRICING FORMULAS OF A BOND 385 vice-versa. There are three diﬀerent quoted yields associated with a bond: (1) Nominal yield is the ratio of annualized coupon payment to par value. For example, 2 coupons per year of $3.50 on a $100 par value bond result in a 7.00% nominal yield. The reader should note that the use of the word “nominal” in this context is diﬀerent than the one used in Section 9. (2) Current yield is the ratio of the annualized coupon payment to the original price of the bond. For example, if you paid $90 per $100 of par value of the bond described above, the current yield would be $7.00 $90 = 7.78% (3) Yield to maturity is the actual annualized yield rate, or internal rate of return. Four formulas for price Like loans, the price of a bond is deﬁned to be the present value of all future payments. We shall consider four diﬀerent ways of determining the present value, or price of a bond. The formulas are derivable one from the other: there are situations where one may be more useful than another for speciﬁc applications. The ﬁrst of these is called the basic formula. It is found by writing the equation of value at time t = 0 of the time diagram given in Figure 43.1 and is given by P = Fran i + Cνn i = Fran i + K. Figure 43.1 This formula has a verbal interpretation: the price of a bond is equal to the present value of future coupons plus the present value of the redemption value. The second formula is the premium/discount formula, and is derived from the basic formula: P =Fran i + Cνn i =Fran i + C(1 −ian i) =C + (Fr −Ci)an i This says that the price of a bond is the sum of the redemption value and the present value of amortization of premium or discount paid at purchase. Amortization of premium and discount will be discussed in the next section. 386 BONDS AND RELATED TOPICS The third, the base amount formula, can also be derived from the ﬁrst formula as follows: P =Fran i + Cνn i =Gian i + Cνn i =G(1 −νn i ) + Cνn i =G + (C −G)νn i This says that the price is the sum of the base amount and the present value of the diﬀerence between base amount and redemption value. The fourth, the Makeham formula, is also obtained from the basic formula: P =Cνn i + Fran i =Cνn i + Cg 1 −νn i i =Cνn i + g i (C −Cνn i ) =K + g i (C −K) Example 43.1 A 10-year 100 par value bond bearing a 10% coupon rate payable semi-annually, and redeemable at 105, is bought to yield 8% convertible semiannually. Find the price. Verify that all four formulas produce the same answer. Solution. We are given the following: F = 100 C = 105 r = 5% Fr = 5 g = Fr C = 1 21 i = 4% n = 20 K = Cνn = 105(1.04)−20 = 47.92 G = Fr i = 125 Using the basic formula we ﬁnd P = Fran + K = 5a20 0.04 + 47.92 = 5(13.59031) + 47.92 = $115.87 43 THE VARIOUS PRICING FORMULAS OF A BOND 387 Using the premium/discount formula we ﬁnd P = C + (Fr −Ci)an = 105 + (5 −4.2)a20 0.04 = $115.87 Using the base amount formula we ﬁnd P = G + (C −G)νn = 125 + (105 −125)(1.04)−20 = $115.87 Using the Makeham formula we ﬁnd P = K + g i (C −K) = 47.92 + 100 84 (105 −47.92) = $115.87 Example 43.2 For the bond of the previous example, determine the following: (a) Nominal yield based on the par value. (b) Nominal yield, based on the redemption value. (c) Current yield. (d) Yield to maturity. Solution. (a) The nominal yield is the annualized coupon value to the par value,i.e. 2Fr F = 10 100 = 10%. (b) This is the annualized modiﬁed coupon rate 2g = 2Fr C = 2 5 105 = 9.52% (c) Current yield is the ratio of annualized coupon to price 10 115.87 = 8.63% (d) Yield to maturity is the annualized actual yield which is 8% In the problems considered thus far, we have assumed that coupon payments are constant. It is possbile to have bonds with varying coupon payments. In this case, the coupons will constitute a varying annuity which can be evaluated using the approach of Sections 26 - 27. Again, the price of the bond is the present value of all future coupon payments plus the present value of the redemption value. Example 43.3 A 1000 par value 20 year bond with annual coupons and redeemable at maturity at 1050 is purchased for P to yield an annual eﬀective rate of 8.25%. The ﬁrst coupon is 75. Each subsequent coupon is 3% greater than the preceding coupon. Determine P. 388 BONDS AND RELATED TOPICS Solution. The coupon payments constitute a annuity that is varying in geometric progression. The present value of this annuity is 75 1 − 1+k 1+i n i −k = 75 1 − 1.03 1.0825 20 0.0825 −0.03 = 900.02. The present value of the redemption value is 1050(1.0825)−20 = 215.09. Thus, the price of the bond is 900.02 + 215.09 = 1115.11 It is possible that the bond yield rate is not constant over the term of the bond. We illustrate this in the following example. Example 43.4 Find the price of a 1000 par value 10-year bond with coupons of 8.4% convertible semi-annually, which will be redeemed at 1050. The bond yield rate is 10% convertible semi-annually for the ﬁrst ﬁve years and 9% convertible semi-annually for the second ﬁve years. Solution. The present value of all future coupons is 42[a10 0.05 + (1.05)−10a10 0.045] = 528.33. The present value of the redemption value is 1050(1.05)−10(1.045)−10 = 415.08. Thus, the price of the bond is 528.33 + 415.08 = 943.41 Next, we consider cases when the yield rate and coupon rate are at diﬀerent frequencies. Consider ﬁrst the case in which each coupon period contains k yield rate conversion periods. Suppose the total number of yield rate conversion periods over the term of the bond is n. There is a coupon payment of Fr paid at the end of each k yield rate conversion periods. The present value of all future coupon payments is Fr an sk and the present value of the redemption value is Cvn. Thus, the price of the bond is P = Fran sk + Cvn. 43 THE VARIOUS PRICING FORMULAS OF A BOND 389 Example 43.5 Find the price of a $1000 par value 10-year bond maturing at par which has coupons at 8% con-vertible semi-annually and is bought to yield 6% convertible quarterly. Solution. The price of the bond is P = 40a40 0.015 s2 0.015 + 1000(1.015)−40 = 1145.12 The second case is when each yield rate conversion period contains m coupon conversion periods. Again, we let n be the total number of yield rate conversion periods. There is a coupon of Fr m paid at the end of each mth of a conversion period. Thus, the present value of the future coupon payments is Fra(m) n and the present value of the redemption value is Cνn. Thus, the price of the bond is P = Fra(m) n + Cνn. Example 43.6 Find the price of a $1000 par value 10-year bond maturing at par which has coupons at 8% con-vertible quarterly and is bought to yield 6% convertible semi-annually. Solution. The price of the bond is P =40a(2) 20 + 1000(1.03)−20 =40 1 −1.03−20 2[(1.03) 1 2 −1] + 1000(1.03)−20 =1153.21 390 BONDS AND RELATED TOPICS Practice Problems Problem 43.1 A 10-year $1,000 par value bond bearing a 8.4% coupon rate convertible semiannually and re-deemable at $1,050 is bought to yield 10% convertible semiannually. Find the price. Use all four formulas. Problem 43.2 For the bond of the previous problem, determine the following: (a) Nominal yield based on the par value. (b) Nominal yield, based on the redemption value. (c) Current yield. (d) Yield to maturity. Problem 43.3 A 5-year $100 par value bond bearing a 8% coupon rate payable semi-annually is selling at par value. If prevailing market rates of interest suddenly go to 10% convertible semiannually, ﬁnd the percentage change in the price of the bond. Problem 43.4 Two 1000 bonds redeemable at par at the end of the same period are bought to yield 4% convertible semiannually. One bond costs $1136.78, and has a coupon rate of 5% payable semiannually. The other bond has a coupon rate of 2.5% payable semiannually. Find the price of the second bond. Problem 43.5 A n−year $1000 par value bond bearing a 9% coupon rate payable semiannually and redeemable at $1125 is bought to yield 10% convertible semiannually. Find the purchase price of the bond if the present value of the redemption value is $225 at the given yield rate. Problem 43.6 A n−year $1000 par value bond maturing at par and with semi-annual coupon of $50 is purchased for $1110. Find the base amount G if the redemption value is $450. Problem 43.7 An investor owns a 1000 par value 10% bond with semiannual coupons. The bond will mature at par at the end of 10 years. The investor decides that an 8-year bond would be preferable. Current yield rates are 7% convertible semiannually. The investor uses the proceeds from the sale of the 10% bond to purchase a 6% bond with semiannual coupons, maturing at par at the end of 8 years. Find the par value of the 8-year bond. 43 THE VARIOUS PRICING FORMULAS OF A BOND 391 Problem 43.8 An n−year 1000 par value bond matures at par and has a coupon rate of 12% convertible semian-nually. It is bought at a price to yield 10% convertible semiannually. If the term of the bound is doubled, the price will increase by 50. Find the price of the n−year bond. Problem 43.9 A 20 year bond with a par value of $10,000 will mature in 20 years for $10,500. The coupon rate is 8% convertible semiannually. Calculate the price that Andrew would pay if he bought the bond to yield 6% convertible twice a year. Problem 43.10 After 5 years (immediately after the 10th coupon is paid), Andrew decides to sell the bond in the previous problem. Interest rates have changed such that the price of the bond at the time of the sale is priced using a yield rate of 9% convertible semiannually. Calculate the selling price. Problem 43.11 A bond matures for its par value of 1000 in one year. The bond pays a annual coupon on 80 each year. The price of the bond is 800. Which of the following are true? i. The nominal yield is 8% per annum. ii. The current yield is 10% per annum. iii. The yield to maturity is 35% per annum Problem 43.12 You paid $950 for a 7 year bond with a face value of $1,000 and semi-annual coupons of $25. What is your current yield on this bond and what is your nominal annual yield to maturity (convertible semi-annually)? Problem 43.13 A 15-year bond with a coupon of $X payable every 6 months has a face (and redemption) value of $10,000. At the nominal annual interest rate, convertible semi-annually, of 6.5%, the price of the bond is $8,576.36. What is X? Problem 43.14 ‡ A ten-year 100 par value bond pays 8% coupons semiannually. The bond is priced at 118.20 to yield an annual nominal rate of 6% convertible semiannually. Calculate the redemption value of the bond. 392 BONDS AND RELATED TOPICS Problem 43.15 ‡ Susan can buy a zero coupon bond that will pay 1000 at the end of 12 years and is currently selling for 624.60 . Instead she purchases a 6% bond with coupons payable semi-annually that will pay 1000 at the end of 10 years. If she pays X she will earn the same annual eﬀective interest rate as the zero coupon bond. Calculate X. Problem 43.16 ‡ A 30-year bond with a par value of 1000 and 12% coupons payable quarterly is selling at 850. Calculate the annual nominal yield rate convertible quarterly. Problem 43.17 ‡ An investor borrows an amount at an annual eﬀective interest rate of 5% and will repay all interest and principal in a lump sum at the end of 10 years. She uses the amount borrowed to purchase a 1000 par value 10-year bond with 8% semiannual coupons bought to yield 6% convertible semiannually. All coupon payments are reinvested at a nominal rate of 4% convertible semiannually. Calculate the net gain to the investor at the end of 10 years after the loan is repaid. Problem 43.18 ‡ Dan purchases a 1000 par value 10-year bond with 9% semiannual coupons for 925. He is able to reinvest his coupon payments at a nominal rate of 7% convertible semiannually. Calculate his nominal annual yield rate convertible semiannually over the ten-year period. Problem 43.19 ‡ You have decided to invest in Bond X, an n−year bond with semi-annual coupons and the following characteristics: • Par value is 1000. • The ratio of the semi-annual coupon rate to the desired semi-annual yield rate, r i , is 1.03125. • The present value of the redemption value is 381.50. Given vn = 0.5889, what is the price of bond X? Problem 43.20 ‡ Bill buys a 10-year 1000 par value 6% bond with semi-annual coupons. The price assumes a nominal yield of 6%, compounded semi-annually. As Bill receives each coupon payment, he immediately puts the money into an account earning interest at an annual eﬀective rate of i. At the end of 10 years, immediately after Bill receives the ﬁnal coupon payment and the redemption value of the bond, Bill has earned an annual eﬀective yield of 7% on his investment in the bond. Calculate i. Problem 43.21 Two $1000 par value bonds redeemable at par at the end of the same period are bought to yield 43 THE VARIOUS PRICING FORMULAS OF A BOND 393 4% convertible quarterly. One bond costs $1098 and has a coupon rate of 5% payable quarterly. The other bond has a coupon rate of 3% payable quarterly. Find the price of the second bond to the nearest dollar. Problem 43.22 A 1000 par value n−year bond maturing at par with annual coupons of 100 is purchased for 1125. The present value of the redemption value is 500. Find n to the nearest integer. Problem 43.23 A 1000 bond with annual coupons is redeemable at par at the end of 10 years. At a purchase price of 870, the yield rate is i. The coupon rate is i −0.02. Calculate i. Problem 43.24 A bond with coupons equal to 40 sells for P. A second bond with the same maturity value and term has coupons equal to 30 and sells for Q. A third bond with the same maturity value and term has coupons equal to 80. All prices are based on the same yield rate, and all coupons are paid at the same frequency. Determine the price of the third bond in terms of P and Q. Problem 43.25 ‡ You have decided to invest in two bonds. Bond X is an n−year bond with semi-annual coupons, while bond Y is an accumulation bond redeemable in n 2 years. The desired yield rate is the same for both bonds. You also have the following information: Bond X • Par value is 1000. • The ratio of the semi-annual bond rate to the desired semi-annual yield rate, r i is 1.03125. • The present value of the redemption value is 381.50. Bond Y • Redemption value is the same as the redemption value of bond X. • Price to yield is 647.80. What is the price of bond X? Problem 43.26 You are given two n−year par value 1,000 bonds. Bond X has 14% semi-annual coupons and a price of 1,407.70 to yield i compounded semi-annually. Bond Y has 12% semi-annual coupons and a price of 1,271.80 to yield the same rate i compounded semi-annually. Calculate the price of Bond X to yield i −1%. 394 BONDS AND RELATED TOPICS Problem 43.27 Henry has a ﬁve year 1,000,000 bond with coupons at 6% convertible semi-annually. Jean buys a 10 year bond with face amount X and coupons at 6% convertible semi-annually. Both bonds are redeemable at par. Henry and Jean both buy their bonds to yield 4% compounded semi-annually and immediately sell them to an investor to yield 2% compounded semi-annually. Jean earns the same amount of proﬁt as Henry. Calculate X. Problem 43.28 On June 1, 1990, an investor buys three 14 year bonds, each with par value 1000, to yield an eﬀective annual interest rate of i on each bond. Each bond is redeemable at par. You are given (1) the ﬁrst bond is an accumulation bond priced at 195.63 (2) the second bond has 9.4% semiannual coupons and is priced at 825.72 (3) the third bond has 10% annual coupons and is priced at P. Calculate P. Problem 43.29 A 10 year bond with par value 1000 and annual coupon rate r is redeemable at 1100. You are given (1) the price to yield an eﬀective annual interest rate of 4% is P (2) the price to yield an eﬀective annual interest rate of 5% is P −81.49 (3) the price to yield an eﬀective annual interest rate of r is X. Calculate X. Problem 43.30 Patrick buys a 28 year bond with a par value of 1200 and annual coupons. The bond is redeemable at par. Patrick pays 1968 for the bond, assuming an annual eﬀective yield rate of i. The coupon rate on the bond is twice the yield rate. At the end of 7 years Patrick sells the bond for P, which produces the same annual eﬀective yield rate of i to the new buyer. Calculate P. Problem 43.31 A 10 year bond with coupons at 8% convertible quarterly will be redeemed at 1600. The bond is bought to yield 12% convertible quarterly. The purchase price is 860.40. Calculate the par value. Problem 43.32 A corporation decides to issue an inﬂation-adjusted bond with a par value of $1000 and with annual coupons at the end of each year for 10 years. The initial coupon rate is 7% and each coupon is 3% greater than the preceding coupon. The bond is redeemed for $1200 at the end of 10 years. Find the price an investor should pay to produce a yield rate of 9% eﬀective. 43 THE VARIOUS PRICING FORMULAS OF A BOND 395 Problem 43.33 A 1000 par value 10 year bond with semiannual coupons and redeemable at 1100 is purchased at 1135 to yield 12% convertible semiannually. The ﬁrst coupon is X. Each subsequent coupon is 4% greater than the preceding coupon. Determine X. Problem 43.34 A 1,000 par value 3 year bond with annual coupons of 50 for the ﬁrst year, 70 for the second year, and 90 for the third year is bought to yield a force of interest δt = 2t −1 2(t2 −t + 1), t ≥0. Calculate the price of this bond. 396 BONDS AND RELATED TOPICS 44 Amortization of Premium or Discount Bonds can be priced at a premium, discount, or at par. If the bond’s price is higher than its par value,i.e. P > C, then the bond is said to sell at a premium and the diﬀerence P −C is called the “premium”. On the other hand, if the bond’s price is less than its par value,i.e. P < C, then the bond is said to sell at a discount and the diﬀerence C −P is called the “discount”. Thus, a discount is merely a negative premium. Using the premium/discount formula of the previous section we see that Premium = P −C = (Fr −Ci)an i = C(g −i)an i if g > i In this case, a loss in that amount incurred when the bond is redeemed. Similarly, Discount = C −P = (Ci −Fr)an i = C(i −g)an i if g < i and a gain in that amount is incurred when the bond is redeemed. Because of this proﬁt or loss at redemption, the amount of each coupon CANNOT be considered as interest income to an investor. It is necessary to divide each coupon into interest earned and principal adjustment portions similar to the separation of payments into interest and principal when discussing loans. In the same sense that a loan has an outstanding balance at any time, we can talk of the book value of a bond at any time t. We deﬁne the book value to be the present value of all future payments. If we let Bt be the book value after the tth coupon has just been paid, then the value of the remaining payments are:n −t coupons and a payment of C at the date of redemption. Hence, Bt = Fran−t + Cνn−t. Example 44.1 Find the book value immediately after the payment of the 14th coupon of a 10-year 1,000 par value bond with semiannual coupons, if r = 0.05 and the yield rate is 12% convertible semiannually. Solution. We have F = C = 1000, n = 20, r = 0.05, and i = 0.06. Thus, B14 = 1000(0.05)a6 0.06 + 1000(1.06)−6 = 950.83 Letting t = 0 in the expression of Bt we ﬁnd B0 = P. Similarly, Bn = C. Thus, on the redemption date, the book value of a bond equals the redemption value of the bond. From this it follows that when a bond is purchased at a premium, the book value of the bond will be written -down(decreased) at each coupon date, so that at the time of redemption its book value 44 AMORTIZATION OF PREMIUM OR DISCOUNT 397 equals the redemption value. This process is called amortization of premium. When it is puchased at a discount, the book value of the bond will be written-up(increased) at each coupon date, so that at the time of redemption its book value equals the redemption value. This process is called accumulation of discount. Book values are very useful in constructing bond amortization schedules. A bond amortization schedule shows the division of each bond coupon into interest earned and principal adjustment portions, together with the book value after each coupon is paid. We will denote the interest earned after the tth coupon has been made by It and the corresponding principal adjustment portion by Pt. The level coupon will continue to be denoted by Fr. Note that It =iBt−1 = i[Fran−t+1 + Cνn−t+1] =Cg(1 −νn−t+1) + iCνn−t+1 = Cg + C(i −g)νn−t+1 and Pt = Fr −It = Cg −It = C(g −i)νn−t+1. where Pt is positive in the case of a premium and negative in the case of a discount. It is easy to see that Bt = B0 − t X k=1 Pk. Example 44.2 A $1000 bond, redeemable at par on December 1, 1998, with 9% coupons paid semiannually. The bond is bought on June 1, 1996. Find the purchase price and construct a bond amortization schedule if the desired yield is 8% compounded semiannually. Solution. Using the premium/discount formula with C = 1000, F = 1000, r = 4.5%, and i = 4%, the purchase price on June 1, 1996 is P = 1000 + (45 −40)a5 0.04 = $1022.26. Thus, the bond is purchased at a premium of $22.26. For the bond amortization schedule, we ﬁrst do the calculation of the ﬁrst line of the schedule. Each successive line of the schedule is calculated in a similar fashion. The interest earned portion of the ﬁrst coupon, i.e. on Dec 1, 1996 is I1 = iB0 = 0.04(1022.26) = 40.89. The principal adjustment portion of the ﬁrst coupon is P1 = Fr −I1 = 45 −40.89 = $4.11. 398 BONDS AND RELATED TOPICS The book value at the end of the ﬁrst period is B1 = B0 −P1 = $1018.15. Amount for Interest amortization of Book Date Coupon earned premium value June 1, 1996 0 0 0 1022.26 Dec 1, 1996 45.00 40.89 4.11 1018.15 June 1, 1997 45.00 40.73 4.27 1013.88 Dec 1, 1997 45.00 40.56 4.44 1009.44 June 1, 1998 45.00 40.38 4.62 1004.82 Dec 1, 1998 45.00 40.19 4.81 1000.01 Total 225.00 202.74 22.25 The one cent error is due to rounding The following observations can be concluded from the above amortization schedule: (1) The sum of the principal adjustment column is equal to the amount of premium/discount. (2) The sum of the interest paid column is equal to the diﬀerence between the sum of the coupons and the sum of the principal adjustment column. (3) The principal adjustment column is a geometric progression with common ration 1 + i. The following is an example of a bond bought at discount. Example 44.3 A $1000 bond, redeemable at par on December 1, 1998, with 9% coupons paid semiannually. The bond is bought on June 1, 1996. Find the purchase price and construct a bond amortization schedule if the desired yield is 10% compounded semiannually. Solution. Using the premium/discount formula with C = 1000, F = 1000, r = 4.5%, and i = 5%, the purchase price on June 1, 1996 is P = 1000 + (45 −50)a5 0.05 = $978.35. Thus, the bond is purchased at a discount of $21.65. 44 AMORTIZATION OF PREMIUM OR DISCOUNT 399 Amount for Interest accumulation of Book Date Coupon earned discount value June 1, 1996 0 0 0 978.35 Dec 1, 1996 45.00 48.92 −3.92 982.27 June 1, 1997 45.00 49.11 −4.11 986.38 Dec 1, 1997 45.00 49.32 −4.32 990.70 June 1, 1998 45.00 49.54 −4.54 995.24 Dec 1, 1998 45.00 49.76 −4.76 1000.00 Total 225.00 246.65 −21.65 Much like dealing with loans, if it is desired to ﬁnd the interest earned or principal adjustment portion of any one coupon, it is not necessary to construct the entire table. Simply ﬁnd the book value at the beginning of the period in question which is equal to the price at that point computed at the original yield rate and can be determined by the methods of Section 41. Then ﬁnd that one line of the table. Example 44.4 Consider a $1000 par value two-year 8% bond with semiannual coupons bought to yield 6% con-vertible semiannually. Compute the interest portion earned and the principal adjustment portion of the 3rd coupon payment. Solution. We have F = C = 1000, n = 4, r = 0.04, and i = 0.03. Thus, the coupon payment is Fr = $40.00. The principal adjustment portion of the 3rd coupon payment is P3 = B3 −B2 = 40a1 0.03 + 1000(1.03)−1 −(40a2 0.03 +1000(1.03)−2) = $9.43. The interest earned portion is I3 = 40.00−9.43 = $30.57 Another method of writing up or writing down the book values of bonds is the straight line method. This does not produce results consistent with compound interest, but it is very simple to apply. In this method, the book values are linear, grading from B0 = P to Bn = C. Thus, the principal adjustment column is constant and is given by Pt = P −C n , t = 1, 2, · · · , n and the interest earned portion is It = Fr −Pt, t = 1, 2, · · · , n Note that Pt > 0 for premium bond and Pt < 0 for discount bond. 400 BONDS AND RELATED TOPICS Example 44.5 Find the book values for the bond in Example 44.2 by the straight line method. Solution. We have Pt = 22.25 5 = 4.45. Thus, B0 =1022.26 B1 =1022.26 −4.45 = 1017.81 B2 =1017.81 −4.45 = 1013.36 B3 =1013.36 −4.45 = 1008.91 B4 =1008.91 −4.45 = 1004.46 B5 =1004.46 −4.45 = 1000.01 44 AMORTIZATION OF PREMIUM OR DISCOUNT 401 Practice Problems Problem 44.1 A $1000 par value two-year 8% bond with semiannual coupons is bought to yield 6% convertible semiannually. (a) Create an amortization schedule for this transaction. (b) Find the book values of the bond using the straight line method. Problem 44.2 A $1000 par value two-year 8% bond with semiannual coupons is bought to yield 10% convertible semiannually. (a) Create an amortization schedule for this transaction. (b) Find the book values of the bond using the straight line method. Problem 44.3 For a bond of face value 1 the coupon rate is 150% of the yield rate, and the premium is p. For another bond of 1 with the same number of coupons and the same yield rate, the coupon rate is 75% of the yield rate. Find the price of the second bond. Problem 44.4 For a certain period a bond amortization schedule shows that the amount for amortization of premium is 5, and that the required interest is 75% of the coupon. Find the amount of the coupon. Problem 44.5 A 10-year bond with semi-annual coupons is bought at a discount to yield 9% convertible semian-nually. If the amount for accumulation of discount in the next-to-last coupon is 8, ﬁnd the total amount for accumulation of discount during the ﬁrst four years in the bond amortization schedule. Problem 44.6 A 1000 par value 5-year bond with a coupon rate of 10% payable semiannually and redeemable at par is bought to yield 12% convertible semiannually. Find the total of the interest paid column in the bond amortization schedule. Problem 44.7 A ﬁve year bond with a par value of 1000 will mature in 5 years for 1000. Annual coupons are payable at a rate of 6%. Create the bond amortization schedule if the bond is bought to yield 8% annually. 402 BONDS AND RELATED TOPICS Problem 44.8 A 40 year bond with a par value of 5000 is redeemable at par and pays semi-annual coupons at a rate of 7% convertible semi-annually. The bond is purchased to yield annual eﬀective rate of 6%. Calculate the amortization of the premium in the 61st coupon. Problem 44.9 A 10 year bond is redeemable at par of 100,000. The bond pays semi-annual coupons of 4000. The bond is bought to yield 6% convertible semi-annually. Calculate the premium paid for the bond. Problem 44.10 A bond is redeemable in eight years at 120% of its par value. The bond pays an annual coupon rate of 6%. Calculate the premium as a percent of the par value if the bond is purchased to yield 6% annually. Problem 44.11 Consider a three-year bond, with a $1,000 face value and a 9% coupon rate paid semi-annually, which was bought to yield 7% convertible semi-annually. Find the amount for amortization of premium during the bonds third half-year (i.e., between the second and third coupon payments). Problem 44.12 A bond of face amount 100 pays semi-annual coupons and is purchased at a premium of 36 to yield annual interest of 7% compounded semiannually. The amount for amortization of premium in the 5th coupon is 1.00. What is the term of the bond? Problem 44.13 ‡ Among a company’s assets and accounting records, an actuary ﬁnds a 15-year bond that was purchased at a premium. From the records, the actuary has determined the following: (i) The bond pays semi-annual interest. (ii) The amount for amortization of the premium in the 2nd coupon payment was 977.19. (iii) The amount for amortization of the premium in the 4th coupon payment was 1046.79. What is the value of the premium? Problem 44.14 ‡ A 10,000 par value 10-year bond with 8% annual coupons is bought at a premium to yield an annual eﬀective rate of 6%. Calculate the interest portion of the 7th coupon. Problem 44.15 ‡ A 1000 par value 5 year bond with 8% semiannual coupons was bought to yield 7.5% convertible semiannually. Determine the amount of premium amortized in the sixth coupon payment. 44 AMORTIZATION OF PREMIUM OR DISCOUNT 403 Problem 44.16 On May 1, 1985, a bond with par value 1000 and annual coupons at 5.375% was purchased to yield an eﬀective annual interest rate of 5%. On May 1, 2000, the bond is redeemable at 1100. The book value of the bond is adjusted each year so that it equals the redemption value on May 1, 2000. Calculate the amount of write-up or write-down in the book value in the year ending May 1, 1991. Problem 44.17 A bond with par value of 1000 and 6% semiannual coupons is redeemable for 1100. You are given (1) the bond is purchased at P to yield 8% convertible semiannually and (2) the amount of principal adjustment for the 16th semiannual period is 5 Calculate P. Problem 44.18 Laura buys two bonds at time 0. Bond X is a 1000 par value 14 year bond with 10% annual coupons. It is bought at a price to yield an annual eﬀective rate of 8%. Bond Y is a 14 year par value bond with 6.75% annual coupons and a face amount of F. Laura pays P for the bond to yield an annual eﬀective rate of 8%. During year 6 the writedown in premium (principal adjustment) on bond X is equal to the writeup in discount (principal adjust-ment) on bond Y. Calculate P. Problem 44.19 A 1000 par value 18 year bond with annual coupons is bought to yield an annual eﬀective rate of 5%. The amount for amortization of premium in the 10th year is 20. The book value of the bond at the end of year 10 is X. Calculate X. Problem 44.20 Thomas buys a bond at a premium of 200 to yield 6% annually. The bond pays annual coupons and is redeemable for its par value of 1000. Calculate the amount of interest in the ﬁrst coupon. Problem 44.21 Megan buys a bond that is redeemable for its par value of 20,000 after 5 years. The bond pays coupons of 800 annually. The bond is bought to yield 8% annually. Calculate the accumulation of discount in the 4th coupon. Problem 44.22 Jenna buys a bond at a premium. The principal amortized in the ﬁrst annual coupon is half that amortized in the 11th annual coupon. Calculate the yield rate used to calculate the purchase price of the bond. 404 BONDS AND RELATED TOPICS Problem 44.23 Betty buys an n−year 1000 par value bond with 6.5% annual coupons at a price of 798.48. The price assumes an annual eﬀective yield rate of i. The total write-up in book value of the bond during the ﬁrst 3 years after purchase is 22.50. Problem 44.24 Ann purchases a 1000 face value 20-year bond, redeemable at 1200, with 5% annual coupons, at a price that yields her an annual eﬀective interest rate of 3%. Immediately upon receiving each coupon, she invests the coupon in an account that earns an annual eﬀective interest rate of 7%. Immediately after receiving the eighth coupon, Ann sells the bond at a price that yields the new purchaser an annual eﬀective interest rate of 4.75%. Determine Ann’s yield rate, as an annual eﬀective interest rate, over the 8-year period that she owns the bond. Problem 44.25 A 1000 face value 20-year 8% bond with semiannual coupons is purchased for 1014. The redemption value is 1000. The coupons are reinvested at a nominal annual rate of 6%, compounded semiannually. Determine the purchaser’s annual eﬀective yield rate over the 20- year period. Problem 44.26 An n−year 1000 par value bond with 8% annual coupons has an annual eﬀective yield of i, i > 0. The book value of the bond at the end of year 3 is 1099.84 and the book value at the end of year 5 is 1082.27. Calculate the purchase price of the bond. Problem 44.27 Bryan buys a 2n−year 1000 par value bond with 7.2% annual coupons at a price P. The price assumes an annual eﬀective yield of 12%. At the end of n years, the book value of the bond, X, is 45.24 greater than the purchase price, P. Assume νn 0.12 < 0.5. Calculate X. 45 VALUATION OF BONDS BETWEEN COUPONS PAYMENT DATES 405 45 Valuation of Bonds Between Coupons Payment Dates Up to now, bond prices and book values have been calculated assuming the coupon has just been paid. In particular, we can ﬁnd a relationship between the book values of two consecutive coupon dates t and t + 1 as follows. Let Bt and Bt+1 be the book values just after the tth and (t + 1)th coupons are paid and Fr the amount of the coupon. Then Bt(1 + i) −Fr =Fr 1 −νn−t i (1 + i) + Cνn−t(1 + i) −Fr =Fr i −Frνn−t i (1 + i) + Cνn−t−1 =Fr i −Frνn−t−1 i + Cνn−t−1 =Fr 1 −νn−t−1 i + Cνn−t−1 =Fran−t−1 + Cνn−t−1 =Bt+1 where we assume a constant yield rate i over the interval. Example 45.1 You buy an n−year 1,000 par value bond with 6.5% annual coupons at a price of 825.44, assuming an annual yield rate of i, i > 0. After the ﬁrst two years, the bond’s book value has changed by 23.76. Calculate i. Solution. We have B2 −B0 = B0(1 + i)2 −Fr(1 + i) −Fr −B0. This implies 23.76 = 825.44(1 + i)2 −65(1 + i) −65 −825.44 which reduces to 825.44(1 + i)2 −65(1 + i) −914.20 = 0 Solving this with the quadratic formula we ﬁnd 1 + i = 1.0925 →i = 9.25% Book values can be computed at times other than immediately after a coupon payment. In this section, we would like to analyze the book value Bt+k for 0 < k < 1 where time k is measured from the last coupon payment. When buying an existing bond between its coupon dates, one must decide how to split up the 406 BONDS AND RELATED TOPICS coupon between the prior owner and the new owner. Consider the price that a buyer would pay for the bond a fractional time k through a coupon period. Assume that the buyer will obtain a yield rate equal to that of the current bond holder. The buyer will receive all of the next coupon. The current holder would expect to receive part of this coupon as interest for the period which we call accrued interest or accrued coupon and is denoted by Frk. Note that Fr0 = 0 and Fr1 = Fr, the coupon payment, where Fr1 is computed just before the coupon is paid. How should the purchase price be allocated between accrued interest (or accrued coupon) and price for the bond? The purchase price for the bond is called the ﬂat price. It is deﬁned to be the money which actually changes hands at the date of sale. It is denoted by Bf t+k. The price for the bond is the book value, which is also called the market price and is denoted by Bm t+k. The market price is the price commonly quoted in the ﬁnancial press, since the market price changes smoothly through time, while the ﬂat price ﬂuctuates due to coupon accrual. From the above deﬁnitions, it is clear that Bf t+k = Bm t+k + Frk. Example 45.2 Consider the amortization schedule of Example 44.2. Amount for Interest amortization of Book Date Coupon earned premium value June 1, 1996 0 0 0 1022.26 Dec 1, 1996 45.00 40.89 4.11 1018.15 June 1, 1997 45.00 40.73 4.27 1013.88 Dec 1, 1997 45.00 40.56 4.44 1009.44 June 1, 1998 45.00 40.38 4.62 1004.82 Dec 1, 1998 45.00 40.19 4.81 1000.01 Total 225.00 202.74 22.25 Sketch a graph describing the relationship between the ﬂat price and the market price. Solution. The graph is shown in Figure 45.1. Note that the accrued coupon at any date is equal to the vertical distance between the solid line and the dotted line 45 VALUATION OF BONDS BETWEEN COUPONS PAYMENT DATES 407 Figure 45.1 There are three methods used to ﬁnd the ﬂat price, the market price, and the accured coupon. Values on the coupon dates are known, so the diﬀerences among the three methods arise only for the interim values between coupon dates. First, we have the theoretical method. This is an exact method, based on compound interest. This method argues that the ﬂat price should be the book value Bt after the preceding coupon, accumulated by (1 + i)k, where i is the yield rate, giving the ﬂat price of Bf t+k = (1 + i)kBt. The accrued coupon is given by (See Problem 15.48) Frk = Fr (1 + i)k −1 i . The market value is then the diﬀerence Bm t+k = (1 + i)kBt −Fr (1 + i)k −1 i The second method is called the practical method. It is an approximate method that uses the linear approximation (1 + i)k ≈1 + ki for 0 < k < 1. Thus, obtaining Bf t+k =(1 + ki)Bt Frk =kFr Bm t+k =(1 + ki)Bt −kFr = Bt + kiBt −kFr =Bt + k[Bt+1 + Fr −Bt] −kFr = (1 −k)Bt + kBt+1 408 BONDS AND RELATED TOPICS where we used the formula Bt(1 + i) −Fr = Bt+1. The third method called the semi-theoretical method is the most widely used method, and has been accepted as the standard method of calculation by the securities industry. The ﬂat price is determined as in the theoretical method, and the accrued coupon is determined as in the practical method. Bf t+k =(1 + i)kBt Frk =kFr Bm t+k =(1 + i)kBt −kFr In computing k, we use an actual count of days: k = number of days since last coupon paid number of days in the coupon period . Both the actual/actual and 30/360 methods (see Section 5) can be used. See Problem 45.12. Example 45.3 Consider a $1000 par value two-year 8% bond with semiannual coupons bought to yield 6% con-vertible semiannually. The price of the bond is computed to be $1037.17. Compute the ﬂat price, accrued interest, and market price ﬁve months after purchase of the bond. Solution. We have Fr = 1000(0.04) = $40.00 and k = 5 6. With the theoretical method we ﬁnd Bf 5 6 =1037.17(1.03) 5 6 = $1063.04 Fr 5 6 =40 " (1.03) 5 6 −1 0.03 # = $33.25 Bm 5 6 =1063.04 −33.25 = $1029.79 With the practical method we ﬁnd Bf 5 6 =1037.17 1 + 5 6(0.03) = $1063.10 Fr 5 6 =5 6(40) = $33.33 Bm 5 6 =1063.10 −33.33 = $1029.77 45 VALUATION OF BONDS BETWEEN COUPONS PAYMENT DATES 409 Finally, with the semi-theoretical method we have Bf 5 6 =1037.17(1.03) 5 6 = $1063.04 Fr 5 6 =5 6(40) = $33.33 Bm 5 6 =1063.04 −33.33 = $1029.71 Example 45.4 A 1000 par value 5 year bond with semi-annual coupons of 60 is purchased to yield 8% convertible semi-annually. Two years and two months after purchase, the bond is sold at the ﬂat price which maintains the yield over the two years and two months. Calculate the ﬂat price using the theoretical method. Solution. The price of the bond after two years is P = 60a6 0.04 + 1000ν−6 0.04 = 1, 104.84. The ﬂat price is Bf 2 6 = 1104.84(1.04) 2 6 = 1, 119.38 We conclude this section by ﬁnding the premium or discount between coupon payment dates. For the premium we have the formula Premium = Bm t+k −C, i < g and for the discount we have Discount = C −Bm t+k, i > g. Thus, premiums and discounts are based on the market value or the book value and not on the ﬂat price. Example 45.5 A n−year $1000 par value bond is selling at $1010 and has a market value of $980 and acrrued coupon of $30. Find premium/discount amount. Solution. Since the premium/discount calculation is based on market value rather than ﬂat price, the bond is a discount bond. The discount amount is $20 410 BONDS AND RELATED TOPICS Practice Problems Problem 45.1 A 10 year bond with semi-annual coupons has a book value immediately after the 5th coupon of 90,000. The ﬂat price 5 months later using the theoretical method is 94,591. Calculate the semi-annual yield on the bond. Problem 45.2 A bond with semi annual coupons of 2500 has a book value immediately after the 6th coupon of 95,000. The market value using the practical method z months after the 6th coupon is 95,137.50. Calculate z if the yield rate is 7% convertible semi-annually. Problem 45.3 A 10 year bond with a par value of 100,000 and semi-annual coupons 2500 is bought at a discount to yield 6% convertible semi-annually. (a) Calculate the book value immediately after the 5th coupon. (b) Calculate the ﬂat price 4 months after the 5th coupon using the theoretical method. (c) Calculate the accrued coupon 4 months after the 5th coupon using the theoretical method. (d) Calculate the market price 4 months after the 5th coupon using the theoretical method. (e) Calculate (b)-(d) using the practical method. (f) Calculate (b)-(d) using the semi-theoretical method Problem 45.4 Which of the following are true for a bond with semi annual coupons? (I) The ﬂat price of the bond under the theoretical method will always equal the ﬂat price of the bond under the practical method. (II) The ﬂat price of the bond under the practical method will always exceed the ﬂat price of the bond under the semi-theoretical method. (III) For a bond purchased at par which matures at par with the yield rate equal to the coupon rate, the ﬂat price at any point in time will equal the par value of the bond under the semi-theoretical method. Problem 45.5 A bond pays coupons of 600 on April 30 and October 31. Calculate the accrued coupon under the practical method on July 1. Problem 45.6 A 10 year bond is redeemable at par of 100,000. The bond has semi-annual coupons of 4000. The bond is bought to yield 6% convertible semi-annually. Four months after purchase, calculate the market price based on the theoretical method less the market price based on the semi-theoretical method. 45 VALUATION OF BONDS BETWEEN COUPONS PAYMENT DATES 411 Problem 45.7 A 10-year 100 par value bond bearing a 10% coupon rate payable semi-annually, and redeemable at 105, is bought to yield 8% convertible semiannually. Using the practical method, compute the ﬂat price, accrued interest, and market price three months after purchase of the bond. Problem 45.8 Show that Bf t+k = (Bt+1 + Fr)v1−k. Problem 45.9 Arrange in increasing order of magnitude for the three interim bond price methods: (a) Flat price. (b) Market price. Problem 45.10 A 10,000 par value bond with 6% semiannual coupons is being sold 3 years and 2 months before the bond matures. The purchase will yield 8% convertible semiannually to the buyer. Determine the price of the bond. Use the practical method. Problem 45.11 ‡ A 10,000 par value bond with coupons at 8%, convertible semiannually, is being sold 3 years and 4 months before the bond matures. The purchase will yield 6% convertible semiannually to the buyer. The price at the most recent coupon date, immediately after the coupon payment, was 5640. Calculate the market price of the bond, assuming compound interest throughout. Problem 45.12 Consider a $1,000 par value bond with coupons at 8% convertible semi-annually and with maturity date 05/06/04. The bond was purchased on 11/06/01. Compute the accrued interest on 12/04/001 taking into account the Actual/Actual day-count basis. Problem 45.13 A 20-year 10% annual coupon bond has a par value of $1,000. When you originally purchased this bond, the eﬀective annual interest rate was 8%. Suppose that ﬁve years after purchase, the eﬀective annual interest rate is 10.5%. What is the absolute diﬀerence between the book and market values of the bond at that point in time? Problem 45.14 ‡ A 1000 bond with semi-annual coupons at i(2) = 6% matures at par on October 15, 2020. The bond is purchased on June 28, 2005 to yield the investor i(2) = 7%. What is the purchase price? Assume simple interest between bond coupon dates and note that: 412 BONDS AND RELATED TOPICS Date Day of the year April 15 105 June 28 179 October 15 288 46 APPROXIMATION METHODS OF BONDS’ YIELD RATES 413 46 Approximation Methods of Bonds’ Yield Rates In this section we consider the question of determining the yield rate to maturity to an investor given the purchase price of a bond. The determination of such yield rates is similar to the determination of an unknown rate of interest for an annuity, discussed in Section 20. We ﬁrst consider the yield rate for a bond purchased on a coupon payment date immediately after the coupon is paid. There are two approaches for this problem. One approach is linear interpolation where bond tables are used. We will not pursue this method any further in this book. Another approach is to develop algebraic estimation formulas for the yield rate. Three main formulas will be discussed. We start with the premium discount formula P =C + (Fr −Ci)an i =C + C(g −i)an i where F, r, g, C, n, and P are known and i is the unknown. Letting k = P−C C we ﬁnd (g −i)an i = k. Solving this equation for i we obtain i = g −k an i which deﬁnes i implicitly. We will develop methods for estimating i in this equation. The ﬁrst is obtained as follows: Writing the power series expansion of 1 an i (see p.209) 1 an i = 1 n 1 + n + 1 2 i + n2 −1 12 i2 + · · · and neglecting terms of higher than the ﬁrst degree in i we obtain the estimation i = g −k an i ≈g −k n 1 + n + 1 2 i . Solving for i we ﬁnd i ≈ g −k n 1 + n+1 2n k. (46.1) Formula (46.1) forms the basis of an even simpler method for calculating approximate yield rates called the bond salesman’s method. Noting that n+1 2n ≈1 2 for large n, this method replaces the ratio n+1 2n in formula (46.1) by 1 2 thus obtaining i ≈g −k n 1 + 1 2k. 414 BONDS AND RELATED TOPICS Error analysis shows that this method produces less accurate results than formula (46.1). The salesman’s formula can be rewritten in the form i ≈n(Cg) + C −P n 2(P + C) = n(Fr) + C −P n 2(P + C) = Total Interest Paid + Capital Gain/Loss Number of Periods × Average Amount Invested Example 46.1 A $100 par value 10-year bond with 8% semiannual coupons is selling for $90. Find the yield rate convertible semi-annually. Solution. We have n = 20, P = 90, C = 100, Fr = 4 so that by the salesman’s formula we have i ≈n(Fr) + C −P n 2(P + C) = 20(4) + 100 −90 0.5(20)(90 + 100) = 0.0474 so the yield rate is 9.48% convertible semi-annually The third method for estimating i is by solving the equation i = g −k an i . by using the Newton-Raphson method. In this case, we are looking for the solutions of the equation f(i) = 0 where f(i) = i −g + k an i = i −g + ki 1 −(1 + i)−n. The Newton-Raphson iterations are given by the formula is+1 = is −f(is) f ′(is). Thus, an expression of the form i −f(i) f′(i) need to be determined. This is done as follows. 46 APPROXIMATION METHODS OF BONDS’ YIELD RATES 415 i −f ′(i) f(i) =i − i −g + ki 1−(1+i)−n 1 + k [1−(1+i)−n]−ni(1+i)−n−1 [1−(1+i)−n]2 =i + g(1 −νn)2 −i(1 −νn)2 −ki(1 −νn) (1 −νn)2 + k[(1 −νn) −niνn+1] =i + (1 −νn)[gian i −i(1 −νn) −ki] ian i(1 −νn) + kian i −nkiνn+1 =i + (1 −νn)[gan i −(1 −νn) − P C −1 ] an i(1 −νn) + P C −1 an i −n P C −1 νn+1 =i + (1 −νn) gan i + νn −P C an i(1 −νn) + (g −i)(an i)2 −n(g −i)an iνn+1 =i + gan i + νn −P C an i + (g −i)(an i)i−1 −n(g −i)i−1νn+1 =i + i gan i + νn −P C ian i + (g −i)(an i) −n(g −i)νn+1 =i " 1 + gan i + νn −P C an i + (g −i)(an i) −n(g −i)νn+1 # =i " 1 + gan i + νn −P C gan i + n(i −g)νn+1 # Thus, the Newton-Raphson iterations are given by is+1 = is " 1 + gan is + (1 + is)−n −P C gan is + n(is −g)(1 + is)−(n+1) # An initial starting guess can be chosen to be i0 = g −k n 1 + n+1 2n k. Example 46.2 A $100 par value 8-year bond with 6% semiannual coupons is selling for $97. Find the yield rate convertible semiannually. 416 BONDS AND RELATED TOPICS Solution. We have g = 3%, n = 16, k = P−C C = 97−100 100 = −0.03. An initial starting guess for the Newton-Raphson method is i0 = 0.03 −(−0.03/16) 1 + 16+1 32 = 0.0323912 and the iterations are given by the formulas is+1 = is 1 + 0.03a16 is + (1 + is)−16 −0.97 0.03a16 is + 16(is −0.03)(1 + is)−17 . We obtain the following values s is 2is 0 0.0323912 0.0647824 1 0.0324329 0.0648658 2 0.0324329 0.0648658 Thus, the yield rate convertible semiannually is 6.48658%. It is worth noting that the selling price is P = 100 0.06 2 a16 3.24329% + 100(1.0324329)−16 = $97 One can also detetermine the yield rate when the purchasing date is between two coupon payment dates. This situation is more complex due to the fractional period prior to the next coupon payment date. In practice, this case involves the use of the semi-theoretical method of Section 45 which leads to an implicit equation in i that can be solved by either linear interpolation or by iterations. We illustrate this in the next example. Example 46.3 A $100 par value 10-year bond with 8% semiannual coupons was issued on March 1, 2001. The market price on May 15, 2003 is $88. Find the yield rate if the bond in bought on that date. Solution. The price of the bond on March 1, 2003 immediately after a coupon payment is 100 + (4 −100i)a16 i. The number of days from March 1, 2003 to May 15, 2003 is 75, while the number of days from March 1, 2003 to September 1, 2003 is 184. Using the semi-theoretical method of Section 45, we have the equation 100 + (4 −100i)a16 i 75 184 −75 184 · 4 = 88. 46 APPROXIMATION METHODS OF BONDS’ YIELD RATES 417 Let’s solve this last equation by using linear interpolation. Let f(i) = 100 + (4 −100i)a16 i 75 184 −75 184 · 4 −88. By trial and error we ﬁnd f(0.05) = 23.43 > 0 and f(0.052) = −2.45 < 0. Thus, using linear interpolation we ﬁnd i ≈0.05 −23.43 × 0.052 −0.05 −2.45 −23.43 = 0.0518. So the nominal yield rate convertible semiannually is 10.36%. The price on May 15,2003 is 100 + (4 −100 × 0.0518)a16 0.0518(1.0518) 75 184 −75 184 · 4 = $87.56 a roundoﬀerror of 0.44 We ﬁnally consider ﬁnding the yield rate of a bond with coupons being reinvested at a diﬀer-ent interest rate. Consider the situation in which a bond is purchased for P, coupons of Fr are paid at the end of each period for n periods, the bond is redeemed for C at the end of n periods, and the coupons are reinvested at rate j. Denoting the yield rate (considering reinvestment) by i′, we would have the equation of value: P(1 + i′)n = Frsn j + C since both sides represent the value of the investment at the end of n periods. Example 46.4 A $100 par value 10-year bond with 8% semiannual coupons is selling for $90 and the coupons can be reinvested at only 6% convertible semi-annually. Find the yield rate taking in consideration reinvestment rates. Solution. We have 90(1 + i′)20 = 4s20 0.03 + 100 →(1 + i′)20 = 4(26.8704)+100 90 = 2.30535 →i′ = 0.04265. Thus, the yield rate is 2(0.04265) = 0.0853 = 8.53% convertible semi-annually. Thus, the yield rate of 9.48% found in Example 44.1 drops to 8.53% if the coupons can be reinvested at 6% 418 BONDS AND RELATED TOPICS Practice Problems Problem 46.1 A $100 par value 12-year bond with 10% semiannual coupons is selling for $110. Find the yield rate convertible semiannually. Problem 46.2 Recompute the overall yield rate to an investor purchasing the bond in the previous problem, if the coupons can be reinvested at only 7% convertible semiannually. Problem 46.3 A $100 bond with annual coupons is redeembale at par at the end of 15 years. At a purchase price of $92 the yield rate is exactly 1% more than the coupon rate. Find the yield rate on the bond. Problem 46.4 An investor buys 20-year bonds, each having semiannual coupons and each maturing at par. For each bond the purchase price produces the same yield rate. One bond has a par value of $500 and a coupon of $45. The other bond has a par value of $1000 and a coupon of $30. The dollar amount of premium on the ﬁrst bond is twice as great as the dollar amount of discount on the second bond. Find the yield rate convertible semiannually. Problem 46.5 A 20 year bond with a 20,000 par value pays semi-annual coupons of 500 and is redeemable at par. Audrey purchases the bond for 21,000. Calculate Audrey’s semi-annual yield to maturity on the bond. Use Salesman’s formula. Problem 46.6 A 10 year bond with a par value of 1000 is redeemable at par and pays annual coupons of 65. If Jamie purchased the bond for 950 and she reinvests the coupons at 4% annual eﬀective rate, calculate the actual yield that Jamie will receive taking into account the reinvestment rate. Problem 46.7 ‡ A 1000 par value 10-year bond with coupons at 5%, convertible semiannually, is selling for 1081.78. Calculate the yield rate convertible semiannually. Problem 46.8 ‡ Dan purchases a 1000 par value 10-year bond with 9% semiannual coupons for 925. He is able to reinvest his coupon payments at a nominal rate of 7% convertible semiannually. Calculate his nominal annual yield rate convertible semiannually over the ten-year period. 46 APPROXIMATION METHODS OF BONDS’ YIELD RATES 419 Problem 46.9 Suppose a 10000 par bond has 6% semi-annual coupons and matures in 10 years. What is the yield rate if the price is 9000? Compute both exactly and using the Salesman’s method. Problem 46.10 A $1000 par value 10-year bond with 7% semiannual coupons was issued on October 1, 2006. The market price on April 28, 2006 is $1020. Find the yield rate if the bond in bought on that date. Problem 46.11 John buys a $1000 bond that pays coupons at a nominal interest rate of 7% compounded semian-nually. The bond is redeemable at par in 15 years. The price he pays will give him a yield of 8% compounded semiannually if held to maturity. After 5 years, he sells the bond to Peter, who desires a yield of 6% on his investment. (a) What price did John pay? (b) What price did Peter pay? (c) What yield did John realize? 420 BONDS AND RELATED TOPICS 47 Callable Bonds and Serial Bonds Callable or redeemable bonds are bonds that can be redeemed or paid oﬀby the issuer prior to the bonds’ maturity date. The earliest such call date will generally be several years after the issue date. Call dates are usually speciﬁed by the bond issuer. When an issuer calls its bonds, it pays investors the call price (usually the face value of the bonds) together with accrued interest to date and, at that point, stops making interest payments. An issuer may choose to redeem a callable bond when current interest rates drop below the interest rate on the bond. That way the issuer can save money by paying oﬀthe bond and issuing another bond at a lower interest rate. This is similar to reﬁnancing the mortgage on your house so you can make lower monthly payments. Callable bonds are more risky for investors than non-callable bonds because an investor whose bond has been called is often faced with reinvesting the money at a lower, less attractive rate. As a result, callable bonds often have a higher annual return to compensate for the risk that the bonds might be called early. Callable bonds present a problem when calculating prices and yields, because the call date of the bond is unknown. Since the issuer has an option whether or not to call the bond, the investor should assume that the issuer will exercise that option to the disadvantage of the investor, and should calculate the yield rate and/or price accordingly. If the redemption value is the same at any call date, including the maturity date, then the following general principle will hold: (i) the call date will most likely be at the earliest date possible if the bond was sold at a premium, which occurs when the yield rate is smaller than the coupon rate (issuer would like to stop repaying the premium via the coupon payments as soon as possible). (ii) the call date will most likely be at the latest date possible if the bond was sold at a discount, which iccurs when the yield rate is larger than the coupon rate (issuer is in no rush to pay out the redemption value). Example 47.1 A 15 year 1000 par bond has 7% semiannual coupons and is callable at par after 10 years. What is the price of the bond to yield 5% for the investor? Solution. Since the yield rate is smaller than the coupan rate, the bond is selling at premium so the earliest redemption date is the most favorable for the issuer and the least favorable for the investor. The price is the smallest of 1000 + (35 −25)an 0.025 for 20 ≤n ≤30, which clearly occurs when n = 20. Thus, the price is 1000 + (35 −25)a20 0.025 = 1, 155.89 47 CALLABLE BONDS AND SERIAL BONDS 421 If the redemption values on all the redemption dates are not equal, this principle is a little harder to apply. In this case one needs to examine all call dates. The most unfavorable date will not necessarily be either the earliest or the latest possible date. The most unfavorable call date ( to the issuer) is the one that produces the smallest purchase price at the investors yield rate as illustrated in the following example. Example 47.2 A 100 par value 4% bond with semi-annual coupons is callable at the following times: 109.00, 5 to 9 years after issue 104.50, 10 to 14 years after issue 100.00, 15 years after issue (maturity date). What price should an investor pay for the callable bond if they wish to realize a yield rate of (1) 5% payable semi-annually and (2) 3% payable semi-annually? Solution. (1) Since the market rate is better than the coupon rate, the bond would have to be sold at a discount and as a result, the issuer will wait until the last possible date to redeem the bond: P = 2.00a30 2.5% + 100.00v30 0.025 = $89.53. (2) Since the coupon rate is better than the market rate, the bond would sell at a premium and as a result, the issuer will redeem at the earliest possible date for each of the three diﬀerent redemption values: P =2.00a10 1.5% + 109.00v10 0.015 = $112.37 P =2.00a20 1.5% + 104.50v20 0.015 = $111.93 P =2.00a30 1.5% + 100.00v30 0.015 = $112.01 In this case, the investor would only be willing to pay $111.93 Serial Bonds Serial bonds are bonds issued at the same time but with diﬀerent maturity dates. The price of any individual bond is found by the methods already discussed earlier in this chapter provided that the redemption date is known. The price of the entire serial bonds is just the sum of the values of the individual bonds. We next consider ﬁnding a formula for pricing an issue of serial bonds with m diﬀerent redemption dates. Let Pt be the purchase price of the tth bond; Ct be the redemption value and Kt the present value of Ct. Then by Makeham’s formula the price of the tth bond is Pt = Kt + g i (Ct −Kt). 422 BONDS AND RELATED TOPICS Summing from t = 1 to t = m to obtain the price of the entire issue of a serial bond m X t=1 Pt = m X t=1 Kt + g i m X t=1 Ct − m X t=1 Kt ! or P ′ = K′ + g i (C′ −K′) where P ′ = Pm t=1 Pt, K′ = Pm t=1 Kt, and C′ = Pm t=1 Ct. Example 47.3 A $10,000 serial bond is to be redeemed in $1000 installments of principal per half-year over the next ﬁve years. Interest at the annual rate of 12% is paid semi-annually on the balance outstanding. How much should an investor pay for this bond in order to produce a yield rate of 8% convertible semi-annually? Solution. We have P ′ = 10 X t=1 1000vt 0.04 + 0.06 0.04 " 10, 000 − 10 X t=1 1000vt 0.04 # =1000a10 0.04 + 1.5[10, 000 −1000a10 0.04] =15, 000 −500a10 0.04 = 10, 944.58 Example 47.4 A 5000 serial bond with 10% annual coupons will be redeemed in ﬁve equal installments of 1100 beginning at the end of the 11th year and continuing through the 15th year. The bond was bought at a price P to yield 9% annual eﬀective. Determine P. Solution. We have: F = 1000, C = 1100, i = 0.09 and r = 0.10. Thus, g = 1000(0.10) 1100 = 1 11. Hence, C′ = 5500 K′ = 15 X t=11 1100vt 0.09 =1100(a15 0.09 −a10 0.09) = 1807.33 The price is P ′ = 1807.33 + 1 0.99(5500 −1807.33) = 5537.30 47 CALLABLE BONDS AND SERIAL BONDS 423 Practice Problems Problem 47.1 A $1000 par value bond has 8% semiannual coupons and is callable at the end of the 10th through the 15th years at par. (a) Find the price to yield 6% convertible semiannually. (b) Find the price to yield 10% convertible semiannually. Problem 47.2 A $1000 par value 8% bond with quarterly coupons is callable ﬁve years after issue. The bond matures for $1000 at the end of ten years and is sold to yield a nominal rate of 6% convertible quarterly under the assumption that the bond will not be called. Find the redemption value at the end of ﬁve years that will provide the purchaser the same yield rate. Problem 47.3 A $1000 par value 4% bond with semiannual coupons matures at the end of 10 years. The bond is callable at $1050 at the ends of years 4 through 6, at $1025 at the ends of years 7 through 9, and at $1000 at the end of year 10. Find the maximum price that an investor can pay and still be certain of a yield rate of 5% convertible semiannually. Problem 47.4 A $1000 par value 6% bond with semiannual coupons is callable at par ﬁve years after issue. It is sold to yield 7% semi-annual under the assumption that the bond will be called. The bond is not called and it matures at the end of 10 years. The bond issuer redeems the bond for 1000 + X without altering the buyer’s yield rate of 7% convertible semiannually. Find X. Problem 47.5 A 15 year bond can be called at the end of any year from 10 through 14. The bond has a par value of 1000 and an annual coupon rate of 5%. The bond is redeemable at par and can also be called at par. (a) Calculate the price an investor would pay to yield 6%. (b) Calculate the price an investor would pay to yield 4%. Problem 47.6 A 15 year bond can be called at the end of any year from 12 through 14. The bond has a par value of 1000 and an annual coupon rate of 5%. The bond is redeemable at par and can be called at 1100. (a) Calculate the price an investor would pay to yield 6%. (b) Calculate the price an investor would pay to yield 4%. 424 BONDS AND RELATED TOPICS Problem 47.7 A callable 20 year bond which matures for 1000 pays annual coupons of 80. The bond is callable in years 10 through 12 at 1100 and in years 13 through 15 at 1050. The bond is bought to yield 4% annually. Calculate the purchase price of the bond. Problem 47.8 ‡ Matt purchased a 20-year par value bond with semiannual coupons at a nominal annual rate of 8% convertible semiannually at a price of 1722.25. The bond can be called at par value X on any coupon date starting at the end of year 15 after the coupon is paid. The price guarantees that Matt will receive a nominal annual rate of interest convertible semiannually of at least 6%. Calculate X. Problem 47.9 ‡ Toby purchased a 20-year par value bond with semiannual coupons at a nominal annual rate of 8% convertible semiannually at a price of 1722.25. The bond can be called at par value 1100 on any coupon date starting at the end of year 15. What is the minimum yield that Toby could receive, expressed as a nominal annual rate of interest convertible semiannually? Problem 47.10 ‡ Sue purchased a 10-year par value bond with semiannual coupons at a nominal annual rate of 4% convertible semiannually at a price of 1021.50. The bond can be called at par value X on any coupon date starting at the end of year 5. The price guarantees that Sue will receive a nominal annual rate of interest convertible semiannually of at least 6%. Calculate X. Problem 47.11 ‡ Mary purchased a 10-year par value bond with semiannual coupons at a nominal annual rate of 4% convertible semiannually at a price of 1021.50. The bond can be called at par value 1100 on any coupon date starting at the end of year 5. What is the minimum yield that Mary could receive, expressed as a nominal annual rate of interest convertible semiannually? Problem 47.12 ‡ A 1000 par value bond with coupons at 9% payable semiannually was called for 1100 prior to maturity. The bond was bought for 918 immediately after a coupon payment and was held to call. The nominal yield rate convertible semiannually was 10%. Calculate the number of years the bond was held. 47 CALLABLE BONDS AND SERIAL BONDS 425 Problem 47.13 ‡ A 1000 par value bond pays annual coupons of 80. The bond is redeemable at par in 30 years, but is callable any time from the end of the 10th year at 1050. Based on her desired yield rate, an investor calculates the following potential purchase prices, P : Assuming the bond is called at the end of the 10th year, P = 957 Assuming the bond is held until maturity, P = 897 The investor buys the bond at the highest price that guarantees she will receive at least her desired yield rate regardless of when the bond is called. The investor holds the bond for 20 years, after which time the bond is called. Calculate the annual yield rate the investor earns. Problem 47.14 Which of the following are true: (I) If a callable bond is called or redeemed at any date other than that used to calculate the purchase price, the buyer’s yield will be higher than that used to calculate the yield rate. (II) If the redemption value of a callable bond is equal at all possible redemption dates and the bond sells at a premium, the price of the bond is calculated by assuming the redemption date is the last possible date. (III) To calculate the price of a callable bond, a buyer should calculate the price at each and every possible redemption date and take the lowest price. Problem 47.15 A 5000 par value 18-year bond has 7% semiannual coupons, and is callable at the end of the 12th through the 17th years at par. (a) Find the price to yield 6% convertible semiannually. (b) Find the price to yield 8% convertible semiannually. (c) Find the price to yield 8% convertible semiannually, if the bond pays a pre- mium of 250 if it is called. (d) Find the price to yield 6% convertible semiannually, if the bond pays a premium of 250 if it is called at the end of years 12 through 14, and a premium of 150 if it is called at the end of years 15 or 16. (It may still be called at the end of year 17 without premium, and otherwise will mature at the end of year 18.) Problem 47.16 Find the price of a $1000 issue of 5.25% bonds with annual coupons which will be redeemed in 10 annual installments at the end of the 11th through the 20th years from the issue date at 105. The bonds are bought to yield 7% eﬀective. 426 BONDS AND RELATED TOPICS Problem 47.17 ABC is issuing a $1000 bond with a maturity of 25 year and annual 10% coupon. The bond is callable with the ﬁrst year call price equal to the oﬀering price plus the coupon; thereafter the call price decreases by equal amounts to equal par at year 20; thereafter the call price is equal to par. If the bond were sold at $950 what would be the call prices for each year? Stocks and Money Market Instruments One of the reasons investors are interested in trading with stocks because of its high rate of return (about 10%). Stocks are examples of ﬁnancial instruments or securities, terms which we deﬁne next. A ﬁnancial instrument is a monetary claim that one party has on another. It is a ﬁnancial asset for the person who buys or holds it, and it is a ﬁnancial liability for the company or institution that issues it. For example, a share of Microsoft stock that you own. The stock gives you a share of Microsoft’s assets and the right to receive a share of dividends (proﬁts), if Microsoft is paying any. To Microsoft owners, a stock is an obligation to include you in their divident payments. A security is a tradeable ﬁnancial instrument, like a bond or a share of stock. A ﬁnancial market is a market in which ﬁnancial assets are traded. In addition to enabling exchange of previously issued ﬁnancial assets, ﬁnancial markets facilitate borrowing and lending by facilitating the sale by newly issued ﬁnancial assets. Examples of ﬁnancial markets include the New York Stock Exchange (resale of previously issued stock shares), the U.S. government bond market (resale of previously issued bonds), and the U.S. Treasury bills auction (sales of newly issued T−bills). A ﬁnancial institution is an institution whose primary source of proﬁts is through ﬁnancial asset transactions. Examples of such ﬁnancial institutions include discount brokers (e.g., Charles Schwab and Associates), banks, insurance companies, and complex multi−function ﬁnancial institutions such as Merrill Lynch. Stocks are ﬁnancial instruments that make the holder a co−owner of the company that issued them. They entitle the holder to a claim on the assets (and implicitly the future proﬁts) of the company. Stocks do not involve the repayment of a debt or the payment of interest. (Some stocks pay dividends, which are shares of the company’s proﬁts, but people typically hold stocks not for the dividends but for the hope of reselling them later at a higher price.) In this chapter, we introduce two types of stocks: preferred and common stocks. We also discuss trading stocks. More speciﬁcally, we will discuss buying stocks by borrowing money (buying on margin) and selling stocks that we don’t own (short sales of socks). We conclude this chapter by introducing some of the ﬁnancial instrument used in today’s ﬁnancial markets. 427 428 STOCKS AND MONEY MARKET INSTRUMENTS 48 Preferred and Common Stocks Three common types of securities which have evolved in the ﬁnancial markets: bonds, preferred stocks, and common stocks. Bonds have been the topic of the previous chapter. In this section, we discuss preferred stocks and common stocks. Preferred Stock This is a type of security which provides a ﬁxed rate of return (similar to bonds). However, unlike bonds, it is an ownership security (bonds are debt security). Generally, preferred stock has no maturity date, although on occasion preferred stock with a redemption provision is issued. The periodic payment is called a dividend, because it’s being paid to an owner. Preferred stock is typically a higher ranking stock than common stock. Preferred stock may or may not carry voting rights, and may have superior voting rights to common stock. Preferred stock may carry a dividend that is paid out prior to any dividends to common stock holders. Preferred stock may have a convertibility feature into common stock. Preferred stock holders will be paid out in assets before common stockholders and after debt holders in bankruptcy. In general, there are four diﬀerent types of preferred stock: • Cumulative preferred stock. Preferred stock on which dividends accrue in the event that the issuer does not make timely dividend payments. Most preferred stock is cumulative preferred. • Non-cumulative preferred stock. Preferred stock for which unpaid dividends do not accrue. • Participating preferred stock. This is a kind of preferred stock that pays a regular dividend and an additional dividend when the dividend on the common stock exceeds a speciﬁc amount. • Convertible preferred stock. Convertible preferred stock has a privilege similar to convertible bonds. Owners have the option to convert their preferred stock to common stock under certain conditions. Since a preferred stock has no redemption date, it pays dividend forever. Their cash ﬂows are, therefore, those of a perpetuity. Thus, the price of a preferred stock is the present value of future dividends of a perpetuity which is given by the formula P = Fr i . Example 48.1 A preferred stock pays a coupon of 50 every six months. Calculate the price of the preferred stock using a yield rate of 5% convertible semi-annually. Solution. The price of the stock is P = Fr i = 50 0.025 = 2000 48 PREFERRED AND COMMON STOCKS 429 Common stock Common stock is another type of ownership security. It does not earn a ﬁxed dividend rate as preferred stock does. Dividends on this type of stock are paid only after interest payments on all bonds and other debt and dividends on preferred stock are paid. The dividend rate is ﬂexible, and is set by the board of directors at its discretion. Value of a share of common stock is being based on the dividend discount model. This model values shares at the discounted value of future dividend payments. That is, the value of a share is the present value of all future dividends. Consider a situation where a dividend D is paid at the end of the current period. Assume that the next dividend payments follow a geometric progression with common ratio 1 + k indeﬁnitely and the stock is purchased at a yied rate of i with −1 < k < i. Then the theoretical price of the stock is P = D lim n→∞ 1 − 1+k 1+i n i −k = D i −k. We point out that the most common frequency of dividend payments on both preferred and commom stock in the US is quarterly. Example 48.2 A common stock is currently earning $1 per share and the dividend is assumed to increase by 5% each year forever. Find the theoretical price to earn an investor an annual eﬀective yield rate of 10%. Solution. The theoretical price is P = D 1 i −k = 1 0.10 −0.05 = 20 It is unrealistic to project constant percentage increases in dividends indeﬁnitely into the future. As corporations increase in size and become more mature, the rate of growth will generally slow down. Example 48.3 A common stock pays annual dividends at the end of each year. The earnings per share for the most recent year were 8 and are assumed to grow at a rate of 10% per year, forever. The dividend will be 0% of earnings for each of the next 10 years, and 50% of earnings thereafter. What is the theoretical price of the stock to yield 12%? 430 STOCKS AND MONEY MARKET INSTRUMENTS Solution. The theoretical price of the stock which is the present value of all future dividends is ∞ X k=11 4(1.1)k(1.12)−k = ∞ X k=11 4 1.1 1.12 k =4 1.1 1.12 11 " 1 1 − 1.1 1.12 # = 183.73 Example 48.4 A common stock is currently earning $4 per share and will pay $2 per share in dividends at the end of the current year. Assuming that the earnings of the corporation increase at the rate of 5% for the ﬁrst ﬁve years, 2.5% for the second ﬁve years and 0% thereafter. Assume that the corporation plans to continue to pay 50% of its earnings as dividends, ﬁnd the theoretical price to earn an investor an annual eﬀective yield rate of 10%. Solution. The theoretical price is 2 " 1 − 1.05 1.10 5 0.10 −0.05 # + 2(1.05)5 (1.10)5 " 1 − 1.025 1.10 5 0.10 −0.025 # + 2(1.05)5(1.025)5 (1.10)10 · 1 0.10 = $25.72 48 PREFERRED AND COMMON STOCKS 431 Practice Problems Problem 48.1 A preferred stock pays a $10 dividend at the end of the ﬁrst year, with each successive annual dividend being 5% greater than the preceding one. What level annual dividend would be equivalent if i = 12%? Problem 48.2 A preferred stock will pay a dividend of 100 today. The stock also provides that future dividends will increase at the rate of inﬂation. The stock is purchased to yield 8% and assuming an inﬂation rate of 3%. Calculate the price of the stock today (before payment of the dividend). Problem 48.3 A common stock pays annual dividends at the end of each year. The earnings per share in the year just ended were $6. Earnings are assumed to grow 8% per year in the future. The percentage of earnings paid out as a dividend will be 0% for the next ﬁve years and 50% thereafter. Find the theoretical price of the stock to yield an investor 15% eﬀective. Problem 48.4 A common stock is purchased at a price equal to 10 times current earnings. During the next 6 years the stock pays no dividends, but earnings increase 60%. At the end of 6 years the stock sold at a price equal 15 times earnings. Find the eﬀective annual yield rate earned on this investment. Problem 48.5 A company’s common stock pays a dividend of $2 each year. The next dividend will be paid in one year. If the dividend is expected to increase at 5% per year, calculate the value of the stock at a 10% yield rate. Problem 48.6 A company’s current earnings are $5 per share. They expect their earnings to increase at 10% per year. Once their earnings are $10 per share, they will payout 75% as a dividend. Calculate the value of the stock to yield 15%. Problem 48.7 Thompson Corporation currently pays a dividend of 3 on earnings of 5 per year. Thompson’s earnings are expected to increase at a rate of 6% per year with 60% of earnings paid as dividends. Calculate the theoretical price of the common stock to yield 10%. 432 STOCKS AND MONEY MARKET INSTRUMENTS Problem 48.8 A common stock is currently earning $4 per share and will pay $2 per share in dividends at the end of the current year. Assuming that the earnings of the corporation increase 5% per year indeﬁnitely and that the corporation plans to continue to pay 50% of its earnings as dividends, ﬁnd the theoretical price to earn an investor an annual eﬀective yield rate of (1) 10%, (2) 8%, and (3) 6% Problem 48.9 A common stock is purchased on January 1, 1992. It is expected to pay a dividend of 15 per share at the end of each year through December 31, 2001. Starting in 2002 dividends are expected to increase K% per year indeﬁnitely, K < 8%. The theoretical price to yield an annual eﬀective rate of 8% is 200.90. Calculate K. Problem 48.10 On January 1 of each year Company ABC declares a dividend to be paid quarterly on its common shares. Currently, 2 per share is paid at the end of each calendar quarter. Future dividends are expected to increase at the rate of 5% per year. On January 1 of this year, an investor purchased some shares at X per share, to yield 12% convertible quarterly. Calculate X. Problem 48.11 ‡ The dividends of a common stock are expected to be 1 at the end of each of the next 5 years and 2 for each of the following 5 years. The dividends are expected to grow at a ﬁxed rate of 2% per year thereafter. Assume an annual eﬀective interest rate of 6%. Calculate the price of this stock using the dividend discount model. Problem 48.12 ‡ A commom stock sells for 75 per share assuming an annual eﬀective interest rate of i. Annual dividends will be paid at the end of each year forever. The ﬁrst dividend is 6, with each subsequent dividend 3% greater than the previous year’s dividend. Calculate i. 49 BUYING STOCKS 433 49 Buying Stocks In this section, we discuss the costs associated with buying stocks. One way to buy stocks is by buying on margin (i.e. buying stocks with borrowed money). When an investor is buying an asset, he is said to have a long position in the asset. If the investor is selling an asset he is said to have a short position in that asset. In the stock and bond markets, if you short an asset, it means that you borrow it, sell the asset, and then later buy it back. Short sales will be discussed in Section 50. Now, depending upon whether an investor is buying or selling a security, he will get diﬀerent price quotes. A bid price is the price at which a broker is ready to purchase the security. Thus, this is the selling price for an investor. The ask price is the price at which a broker is ready to sell the security. The ask is the buying price paid by an investor. The diﬀerence between the two prices is called the bid-ask spread (bid price is lower than the asked price, and the spread is the broker’s proﬁt). Example 49.1 MBF stock has a bid price of $10.35 and an ask price of $11.00. Assume there is a $5 brokerage commission. (a) What is the bid-ask spread? (b) What amount will you pay to buy 50 shares? (c) What amount will you receive for selling 50 shares? (d) Suppose you buy 50 shares, then immediately sell them with the bid and ask price being the same as above. What is your round-trip transaction cost? Solution. (a) The bid-ask spread is 11.00 −10.35 = $0.65. (b) You pay 50 × 11.00 + 5 = $555.00 (c) You receive 50 × 10.35 −5 = $512.50. (d) For each stock, you buy at $11.00 and sell it an instant later for $10.35. The total loss due to the ask-bid spread: 50(11.00 −10.35) = $32.50. In addition, you pay $5 twice. Your total transaction cost is then 32.50 + 2(5) = $42.50 Buying on Margin When buying on margin, the investor borrows part of the purchase price of the stock from a broker or a brokerage ﬁrm and contributes the remaining portion (the initial margin requirement or simply the margin) which is kept in an account with the brokerage ﬁrm. If you buy Q shares at the price of P (per share) with initial margin percentage of IM%, then your initial margin requirement IM is given by the formula IM = Q · P · (IM%). 434 STOCKS AND MONEY MARKET INSTRUMENTS Example 49.2 Suppose you want to buy 100 shares on margin at the price of $50 per share. Assuming that the initial margin percentage is 50%, what was your initial margin requirement? Solution. You initial margin requirement IM is IM = 100(50)(0.5) = $2500 The brokerage ﬁrm is making money on your loan in terms of interest. Moreover, your stock will be hold as the collateral against the loan. If you default, the ﬁrm will take the stock. There are risks associated to buying stock on margin. The price of your stock could always go down. By law, the brokerage ﬁrm will not be allowed to let the value of the collateral (the price of your stock) go down below a certain percentage of the loan value. This percentage is called the maintenance margin. If the stock drops below that set amount, the brokerage ﬁrm will issue a margin call on your stock. The margin call means that you will have to pay the brokerage ﬁrm the amount of money necessary to bring the brokerage ﬁrms risk down to the allowed level. If you don’t have the money, your stock will be sold to pay oﬀthe loan. If there is any money left, it will be sent to you. In most cases, there is little of your original investment remaining after the stock is sold. We deﬁne the actual margin AM by the formula AM = Market Value of Assets−Loan Market Value of Assets If AM < MM, then the investor is undermargined (he or she receives a margin call from his or her broker) and must take corrective action (pay oﬀpart of loan or deposit more collateral). If the investor fails to respond, then the broker will close the investor’s account. If AM > IM, then the investor is overmargined and can withdraw cash or buy more shares. Example 49.3 John wants to purchase 100 stocks of IBM. The current price of a stock is $100. He has $6000 on hand, so he decides to borrow $4000. The maintenance marginal is set to be 50%. (a) What is John’s liability and equity? (b) Suppose the stock falls to $70. What is the actual margin? Would a margin call be in order? (c) How far can the stock price fall before a margin call? Solution. (a) John’s liability is $4000 and his equity is $6000. So John is contributing 6000 10000 = 60% (the margin) of the purchasing price. (b) If the stock falls to $70, then John’s liability is still $4000 but his equity is reduced to $3000 so that the new margin is 3000 7000 = 43% < 50%. Since this is less than the maintenance margin, the 49 BUYING STOCKS 435 broker will issue a margin call. (c) Let P be the original price. Then the margin in this case is 100P −4000 100P . Setting this to 0.5 and solving for P we ﬁnd P = $80. Thus, anytime the stock drops below $80, John will receive a margin call Example 49.4 John wants to purchase 100 stocks of IBM. The current price of a stock is $100. He has $6000 on hand, so he decides to borrow $4000. The broker charges an annual eﬀective rate of interest of 10%. Suppose the stock goes up by 30% in one year period. (a) What is the end-year-value of the shares? (b) Find the total of loan repayment and interest. (c) What is the rate of return? (d) What is the rate of return if not buying on margin? Solution. (a) The end-year-value of the shares is 100[100 + (30%)(100)] = $13, 000. (b) The total of loan repayment and interest is 4000 + 4000(10%) = $4, 400. (c) The rate of return is 13000 −10400 6000 = 43.3% (d) If not buying on margin, the rate of return is 13000 −10000 10000 = 30% The reason why an investor buys on margin is his wish to invest more than what the investor’s money would allow. Buying on margin could mean a huge return when stock prices go up. But there is the risk that you could lose your original investment when stock price goes down. As with any stock purchase there are risks, but when you are using borrowed money, the risk is increased. 436 STOCKS AND MONEY MARKET INSTRUMENTS Practice Problems Problem 49.1 MBF stock has a bid price of $10.35 and an ask price of $11.00. Assume there is a 0.1% brokerage commission on the bid or ask price. (a) What amount will you pay to buy 50 shares? (b) What amount will you receive for selling 50 shares? (c) Suppose you buy 50 shares, then immediately sell them with the bid and ask price being the same as above. What is your round-trip transaction cost? Problem 49.2 MBF stock has a bid price of $10.35 and an ask price of $11.00. At what price can the market-maker purchase a stock? At what price can a market-maker sell a stock? What is the market-maker proﬁt from buying and immediately selling 50 shares? Problem 49.3 Suppose the spot ask exchange rate is $2.10 = £1.00 and the spot bid exchange rate is $2.07 = £1.00. If you were to buy $5,000,000 worth of British pounds and then sell them ﬁve minutes later without the bid or ask changing, how much of your $5,000,000 would be ”eaten” by the bid-ask spread? Problem 49.4 Consider the transaction of Example 49.4 but now assume that there is no change in the stock price in one year period. (a) What is the end-year-value of the shares? (b) Find the total of loan repayment and interest. (c) What is the rate of return? (d) What is the rate of return if not buying on margin? Problem 49.5 Consider the transaction of Example 49.4 but now assume that there is a decrease of 30% in the stock price in one year period. (a) What is the end-year-value of the shares? (b) Find the total of loan repayment and interest. (c) What is the rate of return? (d) What is the rate of return if not buying on margin? Problem 49.6 On July 24, 2002, you bought 100 shares of Microsoft on margin at the price of $45 per share. Assuming that the initial margin percentage is 50%, what was your initial margin requirement? 49 BUYING STOCKS 437 Problem 49.7 Using Problem 49.6, suppose that On July 25, 2002, Microsoft closed at $42.83. What was your actual margin on that day? Problem 49.8 Using Problem 49.6, suppose that the maintenance margin for Microsoft is 40%. Did you receive a margin call on July 25, 2002? Problem 49.9 On December 30, 1998, you decided to bet on the January eﬀect, a well−known empirical regularity in the stock market. On that day, you bought 400 shares of Microsoft on margin at the price of $139 per share. The initial margin requirement is 55% and the maintenance margin is 30%. The annual cost of the margin loan is 6.5%. (a) Determine your initial margin requirement. (b) To what price must Microsoft fall for you to receive a margin call? (c) On January 8, 1999, Microsoft climbed to $151.25. What was the actual margin in your account on that day? (d) On January 9, 1999, you sold your Microsoft shares at the price of $150.50 per share. What was the return on your investment? (e) Recalculate your answer to part (d) assuming that you made the Microsoft stock purchase for cash instead of on margin. 438 STOCKS AND MONEY MARKET INSTRUMENTS 50 Short Sales In this section we discuss trading security over short period of time when the security is likely to decline in value. A short sale is generally a sale of a security by an investor who does not actually own the security. To deliver the security to the purchaser, the short seller will borrow the security. The short seller later ”pays oﬀ” the loan by returning the security to the lender, typically by purchasing the sold security back on the open market (hopefully at a lower price). This process of buying back the security is called covering the short because the seller is no longer at risk to movements in the security. One problem with short sales is determining the yield rate. If the transaction occurs exactly as stated, then technically the yield rate does not exist. We illustrate this in the following example. Example 50.1 An investor sells a stock short for $1000 and buys it back for $800 at the end of one year. What is the yield rate? Solution. A possible equation of value is 1000(1 + i) = 800 which yields i = −20% which is clearly unreason-able since the seller made $200 in proﬁt from the transaction. Now, if we reverse the transaction and solve the equation of value 800(1 + i) = 1000 we obtain i = 25%. But this equation states that there was $200 proﬁt on an $800 investment, but there never was an $800 investment Short sales rarely occur as illustrated above. In practice, the short seller is required to make a deposit (i.e. a collateral) of a percentage of the price at the time the short sale is made. This deposit is called the margin, and cannot be accessed by the short seller until the short sale is covered. However, the short seller will be credited with interest on the margin deposit which will increase the yield rate. Margins are usually expressed as a percentage of the value of the security. For example, govern-mental regulations in the United States require the short seller to make a deposit of 50% of the price at the time the short sale is made. In the example above, the short seller will have to deposit a margin of $500 at the time the short sale. The short seller is not allowed to earn interest on the proceeds from the short sale. Governmental regulations require that these proceeds remain in a non-interest bearing account until the short position is covered, at which time these funds will be used for the purchase necessary to cover the short. Any leftover money is proﬁt on the transaction, while a negative balance would be loss on the transaction. 50 SHORT SALES 439 If coupons or dividends are paid on the security while you are short it, you must pay them to the party you borrowed the security from. Dividend payments are often deducted from the margin account. In order to ﬁnd the yield rate on a short sale we introduce the following notation: • M = Margin deposit when sell was made (t = 0) • S0 = Proceeds from short sale • St = Cost to repurchase stock at time t • dt = Dividend at time t • i = Periodic interest rate of margin account • j = Periodic yield rate of short sale If a short seller repurchases stock after n periods, then the equation of value at time t = n is: M(1 + j)n = M(1 + i)n + S0 −Sn − n X t=1 dt(1 + i)n−t. Example 50.2 ‡ Bill and Jane each sell a diﬀerent stock short for a price of 1000. For both investors, the margin requirement is 50%, and interest on the margin is credited at an annual eﬀective rate of 6%. Bill buys back his stock one year later at a price of P. At the end of the year, the stock paid a dividend of X. Jane also buys back her stock after one year, at a price of (P −25). At the end of the year her stock paid a dividend of 2X. Both investors earned an annual eﬀective yield of 21% on their short sales. Calculate P and X. Solution. The information for Bill gives 500(1.21) = 500(1.06)+1000−P −X and for Jane gives 500(1.21) = 500(1.06) + 1000 −P + 25 −2X. Solving this system of two linear equations in two unknowns we ﬁnd P = 900 and X = 25 Example 50.3 Suppose Eddie sells 10 shares of a company short; the stock price is $87 per share, and margin requirement is 40%. The company pays semiannual dividends of 2 per share. Eddie’s margin account earns 3% nominal annual interest, convertible semiannually, and he buys the stock back after 18 months at a share price of 80. What is his annual eﬀective rate of return? Solution. We have M = 10(87)(0.4) = 348. Thus, 348(1 + j)3 = 348(1.015)3 + 870 −800 −20s3 0.015 Solving for j we ﬁnd j = 2.34%. Hence, the annual eﬀective rate of return is (1.0234)2−1 = 4.73% 440 STOCKS AND MONEY MARKET INSTRUMENTS Example 50.4 On 01/01/2005, Bryan sells a stock short for a price of 800. The margin requirement is 50%, and interest on the margin is credited at an annual eﬀective interest rate of 2%. The stock pays a dividend, D, on 12/31/2005. On 01/01/2006, Bryan buys the stock back at a price of 820 to cover the short. Bryan’s yield on the short sell is −10%. Calculate D. Solution. We are given the following information: M = 400, S0 = 800, S1 = 820, i = 0.02, j = −0.10. The equation of value is 400(0.90) = 400(1.02) + 800 −820 −D. Solving this equation for D we ﬁnd D = 28 Short selling by itself is a speculative endeavor and should not be entered into lightly. The short seller is counting on a signiﬁcant decline in the value of the security in order to come out ahead and such an act is istself a risky one. However, investment strategies have been developed that involve combinations of long and short positions in related securities, which have greatly reduced risk and also have a good prospect to earn a reasonable return on investment. A generic name for these transactions is hedging. Situations of this type in which a proﬁt is certain are called arbitrage. These two concepts will appear later in the book when discussing derivatives markets. Short selling has been blamed (erroneously) for many historic market crashes, and has frequently been harshly regulated or banned. 50 SHORT SALES 441 Practice Problems Problem 50.1 John sells a stock short for 200. A year later he purchases the stock for 160. The margin requirements are 50% with the margin account earning 4%. The stock that John sold paid a dividend of 6 during the year. Calculate John’s return on the stock investment. Problem 50.2 Don and Rob each sell a diﬀerent stock short. Don sells his stock short for a price of 960, and Rob sells his short for 900. Both investors buy back their securities for X at the end of one year. In addition, the required margin is 50% for both investors, and both receive 10% annual eﬀective interest on their margin accounts. No dividends are paid on either stock. Don’s yield rate on his short sale is 50% greater than Rob’s yield rate on his short sale. Calculate Don’s yield rate. Problem 50.3 Brian sells a stock short for 800 and buys it back one year later at a price of 805. The required margin on the sale is 62.5% and interest is credited on the margin deposit at an annual eﬀective rate of i. Dividends of 15 are paid during the year. Brian’s yield rate over the one year period is j. If the interest credited on the margin deposit had been 1.25i, with everything else remaining the same, Brian’s yield rate over the one-year period would have been 1.5j. Calculate i. Problem 50.4 An investor sells short 500 shares of stock at 10 per share and covers the short position one year later when the price of the stock has declined to 7.50. The margin requirement is 50%. Interest on the margin deposit is 8% eﬀective. Four quarterly dividends of 0.15 per share are paid. Calculate the yield rate. Problem 50.5 Andrew sells a stock short for 800. At the end of one year, Andrew purchases the stock for 760. During the year, Andrew’s stock paid a dividend of 50. Amanda sold a diﬀerent stock short 1000. At the end of the year, Amanda purchases the stock she sold short for X. Amanda’s stock paid a dividend of 25 during the year. The margin requirement for both Andrew and Amanda is 60% and they both earn 10% interest on the margin account. Andrew and Amanda both received the same return. Calculate X. Problem 50.6 Thomas sells a stock short for 1000. The margin rate is 60%. The stock does not pay dividends. After one year, Thomas purchases the stock for 880 which results in a yield of 25%. Calculate the interest rate earned on the margin account. 442 STOCKS AND MONEY MARKET INSTRUMENTS Problem 50.7 Jenna sells short a stock for 1000. One year later she buys the stock for 1000. The stock pays a dividend of 10. Jenna earns 10% on her margin requirement of 60%. Jordan sells short a stock for 500. Jordan’s stock does not pay a dividend. Jordan earns 5% on his margin requirement of 60%. When Jordan buys the stock one year later, he earns a yield that is twice that earned by Jenna. Determine the price that Jordan paid to purchase his stock. Problem 50.8 James sells a stock short for 1000. The margin requirement is 60% and James earns 10% on the margin account. One year later he buys the stock for 790. Jack sells a stock short for 600. The margin requirement is 50% and Jack earns 10% on the margin account. One year later she sells the stock for 450. Both stocks pay the same dividend. James and Jack both earn the same yield. Determine the dividend paid on each stock. Problem 50.9 Rachel sells a stock short for 2000. The margin requirement is 50%. During the year, Rachel’s stock pays a dividend of 25. Rachel purchases the stock for 2050. Danielle sells a stock short for 1500. Her margin requirement is 60%. During the year, Danielle’s stock pays a dividend of 10. Danielle purchases the stock one year later for 1535. Danielle earns a return on her transaction that is twice the return that Rachel earns on her trans-action. Both Rachel and Danielle are paid the same interest rate on the margin account. Calculate the interest rate earned on the margin account. Problem 50.10 Adam has a yield on a short sale of the stock of 30%. The interest rate on the margin account is 10%. If the margin requirement had been doubled with no other changes to the transaction, determine the yield on Adam’s transaction. Problem 50.11 On the short sale of a stock, James realizes a yield of 22%. The margin requirement was 50% and he earned 10% on the margin. If the margin had been increased to 60%, but the interest earned on the margin is still 10%, calculate James’ yield. Problem 50.12 ‡ Theo sells a stock short with a current price of 25,000 and buys it back for X at the end of 1 year. Governmental regulations require the short seller to deposit margin of 40% at the time of the short 50 SHORT SALES 443 sale. No dividends incurred. The prevailing interest rate is an 8% annual rate, and Theo earns a 25% yield on the transaction. Calculate X. Problem 50.13 On 01/01/2005, Company XY Z′s stock is trading at $80 per share, and Betty sells short $1000 worth of the stock. (Note: Shares of stock are bought and sold in fractional pieces; you do not have to buy or sell a whole number of shares.) Betty is required to put up a margin of $500, and it is agreed that she will receive interest on the margin at an annual eﬀective interest rate of k%. Company XY Z′s stock pays a dividend of 1.45 per share on 12/31/2005. On 01/01/2006, Company XY Z′s stock is trading at $76 per share, and Betty covers the short at this time for a yield of 9.75%. Calculate k. Problem 50.14 ‡ Eric and Jason each sell a diﬀerent stock short at the beginning of the year for a price of 800. The margin requirement for each investor is 50% and each will earn an annual eﬀective interest rate of 8% on his margin account. Each stock pays a dividend of 16 at the end of the year. Immediately thereafter, Eric buys back his stock at a price of (800 −2X) and Jason buys back his stock at a price of (800 + X). Eric’s annual eﬀective yield, j, on the short sale is twice Jasons annual eﬀective yield. Calculate j. Problem 50.15 ‡ Jose and Chris each sell a diﬀerent stock short for the same price. For each investor, the margin requirement is 50% and interest on the margin debt is paid at an annual eﬀective rate of 6%. Each investor buys back his stock one year later at a price of 760. Jose’s stock paid a dividend of 32 at the end of the year while Chris’s stock paid no dividends. During the 1-year period, Chris’s return on the short sale is i, which is twice the return earned by Jose. Calculate i. Problem 50.16 ‡ On January 1, 2004, Karen sold stock A short for 50 with a margin requirement of 80%. On December 31, 2004, the stock paid a dividend of 2, and an interest amount of 4 was credited to the margin account. On January 1, 2005, Karen covered the short sale at a price of X, earning a 20% return. Calculate X. Problem 50.17 Kathy and Megan each sell short the same stock. Kathy is required to maintain a margin re-quirement of 50% while Megan is required to maintain a margin requirement of 75%. Assume no dividends paid on the stock and the purchase price is the same. Which of the following are true: (I) If no interest is paid on either margin account, Kathy’s yield will be 150% of Megan’s yield. 444 STOCKS AND MONEY MARKET INSTRUMENTS (II) If the same interest rate is paid on both margin accounts, Megan’s yield will exceed 2/3 of Kathy’s yield. (III) If the same interest rate is paid on both margin accounts, Megan’s yield could never exceed Kathy’s yield. 51 MONEY MARKET INSTRUMENTS 445 51 Money Market Instruments In this section we brieﬂy introduce some of the ﬁnancial instruments used in today’s ﬁnancial mar-kets. These instruments can be analyzed with the principles developed in the earlier chapters of this book. The ﬁnancial intruments introduced fall into the two categories: Money market instru-ments and ﬁnancial derivatives. In this section we will cover money markets instruments. Basics of ﬁnancial derivatives will be introduced and discussed at a later sections. Corporations and government organizations are continually borrowing and lending money. One way for raising money is through money market instruments. Money market instruments are cash-equivalent investments that consist of short-term (usually under a year), very low-risk debt instruments. The money market is the market for buying and selling short-term loans and se-curities. The buyer of the money market instrument is the lender of money and the seller is the borrower of money. Below, we introduce few types of money market instruments that are traded in money markets. Money Market Mutual Funds A money market mutual fund is a mutual fund that attempts to keep its share price at $1. Professional money managers will take your cash and invest it in government treasury bills, savings bonds, certiﬁcates of deposit, and other safe and conservative short term commercial paper. They then turn around and pay you, the owner of the shares, your portion of the interest earned on those investments. Money market funds generally allow withdrawals on demand without penalty, making them an ideal place to keep money while considering other investment opportunities. Investment in a money market fund is neither insured nor guaranteed by neither the FDIC or by any other government agency, and there can be no guarantee that the fund will maintain a stable $1 share price. It is possible to lose money by investing in the fund. An example of money market fund investment is a commercial paper security issued by banks and large corporations. An example of a commercial paper is a promissory note. A promissory note is a written promise to repay a loan or debt under speciﬁc terms−usually at a stated time, through a speciﬁed series of payments, or upon demand. Example 51.1 On January 31, Smith borrows $500 from Brown and gives Brown a promissory note. The note states that the loan will be repaid on April 30 of the same year, with interest at 12% per annum. On March 1 Brown sells the promissory note to Jones, who pays Brown a sum of money in return for the right to collect the payment from Smith on April 30. Jones pays Brown an amount such that Jones’ yield (interest rate earned) from March 1 to the maturity date can be stated as an annual rate of interest of 15%. 446 STOCKS AND MONEY MARKET INSTRUMENTS Determine the amount that Jones paid to Brown and the yield rate Brown earned quoted on an annual basis. Assume all calculations are based on simple interest and a 365-day non-leap year. Solution. On April 30, the value of the note is AV = 500 1 + 89 365 (0.12) = $514.63. The amount Jones paid to Brown on March 1 is P 1 + 60 365 (0.15) = $514.63 Solving for P we ﬁnd P = $502.24. The yield rate earned by Brown satisﬁes the equation 502.24 = 500 1 + 29 365 i Solving for i we ﬁnd i = 5.64% Certiﬁcates of Deposit A certiﬁcate of deposit (“CD”) is like a savings account in the sense that it is an FDIC (Federal Deposit Insurance Corporation) insured investment. However, a CD has a ﬁxed term (often three months, six months, or one to ﬁve years). It is intended that the CD be held until maturity, at which time the money may be withdrawn together with the accrued interest. CDs are available at banks and savings and loan institutions. CDs are issued in a range of denom-inations with the higher denominations (e.g. $100,000) carrying the higher rates of interest than the lower denominations. CDs carry a ﬁxed yield rate and in this sense is considered more stable than a money market fund that carries a variable interest rate. However, there is the disadvantage of liquidity since penalties are imposed for early withdrawal. Example 51.2 A two-year certiﬁcate of deposit pays an annual eﬀective rate of 9%. The purchaser is oﬀered two options for prepayment penalties in the event of early withdrawal: A −a reduction in the rate of interest to 7% B −loss of three months interest. 51 MONEY MARKET INSTRUMENTS 447 In order to assist the purchaser in deciding which option to select, compute the ratio of the proceeds under Option A to those under Option B if the certiﬁcate of deposit is surrendered: (a) At the end of six months. (b) At the end of 18 months. Solution. (a) Option’s A proceeds: (1.07)0.5 = 1.03441. Option’s B proceeds: (1.09).25 = 1.02178. The ratio is 1.03441 1.02178 = 1.0124. So Option A is preferable. (b) Option’s A proceeds: (1.07)1.5 = 1.10682. Option’s B proceeds: (1.09)1.25 = 1.11374. The ratio is 1.10682 1.11374 = 0.9938. So Option B is preferable Guaranteed Investment Contracts (GIC) GICs are investment instruments issued by insurance companies, usually to large investors such as pension funds. They guarantee principal and interest over a stated period of time (like CDs), so their market value does not ﬂuctuate with movements in the interest rate. GICs are diﬀerent from CDs in that they allow additional deposits during the ﬁrst 3 months to a year. Like CDs, withdrawals are restricted. GICs pay higher rates of interest than CDs, but they are not insured by the FDIC against default. Interest earned on the initial investment may be reinvested at the guaranteed rate, or it may be reinvested at some lower ﬁxed rate. Example 51.3 An investor deposits $1,000,000 at the beginning of each year for 5 years in a 5-year GIC with an insurance company that guarantees an annual rate 6% on the fund. Determine the lump sum payable by the investor at 5 years (a) if the interest that is generated each year is reinvested at 6%; (b) if the interest that is generated each year is reinvested at 4%; Solution. (a) If the guaranteed interest is reinvested at 6% then the lump sum payable by the investor at 5 years is 448 STOCKS AND MONEY MARKET INSTRUMENTS 1,000,000¨ s5 0.06 = $5, 975, 318.54. (b) If the guaranteed interest is reinvested at 4% then the lump sum payable by the investor at 5 years is 5, 000, 000 + 60, 000(1.04)4 + 2 × 60, 000(1.04)3 + 3 × 60, 000(1.04)2 +4 × 60, 000(1.04) + 5 × 60, 000 = $5, 949, 463.19 Mutual Funds A mutual fund is a professionally-managed form of collective investments that pools money from many investors and invests it in stocks, bonds, short-term money market instruments, and/or other securities. In a mutual fund, the fund manager trades the fund’s underlying securities, realizing capital gains or losses, and collects the dividend or interest income. The investment proceeds are then passed along to the individual investors. The value of a share of the mutual fund, known as the net asset value per share (NAV), is calculated daily based on the total value of the fund divided by the number of shares currently issued and outstanding. Example 51.4 A mutual fund account has the balance $100 on January 1. On July 1 (exactly in the middle of the year) the investor deposits $250 into the account. The balance of the mutual fund at the end of the year is $400. (a) Calculate the exact dollar-weighted annual rate of interest for the year. (b) Suppose that the balance of the fund immediately before the deposit on July 1 is $120. Calculate the time-weighted annual rate of interest for the year. Solution. (a) 100(1 + i) + 250(1 + i) 1 2 = 400 →100(1 + i) + 250(1 + i) 1 2 −400 = 0. By the quadratic formula, (1 + i) 1 2 = 1.108495283 →(1 + i) = 1.228761792 →i = 0.228761792. (b) (1 + j) = (120/100)(400/370) = 1.297297297 so the time-weighted rate is j = 0.297297297 Mortgage Backed Securities (MBS) Mortgage-backed securities(MBS) represent an investment in mortgage loans. An MBS investor owns an interest in a pool of mortgages, which serves as the underlying assets and source of cash ﬂow for the security. Most MBSs are issued by the following three ﬁnancial companies: • Government National Mortgage Association (Ginnie Mae), a U.S. government agency. Ginnie Mae, backed by the full faith and credit of the U.S. government, guarantees that investors receive timely payments. 51 MONEY MARKET INSTRUMENTS 449 • The Federal National Mortgage Association (Fannie Mae) and the Federal Home Loan Mortgage Corporation (Freddie Mac), U.S. government-sponsored enterprises. Fannie Mae and Freddie Mac also provide certain guarantees and, while not backed by the full faith and credit of the U.S. gov-ernment, have special authority to borrow from the U.S. Treasury. Mortgage-backed securities typically carry some of the highest yields of any government or agency security. Example 51.5 A homeowner signs a 30 year mortgage that requires payments of $971.27 at the end of each month. The interest rate on the mortgage is 6% compounded monthly. If the purchase price of the house is $180,000 then what percentage down payment was required? Solution. The purchase price of the home is 971.27a360 0.005 = 162, 000. The amount of down payment is 180, 000 −162, 000 = 18, 000. So the down payment is 10% of the purchasing price 450 STOCKS AND MONEY MARKET INSTRUMENTS Practice Problems Problem 51.1 Consider again Example 51.1. Suppose instead that Jones pays Brown an amount such that Jones’ yield is 12%. Determine the amount that Jones paid. Problem 51.2 You prurchase $10,000 CD with an interest rate of 5% compounded annually and a term of one year. At year’s end, the CD will have grown to what value? Problem 51.3 ‡ A bank oﬀers the following certiﬁcates of deposit: Nominal annual interest rate Term in years (convertible quarterly) 1 4% 3 5% 5 5.65% The bank does not permit early withdrawal. The certiﬁcates mature at the end of the term. During the next six years the bank will continue to oﬀer these certiﬁcates of deposit with the same terms and interest rates. An investor initially deposits $10,000 in the bank and withdraws both principal and interest at the end of six years. Calculate the maximum annual eﬀective rate of interest the investor can earn over the 6-year period. Problem 51.4 An investor deposits $100,000 in a 3-year GIC with an insurance company that guarantees an annual 4% interest rate on the fund. If the GIC speciﬁes that interest is reinvested at 4% within the GIC, determine the lump sum payable to the investor after the three-year period. Problem 51.5 An investor puts $100 into a mutual fund in the ﬁrst year and $50 in the second year. At the end of the ﬁrst year the mutual pays a dividend of $10. The investor sells the holdings in the mutual fund at the end of the second year for $180. Find the dollar weighted return. Problem 51.6 You have a mutual fund. Its balance was $10,000 on 31/12/1998. You made monthly contributions of $100 at the start of each month and the ﬁnal balance was $12,000 at 31/12/1999. What was you approximate return during the year? 51 MONEY MARKET INSTRUMENTS 451 Problem 51.7 An investor purchases 1000 worth of units in a mutual fund whose units are valued at 4.00. The investment dealer takes a 9% “front-end load” from the gross payment. One year later the units have a value of 5.00 and the fund managers claim that the “fund’s unit value has experienced a 25% growth in the past year.” When units of the fund are sold by an investor, there is a redemption fee of 1.5% of the value of the units redeemed. (a) If the investor sells all his units after one year, what is the eﬀective annual rate of interest of his investment? (b) Suppose instead that after one year the units are valued at 3.75. What is the return in this case? Problem 51.8 Julie bought a house with a 100,000 mortgage for 30 years being repaid with payments at the end of each month at an interest rate of 8% compounded monthly. What is the outstanding balance at the end of 10 years immediately after the 120th payment? Problem 51.9 The interest on a 30 year mortgage is 12% compounded monthly. The mortgage is being repaid by monthly payments of 700. Calculate the outstanding balance at the end of 10 years. Problem 51.10 The interest rate on a 30 year mortgage is 12% compounded monthly. Lauren is repaying the mortgage by paying monthly payments of 700. Additionally, to pay oﬀthe loan early, Lauren has made additional payments of 1000 at the end of each year. Calculate the outstanding balance at the end of 10 years. Problem 51.11 A mortgage loan is being repaid with level annual payments of 5000 at the end of the year for 20 years. The interest rate on the mortgage is 10% per year. The borrower pays 10 payments and then is unable to make payments for two years. Calculate the outstanding balance at the end of the 12th year. Problem 51.12 A 20,000 mortgage is being repaid with 20 annual installments at the end of each year. The borrower makes 5 payments, and then is temporarily unable to make payments for the next 2 years. Find an expression for the revised payment to start at the end of the 8th year if the loan is still to be repaid at the end of the original 20 years. 452 STOCKS AND MONEY MARKET INSTRUMENTS Measures of Interest Rate Sensitivity In this chapter we study the sensitivity of the interest rate. We will cover the following topics: (1) The eﬀect of inﬂation on interest rate. We will consider the eﬀect of inﬂation in the calculations of present values and accumulated values. (2) The eﬀect of the investment period on interest rates. That is, interest rate is sensitive to time. This fact is known as the term structure of interest rate. A plot of the interest rate of an investment against time to maturity will be discussed. We call such a plot a yield curve. (3) We want to measure the change in a bond’s price as interest rate change. We call this measure the duration; the two types of duration we will look at are the modiﬁed duration and the Macaulay duration. (4) Immunization is a technique that helps investors shielding their overall ﬁnancial position from exposure to interest rate ﬂuctuations. (5) Absolute matching: The idea here is to structure assets such that the resulting asset inﬂows will exactly match the liability outﬂow. 453 454 MEASURES OF INTEREST RATE SENSITIVITY 52 The Eﬀect of Inﬂation on Interest Rates What is inﬂation? Inﬂation is deﬁned as a sustained increase in the general level of prices for goods and services. It is measured as an annual percentage increase, known as rate of inﬂation. As inﬂation rises, every dollar you own buys a smaller percentage of a good or service. That is, inﬂation represents a loss of purchasing power. The opposite to inﬂation is deﬂation. Experience has shown that inﬂation has a signiﬁcant eﬀect on the rate of interest. As a matter of fact they are positively correlated in the sense that over time both tend to move in the same direction. The following example illustrates how inﬂation aﬀects the purchasing power of money. Suppose that today a gallon of milk costs $4. Then with $100 one can buy 25 gallons of milk. Suppose the $100 is invested for 2 years at an 8% annual eﬀective rate. Then at the end of the two years, you will have 100(1.08)2 = $116.64. Without the presence of inﬂation, that is assuming the cost of milk is still $4, you can now buy 29 gallons of milk. Now suppose that there has been a constant rate of inﬂation during the two years of 5% a year so that milk will cost 4(1.05)2 = $4.41 per gallon. In this case, one can only buy 116.64 4.41 = 26.45 gallons of milk, a reduction of 3 gallons. The real rate of return, i′, is measured by solving the equation 25(1 + i′)2 = 116.64 4.41 = 25(1.08)2 (1.05)2 by taking the square root of both sides we obtain (1 + i′)(1.05) = 1.08 or 1 + ireal rate = 1 + inominal rate 1 + iinflation rate . To study the eﬀect of inﬂation on rates we introduce some notations. The rate of interest after eliminating the eﬀect of inﬂation is called the real rate of interest and will be denoted by i′, while the actual rate of interest charged in the market (i.e. the rate not adjusted for inﬂation) is called the nominal interest rate (not to be confused with the diﬀerent meaning of nominal used in Section 9) and will be denoted by i. Let r denote the rate of inﬂation. Assuming constant rate of inﬂation, the relationship between real and nominal interest rates can be described in the equation: 1 + i = (1 + i′)(1 + r), i > i′ > 0, r > 0. From this we can write 52 THE EFFECT OF INFLATION ON INTEREST RATES 455 1 + i′ = 1 + i 1 + r (52.1) or i′ = i −r 1 + r. A common approximation for the real interest rate is given by the Fisher equation: i′ = i −r which states that the real interest rate is the nominal interest rate minus the expected rate of inﬂation. If inﬂation is positive, which it generally is, then the real interest rate is lower than the nominal interest rate. That’s the reason we are imposing the condition i > i′ whenever r > 0. Formula (52.1) is quite useful in performing calculations involving rates of inﬂation. For example, assume that we wish to ﬁnd the present value of a series of payments at the end of each period for n periods in which the base payment amount at time 0 is R, but each payment is indexed to reﬂect inﬂation. If r is the periodic rate of inﬂation and i is the periodic nominal interest rate, then PV = R " 1 + r 1 + i + 1 + r 1 + i 2 + · · · + 1 + r 1 + i n# = R(1 + r) 1 − 1+r 1+i n i −r . However, if we use Formula (52.1) the above result becomes PV = R 1 1 + i′ + 1 (1 + i′)2 + · · · + 1 (1 + i′)n = Ran i′. The above two formulas provide a guidance in computing present values of future payments as follows: 1. If future payments are not aﬀected by inﬂation, then discount at the nominal rate of interest. 2. If future payments are adjusted to reﬂect the rate of inﬂation and the adjustment is reﬂected in the payment amount, then also discount at the nominal rate of interest. 3. If future payments are adjusted to reﬂect the rate of inﬂation but the adjustment is not reﬂected in the payment amount, the correct procedure is to discount at the real rate of interest. We next consider inﬂation in connection with accumulated values. Consider the common situation in which an investor invests A dollars for n periods at interest rate i. The value of this investment in “nominal dollars” at the end of n periods is AV = P(1 + i)n. 456 MEASURES OF INTEREST RATE SENSITIVITY However, if the rate of inﬂation is accounted for, then the purchasing power of this investment is AV = P 1 + i 1 + r n = P(1 + i′)n. Thus, the value of this investment in “real dollars” is lower since i > i′. Example 52.1 Money is invested for 3 years at an interest rate of 4% eﬀective. If inﬂation is 5% per year over this period, what percentage of purchasing power is lost? Solution. One dollar will accumulate to (1.04)3 = 1.1248 over the period, but the purchasing power will only be (1.04)3 (1.05)3 = 0.9716 due to inﬂation. The loss of purchasing power per dollar of original investment is 1 −0.9716 = 2.84% Example 52.2 The real interest rate is 6% and the inﬂation rate is 4%. Allan receives a payment of 1,000 at time 1, and subsequent payments increase by 50 for 5 more years. Determine the accumulated value of these payments in nominal dollars at time 6 years. Solution. The nominal interest rate is: i = (1.06)(1.04) −1 = 10.24% The accumulated value of the cash ﬂows is: AV =1, 000(1.1024)5 + 1, 050(1.1024)4 + 1, 100(1.1024)3 + 1, 150(1.1024)2 + 1, 200(1.1024) + 1, 250 = 8623.08 52 THE EFFECT OF INFLATION ON INTEREST RATES 457 Practice Problems Problem 52.1 An insurance company is making annual payments under the settlement provisions of a personal injury lawsuit. A payment of $24,000 has just been made and ten more payments are due. Future payments are indexed to the Consumer Price Index which is assumed to increase at 5% per year. Find the present value of the remaining obligation if the assumed interest rate is 8%. Use the exact value of i′. Problem 52.2 The nominal rate of interest is 10% and the rate of inﬂation is 5%. A single deposit is invested for 20 years. Let A denote the value of the investment at the end of 10 years, with the rate of inﬂation accounted for in the calculation. Let B denote the value of the investment at the end of 10 years computed at the real rate of interest. Find the ratio of A B. Problem 52.3 The nominal rate of interest is 8% and the rate of inﬂation is 5%. Calculate the real rate of interest. Problem 52.4 Mindy invests 10,000 for 10 years at a nominal rate of interest of 8%. The rate of inﬂation is 5% over the 10 year period. Calculate the value at the end of 10 years of Mindy’s investment in today’s dollars. Problem 52.5 As a result of sky-high gas prices, the cost of delivery pizza is expected to experience a constant 8% annual rate of inﬂation. In today’s dollars, Obese Oliver spends $2,000 per year on delivery pizza. His delivery pizza-eating habits will not change over the next ten years, so he will pay whatever price increases are necessary to sustain this level of consumption. If the nominal interest rate (note that “nominal” refers to both interest convertibility and its relationship to inﬂation), convertible semiannually, is 10%, ﬁnd the present value of Oliver’s delivery pizza purchases for the next ten years. Problem 52.6 Money is invested for 5 years in a savings account earnings 7% eﬀective. If the rate of inﬂation is 10%, ﬁnd the percentage of purchasing power lost during the period of investment. Problem 52.7 ‡ An insurance company has an obligation to pay the medical costs for a claimant. Average annual claims costs today are $5,000, and medical inﬂation is expected to be 7% per year. The claimant 458 MEASURES OF INTEREST RATE SENSITIVITY is expected to live an additional 20 years. Claim payments are made at yearly intervals, with the ﬁrst claim payment to be made one year from today. Find the present value of the obligation if the annual interest rate is 5%. Problem 52.8 ‡ Seth deposits X in an account today in order to fund his retirement. He would like to receive payments of 50 per year, in real terms, at the end of each year for a total of 12 years, with the ﬁrst payment occurring seven years from now. The inﬂation rate will be 0.0% for the next six years and 1.2% per annum thereafter. The annual eﬀective rate of return is 6.3%. Calculate X. 53 THE TERM STRUCTURE OF INTEREST RATES AND YIELD CURVES 459 53 The Term Structure of Interest Rates and Yield Curves In the previous section we discussed a factor aﬀecting interest rates, i.e. inﬂation. In this section we discuss another factor which is the length of investment period. The term structure of interest rates refers to the way interest rates depend on time. Usually long-term market interest rates are higher than short-term market interest rates. An example of term structure of interest rates is given in the table below. Length of investment Interest rate (or Spot rate) 1 year 3% 2 year 4% 3 year 6% 4 year 7% Table 53.1 We can extend the values in this table to a continuous graph, and the resulting graphical illustration is called the yield curve corresponding to the table. The interest rates on the yield curve (or listed in the table) are often called spot rates. Economists and investors believe that the shape of the yield curve reﬂects the market’s future expectation for interest rates and the conditions for monetary policy. For example, an upward-sloping yield curve implies market’s expectation of future increases in interest rates. In some cases, short-term rates exceed long-term rates, that is the yield curve has negative slope, in this case the curve is called inverted. When all maturities have the same yield the curve is called ﬂat. Now, recall from Section 30 the formula for the net present value of a series of future payments given by NPV (i) = n X t=0 (1 + i)−tct. This formula is based on a single rate of interest i. With the presence of a term structure of interest rates, the net present value can be found by discounting each payment by its associated spot rate. That is, if we denote the spot rate for period t as it then the net present value is given by the equation NPV (i∗) = n X t=0 (1 + it)−tct. This formula can be considered as an application of computing annuity values with varying rates of interest as discussed in Section 21. Example 53.1 You are given the following term structure of interest rates: 460 MEASURES OF INTEREST RATE SENSITIVITY Length of investment Interest Rate 1 7.00% 2 8.00% 3 8.75% 4 9.25% 5 9.50% Find the present value of payments of $1000 at the end of each year for ﬁve years using the spot rates given above. What level yield rate would produce an equivalent value? Solution. The present value of the payments is 1000[(1.07)−1 + (1.08)−2 + (1.0875)−3 + (1.0925)−4 + (1.095)−5] = $3906.63. The equivalent level yield rate is found by solving the equation a5 i = 3.90663 which gives i = 0.0883 = 8.83% using a ﬁnancial calculator A special type of spot rates is the forward interest rate. In order to illustrate this concept, consider a ﬁrm which needs to borrow a sizeable amount of money for two years. The ﬁrm is presented with two options. The ﬁrst option is to borrow money for 2 years at the 2-year spot rate of 10%. The second option is to borrow for one year at the one-year spot rate now of 8%, and to borrow for the second year at the one-year spot rate in eﬀect a year later. This 1-year spot rate for the next year is an example of a forward rate or more speciﬁcally a one year forward interest rate. This is what the one year spot rate will be after one year is passed. Now, if f is the forward rate then the ﬁrm will be indiﬀerent between the two options if (1.10)2 = (1.08)(1 + f) which gives f = 0.12 = 12%. So, if the ﬁrm can predict that the forward rate will be greater than 12%, it should use the ﬁrst option to borrow, but if they expect the forward rate to be less than 12%, it should use the second option to borrow. Example 53.2 Consider the term structure of interest rates given in Table 53.1. Let f n+t n represents the t−year forward rate n years from now. (a) Compute the 1-year, 2-year, and 3-year forward rate, one year from now. (b) Compute the 1-year and 2-year forward rates two years from now. (c) Compute the 1-year forward rate three years from now. 53 THE TERM STRUCTURE OF INTEREST RATES AND YIELD CURVES 461 Solution. (a) The 1-year forward rate one year from now satisﬁes the equation (1.03)(1 + f 2 1) = (1.04)2. Solving this equation we ﬁnd f 2 1 = 5.01%. The 2-year forward rate, one year from now satisﬁes the equation (1.03)(1 + f 3 1)2 = (1.06)3. Solving this equation we ﬁnd f 3 1 = 7.53%. The 3-year forward rate, one year from now satisﬁes the equation (1.03)(1 + f 4 1)3 = (1.07)4. Solving this equation we ﬁnd f 4 1 = 8.37%. (b) The 1-year forward rate, two years from now satisﬁes the equation (1.04)2(1 + f 3 2) = (1.06)3. Solving this equation we ﬁnd f 3 2 = 10.1%. The 2-year forward rate, two years from now satisﬁes the equation (1.04)2(1 + f 4 2)2 = (1.07)4. Solving this equation we ﬁnd f 4 2 = 10.00%. (c) The 1-year forward rate three years from now satisﬁes the equation (1.06)3(1 + f 4 3) = (1.07)4. Solving this equation we ﬁnd f 4 3 = 10.06% From any yield curve, one can calculate a complete set of implied forward rates. In general, the k− year forward rate n years from now satisﬁes the equation (1 + in)n(1 + f n+k n )k = (1 + in+k)n+k where it is the t−year spot rate. Note that f n+k n is the forward interest rate between the periods n and n + k. Example 53.3 Find the present value of the remaining payments in the annuity immediate given in Example 53.1 immediately after two payments have been made assuming that f 2+t 2 = 1% + it where t = 1, 2, 3. 462 MEASURES OF INTEREST RATE SENSITIVITY Solution. The comparison date for this calculation is at the end of two years. At that time there are three remaining annuity payments to be made. The expected forward rates are the current spot rates for 1,2, and 3 years each increased by 1%. Thus, the present value of the remaining three payments is 1000[(1.08)−1 + (1.09)−2 + (1.0975)−3] = $2524.07 Example 53.4 You are given the following term of structure of spot interest rates: Term (in Years) Spot Interest Rate 1 7.00% 2 8.00% 3 8.75% 4 9.25% 5 9.50% (a) Find the price of a $1000 two-year bond with annual 5% coupon using the spot rates given above. (b) Compute the yield to maturity as deﬁned in Section 43. Solution. (a) The coupon is $50 each. The price of the bond is P = 50(1.07)−1 + 1050(1.08)−2 = 946.93 (b) The yield to maturity i satisﬁes the equation 50(1 + i)−1 + 1050(1 + i)−2 = 946.93. Solving this equation we ﬁnd i = 7.975% Example 53.5 Yield rates for 6% annual coupon bonds are given to be 3.5% for 1-year bonds and 4% for 2-year bonds. Find the implied 2-year spot rate from these yield rates. Solution. Per 100 of par value we have 6a2 0.04 + 100v2 0.04 = 6 1.035 + 106 (1 + i)2. Solving this equation for i we ﬁnd i = 4.01% 53 THE TERM STRUCTURE OF INTEREST RATES AND YIELD CURVES 463 Practice Problems Problem 53.1 You are given the following term structure of interest rates: Length of investment Interest Rate 1 5.00 2 6.00 3 6.75 4 7.25 5 7.50 Table 53.2 Which of the following are true: (I) The yield curve has a positive slope. (II) The yield curve is inverted. (III) The interest rates in the table are called forward rates. Problem 53.2 Using Table 53.2, calculate the present value of a ﬁve year annuity due of 100 per year. Problem 53.3 Using Table 53.2, determine the three year deferred two year forward rate. Problem 53.4 Using Table 53.2, determine the accumulated value of a three year annuity immediate of 100 at the end of each year. Problem 53.5 You are given the following yield curve Length of investment Interest Rate 1 0.040 2 0.045 3 0.048 4 0.050 5 0.051 464 MEASURES OF INTEREST RATE SENSITIVITY (a) Calculate the present value of an annuity immediate of 10 at the end of each year for 5 years using the spot rates. (b) Calculate the equivalent level rate. Problem 53.6 You are given the following yield curve Length of investment Interest Rate 1 0.040 2 0.045 3 0.048 4 0.050 5 0.051 (a) Calculate the present value of an annuity due of 10 at the beginning of each year for 5 years using the spot rates. (b) Calculate the equivalent level rate. Problem 53.7 You are given the following yield curve Length of investment Interest Rate 1 0.040 2 0.045 3 0.048 4 0.050 5 0.051 (a) Calculate the 2 year deferred 3 year forward rate less the three year spot rate. (b) Calculate what the 3 year spot rate is expected to be in 2 years. Problem 53.8 You are given the following term structure of spot interest rates: Term (in Years) Spot Interest Rate 1 7% 2 8% 3 9% What is the one-year forward rate beginning two years from now? 53 THE TERM STRUCTURE OF INTEREST RATES AND YIELD CURVES 465 Problem 53.9 ‡ Using the table of the previous problem, ﬁnd (a) the current price of a $1000 3-year bond with coupon rate 6% payable annually. (b) the annual eﬀective yield rate for the bond if the bond is sold at a price equal to its value. Problem 53.10 ‡ You are given the following term structure of spot interest rates: Term (in Years) Spot Interest Rate 1 5.00% 2 5.75% 3 6.25% 4 6.50% A three-year annuity-immediate will be issued a year from now with annual payments of 5000. Using the forward rates, calculate the present value of this annuity a year from now. Problem 53.11 ‡ Consider a yield curve deﬁned by the following equation: ik = 0.09 + 0.002k −0.001k2 where ik is the annual eﬀective rate of return for zero coupon bonds with maturity of k years. Let j be the one-year eﬀective rate during year 5 that is implied by this yield curve. Calculate j. Problem 53.12 ‡ Yield rates to maturity for zero coupon bonds are currently quoted at 8.5% for one-year maturity, 9.5% for two-year maturity, and 10.5% for three-year maturity. Let i be the one-year forward rate for year two implied by current yields of these bonds. Calculate i. Problem 53.13 You are given the following term of structure of spot interest rates: Term (in Years) Spot Interest Rate 1 7.00% 2 8.00% 3 8.75% 4 9.25% 5 9.50% 466 MEASURES OF INTEREST RATE SENSITIVITY Find the following expected forward rates: (a) 1-year deferred 2-year forward rate. (b) 2-year deferred 3-year forward rate. Problem 53.14 You are given the following term of structure of spot interest rates: Term (in Years) Spot Interest Rate 1 7.00% 2 8.00% 3 8.75% 4 9.25% 5 9.50% (a) Find the price of a $1000 two-year bond with annual 10% coupon using the spot rates given above. (b) Compute the yield to maturity as deﬁned in Section 41. Problem 53.15 You are given 1−year, 2−year, and 3−year spot rates of 4%, 5%, and 6%, respectively. Calculate the annual yield rate for 3−year 5% annual coupon bonds implied by these spot rates. Problem 53.16 Consider a yield curve deﬁned by the following equation: ik = 0.09 + 0.002k −0.001k2 where ik is the annual eﬀective rate of return for zero coupon bonds with maturity of k years. (a) Calculate the 2-year spot rate implied by this yield curve. (b) Calculate the 2-year deferred 3-year forward rate implied by this yield curve. Problem 53.17 A two-year bond with annual coupons of 100 and redemption value of 1000 is priced at 1,037.41. The current two-year spot rate is 8%. Determine the current one-year spot rate that is consistent with the pricing of the bond. Problem 53.18 A 5 year bond with 6% annual coupons has a yield rate of 10% eﬀective and a 5 year bond with 8% annual coupons has a yield rate of 9% eﬀective. What is the 5 year spot rate? 53 THE TERM STRUCTURE OF INTEREST RATES AND YIELD CURVES 467 Problem 53.19 ‡ Which of the following statements about zero-coupon bonds are true? (I) Zero-coupon bonds may be created by separating the coupon payments and redemption values from bonds and selling each of them separately. (II) The yield rates on stripped Treasuries at any point in time provide an immediate reading of the risk-free yield curve. (III) The interest rates on the risk-free yield curve are called forward rates. 468 MEASURES OF INTEREST RATE SENSITIVITY 54 Macaulay and Modiﬁed Durations By now, it should be clear to you that the timing of cash ﬂows is a signiﬁcant factor in the analysis of ﬁnancial instruments. In this section we develop some “indices” to measure the timing of future payments. One index known as the duration is useful as a measure of the sensitivity of a bond’s price to interest rate movements. The two types of duration we will look at are the modiﬁed duration and the Macaulay duration. The most basic index of time measuring is the term-to-maturity. For a given price and par value, among zero coupon bonds, one wishes to purchase the bond with the earliest redemption time. In the case of bonds carrying coupons, the redemption time is not enough for helping choose the bond to purchase. A better index than the term-to-maturity is the method of equated time, which was deﬁned in Section 13: Let R1, R2, · · · , Rn be a series of payments made at times 1, 2, · · · , n. Then the weighted average of the various times of payments, where the weights are the various amounts paid is t = Pn t=1 tRt Pn t=1 Rt . This is also known as the average term-to-maturity. Example 54.1 Consider two 10-year par value 100 bonds one with 5% coupons and the other with 10% coupons. Find the equated time for each coupon. Solution. The average term to maturity of the 5% bond is t = 1 · 5 + 2 · 5 + · · · + 10 · 5 + 10 · 100 5 + 5 + · · · + 5 + 100 = 8.50. The average term to maturity of the 10% bond is t = 1 · 10 + 2 · 10 + · · · + 10 · 10 + 10 · 100 10 + 10 + · · · + 10 + 100 = 7.75. This shows that the 5% bond is a longer term bond than the 10% bond An even better index that is similar to the method of equated time is obtained by replacing each payment in the above formula by its present value thus obtaining d = Pn t=1 tνtRt Pn t=1 νtRt . 54 MACAULAY AND MODIFIED DURATIONS 469 We call d the Macaulay duration or simply duration. Example 54.2 Suppose payments of 2000, 4000, and 10000 are to be made at times 1, 2, and 4, respectively. Assume, an annual yield of 25%. (a) Find the average term to maturity using the method of equated time. (b) Find the duration of the investment. Solution. (a) The method of equated time produces the value t = 1 · 2000 + 2 · 4000 + 4 · 10000 2000 + 4000 + 10000 = 3.125. (b) The duration is given by d = 1 · (1.25)−1 · 2000 + 2 · (1.25)−2 · 4000 + 4 · (1.25)−4 · 10000 (1.25)−1(2000) + (1.25)−2(4000) + (1.25)−4(10000) = 2.798 Example 54.3 Consider two bonds purchased at the redemption value of 1000, and due in 5 years. One bond has 5% annual coupon rate payable semi-annually and the other has 10% annual coupon rate payable semi-annually. Find the duration of each bond if both bonds were purchased to yield 7% compounded semi-annually. Solution. The duration of the 5% bond is d =0.5[1 · (1.035)−1(25) + 2 · (1.035)−2(25) + · · · + 10 · (1.035)−10(25)] + 5000(1.035)−10 (1.035)−1(25) + (1.035)−2(25) + · · · + (1.035)−10(25) + 1000(1.035)−10 =12.5(Ia)10 0.035 + 5000(1.035)−10 25a10 0.035 + 1000(1.035)−10 =4.4576 The duration of the 10% bond is d =0.5[1 · (1.035)−1(50) + 2 · (1.035)−2(50) + · · · + 10 · (1.035)−10(50)] + 5000(1.035)−10 (1.035)−1(50) + (1.035)−2(50) + · · · + (1.035)−10(50) + 1000(1.035)−10 =4.1158 470 MEASURES OF INTEREST RATE SENSITIVITY Note that d is a function of i. If i = 0 we have d = t so that the method of equated time is a special case of duration which ignores interest. Also, note that in the case of only one future payment, duration is the point in time at which that payment is made. This is clear since the summations in the numerator and the denominator have only one term each and everything cancels except the time of payment. Example 54.4 Find the duration of a 10-year zero coupon bond assuming a yield 8% eﬀective. Solution. Only one payment is involved so that d = 10. Notice that this answer is independent of the yield rate Another important fact is that d is a decreasing function of i. To see this, we have d di(d) = d di Pn t=1 tνtRt Pn t=1 νtRt = −ν (Pn t=1 νtRt) (Pn t=1 t2νtRt) −(Pn t=1 tνtRt)2 (Pn t=1 νtRt)2 = −ν "Pn t=1 t2νtRt Pn t=1 νtRt − Pn t=1 tνtRt Pn t=1 νtRt 2# Now using Cauchy-Schwartz’s inequality ( n X t=1 atbt)2 ≤( n X t=1 a2 t)( n X t=1 b2 t) with at = ν t 2R 1 2 t and bt = tν t 2R 1 2 t we ﬁnd that Pn t=1 t2νtRt Pn t=1 νtRt − Pn t=1 tνtRt Pn t=1 νtRt 2 ≥0 and therefore the derivative of d with respect to i is negative. We next describe a measure of how rapidly the present value of a series of future payments changes 54 MACAULAY AND MODIFIED DURATIONS 471 as the rate of interest changes. Let P(i) denote the present value of all future payments. Then by Taylor series we can write P(i + ϵ) = P(i) + ϵ 1!P ′(i) + ϵ2 2!P ′′(i) + · · · . Using the ﬁrst two terms we can write P(i + ϵ) ≈P(i) + ϵP ′(i). Hence, the percentage change in P is given by P(i + ϵ) −P(i) P(i) ≈ϵP ′(i) P(i) . Thus, the percentage change of P(i) is a function of P ′(i) P(i) . We deﬁne the volatility of the present value by ν = −P ′(i) P(i) = −d di[ln P(i)]. The minus sign is necessary to make ν positive since P ′(i) is negative ( increasing the yield rate results in a decreasing present value). Now, since P(i) = Pn t=1(1 + i)−tRt, by taking the derivative we ﬁnd P ′(i) = −Pn t=1 t(1 + i)−t−1Rt. Hence, ν = Pn t=1 t(1 + i)−t−1Rt Pn t=1(1 + i)−tRt = 1 1 + i · Pn t=1 t(1 + i)−tRt Pn t=1(1 + i)−tRt = d 1 + i = νd. Because of this close relationship, volatility is often called modiﬁed duration. Remark 54.1 None of the results in this section are valid if the payment amounts vary depending on the interest rate. Example 54.5 (a) Find the Macaulay duration of a ten-year, $1,000 face value, 8% annual coupon bond. Assume an eﬀective annual interest rate of 7%. (b) Find the modiﬁed duration for the bond Solution. (a) We have d =80ν + 2(80ν2) + · · · + 10(80ν10) + 10(1000ν10) 80ν + (80ν2) + · · · + (80ν10) + (1000ν10) =80(Ia)10 0.07 + 10000ν10 0.07 80a10 0.07 + 1000ν10 0.07 472 MEASURES OF INTEREST RATE SENSITIVITY But ν10 0.07 =(1.07)−10 = 0.508349 a10 0.07 =1 −(1.07)−10 0.07 = 7.02358 (Ia)10 0.07 =1.07a10 0.07 −10(1.07)−10 0.07 = 34.7391 Hence, d = 80(34.7392)+10000(0.508349) 80(7.02358)+1000(0.508349) = 7.3466 years (b) The modiﬁed duration is ν = νd = (1.07)−1(7.3466) = 6.8660 Example 54.6 A 10-year $100,000 mortgage will be repaid with level semi-annual payments of interest and prin-cipal. Assume that the interest rate on the mortgage is 8%, convertible semiannually, and that payments are made at the end of each half-year. Find the Macaulay duration of this mortgage. Solution. Per dollar of mortgage payment, the Macaulay duration is d =0.5ν0.04 + 2ν2 0.04 + 3ν3 0.04 + · · · + 20ν20 0.04 ν0.04 + ν2 0.04 + ν3 0.04 + · · · + ν20 0.04 =0.5(Ia)20 0.04 a20 0.04 = 4.6046 years Note that the answer is independent of the rate of interest being paid on the mortgage. Duration depends on the pattern of the level payments and not their amount Example 54.7 Find the duration of a preferred stock paying level dividends into perpetuity assuming eﬀective 8%. Solution. Per dollar dividend we have d = (Ia)∞ a∞ = 1.08/.082 1/.08 = 13.5 54 MACAULAY AND MODIFIED DURATIONS 473 Example 54.8 Find the modiﬁed duration for a share of common stock. Assume that the stock pays annual dividends, with the ﬁrst dividend of $2 payable 12 months from now, and that subsequent dividends will grow at an annual rate of 4%. Assume that the eﬀective annual interest rate is 9%. Solution. We have P(i) = D1 i −g P ′(i) = −D1(i −g)−2 v = −P ′(i) P(i) = 1 i −g = 1 0.09 −0.04 = 20 Example 54.9 You are the actuary for an insurance company. Your company’s liabilities include loss reserves, which are liability reserves set aside to make future claim payments on policies which the company has already sold. You believe that these liabilities, totaling $100 million on December 31, 2006, will be paid out according to the following schedule: Proportionl of Calendar year reserves paid out 2007 40% 2008 30% 2009 20% 2010 10% Find the modiﬁed duration of your company’s loss reserves. Assume that the annual interest rate is 10%, and that all losses paid during a given calendar year are paid at the mid-point of that calendar year. Solution. We have ν = ν 0.5(4ν0.5) + 1.5(3ν1.5) + 2.5(2ν2.5) + 3.5(ν3.5) 4ν0.5 + 3ν1.5 + 2ν2.5 + ν3.5 = 1.2796 Remark 54.2 The equation P(i + ϵ) −P(i) P(i) = −ϵν is valid if P(i) stands for the current price of a bond. 474 MEASURES OF INTEREST RATE SENSITIVITY Example 54.10 The current price of a bond is $110 and its yield is 7%. The modiﬁed duration is 5. Estimate the price of the bond if its yield falls to 6%. Solution. The approximate percentage change in the price is −(−0.01)(5) = 5%. That is, the price of the bond increased by approximately 5%. Thus, the new price of the bond is 110 × 1.05 = $115.50 Portfolio Modiﬁed Duration Consider a portfolio consisting of n bonds. Let bond k have current price Pk(i) and modiﬁed duration νk(i). Then the current value of the portfolio is P(i) = P1(i) + P2(i) + · · · + Pn(i). Let ν denote the modiﬁed duration of the portfolio. Then, we have ν = −P ′(i) P(i) = −P ′ 1(i) P(i) + −P ′ 2(i) P(i) + · · · + −P ′ n(i) P(i) =P1(i) P(i) −P ′ 1(i) P1(i) + P2(i) P(i) −P ′ 2(i) P2(i) + · · · + Pn(i) P(i) −P ′ n(i) Pn(i) =P1(i) P(i) ν1 + P2(i) P(i) ν2 + · · · + Pn(i) P(i) νn Thus, the modiﬁed duration of a portfolio can be calculated as the weighted average of the bonds’ modiﬁed durations, using the market values of the bonds as the weights. Example 54.11 Megan buys the following bonds: (I) Bond A with a modiﬁed duration of 7 for 1000; (II) Bond B with a modiﬁed duration of 5 for 2000; and (III) Bond C with a modiﬁed duration of 10 for 500. Calculate the modiﬁed duration of the portfolio. Solution. The modiﬁed duration of the portfolio is weighted average of the modiﬁed durations weighted by the purchase price ν = 7 · 1000 + 5 · 2000 + 10 · 500 1000 + 2000 + 500 = 6.286 54 MACAULAY AND MODIFIED DURATIONS 475 Example 54.12 An investment portfolio consists of two bonds: a two-year zero-coupon bond, and a four-year zero-coupon bond. Both bonds have redemption values of $1,000. Assume an eﬀective annual interest rate of 8%. Find the Macaulay duration of the investment portfolio. Solution. The Macaulay duration of the two-year zero-coupon is 2 with current value 1000v2. The Macaulay duration of the four-year zero-coupon is 4 with current value 1000v4. The Macaulay duration of the portfolio is d = 2(1000ν2) + 4(1000ν4) 1000ν2 + 1000ν4 = 2.9232 years 476 MEASURES OF INTEREST RATE SENSITIVITY Practice Problems Problem 54.1 Find the modiﬁed duration of a 30-year, $1,000 face value, 6% annual coupon bond. Assume an eﬀective annual interest rate of 9%. Problem 54.2 Calculate the duration of a payment to be made in 5 years. Problem 54.3 A 10 year bond has annual coupons of 10 and matures for 100. Which of the following are true: (I) The term to maturity is 10 years. (II) The average term to maturity using the method of equated time is 7.5 years. (III) The Macaulay duration at 8% interest is 6.97. (IV) The modiﬁed duration at 8% interest is 6.45. Problem 54.4 Which of the following are true: (I) The average term to maturity under the method of equated time is always greater than the volatility. (II) The duration is a decreasing function of i. (III) A zero coupon bond will have a Macaulay duration equal to the term to maturity. Problem 54.5 (a) Calculate the duration of a 12-year annuity immediate payable using an interest rate of 5%. (b) Calculate the modiﬁed duration of the annuity. Problem 54.6 A perpetuity pays 100 immediately. Each subsequent payment is increased by inﬂation. Calculate the duration of the perpetuity using 10.25% assuming that inﬂation will be 5% annually. Problem 54.7 Calculate the duration of a perpetuity immediate of 1 less the duration of a perpetuity due of 1 at an interest rate of 10%. Problem 54.8 Calculate the modiﬁed duration of an annuity due with payments of 100 for 10 years using an interest rate of 8%. 54 MACAULAY AND MODIFIED DURATIONS 477 Problem 54.9 ‡ A bond will pay a coupon of 100 at the end of each of the next three years and will pay the face value of 1000 at the end of the three-year period. The bond’s duration (Macaulay duration) when valued using an annual eﬀective interest rate of 20% is X. Calculate X. Problem 54.10 Suppose that a 3-year ﬁnancial instrument is expected to make increasing payments to you at the end of each of the next three years. Speciﬁcally, the payments will be CF(t) = 1, 000t , for t = 1, 2, and 3. Assume that you purchase this ﬁnancial instrument, at time 0, at a price which provides an annual eﬀective yield of 8%. Calculate the Macaulay duration, and the modiﬁed duration of this ﬁnancial instrument. Problem 54.11 Calculate the Macaulay duration and the modiﬁed duration of a 30-year zero-coupon bond with a face value of $ 1,000. Assume that the annual yield-to-maturity is 8%. Problem 54.12 A 10-year $200,000 mortgage will be paid oﬀwith level quarterly amortization payments. Assume that the interest rate on the mortgage is 6%, convertible quarterly, and that payments are made at the end of each quarter. Find the modiﬁed duration of this mortgage. Problem 54.13 ‡ Calculate the duration of a common stock that pays dividends at the end of each year into perpetuity. Assume that the dividend increases by 2% each year and that the eﬀective rate of interest is 5%. Problem 54.14 ‡ Calculate the duration of a common stock that pays dividends at the end of each year into perpetuity. Assume that the dividend is constant, and that the eﬀective rate of interest is 10%. Problem 54.15 ‡ The current price of an annual coupon bond is 100. The derivative of the price of the bond with respect to the yield to maturity is −700. The yield to maturity is an annual eﬀective rate of 8%. Calculate the duration of the bond. Problem 54.16 ‡ Calculate the Macaulay duration of an eight-year 100 par value bond with 10% annual coupons and an eﬀective rate of interest equal to 8%. 478 MEASURES OF INTEREST RATE SENSITIVITY Problem 54.17 ‡ John purchased three bonds to form a portfolio as follows: Bond A has semiannual coupons at 4%, a duration of 21.46 years, and was purchased for 980. Bond B is a 15-year bond with a duration of 12.35 years and was purchased for 1015. Bond C has a duration of 16.67 years and was purchased for 1000. Calculate the duration of the portfolio at the time of purchase. Problem 54.18 A 10-year 1000 face value 6% annual coupon bond with redemption value C has duration equal to 6.06 using an annual eﬀective interest rate of 6%. Calculate C. Problem 54.19 You are given: (i) the annual yield rate on a zero-coupon bond with duration of 6 months is 3% (ii) the annual yield rate on a zero-coupon bond with duration of 12 months is 3.25% (iii) the annual yield rate on a zero-coupon bond with duration of 18 months is 3.5% (iv) the annual yield rate on a zero-coupon bond with duration of 24 months is 3.75% Determine the semiannual eﬀective yield rate on a 2-year 100 par value bond with 4% semiannual coupons. Problem 54.20 John purchased three bonds to form a portfolio as follows: Bond A is purchased for X and has a duration of 20 years. Bond B is purchased for Y and has a duration of 20 years. Bond C is purchased for X + Y. The duration (in years) of the portfolio at the time of purchase is 18. Determine the duration (in years) of Bond C at the time of purchase. 55 REDINGTON IMMUNIZATION AND CONVEXITY 479 55 Redington Immunization and Convexity Most ﬁnancial institutions try to insulate their portfolios from interest rate risk movements. Some institutions (such as banks) are concerned with protecting their current net worth against interest rate movements. Other institutions (such as pension funds) may face an obligation to make pay-ments after a given number of years. These institutions are more concerned with protecting the future values of their portfolios. The common factor amongst diﬀerent investors and ﬁnancial institutions is the presence of interest rate risk because net worth ﬂuctuates with interest rate. Financial institutions try to structure their assets and liabilities in such a way that they are protected against small changes in interest rates. There are three approaches that all work toward the same goal of minimizing (or even eliminating) the risk caused by ﬂuctuating interest rates: Immunizatiom, full immunization, and dedication. In this section, we consider the ﬁrst strategy. The basic idea of immunization is that ﬁnancial institutions (i.e., banks) are vulnerable to incur-ring losses if interest rates change. As a result, they “immunize” themselves by structuring their assets and liabilities such that the modiﬁed duration of the assets equals the modiﬁed duration of the liabilities. When this occurs, a rise in the interest rate would cause the present value of assets to decline, but there would be a matching decrease in the present value of liabilities. Thus, the company is protected against small changes in interest rates. We next discuss how immunization achieves the protection required. Consider cash inﬂows A1, A2, · · · , An generated by the assets at times 1, 2, · · · , n. Similarly, let L1, L2, · · · , Ln be the cash outﬂows gen-erated by the liabilities at times 1, 2, · · · , n. Let the net cash ﬂows at time t be Rt = At −Lt, t = 1, 2, · · · , n. In general, banks assets and liabiltiy are roughly equal in size so that we can assume that the present value of the cash inﬂows from the assets is equal to the present value of the cash outﬂows from the liabilities. This leads to P(i) = n X t=1 νtRt = 0. Next, we would like P(i) to have a local minimum at i, so that small changes in the interest rate in either direction will increase the present value of the cash ﬂows. But for the function P(i) to have a minimum requires two conditions to hold. The ﬁrst is that P ′(i) = 0 (i.e. the modiﬁed duration of the assets is equal to the modiﬁed duration of the liabilities ). The second condition is P ′′(i) > 0. The second derivative of P(i) is used to deﬁne the convexity of a series of cash ﬂows given by c = P ′′(i) P(i) . 480 MEASURES OF INTEREST RATE SENSITIVITY To summarize, immunization is achieved when the following three conditions are met: (1) P(i) = 0 : The present value of cash inﬂows (assets) should be equal to the present value of cash outﬂows (liabilities). (2) P ′(i) = 0 : The modiﬁed duration of the assets is equal to the modiﬁed duration of the liabilities. (3) P ′′(i) > 0 : The convexity of the present value of cash inﬂows (assets) should be greater than the convexity of the present value of cash outﬂows(liabilities). In other words, asset growth (decline) should be greater (less) than liability growth(decline). In practice there are some diﬃculties and limitations in implementing the immunization strategy: (a) choice of i is not always clear. (b) doesn’t work well for large changes in i. For large changes in i a technique, known as full immunization, was developed for that purpose. This concept will be covered in Section 56. (c) yield curve is assumed to change with the change in i; actually, short-term rates are more volatile than long-term rates. (d) frequent rebalancing is required in order to keep the modiﬁed duration of the assets and liabil-ities equal. (e) exact cash ﬂows may not be known and may have to be estimated. (f) assets may not have long enough maturities of duration to match liabilities. We next express the percentage change of P(i) in terms of both the modiﬁed duration and convexity. Expanding P(i + ϵ) as a Taylor series as far as second derivatives we ﬁnd P(i + ϵ) ≈P(i) + ϵP ′(i) + ϵ2 2 P ′′(i). Thus, P(i + ϵ) −P(i) P(i) ≈ϵP ′(i) P(i) + ϵ2 2 P ′′(i) P(i) = −ϵν + ϵ2 2 c. Example 55.1 A client deposits 100,000 in a bank, with the bank agreeing to pay 8% eﬀective for two years. The client indicates that half of the account balance will be withdrawn at the end of the ﬁrst year. The bank can invest in either one year or two year zero coupon bonds. The one year bonds yield 9% and the two year bonds yield 10%. Develop an investment program based on immunization. Solution. The cash outﬂows are 50000(1.08) at the end of the ﬁrst year and 50000(1.08)2 at the end of the second year. 55 REDINGTON IMMUNIZATION AND CONVEXITY 481 Suppose the bank invests x in the one year bonds, and y in two year bonds. Then P(i) =x(1.09)(1 + i)−1 + y(1.10)2(1 + i)−2 −50000(1.08)(1 + i)−1 −50000(1.08)2(1 + i)−2 P ′(i) = −x(1.09)(1 + i)−2 −2(1.10)2y(1 + i)−3 + 50000(1.08)(1 + i)−2 +2(50000)(1.08)2(1 + i)−3 P ′′(i) =2(1.09)x(1 + i)−3 + 6(1.10)2y(1 + i)−4 −100000(1.08)(1 + i)−3 −300000(1.08)2(1 + i)−4 To immunize against interest rate risk at i = 0.08, the bank should have P(0.08) = 0, P ′(0.08) = 0 and P ′′(0.08) > 0. The ﬁrst two conditions lead to a system in the unknowns x and y. Solving the system gives x = 49541.28 and y = 48198.35. Moreover, P ′′(0.08) = 128, 773.89 > 0. Thus, the bank is well protected against interest rate ﬂuctuations Example 55.2 For the assets in Example 55.1, ﬁnd the modiﬁed duration and convexity. Solution. We have PA(i) = x(1.09)(1 + i)−1 + (1.10)2y(1 + i)−2 →PA(0.08) = 50, 000.00 + 50, 000.00 = 100, 000.00. Also, P ′ A(i) = −x(1.09)(1 + i)−2 −2(1.10)2y(1 + i)−3 →P ′ A(0.08) = −138, 888.89. The modiﬁed duration is v(0.08) = −P ′ A(0.08) PA(0.08) = 138, 888.89 100, 000.00 = 1.39. The convexity is c(0.08) = P ′′ A(0.08) PA(0.08) = 342, 935.54 100, 000.00 = 3.43 since P ′′ A(i) = 2(1.09)x(1 + i)−3 + 6(1.10)2y(1 + i)−4 →P ′′ A(0.08) = 342, 935.54 Example 55.3 For a 30-year home mortgage with level payments and an interest rate of 10.2% convertible monthly, ﬁnd the modiﬁed duration and the convexity of the payments. Solution. The monthly rate of interest is 0.102/12 = 0.0085. Then for a dollar of monthly payment, we have P(0.0085) = 360 X t=1 (1.0085)−t = (1.0085)−1 1 −(1.0085)−360 1 −(1.0085)−1 = 112.0591 482 MEASURES OF INTEREST RATE SENSITIVITY and P ′(0.0085) = − 360 X t=1 t(1.0085)−t−1 = −(1.0085)−1(Ia)360 0.0085 = −11, 188.69608 Thus, the modiﬁed duration is ν(0.0085) = 11, 188.69608 112.0591 = 99.85. Thus, the modiﬁed duration of the payments is just under 100 months of the 360-month term of the mortgage. Now, P ′′(0.0085) = 360 X t=1 t(t + 1)(1.0085)−t−2 = ν2 360 X t=1 (t2 + t)νt =ν2[ 360 X t=1 t2νt + 360 X t=1 tνt] =ν2[ 360 X t=1 t2νt + (Ia)360 0.0085] Hence, using a calculator we ﬁnd c(0.0085) = 1, 940, 079 + 11, 283.80 (1.0085)2(112.0591) = 17, 121 Example 55.4 A bank agrees to pay 5% compounded annually on a deposit of 100,000 made with the bank. The depositor agrees to leave the funds on deposit on these terms for 8 years. The bank can either buy 4 year zero coupon bonds or preferred stock, both yielding 5% eﬀective. How should the bank apportion its investment in order to immunize itself against interest rate risk? Solution. The present value of the cash ﬂows in terms of the amount B invested in the bonds is P(i) = B(1.05)4(1 + i)−4 + (100000 −B)(0.05) i −100000(1.05)8(1 + i)−8 since the preferred stock is assumed to pay 5% forever and the bank must repay the deposit with interest at the end of 8 years. Substitution veriﬁes that P(.05) = 0. On the other hand, P ′(0.05) = −4B(1.05)4(1.05)−5 −(100000 −B)(0.05) 0.052 + 800, 000(1.05)8(1.05)−9 55 REDINGTON IMMUNIZATION AND CONVEXITY 483 So setting P ′(0.05) = 0 and solving for B gives B = 76470.58 as the amount to be invested in bonds with the remainder in stock. With this allocation, one ﬁnds P ′′(.05) = 20(76470.58)(1.05)−2 + 2(100000 −7647.58)(.05)−2 −720, 000(1.05)−9 = 965786.7413 > 0, so the portfolio is optimal Example 55.5 Find the duration and convexity of a 20 year zero coupon bond assuming that the interest rate is 7% eﬀective. Solution. The duration is 20 since the par value is repaid at the end of 20 years and this is the only payment made by the bond. Now, per dollar of par value, P(i) = (1 + i)−20, so that the convexity P ′′(i) P(i) = 420 (1 + i)2. Substituting i = 0.07 gives the convexity as 366.844 Convexity of a Portfolio Consider a portfolio consisting of n bonds. Let bond k have current price Pk(i) and convexity ck(i). Then the current value of the portfolio is P(i) = P1(i) + P2(i) + · · · + Pn(i). Let c denote the convexity of the portfolio. Then, we have c =P ′′(i) P(i) = P ′′ 1 (i) P(i) + P ′′ 2 (i) P(i) + · · · + P ′′ n(i) P(i) =P1(i) P(i) P ′′ 1 (i) P1(i) + P2(i) P(i) P ′′ 2 (i) P2(i) + · · · + Pn(i) P(i) P ′′ n(i) Pn(i) =P1(i) P(i) c1 + P2(i) P(i) c2 + · · · + Pn(i) P(i) cn Thus, the convexity of a portfolio can be calculated as the weighted average of the bonds’ convexities, using the market values of the bonds as the weights. 484 MEASURES OF INTEREST RATE SENSITIVITY Practice Problems Problem 55.1 ‡ Which of the following statements about immunization strategies are true? I. To achieve immunization, the convexity of the assets must equal the convexity of the liabilities. II. The full immunization technique is designed to work for any change in the interest rate. III. The theory of immunization was developed to protect against adverse eﬀects created by changes in interest rates. Problem 55.2 Nova Inc. has liabilities of 10 due in 1, 4, and 7 years. What asset incomes must the company arrange for in years 1 and 6 to immunize their cash ﬂow, assuming an annual interest rate of 10% on all transactions? Problem 55.3 A bank is required to pay 1,100 in one year. There are two investment options available with respect to how money can be invested now in order to provide for the 1,100 payback: (i) a non-interest bearing cash fund, for which x will be invested, and (ii) a two-year zero-coupon bond earning 10% per year, for which y will be invested. Based on immunization theory, develop an asset portfolio that will minimize the risk that liability cash ﬂows will exceed asset cash ﬂows. Assume the eﬀective rate of interest is equal to 10% in all calculations. Problem 55.4 A $5,000 payment is owed from Abby to Ben two years from now. Abby wants to set up an investment fund to meet this obligation, but the only investments she has available are a money market fund (currently earning 8%, but the rate changes daily), and a ﬁve-year zero-coupon bond earning 8%. Use an immunization framework to determine the amount of money Abby should invest now in each of the two investment vehicles. Assume an eﬀective annual interest rate of 8% for present value calculations. Problem 55.5 For the assets in Problem 55.4, ﬁnd the modiﬁed duration and convexity. Problem 55.6 Find the convexity of the following investments,assuming the eﬀective rate of interest is 8%: (a) A money market fund. (b) A 10-year zero coupon bond. (c) A preferred stock paying level dividends into perpetuity. 55 REDINGTON IMMUNIZATION AND CONVEXITY 485 Problem 55.7 Find the convexity of a loan repaid with equal installments over n periods if i = 0. Problem 55.8 A common stock pays dividends at the end of each year. It is assumed that each dividend is 4% greater than the prior dividend and the eﬀective rate of interest is 8%. Find the convexity of this common stock. Problem 55.9 Derive the following relationship between modiﬁed duration and convexity: d diν = ν2 −c. Problem 55.10 Suppose a company must make payments of 300 at time t = 3 and 500 at time t = 5, and the annual interest rate is 4%. Find asset income at times t = 0 and t = 6 so that the portfolio will be immunized. 486 MEASURES OF INTEREST RATE SENSITIVITY 56 Full Immunization and Dedication In this section we discuss the two other strategies of immunization: Full immunization and dedica-tion. Full Immunization Full immunization is an extension to the Redington immunization discussed in Section 50. While Redington immunization works for small changes in i, full immunization can be applied for all changes of i. We will say that a portfolio is fully immunized if PVA(i + ϵ) > PVL(i + ϵ) for all ϵ > 0 or equivalently PVA(i) ≥PVL(i) for all positive i. Suppose we have one liability outﬂow Lk at time k. The concept is to hold two assets, one that will produce a cash inﬂow, A, prior to the liability outﬂow (say at time k −a) , and another that will produce a cash inﬂow subsequent to the liability outﬂow (say at time k + b). We use the force of interest δ that is equivalent to i. The full immunization conditions for this single liability cash ﬂow are: (1) Present value of assets = Present value of liability. (2) Modiﬁed duration of assets = Modiﬁed duration of liability. (3) The asset cash ﬂows occur before and after the liability cash ﬂow. Example 56.1 An insurance company has committed to make a payment of $100,000 in 10 years. In order to fund this liability, the company has invested $27,919.74 in a 5-year zero-coupon bond and $27,919.74 in a 15-year zero-coupon bond. The annual eﬀective yield on all assets and liabilities is 6%. Determine whether the company’s position is fully immunized. Solution. Condition 1: Present value of assets = Present value of liability. We have PVL = 100, 000(1.06)−10 = 55, 839.48 and PVA = 27, 919.74 × 2 = 55839.48, so condition (1) is met. Condition 2: Modiﬁed duration of assets = Modiﬁed duration of liability. We have Modiﬁed duration of liabilities = 10(100,000)(1.06)−10 100,000(1.06)−10 = 10 and Modiﬁed duration of assets = 5(27,919.74)+15(27,919.74) 55,839.48 = 10. 56 FULL IMMUNIZATION AND DEDICATION 487 Thus, condition (2) is true. Condition 3: The asset cash ﬂows occur before and after the liability cash ﬂow. Thus, the po-sition is fully immunized Now, conditions (1) and (2) leads to the following system P(δ) =Ae−(k−a)δ + Be−(k+b)δ −Lke−kδ = 0 P ′(δ) = −A(k −a)e−(k−a)δ −B(k + b)e−(k+b)δ + kLke−kδ = 0. The ﬁrst condition implies Lk = Aeaδ + Be−bδ. The second condition implies −A(k −a)e−(k−a)δ −B(k + b)e−(k+b)δ + kLke−kδ =0 −Ake−(k−a)δ + Aae−(k−a)δ −Bke−(k+b)δ −Bbe−(k+b)δ + kLke−kδ =0 −k Ae−(k−a)δ + Be−(k+b)δ −Lke−kδ + [Aae−(k−a)δ −Bbe−(k+b)δ] =0 Aae−(k−a)δ −Bbe−(k+b)δ =0 Aaeaδ −Bbe−bδ =0 Thus, we obtain the system Aeaδ + Be−bδ =Lk Aaeaδ =Bbe−bδ There are exceptional cases when the system has no solutions, but if the known quantities are (i) a, b (ii) B, b (iii) A, a or (iv) A, b then unique values of the other two quantities can be found. Solving the last equation for B we ﬁnd B = A a b e(a+b)δ. We next show that conditions (1)-(3) achieve the required immunization. That is, we will show that for any δ′ ̸= δ there is a surplus, i.e. PVA(δ′) > PVL(δ′). Let S(δ′) denote the present value, at force of interest δ′ per annum, of the total-assets less the present value of the liability, i.e. the surplus function S(δ′) = PVA(δ′) −PVL(δ′) = e−kδ′(Aeaδ′ + Be−bδ′ −Lk). 488 MEASURES OF INTEREST RATE SENSITIVITY Then we have S(δ′) =e−kδ′[Aeaδ′ + Be−bδ′ −Lk] =e−kδ′[Aeaδ′ + Be−bδ′ −(Aeaδ + Be−bδ)] =e−kδ′Aeab h ea(δ′−δ) + a be−b(δ′−δ) − 1 + a b i Next, consider the function f(x) = eax + a be−bx − 1 + a b . Clearly, f(0) = 0. Moreover, since f ′(x) = a(eax −e−bx), one can easily check the following f ′(0) =0 f ′(x) >0 for x > 0 f ′(x) <0 for x < 0 Thus, f(x) > 0 for all x ̸= 0. This shows that S(δ′) > 0 for all δ′ ̸= δ as required. For a portfolio with multiple liabilities, repeat the above process for each liability outﬂow. There will be two asset inﬂows for each liability outﬂow. Full immunization is like Redington immunization in the sense that portfolio must be rebalanced periodically so that to ensure that modiﬁed duration of assets is equal to modiﬁed duration of liabilities. Example 56.2 An investor has a single liability of $1000 due in 15 years’ time. The yield on zero coupon bonds of any term is currently 4% per annum, and the investor possesses cash equal to the present value of his liability, i.e. 1000(1.04)−15 = 555.26. He wishes to invest in 10-year and 20-year zero-coupon bonds in such a way that he will make a proﬁt on any immediate change in the force of interest. How much of each security should he buy, and how large a proﬁt will he make if the rate of interest per annum immediately becomes 0.01, 0.02, 0.03, 0.05, 0.06, 0.07,or 0.08? Solution. Letting δ = ln 1.04 we ﬁnd Ae5δ + Be−5δ =1000 5Ae5δ −5Be−5δ =0 56 FULL IMMUNIZATION AND DEDICATION 489 or equivalently, we have the system of two equations A(1.04)5 + B(1.04)−5 =1000 A(1.04)5 −B(1.04)−5 =0. These equations may easily be solved, giving A = 410.96 and B = 608.32. The quantities of zero-coupon bonds which he should buy are therefore those providing $410.96 at time 10 years and $608.32 at time 20 years. The proﬁt on an immediate change in the rate of interest to i per annum is, S(i) = (1 + i)−15[410.96(1 + i)5 + 608.32(1 + i)−5 −1000]. In the following table we give the values of the liability ($1000(1 + i)−15), the assets (410.96(1 + i)−10 + 608.32(1 + i)−20) and the proﬁt to the investor for each of the speciﬁed rates. Annual interest rate PVL PVA Present value of proﬁt 0.00 1000.00 1019.28 19.28 0.01 861.35 870.58 9.23 0.02 743.01 746.51 3.50 0.03 641.86 642.61 0.75 0.04 555.26 555.26 0.00 0.05 481.02 481.56 0.54 0.06 417.27 419.16 1.89 0.07 362.45 366.11 3.66 0.08 315.24 320.87 5.63 Dedication In this section we discuss an approach for matching liabilities and assets. This approach is called dedication or absolute matching. In this approach, an company structures an asset portfolio in such a fashion that the cash inﬂow that will be generated from assets will exactly match the cash outﬂow from the liabilities. If the above strategy is achieved, the company has full protection against any movement in interest rates. However, the following problems exist with implementing this strategy. 1. Cash ﬂows are usually not that predictable on either the asset or the liability side. 2. If the liabilities are long term in nature, it might be impossible to ﬁnd assets to exactly match the liabilities without creating reinvestment risk. 3. The yield rate on a fund structured with the major restrictions imposed by absolute matching may be less than that on a fund for which more ﬂexibility is available, and the extra return may overshadow the advantages of absolute matching in importance. 490 MEASURES OF INTEREST RATE SENSITIVITY Example 56.3 An insurance company accepts an obligation to pay 10,000 at the end of each year for 2 years. The insurance company purchases a combination of the following two bonds (both with $1,000 par and redemption values) in order to exactly match its obligation: Bond A: A 1-year 5% annual coupon bond with a yield rate of 6%. Bond B: A 2-year 8% annual coupon bond with a yield rate of 7%. (a) Find the numbers (which need not be integers) of each bond the insurer must purchase to exactly match its obligations. (b) Find the total cost to the insurer of purchasing the needed numbers of bonds. Solution. (a) We have the following chart Period Bond A Bond B Liabilities 1 1050 80 10,000 2 1080 10,000 Let a be the number of bonds A purchased and b that of bonds B. Then we have the following system of two equations in two unknowns 1050a + 80b =10, 000 1080b =10, 000 Solving for a and b we ﬁnd a = 8.8183 bonds and b = 9.2593 bonds. (b) The answer is 8.8183(1050ν0.06) + 9.2593(80ν0.07 + 1080ν2 0.07) = $18161.82 56 FULL IMMUNIZATION AND DEDICATION 491 Practice Problems Problem 56.1 An insurance company has an obligation to pay $1,000,000 at the end of 10 years. It has a zero-coupon bond that matures for $413,947.55 in 5 years, and it has a zero-coupon bond that matures for $864,580.82 in 20 years. The eﬀective yield for assets and liability is 10%. (a) Determine whether the company’s position is fully immunized. (b) What is the level of surplus if the interest rate falls to 0%? (c) What is the level of surplus if the interest rate rises to 80%? Problem 56.2 ‡ A company must pay liabilities of 1000 and 2000 at the end of years 1 and 2, respectively. The only investments available to the company are the following two zero-coupon bonds: Maturity(years) Eﬀective annual yield Par 1 10% 1000 2 12% 1000 Determine the cost to the company today to match its liabilities exactly. Problem 56.3 ‡ An insurance company accepts an obligation to pay 10,000 at the end of each year for 2 years. The insurance company purchases a combination of the following two bonds at a total cost of X in order to exactly match its obligation: (i) 1-year 4% annual coupon bond with a yield rate of 5% (ii) 2-year 6% annual coupon bond with a yield rate of 5%. Calculate X. The following information applies to Problems 56.4 - 56.6 Joe must pay liabilities of 1,000 due 6 months from now and another 1,000 due one year from now. There are two available investments: • a 6-month bond with face amount of 1,000, a 8% nominal annual coupon rate convertible semi-annually, and a 6% nominal annual yield rate convertible semiannually; and • a one-year bond with face amount of 1,000, a 5% nominal annual coupon rate convertible semi-annually, and a 7% nominal annual yield rate convertible semiannually. Problem 56.4 ‡ How much of each bond should Joe purchase in order to exactly (absolutely) match the liabilities? 492 MEASURES OF INTEREST RATE SENSITIVITY Problem 56.5 ‡ What is Joe’s total cost of purchasing the bonds required to exactly (absolutely) match the liabili-ties? Problem 56.6 ‡ What is the annual eﬀective yield rate for investment in the bonds required to exactly (absolutely) match the liabilities? Problem 56.7 John wants to absolutely match a debt that he owes under which he must make payment of 1000 in one year and 2000 in two years. He can purchase the following bonds: (1) Bond A is a two year bond with annual coupons of 100 and a maturity value of 1000. (2) Bond B matures in one year for 1000 and pays a coupon of 80. Calculate the amount of Bond B that John should purchase. An Introduction to the Mathematics of Financial Derivatives What is a ﬁnancial derivative? A ﬁnancial derivative is a ﬁnancial contract (between two parties) that derives its value from the value of some underlying asset. For example, a homeowner’s insurance policy promises that in the event of a damage to your house, the insurance company will compensate you for at least part of the damage. The greater the damage, the more the insurance company will pay. Your insurance policy thus derives its value from the value of your house and therefore is a derivative. In this chapter, we introduce four ﬁnancial derivatives: forward contracts, future contracts, options (call and put) and swaps and we discuss some of the related topics. 493 494 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES 57 Financial Derivatives and Related Issues In this section we introduce the concept of ﬁnancial derivatives and discuss some of the relevant topics of derivatives. What is a Financial Derivative? A ﬁnancial derivative is a contract (between two parties) that derives its value from the value of some underlying asset(s). Examples of underlying assets include ﬁnancial assets (such as stocks or bonds), commodities (such as gold or oil), market indices (such as S&P 500 or FTSE 100), interest rates (LIBOR rates) and many others. An example of a derivative is the following: Party A makes an agreement with Party B regarding the price of gold. If the price of one ounce of gold in one year is above $300, Party A agrees to pay Party B the amount of $1. Otherwise, Party B will pay Party A $1. The agreement is a derivative since the outcome depends on the price of gold. Note that this agreement is a kind of a “bet” on the price moves of gold. In general derivatives can be viewed as bets on the price of something. The most common types of ﬁnancial derivatives that we will discuss in this book are futures, for-wards, options, and swaps. Derivatives Trading Markets Derivatives can be traded through organized exchanges such as The Chicago Board of Trade(CBOT), the New York Stock Exchange (NYSE), the New York Mercantile Exhange (NYMEX) and many oth-ers, or through contracts negotiated privately between two parties, referred to as over−the−counter (abbreviated OTC). Financial markets are usually regulated by the Securities and Exchange Com-mission (SEC) and the Commodity Futures Trading Commission (CFTC). Financial markets serve six basic functions. These functions are brieﬂy listed below: • Borrowing and Lending: Financial markets permit the transfer of funds (purchasing power) from one agent to another for either investment or consumption purposes. • Price Determination: Financial markets provide vehicles by which prices are set both for newly issued ﬁnancial assets and for the existing stock of ﬁnancial assets. • Information Aggregation and Coordination: Financial markets act as collectors and aggregators of information about ﬁnancial asset values and the ﬂow of funds from lenders to borrowers. • Risk Sharing: Financial markets allow a transfer of risk from those who undertake investments to those who provide funds for those investments. • Liquidity: Financial markets provide the holders of ﬁnancial assets with a chance to resell or liquidate these assets. • Eﬃciency: Financial markets reduce transaction costs and information costs. 57 FINANCIAL DERIVATIVES AND RELATED ISSUES 495 Most Common Uses of Derivatives Below are some of the motives for someone to use derivatives: (1) to reduce (or hedge) exposure to risk. For example, a wheat farmer and a wheat miller could enter into a futures contract to exchange cash for wheat in the future. Both parties have reduced a future risk: for the wheat farmer, the uncertainty of the price, and for the wheat miller, the availability of wheat. (2) to speculate expected changes in future prices with the hope of making proﬁt. In this case, speculation increases the exposure to risk. The potential gain or loss can be leveraged (i.e. mag-niﬁed) relative to the initial investment. (3) to reduce transaction costs, such as commissions and other trading costs. (4) to maximize return on investments through asset management activities, tax loopholes, and regulatory restrictions. For example, a company can use derivatives to produce temporary losses to lower its taxes. We refer to this motive as regulatory arbitrage. Example 57.1 Weather derivatives are ﬁnancial instruments that can be used by organizations or individuals as part of a risk management strategy to reduce risk associated with adverse or unexpected weather conditions (rain/temperature/snow). Two types of weather derivatives are heating degree-day (HDD) and cooling degree-day (CDD) future contracts. These future contracts make payments based on whether the temperature is abnormally hot or cold. (a) Explain why a soft-drink manufacturer might be interested in such a contract. (b) The business buys these future contracts to hedge against temperature-related risk. Who is the other party accepting the risk? Solution. (a) Soft drink sales greatly depend on weather. Generally, warm weather boosts soft drink sales and cold weather reduces sales. A soft drink producer can use weather futures contracts to reduce the revenue swing caused by weather and smooth its earnings. Shareholders of a company generally want the earnings to be steady. They don’t want the management to use weather as an excuse for poor earnings or wild ﬂuctuations of earnings. (b) Anyone who can predict a weather index can enter into weather futures and make a proﬁt. However, since predicting weather is not usually easy and accurate a loss may occur Perspectives on Derivatives There are three diﬀerent user perspectives on derivatives: • The end-user perspective. End-users include corporations, investment managers, and in-vestors. End-users use derivatives in order to achieve a speciﬁc goal or goals such as managing risk, speculating, reducing costs, or avoiding regulations. 496 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES • The market-maker perspective. These are traders or intermediaries between diﬀerent end-users. They buy from end-users who want to sell (usually at a low price) and sell to end-users who want to buy (usually at a higher price.) Also, market-users might charge commissions for trading transactions. • The economic observer such as a regulator or a research economist, whose role is to watch and even sometimes regulate the markets. Traders of Derivatives There are three main traders of derivatives: hedgers, speculators and arbitrageurs. • Hedgers use derivatives to reduce risk that they face from potential future movements in market variables. Example 57.2 Suppose that it is August 16 2007 and, ﬁrm A, a company based in the US knows that it will pay £10 million on November 16, 2007 for goods it has purchased from a British ﬁrm. Firm A can buy pounds from a bank in a three-month forward contract and thus hedge against foreign exchange risk. Suppose the agreement is to buy £1 for $1.55. (a) What would the ﬁrm A pay on November 16, 2007? (b) Suppose that on November 16, 2007 the exchange rate is $3 for £1. What would ﬁrm A pay on Novemebr 16, 2007 in the absence of hedging? Solution. (a) The ﬁrm has hedged against exchange rate volatility. The amount to be paid in US dollars on November 16, 2007 is 1.55 × 10, 000, 000 = $15.5 millions. (b) The ﬁrm would have to pay 3 × 10, 000, 000 = $30 millions • Speculators use derivatives to bet on the future direction of a market variable. They can, for example, buy a put option on a stock if they think it will go down. If they think that the price of the stock will go up they will buy a call option. Another example of a speculator is an investor in weather derivatives who buys temperature-related future contracts. • Arbitrageurs take oﬀsetting positions in two or more instruments to lock in a riskless proﬁt if securities are inconsistently priced. Example 57.3 Suppose that ﬁrm A is dually listed on both the NYSE and LSE. On the NYSE its share is trading at $152 and on the LSE its share is trading at £100. The exchange rate is $1.55 per pound. An arbitrageur makes a riskless proﬁt by simultaneously buying 100 shares on the NYSE and sell them on the LSE. What is the amount of this proﬁt? 57 FINANCIAL DERIVATIVES AND RELATED ISSUES 497 Solution. He buys the shares on the NYSE for $152 a share. He sells the shares on the LSE for $155 a share. So the proﬁt per share is $3. Total proﬁt from the 100 shares is then $300 498 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES Practice Problems Problem 57.1 Which of the following is a derivative? (I) A bushel of corn (II) A contract to sell 100 bushels of corn (III) Short sale of a stock. Problem 57.2 The Security and Exchange Commission (SEC) holds primary responsibility for enforcing the fed-eral securities laws and regulating the securities industry. Which of the following describes the perspective of SEC on derivatives? (I) The end-user perspective (II) The market-maker perspective (III) The economic observer. Problem 57.3 List four motives for using derivatives. Problem 57.4 Company A based in the US buys goods from a British company and is required to pay in British pounds for the cost of the goods. Company A buys a derivative that allows it to use the current exhange rate after six months from today. Which of the following characterizes company A uses of the derivative? (I) Hedging (II) Speculation (III) Reduced transaction costs (IV) Regulatory loop holes. Problem 57.5 You buy 100 shares of a stock through a brokerage ﬁrm. Which of the following describes your perspective of using derivatives? (I) The end-user perspective (II) The market-maker perspective (III) The economic observer. Problem 57.6 You buy 100 shares of a stock through a brokerage ﬁrm. Which of the following describes brokerage ﬁrm perspective of using derivatives? 57 FINANCIAL DERIVATIVES AND RELATED ISSUES 499 (I) The end-user perspective (II) The market-maker perspective (III) The economic observer. Problem 57.7 You work for a ﬁrm who has a reimbursement program for college credits. The program reimburses 100% of costs for an “A”, 75% of costs for a “B”, 50% for a “C” and 0% for anything less. Would this program be considered an example of a derivative? Explain. Problem 57.8 Explain why the following ﬁrms might be interested in buying HDD or DCC future contracts. (a) Ski-resort owner. (b) Electric utilities. (c) Amusement park owner. Problem 57.9 Match each of the following market traders: (a) hedgers (b) speculators (c) arbitrageurs with one of the following roles: (I) proﬁt from perceived superior expectations (II) trade to eliminate or reduce future price risk (III) attempt to proﬁt when the same security or commodity is trading at diﬀerent prices in two or more markets. Problem 57.10 Explain carefully the diﬀerence between (a) hedging (b) speculation and (c) arbitrage. 500 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES 58 Derivatives Markets and Risk Sharing As pointed out in the previous section, one of the basic function of derivatives markets is risk shar-ing. Risk is a major feature of economic activity. In this section we discuss the role of ﬁnancial markets as a vehicle of sharing risks. Risk-sharing is one of the most important functions of ﬁnancial markets. To illustrate this con-cept, consider auto insurance companies. These companies collect premiums for auto insurance policies. The total premiums collected are then being available to help those who get involved into car wrecks. Thus, those policyholders who had no wrecks in their records have basically lost their premiums. However, their premiums went to help those who needed it. A similar scenario occurs in the business world. Some companies proﬁt and others suﬀer. Thus, it makes sense to have a mechanism enabling companies to exchange ﬁnancial risks. Share risking mechanisms should beneﬁt everyone. Another example of risk sharing is the use of the so-called catastrophe bonds. For example, an insurance company that provides hurricane insurance for Florida residents will usually face large insurance claims in the case of a devastating hurricane happening. For that reason, the insurance company usually either issues or buys from reinsurers cat bonds enabling the company to share risks with the bondholders. Catastrophe bondholders receive interest rates at levels commensurate with the risk that they may lose all of their principal on the occurrence of a major hurricane. The sponsor issues cat bonds and typically invests the proceeds from the bond issuance in low-risk securities. The earnings on these low-risk securities, as well as insurance premiums paid to the sponsor, are used to make periodic, variable rate interest payments to investors. Sponsors of cat bonds include insurers, reinsurers, corporations, and government agencies. Diversiﬁable Risk Versus Non-Diversiﬁable Risk Risk management is one of the topics discussed in portfolio theory. The risk level of an asset can be categorized into two groups: the diversiﬁable risk and the non-diversiﬁable risk. Before we discuss the diﬀerence between diversiﬁable and non-diversiﬁable risks, we need to ﬁrst understand the term diversiﬁcation. The act of diversiﬁcation implies that an individual (or a ﬁrm) allocates his/her wealth among several types of investments rather than just one investment. In other words, diversi-ﬁcation means spreading the risk of a portfolio by investing in several diﬀerent types of investments rather than putting all the money in one investment. Diversiﬁable risk (also known as non-systematic risk) is a risk that can be reduced or eliminated by combining several diverse investments in a portfolio. Example 58.1 Suppose you are an avid Apple computer user, and you decide to invest all your money in Apple stocks. Are you making a sound decision? If not, then what is a good investment strategy? 58 DERIVATIVES MARKETS AND RISK SHARING 501 Solution. The answer is no. If you have done so, you will be very vulnerable in the last few months when Apple announced losses and a major change in management. However, you can reduce the risk of your portfolio if you invest in other computer companies such IBM, Compaq, etc. However, this is not a very well diversiﬁed portfolio because you have invested your money in the computer industry. As a result, you will still be vulnerable to changes in the computer industry. It is wise to have a portfolio consisting of shares in ﬁrms from various industries such as real estate, electronics, biomedical and so on Non-diversiﬁable risk is the part of an asset’s risk that cannot be eliminated through diver-siﬁcation. This type of risk is also known as market risk or systematic risk. Examples of non-diversiﬁable risks are natural disasters, wars, and economic factors such as inﬂation or unem-ployment levels events that can inﬂuence the entire market and usually aﬀects your entire portfolio. Example 58.2 Which of the following can be categorized as a non-diversiﬁable risk? (a) Recession. (b) The fall of a stock share. (c) Government defaults in paying bonds. Solution. Obviously only (a) and (c) might have major impact of one’s entire portfolio 502 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES Practice Problems Problem 58.1 Explain why a health insurance policy is a risk sharing derivative. Problem 58.2 List the two types of risk involved in a portfolio. Problem 58.3 The act of diversiﬁcation implies that an individual (or a ﬁrm) allocates his/her wealth (I) among several types of assets (II) in just one asset Problem 58.4 Determine whether each of the following events is considered a diversiﬁable or non-diversiﬁable risk for a portfolio: (a) A software ﬁrm’s best programmer quits. (b) Oil-prices on international markets suddenly increase. (c) Congress votes for a massive tax cut. (d) A low-cost foreign competitor unexpectedly enters a ﬁrm’s market. Problem 58.5 Determine whether each of the following events is considered a diversiﬁable or non-diversiﬁable risk for a portfolio: (a) A wildcat strike is declared. (b) Oil is discovered on a ﬁrm’s property. (c) The Federal Reserve institutes a restrictive monetary policy. (d) Long term interest rates rise sharply. Problem 58.6 An earthquake linked bond is called (a) callable bond (b) perpetual bond (c) cat bond (d) serial bond Problem 58.7 Decide whether the risk is a diversiﬁable risk or non-diversiﬁable risk: 58 DERIVATIVES MARKETS AND RISK SHARING 503 (a) Management risk (b) Inﬂation risk (c) Sociopolitical risk (d) Credit risk (e) Currency risk (f) Interest rate risk (g) Liquidity risk. Problem 58.8 True or false: Cat bonds are used by sponsors as a hedge against natural disasters. Problem 58.9 Which of the following is not a term that refers to non-systematic risk? (A) Non-diversiﬁable risk (B) Market risk (C) Diversiﬁable risk (D) Both (A) and (B) Problem 58.10 In 9/11, immediately after the terrorist attacks the stock market fell signiﬁcantly. This is an example of (A) Systematic risk (B) Market risk (C) Non-diversiﬁable risk (D) All of the above Problem 58.11 Recently in California the workers at Kroger supermarkets had been on strike. This strike would most likely have been seen as an example of (A) Non-systematic risk (B) Non-diversiﬁable risk (C) Diversiﬁable risk (D) None of the above Problem 58.12 Explain how cat bonds work. 504 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES 59 Forward and Futures Contracts: Payoﬀand Proﬁt Dia-grams In this section we introduce two examples of derivatives: forward and futures contracts. We also analyze the possible payoﬀs and proﬁts of these two securities. Forwards and futures are very similar as contracts but diﬀer in pricing and trading mechanisms. For example, forwards are traded over the counter (i.e. contract speciﬁcation can be customized) whereas futures are traded on exchange markets (where contracts are standardized). For the purpose of this section, we think of them as interchangeable. A more detailed discussion of futures will be covered in Section 72. A forward contract (or a futures contract) is a commitment to purchase at a future date, known as the expiration date, the delivery date or the maturity date, a given amount of a commodity or an asset at a price agreed on today. The price ﬁxed now for future exchange is called the forward price or the delivery price. The asset or commodity on which the forward contract is based is called the underlying asset. Apart from commissions and bid-ask spreads, a forward contract requires no initial payment or premium between the two parties. At the delivery date, cash is exchanged for the asset. Examples of forward contracts include • A bond forward is an obligation to buy or sell a bond at a predetermined price and time. • An equity forward is an obligation to buy or sell an equity at a predetermined price and time. • A gold forward is an obligation to deliver a speciﬁed quantity of gold on a ﬁxed date and receive a ﬁxed delivery price. • A foreign exchange forward is an obligation to buy or sell a currency on a future date for a predetermined ﬁxed exchange rate. This type of forwards is used by ﬁrms as a protection against ﬂuctuation in foreign currency exchange rates. • An interest rate forward is used to lock-in future interest rates. For example, a ﬁrm who is planning for a loan in six months can buy forward contract that lock-in the rate at the present. This rate will be the rate that will apply when the loan is exercised in six months. The risk here is that the ﬁrm will stuck with borrowing money at the stated rate so any decline of interest rates before the six months period will cause losses for the ﬁrm. Example 59.1 Explain in words the meaning of the following: A 3-month forward contract for 1,000 tons of soybean at a forward price of $165/ton. Solution. In this contract, the buyer is committed to buy 1000 tons of soybean from the seller in three months for a price of $165 a ton 59 FORWARD AND FUTURES CONTRACTS: PAYOFF AND PROFIT DIAGRAMS 505 Payoﬀon a Forward Contract Forward contracts are privately executed between two parties. The buyer of the underlying com-modity or asset is referred to as the long side whereas the seller is the short side. The obligation to buy the asset at the agreed price on the speciﬁed future date is referred to as the long position. A long position proﬁts when prices rise. The obligation to sell the asset at the agreed price on the speciﬁed future date is referred to as the short position. A short position proﬁts when prices go down. What is the payoﬀof a forward contract on the delivery date? Let T denote the expiration date, K denote the forward price, and PT denote the spot price (or market price) at the delivery date. Then • for the long position: the payoﬀof a forward contract on the delivery date is PT −K; • for the short position: the payoﬀof a forward contract on the delivery date is K −PT. Figure 59.1 shows a payoﬀdiagram on a contract forward. Note that both the long and short forward payoﬀpositions break even when the spot price is equal to the forward price. Also note that a long forward’s maximum loss is the forward price whereas the maximum gain is unlimited. For a short forward, the maximum gain is the forward price and the maximum loss is unlimited. Figure 59.1 Payoﬀdiagrams show the payoﬀof a position at expiration. These payoﬀs do not include any costs or gains earned when purchasing the assets today. Payoﬀdiagrams are widely used because they summarize the risk of the position at a glance. We have pointed out earlier that the long makes money when the price rises and the short makes money when the price falls. Example 59.2 An investor sells 20 million yen forward at a forward price of $0.0090 per yen. At expiration, the spot price is $0.0083 per yen. 506 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES (a) What is the long position payoﬀ? (b) What is the short position payoﬀ? Solution. (a) At the expiration date, the long position’s payoﬀis (0.0083 −0.009) × 20 · 106 = −$14, 000, a loss of $14,000 (b) The short position’s payoﬀis $14, 000 that is a proﬁt of $14,000 Example 59.3 Consider a forward contract on a stock with spot price of $25 and maturity date of 3 months from now. The forward price is $25.375. Draw the payoﬀdiagram for both the long and short forward positions after 3 months. Solution. The diagram is given in Figure 59.2. Figure 59.2 Note that both the long and short forward payoﬀpositions break even when the price of the stock at maturity is equal to the forward price, i.e., at $25.375 Example 59.4 Consider the forward contract of the previous example. Suppose the investor decides to purchase the stock outright today for $25. Is there any advantage of using the forward contract to buy the stock as opposed to buying it outright? (a) Draw the diagram for the payoﬀto the long physical position as well as the payoﬀdiagram to the long forward. (b) If the spot price is $0 in three months then ﬁnd the long forward position’s payoﬀ. 59 FORWARD AND FUTURES CONTRACTS: PAYOFF AND PROFIT DIAGRAMS 507 (c) If the spot price of the stock in three months is $25.372 then ﬁnd the long physical position’s payoﬀand the long forward’s payoﬀ. (d) Describe the investement for the outright purchase as well as the long forward in order to own the stock. Solution. (a) The diagram is shown in Figure 59.3. Note that the value of the payoﬀof the physical position is equal to the spot price of the stock. The payoﬀdiagram does not include costs when the stock is purchased outright. Figure 59.3 (b) The long forward’s payoﬀis 0 −25.372 = −$25.372. (c) The payoﬀof the long physical position is $25.372. The payoﬀof the long forward is 25.372 − 25.372 = $0. (d) With the outright purchase, the investor invests $25 now and owns the stock. With the forward contract, the investor invests $0 now and $25.372 after three months and owns the stock Note that the payoﬀof the outright purchase in Figure 59.3 tells us how much money we end up with after 3 months, but does not account for the initial $25 investment for purchasing the stock outright. Thus, from the graph one can not tell whether there is an advantage to either a forward purchase or an outright purchase. In order to have a fair comparison, the initial investments for both must be the same and then account for interest rate earned over the 3 months. We illustrate this point in the next two examples. Example 59.5 Suppose the investor invests $25 in a zero-coupon bond with par value $25.372 and maturity date in three months along with the forward contract of the previous example. Note that this combined position and the outright purchase position each initially costs $25 at time 0. Suppose the bond yield rate is 5.952% payable quarterly. Show that the combined position mimics the eﬀect of buying the stock outright. 508 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES Solution. The investor pays $25 now for the bond. After three months the zero-coupon bond return is $25.372. He then uses the proceeds to pay for the forward price of $25.372 and owns the stock. This alternative is equivalent to outright purchase of the stock today for $25 Example 59.6 Suppose the investor borrows $25 to buy the physical stock which costs $25. Interest rate on the loan is 5.952% compounded quarterly. In this case, this position and the long forward position each initially costs $0 at time 0. Show that this act of borrowing mimics the eﬀect of entering into a long forward contract. Solution. The investor uses the borrowed money to buy the stock today. After three months, the investor repays $25 × 1.01488 = $25.372 for the borrowed money. Thus, the investor invested nothing ini-tially, and after three months he paid $25.372 and owned the stock. This shows that borrowing to buy the stock mimics the eﬀect of entering into a long forward contract In the above two examples, we conclude that the forward contract and the cash position are equiv-alent investments, diﬀering only in the timing of the cash ﬂows. Besides payoﬀdiagrams, one deﬁnes a proﬁt diagram. Recall that a payoﬀdiagram graphs the cash value of a position at maturity. A proﬁt diagram subtracts from the payoﬀthe future value of the initial investment in the position. Geometrically, the proﬁt diagram is a vertical shift of the payoﬀdiagram by the amount of the future value of the initial investment. Since forward contracts require no initial investment, the payoﬀand proﬁt diagrams coincide. Example 59.7 A stock is priced at $50 and pays no dividends. The eﬀective annual interest rate is 10%. Draw the payoﬀand proﬁt diagrams for a one year long position of the stock. Solution. The payoﬀof a stock you buy long (and therefore you own) is just the stock market value. Let’s assume you want to ﬁnd the proﬁt after one year. Your proﬁt is the payoﬀ(market price) minus the future value of the initial investment over one year. If we denote the stock price by S then your proﬁt will be S −1.1(50) = S −55. Figure 59.4 shows both the payoﬀand proﬁt diagrams. Notice that the proﬁt is 0 when the stock price is $55 59 FORWARD AND FUTURES CONTRACTS: PAYOFF AND PROFIT DIAGRAMS 509 Figure 59.4 Example 59.8 Suppose we own an index which we want to be valued at $1020 six months from today. For that we buy a forward contract with forward price of $1020 and maturity date of six months. Consider investing $1000 in a zero-coupon bond with six-month interest rate of 2%. (a) Compare the payoﬀdiagram of the combined position of the forward contract and the bond with the payoﬀdiagram of the index after six months. (b) Answer the same question for the proﬁt diagrams. Solution. (a) The payoﬀof the forward contract and the bond in six months is forward + bond = spot price at expiration - 1020 + (1000 + 1000(0.02)) = spot price at expiration Thus, the payoﬀdiagram of the combined position is the same as the payoﬀdiagram of the index. Adding the bond shifts the payoﬀdiagram of the forward vertically upward by $1020. (b) The proﬁt diagram of the forward contract coincides with its payoﬀ’s diagram since there is no initial investment. The proﬁt of the bond is the payoﬀminus the future value of the investment which comes out to 0. Thus, the proﬁt diagram of the combined forward and bond is the same as the proﬁt diagram of the forward Cash-Settled Forward Contracts A forward may be cash-settled, in which case the underlying asset and payment never exchange hands. Instead, the contract settles with a single payment in the amount spot price minus forward 510 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES price at delivery date. If the diﬀerence is positive, the short party pays the long party the diﬀerence. If it is negative, the long party pays the short party. Example 59.9 Two parties agree today to exchange 500,000 barrels of crude oil for $42.08 a barrel three months from today. Describe this forward when settled for cash (a) if the spot price at the delivery date is $47.36 a barrel; (b) if the spot price at the delivery date is $40.17 a barrel. Solution. If the forward were cash-settled then on the delivery date three months from today, no oil would change hands, and there would be no payment of 500, 000 × 42.08 = $21.04 millions. (a) If the spot price at the delivery date was $47.36 a barrel then the forward would settle with a single payment of 500, 000(47.36 −42.08) = $2, 640, 000 made by the short party to the long party. (b) If the spot price at the delivery date was $40.17 a barrel then the forward would settle with a single payment of 500, 000(42.08 −40.17) = $955, 000 made by the long party to the short party Credit Risk We conclude this section by mentioning the credit risk (i.e. the risk that a party will not meet its contractual obligations) associated with forward contracts. Exchange-traded contracts typically require collateral in order to minimize this risk . In over-the-counter contracts, each party bears the other’s credit risk. Example 59.10 Creditwise, which is riskier a forward contract or a futures contract? Solution. Since forward contracts are traded on over-the-counter and futures contracts are traded on ex-changes, forward contracts are more riskier than futures contracts 59 FORWARD AND FUTURES CONTRACTS: PAYOFF AND PROFIT DIAGRAMS 511 Practice Problems Problem 59.1 What is the diﬀerence between a long forward position and a short forward position? Problem 59.2 What is the diﬀerence between a long futures position and a short futures position? Problem 59.3 A trader enters into a short forward contract on 100 million yen. The forward exchange rate is 189.4 yen per £. How much does the trader gain or lose if the exchange rate at the end of the contract is (a) 192.0 yen per £; (b) 183.0 yen per £? Problem 59.4 A stock with no dividends has a current price of $50. The forward price for delivery in one year is $53. If there is no advantage of buying either the stock or the forward contract, what is the 1-year eﬀective interest rate? Problem 59.5 A default-free zero-coupon bond costs $91 and will pay $100 at delivery date in one year. What is the eﬀective annual interest rate? What is the payoﬀdiagram for the bond? What is the proﬁt diagram for the bond? Problem 59.6 Suppose that you enter into a long six-month forward position at a forward price of $50. What is the payoﬀin 6 months for prices of $40, $45, $50, $55, and $60? Problem 59.7 Suppose that you enter into a short six-month forward position at a forward price of $50. What is the payoﬀin 6 months for prices of $40, $45, $50, $55, and $60? Problem 59.8 Suppose shares of XY Z corporation are trading at $50 and currently pay no dividends. The forward price for delivery in 1 year is $55. Suppose the 1-year eﬀective annual interest rate is 10%. Draw the payoﬀand proﬁt diagrams of this forward contract for a long position. Problem 59.9 A company agrees to buy 2,000 barrels of oil in two years; the current price is $63.50 per barrel. What is reasonable price to pay per barrel in two years? 512 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES Problem 59.10 Consider a forward contract of a stock index with maturity date six months from now and forward price of $1020. Answer the following questions if this forward contract is settled for cash. (a) Find the long and short positions payoﬀs if the spot price of the index at maturity is $1040. (b) What if the spot sprice is $960? Problem 59.11 You are a US exporter who will receive 5 millions Euro in six months time from the sale of your product in Spain. How can you hedge your foreign exchange risk ? Problem 59.12 Consider a forward contract of a stock index with maturity date three months from now and forward price of $930. Assume that the spot price of the index today is $900. What is the proﬁt or loss to a short position if the spot price of the market index rises to $920 by the expiration date? Problem 59.13 One ounce of gold today is being sold at the price of $300 per ounce. An investor borrows $300 with annual interest rate of 5% and buys one ounce of gold. The investor enters into a short forward contract to sell 1 oz. of gold in one year at $340. What is the investor’s proﬁt in this transaction? What would be his proﬁt if the original price of 1 oz. of gold is $323.81 and the loan is $323.81? Problem 59.14 An investor enters into a short forward contract to sell 100,000 British pounds for US dollars at an exchange rate of 1.50 US dollars per pound. How much does the investor gain or lose if the exchange rate at the end of the contract is (a) 1.49 and (b) 1.52? Problem 59.15 A forward contract is made for the delivery of a commodity 10 months from today at a forward price of $200 per unit of the commodity. (A) The purchaser of the forward contract pays the forward price, $200, immediately but must wait for 10 months to receive the commodity. (B) The holder of a forward contract always has the right to exchange it for an identical futures contract at any time before the delivery date. (C) The seller of the forward contract has an obligation to deliver the commodity 10 months from today in return for the spot price on the delivery date. (D) The seller of the forward contract has an obligation to deliver the commodity 10 months from today in return for $200 per unit, payable on delivery. 59 FORWARD AND FUTURES CONTRACTS: PAYOFF AND PROFIT DIAGRAMS 513 Problem 59.16 The main diﬀerence between forward and futures contracts is: (A) Futures contracts are traded every day (i.e., frequently) in organized exchanges prior to the delivery date speciﬁed in the contract. (B) For a futures contract there is never any obligation to deliver, or to receive, the underlying asset. (C) Forward contracts always have delivery dates further from the present than futures contracts. (D) The underlying assets in forward contracts are always physical commodities (e.g., wheat, oil or silver), while for futures contracts the underlying assets could be anything (e.g. weather indices or notional bank deposits). 514 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES 60 Call Options: Payoﬀand Proﬁt Diagrams One problem with a forward contract is the obligation of the buyer to pay the forward price at the expiration date which causes a loss if the spot price is below the forward price. So it is natural to wonder if there is a type of contract where the buyer has the option not to buy the underlying asset. The answer is aﬃrmative thanks to call options. An option is a contract to buy or sell a speciﬁc ﬁnancial product such as a security or a commodity. The contract itself is very precise. It establishes a speciﬁc price, called the strike price or exercise price at which the contract may be exercised, or acted on, and it has an expiration date. When an option expires, it no longer has value and no longer exists. There are two types of options: calls and puts. In this section we discuss call options and in the next section we discuss the put options. In a call option the call buyer expects that the price of an asset may go up. The buyer pays an upfront premium that he will never get back. He has the right to exercise the option at or before the expiration date depending on the option’s style(to be introduced below). The call seller, also known as the option’s writer, receives the premium. If the buyer decides to exercise the option, then the seller has to sell the asset at the strike price. If the buyer does not exercise the option, then the seller proﬁt is just the premium. Thus, a call option is a contract where the buyer has the right to buy but not the obligation to buy. An option style governs the time at which exercise can occur. A European call option allows the holder to exercise the option (i.e., to buy) only on the option expiration date. An American call option allows exercise at any time during the life of the option. A Bermuda call option can be exercised at certain pre-speciﬁed dates before or at the expiration date. Unless otherwise speciﬁed, we will limit our discussions to European style. For a call option, the buyer has a long position whereas the seller has a short position. Example 60.1 Trader A (call buyer) purchases a call contract to buy 100 shares of XY Z Corp from Trader B (call seller) at $50 a share. The current price is $45 a share, and Trader A pays a premium of $5 a share. (a) If the share price of XY Z stock rises to $60 a share right before expiration, then what proﬁt does Trader A make if he exercised the call by buying 100 shares from Trader B and sell them in the stock market? (b) If, however, the price of XY Z drops to $40 a share below the strike price, then Trader A would not exercise the option. What will be his losses in this case? Solution. (a) Trader A total cost for owning the 100 shares before or on the expiration date is 50(100) + 5(100) = 5, 500. The revenue from selling the shares is 60(100) = 6, 000. Thus, Trader A makes a 60 CALL OPTIONS: PAYOFF AND PROFIT DIAGRAMS 515 proﬁt of $500 in this transaction. (b) The losses will consist basically of the future value of premium paid, i.e. the future value of 5(100) = $500 Since a buyer of a call option will exercise the option only when the spot price is higher than the strike price, the buyer’s payoﬀis then deﬁned by the formula Buyer′s call payoﬀ= max{0, spot price at expiration −strike price}. Example 60.2 Consider a call option with a strike price of $500. (a) What would be the payoﬀto the buyer if the spot price at the expiration date is $550? (b) What would be the payoﬀto the buyer if the spot price at the expiration date is $450? Solution. (a) Since the spot price is larger than the strike price, the buyer will most likely exercise his option. The payoﬀin this case is max{0, 550 −500} = $50. (b) In this case, the buyer will not exercise his option and therefore the payoﬀis max{0, 450 −500} = $0 Notice that the payoﬀdoes not take into consideration the cost of acquiring the position, i.e. the premium which is paid at the time the option is acquired. The payoﬀof an option is not the money earned (or lost). The proﬁt earned by the buyer is given by the formula Buyer′s call proﬁt = Buyer′s call payoﬀ−future value of premium Example 60.3 Consider a 3-month call option with strike price of $500 and premium of $46.90. If the spot price at expiration date is $550, would you exercise the call option? How much would you make or lose? The risk-free interest rate is 1% over three months. Solution. Since spot price is higher than the strike price, you expect to exercise the call option. The future value of the premium is 46.90(1.01) = $47.37. The buyer’s proﬁt is then max{0, 550 −500} −47.37 = $2.63 516 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES A payoﬀdiagram and a proﬁt diagram of a long call with strike price K, spot price PT, and future value of premium Pc is shown in Figure 60.1. Figure 60.1 Notice that for a long call the maximum loss is the future value of the premium whereas the maximum gain is unlimited. Example 60.4 Consider a long call option (with strike price K) and a long forward contract (with a forward price K) with underlying asset a stock index. Draw the proﬁt diagram of both on the same window. (a) If the stock index is rising past K,, which of the two is more proﬁtable? (b) If the stock index falls considerably, which is more proﬁtable? Solution. Figure 60.2 plots the proﬁt for both the long forward and the long call. Figure 60.2 (a) The forward contract is more proﬁtable than the call option. (b) The call option is more proﬁtable (i.e. less loses) because the most you lose is the future 60 CALL OPTIONS: PAYOFF AND PROFIT DIAGRAMS 517 value of your premium. Notice that a call option can be thought as an insured position in the in-dex. It protects the buyer against substantial loses. The cost of this insurance is the premium paid Thus far we have considered the payoﬀand the proﬁt from the buyer’s perspective. Next, we consider these issues from the seller’s (i.e. the writer) perspective. The writer’s payoﬀand proﬁt are the negative of those of the buyer’s and thus they are given by Writer′s call payoﬀ= −max{0, spot price at expiration −strike price} = min{0, strike price −spot price at expiration} and Writer′s call proﬁt = Writer′s call payoﬀ+ future value of premium. Example 60.5 Again, consider a 3-month call option with a strike price of $500 and a premium $46.90. Find the writer’s payoﬀand the proﬁt if the spot price at the expiration date is $550. Assume a risk-free 3-month interest rate of 1%. Solution. Since the spot price is higher than the strike price, the writer has to sell the option. His/her payoﬀ is min{0, 500 −550} = −$50 and the proﬁt is min{0, 500 −550} + 46.90(1.01) = −$2.63 The payoﬀand proﬁt diagrams for a short call with strike price K, future value of premium Pc are shown in Figure 60.3. 518 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES Figure 60.3 Note that for a short call, the maximum gain is the future value of the premium whereas the maximum loss can be unlimited. Example 60.6 Consider a short call option (with strike price K) and a short forward contract (with a forward price K) with underlying asset a stock index. Draw the proﬁt diagram of both on the same window. (a) If the stock index is rising, which of the two is more proﬁtable? (b) If the stock index falls considerably, which is more proﬁtable? Solution. Figure 60.4 plots the proﬁt on both the short forward and the short call. Figure 60.4 (a) Both derivatives loss when stock price rises. However, the loss with the short forward is higher than that with the short call. (b) The short forward is more proﬁtable than the short call 60 CALL OPTIONS: PAYOFF AND PROFIT DIAGRAMS 519 Practice Problems Problem 60.1 A person who buys an option may do any of the following except (A) extend it. (B) exercise it. (C) sell it. (D) allow it to expire. Problem 60.2 Deﬁne the following terms associated with options: (a) Option. (b) Exercise. (c) Strike price. (d) Expiration date. (e) Call option. Problem 60.3 What is the diﬀerence between European options and American options? Problem 60.4 An investor buys a call option whose underlying asset is a stock index. The strike price is $1020 and the expiration date is six months. (a) If the spot price at the expiration date is $1100, would the buyer exercise his option? What is his payoﬀin this case? (b) What if the spot price is $900? Problem 60.5 An investor buys a call option whose underlying asset is a stock index. The strike price is $1020 and the expiration date is six months. (a) If the spot price at the expiration date is $1100, what would the seller’s payoﬀbe? (b) If the spot price at the expiration date is $900, what would the seller’s payoﬀbe? Problem 60.6 Consider a call option (whose underlying asset is a stock index) with strike price of $1000 and expiration date in six months. Suppose the risk-free 6-month interest rate is 2% and the call premium is $93.81 (a) What the future value of the premium? (b) What is the buyer’s proﬁt if the spot price at the expiration date is $1100? (c) What if the spot price at the expiration date is $900? 520 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES Problem 60.7 Graphically, the proﬁt diagram of a call option is a vertical shift of the payoﬀdiagram by what quantity? Problem 60.8 Suppose you buy a 6-month call option with a strike price of $50. What is the payoﬀin 6 months for prices $40, $45, $50, $55, and $60? Problem 60.9 A call option on ABC Corp stock currently trades for $6. The expiration date is December 17,2005. The strike price of the option is $95. (a) If this is an American option, on what dates can the option be exercised? (b) If this is a European option, on what dates can the option be exercised? (c) Suppose the stock price on Nov 14, 2005 is $80. Is this option worthless? Problem 60.10 The strike price of a call option on Sony Corporation common stock is $40. (a) What is the payoﬀat expiration on this call if, on the expiration date, Sony stock sells for $35? (b) What is the payoﬀat expiration of this call if, on the expiration date, Sony stock sells for $45? (c) Draw the payoﬀdiagram for this option. Problem 60.11 You hold a European call option contract (i.e. 100 shares) on Coca-Cola stock. The exercise price of the call is $50. The option will expire in moments. Assume there are no transactions costs or taxes associated with this contract. (a) What is your proﬁt on this contract if the stock is selling for $51? (b) If Coca-Cola stock is selling for $49, what will you do? Problem 60.12 Suppose the stock price is $40 and the eﬀective annual interest rate is 8%. (a) Draw the payoﬀdiagrams on the same window for the options below. (b) Draw the proﬁt diagrams on the same window for the options below. (i) 35-strike call with a premium of $9.12. (ii) 40-strike call with a premium of $6.22. (iii) 45-strike call with a premium of $4.08. Problem 60.13 What will be the value of adding a long call payoﬀwith a short call payoﬀ? 60 CALL OPTIONS: PAYOFF AND PROFIT DIAGRAMS 521 Problem 60.14 What position is the opposite of a purchased call option? Problem 60.15 What is the diﬀerence between entering into a long forward contract when the forward price is $55 and taking a long position in a call option with a strike price of $55? Problem 60.16 The current spot price of a market index is $900. An investor buys a 3-month long call option on the index with strike price at $930. If the spot price at the expiration date is $920, what is the investor’s payoﬀ? Problem 60.17 You would like to speculate on a rise in the price of a certain stock. The current stock price is $29, and a 3-month call with a strike price of $30 costs $2.90. You have $5,800 to invest. Identify two alternative investment strategies, one in the stock and the other in an option on the stock. What are the potential gains and losses from each? Problem 60.18 An American call option allows the holder to: (A) buy the underlying asset at the strike price on or before the expiration date. (B) sell the underlying asset at the strike price on or before the expiration date. (C) buy or sell the underlying asset at the strike price only on the expiration date. (D) None of the above. Problem 60.19 Which of the following is TRUE regarding the purchaser of a call option? (A) The yield on the purchasers portfolio would decrease by purchasing the option (B) The purchaser would limit the amount of money he could lose if the underlying stock declined (C) The purchaser would beneﬁt if the underlying stock declined (D) The purchaser would exercise the option if the stock declined Problem 60.20 Which of the following statements about the payoﬀof a call option at expiration is false? (a) A short position in a call option will result in a loss if the stock price exceeds the exercise price. (b) The payoﬀof a long position equals zero or the stock price minus the exercise price, whichever is higher. (c) The payoﬀof a long position equals zero or the exercise price minus the stock price, whichever is higher. (d) A short position in a call option has a zero payoﬀfor all stock prices equal to or less than the exercise price. 522 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES Problem 60.21 Suppose that the strike price of a call option is $80 and future value of premium at expiration is $7. Complete the following table. Spot price Payoﬀto Payoﬀto Proﬁt/Loss Proﬁt/Loss at expiration buyer seller to buyer to seller 50 60 70 80 90 100 110 Problem 60.22 The strike price of an European call option is 88 and the cost (including interest) is $3.50. For what value of the stock at maturity will the option buyer exercise the option? For what spot prices at maturity the option buyer makes a proﬁt? Problem 60.23 A European call option on Intel ordinary shares with exercise price $10.50 per share expires on June 30: (A) Permits the option writer to sell Intel shares for $10.50 on June 30. (B) Requires the option writer to buy Intel shares for $10.50 any time before or on June 30, at the discretion of the option holder. (C) Permits the option holder to buy Intel shares for $10.50 on June 30. (D) Permits the option holder to sell Intel shares for $10.50 on June 30. 61 PUT OPTIONS: PAYOFF AND PROFIT DIAGRAMS 523 61 Put Options: Payoﬀand Proﬁt Diagrams A call option gives the right to the option holder to buy or walk away. A put option gives the right, but not the obligation, to the option holder (the put buyer) to sell the underlying asset at the strike price to the option writer. Note that the buyer of a put option is the seller of the underlying asset. The put seller is obligated to buy the underlying asset if the put is exercised. In a put option, the put buyer thinks price of an asset will decrease. The buyer pays an upfront premium which he will never get back. The buyer has the right to sell the asset (to the put seller) at strike price. The put seller (or option writer) receives a premium. If buyer exercises the option, the option writer will buy the asset at strike price. If buyer does not exercise the option, the option writer’s proﬁt is just the premium. For a put option, the put buyer has a short position (since he has the right to sell the underlying asset) whereas the put seller has a long position (since he has the obligation to buy the underlying asset from the put holder). Example 61.1 Consider a put option with underlying asset a stock index and strike price of $500 in 3 months. (a) If in three months the stock index is $550, would the option holder exercise the option? (b) What if the stock index is $450? Solution. (a) If the stock index is rising the option holder will not exercise the option and he will walk away. (b) Since the stock index is below $500, the option holder will sell the stock and earn $50 Now, since a buyer of a put option will exercise the option only when the spot price is lower than the strike price, the buyer’s payoﬀis then deﬁned by the formula Buyer′s put payoﬀ= max{0, strike price −spot price at expiration}. Example 61.2 Consider a put option with a strike price of $500. (a) What would be the payoﬀto the buyer if the spot price at the expiration date is $550? (b) What would be the payoﬀto the buyer if the spot price at the expiration date is $450? Solution. (a) Since the spot price is larger than the strike price, the buyer will not exercise his option. The payoﬀin this case is max{0, 500 −550} = $0. 524 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES (b) In this case, the buyer will exercise his option and therefore the payoﬀis max{0, 500 −450} = $50 Note that the payoﬀdoes not take into consideration the cost of acquiring the position, i.e. the premium which is paid at the time the option is acquired. The proﬁt earned by the buyer is found by the formula Buyer′s put proﬁt = Buyer′s put payoﬀ−future value of premium Example 61.3 Consider a put option with strike price of $500, spot price at expiration date of $550 and premium $41.95. Assume the risk-free interest rate for the three months is 1%. Would you exercise the put option? How much would you make or lose? Solution. Since spot price is higher than the strike price, you expect not to exercise the put option. In this case your proﬁt is max{0, 500 −550} −41.95(1.01) = −$42.37. That is you will lose −$42.37 The payoﬀdiagram and the proﬁt diagram of a long put with strike price K and future value of premium Pp are given in Figure 61.1. Note that a call proﬁt increases as the value of the un-derlying asset increases whereas put proﬁt increases as the value of the underlying asset decreases. Also note that for a purchased put the maximum gain is the strike price minus the future value of the premium whereas the maximum loss is the future value of the premium. Figure 61.1 Example 61.4 Draw the proﬁt diagram of the put option of the previous example and that of a short forward with delivery price of $500 and expiration date that of the put option. Which one has higher proﬁt if the index price goes below $500? What if the index price goes up suﬃciently? 61 PUT OPTIONS: PAYOFF AND PROFIT DIAGRAMS 525 Solution. The proﬁt diagrams for both positions are shown in Figure 61.2. If the index price goes down, the short forward, which has no premium, has a higher proﬁt than the purchased put. If the index goes up suﬃciently, the put outperforms the short forward Figure 61.2 A put option is like an insured short forward. When prices go up, the losses from a short forward can be potentially unlimited. With a put the losses are limited. Up to this point we have considered the payoﬀand the proﬁt from the buyer’s perspective. Next, we consider these issues from the seller’s (i.e. the writer) perspective. The writer’s payoﬀand proﬁt are the negative of those of the buyer’s and thus they are given by Writer′s put payoﬀ= min{0, spot price at expiration −strike price} and Writer′s put proﬁt = Writer′s put payoﬀ+ future value of premium. Example 61.5 Again, consider a put option with a strike price of $500 and a premium valued at expiration date for $41.95. Find the writer’s payoﬀand the proﬁt if the spot price at the expiration date is $550. Solution. Since the spot price is higher than the strike price, the writer does not have to buy the underlying asset. His/her payoﬀis min{0, 550 −500} = $0 and the proﬁt is min{0, 550 −500} + 41.95(1.01) = $42.37 526 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES The payoﬀand proﬁt diagrams of a short put are shown in Figure 61.3. Figure 61.3 Note that for a written put the maximum gain is the future value of the premium whereas the maximum loss is the future value minus the strike price. Options as Insurance Policies Options can be used very much like insurance policies against moving prices and a way to protect physical assets already owned. For example, a call option can provide insurance against the rise in the price of something we plan to buy in the future. The long position is buying the insurance and the short position is selling the insurance. A put option can provide insurance against the fall in the price of an underlying asset. The long position is selling the insurance and the short position is buying the insurance. A homeowner insurance policy is an example of a put option. For example, suppose your house is valued at $200,000. You buy a homeowner insurance policy for the premium of $15,000 that expires in one year. The policy requires a deductible for $25,000. This means that if the damage of the house is less than $25,000 you are fully responsible for the cost. If the damage is greater than $25,000 then the insurance company will cover the portion above the $25,000. The insurance policy acts as a put option. The premium $15,000 is like the premium of a put, the expiration date is one year, and the striking price of the put is $175,000. The Moneyness of an Option In ﬁnance, “moneyness” is a measure of the degree to which a derivative is likely to have positive monetary value at its expiration. From the buyer perspective and neglecting the premium for buying an option, an option is said to be in-the-money if the buyer proﬁts when the option is exercised immediately. For example, a call option with spot price larger than the strike and a put option with strike price larger than a 61 PUT OPTIONS: PAYOFF AND PROFIT DIAGRAMS 527 spot price are both in-the-money. On the other hand, an option is said to be out-of-the-money option if the buyer loses when the option is exercised immediately. A call option with spot price less than the strike price and a put option with strike price less than the spot price are both out-of-the money. An option is said to be at-the-money option if the buyer does not lose or proﬁt when the option is exercised immediately. This occurs when the spot price is approximately equal to the strike price. Figure 61.4 illustrates these concepts. Figure 61.4 Example 61.6 If the underlying stock price is $25, indicate whether each of the options below is in the money, at the money, or out of the money. Strike Call Put $20 $25 $30 Solution. Strike Call Put $20 In the money out of the money $25 At the money At the money $30 Out of the money In the money 528 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES Practice Problems Problem 61.1 Suppose you have 5,000 shares worth $25 each. How can put options be used to provide you with insurance against a decline in the value of your holdings over the next four months? Problem 61.2 Consider a put option whose underlying asset is a stock index with 6 months to expiration and a strike price of $1000. (a) What is the buyer’s payoﬀif the index price is $1100 in 6 months? (b) What is the buyer’s payoﬀif the index price is $900 in 6 months? Problem 61.3 Consider a put option whose underlying asset is a stock index with 6 months to expiration and a strike price of $1000. Suppose the risk-free interest rate for the six months is 2% and that the option’s premium is $74.20. (a) Find the future premium value in six months. (b) What is the buyer’s proﬁt is the index spot price is $1100? (c) What is the buyer’s proﬁt is the index spot price is $900? Problem 61.4 A put is trading on Sony Corporation stock. It has a strike price of $40. (a) What is the payoﬀat expiration of this put if, on the expiration date, Sony stock sells for $45? (b) What is the payoﬀat expiration of this call if, on the expiration date, Sony stock sells for $25? (c) Draw the payoﬀdiagram for this option. Problem 61.5 A put option in which the stock price is $60 and the exercise price is $65 is said to be (a) in-the-money (b) out-of-the-money (c) at-the-money (d) none of the above Problem 61.6 What position is the opposite of a purchased put option? Problem 61.7 Suppose that you own 5000 shares worth $30 each.How can put options be used to provide you with insurance against a decline in the value of your holding over the next four months? 61 PUT OPTIONS: PAYOFF AND PROFIT DIAGRAMS 529 Problem 61.8 A Swiss company knows that it is due to receive $1,400,000 in (exactly) one year. What type of option contract is appropriate for hedging? Problem 61.9 A trader buys a European call option and sells a European put option. The options have the same underlying asset, strike price, and maturity date. Describe the trader’s position. Problem 61.10 Explain carefully the diﬀerence between selling a call option and buying a put option. Assume European options. Problem 61.11 Suppose that you write a put contract with a strike price of $40 and an expiration date in 3 months. The current stock price is $41 and the contract is on 100 shares. What have you committed yourself to? How much could you gain or lose? Problem 61.12 Describe the moneyness of the following options, i.e. as in-the-money, out-of-the-money or at-the-money: (a) A call or a put option with spot price of $100 at expiration and strike price of $100. (b) A call option with a spot price of $100 at expiration and strike price of $80. (c) A put option with a spot price of $100 at expiration and strike price of $80. (d) A call option with a spot price of $100 at expiration and strike price of $120. (e) A put option with a spot price of $100 at expiration and strike price of $120. Problem 61.13 Suppose the stock price is $40 and the eﬀective annual interest rate is 8%. (a) Draw the payoﬀdiagrams on the same window for the options below. (b) Draw the proﬁt diagrams on the same window for the options below. (i) 35-strike put with a premium of $1.53. (ii) 40-strike put with a premium of $3.26. (iii) 45-strike put with a premium of $5.75. Problem 61.14 You are a speculator and you think stock prices will increase. Should you buy a call or a put option ? 530 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES Problem 61.15 A call option and a put option of a stock index have an exercise price of $60. Complete the following table of payoﬀs. stock price Buy call Write call Buy put Write put $30 $40 $70 $900 Problem 61.16 A trader writes a December put option with a strike price of $30. The premium of the option is $4. Under what circumstances does the trader make a gain? Ignore the time value of money. Problem 61.17 Complete the following table. Derivative Maximum Minimum Position with Respect Strategy Position Loss Gain to Underlying Asset Long Forward Short Forward Long Call Short Call Long Put Short Put Problem 61.18 The payoﬀof an out-of-the-money long put is equal to A. the stock price minus the exercise price. B. the put premium. C. zero. D. the exercise price minus the stock price. E. none of the above. Problem 61.19 An option whose striking price is above the stock price is A. out-of-the-money. B. in-the-money. C. at-the-money. D. cannot be determined. 61 PUT OPTIONS: PAYOFF AND PROFIT DIAGRAMS 531 Problem 61.20 Options are not normally A. exercised early. B. exercised after expiration. C. exercised when out-of-the-money. D. All of the above Problem 61.21 If you write a put option, A. the maximum proﬁt is unlimited. B. the maximum loss is unlimited. C. the maximum gain equals the premium. D. the maximum gain equals the stock price minus the striking price. Problem 61.22 If someone writes a put, they usually want the market to A. go up. B. go down. C. stay unchanged. D. ﬂuctuate. Problem 61.23 An American put option on Intel ordinary shares with exercise price $10.50 per share expires on June 30: (A) Permits the option holder to sell Intel shares for $10.50 any time before or on June 30. (B) Permits the option writer to sell Intel shares for $10.50 on June 30. (C) Requires the option writer to sell Intel shares for $10.50 on June 30, at the discretion of the option holder. (D) Requires the option holder to buy Intel shares for $10.50 any time before or on June 30, at the discretion of the option writer. Problem 61.24 A stock is trading at a price of $20 at expiration date. Which of the following statement about options would be correct? (i) A put option with a strike of $17.50 would be in-the-money. (ii) A put option with a strike of $20 would be at-the-money. (iii) A put option with a strike price of $25 would be in-the-money. 532 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES (iv) A call option with a strike price of $20 would be in-the-money. (v) A call option with a strike price of $17.50 would be out-of-the-money. (A) i, iv and v only. (B) i, ii, iv and v only. (C) ii, iv and v only. (D) ii and iii only. 62 STOCK OPTIONS 533 62 Stock Options Options can have various underlying securities, namely: • Index Options: An index option contract is where the underlying security is not an individual share but an index (i.e. a collection of stocks) such as the S& P 500 index or the Nasdaq Compsoite index. Thus the buyer of a call option has a right to buy the index at a predetermined price on or before a future date. All index option contracts are cash settled. • Futures Options: An option with underlying asset a futures contract. • Foreign currency options: An option which gives the owner the right to buy or sell the indicated amount of foreign currency at a speciﬁed price before or on a speciﬁc date. • Stock options. In this section we consider stock options and the main factors that an investor should be aware oﬀwhen trading with stock options. A stock option is the right to sell or buy a speciﬁed number of shares of a stock at a speciﬁc price (the strike price) and time (the expiration date.) Stock options are the most popular long-term incentive compensation approach used in U.S. companies. A stock option contract is a contract representing 100 shares in the underlying stock. For example, if you want to own 1000 shares of Microsoft then you either can buy the 1000 shares at the stock exchange or you buy 10 Microsoft option contracts at the options exchange. A stock option’s expiration date can be up to nine months from the date the option is ﬁrst listed for trading. Longer-term option contracts, called LEAPS (Long-Term Equity AnticiPation Securities), are also available on many stocks, and these can have expiration dates up to three years from the listing date. In practice, stock options expire on the third Saturday of the expiration month but the last day of trading is the third Friday of the expiration month. Buying and selling stock options are not made directly between the buyer (holder) and the seller (writer) of the stock option but rather through a brokerage ﬁrm. In reality, an organization called the OCC or Options Clearing Corporation steps in between the two sides. The OCC buys from the seller and sells to the buyer. This makes the OCC neutral, and it allows both the buyer and the seller to trade out of a position without involving the other party. Example 62.1 The following is listed in an edition of the Wall Street Journal: IBM Oct 90 Call at $2.00. Suppose you buy one contract. (a) What type of option is considered in the listing? (b) What is the strike price of a share of the stock? (c) When does the option expires? (d) If you own the option and you want to sell, how much will you receive? (e) If you are buying the option, how much will it cost you? (f) Suppose you hold the option and you decide to exercise, how much will you pay? 534 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES Solution. (a) This is a call stock option. (b) The strike price is $90 per share. (c) It expires on the Saturday following the third Friday of October in the year it was purchased. (d) You will receive $2 × 100 = $200−commission. (e) The cost of buying the option is $2 × 100+ brokerage commission. (f) You have to pay $90 × 100+ brokerage commission In this section we consider the basic issues that an investor must be aware of when considering buying stock options, namely, dividends, exercise, margins, and taxes. The Dividend Eﬀect Dividends that might be paid during the life of a stock option can aﬀect the stock price. We may regard dividends as a cash return to the investors. The company has the choice to either paying dividends to the shareholders or reinvesting that money in the business. The reinvestment of that cash could create more proﬁt for the business and therefore leading to an increase in stock price. On the other hand, paying dividends eﬀectively reduce the stock price by the amount of the dividend payment. Such a drop has an adverse eﬀect on the price of a call option and a beneﬁcial eﬀect on the price of a put option. Since the payoﬀof a call option is the maximum of zero or the stock price minus the exercise price, a drop in the stock price will lower the payoﬀof a call option at expiration date. On the contrary, since the payoﬀof a put option is the maximum of zero or the exercise price minus the stock price, lowering the stock price will increase the return on a put option. Example 62.2 Consider a call option and a put option with the same expiration date and the same strike price of $100. (a) Assume no dividends occurred during the life of the options, ﬁnd the payoﬀof both options if the stock price is $102 at expiration. (b) Suppose that a dividend of $3 was paid just before expiration. What will be the stock price at expiration? (c) Find the payoﬀof both options after the dividend was paid. Solution. (a) The payoﬀof the call option is max{0, 102 −100} = $2 while that of the put option is max{0, 100 −102} = $0. 62 STOCK OPTIONS 535 (b) The stock price will decline by $3 so that the stock price at expiration is $99. (c) The call option’s payoﬀis max{0, 99 −100} = $0 and the put’s option payoﬀis max{0, 100 −99} = $1 The Exercising Eﬀect After buying an option, an investor must provide exercise instructions to the broker’s deadline. Failing to do so, the option will expire worthless. However, this is not required for cash-settled options where the option is automatically exercised at the expiration date. Exercising an option usually results in a commission paid to the broker. Exercising a put option results in an additional commission for selling the underlying shares. Therefore, if a holder of a stock option does not wish to own the shares of the stock he is better of selling the option instead of exercising it. When an option is exercised the writer of the option is obligated to sell the underlying assets (in the case of a call option) or buy the underlying assets (in the case of a put option). He is said to have been assigned. Assignment results in a commision to be paid by the option’s writer. Dividends are one factor that can aﬀect the exercise decision. Since American stock options can be exercised at anytime before the expiration date, dividends make early exercise more likely for American calls and less likely for American puts. Example 62.3 An investor buys a European put on a share for $3. The strike price is $40. (a) Under what circumstances will the option be exercised? (b) Under what circumstances does the investor make a proﬁt? (c) Draw a diagram showing the variation of the investor’s proﬁt with the stock price at the maturity of the option. Solution. (a) The option will be exercised if the stock price is less than $40 on the expiration date. (b) The investor makes a proﬁt if the stock price on the expiration date is less than $37, because the gain from exercising the option is greater than $3. Taking into account the initial cost of the option ($3), the proﬁt will be positive. (c) The graph of the investor’s proﬁt as a function of the stock price at the maturity of the option is shown in Figure 62.1. 536 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES Figure 62.1 Margins for Written Options By a margin we mean a collateral deposited into a margin account to insure against the possibility of default. The writer of an option is in a short position. Thus, an investor is required to post a collateral with his brokerage ﬁrm to insure against the possibility of default (i.e. not being able to fulﬁll his obligations toward the option buyer). Writers of options should determine the applicable margin requirements from their brokerage ﬁrms and be sure that they are able to meet those requirements in case the market turns against them. Example 62.4 Explain why brokers require margins when the clients write options but not when they buy options. Solution. When an investor buys an option, he must pay it upfront. There are no future liabilities and, therefore, no need for a margin account. When an investor writes an option, there are potential future liabilities. To protect against the risk of a default, margins are required Tax Considerations when Trading Stock Options Tax rules for derivatives are very complicated and change very frequently. Brieﬂy put, under current IRS rules, both stock and stock options are capital assets, and any net gain (i.e. gross gain −costs) on their sale is taxable. If the stock or option was held for less than a year, then ordinary income tax rates apply to the gain. But if the stock or option has been held for more than a year, the holder gets a tax break on the sale, because the gain is taxed at the long-term capital gains rate. Since all stock options (except LEAPS options) have a duration of 9 months or less you cannot hold them for a year or more, and so tax on option gains will usually be at the short-term (ordinary income tax) rates. All short-term gains and losses are combined on the investor tax return. One category of tax rules that can apply when trading with stocks is the one that governs con-structive sales. If you own a stock and you have entered into a transaction such as a short sale or a 62 STOCK OPTIONS 537 forward contract hoping to reduce your risk from a ﬂuctuating market, the IRS consider that you have made a sale referred to as a constructive sale and as a result your gain or loss are subject to taxes. We illustrate this concept in the following example. Example 62.5 Suppose you own 100 shares of ABC stocks that you bought for $20 per share. So, you have a $2,000 basis in your stock. Currently, ABC stock has reached a peak of $50 a share. So you decide to make a short sale. You borrow 100 shares from a broker and sell them for $50 per share, getting a total of $5,000. Now, you have to give 100 shares back to your broker to close the transaction. So, if the stock price goes up between the time that you sold the borrowed shares and the time that you must give the broker back 100 shares, you just give the broker the 100 shares that you originally own and you are done with it. You have made $3000. If, instead, the stock price goes down in between the time you sold the borrowed shares and the day you must give the broker back his 100 shares you would not give him your original 100 shares, but instead go out to the open market and buy 100 shares and give those new ones to the broker. In this scenario, then, you would make a proﬁt and he would still have your shares to hold onto for long-term. In both scenarios, your gain is taxable. 538 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES Practice Problems Problem 62.1 Assuming you own XY Z company shares trading at $40 right now. You bought a put stock options contract that allows you to sell your XY Z shares at $40 anytime before it expires in 2 months. 1 month later, XY Z company shares are trading at $30. If you decide to exercise the option, at what price will you be selling each share? Problem 62.2 Explain the meaning of “XYZ April 25 Call.” Problem 62.3 Which is more expensive a paying dividend call option or a paying dividend put option? Problem 62.4 An investor buys a European call on a share for $4. The strike price is $50. (a) Under what circumstances will the option be exercised? (b) Under what circumstances does the investor make a proﬁt? (c) Draw a diagram showing the variation of the investor’s proﬁt with the stock price at the maturity of the option. Problem 62.5 A stock price is S just before a dividend D is paid. What is the price immediately after the payment? (Assume that income tax on dividend income is zero.) Problem 62.6 When would be the best time to exercise an American call on a dividend-paying stock? Problem 62.7 A company has 400,000 shares and the stock price is currently $40 per share. If the company issues 25% stock dividend (i.e. for every 100 shares of stock owned, 25% stock dividend will yield 25 extra shares), what would you expect the stock price to be after the dividend is paid? Problem 62.8 In January, 1995, you bought 100 shares of XYZ stock at $5 a share. One day in March, 1995, when the stock was worth $10 a share, you borrowed 100 shares from your broker, sold those, and pocketed the $1000. You had, so far, made $500. But eventually you would have to give the stocks back to the broker. (a) Early in 1996, the stock went up to $15 a share. What would you do to give the broker back his shares? (b) Are you supposed to report the gain to the IRS? 62 STOCK OPTIONS 539 Problem 62.9 A LEAP is a A. commodity option. B. gold or silver option. C. long-term option. D. option with a high striking price. Problem 62.10 The guarantor of option trades is the A. SEC B. OCC C. CFTC D. FDIC Problem 62.11 LEAPS are issued with durations of all of the following except A. 1 year B. 2 years C. 3 years D. 10 years Problem 62.12 When is the last trading day for an option? (A) The last business day of the expiry month. (B) The last Friday of the expiry month. (C) The ﬁrst business day of the expiry month. (D) The third Friday of the expiry month. 540 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES 63 Options Strategies: Floors and Caps In this section we discuss the strategy of using options to insure assets we own (or purchase) or assets we short sale. Insuring Owned Assets with a Purchased Put: Floors An investor who owns an asset (i.e. being long an asset) and wants to be protected from the fall of the asset’s value can insure his asset by buying a put option with a desired strike price. This combination of owning an asset and owning a put option on that asset is called a ﬂoor. The put option guarantees a minimum sale price of the asset equals the strike price of the put. To examine the performance of this strategy, we look at the combined payoﬀand proﬁt of the asset position and the put as illustrated in the example below. Example 63.1 Suppose you buy an index valued at $500. To insure your index from value decline you buy a 500-strike 3-month put with risk-free 3-month rate of 1% and a premium of $41.95. (a) Complete the table below where the values are assumed at the expiration date. Index Payoﬀ Put Payoﬀ Combined Payoﬀ Combined Proﬁt 400 450 500 550 600 650 700 (b) Graph the payoﬀand proﬁt diagrams for the combined position. (c) Compare the payoﬀdiagram of the combined position with the payoﬀof a long 500-strike 3-month call with risk-free 3-month interest of 1% and a premium of $46.90. (d) Compare the proﬁt diagram of the combined position with the proﬁt diagram of a long 500-strike 3-month call with risk-free 3-month rate of 1% and a premium of $46.90. (e) Compare the payoﬀand proﬁt diagrams of the combined position with the payoﬀand proﬁt diagrams of the combined position of a 3-month 500-strike call (with 3-month risk-free interest of 1% and premium 0f $46.90) and a zero-coupon bond that pays $500 in three months (with present cost of 500(1.01)−1 = $495.05). Solution. Note that the proﬁt of the combined position is the proﬁt from the long index plus the proﬁt from the long put. That is, the combined proﬁt is 63 OPTIONS STRATEGIES: FLOORS AND CAPS 541 Index payoﬀ−500 × 1.01 + Put payoﬀ−41.95 × 1.01 = Combined payoﬀ−$547.37. (a) Recall that the (long) index payoﬀis the spot price of the index and the (long) put payoﬀis equal to max{0, strike price −spot price}. Index Payoﬀ Put Payoﬀ Combined Payoﬀ Combined Proﬁt 400 100 500 −47.37 450 50 500 −47.37 500 0 500 −47.37 550 0 550 2.63 600 0 600 52.63 650 0 650 102.63 700 0 700 152.63 (b) Figure 63.1 shows the payoﬀand the proﬁt diagrams of the combined position. Figure 63.1 Notice that the level of the ﬂoor is −$47.37, which is the lowest possible proﬁt (a loss in this case). (c) We notice from Figure 63.1(a) that the payoﬀdiagram of the combined position is a vertical shift of the long call (compare with Figure 60.1). (d) The proﬁt diagram in Figure 63.1(b) is identical to the proﬁt diagram of a 3-month 500-strike call with 3-month risk-free interest of 1% and premium of $46.90 (See Figure 60.1). It follows that the cash ﬂows in purchasing a call are diﬀerent from the cash ﬂows of buying an asset and insuring it with a put, but the proﬁt for the two positions is the same. (e) Recall that a long call payoﬀis equal to max{0, spot price −strike price}. Also, the combined proﬁt is Call payoﬀ−46.90 × 1.01 + Bond payoﬀ−495.05 × 1.01 = Combined payoﬀ−$547.37 542 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES We have the following table for the combined position of the 3-month 500-strike call with a zero-coupon bond that pays $500 in three months. spot Price Call Payoﬀ Bond’s Payoﬀ Combined Payoﬀ Combined Proﬁt 400 0 500 500 −47.37 450 0 500 500 −47.37 500 0 500 500 −47.37 550 50 500 550 2.63 600 100 500 600 52.63 650 150 500 650 102.63 700 200 500 700 152.63 Thus, the combined position of index plus put is equivalent to the combined position of a zero-coupon bond plus call Insuring a Short Sale: Caps When you short an asset, you borrow the asset and sell, hoping to replace them at a lower price and proﬁt from the decline. Thus, a short seller will experience loss if the price rises. He can insure his position by purchasing a call option to protect against a higher price of repurchasing the asset. This combination of short sale and call option purchase is called a cap. To examine the performance of this strategy, we look at the combined payoﬀand proﬁt of the asset position and the call as illustrated in the example below. Example 63.2 Suppose you short an index valued at $500. To insure your index from value increase you buy a 500-strike 3-month call with risk-free 3-month rate of 1% and a premium of $46.90. (a) Complete the table below where the values are assumed at the expiration date. Short Index Payoﬀ Call Payoﬀ Payoﬀ Accumulated Cost Proﬁt −400 −450 −500 −550 −600 −650 −700 (b) Graph the payoﬀand proﬁt diagrams for the combined position. (c) Compare the payoﬀdiagram of the combined position with the payoﬀdiagram of a purchased 500-strike put option with expiration date of three months, risk-free 3-month interest of 1% and 63 OPTIONS STRATEGIES: FLOORS AND CAPS 543 premium $41.95. (d) Compare the proﬁt diagram of the combined position to the proﬁt diagram of a 500-strike put with expiration date of three months, risk-free 3-month interest of 1% and premium of $41.95. (e) Compare the payoﬀand proﬁt diagram of the combined position with the payoﬀand proﬁt dia-grams of a purchased 500-strike put option with expiration date of three months, risk-free 3-month interest of 1% and premium $41.95 coupled with borrowing 500(1.01)−1 = $495.05. Solution. (a) Recall that the short index payoﬀis the negative of the long index position. Also, recall that (long) call payoﬀis max{0, spot price −strike price}. The proﬁt of the combined position is the proﬁt from the short index plus the proﬁt from the long call. That is, the combined proﬁt is −long index payoﬀ+ 500 × 1.01 + long call payoﬀ−46.90 × 1.01 = Combined payoﬀ+ $457.63 Short Index Payoﬀ Call Payoﬀ Combined Payoﬀ Combined Proﬁt −400 0 −400 57.63 −450 0 −450 7.63 −500 0 −500 −42.37 −550 50 −500 −42.37 −600 100 −500 −42.37 −650 150 −500 −42.37 −700 200 −500 −42.37 (b) Figure 63.2 shows the payoﬀand the proﬁt diagrams of the combined position. Figure 63.2 544 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES (c) By comparing Figure 63.2 with Figure 61.1, we note that the payoﬀdiagram of the combined position generates a position that looks like a long put. (d) The proﬁt diagram in Figure 63.2(b) is identical to the proﬁt diagram of a 3-month 500-strike put with 3-month risk-free interest of 1% and premium of $41.95 (See Figure 61.2). (e) Recall that a long put payoﬀis equal to max{0, strike price −spot price} and the loan payoﬀis −$500 so that the combined payoﬀis max{0, strike price −spot price} −500. Since borrowing has no eﬀect on the proﬁt, the combined proﬁt is just Put payoﬀ−41.95 × 1.01 = Put payoﬀ−$42.37 We have the following table for the combined position of the 3-month 500-strike put with borrowing $495.05. Spot Price Put Payoﬀ Combined Payoﬀ Combined Proﬁt 400 100 −400 57.63 450 50 −450 7.63 500 0 −500 −42.37 550 0 −500 −42.37 600 0 −500 −42.37 650 0 −500 −42.37 700 0 −500 −42.37 Thus, the combined position of short sale plus call is equivalent to the combined position of a purchased put coupled with borrowing 63 OPTIONS STRATEGIES: FLOORS AND CAPS 545 Practice Problems Problem 63.1 Suppose a speculator is short shares of stock. Which of the following would be the best hedge against the short position? A. buying a put B. writing a put C. buying a call D. writing a call Problem 63.2 Buying a long position in the underlying asset and a long put resembles (A) Buying a long position in the underlying asset and a long call (B) Buying a long call (C) Buying a short call (D) Buying a long position in the underlying asset and short call. Problem 63.3 Shorting an index and a long call resembles (A) Buying a long position in the underlying asset and a long call (B) Buying a long call (C) Buying a long put (D) Shorting the index and buying and long put. Problem 63.4 Consider the following combined position: • buy an index for $500 • buy a 500-strike put with expiration date in 3 months, 3-month risk free rate of 1% and premium of $41.95 • borrow $495.05 with 3-month interest rate of 1%. Graph the payoﬀdiagram and proﬁt diagram of this position. Problem 63.5 Suppose you buy an index valued at $1000. To insure your index from value decline you buy a 1000-strike 6-month put with risk-free 6-month rate of 2% and a premium of $74.20. (a) Construct the payoﬀdiagram and the proﬁt diagram of this combined position. (b) Verify that you obtain the same payoﬀdiagram and proﬁt diagram by investing $980.39 in zero coupon bond and buying a 1000-strike 6-month call with risk-free 6-month rate of 2% and a premium of $93.81. 546 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES Problem 63.6 Suppose you short an index valued for $1000. To insure your index from value increase you buy a 1000-strike 6-month call with risk-free 6-month rate of 2% and a premium of $93.81. (a) Graph the payoﬀand proﬁt diagrams for the combined position. (b) Verify that you obtain the same payoﬀdiagram and proﬁt diagram by purchasing a 1000-strike put option with expiration date of six months, risk-free 6-month interest of 2% and premium $74.20 coupled with borrowing $980.39. 64 COVERED WRITINGS: COVERED CALLS AND COVERED PUTS 547 64 Covered Writings: Covered Calls and Covered Puts Writing an option backed or covered by the underlying asset (such as owning the asset in the case of a call or shorting the asset in the case of a put) is referred to as covered writing or option overwriting. The most common motivation for covered writing is to generate additional income by means of premium. In contrast to covered writing, when the writer of an option has no position in the underlying asset we refer to this as naked writing. In this section we discuss two types of cov-ered writings: Covered calls and covered puts. Note that covered writing resembles selling insurance. Writing a covered call A covered call is a call option which is sold by an investor who owns the underlying assets. An investor’s risk is limited when selling a covered call since the investor already owns the underlying asset to cover the option if the covered call is exercised. By selling a covered call an investor is attempting to capitalize on a neutral or declining price in the underlying stock. When a covered call expires without being exercised (as would be the case in a declining or neutral market), the investor keeps the premium generated by selling the covered call. The opposite of a covered call is a naked call, where a call is written without owned assets to cover the call if it is exercised. Example 64.1 Suppose you buy an index valued for $500 and sell a 500-strike 3-month call with risk-free 3-month rate of 1% and a premium of $46.90. (a) Construct the payoﬀdiagram and the proﬁt diagram of this covered call. (b) Verify that you obtain the same proﬁt diagram by selling a 500-strike 3-month put with risk-free 3-month rate of 1% and a premium of $41.95. Solution. (a) Recall that the long index payoﬀis the same as the spot index at expiration. Also, recall that the written call payoﬀis −max{0, spot price −strike price}. The proﬁt of the combined position is the proﬁt from the long index plus the proﬁt from the written call. That is, the combined proﬁt is Long index payoﬀ−500 × 1.01 −max{0, spot price −strike price} + 46.90 × 1.01 = Combined payoﬀ- $457.63 Long Index Payoﬀ Written Call Payoﬀ Combined Payoﬀ Combined Proﬁt 400 0 400 −57.63 450 0 450 −7.63 500 0 500 42.37 550 −50 500 42.37 600 −100 500 42.37 650 −150 500 42.37 700 −200 500 42.37 548 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES Figure 64.1 shows the payoﬀand the proﬁt diagrams of the combined position. Figure 64.1 (b) Recall that the proﬁt of a written put is −max{0, strike price −spot price} + 41.95 × 1.01. The proﬁt diagram is given in Figure 64.2. Figure 64.2 Thus selling a covered call has the same proﬁt as selling the put in which we don’t own the under-lying asset . Writing a covered put A covered put is a put option which is sold by an investor and which is covered (backed) by a short sale of the underlying assets. A covered put may also be covered by deposited cash or cash equivalent equal to the exercise price of the covered put. Since a covered put is covered (or backed) in advance by assets or cash, a covered put represents a known and limited risk should the holder of the put choose to exercise the option of the covered put. The opposite of a covered put would be an uncovered or naked put. Example 64.2 An investor shorts an index for $500 and sell a 500-strike put with expiration date in 3 months, 64 COVERED WRITINGS: COVERED CALLS AND COVERED PUTS 549 3-month risk free rate of 1% and premium of $41.95 (a) Graph the payoﬀdiagram and proﬁt diagram of this covered put. (b) Verify that the proﬁt diagram coincides with the proﬁt diagram of a 500-strike written call with expiration date in 3 months, 3-month risk free rate of 1% and premium of $46.90. Solution. (a) The combined payoﬀis −Long index payoﬀ−max{0, strike price −spot price} The combined proﬁt is −Long index payoﬀ+500 × 1.01 −max{0, 500 −spot price} + 41.95 × 1.01 or −Long index payoﬀ−max{0, 500 −spot price} + $547.37 The following table summarizes the combined payoﬀand the combined proﬁt. Short Index Payoﬀ Written Put Payoﬀ Combined Payoﬀ Combined Proﬁt −400 −100 −500 47.37 −450 −50 −500 47.37 −500 0 −500 47.37 −550 0 −550 −2.63 −600 0 −600 −52.63 −650 0 −650 −102.63 700 0 −700 −152.63 The payoﬀdiagram and the proﬁt diagram are shown in Figure 64.3. Figure 64.3 550 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES (b) Recall that the proﬁt of a written call is given by −max{0, spot price at expiration−strike price}+ 46.90(1.01) = −max{0, spot price at expiration−strike price}+47.37. The proﬁt diagram coincides with the one in Figure 64.3 Example 64.3 An investor sells short 100 shares of ABC stock at $9.25 a share. He sells one put contract (100 shares) with a striking price of $10 and a premium of $1.50 a share. (a) What is the investor’s proﬁt if the spot price is below $10 at expiration? (b) The investor’s will be making proﬁt as long as the spot price is less than what value at expiration? Solution. (a) If the spot price at expiration is below $10 the put holder will exercise the option. In this case, the investor will buy the shares for $10 a share and give it back to the lender. The investor’s proﬁt is then 9.25 × 100 + 100 × 1.50 −10 × 100 = $75. (b) The investor will be making proﬁt as long as the spot price at expiration is less than $10.75 64 COVERED WRITINGS: COVERED CALLS AND COVERED PUTS 551 Practice Problems Problem 64.1 If someone writes a call while owning the underlying asset, the call is (A) covered (B) long (C) naked (D) cash-secured Problem 64.2 A covered call position is (A) the simultaneous purchase of the call and the underlying asset. (B) the purchase of a share of stock with a simultaneous sale of a put on that stock. (C) the short sale of a share of stock with a simultaneous sale of a call on that stock. (D) the purchase of a share of stock with a simultaneous sale of a call on that stock. (E) the simultaneous purchase of a call and sale of a put on the same stock. Problem 64.3 A covered put position is (A) the simultaneous purchase of the put and the underlying asset. (B) the short sale of a share of stock with a simultaneous sale of a put on that stock. (C) the purchase of a share of stock with a simultaneous sale of a call on that stock. (D) the simultaneous purchase of a call and sale of a put on the same stock. Problem 64.4 If a person writes a covered call with a striking price of $45 and receives $3 in premium, exercise will occur if the stock price is above on expiration day. (A) $42 (B) $45 (C) $48 (D) $50 Problem 64.5 If a stock is purchased at $50 and a $55 call is written for a premium of $2, the maximum possible gain per share is (neglecting interest) (A) $2 (B) $5 (C) $7 (D) $10 552 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES Problem 64.6 If someone writes a covered call, they usually want the market to (A) go up. (B) go down or stay unchanged. (C) ﬂuctuate. Problem 64.7 The most common motivation for option overwriting is (A) risk management. (B) tax reduction. (C) leverage. (D) income generation. Problem 64.8 An investor buys 500 shares of ABC stock at $17 a share. He sells 5 call contracts (100 shares each) with a striking price of $17.50 and premium of $2 a share. (a) What is the investor’s initial investment? (b) What will be the investor’s proﬁt at expiration if the spot price of the stock is $17? Ignore interest. (c) What will be the investor’s proﬁt at expiration if the spot price of the stock is $18 if the calls are exercised? (d) What will be the investor’s proﬁt at expiration if the spot price of the stock is $16 and the investor sells his shares? Problem 64.9 An investor sells short 200 shares of ABC stock at $5.25 a share. He sells two put contracts (100 shares each) with a striking price of $5 and a premium of $0.50 a share. (a) What is the investor’s proﬁt if the spot price is below $50 at expiration? (b) The investor’s will be making proﬁt as long as the spot price is less than what value at expiration? 65 SYNTHETIC FORWARD AND PUT-CALL PARITY 553 65 Synthetic Forward and Put-Call Parity A synthetic forward is a combination of a long call and a short put with the same expiration date and strike price. Example 65.1 An investor buys a 500-strike call option with expiration date in three months, risk-free 3-month interest of 1% and a premium of $46.90. He also sells a 500-strike put option with expiration date in three months, risk-free 3-month interest of 1% and a premium of $41.95. (a) Show that the investor is obliged to buy the index for $500 at expiration date. (b) Draw the proﬁt diagram of the long call, the short put, and the combined position of the long call and the written put. (c) State two diﬀerences between the synthetic long forward and the actual long forward. Solution. (a) If the spot price at expiration is greater than $500, the put will not be exercised (and thus expires worthless) but the investor will exercise the call. So the investor will buy the index for $500. If the spot price at expiration is smaller than $500, the call will not be exercised and the investor will be assigned on the short put, if the owner of the put wishes to sell then the investor is obliged to buy the index for $500. Either way, the investor is obliged to buy the index for $500 and the long call-short put combination induces a long forward contract that is synthetic since it was fabricated from options. (b) The payoﬀof the combined position is max{0, spot price −strike price}−max{0, strike price −spot} = spot price −strike price. But this is the payoﬀof a long forward contract expiring at time T and with forward price of $500. The proﬁt for the combined position is max{0, spot −strike}−46.90(1.01)−max{0, strike −spot}+ 41.95(1.01). The proﬁt diagram of each position is given in Figure 65.1. Figure 65.1 554 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES Thus, a synthetic long forward is equivalent to a long forward with forward price of $505 (= 500(1.01)). (c) The actual forward (i.e. the long forward with forward price $505) premium is zero whereas there is a net option premium of 46.90 −41.95 = $4.95 with the synthetic forward. At expiration, we pay the forward price (i.e. $505) with the actual forward whereas with the synthetic forward we pay the strike price of the options (i.e. $500) Let F0,T denote the no-arbitrage (i.e. zero cost) forward price . This means we pay $0 today and at time T we are obliged to buy the asset at the price F0,T. The cost of the contract today is the present value of F0,T which we will denote by PV (F0,T). Note that PV (F0,T) = S0, where S0 is the current asset price. Let Call(K, T) and Put(K, T) denote the premiums of the purchased call and the written put respectively. We pay Call(K, T) −Put(K, T) today to buy the assets for K at time T which gives a cost at time 0 of Call(K, T) −Put(K, T) + PV (K). Using the no-arbitrage pricing, the net cost of asset must be the same whether through options or forward contract, that is, Call(K, T) −Put(K, T) + PV (K) = PV (F0,T) or Call(K, T) −Put(K, T) = PV (F0,T −K) (65.2) Equation (65.2) is known as put-call parity. This is an important relationship; it “ties” the option and forward markets together. Note that if the strike price of a synthetic forward equals the no-arbitrage forward price, then the net option premium is zero. In this case, the premium paid for the purchased call equals the premium received for the written put. Note that a synthetic long forward is a forward contract with a premium as opposed to an actual forward contract which does not require a premium. A forward contract with a premium is called an oﬀ-market forward. Thus, unless the strike price and the no-arbitrage forward price are equal, buying a call and selling a put creates an oﬀmarket forward. Example 65.2 The premium of a 6-month oﬀ-market (synthetic) forward contract with a forward price of $1000 is $19.61. The premium of a 6-month 1000-strike call is $X and that of a 6-month 1000-strike put is $74.20, determine X. Solution. Using Equation (65.2) we obtain X −74.20 = 19.61. 65 SYNTHETIC FORWARD AND PUT-CALL PARITY 555 Solving for X we ﬁnd X = $93.81 Equation (65.2) can be rearranged to show the equivalence of the prices (and payoﬀs and prof-its) of a variety of diﬀerent combinations of positions. Example 65.3 Show that buying an index plus a put option with strike price K is equivalent to buying a call option with strike price K and a zero-coupon bond with par value of K. Solution. The cost of the index is PV (F0,T) and that of the put option is Put(K, T). The cost of the call option is Call(K, T) and that of the zero-coupon bond is PV (K). From Equation (65.2) we can write PV (F0,T) + Put(K, T) = Call(K, T) + PV (K) which establishes the equivalence of the two positions Example 65.4 Show that shorting an index plus buying a call option with strike price K is equivalent to buying a put option with strike price K and taking out a loan with maturity value of K. Solution. The cost of the short index is −PV (F0,T) and that of the call option is Call(K, T). The cost of the put option is Put(K, T) and the loan is −PV (K). From Equation (65.2) we can write −PV (F0,T) + Call(K, T) = Put(K, T) −PV (K) which establishes the equivalence of the two positions Remark 65.1 Recall that neither a zero-coupon bond nor borrowing aﬀect the proﬁt function. Example 65.5 A call option on XYZ stock with an exercise price of $75 and an expiration date one year from now is worth $5.00 today. A put option on XYZ stock with an exercise price of $75 and an expiration date one year from now is worth $2.75 today. The annual risk-free rate of return is 8% and XYZ stock pays no dividends. Find the current price of the stock. Solution. Using the Put-Call parity we have PV (F0,T) = Call(K, T) −Put(K, T) + 75(1.08)−1 = $71.69 556 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES Example 65.6 A short synthetic forward contract can be created by reversing the long synthetic forward. This is done by combining a purchased put with a written call. Let K be the strike price and T the expiration time on both options. Show that the payoﬀfor the short synthetic forward contract is given by K −PT. Solution. The payoﬀof the short synthetic forward is the payoﬀof the purchased put plus the payoﬀof the written call which is given by the expression max{K −PT, 0} −max{K −Pt, 0}. Whether PT ≥K or PT < K one can easily check that the payoﬀis K −PT 65 SYNTHETIC FORWARD AND PUT-CALL PARITY 557 Practice Problems Problem 65.1 If a synthetic forward contract has no initial premium then (A) The premium you pay for the call is larger than the premium you receive from the put (B) The premium you pay for the call is smaller than the premium you receive from the put (C) The premium you pay for the call is equal to the premium you receive from the put (D) None of the above. Problem 65.2 In words, the Put-Call parity equation says that (A) The cost of buying the asset using options must equal the cost of buying the asset using a forward. (B) The cost of buying the asset using options must be greater than the cost of buying the asset using a forward. (C) The cost of buying the asset using options must be smaller than the cost of buying the asset using a forward. (D) None of the above. Problem 65.3 According to the put-call parity, the payoﬀs associated with ownership of a call option can be replicated by (A) shorting the underlying stock, borrowing the present value of the exercise price, and writing a put on the same underlying stock and with same exercise price (B) buying the underlying stock, borrowing the present value of the exercise price, and buying a put on the same underlying stock and with same exercise price (C) buying the underlying stock, borrowing the present value of the exercise price, and writing a put on the same underlying stock and with same exercise price (D) None of the above Problem 65.4 State two features that diﬀerentiate a synthetic forward contract from a no-arbitrage forward con-tract. Problem 65.5 Show that buying an index plus selling a call option with strike price K (i.e. selling a covered call) is equivalent to selling a put option with strike price K and buying a zero-coupon bond with par value K. 558 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES Problem 65.6 Show that short selling an index plus selling a put option with strike price K (i.e. selling a covered put) is equivalent to selling a call option with strike price K and taking out a loan with maturity value of K. Problem 65.7 ‡ You are given the following information: • The current price to buy one share of XYZ stock is 500. • The stock does not pay dividends. • The risk-free interest rate, compounded continuously, is 6%. • A European call option on one share of XYZ stock with a strike price of K that expires in one year costs $66.59. • A European put option on one share of XYZ stock with a strike price of K that expires in one year costs $18.64. Using put-call parity, determine the strike price, K. Problem 65.8 The current price of an stock is $1000 and the stock pays no dividends in the coming year. The premium for a one-year European call is $93.809 and the premium for the corresponding put is $74.201. The annual risk-free interest rate is 4%. Determine the strike price of the synthetic forward. Problem 65.9 The actual forward price of a stock is $1020 and the stock pays no dividends in the coming year. The premium for a one-year European call is $93.809 and the premium for the corresponding put is $74.201. The strike price of the synthetic forward contract is $1000. Determine the annual risk-free eﬀective rate r. Problem 65.10 Describe how you can ﬁnancially engineer the payoﬀto holding a stock by a combination of a put, call and a zero-coupon bond. Assume that the put and call have the same maturity date T and exercise price K, and that the bond has also has maturity date T and face value K. Problem 65.11 A stock currently sell for $100 and the stock pays no dividends. You buy a call with strike price $105 that expires in one year for a premium $3.46. You sell a put with the same expiration date and the strike price. Assuming, a risk-free eﬀective annual rate of 6%, determine the put’s premium. 65 SYNTHETIC FORWARD AND PUT-CALL PARITY 559 Problem 65.12 A $50, one-year call sells for $4; a $50 one-year put sells for $6.45. If the one-year interest rate is 8%, calculate the implied stock price. Problem 65.13 XYZ Corporation sells for $35 per share; the AUG option series has exactly six months until expiration. At the moment, the AUG 35 call sells for $3, and the AUG 35 put sells for $1.375. Using this information, what annual interest rate is implied in the prices? Problem 65.14 ‡ You are given the following information: • One share of the PS index currently sells for 1,000. • The PS index does not pay dividends. • The eﬀective annual risk-free interest rate is 5%. You want to lock in the ability to buy this index in one year for a price of 1,025. You can do this by buying or selling European put and call options with a strike price of 1,025. Which of the following will achieve your objective and also gives the cost today of establishing this position. (A) Buy the put and sell the call, receive 23.81 (B) Buy the put and sell the call, spend 23.81 (C) Buy the put and sell the call, no cost (D) Buy the call and sell the put, receive 23.81 (E) Buy the call and sell the put, spend 23.81 560 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES 66 Spread Strategies Forming a spread strategy with options means creating a position consisting only calls or only puts, in which some options are purchased and some are sold. In this section, we discuss some of the typical spread strategies. Bull Spreads A position that consists of buying a call with strike price K1 and expiration T and selling a call with strike price K2 > K1 and same expiration date is called a bull call spread. In contrast, buying a put with strike price K1 and expiration T and selling a put with strike price K2 > K1 and same expiration date is called a bull put spread. An investor who enters a bull spread is speculating that the stock price will increase. Example 66.1 Find a formula of the payoﬀof a bull call spread. Solution. The payoﬀfor buying a K1−strike call with expiration date T is max{0, PT −K1}. The payoﬀfor selling a K2−strike call (where K2 > K1) with the same expiration date is −max{0, PT −K2}. Thus, the payoﬀfor the call bull spread is max{0, PT −K1} −max{0, PT −K2}. Now, if PT ≤K1 then the call bull payoﬀis 0. If K1 < PT < K2 then the call bull payoﬀis PT −K1. Finally, if K2 ≤PT then the call bull payoﬀis K2 −K1. The payoﬀdiagram for the bull call spread is given in Figure 66.1 Figure 66.1 Example 66.2 Find a formula of the proﬁt of a bull call spread. Solution. The proﬁt of the position is the proﬁt of the long call plus the proﬁt of the short call. Thus, 66 SPREAD STRATEGIES 561 the proﬁt is the payoﬀof the combined position minus the future value of the net premium. If PT ≤K1 then the proﬁt is −FV (Call(K1, T) −Call(K2, T)). If K1 < PT < K2 then the proﬁt is PT −K1−FV (Call(K1, T)−Call(K2, T)). If K2 ≤PT then the proﬁt is K2−K1−FV (Call(K1, T)− Call(K2, T)). Figure 66.2 shows a graph of the proﬁt of a bull call spread. Note that from the graph, a bull spread limits the investor’s risk but also limits the proﬁt potential Figure 66.2 Example 66.3 An investor buys a $70-strike call and sells a $85-strike call of a stock. Both have expiration date one year from now. The risk free annual eﬀective rate of interest is 5%. The premiums of the $70-strike and $85-strike calls are $10.76 and $3.68 respectively. (a) Find the proﬁt at expiration as a function of the spot price. (b) Complete the following table Stock Price Total at Expiration Proﬁt 65 70 75 80 85 90 (c) Draw the proﬁt diagram. Solution. (a) The future value of the premium is (10.76−3.68)(1.05) = 7.43. The proﬁt function as a function of PT is given by    −7.43 if PT ≤70 PT −70 −7.43 if 70 < PT ≤< 85 85 −70 −7.43 if 85 ≤PT 562 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES (b) Stock Price Total at Expiration Proﬁt 65 −7.43 70 −7.43 75 −2.43 80 2.57 85 7.57 90 7.57 (c) The proﬁt diagram is similar Figure 66.2 Example 66.4 Show that the proﬁt diagram of bull call spread coincides with the proﬁt diagram of the bull put spread. Solution. As a function of PT the proﬁt from buying a K1−strike call and selling a K2−strike call with the same expiration date is given by    −FV (Call(K1, T) −Call(K2, T)) if PT ≤K1 PT −K1 −FV (Call(K1, T) −Call(K2, T)) if K1 < PT ≤< K2 K2 −K1 −FV (Call(K1, T) −Call(K2, T)) if K2 ≤PT The proﬁt from buying a K1−strike put with expiration date T and premium Put(K1, T) is max{0, K1 −PT} −FV (Put(K1, T)). The proﬁt from selling a K2−strike put (where K2 > K1) with the same expiration date is −max{0, K2 −PT} + FV (Put(K2, T)). Thus, the proﬁt for the put bull spread is max{0, K1 −PT} −max{0, K2 −PT} −FV (Put(K1, T) −Put(K2, T)). The proﬁt as a funtion of PT is given by    K1 −K2 −FV (Put(K1, T) −Put(K2, T)) if PT ≤K1 PT −K2 −FV (Put(K1, T) −Put(K2, T)) if K1 < PT ≤< K2 −FV (Put(K1, T) −Put(K2, T)) if K2 ≤PT Using the Put-Call parity we obtain the equation FV ([Call(K1, T) −Call(K2, T)] −[Put(K1, T) −Put(K2, T)]) = K2 −K1. This equation shows that the two proﬁts coincide. Hence, we can form a bull spread either buying a K1−strike call and selling a K2−strike call, or buying a K1−strike put and selling a K2−strike put 66 SPREAD STRATEGIES 563 Bear Spread A bear spread is precisely the opposite of a bull spread. An investor who enters a bull spread is hoping that the stock price will increase. By contrast, an investor who enters a bear spread is hoping that the stock price will decline. Let 0 < K1 < K2. A bear spread can be created by either selling a K1−strike call and buying a K2−strike call, both with the same expiration date (bear call spread), or by selling a K1−strike put and buying a K2−strike put, both with the same expiration date (bear put spread). Example 66.5 Find a formula of the proﬁt of a bear spread created by selling a K1−strike call and buying a K2−strike call, both with the same expiration date. Graph the proﬁt diagram. Solution. The proﬁt of the position is the proﬁt of the short call plus the proﬁt of the long call. Thus, the proﬁt is the payoﬀof the combined position minus the future value of the net premium. That is, max{0, PT −K2} −max{0, PT −K1} + FV (Call(K1, T) −Call(K2, T)). The proﬁt function is given by    FV (Call(K1, T) −Call(K2, T)) if PT ≤K1 K1 −PT + FV (Call(K1, T) −Call(K2, T)) if K1 < PT ≤< K2 K1 −K2 + FV (Call(K1, T) −Call(K2, T)) if K2 ≤PT Figure 66.3 shows a graph of the proﬁt of a bear spread. Note that the graph is the reﬂection of the graph in Figure 66.2 about the horizontal axis Figure 66.3 Combination of a Bull Spread and a Bear Spread: Box Spread A long box spread comprises four options, on the same underlying asset with the same expiration date. They can be paired in two ways as shown in the following table 564 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES Bull Call Spread Bear Put Spread Synthetic Long Forward Buy Call at K1 Sell Put at K1 Synthetic Short Forward Sell Call at K2 Buy Put at K2 Reading the table horizontally and vertically we obtain two views of a long box spread. • A long box spread can be viewed as a long synthetic stock at a price plus a short synthetic stock at a higher price. • A long box spread can be viewed as a long bull call spread at one pair of strike prices plus a long bear put spread at the same pair of strike prices. No matter what the spot price of the stock is at expiration, the box spread strategy guarantees a cash ﬂow of K2 −K1 in the future. Thus, this strategy has no stock price risk. Using Put-Call parity (assuming no-arbitrage), the net premium of acquiring this position is given by (Call(K1, T) −Put(K1, T)) −(Call(K2, T) −Put(K2, T)) = PV (K2 −K1). If K1 < K2, a box spread is a way to lend money. An investment of PV (K2 −K1) is made at time zero and a return of K2 −K1 per share is obtained at time T. If K1 > K2, a box spread is a way to borrow money. A return of PV (K1 −K2) is received at time zero and a loan payment of K1 −K2 is made at time T. Example 66.6 Find an expression for the payoﬀat expiration of a (long) box spread. Solution. The payoﬀis the payoﬀof a synthetic long forward (PT −K1) plus the payoﬀof a synthetic short forward (K2 −PT). That is, K2 −K1. Hence, the payoﬀat expiration is just the diﬀerence between the strike prices of the options involved. The proﬁt will be the amount by which the discounted payoﬀexceeds the net premium. Under no-arbitrage, the proﬁt is zero. Normally, the discounted payoﬀwould diﬀer little from the net premium, and any nominal proﬁt would be consumed by transaction costs. Example 66.7 Consider a three-month option on a stock whose current price is $40. If the eﬀective annual risk-free interest rate is 8.33% then price for the options might be K Call Put 40 2.78 1.99 45 0.97 5.08 66 SPREAD STRATEGIES 565 Find the cost at time zero and the payoﬀat expiration of the box spread obtained by • buying a 40-strike call and selling a 40-strike put • selling a 45-strike call and buying a 45-strike put Solution. The cost at time zero is the discounted value of the payoﬀ45 −40 = 5 which is 5 × (1.0833)0.25 = $4.90 Ratio Spreads A (call) ratio spread is achieved by buying a certain number of calls with a low strike and selling a diﬀerent number of calls at a higher strike. By replacing the calls with puts one gets a (put) ratio spread. All options under considerations have the same expiration date and same underlying asset. If m calls were bought and n calls were sold we say that the ratio is m n . Example 66.8 An investor buys one $70-strike call and sells two $85-strike call of a stock. All the calls have expiration date one year from now. The risk free annual eﬀective rate of interest is 5%. The premiums of the $70-strike and $85-strike calls are $10.76 and $3.68 respectively. Draw the proﬁt diagram of this position. Solution. The proﬁt of the 1:2 ratio is given by max{0, spot −70} −2 max{0, spot −85} + (2 × 3.68 −10.76) × 1.05. The proﬁt diagram is given in Figure 66.4 Figure 66.4 566 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES Practice Problems Problem 66.1 Explain two ways in which a bull spread can be created. Problem 66.2 The premium of a 3-month 40-strike call option on a stock currently selling for $40 is $2.78, and the premium for a 45-strike call option on the same stock is $0.97. Consider the bull spread achieved by purchasing the 40-strike call and selling the 45-strike call. The risk-free eﬀective annual interest rate is 8.33%. (a) Find the future value of the net premium. (b) Graph the proﬁt function of the bull spread. Problem 66.3 The premium of a 3-month 40-strike put option on a stock currently selling for $40 is $1.99, and the premium for a 45-strike put option on the same stock is $5.08. Consider the bull spread achieved by purchasing the 40-strike put and selling the 45-strike put. The risk-free eﬀective annual interest rate is 8.33%. (a) Find the future value of the net premium. (b) Graph the proﬁt function of the bull spread. Problem 66.4 A 3-month call options with strike prices $35 and $40 have premiums $6.13 and $2.78, respectively. The risk-free eﬀective annual interest rate is 8.33%. (a) What is the maximum gain when a bull spread is created from calls? (b) What is the maximum loss when a bull spread is created from calls? Problem 66.5 True or false:For a bull put spread I would sell a put option with a low strike price, and buy a put option with a higher strike price. Problem 66.6 Explain two ways in which a bear spread can be created. Problem 66.7 A 3-month call options with strike prices $35 and $40 have premiums $6.13 and $2.78, respectively. The risk-free eﬀective annual interest rate is 8.33%. (a) What is the maximum gain when a bear spread is created from calls? (b) What is the maximum loss when a bear spread is created from calls? 66 SPREAD STRATEGIES 567 Problem 66.8 Suppose you buy a 100-strike call, sell a 120-strike call, sell a 100-strike put, and buy a 120-strike put. Assume eﬀective annual risk-free interest rate of 8.5% and the expiration date of the options is one year. (a) Verify that there is no price risk in this transaction. (b) What is the initial cost of the position? (c) What is the value of the position at expiration? Problem 66.9 An investor buys two $70-strike call and sells one $85-strike call of a stock. All the calls have expi-ration date one year from now. The risk free annual eﬀective rate of interest is 5%. The premiums of the $70-strike and $85-strike calls are $10.76 and $3.68 respectively. Draw the proﬁt diagram of this position. Problem 66.10 Which of the following will create a bull spread (A) Buy a put with strike price $50, sell a put with strike price 55 (B) Buy a put with strike price $55, sell a put with strike price 50 (C) Buy a call with premium of $5, sell a call with premium $7 (D) Buy a call with strike price $50, sell a put with strike $55. 568 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES 67 Collars A collar is achieved with the purchase of an at-the-money put option with strike price K1 and expiration date T and the selling of an out-of-the-money call option with strike price K2 > K1 and expiration date T. Both options use the same underlying asset. A collar can be used to speculate on the decrease of price of an asset. The diﬀerence K2 −K1 is called the collar width. A written collar is the reverse of collar (sale of a put and purchase of a call). Example 67.1 Find the proﬁt function of a collar as a function of PT. Solution. The proﬁt function is max{0, K1 −PT} −max{0, PT −K2} + FV (Call(K2, T) −Put(K1, T)) which is    K1 −PT + FV (Call(K2, T) −Put(K1, T)) for PT ≤K1 FV (Call(K2, T) −Put(K1, T)) for K1 < PT < K2 K2 −PT + FV (Call(K2, T) −Put(K1, T)) for K2 ≤PT Example 67.2 An investor buys a 65-strike put with premium $1.22 and sells an 80-strike call with premium $5.44. Both have expiration date one year from now. The current price of the stock is $65. The risk free annual eﬀective rate of interest is 5%. Draw the proﬁt diagram of this collar. Solution. Using the function from the previous example we can create the following table PT 55 60 65 70 75 80 85 90 Proﬁt 14.43 9.43 4.43 4.43 4.43 4.43 −0.57 −5.57 The proﬁt diagram is shown in Figure 67.1 67 COLLARS 569 Figure 67.1 Note that the ﬁgure depicts a short position. The position beneﬁts from declining price asset and suﬀers losses from price asset appreciation. It resembles a short forward with the only exception that with a collar there a range between the strikes in which the proﬁt is unaﬀected by changes in the value of the underlying asset Collars can be used to insure assets we own. This position is called a collared stock. A col-lared stock involves buying the index, buy an at-the-money K1−strike put option (which insures the index) and selling an out-of-the-money K2−strike call option (to reduce the cost of the insur-ance), where K1 < K2. Example 67.3 Find the proﬁt function of a collared stock as a function of PT. Solution. The proﬁt function at expiration is PT + max{0, K1 −PT} −max{0, PT −K2} + FV (Call(K2, T) −Put(K1, T) −P0) which is    K1 + FV (Call(K2, T) −Put(K1, T) −P0) for PT ≤K1 PT + FV (Call(K2, T) −Put(K1, T) −P0) for K1 < PT < K2 K2 + FV (Call(K2, T) −Put(K1, T) −P0) for K2 ≤PT Example 67.4 Amin buys an index for $65. To insure the index he buys a 65-strike put with premium $1.22. In order to reduce the cost of this insurance he sells an 80-strike call with premium $5.44. Both options have expiration date one year from now. The risk free annual eﬀective rate of interest is 5%. Draw the proﬁt diagram of this collar. Solution. Using the function from the previous example we can create the following table PT 55 60 65 70 75 80 85 90 Proﬁt 1.181 1.181 1.181 6.18 11.18 16.18 16.18 16.18 570 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES The proﬁt diagram is shown in Figure 67.2 Figure 67.2 Note that the proﬁt diagram resembles the one of a bull spread Zero-cost Collars A collar with zero cost at time 0, i.e. with zero net premium, is called a zero-cost collar (also known as a costless collar). In this strategy, one buys an at-the-money put and selling an out-of-the-money call with both having the same premium. Example 67.5 Suppose that a zero-cost collar consists of buying a K1−strike put option and selling a K2−strike call option, where K1 < K2. Find the proﬁt function as a function of PT. Solution. The proﬁt coincides with the payoﬀsince the net premium is zero at time 0. Hence, the cost function is max{0, K1 −PT} −max{0, PT −K2} which is    K1 −PT for PT ≤K1 0 for K1 < PT < K2 K2 −PT for K2 ≤PT Example 67.6 A zero-cost collar on an index is created by buying the index for $60, buying a 55-strike put and selling a 65-strike call. Both options have expiration date one year from now. The risk free annual eﬀective rate of interest is 5%. Draw the proﬁt diagram of this position. 67 COLLARS 571 Solution. The proﬁt function is    55 −60(1.05) for PT ≤55 PT −60(1.05) for 55 < PT < 65 65 −60(1.05) for 65 ≤PT We have the following table PT 45 50 55 60 65 70 75 Proﬁt −8 −8 −8 −3 2 2 2 The proﬁt diagram is shown in Figure 67.3 Figure 67.3 Remark 67.1 (1) For a given stock, it can be shown that there are an inﬁnite number of possible combinations of strike prices that will produce a zero-cost collar. That is, there is an inﬁnite number of zero-cost collars. (2) If K1 = K2 = F0,T (i.e. you will receive F0,T at time T) then by the Put-Call parity the cost of the collar is zero and the collar is identical to a forward contract (a short forward in the case of a purchased collar (See Figure 67.1), or a long forward in the case of a written collar (See Problem 67.6). (c) In (b), it sounds that you are insuring your asset for free. That is not the case. In fact, you have lost money because you have forgone interest on the asset. 572 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES Practice Problems Problem 67.1 A purchased collar consists of a 40-strike put and a 65-strike call. What is the collar width? Problem 67.2 A purchased collar is achieved by (A) Buying a K1−strike put and selling a K2−strike call with K1 < K2 (B) Selling a K1−strike put and buying a K2−strike call with K1 < K2 (A) Buying a K1−strike call and selling a K2−strike put with K1 < K2 (A) Selling a K1−strike call and buying a K2−strike put with K1 < K2 Problem 67.3 A written collar is achieved by (A) Buying a K1−strike put and selling a K2−strike call with K1 < K2 (B) Selling a K1−strike put and buying a K2−strike call with K1 < K2 (A) Buying a K1−strike call and selling a K2−strike put with K1 < K2 (A) Selling a K1−strike call and buying a K2−strike put with K1 < K2 Problem 67.4 ‡ Which statement about zero-cost purchased collars is FALSE? (A) A zero-width, zero-cost collar can be created by setting both the put and call strike prices at the forward price. (B) There are an inﬁnite number of zero-cost collars. (C) The put option can be at-the-money. (D) The call option can be at-the-money. (E) The strike price on the put option must be at or below the forward price. Problem 67.5 Find the proﬁt function of a written collar as a function of PT. Problem 67.6 An investor sells a 65-strike put with premium $1.22 and buys an 80-strike call with premium $5.44. Both have expiration date one year from now. The current price of the stock is $65. The risk free annual eﬀective rate of interest is 5%. Draw the proﬁt diagram of this written collar. Problem 67.7 To insure a short sale of a stock, an investor buys a K1−strike call option (which insures the index) 67 COLLARS 573 and sells a K2−strike put option (to reduce the cost of the insurance), where K1 > K2. Find the proﬁt function of this position as a function of PT. What spread does the diagram resemble to? Problem 67.8 Amin short sale an index for $75. To insure the index he buys a 80-strike call with premium $5.44. In order to reduce the cost of this insurance he sells a 65-strike put with premium $1.22. Both options have expiration date one year from now. The risk free annual eﬀective rate of interest is 5%. Draw the proﬁt diagram of this collar. What spread strategy does the diagram resemble to? Problem 67.9 ‡ Happy Jalapenos, LLC has an exclusive contract to supply jalapeno peppers to the organizers of the annual jalapeno eating contest. The contract states that the contest organizers will take delivery of 10,000 jalapenos in one year at the market price. It will cost Happy Jalapenos 1,000 to provide 10,000 jalapenos and today’s market price is 0.12 for one jalapeno. The continuously compounded risk-free interest rate is 6%. Happy Jalapenos has decided to hedge as follows (both options are one-year, European): Buy 10,000 0.12-strike put options for 84.30 and sell 10,000 0.14-strike call options for 74.80. Happy Jalapenos believes the market price in one year will be somewhere between 0.10 and 0.15 per pepper. Which interval represents the range of possible proﬁt one year from now for Happy Jalapenos? (A) −200 to 100 (B) −110 to 190 (C) −100 to 200 (D) 190 to 390 (E) 200 to 400 574 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES 68 Volatility Speculation: Straddles, Strangles, and Butter-ﬂy Spreads In this section we consider strategies that depend on the volatility of a ﬁnancial market rather than on the direction of the asset price change. For example, straddles (to be deﬁned below) create positive payoﬀs when the market price is way diﬀerent from the options strike price in either direction (i.e. up or down). That is, the investor with these strategies does not care whether the stock price goes up or down, but only how much it moves. The ﬁrst strategy that we examine is that of a straddle. A long straddle or simply a straddle is an option strategy that is achieved by buying a K−strike call and a K−strike put with the same expiration time T and same underlying asset. This strategy has a high premium since it requires purchasing two options. However, there is a guaranteed positive payoﬀwith this strategy as long as the stock price at expiration is diﬀerent from the strike price (whether higher or lower). Indeed, the investor beneﬁts from the call if the stock price appreciates and from the put if the stock price declines. If the spot price at expiration is almost equal to the strike price the premiums from the two options are lost. The owner of a straddle is betting on high volatile market, regardless of the direction of price movement (up or down). Premiums of straddles become greater when the market’s perception is that volatility is greater. The payoﬀof a straddle is the sum of the payoﬀof a purchased call and a purchased put. That is, max{0, K −PT} + max{0, PT −K} = \|PT −K\|. The proﬁt function is given by \|PT −K\| −FV (Call(K, T) + Put(K, T)). The proﬁt diagram is shown in Figure 68.1. Note that a straddle has a limited risk, since the most a purchaser may lose is the cost of both options. At the same time, there is unlimited proﬁt potential. 68 VOLATILITY SPECULATION: STRADDLES, STRANGLES, AND BUTTERFLY SPREADS575 Figure 68.1 Example 68.1 Tess buys a 40-strike call option and a 40-strike put of an index. Both have expiration date three months from now. The current price of the index is $40. The risk free annual eﬀective rate of interest is 8.33%. The premium of the call option is $2.78 and that of the put option is $1.99. (a) Determine the future value of the initial cost. (b) Find the proﬁt as a function of the spot price at expiration PT. (c) Find the values of the spot price at expiration at which Tess makes a proﬁt. Solution. (a) The future value of the initial cost is (2.78 + 1.99)(1.0833)0.25 = $4.87. (b) The proﬁt is given by \|PT −40\| −4.87. (c) From Figure 68.1, a proﬁt occurs when 0 < PT < 35.13 or PT > 44.87 As pointed out above, a straddle has a high premium cost. One way to reduce the cost is to buy out-of-the-money options. Such a strategy is called a strangle. More formally, a strangle is a strategy that invovles buying an out-of-the-money call and an out-of-the-money put option with diﬀerent strike prices but with the same maturity and underlying asset. For example, buy a K1−strike put and a K2−strike call with the same expiration date and same underlying asset and such that K1 < K2. The proﬁt of such a strangle is max{0, K1 −PT} + max{0, PT −K2} −FV (Put(K1, T) + Call(K2, T)) or    K1 −PT −FV (Put(K1, T) + Call(K2, T)) if PT ≤K1 −FV (Put(K1, T) + Call(K2, T)) if K1 < PT < K2 PT −K2 −FV (Put(K1, T) + Call(K2, T)) if K2 ≤PT. In Poblem 68.10 we consider the proﬁt of a strangle with K2 < K1. Example 68.2 Drew buys a 35-strike put option and a 45-strike call of an index. Both have expiration date three months from now. The current price of the index is $40. The risk free annual eﬀective rate of interest is 8.33%. The premium of the call option is $0.97 and that of the put option is $0.44. (a) Determine the future value of the initial cost. (b) Find the proﬁt as a function of the spot price at expiration PT. (c) Find the break-even prices. (d) Draw the proﬁt diagram of this position. (e) On the same window, draw the proﬁt diagram of the straddle position of Example 68.1. 576 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES Solution. (a) The future value of the initial cost is (0.97 + 0.44)(1.0833)0.25 = $1.44. (b) The proﬁt is given by    35 −PT −1.44 if PT ≤35 −1.44 if 35 < PT < 45 PT −45 −1.44 if 45 ≤PT. (c) The break-even prices occur at PT = 33.56 and PT = 46.44 (d) The proﬁt diagram is shown in Figure 68.2 Figure 68.2 (e) The proﬁt diagram of the straddle as well as the one for the strangle are shown in Figure 68.3 Figure 68.3 Some observations are in order. Even though these options reduce the initial cost (and therefore the maximum loss), they increase the stock-price move required to make a proﬁt. The proﬁt diagrams of the two positions interesect at the points where −PT +40−4.87 = −1.44 and PT −40−4.87 = −1.44. That is, at the points PT = 36.57 and PT = 43.43. Only, in the interval 36.57 < PT < 43.43, the 68 VOLATILITY SPECULATION: STRADDLES, STRANGLES, AND BUTTERFLY SPREADS577 strangle outperforms the straddle. A straddle is a bet on high volatility. A corresponding position that bets on low volatility is a written straddle. This position is achieved by selling a K−strike call and a K−strike put with the same expiration date and same underlying asset. Similarly, a written strangle is a bet on low volatility. It is achieved by selling a K1−strike call and a K2−strike put (K1 < K2) with the same expiration date and same underlying asset. The proﬁt of a written straddle is given by −\|PT −K\| + FV (Call(K, T) + Put(K, T)). The proﬁt diagram is shown in Figure 68.4. Note that a written straddle is most proﬁtable if the stock price is K at expiration, and in this sense it is a bet on low volatility. However, a large change in the direction of the stock price results in a potentially unlimited loss. Figure 68.4 One way to limit the losses that can occur with a written straddle is by buying options to insure against losses whether is stock price is falling or rising. Buying an out-of-the-money put protects against losses that can occur by a sharp decline in the at-the-money written put. Buying an out-of-the-money call protects against losses that can occur by a sharp increase in the at-the-money written call. An insured written straddle strategy is referred to as a butterﬂy spread. Suppose that K1 < K2 < K3. One way to create a butterﬂy spread is by using the following combination: (1) Create a written straddle by selling a K2−strike call and a K2−strike put. (2) Create a long strangle by buying a K1-strike call and a K3-strike put. The purchased strangle provides insurance against the written straddle. All the options have the same expiration date T and the same underlying asset. The initial cost of the position is Call(K1, T) −Call(K2, T) −Put(K2, T) + Put(K3, T). 578 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES Letting FV denote the future value of the net premium, the proﬁt is given by max{0, PT −K1} + max{0, K3 −PT} −max{0, PT −K2} −max{0, K2 −PT} −FV or        K3 −K2 −FV if PT ≤K1 PT −K1 −K2 + K3 −FV if K1 < PT < K2 −PT −K1 + K2 + K3 −FV if K2 ≤PT < K3 K2 −K1 −FV if K3 ≤PT. Example 68.3 Sophia sells a 40-strike call option (with premium $2.78) and a 40-strike put (with premium $1.99) of an index. To insure against unlimited losses she buys a 35strike call option (with premium $6.13) and a 45strike put (with premium $5.08). All the options have expiration date three months from now. The current price of the index is $40. The risk free annual eﬀective rate of interest is 8.33%. Draw the proﬁt diagram of this position. Solution. The future value of the net premium is (6.13 −2.78 −1.99 + 5.08)(1.0833)0.25 = $6.57. The proﬁt function is given by        45 −40 −6.57 if PT ≤35 PT −35 −40 + 45 −6.57 if 35 < PT < 40 −PT −35 + 40 + 45 −6.57 if 40 ≤PT < 45 40 −35 −6.57 if 45 ≤PT. The proﬁt diagram is shown in Figure 68.5 Figure 68.5 A butterﬂy spread may be created by using only calls, only puts, or by a long (or a short) position combined with both calls and puts. See Problems 68.3 - Problems 68.5. Given 0 < K1 < K2 < K3. When a symmetric butterﬂy is created using these strike prices the 68 VOLATILITY SPECULATION: STRADDLES, STRANGLES, AND BUTTERFLY SPREADS579 number K2 is the midpoint of the interval with endpoints K1 and K3. What if K2 is not midaway between K1 and K3? In this case, one can create a butterﬂy-like spread with the peak tilted either to the left or to the right as follows: Deﬁne a number λ by the formula λ = K3 −K2 K3 −K1 . Then λ satisﬁes the equation K2 = λK1 + (1 −λ)K3. Thus, for every written K2−strike call, a butterﬂy-like spread can be constructed by buying λ K1−strike calls and (1−λ) K3−strike calls. The resulting spread is an example of an asymmetric butterﬂy spread. Example 68.4 Construct an an asymmetric butterﬂy spread using the 35-strike call (with premium $6.13), 43-strike call (with premium $1.525 ) and 45-strike call (with premium $0.97). All options expire 3 months from now. The risk free annual eﬀective rate of interest is 8.33%. Draw the proﬁt diagram of this position. Solution. With the given strike prices we ﬁnd Then λ = 0.2. Thus, an asymmetric butterﬂy spread is created by selling ten 43-strike calls, buying two 35-strike calls and eight 45-strike calls. The future value of the net premium is (2 × 6.13 + 8 × 0.97 −10 × 1.525)(1.0833)0.25 = $4.87. The proﬁt is 2 max{0, PT −35} + 8 max{0, PT −45} −10 max{0, PT −43} −4.87. or        −4.87 if PT ≤35 2PT −74.87 if 35 < PT < 43 −8PT + 355.13 if 43 ≤PT < 45 −4.87 if 45 ≤PT. The proﬁt diagram is shown in Figure 68.6 580 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES Figure 68.6 As with the symmetric butterﬂy spread, an investor can create an asymmetric butterﬂy spread by using only calls, using only puts, or by using a long or short position in the stock, combined with both puts and calls. See Problem 68.12. 68 VOLATILITY SPECULATION: STRADDLES, STRANGLES, AND BUTTERFLY SPREADS581 Practice Problems Problem 68.1 Which of the following is most equivalent to writing a straddle? (A) buy stock, write two calls (B) buy stock, buy one put (C) short stock, buy one call (D) short stock, buy one put. Problem 68.2 A butterﬂy spread of an index is created as follows: (1) buying a 35-strike call with premium $6.13 and selling a 40-strike call with premium $2.78 (that is buying a bull call spread). (2) selling a 40-strike call with premium $2.78 and buyin a 45-strike call with premium $0.97 ( that is buying a bear call spread). All options expire in three months from now. The current price of the index is $40. The risk free annual eﬀective rate of interest is 8.33%. Show that the proﬁt diagram of this position coincides with the one in Example 68.3. Problem 68.3 A butterﬂy spread of an index is created by buying a 35-strike call (with premium $6.13), selling two 40-strike calls (with premium $2.78 each) and buying a 45-strike call (with premium $0.97). All options expire in three months from now. The current price of the index is $40. The risk free annual eﬀective rate of interest is 8.33%. Show that the proﬁt diagram of this position coincides with the one in Example 68.3. Problem 68.4 A butterﬂy spread of an index is created by buying a 35-strike put (with premium $0.44), selling two 40-strike puts (with premium $1.99 each) and buying a 45-strike put (with premium $5.08). All options expire in three months from now. The current price of the index is $40. The risk free annual eﬀective rate of interest is 8.33%. Show that the proﬁt diagram of this position coincides with the one in Example 68.3. Problem 68.5 A butterﬂy spread of an index is created by buying the stock, buying a 35-strike put (with premium $0.44), selling two 40-strike calls (with premium $2.78 each) and buying a 45-strike call (with premium $0.97). All options expire in three months from now. The current price of the index is $40. The risk free annual eﬀective rate of interest is 8.33%. Show that the proﬁt diagram of this position coincides with the one in Example 68.3. 582 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES Problem 68.6 ‡ You believe that the volatility of a stock is higher than indicated by market prices for options on that stock. You want to speculate on that belief by buying or selling at-the-money options. What should you do? (A) Buy a strangle (B) Buy a straddle (C) Sell a straddle (D) Buy a butterﬂy spread (E) Sell a butterﬂy spread Problem 68.7 ‡ You are given the following information: • The current price to buy one share of ABC stock is 100 • The stock does not pay dividends • The risk-free rate, compounded continuously, is 5% • European options on one share of ABC stock expiring in one year have the following prices: Strike Price Call Premium Put Premium 90 14.63 0.24 100 6.80 1.93 110 2.17 6.81 A butterﬂy spread on this stock has the following proﬁt diagram. Which of the following will NOT produce this proﬁt diagram? (A) Buy a 90 put, buy a 110 put, sell two 100 puts (B) Buy a 90 call, buy a 110 call, sell two 100 calls (C) Buy a 90 put, sell a 100 put, sell a 100 call, buy a 110 call (D) Buy one share of the stock, buy a 90 call, buy a 110 put, sell two 100 puts (E) Buy one share of the stock, buy a 90 put, buy a 110 call, sell two 100 calls. 68 VOLATILITY SPECULATION: STRADDLES, STRANGLES, AND BUTTERFLY SPREADS583 Problem 68.8 Which of the following has the potential to lose the most money? (A) a long butterﬂy spread (B) a short strangle (C) a long strangle (D) an asymetric butterﬂy spread Problem 68.9 Three month European put options with strike prices of $50, $55, and $60 cost $2, $4, and $7, respectively. A butterﬂy is created by buying a 50-strike put, selling two 55-strike put, and buying a 60-strike put. The risk-free 3-month interest rate is 5%. (a) What is the maximum gain when a butterﬂy spread is created from the put options? (b) What is the maximum loss when a butterﬂy spread is created from the put options? (c) For what two values of PT does the holder of the butterﬂy spread breakeven, where PT is the stock price in three months? Problem 68.10 Consider a strangle with the strike price of the put is higher than the strike price of the call. Draw the proﬁt diagram. Problem 68.11 An asymmetric butterﬂy spread has the following payoﬀdiagram: This position was created using calls that are priced as follows: Strike Price Premium 80 4 84 2 90 0.50 Ignoring commissions and bid-ask spreads, what was the cost to establish this asymmetric butterﬂy spread? 584 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES Problem 68.12 Show that the asymmetric butterﬂy of Figure 68.6 can be duplicated by buying two 35-strike puts (with premium $0.44 each), selling ten 43-strike puts (with premium $3.674 each ) and buying eight 45-strike puts (with premium $5.08 each). All options expire 3 months from now. The risk free annual eﬀective rate of interest is 8.33%. 69 EQUITY LINKED CDS 585 69 Equity Linked CDs An equity-linked CD (ELCD) (also known as index-linked CD) is an FDIC-insured certiﬁcate of deposit that ties the rate of return to the performance of the market indices such as the S&P 500 Composite Stock Price Index, the Dow Jones Industrial Average, or the NASDAQ 100 composite index. Equity-linked CDs are usually issued by a bank. The CDs may vary by maturity (ﬁve years, six years, etc.), participation rate (some oﬀer 100% of the increase in the market index, others only 90% or 70%), and the method of computing the return of the stock index (some use an averaging method to determine the ending value of the index level, while others use the ﬁnal index value). For example, an equity-linked CD can have the following structure: A CD with a return linked to the S&P 500 index is guaranteed to repay the invested amount plus 70% of the simple appreciation in the S&P 500 at the maturity date of 5.5 years. If the invested amount is $10,000 and the index is 1300 initially then a decline in the index after 5.5 years will result in a return of the $10,000 to the investor. If on the other hand the index rises to 2200 at the maturity date then the return to the investor is the original investment plus 70% of the precentage gain on the index. That is, 10, 000 1 + 0.70 2200 1300 −1 = $14, 846. More generally, the payoﬀis given by the formula 10, 000 1 + 0.70 max 0, S5.5 1300 −1 where S5.5 is the value of the index after 5.5 years. Example 69.1 A 5.5-year ELCD promises to repay $10,000 and 70% of the gain in S&P 500 index at expiration date. Assume investment occurs when S&P 500 index was 1300. (a) Complete the following table Index after 5.5 years Payoﬀ 500 1000 1500 2000 2500 3000 586 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES (b) Draw the payoﬀdiagram. Solution. (a) Index after 5.5 years Payoﬀ 500 10,000 1000 10,000 1500 11,076.92 2000 13,769.23 2500 16,461.54 3000 19,153.85 (b) The payoﬀdiagram is shown in Figure 69.1 Figure 69.1 Building an ELCD Using a Zero-Coupon Bond and a Call Option We next use ﬁnancial engineering to create the building structure of an ELCD by buying a zero-coupon bond and a call option. Consider an investor who wants to invest $1,000 in a S&P 500 index with no risk of losing principal and the opportunity to earn market-like returns. This investor might purchase a zero-coupon bond and an at-the-money call option on the S&P 500 market index. The investor buys the zero-coupon 69 EQUITY LINKED CDS 587 bond at a discount and receives the $1,000 bond par value at maturity. She then uses the diﬀerence between the $1,000 investment and the price of the zero-coupon bond to pay for the call option premium. The call option has expiration date as the time to maturity of the bond. If, at maturity, the S&P 500 has increased, the investor exercises the call option and earns a return on the capital gains portion of the S&P 500. If the S&P 500 decreases, the investor does not exercise the option and receives nothing for it. However, the investor still receives the face value of the bond thereby insuring the original $1,000 investment. A setback of the ELCD in the above example is that in the case of a decline in the index the investor will get back his $1,000 at maturity date. If, say, the maturity date is 5 years and if the annual eﬀective interest rate during the investment time is 6%, the investor has lost in interest 1, 000(1.06)5 −1, 000 = $338.22. The present value of this interest is 338.22(1.06)−5 = $252.74. Thus as the investor invests today he has foregone interest with a present value of $252.74, and that is the implied cost today of investing in the CD. Summarizing, an investor is attracted to ELCD’s because it has the potential for market appreciation without risking capital. One disadvantage is the possible loss of interest on the invested principal. In general, the payoﬀof an ELCD with invested principal P and rate of participation r is given by the formula P 1 + r max{PT P0 −1, 0} = P + Pr max{PT P0 −1, 0}. That is, the payoﬀof a typical equity-linked CD at expiration is equal to the par amount of the CD plus an equity-linked coupon. In general, the equity-linked coupon is equal to either: (a) zero, if the underlying equity has depreciated from an agreed upon strike level (usually the index level on the issue date of the CD), or (b) the participation rate times the percentage change in the underlying equity times the par amount of the note, if the underlying equity appreciated. Example 69.2 An investor buys a ﬁve-year ELCD with 100% participation in the simple appreciation of the S&P 500 index for $1,000. The starting index-level is 1,400. (a) Find the payoﬀif after ﬁve years the index-level is 2,100. (b) Find the payoﬀif after ﬁve years the index-level is below 1,400. (c) Create the payoﬀdiagram. Solution. (a) The payoﬀis 1, 000 1 + 2100 1400 −1 × 100% = $1, 500. 588 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES (b) The total payoﬀis just the original investment of $,1000. (c) The proﬁt diagram is shown in Figure 69.2 Figure 69.2 Example 69.3 Create a synthetic equivalent to the ELCD in Example 69.2. Solution. A synthetic equivalent can be created as follows: (1) invest in a ﬁve-year 6% discount bond with par value of $1,000 for 1, 000(1.06)−5 = $747.26 and (2) buy a ﬁve-year, S&P 500 at-the-money call option with a $1,400 strike price for a premium $252.74 We conclude this section by looking at how the three perspectives the end-user, the market-maker, and the economic observer view an ELCD: • The end-user is interested in the product and whether it meets a ﬁnancial need at a fair cost. • The market-maker (i.e. the bank issuer) is interested in making proﬁt without bearing risk from having issued the CD. Some of the proﬁt come from fees for issuing the CD and from penalities of early withdrawals since ELCDs cannot be withdrawn before maturity date. • The economic observer studies the advantage of having ELCDs. 69 EQUITY LINKED CDS 589 Practice Problems Problem 69.1 The payoﬀin Example 69.1 can be regarded as the payoﬀof $10,000 and the payoﬀof a certain unit of European call options. What is that number of units? Problem 69.2 Suppose that an ELCD promises to repay the principal investment plus 80% of the gain of a linked equity. If the gain on the equity is 50% during the investment period then what is the gain for each dollar invested? Problem 69.3 In a ELCD, an investor will receive a return consisting of his orginal investment only when (a) the underlying index has increased from the starting index value (b) the underlying index has remained static or have fallen. Problem 69.4 An investor has an equity linked CD with a nominal value of $10,000 and a participation of %110. If the positive movement of the equity markets at maturity is 35% then ﬁnd the amount due at maturity to the investor. Problem 69.5 Which of the following statements are true about equity linked CDs: (I) pays a ﬁxed interest rate (II) return depends on the performance of the underlying assets (III) Miniumum return is the initially invested money (IV) can be viewed as a combination of a zero-coupon bond and a call option Problem 69.6 A 5-year CD promises to repay $1,000,000 and 115% of the gain in S&P 500 index at expiration date. Assume investment occurs when S&P 500 index was 500. (a) Complete the following table Index after 5 years Payoﬀ 400 450 500 550 600 700 800 590 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES (b) Draw the payoﬀdiagram. Problem 69.7 What is the minimum value of a call option? An equity-linked CD? Problem 69.8 A 5-year CD promises to repay $1,000,000 and 115% of the gain in S&P 500 index at expiration date. Assume investment occurs when S&P 500 index was 500. Assume annual eﬀective interest rate of 8% during the 5-year period. What is the implied cost today of investing in the equity-linked index ? Problem 69.9 A 5-year CD promises to repay $1,000,000 and 115% of the gain in S&P 500 index at expiration date. Assume investment occurs when S&P 500 index was 500. Create a synthetic ELCD for this investment. 70 PREPAID FORWARD CONTRACTS ON STOCK 591 70 Prepaid Forward Contracts On Stock There are four alternative ways for buying shares of a stock. Namely, • Outright purchase: An investor simultaneously pays S0 dollars in cash at time 0 and owns the stock. • Full leveraged purchase: An investor borrows S0 dollars and receives the ownership of the stock at time 0. Letting r denote the annual continuous compound interest, the investor has to repay S0ert at time T. • Forward contract: An arrangement for an agreed-upon future price (the forward price) and date in which the buyer pays for the stock at time T and the seller transfers the ownership of the stock to the buyer at that time. • Prepaid forward contract or prepay: This is like a forward contract but with the (prepaid) forward price paid at time 0. That is, the contract entails the buyer to pay today for the stock and owns it in the future. The price one pays is not necessarily the stock price S0. In this section, we will discuss the question of pricing a prepaid forward contract. Pricing of Prepaid Forward Contracts with no Dividends Let F P 0,T denote the prepaid forward price for an asset bought at time 0 and delivered at time T. This price depends on the dividends paid to the stockholders. In the absence of dividends, the timing of delivery is irrelevant. Therefore, the prepaid forward price is just the stock price at time 0. That is, F P 0,T = S0. Example 70.1 The current price of a stock is $50. The free-risk interest rate is 4% compounded continuously. Assuming no dividends, what is the prepaid forward price? Solution. The price of a prepaid forward contract on a stock that pays no dividends is the initial stock price. So the answer is $50 Now, arbitrage opportunities (positive cash ﬂows either today or in the future and with no risk) can be achieved with prepaid forward price as illustrated in the following two examples. Example 70.2 Suppose that the prepaid forward price exceeds the stock price. An arbitrageur will buy the stock at the low price of S0 and sell the prepaid forward contract for F P 0,T. Assuming no dividends, show that this transaction has a positive cash ﬂow at time 0 and is risk-free. 592 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES Solution. The net cash ﬂow at time 0 is F P 0,T −S0 > 0 since the cash ﬂow (outﬂow) from buying the stock is −S0 and from selling the prepaid contract (inﬂow) is F P 0,T. This transaction is risk-free: Selling the prepaid forward contract obliges the seller to deliver the stock at time T and buying the stock today ensures that the stock will be delivered at that time Example 70.3 Suppose that the stock price exceeds the prepaid forward. An arbitrageur will short the stock at the price of S0 and buy the prepaid forward contract for F P 0,T. Assuming no dividends, show that this transaction has a positive cash ﬂow at time 0 and is risk-free. Solution. The net cash ﬂow at time 0 is S0 −F P 0,T > 0 since the cash ﬂow (outﬂow) from buying the prepaid forward is −F0,T and from selling the stock (inﬂow) is S0. This transaction is risk-free: By buying the prepaid forward contract we are entitled to acquire the stock at time T which we use to close the short position Throughout the book all prices are assumed to be free arbitrage prices. Pricing of Prepaid Forward Contracts with Dividends Stocks that pay dividends have prepaid forward price lower than the stock price. Only the owner of the stock receives the dividends but not the holder of the prepaid forward. When dividend is paid the stock price is reduced by the amount of the dividend. Suppose that over the life of a forward contract, a stock receives dividends Di at time ti where i = 1, 2, · · · , n. Then the prepaid forward price is lower than the stock price by the present value of those dividends: F P 0,T = S0 − n X i=1 PV (Di). (70.1) Example 70.4 Suppose ABC stock costs $75 today and is expected to pay semi-annual dividend of $1 with the ﬁrst coming in 4 months from today and the last just prior to the delivery of the stock. Suppose that the annual continuously compounded risk-free rate is 8%. Find the cost of a 1-year prepaid forward contract. Solution. The cost is F P 0,1 = 75 −e−0.08× 4 12 −e−0.08× 10 12 = $73.09 70 PREPAID FORWARD CONTRACTS ON STOCK 593 Formula (70.1) applies for the case of discrete dividends. What about a stock index where dividends are given almost daily? In this case, it is common to model the dividends as being continuously paid at a constant continuoulsy compounded dividend yield rate δ (which is deﬁned to be the annualized dividend payment divided by the stock price). Note that δ is used here as a continuously compounded dividend rate, rather than a force of interest. Like an interest rate, the dividend rate is a rate per year. Divide the interval [0, T] into n equal subintervals using the mesh point ti = i nT where 1 ≤i ≤n. The dollar dividend over each interval is Dividend = δT n S0. Suppose that dividends payments are reinvested into stock. Let Ai be the total number of shares at time ti. Then Ai+1 = Ai 1 + δT n . Hence, the total amount of shares multiplies by 1 + δT n so that one share at time 0 grows to 1 + δT n n at time T. Letting n →∞we get that one share at time 0 grows to eδT shares at time T. Example 70.5 You buy one share of ABC stock and hold it for two years. The dividends are paid at continuously compounded rate of 3% and you reinvest all the dividends when they are paid. The current stock price is $125. (a) Find the daily dollar dividend. (b) How many shares do you have at the end of two years? Solution. (a) The daily dollar dividend is Dividend= 0.03 365 × 125 = $0.01027. (b) At the end of two years, the number of shares increased from one share to e0.03×2 = 1.061836 shares Now, if one share of the index is desired at time T, then one has to buy e−δT shares today. Due to dividend reinvestment, at time T we will have e−δTeδT = 1 share. This process of buying just enough shares today so that when dividends are reinvested we accumulate exactly one share at time T is called tailing the position. The cost per share at time 0 of a tailed position is e−δTS0. This position can be achieved as well by buying a T−year prepaid forward contract on a stock (or 594 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES index) that pays dividends at a continuous rate of δ for delivery of one full share of the index at time T. If there exists no arbitrage, the prepaid forward price must be equal to the cost of the tailing position at time 0. That is, F P 0,T = S0e−δT. Remark 70.1 In either the case of a tailed position or a prepaid forward contratct, the buyer of the share does not receive (in cash) any dividends paid prior to time T. In the case of the tailed position, those dividends are used to buy additional stock to bring the total to one full share; in the case of the prepaid forward contract, dividends are not payable to the holder of a long forward prior to the expiration (delivery) date, time T (at which time the investor receives one full share). Example 70.6 Find the 1-year prepaid forward price of the index in the previous example. Solution. The prepaid forward price is F P 0,1 = 125e−0.03 = $121.306 70 PREPAID FORWARD CONTRACTS ON STOCK 595 Practice Problems Problem 70.1 Complete the following charts that descibe the cash ﬂows and transactions to undertake arbitrage (stocks pay no dividends). F0,T > S0 Cash Flows t = 0 t = T Buy Stock Sell Prepaid Forward Total F0,T < S0 Cash Flows t = 0 t = T Short Stock Buy Prepaid Forward Total Problem 70.2 Suppose that the current price of a stock is $1100. The stock does not pay dividends and the risk-free annual rate is 5%. (a) Suppose that an investor is willing to buy a prepaid forward contract at a prepaid forward price of $1135 on a one year prepaid forward contract. Show how to make an arbitrage gain under these circumstances. (b) Suppose that an investor is willing to sell a prepaid forward contract at a prepaid forward price of $1050 on a one year prepaid forward contract. Show how to make an arbitrage gain under these circumstances. Problem 70.3 Suppose ABC stock costs $X today. It is expected that 4 quarterly dividends of $1.25 each will be paid on the stock with the ﬁrst coming 3 months from now. The 4th dividend will be paid one day before expiration of the forward contract. Suppose the annual continuously compounded risk-free rate is 10%. Find X if the cost of the forward contract is $95.30. Round your answer to the nearest dollar. Problem 70.4 Suppose ABC stock costs $75 today and is expected to pay semi-annual dividend of $X with the 596 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES ﬁrst coming in 4 months from today and the last just prior to the delivery of the stock. Suppose that the annual continuously compounded risk-free rate is 8%. Find X if the cost of a 1-year prepaid forward contract is $73.09. Round your answer to the nearest dollar. Problem 70.5 You buy one share of Ford stock and hold it for 2 years. The dividends are paid at the annualized daily continuously compounded rate of 3.98% and you reinvest all the dividends when they are paid. How many shares do you have at the end of two years? Problem 70.6 Suppose that annual dividend of 30 on the stocks of an index is valued at $1500. What is the continuously compounded dividend yield? Problem 70.7 Suppose XYZ stock costs $50 today and is expected to pay quarterly divident of $1 with the ﬁrst coming in 3 months from today and the last just prior to the delivery of the stock. Suppose that the annual continuously compounded risk-free rate is 6%. What is the price of a prepaid forward contract that expires 1 year from today, immediately after the fourth-quarter dividend? Problem 70.8 Suppose XYZ stock costs $50 today and is expected to pay 8% continuous risk-free dividend. What is the price of a prepaid forward contract that expires 1 year from today, immediately after the fourth-quarter dividend? Problem 70.9 LEPOs (Low Exercised Price Options) exist in order to avoid taxes and transaction fees associated with trading with stocks. These stocks have zero-dividends and are purchased outright. These options are certain to be exercised due to the fact that the strike price K is so low. The payoﬀat expiration time T is PT −K. Find the cost of this option at time 0. Show that a LEPO is essentially a prepaid forward contract with no dividends. Problem 70.10 An investor is interested in buying XYZ stock. The current price of stock is $45 per share. This stock pays dividends at an annual continuous risk-free rate of 5%. Calculate the price of a prepaid forward contract which expires in 18 months. Problem 70.11 ‡ A non-dividend paying stock currently sells for 100. One year from now the stock sells for 110. The risk-free rate, compounded continuously, is 6%. The stock is purchased in the following manner: 70 PREPAID FORWARD CONTRACTS ON STOCK 597 • You pay 100 today • You take possession of the security in one year Which of the following describes this arrangement? (A) Outright purchase (B) Fully leveraged purchase (C) Prepaid forward contract (D) Forward contract (E) This arrangement is not possible due to arbitrage opportunities 598 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES 71 Forward Contracts on Stock Recall that prepaid forward contracts are forward contracts with payment made at time 0. The forward price is the future value of the prepaid forward price and this is true regardless of whether there are discrete dividends, continuous dividends, or no dividends If the expiration time is T and the continuous risk-free rate is r then for a stock with no dividends (i.e. F P 0,T = S0) the future value of the prepaid forward price is given by the formula F0,T = S0erT. In the case of a stock with discrete dividends D1, D2, · · · , Dn made at time t1, t2, · · · , tn the forward price is F0,T =FV (F P 0,T) = S0 − n X i=1 Die−rti ! erT =S0erT − n X i=1 Dier(T−ti) In the case of a stock with continuous dividends with continuously compounded yield δ we have F0,T = (S0e−δT)erT = S0e(r−δ)T. The forward premium is the ratio of the forward price to the current spot price deﬁned by Forward premium= F0,T S0 . The expression 1 −F0,T S0 is the percentage by which the forward price exceeds (or falls below) the spot price. A number α that satisﬁes the equation F0,T = S0eαT is called the annualized forward premium. Solving this equation for α we ﬁnd α = 1 T ln F0,T S0 . 71 FORWARD CONTRACTS ON STOCK 599 If a stock pays no dividends, the annualized forward premium is just the annual risk-free interest rate α = r. For the case of continuous dividends, the annualized forward premium reduces to α = 1 T ln S0e(r−δ)T S0 = r −δ. Remark 71.1 In the literature, the forward premium of an exchange rate forward contract is deﬁned to be the percentage by which today’s forward rate exceeds (falls below) today’s spot rate and is given by the formula ln F0,T S0 . Usually these premia are quoted as annualized rates given by the formula 1 T ln F0,T S0 . Example 71.1 Suppose the stock price is $90 and the continuously compounded interest rate is 10%. (a) What is the 9-month forward price assuming dividends are zero? What is the forward premium? (b) If the 9-month forward price is $92, what is the annualized forward premium? (c) If the 9-month forward price is $92, what is the annualized continuous dividend yield? Solution. (a) F0, 9 12 = 90e0.1× 9 12 = $97.01. The forward premium is 97.01 90 = 1.07789. (b) The annualized forward premium is α = 1 T ln F0,9/12 S0 = 12 9 ln 92 90 = 0.0293. (c) We have 0.0293 = 0.10 −δ. Solving for δ we ﬁnd δ = 0.0707 = 7.07% In Section 65, we have seen how to create a synthetic forward contract by buying a call and selling a put with both having the same strike price. We next discuss another way for creating a synthetic forward. In this discussion we consider a stock that pays continuous dividends at the rate δ. We buy a tailed position by buying e−δT shares of the stock which cost S0e−δT at time 0 so that at time T we have one share of the stock. We borrow S0e−δT to pay for the shares of stock. In this case, we invest 0 at time 0. At time T we must repay S0e(r−δ)T and we sell the one share of stock at price PT. The payoﬀfor this position is 600 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES Payoﬀat expiration= PT −S0e(r−δ)T = PT −F0,T which is the payoﬀof a long forward contract with forward price F0,T and expiration time T. In terms of a word equation we have Forward = Stock −Zero coupon bond (71.1) By rearranging the terms of this equation we can derive a synthetic stock Stock = Forward + Zero-coupon bond This position is created by buying a forward contract with forward price F0,T = S0e(r−δ)T and expiration T and lending S0e−δT ( the present value of the forward price) at time 0. The payoﬀat time T is Payoﬀat expiration= (PT −F0,T) + F0,T = PT which is the payoﬀof buying e−δT shares of the stock. We can use (71.1) to create a synthetic bond Zero-coupon bond = Stock −Forward. This position is created by buying e−δT shares of the index at the cost of S0e−δT at time 0 and short one forward contract with forward price F0,T and expiration time T. The payoﬀof this position at time T is Payoﬀat expiration= PT + (F0,T −PT) = F0,T which is the payoﬀof a long risk-free bond with par value of F0,T and expiration time T. Example 71.2 Consider the position of selling e−δT shares for S0e−δT at time 0 and buying a forward with forward price F0,T = S0e(r−δ)T. Complete the following table Cash Flows t = 0 t = T Selling e−δT shares Long one forward Total Solution. 71 FORWARD CONTRACTS ON STOCK 601 Cash Flows t = 0 t = T Selling e−δT shares S0e−δT −PT Long one forward 0 PT −F0,T Total S0e−δT −F0,T Thus going short on a tailed position and buying a forward create cash ﬂows like those of a risk-free short zero-coupon bond (i.e. borrowing) If i is the rate of return on the synthetic bond then i satisﬁes the equation S0e(i−δ)T = F0,T. Solving for i we ﬁnd i = 1 T ln F0,T S0e−δT = δ + 1 T ln F0,T S0 . This rate is called the implied repo rate. Remark 71.2 “Repo” is short for “repurchase.” In the synthetic short bond, if we sell the stock short and enter into a long forward contract, then we have sold the stock and agreed to repurchase it. We have in eﬀect taken a loan (shorted a bond), which must be repaid with interest at the implied repo rate. Example 71.3 A $124 index pays a 1.5% continuous dividend. Suppose you observe a 2-year forward price of $135.7. Calculate the annual continuous rate of interest which you earn by buying stock and entering into a short forward contract, both positions for the same nominal amount. Solution. The implies repo rate is i = δ + 1 T ln F0,T S0 = 0.015 + 1 2 ln 135.7 124 = 6% We next discuss how a market-maker uses the synthetic positions strategies to hedge his clients positions. Suppose that a client enters into a long forward position with forward price F0,T and expiration time T. In this case, the conterparty represented by the market-maker is holding a short forward position. To oﬀset this risk, the market-maker borrows S0e−δT and use this amount to buy a tailed position in the index, paying S0e−δT and thus creating a synthetic long forward position. These transactions are summarized in the table below. 602 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES Table 71.1 Cash Flows t = 0 t = T Buy tailed position in stock, paying −S0e−δT −S0e−δT PT Borrow S0e−δT S0e−δT −S0e(r−δ)T Short forward 0 F0,T −PT Total 0 F0,T −S0e(r−δ)T A transaction in which you buy the asset and short the forward contract is called cash-and-carry (or cash-and-carry hedge). It is called cash-and-carry, because the cash is used to buy the asset and the asset is kept. A cash-and-carry has no risk: You have obligation to deliver the asset, but you also own the asset. An arbitrage that involves buying the asset and selling it forward is called cash-and-carry arbitrage. Now suppose that a client wishes to enter into a short forward contract with forward price F0,T and expiration time T. In this case, the market-maker is holding a long forward position. To oﬀset this risk, the market-maker short a tailed position in the index receiving S0e−δT a tailed position in the index, paying S0e−δT and thus creating a synthetic long forward position. Example 71.4 A transaction in which you short the asset and long the forward contract is called reverse cash-and-carry (or cash-and-carry hedge). Complete the following table: Cash Flows t = 0 t = T Short tailed position in stock, receiving S0e−δT Lent S0e−δT Long forward Total Solution. We have Table 71.2 Cash Flows t = 0 t = T Short tailed position in stock, receiving S0e−δT S0e−δT −PT Lent S0e−δT −S0e−δT S0e(r−δ)T Long forward 0 PT −F0,T Total 0 S0e(r−δ)T −F0,T Note that with a reverse cash-and-carry the market-maker is long a forward and short a synthetic forward contract 71 FORWARD CONTRACTS ON STOCK 603 Note that if the forward contract is priced as F0,T = S0e(r−δ)T, then proﬁts on a cash-and-carry or reverse cash-and-carry are zero. If F0,T ̸= S0e(r−δ)T, an arbitrageur can do arbitrage. For example, if the forward contract is overpriced, that is F0,T > S0e(r−δ)T, than an arbitrageur can use the strategy described in Table 71.1 to make a risk-free proﬁt. In contrast, if the forward contract is underpriced, that is F0,T < S0e(r−δ)T, than an arbitrageur can use the strategy described in Table 71.2 to make a risk-free proﬁt. Example 71.5 Suppose you are a market-maker in index forward contracts. The current price of an index is $42.50. The index pays no dividends. You observe a 6-month long forward contract with forward price of $43. The risk-free annual continuously compounded interest rate is 5%. Describe a strategy for creating an arbitrage proﬁt and determine the amount of the proﬁt. Solution. The no-arbitrage forward price of the long forward contract is 42.50e0.05(0.5) = $43.58 which is larger than the observed long forward contract. We use the strategy of reverse cash-and-carry: We sell the index for $42.50 and deposit the money into a savings account for six months. We also buy the forward contract. At the expiration date, the money in the bank grew to 42.50e0.05(0.5) = 43.58 and the cost of the forward is $43. Thus, we make a proﬁt of 43.58 −43 = $0.58 According to either Table 71.1 or Table 71.2 we have seen that an arbitrageur makes a costless proﬁt only when F0,T ̸= S0e(r−δ)T. This says that there is only one single no arbitrage forward price. In practice that is not the case. That is, there is a range of prices that preclude arbitrage opportunities. In fact, there is a no-arbitrage lower bound and upper bound between which there arbitrage can not be practiced. Transactions costs that determine these bounds are: • Index ask and bid prices denoted by Sa 0 and Sb 0 with Sb 0 < Sa 0. • Forward contract ask and bid prices denoted by F a 0,T and F b 0,T with F b 0,T < F a 0,T. • Index and forward transaction fees (commissions) denoted by CS and CF respectively. • The diﬀerence between borrowing and lending rates rb > rl. Let F + and F −be the upper and lower bound respecively. Suppose that an arbitrageur believe that the forward price F0,T is too high. Assume that the index pays no dividends and that the arbitrageur applies the cash-and-carry strategy as illustrated in the table below. 604 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES Cash Flows t = 0 t = T Buy tailed position in stock −Sa 0e−δT −CI PT Borrow Sa 0e−δT + CI + CF [Sa 0e−δT + CI + CF] −[Sa 0e−δT + CI + CF]erbT Short forward −CF F b 0,T −PT Total 0 F b 0,T −[Sa 0e−δT + CI + CF]erbT An arbitrageur makes costless proﬁt if F b 0,T > F + = [Sa 0e−δT + CI + CF]erbT. This upper bound reﬂects the fact that we pay a high price for the index (the ask price), pay transaction costs on both the index and the forward, and borrow at a high rate. In our discussion, we assume no transaction fees at time T. In Problem 71.14 we ﬁnd the lower bound to be F −= (Sb 0 −CI −CF)erlT. Example 71.6 Consider an index with current price of $800 and with zero dividends. Suppose that an arbitrageur can borrow at the continuously compounded rate of 5.5% and lend at the continuously compounded rate of 5%. Suppose there is a $1 transaction fee, paid at time 0, only for going either long or short a forward. Find the lower and upper bound between which the arbitrage is not proﬁtable. Solution. We have F −= (800 −1)e0.05 = $839.97 and F + = (800 + 1)e0.055 = $846.29 Expected Future Price Consider a forward contract with underlying asset a non-dividend paying stock. Suppose that the current price is S0, the risk-free rate is r, and the delivery time is T. A forward price in this case is found to be S0erT. Does this mean that the stock price at time T is S0erT? The answer is no. That is, the forward price does not conveys any information about the future price of the stock. However, there is (conceptually) an expected rate of accumulation for the stock, call it α, that can be used to project an expected price for the stock at time T : S0eαT. Since α = r + (α −r) the expected rate α has two components: the a risk-free component, r, representing compensation for the passage of time, and a risk component, α −r, which represents compensation for assuming the risk of loss on the stock (risk premium). If α −r > 0, then you must expect a positive return from the forward contract. The only way for this to happen is if the forward price predicts too low a stock price. In other words, the forward price a biased predictor of the future stock price. 71 FORWARD CONTRACTS ON STOCK 605 Example 71.7 ‡ The current price of one share of XYZ stock is 100. The forward price for delivery of one share of XYZ stock in one year is 105. Which of the following statements about the expected price of one share of XYZ stock in one year is TRUE? A. It will be less than 100 B. It will be equal to 100 C. It will be strictly between 100 and 105 D. It will be equal to 105 E. It will be greater than 105. Solution. In general, an investor should be compensated for time and risk. A forward contract has no invest-ment, so the extra 5 represents the risk premium. Those who buy the stock expect to earn both the risk premium and the time value of their purchase and thus the expected stock value is greater than 100 + 5 = 105 and the answer is (E) Cost of Carry and Lease Rate Suppose you borrow S0 at the continuously compounded risk-free rate r to purchase a share of stock that pays continuous dividends at the rate δ. At the end of one year, you will repay S0er. However, the dividends will provide an oﬀsetting income of S0eδ. Hence, at the end of one year you will have to pay S0er−δ. This is the net cost of carrying a long position in the stock. We call r −δ the cost of carry. Now suppose you short the index and depositing the proceeds at a savings account paying contin-uously compounded annual rate r. Then at the end of one year you will earn S0er. However, you would have to pay S0eδ to the index lender. Note that the index dividends are what the lender gives up by allowing the stock to be borrowed. We call δ the lease rate of the index: It is the annualized cash payment that the borrower must make to the lender. With the deﬁnitions of the cost of carry and the lease rate, the forward pricing formula F0,T = S0e(r−δ)T has the following word interpretation Forward price = Spot price + Interest to carry the asset - Lease rate or Forward price = Spot price + Cost of carry. 606 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES Example 71.8 What is the lease rate for a non-dividend paying stock? What is the cost of carry of a forward contract? Solution. For a non-dividend paying stock the lease rate is δ = 0. A forward contract requires no investment and makes no payouts and therefore has a zero cost of carry 71 FORWARD CONTRACTS ON STOCK 607 Practice Problems Problem 71.1 A stock has a spot price of $35. The continuously compounded risk free rate of interest is 6%. Find the prepaid forward price, the forward price and the annualized forward premium for a one-year forward contract on one share of the stock in each of the following situations. (a) The stock pays no dividends. (b) The stock pays a dividend of 1 in 6 months and 0.5 in one year just before delivery. (c) The stock pays continuous dividends at a rate of 3.5%. Problem 71.2 Suppose the stock price is $55 and the 4-month forward price is $57.50. Assuming the stock pays continuous dividends, ﬁnd (a) the 8-month forward premium (b) the 8-month forward price. Problem 71.3 Suppose the stock price is $55 and the 4-month forward price is $57.50. Assuming the stock pays continuous dividends, ﬁnd (a) Calculate the annualized forward premium. (b) Calculate the 1-year forward price. Problem 71.4 Consider the position of borrowing S0e−δT at time 0 and selling a forward with forward price F0,T = S0e(r−δ)T. Complete the following table Cash Flows t = 0 t = T Short one forward Borrow S0e−δT Total Problem 71.5 Consider the position of selling e−δT shares of the index for S0e−δT at time 0 and lending this amount for a period of T with interest rate r. Complete the following table Cash Flows t = 0 t = T Short tailed position in Stock Lend S0e−δT Total 608 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES Problem 71.6 XYZ stock costs $123.118 per share. This stock pays dividends at an annual continuous rate of 2.5%. A 18 month forward has a price of $130.242. You own 10000 shares of XYZ stock. Calculate the annual continuous rate of interest at which you can borrow by shorting your stock. Problem 71.7 You are a market-maker in S&R index forward contracts. The current price of the index is $1100. The index pays no dividends. The risk-free annual continuously compounded interest rate is 5%. (a) What is the no-arbitrage forward price of a 9-month forward contract. (b) Suppose a customer wishes to enter into a long index forward position. If you take the opposite position, describe the strategy you would use to hedge the resulting short position. (c) Suppose a customer wishes to enter into a short index forward position. If you take the opposite position, describe the strategy you would use to hedge the resulting long position. Problem 71.8 Repeat the previous problem that index pays an 1.5% continuous dividend. Problem 71.9 Suppose you are a market-maker in index forward contracts. The current price of an index is $1100. The index pays no dividends. The risk-free annual continuously compounded interest rate is 5%. (a) You observe a 6-month forward contract with forward price of $1135. Describe a strategy for creating an arbitrage proﬁt and determine the amount of the proﬁt. (b) You observe a 6-month forward contract with forward price of $1115. Describe a strategy for creating an arbitrage proﬁt and determine the amount of the proﬁt. Problem 71.10 Suppose you are a market-maker in index forward contracts. The current price of an index is $1100. The index pays continuous dividends at 2%. The risk-free annual continuously compounded interest rate is 5%. (a) You observe a 6-month forward contract with forward price of $1120. Describe a strategy for creating an arbitrage proﬁt and determine the amount of the proﬁt. (b) You observe a 6-month forward contract with forward price of $1110. Describe a strategy for creating an arbitrage proﬁt and determine the amount of the proﬁt. Problem 71.11 An ABC index has a current price of $124 and the continuously compounded risk-free rate is 6%. Suppose you observe a 2-year forward price of $135.7. What dividend yield is implied by this forward price? 71 FORWARD CONTRACTS ON STOCK 609 Problem 71.12 Using the previous problem (a) Suppose you believe that the dividend yield over the next two years will be 0.5%. What arbitrage would you take? (b) Suppose you believe that the dividend yield over the next two years will be 3%. What arbitrage would you take? Problem 71.13 Suppose the current stock price is $45.34 and the continuously compounded intrest rate is 5%. The stock pays dividend of $1.20 in three months. You observe a 9-month forward contract with forward price $47.56. Is there an arbitrage opportunity on the forward contract? If so, describe the strategy to realize proﬁt and ﬁnd the arbitrage proﬁt. Problem 71.14 Suppose that an arbitrageur believes that the observed forward price F0,T is too low. Show that, for the arbitrageur to make costless proﬁt we must have F0,Ta < F −= [Sa 0e−δT −CI −CF]erlT. Problem 71.15 Suppose that an arbitrageur would like to enter a cash-and-carry for 10000 barrels of oil for delivery in six months. Suppose that he can borrow at an annual eﬀective rate of interest of 4.5%. The current price of a barrel of oil is $55. (a) What is the minimum forward price at which he would make a proﬁt? (b) What is his proﬁt if the forward price is $57? Problem 71.16 Consider an index with current price of $800 and with zero dividends. Suppose that an arbitrageur can borrow at the continuously compounded rate of 5.5% and lend at the continuously compounded rate of 5%. Suppose that there are no transaction fees. Then an arbitrage is not proﬁtable if the forward price is between F −and F +. Determine the values of F −and F +. Problem 71.17 Consider an index with current price of $800 and with zero dividends. Suppose that an arbitrageur can borrow at the continuously compounded rate of 5.5% and lend at the continuously compounded rate of 5%. Suppose the transaction fee for going long or short a forward is $1 and $2.40 for going long or short an index. Find the lower and upper no-arbitrage bounds. 610 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES 72 Futures Contracts A futures contract is an agreement between two counterparts to buy/sell an asset for a speciﬁed price (the future price) at a speciﬁed date (the delivery date). Like the case of a forward con-tract, the holder of the contract is called the long future and the seller is called the short future. Futures are exchange-traded forward contracts. That is, they are bought and sold in organized future exhanges. Examples of exchanges the Chicago Mercantile Exchange, the Chicago Board of Trade, the International Petroleum Exchange of London, the New York Mercantile Exchange (NYMEX), the London Metal Exchange and the Tokyo Commodity Exchange. In the US, futures transactions are regulated by a government agency, the Commodity Futures Trading Commission (CFTC). Each exchange has a clearinghouse. The role of the clearinghouse is to match the purchases and the sales which take place during the trading day. By matching trades, the clearinghouse never takes market risk because it always has oﬀsetting positions with diﬀerent counterparts. By having the clearinghouse as counterpart, an individual entering a futures contract does not face the possible credit risk of its counterpart. Forwards and futures are similar in many aspects. However, there are still some diﬀerences which we list next. • Forward contracts are privately negotiated and not standardized. They are traded mainly over the counter and can be customized to suit the buyer or the seller. In contrast, futures contracts are standardized and have speciﬁed delivery dates, locations, and procedures. • Forward contracts are settled at the delivery date. Futures contracts are settled daily. As a result of daily settlements, value of a future contract is calculated according with the current value of the the underlying asset. We refer to this process as marking-to-market. Frequent marking-to-market can lead to pricing diﬀerences between futures contracts and identical forward contracts. • Futures contracts are liquid. Unlike a forward contract, which requires settling with the coun-terparty by actually buying or selling the asset on the expiration date, a futures contract can be cancelled by entering into the opposite position. For example, if you are short a futures contract, you can cancel your obligation to sell by entering an oﬀsetting obligation to buy the contract. This is done through a brokerage ﬁrm. • In the presence of clearinghouse, futures contracts are structured so as to minimize the eﬀects of credit risks. In contrast, in over-the-counter forward contracts, each party bears the other’s credit risk. • Because futures contracts are traded on an exchange, trades are subject to the exchange’s rules, which typically include a limit on the amount of daily price changes by imposing temporary halts in trading if the price change exceeds a deﬁned amount. • Futures contracts are commonly cash-settled, rather than requiring delivery of the underlying 72 FUTURES CONTRACTS 611 asset. We next illustrate an example of futures contracts with the S&P 500 index futures contract. The underlying asset for such a contract is the S&P 500 stock index. The contract has the following speciﬁcations: • The futures contract is traded at the Chicago Mercantile Exchange. • Notional Value: One contract has a value that is deﬁned as the level of the index (let’s say it is currently 1300) times $250 (so an S&P 500 futures contract currently has a notional value of $325,000). • The contract has delivery months March, June, September and December. • Trading ends on each business day prior to determination of settlement price. The settlement price is the average of the prices at which the contract traded immediately before the bell signaling the end of trading for the day. • The contracts are cash-settled because it is inconvenient or impossible to deliver the underlying asset which in this case consists of a basket of 500 stocks. The cash settlement is based on the opening price of the index on the third Friday of delivery month. Example 72.1 Suppose you wish to acquire $2.2 million worth of S&P 500 index with futures price 1100. (a) What is the notional value of one contract and how many futures contracts can you long for the invested amount? (b) The open interest of a futures contract is deﬁned to be the number of all futures outstanding contracts. It is the number of long positions. The activities of ﬁve traders are listed below. Complete the following table: Date Trading Activity Open interest Jan 1 A buys 1 contract and B sells 1 contract Jan 2 C buys 5 contracts and D sells 5 contracts Jan 3 A sells his contract and D buys one contract Jan 4 E buys 5 contracts from C who sells his 5 contracts Solution. (a) The notional value of one futures contract is 250 × 1100 = $275, 000. With $2.2 million you can long 2200000 275000 = 8 futures contract. (b) 612 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES Date Trading Activity Open interest Jan 1 A buys 1 contract and B sells 1 contract 1 Jan 2 C buys 5 contracts and D sells 5 contracts 6 Jan 3 A sells his contract and D buys one contract 6 Jan 4 E buys 5 contracts from C who sells his 5 contracts 6 Margins and Marking to Market One of the advantages of the futures contracts over the forward contracts is the dealing with credit risk. For example, forward contracts can be traded between two parties directly. This creates a credit risk related with the solvency of each party: One of the investors may regret the deal and try to back out or simply might not have the ﬁnancial resources to honor the agreement. In contrast, in the case of futures contracts the exchange tries to organize trading in such a way to avoid credit defaults. This is when margins come in. To avoid credit risk, a brokerage ﬁrm requires investors entering a futures contract to make deposit into an account called the margin account (also known as perfomance bond) which earns interest. The amount to be deposited at the time the contract is entered into is called the initial margin and is determined by the exchange. It is usually a fraction of the market value of the futures’ underlying asset. The margin account is mainly intended to protect the counterparty of the contract. Margin accounts are adjusted on a daily basis to reﬂect the investor’s gain or loss. This practice is referred to as marking to market the account. A trade is marked to market at the close of the day on which it takes place. It is then marked to market at the close of the trading on each subsequent day. If there exists a loss, the investor’s broker transfers that amount from the investor’s margin account to the clearinghouse. If a proﬁt, the clearinghouse transfers that amount to investor’s broker who then deposits it into the investor’s margin account. We illustrate the role of margins in the following example. Example 72.2 On July 5, 2007, John instructs his broker to buy two gold futures contracts (each contract with nominal amount of 100 ounces of gold) on the New York Commodity Exchange (COMEX) with futures’ price of $400 per ounce. The annual continuously compounded rate of return is 1.5%. (a) How many ounces of gold has John contracted to buy at the price of $400 per ounce? (b) What is the notional value of the two contracts? (c) Suppose that the exchange requires 5% margin with daily settlement. What is the initial margin? (d) Suppose that on July 6,2007, the futures price of gold has dropped from $400 to $397. What is the balance in John’s margin account after settlement? (e) Suppose that on July 7,2007, the futures price of gold has risen to $400. What is the balance in John’s margin account after settlement? 72 FUTURES CONTRACTS 613 Solution. (a) Since each contract consists of 100 ounces of gold, John has contracted to buy 200 ounces at the price of $400 per ounce. (b) The notional value of the two contracts is 2 × 100 × 400 = $80, 000. (c) The initial margin is 5% × 80, 000 = $4, 000. (d) The balance in John’s margin account after settlement is 4000e 0.015 365 + 200(397 −400) = $3400.16. (e) The balance in John’s margin account after settlement is 34, 000.16e 0.015 365 + 200(400 −397) = $4, 000.30 An investor is entitled any balance in the margin account in excess of the initial margin. However, if the balance in an investors margin account falls, the broker (or clearinghouse) has less protection against default. Investors are required to keep the margin account to a minimum level which is a percentage of the initial margin. This minimum level is called the maintenance margin. When the balance in the margin account drops below the maintenance level the investor receives a margin call and is required to make a deposit to return the balance to the initial margin level. If the investor does not meet the margin call, the position is closed out and the investor receives any remaining balance in the margin account. Example 72.3 A farmer enters into a short futures contract to sell 100,000 pounds of wheat for $1.4 per pound. There is 30% margin and weekly settlement. The maintenance margin is 20% and the annual eﬀective rate of interest is 4.5%. What is the minimum next week price which would lead to a margin call? Solution. The initial margin is (0.30)(100000)(1.4) = $42000. The maintenance margin is (0.20)(100000)(1.4) = $28000. After settlement, next week’s balance is 42000(1.045) 1 52 + 100000(1.4 −P 1 52). A margin call happens if 28000 > 42000(1.045) 1 52 + 100000(1.4 −P 1 52). 614 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES or P 1 52 > 1.4 −28000 −42000(1.045) 1 52 100000 = 1.540355672 72 FUTURES CONTRACTS 615 Practice Problems Problem 72.1 The open interest on silver futures at a particular time is the number of (A) all silver futures outstanding contracts (B) outstanding silver futures contracts for a particular delivery month (C) silver futures contracts traded during the day (D) silver futures contracts traded the previous day Problem 72.2 The fact that the exchange is the counter-party to every futures contract issued is important because it eliminates (A) market risk (B) credit risk. (C) interest rate risk Problem 72.3 Margin must be posted by (A) buyers of futures contracts (B) sellers of futures contracts (C) both buyers and sellers of futures contracts (D) speculators only Problem 72.4 The daily settlement of obligations on futures positions is called (A) a margin call (B) marking to market (C) a variation margin check (D) None of the above Problem 72.5 All of the following are false except (A) A margin deposit can only be met by cash (B) All futures contracts require the same margin deposit (C) The maintenance margin is the amount of money you post with your broker when you buy or sell a futures contract (D) The maintenance margin is the value of the margin account below which the holder of a futures contract receives a margin call. 616 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES Problem 72.6 You are currently long in a futures contract. You then instruct a broker to enter the short side of a futures contract to close your position. This is called (A) a cross hedge (B) a reversing trade (C) a speculation (D) marking to market Problem 72.7 A silver futures contract requires the seller to deliver 5,000 Troy ounces of silver. An investor sells one July silver futures contract at a price of $8 per ounce, posting a $2,025 initial margin. Neglecting interest, if the required maintenance margin is $1,500, the price per ounce at which the investor would ﬁrst receive a maintenance margin call is closest to: (A) $5.92 (B) $7.89 (C) $8.11 (D) $10.80 Problem 72.8 If the initial margin on a futures contract I have sold increases, it likely means: (A) The market has gotten less liquid. (B) I am not making money in my futures positions. (C) Prices have gotten more volatile. (D) The margins never change once you have a futures position. Problem 72.9 Complete the following statement: At the end of each trading day, the margin account is adjusted to reﬂect the futures trader’s gain or loss. This process is referred to as the account. Problem 72.10 What is the diﬀerence between the initial margin and the maintenance margin? Problem 72.11 Explain some diﬀerences between a futures contract and a forward contract. Problem 72.12 Crude oil futures trade in units of 1,000 U.S. barrels (42,000 gallons). The current value is $50/bar-rel. Find the notional value of this contract. 72 FUTURES CONTRACTS 617 Problem 72.13 Assume today’s settlement price on a CME EUR futures contract is $1.3140/EUR. The nominal amount of one contract is 125,000. You have a short position in one contract. Your margin account currently has a balance of $1,700. The next three days’ settlement prices are $1.3126, $1.3133, and $1.3049. Calculate the changes in the margin account from daily marking-to-market and the balance in the account after the third day. Problem 72.14 Do Problem 72.13 again assuming you have a long position in the futures contract. Assume 40% maintenance margin. Would you receive a margin call? Problem 72.15 Suppose the S&P 500 index futures price is currently 1200. You wish to purchase four futures contracts on margin. (a) What is the notional value of your position? (b) Assuming a 10% initial margin, what is the value of the initial margin? Problem 72.16 Suppose the S&P 500 index futures price is currently 950 and the initial margin is 10%. You wish to enter into 10 S&P 500 futures contracts. (a) What is the notional value of your position? What is the margin? (b) Suppose you earn a continuously compound rate of 6% on your margin balance, your position is marked to market weekly, and the maintenance margin is 80% of the initial margin. What is the greatest S&P 500 index futures price 1 week from today at which you will receive a margin call? 618 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES 73 Understanding the Economy of Swaps: A Simple Com-modity Swap A swap is an agreement to exchange cash ﬂows at speciﬁed future times according to certain spec-iﬁed rules. For example, a forward contract is a single-payment swap. To see this, suppose you enter a forward contract to buy 50 ounces of gold for $300 an ounce in one year. You can sell the gold as soon you receive it. The forward contract acts like a single-payment swap because at the end of the year you will pay 50 × 300 = $15000 and receive 50 × P1 where P1 is the market price of one ounce of gold on that date. While a forward contract leads to an exchange of cash on a single future dates, swaps typically lead to cash ﬂow exchanges taking place on several future dates. In some sense, a swap is a series of forward contracts combined into one contract or put it diﬀerently it can be regarded as a portfolio of forward contracts. There are various types of swaps, namely, interest rates swap, currency swaps, commodity swaps, and equity swaps. In this section we will try to understand the economy of swaps. Our discussion will be centered around commodity swaps. A commodity swap is a swap in which one party is selling a commodity (or the cash equivalent) and the other is paying cash for the commodity (the buyer). The contract needs to specify the type and quality of the commodity, how to settle the contract, etc. Commodity swaps are predominantly associated with energy-related products such as oil. Suppose that quantity Qi of the commodity is to be delivered at time ti where i = 1, 2, · · · , n. The buyer can either make one payment of C0 at time 0 for the swap (a prepaid swap) or make a payment of Ci at time ti. Let F0,ti be the forward price of the commodity with delivery at time ti where 1 ≤i ≤n. Let ri be the annual interest rate applied over the interval of time [0, ti]. Then the present value of the commodity delivered over the life of the swap is n X i=1 QiPVi(F0,ti). In a prepaid swap, the buyer makes a single payment today in the amount of n X i=1 QiPVi(F0,ti) to obtain multiple deliveries in the future. Example 73.1 A manufacturer is going to buy 100,000 barrels of oil 1 year from today and 2 years from today. 73 UNDERSTANDING THE ECONOMY OF SWAPS: A SIMPLE COMMODITY SWAP 619 Suppose that the forward price for delivery in 1 year is $20/barrel and in 2 years $21/barrel. The 1-year spot interest rate is 6% and the 2-year spot interest rate is 6.5%. The manufacturer enters into long forward contracts for 100,000 barrels for each of the next 2 years, committing to pay $20/barrel in 1 year and $21/barrel in 2 years. Alternatively, the manufacturer can make a single payment P now and the supplier would commit to deliver 1 barrel in year 1 and 1 barrel in year 2. Find P. Solution. P is just the cost of a prepaid swap given by 20 1.06 + 21 1.0652 = $37.383 Now, with multiple deliveries and multiple payments, the buyer makes swap payments at the time the commodity is delivered. Usually each swap payment per unit of commodity is a ﬁxed amount say R. That is, at time ti the swap payment is QiR. Thus, the present value of all the payments is given by the sum n X i=1 QiPVi(R). We thus obtain the equation n X i=1 QiPVi(F0,ti) = n X i=1 QiPVi(R) From this equation, one ﬁnds R given by R = Pn i=1 QiPVi(F0,ti) Pn i=1 QiPVi(1) . Suppose that at time of delivery, the buyer pays a level payment of R at each of the delivery times. Then, the present value of the cashﬂow of payments is n X i=1 PVi(R) and R is given by R = Pn i=1 QiPVi(F0,ti) Pn i=1 PVi(1) . 620 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES Example 73.2 Consider the information of Example 73.1. Suppose a swap calls for equal payments each year. Let x be the payment per year per barrel. This is the swap price. Find x. Solution. We must have x 1.06 + x 1.0652 = 37.383. Solving this equation for x we ﬁnd x = $20.483. This means, that each year, the manufacturer will pay $20.483 and receives a barrel of oil Remark 73.1 The payments need not be equal. Any payments that have a present value of $37.383 are acceptable. That is, any two payments that satisfy the equation P1 1.06 + P2 1.0652 = 20 1.06 + 21 1.0652. A swap can be settled either by physical settlement or by cash settlement. If a swap is settled physically, the party with the commodity delivers the stipulated notional amount to the buyer in exchange of cash as shown in the previous example. If a swap is cashed settled, then the commodity is valued at the current spot price. If the current value of the commodity is larger than the value of the cash payment, then the buyer pays the swap counterparty the diﬀerence and then buys the commodity at the spot price. Reciprocally, if the current value of the commodity is smaller than the value of the cash payment, the counterparty pays this diﬀerence to the buyer and the buyer buys the commodity. Example 73.3 Again, we consider the information of both previous examples. Suppose the swap is cash settled. (a) Suppose that the market price is $25. Calculate the payment which the buyer receives. (b) Suppose that the market price is $18. Calculate the payment which the counterparty receives. Solution. (a) The swap counterparty pays the buyer 25 −20.483 = $4.517 per barrel. (b) The buyer pays the swap counterparty 18 −20.483 = −$2.483 per barrel Notice that whatever the market price of oil, the net cost to the buyer is the swap price $20.483: spot price −(spot price −swap price) = swap price. 73 UNDERSTANDING THE ECONOMY OF SWAPS: A SIMPLE COMMODITY SWAP 621 This formula describes the payments per one barrel of oil. For a swap agreement involving 100,000 barrels of oil, the values in the formula are multiplied by the notional amount of the swap (i.e. 100,000 barrels). Remark 73.2 Notice that the results for the buyer are the same whether the swap is settled physically or ﬁnancially. In both cases, the net cost to the buyer is $20.483 per barrel. A swap can be regarded as a series of forward contracts, combined with borrowing and/or lending money. In the above oil swap example, the buyer under the swap agreement would pay (and the counterparty would receive) the swap price of $20.483 on each of the two dates. Under a pair of forward contracts, the forward prices of $20 and $21 would have been paid on the ﬁrst and second dates, respectively. Relative to the forward prices, the buyer lends to the counterparty an amount of $0.483 (= 20.483 −20) on the ﬁrst date by paying that much more than the forward rate; then the counterparty repays the loan by accepting $20.483 on the second date, which is less than the forward price by $0.517 (= 21 −20.483). This loan has the eﬀect of equalizing the net cash ﬂow on the two dates. The interest rate on the loan is 0.517 0.483 −1 = 7% This is the 1-year implied forward rate from year 1 to year 2. A commodity swap allows to lock the price of a sale. It can be used by a producer of a com-modity to hedge by ﬁxing the price that he will get in the future for this commodity. It also can be used by a manufacturer to hedge by ﬁxing the price that he will pay in the future for a commodity. Swap Dealer A dealer is an agent who makes a market buying and selling swaps to parties who have natural exposure to price risk. He earns his keep on the bid/ask spread. The dealer buys a swap from the producer, at ﬁxed price S (the swap price). The seller receives S −PT, where PT is the spot price. The dealer sells a swap to the consumer, at ﬁxed price S + ∆, where δ is the ask-bid spread. The buyer pays S + ∆−PT. The dealer earns diﬀerence between what buyer gives and seller gets, i.e. ∆. The process of matching a buyer to the seller is called a back-to-back transaction. As you can see from the above discussion, the dealer is subject to each party’s credit risk, but is not exposed to price risk. Figure 73.1 illustrates the role of the swap dealer. 622 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES Figure 73.1 We next consider the case when the dealer serves as the counterparty. A dealer enters the swap as the counterparty. Thus, the dealer has the obligation to pay the spot price and receives the swap price. The dealer has a short position in 1- and 2- year oil. The dealer can lose money if the spot price increases. The dealer hedges his short position by going long year 1 and year 2 gold forwards. The table below illustrates how this strategy works. Year Payment from Gold Buyer Long Forward Net 1 $20.483 −P1 P1 −$20 $0.483 2 $20.483 −P2 P2 −$21 −$0.517 The net cash ﬂow for the hedged position is a loan where the dealer receives $0.483 in year 1 and repays $0.517 in year two. The dealer has interest rate exposure. In interest rate fall, the dealer will not be able to earn suﬃcient return from investing $0.483 in year 1 to repay $0.517 in year 2. This shows that hedging oil price risk does not fully hedge the position. In summary, the dealer can eliminate its price risk by hedging the swap contract through the use of multiple forward contracts. The dealer must also enter into forward rate agreements in order to hedge the interest rate exposure resulting from the timing diﬀerence between the forward contract’s payments and the payments under the swap agreement. The Market Value of a Swap When a swap agreement is created, its market value is zero. In our oil example, the swap consists of two forward contracts and an agreement to lend money at the implied forward rate of 7%. These forwards have zero values at time 0 so the swap does as well. As time passes the market value of the swap will no longer be zero due to oil prices and interest rates change. So over time, the swap contract develops a non-zero value. Even if prices and interest rates do not change, the swap agreement will have a non-zero value once the ﬁrst payment has been made. Let us ﬁnd the market value of the oil swap example. Suppose that right after entering the swap the oil prices go up by $2 in year 1 and year 2. The original swap in this case has a non-zero market value. Assuming interest rate are the same, the new swap price x satisﬁes the equation x 1.06 + x 1.0652 = 22 1.06 + 23 1.0652 73 UNDERSTANDING THE ECONOMY OF SWAPS: A SIMPLE COMMODITY SWAP 623 which gives x = $22.483. In this case, the buyer buys a barrel of oil for $20.483 and can sell it for $22.483 making a net proﬁt of $2 per barrel each year. The present value of this diﬀerence of prices is 2 1.06 + 2 1.0652 = $3.65. The buyer can receive a stream of payments worth $3.65 by oﬀsetting a new swap. Thus, $3.65 is the market value of the swap. If interest rates had changes, we would have used the new interest rate in computing the new swap price. The swap market is very large and has no major governmental oversight. Swap markets have none of the provisions as the futures and options markets do, thus resulting in a default risk and the need to assess the creditworthiness of each counterparty. Traders must carefully assess their trade partners. On the other side of this, swaps have a measure of privacy and can be negotiated for unique quan-tities, qualities, and contract provisions. Example 73.4 Suppose that oil forward prices for 1 year, 2 years, and 3 years are $20, $21, and $22. The eﬀective annual interest rates are shown in the table below Maturity(in years) yield % 1 6 2 6.5 3 7 (a) What is the swap price, assuming level payments? (b) Suppose you are a dealer who is paying the swap price and receiving the spot price. Suppose you enter into the swap and immediately thereafter all interest rates rise by 0.5% with the oil forward prices remain unchanged. What happens to the value of your swap position? (c) What if interest rates fall by 0.5%? (d) What hedging instrument would have protected you against interest rate risk in this position? Solution. (a) The swap price x satisﬁes the equation x 1.06 + x 1.0652 + x 1.073 = 20 1.06 + 21 1.0652 + 22 1.073. Solving for x we ﬁnd x = $20.952. (b) You should short 3 forward contracts. This strategy is illustrated in the table below. 624 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES Year Payment from Gold Buyer Short Forward Net 1 P1 −$20.952 $20 −p1 −$0.952 2 P2 −$20.952 $21 −p2 $0.048 3 P3 −$20.952 $22 −p2 $1.048 With the 0.5% increase in rates we ﬁnd that the net present value of the loan is −0.952 1.065 + 0.048 1.072 + 1.048 1.0753 = −$0.0081. The dealer never recovers from the increased interest rate he faces on the overpayment of the ﬁrst swap payment. (c) If interest rates fall by 0.5% then the net present value of the loan is −0.952 1.055 + 0.048 1.062 + 1.048 1.0553 = $0.0083. The dealer makes money, because he gets a favorable interest rate on the loan he needs to take to ﬁnance the ﬁrst overpayment. (d) The dealer could have tried to hedge his exposure with a forward rate agreement or any other derivative protecting against interest rate risk 73 UNDERSTANDING THE ECONOMY OF SWAPS: A SIMPLE COMMODITY SWAP 625 Practice Problems Problem 73.1 Match (a)-(d) with (I)-(IV) (a) Notional amount is (b) The contracting parties in a swap are called (c) What the swap dealers receive for their services is (d) The ask-spread commission is (I) ask-spread bid (II) counterparties (III) the diﬀerence between the cash ﬂow one counterparty pays to the dealer and what dealer pays out to the other counterparty (IV) used to calculate the payments that will be exchanged between the two parties. Problem 73.2 ‡ Zero-coupon risk-free bonds are available with the following maturities and yield rates (eﬀective, annual): Maturity(in years) yield % 1 6 2 6.5 3 7 You need to buy corn for producing ethanol. You want to purchase 10,000 bushels one year from now, 15,000 bushels two years from now, and 20,000 bushels three years from now. The current forward prices, per bushel, are $3.89, $4.11, and $4.16 for one, two, and three years respectively. You want to enter into a commodity swap to lock in these prices. Which of the following sequences of payments at times one, two, and three will NOT be acceptable to you and to the corn supplier? (A) $38,900; $61,650; $83,200 (B) $39,083; $61,650; $82,039 (C) $40,777; $61,166; $81,554 (D) $41,892; $62,340; $78,997 (E) $60,184; $60,184; $60,184 Problem 73.3 A manufacturer uses oil in his profession. He needs 10,000 barrels in two months, 12,000 barrels in four months, and 15,000 barrels in six months. The manufacturer enters into a long oil swap. The 626 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES payment of the swap will be made at the delivery times. The current forward prices, per barrel, are $55, $56, and $58 for two, four, and six months re-spectively. Zero-coupon risk-free bonds are available with the following maturities and yield rates (nominal, convertible monthly): Maturity(in months) yield % 2 4.5 4 4.55 6 4.65 (a) Suppose level payments of R are made at each of the delivery times. Calculate R. (b) Suppose that the cash swap payment consists of a unique payment per barrel made at each of the delivery times. Calculate the no arbitrage swap price per barrel. Problem 73.4 Suppose that a transportation company must buy 1000 barrels of oil every six months, for 3 years, starting 6 months from now. Instead of buying six separate long forward contracts, the company enters into a long swap contract. The payment of the swap will be made at the delivery times. The current forward prices, per barrel, are $55, $57, $57, $60, $62, and $64 for 6-, 12-, 18-, 24-, 30-, and 36- months respectively. Zero-coupon risk-free bonds are available with the following maturities and yield rates (nominal, convertible semiannually): Maturity(in months) yield % 6 5.5 12 5.6 18 5.65 24 5.7 30 5.7 36 5.75 (a) With level payments, ﬁnd the price per barrel of oil using the swap. (b) Suppose the swap is settled in cash. Assume that the spot rate for oil in 18 months is $57. Calculate the payment which the counterparty receives. (c) Suppose that immediately after the swap is signed up, the future prices of oil are given by the table Forward Price $55 $58 $59 $61 $62 $63 Expiration (in months) 6 12 18 24 30 36 Find the market value of the swap for the company. 73 UNDERSTANDING THE ECONOMY OF SWAPS: A SIMPLE COMMODITY SWAP 627 Problem 73.5 Suppose that a jewelry manufacturer needs to purchase 100 ounces of gold one year from now and another 100 ounces in two years, and that the forward price for gold is $750 per ounce for delivery in one year and $800 for delivery in two years. The manufacturer could enter into two long forward contracts for these two dates at these two prices, and eliminate the risk of price variation. Alternatively, a single swap agreement could be used to ﬁx a single price for the purchases on these two dates. Suppose that the 1-year and 2-year spot rates are 5% and 6% respectively. (a) With level payments, ﬁnd the price per ounce of gold using the swap. (b) Suppose that the payments are not leveled, say with payment P1 per ounce for year 1 and P2 for year 2. Set the equation that P1 and P2 must satisfy. (c) What is the notional amount of the swap? (d) The buyer pays an amount larger than the forward price in 1 year. The diﬀerence is considered as a loan for the counterparty. What is the value of this amount per ounce? (e) The buyer pays an amount smaller than the forward price in 2 years. The diﬀerence is considered as the counterparty repayment of the loan. What is the value of this amount per ounce? (f) What is the implied forward rate from year 1 to year 2 of the loan? Problem 73.6 Consider the information of the previous problem. Suppose the forward prices are $780 and $830. The 1- and 2-year interest rates are 5% and 6% respectively. Find the new swap price. Problem 73.7 Suppose that gold forward prices for 1 year, 2 years, and 3 years are $750, $800, and $850. The eﬀective annual interest rates are shown in the table below Maturity(in years) yield % 1 6 2 6.5 3 7 (a) What is the swap price? (b) What is the price of a 2-year swap with the ﬁrst settlement two years from now and the second in 3 years? Problem 73.8 Consider the 2-year gold swap of Problem 73.5. Suppose a dealer is paying the spot price and receiving the swap price. What position in gold forward contracts will hedge gold price risk in this position? 628 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES Problem 73.9 Consider the 2-year gold swap of Problem 73.5. Suppose a dealer is paying the swap price and receiving the spot price. What position in gold forward contracts will hedge gold price risk in this position? 74 INTEREST RATE SWAPS 629 74 Interest Rate Swaps In this section we consider another type of swaps: the interest rate swaps. Companies use interest rate swap as a means of converting a series of future interest payments that vary with changes in interest rates (ﬂoating payments) to a series of ﬁxed level payments (i.e. swap rate), or vice-versa. The ﬁxed stream of payments are computed with respect to a rate determined by the contract. The ﬂoating stream of payments are determined using a benchmark, such as the LIBOR1 The amount in which the interest payments is based is called the notional principal. The life of a swap is called the swap term or swap tenor. Interest rate swaps are used to hedge against or speculate on changes in interest rates: They are often used by ﬁrms to alter their exposure to interest-rate ﬂuctuations, by swapping ﬁxed-rate obligations for ﬂoating rate obligations, or vice versa. By swapping interest rates, a ﬁrm is able to alter its interest rate exposures and bring them in line with management’s appetite for interest rate risk. They are also used speculatively by hedge funds or other investors who expect a change in interest rates or the relationships between them. Interest rate swaps are also very popular due to the arbitrage opportunities they provide. The interest rate swap market is closely linked to the Eurodollar2 futures market which trades at the Chicago Mercantile Exchange. All interest rate swaps payments are settled in net: If the ﬁxed rate is greater than the ﬂoating rate, the buyer pays the counterparty the diﬀerence. In contrast, if the ﬁxed rate is smaller than the ﬂoating rate, the counterparty pays the buyer the diﬀerence. The interest rate determination date for the ﬂoating interest payment occurs at the beginning of the period with payment due at the end of the period. It is important to keep in mind that there is no exchange of the notional principal in a swap. Example 74.1 On December 31, 2006, ﬁrm ABC enters into an interest rate swap with ﬁrm XYZ. The notional principal is $100 m. Firm ABC pays the ﬁxed rate of 5% compounded annually and receives ﬂoating rate from ﬁrm XYZ at a 12-month LIBOR. The swap term is 5 years. That is, the two parties exchange payments annually on December 31, beginning in 2007 and concluding in 2011. This swap is illustrated in Figure 74.1 1The (London Interbank oﬀer rate) LIBOR is the most widely used reference rate for short term interest rates world-wide. The LIBOR is published daily the (British Bankers Association) BBA. It is based on rates that large international banks in London oﬀer each other for interbank deposits. Rates are quoted for 1-month, 3-month, 6-month and 12-month deposits. 2Eurodollars are US dollars deposited at banks outside the United States, primarily in Europe. The interest rate paid on Eurodollars is largely determined by LIBOR. Eurodollar futures provide a way of betting on or hedging against future interest rate changes. 630 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES Figure 74.1 (a) On December 31,2006 the 12-month LIBOR rate was 4.5%. Determine the amount paid by ABC to XYZ on December 31,2007. (b) On December 31,2006 the 12-month LIBOR rate was 5.5%. Determine the amount received by ABC from XYZ on December 31,2007. Solution. (a) On December 31,2007 ABC will pay XYZ the amount $100m×0.05 = $5, 000, 000 and receives from XYZ the amount $100m×0.045 = $4, 500, 000. Thus, on December 31,2007, ABC pays XYZ the net payment of $500,000. (a) On December 31,2007 ABC will pay XYZ the amount $100m×0.05 = $5, 000, 000 and receives from XYZ the amount $100m×0.055 = $5, 500, 000. Thus, on December 31,2007, ABC receives from XYZ the net amount of $500,000. On net, ABC is paying At no point does the principal change hands, which is why it is referred to as a notional amount Remark 74.1 A ﬁxed-for-ﬂoating interest rate swap is often referred to as a plain vanilla swap because it is the most commonly encountered structure. Counterparty Risk of a Swap Unlike future markets, swaps are over-the-counter instruments and so they are not backed by the guarantee of a clearing house or an exchange. If one party defaults, they owe to the other party at most the present value of net swap payments that they are obligated to make at current market prices. Asset Swap An asset swap is an interest rate swap used to convert the cash ﬂows from and underlying security (such a bond) from ﬁxed coupon to ﬂoating coupon. Thus, an asset swap converts a ﬁxed-rate bond 74 INTEREST RATE SWAPS 631 to a synthetic ﬂoating-rate bond. The asset swap transaction is illustrated in Figure 74.2. Figure 74.2 Asset swaps are commonly used by fund managers who own ﬁxed-rate bonds and wish to have ﬂoating-rate exposure while continuing to own the bonds. Computing the Swap Rate Consider an interest rate swap with n settlements occurring on dates ti, i = 1, 2, · · · , n. Let P(0, ti) be the price of a $1 face value zero-coupon bond maturing on date ti. Notice that P(0, ti) is the discount factor from time zero to time ti based on the ti−year spot rate. That is, if it is the t−year spot rate then P(0, t) = (1 + it)−t. Let r0(ti−1, ti) be the implied forward rate between the times ti−1 and ti. From Section 50, we have r0(ti−1, ti) = P(0, ti−1) P(0, ti) −1. An interest rate swap acts much like the oil price swap described in the previous section. When initiated, the swap has a value of zero, i.e., the ﬁxed rate payments and the variable rate payments have equal present values. In terms of a formula, we have: R n X i=1 P(0, ti) = n X i=1 P(0, ti)r0(ti−1, ti) where R denote the ﬁxed swap rate. Solving this last equation for R we obtain R = Pn i=1 P(0, ti)r0(ti−1, ti) Pn i=1 P(0, ti) 632 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES where Pn i=1 P(0, ti)r0(ti−1, ti) is the present value of the variable payments and Pn i=1 P(0, ti) is the present value of a $1 annuity when interest rates vary over time. Note that the swap rate can be interpreted as a weighted average of the implied forward rates, where zero-coupon bond prices are used to determine the weights. Now, since r0(ti−1, ti) = P(0, ti−1) P(0, ti) −1 we can write n X i=1 P(0, ti)r0(ti−1, ti) = n X i=1 [P(0, ti−1) −P(0, ti) = 1 −P(0, tn) so that R = 1 −P(0, tn) Pn i=1 P(0, ti) This formula can be rearranged to obtain P(0, tn) + R n X i=1 P(0, ti) = 1. This formula states that a payment of 1 that will be received n years from now, plus a payment of R at the end of each of the next n years, has a total present value of 1. This is the valuation equation for a bond priced at par (i.e. face value $1) with a coupon rate R. Thus, the swap rate is the coupon rate for a bond that is selling at par. Example 74.2 The following table lists prices of zerocoupon $1face value bonds with their respective maturities: Years to Maturity Price 1 $0.956938 2 $0.907029 3 $0.863838 (a) Calculate the 1-, 2-, and 3-year spot rates of interest. (b) Calculate the 1- and 2-year forward rates of interest. (c) Calculate the coupon rate R for a 3-year bond with annual coupons whose face value, redemption value and price are all one. Solution. (a) Let i1 be the spot rate of interest. Then 1 + i1 = P(0, 1)−1 = 0.956938−1 →i1 = 0.956938−1 − 74 INTEREST RATE SWAPS 633 1 = 4.499967135%. Likewise, i2 = 0.907029 −1 2 −1 = 5.000027694% and i3 = 0.863838 −1 3 −1 = 4.99991613% (b) The 1-year forward rate is given by r0(1, 2) = P(0, 1) P(0, 2) −1 = 0.956938 0.907029 −1 = 5.502470153%. Likewise, the 2-year forward rate is given by r0(2, 3) = P(0, 2) P(0, 3) −1 = 0.907029 0.863838 −1 = 4.999895814%. (c) The coupon rate is given by R = 1 −P(0, 3) P(0, 1) + P(0, 2) + P(0, 3) = 1 −0.863838 0.956938 + 0.907029 + 0.863838 = 4.991632466% Deferred Swaps By a k-deferred swap we mean a swap whose exchange of payments start k periods from now. The (deferred) swap rate is applied in the future but it is agreed upon today. It is computed as follows R = Pn j=k P(0, tj)r0(tj−1, tj) Pn j=k P(0, tj) = Pn j=k P(0, tj) P(0,tj−1) P(0,tj) −1 Pn j=k P(0, tj) = Pn j=k(P(0, tj−1) −P(0, tj) Pn j=k P(0, tj) =P(0, tk−1) −P(0, tn) Pn j=k P(0, tj) Example 74.3 Given the following zero-coupon bond prices: Quarter 1 2 3 4 5 6 7 8 P(0, t) 0.9852 0.9701 0.9546 0.9388 0.9231 0.9075 0.8919 0.8763 What is the ﬁxed-rate (deferred swap rate) in a 6-quarter interest rate swap with the ﬁrst settlement in quarter 3? 634 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES Solution. We have R =P(0, 2) −P(0, 8) P8 j=3 P(0, j) = 0.9701 −0.8763 0.9546 + 0.9388 + 0.9231 + 0.9075 + 0.8919 + 0.8763 =1.71% The Value of a Swap At initiation of the swap, the value of the swap is set to be zero to both parties. The ﬁxed rate of the swap is then calculated such that the present value of the ﬁxed payments equals that of the ﬂoating payments. As time evolves, the interest rate may move upward or downward so that the value of the ﬂoating payments changes. Therefore, the value of the swap changes in later times, and its value at each time is given by the diﬀerence in the present values of the remaining cash ﬂows from the two parties. Even in the absence of changes in interest rates, as soon as the ﬁrst swap payment occurs, the swap will have a value. The swap will have a positive value for the party who paid too much (e.g., the ﬁxed rate payer if the ﬁxed rate exceeds the variable rate on the ﬁrst payment date), and a negative value for the party who received the net payment. As with the commodity swap, an interest rate swap is equivalent to entering into a series of forward contracts and also undertaking some borrowing and lending. 74 INTEREST RATE SWAPS 635 Practice Problems Problem 74.1 Interest rate swaps allow one party to exchange a: (A) ﬂoating interest for a ﬁxed rate over the contract term. (B) ﬁxed interest rate for a lower ﬁxed rate over the contract term. (C) ﬂoating interest rate for a lower ﬂoating rate over the contract term. Problem 74.2 All of the following are types of ﬁnancial instruments except for: (A) currency forward contracts (B) currency futures contracts (C) currency options (D) swap agreements (E) money-market hedge Problem 74.3 On March 5,2004, an agreement by Microsoft to receive 6-month LIBOR and pay a ﬁxed rate of 5% per annum every 6 months for 3 years on a notional principal of $100 million. Complete the following table: Date 6-month Fixed Floating Net LIBOR rate Payment Payment Cash Flow March 5,2004 4.2% Sept 5,2004 4.8% March 5, 2005 5.3% Sept 5, 2005 5.5% March 5, 2006 5.6% Sept 5, 2006 5.9% March 5,2007 6.4% Problem 74.4 Firm ABC is paying $750,000 in interest payments a year while Firm XYZ is paying LIBOR plus 75 basis points (i.e. 0.75%) on $10,000,000 loans. The current LIBOR rate is 6.5%. Firm ABC and XYZ have agreed to swap interest payments, how much will paid to which Firm this year? Problem 74.5 Given the following information 636 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES t (in months) P(0, t) 3 0.986923 6 0.973921 9 0.961067 12 0.948242 Calculate the annual nominal interest rate compounded quarterly for a loan with the following maturity dates: 3, 6, 9 and 12 months. Problem 74.6 Given the following information t (in years) P(0, t) 1 0.95785 2 0.91049 3 0.85404 Calculate the 3-year interest rate swap. Problem 74.7 The ﬁxed rate for a 2-year interest rate swap is 6.06%. The 2-year spot rate is 6.10%. What is the price of a 1-year zero-coupon bond with a maturity value of 1,000? Problem 74.8 Given the following zero-coupon bond prices: Quarter 1 2 3 4 5 6 7 8 P(0, t) 0.9852 0.9701 0.9546 0.9388 0.9231 0.9075 0.8919 0.8763 What is the ﬁxed-rate (deferred swap rate) in a 5-quarter interest rate swap with the ﬁrst settlement in quarter 2? Problem 74.9 Given the following zero-coupon bond prices: Quarter 1 2 3 4 5 6 7 8 P(0, t) 0.9852 0.9701 0.9546 0.9388 0.9231 0.9075 0.8919 0.8763 What is the swap rate in an 8-quarter interest rate swap? Problem 74.10 The following table lists prices of zerocoupon $1face value bonds with their respective maturities: 74 INTEREST RATE SWAPS 637 Years to Maturity Price 1 $0.95238 2 $0.89845 3 $0.842.00 Complete the following table: Years to Maturity Price Spot Rate in r0(n −1, n) 1 $0.95238 2 $0.89845 3 $0.842.00 Problem 74.11 Using the data in the previous table, ﬁnd the ﬁxed rate (i.e. the swap rate R). 638 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES 75 Risk Management Risk management consists of identifying the sources of risk, choosing the ones to be hedged, and choosing the way of hedging. A ﬁrm’s risk management consists of the use of ﬁnancial derivatives such as forwards, calls, and puts to alter its exposure to risk and protect its proﬁtability. We will next discuss risk management from both the producer’s perspective as well as the buyer’s perspective. The Producer’s Perspective A producer normally has a long position that consists of its product. In order to protect its future proﬁts from price ﬂuctuations, the producer can oﬀset his long position by using forward contracts, puts options, call options, or some combination of these derivatives. This process is also referred to as hedging or insuring the long position. Some of the risk management strategies for a producer include the following: (1) Hedging with a forward contract: In order for the producer to lock in the price of his product, the producer can enter into a short forward contract that ﬁxes the future sale price at the current forward price. There is no initial cost for the producer. For example, ABC Inc, an oil producing ﬁrm, enters into a short forward contract, agreeing to sell oil at today’s current price of $65 a barrel in one year. (2) Hedging with a put option: Creating a ﬂoor by purchasing a put option which limits the losses if the price declines, but allows unlimited proﬁt if the price rises. Note that the option’s premium is a cost to the producer for creating the ﬂoor. For example, ABC Inc purchases 65-strike put at a premium of $3.50 per barrel. In this case, the put option plays the role of an insurance that guarantees a minimum price of $65 for a barrel of oil. (3) Insuring by selling a call: A written call (selling a cap) sets a maximum price (and therefore limit proﬁt) for the product if the price rises, but does not limit losses if the price declines. The premium received by the producer helps reduce the losses. For example, ABC Inc sells a 65-strike call for a premium of $3.50 per barrel. (4) Creating a collar by purchasing a put at one strike price and writing a call at a higher strike price. A collar sets maximum and minimum prices (the call strike and the put strike, re-spectively) that the producer may realize for its product. The producer is exposed to the risk of variation between these two prices, but is not aﬀected by price variation above or below this range. The put and call options constitute the collar, but in combination with the producer’s long position in the product, they form a bull spread. 75 RISK MANAGEMENT 639 Example 75.1 ABC is a coppermining company with ﬁxed costs $0.75/lb and variable costs of $2.25/lb. (a) If ABC does nothing to hedge price risk, what is its proﬁt 1 year from now, per pound of copper? (b) Suppose that ABC can enter a short forward contract agreeing to sell its copper production one year from now. The 1-year forward price is $3.5/lb. What is the estimated proﬁt one year from now? (c) Suppose that ABC would like to beneﬁt if the price of the copper goes higher than $3.5/lb. It could use options. One year option prices for copper are given below Strike 3.3 3.4 3.5 3.6 3.7 3.8 3.9 Call 0.42567 0.3762 0.33119 0.29048 0.25386 0.22111 0.19196 Put 0.23542 0.28107 0.33119 0.3856 0.44411 0.50648 0.57245 The 1-year continuously compounded interest rate is 5%. Compute the estimated proﬁt in 1 year if ABC buys put options with a strike of $3.7/lb and with a strike of $3.4/lb. (d) Compute the estimated proﬁt in 1 year if ABC sells call options with a strike of $3.7/lb and with a strike of $3.4/lb. (e) Compute the proﬁt in 1 year if ABC buys a 3.6-strike put and sells a 3.9-strike call. Solution. (a) The proﬁt is the one-year spot price minus total cost. That is, P1 −3. A graph of the unhedged position is shown in Figure 75.1. (b) In this case, the proﬁt is P = P1 −3 + (3.5 −P1) = $0.5. A graph of the hedged position is shown in Figure 75.1. For an uninsured position, the possible losses can be very high. However, by entering the forward, the company has a ﬁxed beneﬁt. Figure 75.1 (c) The proﬁt per pound that results from buying the 3.7-strike put is P1 −3 + max{3.7 −P1, 0} −0.44411e0.05. 640 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES The graph of the proﬁt is given in Figure 75.2. Figure 75.2 The proﬁt per pound that results from buying the 3.4-strike put is P1 −3 + max{3.4 −P1, 0} −0.28107e0.05. The graph of the proﬁt is given in Figure 75.2. With the 3.4-strike put, the company makes $0.17139924/lb more than the 3.7-strike put if the price of copper is over $3.7/lb. However, its guaranteed proﬁt is $0.104519233, which is smaller than the guaranteed proﬁt under a 3.7-strike put. Buying a put is like buying insurance against small spot prices. The larger the strike price the larger the price of the insurance. A producer company needs to buy a put with a strike large enough to cover low spot prices. If the strike put is too large, the company will be wasting money in insurance which it does not need. (d) The proﬁt per pound that results from selling the 3.7-strike call is P1 −3 −max{P1 −3.7, 0} + 0.25386e0.05. The graph of the proﬁt is given in Figure 75.3. 75 RISK MANAGEMENT 641 Figure 75.3 The proﬁt per pound that results from selling the 3.4-strike call is P1 −3 −max{P1 −3.4, 0} + 0.3762e0.05. The graph of the proﬁt is given in Figure 75.3. (e) The proﬁt is S1 −3 + max{3.6 −S1, 0} −max{3.4 −S1, 0} −(0.3856 −0.19196)e0.05. Under this strategy the proﬁt of the company is very close to that of a forward contract. But, instead of winning a constant of $0.5/lb. in the forward contract, the company’s proﬁt varies with the future price of copper, although not much The Buyer’s Perspective A buyer of the producer’s product is in the opposite position from the producer. The buyer can engage in any of the above risk management strategies, but would do the opposite of what the producer does. The buyer’s possible strategies are therefore as follows: • Enter into a long forward contract. • Sell a put option (ﬂoor). • Buy a call option (cap). Example 75.2 MBF is a wire manufacturer. It buys copper and manufactures wires. Suppose the ﬁxed cost of a unit of wire is $2 and the noncopper variable cost is $1.25. A unit of wire sells for $5 plus the price of copper. One pound of copper is needed to manufacture one unit of wire. (a) What is the unhedged proﬁt of MBF? (b) Suppose that MBF buys a forward contract with forward price of $3.5/lb. What is the proﬁt one year from now? Solution. (a) MBF needs copper to manufacture its product. Thus, an increase in copper price results in an increase in cost of making the wire. So one may think that the company needs to hedge for the price ﬂuctuations. But since the company will adjust its selling price to reﬂect the change in the copper price, this chnage will cancel the copper price risk and as a result there is no need for hedging. The unhedged proﬁt per pound of copper in one year from today is Revenue −Cost = P1 + 5 −(2 + 1.25 + P1) = 1.75. 642 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES We see that the proﬁts of MBF do not depend on the price of copper. Cost and revenue copper price risk cancel each other out. (b) In this case, the proﬁt per pound of copper one year from today is given by P1 + 5 −(2 + 1.25 + P1) + P1 −3.5 = P1 −1.75. Note that buying a forward introduces price risk Example 75.3 Alumco is a company that manufactures dishwashers. It uses aluminum to make their product. Each dishwasher requires 5 pounds of aluminum. The ﬁxed cost is $100 and the selling price is $350. (a) What is the unhedged proﬁt one year from now? (b) Alumco could faces severe loses if the price of the aluminum goes very high. To hedge risk, the company buys forward contract with the current forward price of $40 per pound. What is the proﬁt one year from now per dishwasher? (c) Another alternative to hedge the aluminum price risk is to buy a call. Suppose that Alumco buys a 35-strike call with an expiration date one year from now, nominal amount 5 lbs, and for a premium of $7.609104. The one year continuously compounded interest rate is 6%. Find the proﬁt one year from now per dishwasher. (d) Another alternative to hedge the aluminum price risk is to sell a put. Suppose that Alumco sells a 35-strike put with an expiration date one year from now, nominal amount 5 lbs, and for a premium of $7.609104. Find the proﬁt one year from now per dishwasher. Solution. (a) Let P1 be the price of one pound of aluminum one year from now. The proﬁt from selling one dishwasher one year from now is given by 350 −(5P1 + 100) = 250 −5P1. The unisured proﬁt is shown in Figure 75.4. (b) The proﬁt of entering a long forward is 350 −(5 × 40 + 100) = $50. The graph of this proﬁt is shown in Figure 75.4. (c) The proﬁt that results from buying the call is 250 −5S1 + 5 max{S1 −35, 0} −5(7.609104)e0.06 = 209.602 −5S1 + 5 max{S1 −35, 0}. 75 RISK MANAGEMENT 643 The graph of the proﬁt is shown in Figure 75.4. (d) The proﬁt that results from selling the put is 250 −5S1 −5 max{35 −S1, 0} + 5(7.609104)e0.06 = 290.40 −5S1 −5 max{35 −S1, 0} Figure 75.4 Why do ﬁrms manage risks through derivatives? Firms engage in hedging for a variety of reasons namely: • To lower expected taxes • To lower ﬁnancial distress costs: A large loss can threaten the survival of a ﬁrm. For example, a money-losing ﬁrm may be unable to meet ﬁxed obligations (such as, debt payments and wages). If a ﬁrm appears in ﬁnancial distress, customers may be less willing to purchase its goods (for the fear that the ﬁrm may go out of business and therefore will not provide warranty for its goods). A ﬁrm uses derivatives to transfer income from proﬁt states to loss states. Hedging allows a ﬁrm to reduce the probability of bankruptcy or ﬁnancial distress. • To lower costly external ﬁnance: Usually when a form is in a loss state they have to oﬀset that by either using cash reserve or raising funds (by borrowing or issuing bonds). Borrowing entails costs such as underwriting fees and high interest rates (because the lender fears for his money when the borrower is in s loss state). If a ﬁrm uses its cash reserve to pay for the losses, this reduction in cash can increase the probability of costly external ﬁnancing and can lead a ﬁrm to forego invest-ment projects it would have taken had cash been available to use for ﬁnancing. Thus, hedging with derivatives can safeguard cash reserves and reduce the probability of raising funds externally. • Increasing debt capacity: That is the amount a company can borrow. • Managerial risk aversion: Firm managers are typically not well-diversiﬁed. Salary, bonus, 644 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES and compensation are tied to the performance of the ﬁrm. Poor diversiﬁcation makes managers risk-averse, i.e., they are harmed by a dollar of loss more than they are helped by a dollar of gain. Managers have incentives to reduce uncertainty through hedging. As there are reasons for hedging, there are also reasons why a company may choose not to hedge, namely: • Transaction costs of engaging in hedges (such as commissions and the bid-ask spread) • The cost of expertise required to analyze a hedging strategy • The cost of monitoring and controlling the hedging transactions. • Potential collateral requirements associated with some types of hedging. • The tax and accounting consequences of hedges. In the real world, small companies are discouraged to do derivatives because the reasons above. However, large companies have ﬁnancial, accounting and legal departments which allow them to take advantage of the opportunities on market derivatives. The ﬁnancial department of a large company can asses derivatives as well or better than the market does. Their legal and accounting departments allow them to take advantage of the current tax laws. Answer Key The answer key to the book can be requested directly from the author through email: mﬁ-nan@atu.edu 645 646 ANSWER KEY Bibliography S. Kellison , The Theory of Interest, 2nd Edition (1991), McGraw-Hill. R.L. McDonald , Derivatives Markets, 2nd Edition (2006), Pearson. SOA/CAS, Exam FM Sample Questions. 647
15701	https://pubmed.ncbi.nlm.nih.gov/15474239/	Major placenta praevia should not preclude out-patient management - PubMed Clipboard, Search History, and several other advanced features are temporarily unavailable. Skip to main page content An official website of the United States government Here's how you know The .gov means it’s official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you’re on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. 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Major placenta praevia should not preclude out-patient management Corinne D B Love1,Kevin J Fernando,Liz Sargent,Rhona G Hughes Affiliations Expand Affiliation 1 Simpson Centre for Reproductive Health, The Royal Infirmary Little France, Edinburgh EH16 45A, UK. clove@htbs.org.uk PMID: 15474239 DOI: 10.1016/j.ejogrb.2003.10.039 Item in Clipboard Comparative Study Major placenta praevia should not preclude out-patient management Corinne D B Love et al. Eur J Obstet Gynecol Reprod Biol.2004. Show details Display options Display options Format Eur J Obstet Gynecol Reprod Biol Actions Search in PubMed Search in NLM Catalog Add to Search . 2004 Nov 10;117(1):24-9. doi: 10.1016/j.ejogrb.2003.10.039. Authors Corinne D B Love1,Kevin J Fernando,Liz Sargent,Rhona G Hughes Affiliation 1 Simpson Centre for Reproductive Health, The Royal Infirmary Little France, Edinburgh EH16 45A, UK. clove@htbs.org.uk PMID: 15474239 DOI: 10.1016/j.ejogrb.2003.10.039 Item in Clipboard Full text links Cite Display options Display options Format Abstract Objective: To review current management of women with major and minor placenta praevia in view of the recommendations made in the RCOG guideline 2001. To assess whether out-patient care was detrimental to pregnancy outcome. Study design: Retrospective observational study at the Simpson Memorial Maternity Pavilion, Edinburgh (a tertiary referral centre). One hundred and sixty-one women with major and minor placenta praevia between 1994 and 2000 were separated into those who experienced bleeding (antepartum haemorrhage (APH)) and those who had no bleeding during pregnancy (non-APH). Statistical analysis was carried out using SPSS. Results: There were 129 women (80%) in the APH group. Forty-three were out-patients at the time of delivery and 63% had a major degree of praevia. Thirty-two women were in the non-APH group. Sixty-eight were managed as out-patients and 50% had a major degree of praevia. Women with a major degree of praevia were not significantly more likely to experience bleeding. Women with APH were significantly more likely to be delivered early, by emergency caesarean section (C/S), of lower birthweight babies who required neonatal admission than the non-APH group. Conclusion: There is a place for out-patient management of women with placenta praevia. Caution is required with increasing number of bleeds but not degree of praevia. PubMed Disclaimer Similar articles Do women with placenta praevia without antepartum haemorrhage require hospitalization?Rosen DM, Peek MJ.Rosen DM, et al.Aust N Z J Obstet Gynaecol. 1994 May;34(2):130-4. doi: 10.1111/j.1479-828x.1994.tb02674.x.Aust N Z J Obstet Gynaecol. 1994.PMID: 7980298 Placenta praevia: maternal morbidity and place of birth.Olive EC, Roberts CL, Algert CS, Morris JM.Olive EC, et al.Aust N Z J Obstet Gynaecol. 2005 Dec;45(6):499-504. doi: 10.1111/j.1479-828X.2005.00485.x.Aust N Z J Obstet Gynaecol. 2005.PMID: 16401216 Women with placenta praevia and antepartum haemorrhage have a worse outcome than those who do not bleed before delivery.Lam CM, Wong SF, Chow KM, Ho LC.Lam CM, et al.J Obstet Gynaecol. 2000 Jan;20(1):27-31. doi: 10.1080/01443610063417.J Obstet Gynaecol. 2000.PMID: 15512459 Ante-partum haemorrhage: an update.Sinha P, Kuruba N.Sinha P, et al.J Obstet Gynaecol. 2008 May;28(4):377-81. doi: 10.1080/01443610802091487.J Obstet Gynaecol. 2008.PMID: 18604667 Review. [Placenta praevia--a diagnostic problem in hemorrhage in pregnancy].Loch EG.Loch EG.Gynakologe. 1989 Jun;22(3):185-9.Gynakologe. 1989.PMID: 2670700 Review.German.No abstract available. See all similar articles Cited by Evaluation of Antepartum Factors for Predicting the Risk of Emergency Cesarean Delivery in Pregnancies Complicated With Placenta Previa.Oğlak SC, Ölmez F, Tunç Ş.Oğlak SC, et al.Ochsner J. 2022 Summer;22(2):146-153. doi: 10.31486/toj.21.0138.Ochsner J. 2022.PMID: 35756596 Free PMC article. Prevalence of antepartum hemorrhage in women with placenta previa: a systematic review and meta-analysis.Fan D, Wu S, Liu L, Xia Q, Wang W, Guo X, Liu Z.Fan D, et al.Sci Rep. 2017 Jan 9;7:40320. doi: 10.1038/srep40320.Sci Rep. 2017.PMID: 28067303 Free PMC article. Publication types Comparative Study Actions Search in PubMed Search in MeSH Add to Search MeSH terms Adult Actions Search in PubMed Search in MeSH Add to Search Ambulatory Care / methods Actions Search in PubMed Search in MeSH Add to Search Female Actions Search in PubMed Search in MeSH Add to Search Gestational Age Actions Search in PubMed Search in MeSH Add to Search Hospitalization Actions Search in PubMed Search in MeSH Add to Search Humans Actions Search in PubMed Search in MeSH Add to Search Incidence Actions Search in PubMed Search in MeSH Add to Search Outpatients Actions Search in PubMed Search in MeSH Add to Search Placenta Previa / complications Actions Search in PubMed Search in MeSH Add to Search Placenta Previa / epidemiology Actions Search in PubMed Search in MeSH Add to Search Placenta Previa / therapy Actions Search in PubMed Search in MeSH Add to Search Pregnancy Actions Search in PubMed Search in MeSH Add to Search Pregnancy Outcome Actions Search in PubMed Search in MeSH Add to Search Pregnancy Trimester, Third Actions Search in PubMed Search in MeSH Add to Search Retrospective Studies Actions Search in PubMed Search in MeSH Add to Search Uterine Hemorrhage / epidemiology Actions Search in PubMed Search in MeSH Add to Search Uterine Hemorrhage / etiology Actions Search in PubMed Search in MeSH Add to Search Uterine Hemorrhage / therapy Actions Search in PubMed Search in MeSH Add to Search Related information MedGen LinkOut - more resources Full Text Sources Elsevier Science Full text links[x] Elsevier Science [x] Cite Copy Download .nbib.nbib Format: Send To Clipboard Email Save My Bibliography Collections Citation Manager [x] NCBI Literature Resources MeSHPMCBookshelfDisclaimer The PubMed wordmark and PubMed logo are registered trademarks of the U.S. Department of Health and Human Services (HHS). 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15702	https://en.wikisource.org/wiki/Page:1902_Encyclop%C3%A6dia_Britannica_-_Volume_25_-_A-AUS.pdf/60	Page:1902 Encyclopædia Britannica - Volume 25 - A-AUS.pdf/60 - Wikisource, the free online library Jump to content [x] Main menu Main menu move to sidebar hide Navigation Main Page Community portal Central discussion Recent changes Subject index Authors Random work Random author Random transcription Help Special pages Search Search [x] Appearance Appearance move to sidebar hide Text Small Standard Large This page always uses small font size Width Standard Wide The content is as wide as possible for your browser window. Donate Create account Log in [x] Personal tools Donate Create account Log in Page:1902 Encyclopædia Britannica - Volume 25 - A-AUS.pdf/60 [x] Add languages Previous page Next page Page Discussion Image Index [x] English Read Edit View history [x] Page Page logs… All logs AbuseFilter log Deletion log Move log Protection log Spam blacklist log Analysis… Analysis – XTools Analysis – Σ Basic statistics Copyright vio detector Link count Traffic report Search… Latest diff Search by contributor Search history – WikiBlame Search subpages Purge cache Subpages [x] Tools Tools move to sidebar hide Actions Read Edit View history Purge Hard purge Null edit General What links here Related changes Permanent link Page information Cite this page Get shortened URL Download QR code Print/export Printable version Download EPUB Download MOBI Download PDF Other formats In other projects From Wikisource This page needs to be proofread. 40 ACHILL —ACHROMATIC out into a butterfly flame. Soon afterwards, Billwiller introduced the idea of sucking air into the flame at or just below the burner tip, and at this juncture the Naphey or Dolan burner was introduced in America, the principle (employed being to use two small and widely separated jets instead of the two openings of the union jet burner, and to make each a minute Bunsen, the acetylene dragging in from the base of the nipple enough air to surround and protect it while burning from contact with the steatite. This class of burner has been very successful, and its introduction, together with the realization of the importance of purifying the gas before combustion, has removed perhaps the most important obstacle to the use of this beautiful illuminant. Authorities.—Dommer, L'Acetylene et ses applications. Paris, 1896.—Lewes, Acetylene. London, 1900.—Liebetanz, Calcium£arbid und Acetylen. Leipzig, 1899.—Pellxssier, L'eclairage a, 1'acetylene. Paris, 1897.—Perrodil, Le carbure de calcium et I'acitylene. Paris, 1897.—For a complete list of the various papers and memoirs on Acetylene, see Ludwig’s Fiihrer durch die gesammte Calciumcarbid- und Acchjlen-Literatur. Berlin, 1899. (v. B. L.) Achill Island, off the west coast of Ireland, part of the county of Mayo. How under the control of the Congested Districts Board, who have made efforts to improve the condition of the people. There is now a station at Achill Sound, which is crossed by a swivel bridge, opened in 1888. Population, 4677. Achin (Dutch Atjeh) and its dependencies form a government of Northern Sumatra, extending from 2° 53' N. on the W. coast to 4° 32' N. on the E. coast. The area of Achin is estimated at 20,520 square miles. Since 1874 the valley of the Achin river has been subjugated by the Dutch. The restriction of export and import to Achin (1888) and further regulation of the ports (1892), the death of the traitor Tuku Umar, and the successful expeditions of General van Heutsz (1898-99) on both coasts and in the valley, have broken resistance and firmly established Dutch government. A scheme to unite the coasts by a railway is under consideration. The administrative divisions are as follows—1. Great Achin (the nine districts within and beyond the military posts) with Poeloe Wai (isle); 2. dependencies (west coast, with the island of Simalu [Babi or Hog], north coast, east coast, and the southern settlements of Great Achin). Under the military and civil chief is a resident (for the regulation of shipping ,and Achin affairs), and under him again assistant-residents for the dependencies. Geographical knowledge of the Achin valley, river, and coast has been considerably advanced since 1874. In its upper part, near Selimun, the valley is 3 miles broad, the river having a breadth of 99 feet and a depth of l]? feet; but in its lower course, north ,of its junction with the Krung Daru, the valley broadens to 121 miles. The marshy soil is covered by rice-fields, and on higher ground by kampongs full of trees. The river at its mouth is 327 feet broad and 20-33 feet deep, but before it lies a sandbank covered at low water by a depth of only 4 feet. The coasts are low and the rivers insignificant, rising in the coast ranges and flowing through the coast states (the chief of which are Pedir, Gighen, and Samalanga on the N., Edi, Perlak, and Langsar on the E., Kluwah, Rigas, and Melabuh on the W.). The chief ports are Olehleh, the port of Kotaraja or Achin (formerly Kraton, now the seat of the Dutch Government), Segli on the N., Edi on the E., and Analabu or Melabuh on the W., all visited by steamers of the Royal Packet Company. The relief of the soil of Achin is imperfectly understood. With regard to the west coast, Resident van Langen has spoken of the Barisan and other parallel ranges which are characteristic of the island of Sumatra, and a geological description of a small portion of the same coast has been given by the mining engineer Renaud, but the interior, possibly a continuation of the Batta plateau, is unexplored. The population of Achin in 1898 was estimated at 535,432, of whom 328 were Europeans, 3933 Chinese, 30 Arabs, and 372 other foreign Asiatics. The natives of this commercial state are of very mixed origin (Hindu, Klings, Malay, Arab). They live in kampongs, collections of houses and gardens, which combine to form mukims or districts, which again combine to form sagis, of which there are three. The chief of a mukim is called an imeum, of a sagi a panglima sagi. The people of the highlands {prang tunong) differ in many respects from those of the lowlands (prang baroh). The means of subsistence are furnished by the culture of rice, betel (penang), tobacco, and pepper, but agriculture and stock-raising both declined during the war. The following 'industries are of some importance—^gold-working, weapon-making, silk-weaving, the making of pottery, fishing and coasting trade. The value of the exports (chiefly pepper) has of late years been about £58,000, of the imports from £165,000 to £250,000. Kruijt. Atjeh en de Atjehers. Leiden, 1877.—Van Langen. Atjeh’s Wesskud, Tijdschrift Aardrijko. Genotktsch. Amsterdam, 1888, p. 226.—Renaud. Jaarboek van het Mynwczen. 1882.— Jacobs Het famille-en Kampongleven op Groot Atjeh. Leiden, 1894. —Snouck Hurgronje. He Atjehers. Batavia, 1894. (C. M. K.) Achinsk, a district town of Russia, East Siberia, government of Yeniseisk, 110 miles by rail W. of Krasnoyarsk, and on the Chulym river. It was founded in 1642, and remained quite insignificant till lately, when steamers began to ply on the Chulym to the gold mines. There are tanneries and soap and candle works. Population (1860), 2501; (1897), 6714. Achromatic Objectives.—The general equation for two lenses in contact and of negligible thicknesses is, P = A(™-1) + B(w'-1), where P is the power of the combined system, or the reciprocal of its focal length, A and B are the sums of the reciprocals of the radii of curvatures of the first and second lenses respectively, and n and n are the indices of refraction of the materials of which these two lenses are made. It is obvious that in general the value of P varies with the refrangibility of light, i.e., with its wave-length. The mathematical condition which must obtain in order that the power shall be invariable is that the derivative of this equation shall vanish, or dP -r-da' - - = 0n = A. + B t-. dn dn Lenses, whether binary or multiple, subject to this condition and employed for the formation of real images are called achromatic objectives. These two equations would serve to determine the values of A and B, when P is assumed, and thus completely solve the problem in its elementary form, were it not unfortunately true that the value of the coefficient (called the dispersive ratio), for all practicable materials is far from constant throughout the range of wave-lengths which are involved in optical images. For example, for the kinds of glass most generally employed for large telescopes, this quantity varies from D80 in the extreme red to 2'20 in the extreme violet. The method of fixing upon the most advantageous value to substitute in the second equation was not obvious to the earlier opticians, even to those who produced some of the finest telescopes now in use; but it is easily demonstrated that the value should be that which obtains for light of Retrieved from " Category: Not proofread This page was last edited on 20 May 2020, at 17:00. 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15703	https://math.stackexchange.com/questions/2154998/finding-the-general-term-of-a-partial-sum-series	calculus - Finding the general term of a partial sum series? - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Finding the general term of a partial sum series? Ask Question Asked 8 years, 7 months ago Modified8 years, 7 months ago Viewed 4k times This question shows research effort; it is useful and clear -2 Save this question. Show activity on this post. If s n=5−n 3 n s n=5−n 3 n is the partial sum of the series of ∑∞k=1 a k∑k=1∞a k, can we find, a-) the general term of the series b-) the sum of the series? calculus sequences-and-series convergence-divergence Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Feb 21, 2017 at 19:59 Mauro ALLEGRANZA 100k 8 8 gold badges 75 75 silver badges 160 160 bronze badges asked Feb 21, 2017 at 18:21 OnurOnur 85 1 1 silver badge 9 9 bronze badges Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 1 Save this answer. Show activity on this post. Hint: a n=s n−s n−1 a n=s n−s n−1 ∑k=1∞a k=lim k→∞s k∑k=1∞a k=lim k→∞s k Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Feb 21, 2017 at 18:23 Simply Beautiful ArtSimply Beautiful Art 76.8k 13 13 gold badges 134 134 silver badges 301 301 bronze badges 11 So the sum is 5. What about the general term?Onur –Onur 2017-02-21 18:25:04 +00:00 Commented Feb 21, 2017 at 18:25 1 Have you tried my first hint? It's pretty direct.Simply Beautiful Art –Simply Beautiful Art 2017-02-21 18:25:53 +00:00 Commented Feb 21, 2017 at 18:25 s n=5−n/3 n s n=5−n/3 n s(n−1)=5−(n−1)/3(n−1)s(n−1)=5−(n−1)/3(n−1) Onur –Onur 2017-02-21 18:29:21 +00:00 Commented Feb 21, 2017 at 18:29 1 You actually want s n−1 s n−1.Simply Beautiful Art –Simply Beautiful Art 2017-02-21 18:32:04 +00:00 Commented Feb 21, 2017 at 18:32 1 Well, there shouldn't be a summation, and then it's all good and well.Simply Beautiful Art –Simply Beautiful Art 2017-02-21 18:51:06 +00:00 Commented Feb 21, 2017 at 18:51 \|Show 6 more comments This answer is useful 1 Save this answer. Show activity on this post. s n+1=10 3+1 3 s n−1 3 n+1 s n+1=10 3+1 3 s n−1 3 n+1 with lim∞s n=ℓ lim∞s n=ℓ then ℓ=10 3+1 3 ℓ ℓ=10 3+1 3 ℓ so ℓ=5 ℓ=5. a n=s n−s n−1=2 n−3 3 n a n=s n−s n−1=2 n−3 3 n. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Feb 21, 2017 at 18:41 answered Feb 21, 2017 at 18:25 NosratiNosrati 30.6k 7 7 gold badges 35 35 silver badges 65 65 bronze badges 4 Thanks. What about the general term?Onur –Onur 2017-02-21 18:36:19 +00:00 Commented Feb 21, 2017 at 18:36 Hm, a strange way of doing. If I may point out, you quietly assume the limit exists.Simply Beautiful Art –Simply Beautiful Art 2017-02-21 18:37:44 +00:00 Commented Feb 21, 2017 at 18:37 @MyGlasses lol, well, see if you can figure out why my hints work.Simply Beautiful Art –Simply Beautiful Art 2017-02-21 18:38:30 +00:00 Commented Feb 21, 2017 at 18:38 Well, it looks good now.Simply Beautiful Art –Simply Beautiful Art 2017-02-21 18:46:58 +00:00 Commented Feb 21, 2017 at 18:46 Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. 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15704	https://phys.libretexts.org/Bookshelves/Astronomy__Cosmology/Celestial_Mechanics_(Tatum)/02%3A_Conic_Sections/2.04%3A_The_Hyperbola	Skip to main content 2.4: The Hyperbola Last updated : Nov 1, 2022 Save as PDF 2.3: The Parabola 2.5: Conic Sections Page ID : 6795 Jeremy Tatum University of Victoria ( \newcommand{\kernel}{\mathrm{null}\,}) A hyperbola is the locus of a point that moves such that the difference between its distances from two fixed points called the foci is constant. We shall call the difference between these two distances 2a2a and the distance between the foci 2ae2ae, where ee is the eccentricity of the hyperbola, and is a number greater than 1. See figure II.28II.28. FIGURE II.28FIGURE II.28 For example, in a Young's double-slit interference experiment, the mmth bright fringe is located at a point on the screen such that the path difference for the rays from the two slits is mm wavelengths. As the screen is moved forward or backwards, this relation continues to hold for the mmth bright fringe, whose locus between the slits and the screen is therefore a hyperbola. The "Decca" system of radar navigation, first used at the D-Day landings in the Second World War and decommissioned only as late as 2000 on account of being rendered obsolete by the "GPS" (Global Positioning Satellite) system, depended on this property of the hyperbola. (Since writing this, part of the Decca system has been re-commissioned as a back-up in case of problems with GPS.) Two radar transmitters some distance apart would simultaneously transmit radar pulses. A ship would receive the two signals separated by a short time interval, depending on the difference between the distances from the ship to the two transmitters. This placed the ship on a particular hyperbola. The ship would also listen in to another pair of transmitters, and this would place the ship on a second hyperbola. This then placed the ship at one of the four points where the two hyperbolas intersected. It would usually be obvious which of the four points was the correct one, but any ambiguity could be resolved by the signals from a third pair of transmitters. In figure II.28II.28, the coordinates of F1F1 and F2F2 are, respectively, (−ae,0)(−ae,0) and (ae,0)(ae,0). The condition PF1−PF2=2aPF1−PF2=2a requires that [(x+ae)2+y2]12−[(x−ae)2+y2]12=2a, [(x+ae)2+y2]12−[(x−ae)2+y2]12=2a,(2.5.1) and this is the Equation to the hyperbola. After some arrangement, this can be written x2a2−y2a2(e2−1)=1, x2a2−y2a2(e2−1)=1,(2.5.2) which is a more familiar form for the Equation to the hyperbola. Let us define a length bb by b2=a2(e2−1). b2=a2(e2−1).(2.5.3) The Equation then becomes x2a2−y2b2=1, x2a2−y2b2=1,(2.5.4) which is the most familiar form for the Equation to a hyperbola. Example 1 When a meteor streaks across the sky, it can be tracked by radar. The radar instrumentation can determine the range (distance) of the meteoroid as a function of time. Show that, if the meteoroid is moving at constant speed (a questionable assumption, because it must be decelerating, but perhaps we can assume the decrease in speed is negligible during the course of the observation), and if the range rr is plotted against the time, the graph will be a hyperbola. Show also that, if r2r2 is plotted against tt, the graph will be a parabola of the form r2=at2+bt+c, r2=at2+bt+c,(2.4.1) where a=V2, b=−2V2t0, c=V2t20+r20, V=speed of the meteoroid, t0=time of closest approach, r0=distance of closest approacha=V2, b=−2V2t0, c=V2t20+r20, V=speed of the meteoroid, t0=time of closest approach, r0=distance of closest approach. Radar observation of a meteor yields the following range-time data: t(s)r(km)0.0101.4 ∗0.1103.00.2105.80.3107.80.4111.10.5112.60.6116.70.7119.30.8123.8 ∗0.9126.41.0130.61.1133.31.2138.11.3141.3 ∗ t(s)0.00.10.20.30.40.50.60.70.80.91.01.11.21.3r(km)101.4 ∗103.0105.8107.8111.1112.6116.7119.3123.8 ∗126.4130.6133.3138.1141.3 ∗(2.4.2) Assume that the velocity of the meteor is constant. Determine i. The time of closest approach (to 0.01 ss) ii. The distance of closest approach (to 0.1 kmkm) iii. The speed (to 1.0 km s−1km s−1) If you wish, just use the three asterisked data to determine aa, bb and cc. If you are more energetic, use all the data, and determine aa, bb and cc by least squares, and the probable errors of VV, t0t0 and r0r0. The distance between the two vertices of the hyperbola is its transverse axis, and the length of the semi transverse axis is aa − but what is the geometric meaning of the length bb? This is discussed below in the next subsection (on the conjugate hyperbola) and again in a later section on the impact parameter. The lines perpendicular to the xx-axis and passing through the foci are the two latera recta. Since the foci are at (±ae, 0)(±ae, 0), the points where the latera recta intersect the hyperbola can be found by putting x=aex=ae into the Equation to the hyperbola, and it is then found that the length ll of a semi latus rectum is l=a(e2−1). l=a(e2−1).(2.5.5) Definition: The Conjugate Hyperbola The Equation x2a2−y2b2=−1 x2a2−y2b2=−1(2.5.6) is the Equation to the conjugate hyperbola. The conjugate hyperbola is drawn dashed in figure II.28II.28, and it is seen that the geometric meaning of bb is that it is the length of the semi transverse axis of the conjugate hyperbola. It is a simple matter to show that the eccentricity of the conjugate hyperbola is e/√e2−1e/e2−1−−−−−√. Definition: The Asymptotes The lines y=±bxa y=±bxa(2.5.7) are the asymptotes of the hyperbola. Equation 2.5.72.5.7 can also be written x2a2−y2b2=0. x2a2−y2b2=0.(2.5.8) Thus x2a2−y2b2=c x2a2−y2b2=c(2.5.9) is the hyperbola, the asymptotes, or the conjugate hyperbola, if c=+1c=+1, 00 or −1−1 respectively. The asymptotes are drawn as dotted lines in figure II.28II.28. The semi angle ψ between the asymptotes is given by tanψ=b/a. tanψ=b/a.(2.5.10) Exercise 1 If the eccentricity of a hyperbola is ee, show that the eccentricity of its conjugate is e√e2−1ee2−1−−−−−√. Exercise 1: Corollary No one will be surprised to note that this implies that, if the eccentricities of a hyperbola and its conjugate are equal, then each is equal to √22–√. The Directrices The lines y=±a/ey=±a/e are the directrices, and, as with the ellipse (and with a similar proof), the hyperbola has the property that the ratio of the distance PF2PF2 to a focus to the distance PNPN to the directrix is constant and is equal to the eccentricity of the hyperbola. This ratio (i.e. the eccentricity) is less than one for the ellipse, equal to one for the parabola, and greater than one for the hyperbola. It is not a property that will be of great importance for our purposes, but is worth mentioning because it is a property that is sometimes used to define a hyperbola. I leave it to the reader to draw the directrices in their correct positions in figure II.28II.28. Parametric Equations to the Hyperbola. The reader will recall that the point (acosE,bsinE)(acosE,bsinE) is on the ellipse (x2/a2)+(y2/b2)=1(x2/a2)+(y2/b2)=1 and that this is evident because this Equation is the EE-eliminant of x=acosEx=acosE and y=bsinEy=bsinE. The angle EE has a geometric interpretation as the eccentric anomaly. Likewise, recalling the relation cosh2ϕ−sinh2ϕ=1cosh2ϕ−sinh2ϕ=1, it will be evident that (x2/a2)−(y2/b2)=1(x2/a2)−(y2/b2)=1 can also be obtained as the ϕϕ−eliminant of the Equations x=acoshϕ,y=bsinhϕ x=acoshϕ,y=bsinhϕ(2.5.11) These two Equations are therefore the parametric Equations to the hyperbola, and any point satisfying these two Equations lies on the hyperbola. The variable ϕϕ is not an angle, and has no geometric interpretation analogous to the eccentric anomaly of an ellipse. The Equations x=asecE,y=btanE x=asecE,y=btanE(2.5.12) can also be used as parametric Equations to the hyperbola, on account of the trigonometric identity 1+tan2E=sec2E1+tan2E=sec2E. In that case, the angle EE does have a geometric interpretation (albeit not a particularly interesting one) in relation to the auxiliary circle, which is the circle of radius a centred at the origin. The meaning of the angle should be evident from figure II.29II.29, in which EE is the eccentric angle corresponding to the point PP. FIGURE II.29FIGURE II.29 Impact Parameter A particle travelling very fast under the action of an inverse square attractive force (such as an interstellar meteoroid or comet - if there are such things - passing by the Sun, or an electron in the vicinity of a positively charged atomic nucleus) will move in a hyperbolic path. We prove this in a later chapter, as well as discussing the necessary speed. We may imagine the particle initially approaching from a great distance along the asymptote at the bottom right hand corner of figure II.30II.30. As it approaches the focus, it no longer moves along the asymptote but along an arm of the hyperbola. FIGURE II.30FIGURE II.30 The distance K2 F2K2 F2, which is the distance by which the particle would have missed F2F2 in the absence of an attractive force, is commonly called the impact parameter. Likewise, if the force had been a repulsive force (for example, suppose the moving particle were a positively charged particle and there were a centre of repulsion at F1F1, F1K1F1K1 would be the impact parameter. Clearly, F1K1F1K1 and F2K2F2K2 are equal in length. The symbol that is often used in scattering theory, whether in celestial mechanics or in particle physics, is bb - but is this bb the same bb that goes into the Equation to the hyperbola and which is equal to the semi major axis of the conjugate hyperbola? OF2=aeOF2=ae, and therefore K2F2=aesinψK2F2=aesinψ. This, in conjunction with tanψ=b/atanψ=b/a and b2=a2(e2−1)b2=a2(e2−1), will soon show that the impact parameter is indeed the same bb that we are familiar with, and that bb is therefore a very suitable symbol to use for impact parameter. Tangents to the Hyperbola Using the same arguments as for the ellipse, the reader should easily find that lines of the form y=mx±√a2m2−b2 y=mx±a2m2−b2−−−−−−−−√(2.5.13) are tangent to the hyperbola. This is illustrated in figure II.33II.33 for a hyperbola with b=a/2b=a/2, with tangents drawn with slopes 30∘30∘ to 150∘150∘ in steps of 5∘5∘. (The asymptotes have ψ=26∘34′ψ=26∘34′.) (Sorry, but there are no figures II.31II.31 or II.32II.32 - computer problems!) FIGURE II.31FIGURE II.31 Likewise, from similar arguments used for the ellipse, the tangent to the hyperbola at the point (x,y)(x,y) is found to be x1xa2−y1yb2=1. x1xa2−y1yb2=1.(2.5.14) Director Circle As for the ellipse, and with a similar derivation, the locus of the points of intersection of perpendicular tangents is a circle, called the director circle, which is of radius √(a2−b2)√(a2−b2). This is not of particular importance for our purposes, but the reader who is interested might like to prove this by the same method as was done for the director circle of the ellipse, and might like to try drawing the circle and some tangents. If b>ab>a, that is to say if ψ>45∘ψ>45∘ and the angle between the asymptotes is greater than 90∘90∘, the director circle is not real and it is of course not possible to draw perpendicular tangents. Rectangular Hyperbola If the angle between the asymptotes is 90∘90∘, the hyperbola is called a rectangular hyperbola. For such a hyperbola, b=ab=a, the eccentricity is √2√2, the director circle is a point, namely the origin, and perpendicular tangents can be drawn only from the asymptotes. The Equation to a rectangular hyperbola is x2−y2=a2 x2−y2=a2(2.5.15) and the asymptotes are at 45∘45∘ to the xx axis. Let Ox′, Oy′Ox′, Oy′ be a set of axes at 45∘45∘ to the xx axis. (That is to say, they are the asymptotes of the rectangular hyperbola.) Then the Equation to the rectangular hyperbola referred to its asymptotes as coordinate axes is found by the substitutions (xy)=(cos45∘sin45∘−sin45∘cos45∘)(x′y′) into x2−y2=a2. This results in the Equation x′y′=12a2=c2,wherec=a/√2, for the Equation to the rectangular hyperbola referred to its asymptotes as coordinate axes. The geometric interpretation of c is shown in figure II.32, which is drawn for c=1, and we have called the coordinate axes x and y. The length of the semi transverse axis is c√2. FIGURE II.32 The simple Equation y=1/x is a rectangular hyperbola and indeed it is this Equation that is shown in figure II.32. It is left to the reader to show that the parametric Equations to the rectangular hyperbola xy=c2 (we have dropped the primes) are x=ct, y=c/t, that lines of the form y=mx±2c√−m are tangent to xy=c2 (figure II.35, drawn with slopes from 90∘ to 180∘ in steps of 5∘ ), and that the tangent at (x1,y1) is x1y+y1x=2c. FIGURE II.33 Equation of a Hyperbola Referred to its Asymptotes as Axes of Coordinates We have shown that the Equation to a rectangular hyperbola referred to its asymptotes as axes of coordinates is x′y′=12a2=c2. In fact the Equation x′y′=c2 is the Equation to any hyperbola (centred at (0, 0)), not necessarily rectangular, when referred to its asymptotes as axes of coordinates, where c2=14(a2+b2) In the figure below I have drawn a hyperbola and a point on the hyperbola whose coordinates with respect to the horizontal and vertical axes are (x, y), and whose coordinates with respect to the asymptotes are (x′,y′). I have shown the distances x and y with blue dashed lines, and the distances x′ and y′ with red dashed lines. The semiangle between the asymptotes is ψ. The Equation to the hyperbola referred to the horizontal and vertical axes is x2a2−y2b2=1. From the drawing, we see that x=(x′+y′)cosψ,y=(y′−x′)sinψ. If we substitute these into Equation 2.5.18, and also make use of the relation tanψ=b/a (Equation 2.5.10), we arrive at the Equation to the hyperbola referred to the asymptotes as axes of coordinates: x′y′=14(a2+b2)=c2. Polar Equation to the Hyperbola We found the polar Equations to the ellipse and the parabola in different ways. Now go back and look at both methods and use either (or both) to show that the polar Equation to the hyperbola (focus as pole) is r=l1+ecosθ. This is the polar Equation to any conic section - which one being determined solely by the value of e. You should also ask yourself what is represented by the Equation r=l1−ecosθ. Try sketching it for different values of e. 2.3: The Parabola 2.5: Conic Sections
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Discover how to analyze a regression line, and work through examples of interpreting the slope and y-intercept. Updated: 11/21/2023 Table of Contents What Does It Mean to Interpret the Slope and Intercept? How to Interpret Slope How to Interpret Intercept Examples of Interpreting Slope and Y Intercept Lesson Summary Show Create an account LessonTranscript VideoQuizCourse Click for sound 8:05 You must c C reate an account to continue watching Register to view this lesson Are you a student or a teacher? I am a student I am a teacher Create Your Account To Continue Watching As a member, you'll also get unlimited access to over 88,000 lessons in math, English, science, history, and more. Plus, get practice tests, quizzes, and personalized coaching to help you succeed. Get unlimited access to over 88,000 lessons. Try it now Already registered? Log in here for access Go back Resources created by teachers for teachers Over 30,000 video lessons & teaching resources—all in one place. 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Your next lesson will play in 10 seconds 0:00 Interpreting Slope and… 0:25 Identifying Slope and… 1:38 Interpreting Slope 4:18 Interpreting Intercept 5:34 Practice Problem 6:56 Lesson Summary QuizCourseView Video Only Save Timeline 184K views Recommended lessons and courses for you Related LessonsRelated Courses ##### Line of Best Fit \| Definition, Formula & Examples 6:00 ##### Line of Best Fit \| Definition, Formula & Equation 6:05 ##### Least-Squares Regression \| Line Formula, Method & Examples 8:26 ##### Residual Plot \| Definition, Interpretation & Examples 5:19 ##### How to Draw a Trend Line 3:43 ##### How to Find the Equation of a Trend Line 6:31 ##### Trend Line in Math \| Definition, Formula & Graphs 7:00 ##### Understanding Simple Linear Regression \| Graphing & Examples 9:52 ##### Making Estimates and Predictions using Quantitative Data 4:07 ##### Predictor Variable Overview, Definition & Examples 4:34 ##### Interpolation in Statistics \| Definition, Methods & Formula 4:11 ##### Mean Squared Error \| Definition, Formula & Examples 5:02 ##### Response Variable in Statistics \| Definition & Examples 3:59 ##### Statistical Modeling Purpose, Types & Examples 10:20 ##### Center, Shape & Spread of a Distribution \| Overview & Examples 6:11 ##### Line of Best Fit \| Definition, Formula & Equation 6:05 ##### The Correlation Coefficient \| Definition, Formula & Calculation 9:57 ##### DSST Principles of Statistics Study Guide and Test Prep ##### TECEP Principles of Statistics: Study Guide & Test Prep ##### UExcel Statistics: Study Guide & Test Prep ##### Math 99: Essentials of Algebra and Statistics ##### 8th Grade Math ##### Calculus-Based Probability & Statistics ##### Contemporary Math ##### ORELA Mathematics: Practice & Study Guide ##### OAE Mathematics (027) Study Guide and Test Prep ##### Math 107: Quantitative Literacy ##### Intro to Public Relations ##### NYSTCE Multi-Subject - Teachers of Early Childhood (Birth-Grade 2)(211/246/245) Study Guide and Test Prep ##### Auditing & Assurance Services ##### Common Core Math - Statistics & Probability: High School Standards ##### Common Core Math Grade 7 - Statistics & Probability: Standards ##### ORELA Business Education Study Guide and Test Prep ##### OSAT Advanced Mathematics (111) Study Guide and Test Prep ##### MTTC Mathematics (Elementary) (089) Study Guide and Test Prep ##### AP Statistics Study Guide and Exam Prep ##### 7th Grade Math What Does It Mean to Interpret the Slope and Intercept? ------------------------------------------------------- When modeling linear data, the slope and intercept of the graph provide useful information about the initial conditions and rate of change of what is being studied. First, the slope of a line is a measure of its steepness. In a line, slope is a ratio of the change in one variable to the change in the other. Usually, this refers to the change in y for each unit change in x, but sometimes other variables may be used. The formula for slope is the change in y over the change in x. The triangle is the Greek letter delta, which represents change. Slope is usually represented by the variable m=Δ y Δ x=y 2−y 1 x 2−x 1=change in y change in x The intercept refers to the y-intercept, which is where the line intersects the y-axis. Again, other variables may be used, but the intercept generally refers to the independent variable and the vertical axis. When data appears to form a straight line, statisticians will use a linear model. A linear model is a comparison of two variables with a constant rate of change or slope. When graphing data that appears to be linear, a regression line is usually graphed to show the closest linear approximation to the data. A regression line is a straight line that approximates the relationship between data points. Interpreting the slope and intercept using a linear model means explaining what the slope and intercept represent for the data and the situation. Slope-Intercept Formula The slope-intercept form of a linear equation is y = mx + b, where m represents the slope and b represents the y-intercept. This form of the equation is useful because the slope and the intercept are clearly visible in the equation. When looking at a linear equation in slope-intercept form, the number next to the x is the slope m and the number added on at the end is the intercept b. Equations are not always in this form, and terms are not always in this order. The terms may need to be rearranged. The equation y = mx + b is often referred to as the slope-intercept formula. Slope-Intercept Form How to Interpret Slope ---------------------- To interpret the slope of the line, identify the variables in the situation. Since slope is change in y divided by change in x, divide the y-variable by the x-variable to get the units for the slope. Then, write a sentence to connect this value and its units back to the scenario in the problem. Consider the following scenario. Data is collected on the cost of flights from Baltimore, Maryland, to different destinations. The data is graphed in the scatter plot below. Distance to Destination (from Baltimore, MD) vs. Cost of Airfare The regression line is also shown on the graph and is labeled with its equation in slope-intercept form: y = 0.117 x + 83.267. So, the slope is 0.117. To identify the units of the slope, divide the units of y by the units of x. In this example, y is measured in dollars and x is measured in miles. So, the units of the slope are dollars divided by miles, or dollars per mile. Then, interpret the slope of the regression line by writing a sentence relating the answer to the original problem scenario. One way to do this is to state that y increases/decreases by approximately m units per 1 unit of x. The cost of airfare increases by 0.117 dollars per mile that the destination is from Baltimore, MD. The word decreases would be used if the slope is negative, depending on the scenario. Another way to interpret the slope is to state that the rate of change of y is m units. The rate of change in the cost of airfare is 0.117 dollars per mile. How to Interpret Intercept -------------------------- The y-intercept generally represents the initial condition for a set of data. It always represents the amount of y when the value of x is zero. This means that the units will be the same as the units for the y-axis. In the previous example, the equation of the regression line was y = 0.117 x + 83.267. The intercept, b, is 83.267 dollars. This means that for a flight of zero miles, the cost of airfare would be $83.27. When written like that, it doesn't make sense. It could also represent the baseline cost for any flight from Baltimore before the distance of the flight is determined, which makes a little more sense. It is important to note that sometimes the slope and intercept are not realistic for the scenario. That is because a linear model is just an approximation and is not perfectly accurate. Examples of Interpreting Slope and Y Intercept ---------------------------------------------- Suppose that some doctors are conducting a study. They ask their patients how many servings of fruit or vegetables they consume per day. They also keep track of whether those patients are obese. They record the percentage of their patients who consume 5 or more servings of fruits and vegetables per day and the percentage of their patients who are obese. The data is shown in the graph below. Percent of Adults Getting 5+ Servings of Fruits and Vegetables vs. Percentage of Adults that are Obese The equation of the regression line is given to be y = -0.91 x + 38.42. The slope is -0.91. The units of y are % of adults and the units of x are also % of adults. The units of y and the units of x will cancel out when they are divided. Instead, to interpret the slope, state that the percentage of obese adults decreases by 0.91% for every 1% increase in the number of adults who are getting 5 or more servings of fruits and vegetables. The y-intercept is 38.42. To interpret the intercept, relate it to what happens in the scenario when x is zero. In this scenario, that means that if 0% of adults are getting five or more servings of fruits and vegetables, it would be expected for 38.42% of the adults to be obese. Suppose a company collects data on the salaries of their employees (y) and the length of time that their employees have been with the company (x). From this data, they obtain the regression line y = 0.536 x + 40, where x is measured in years and y is measured in thousands of dollars. First, interpret the slope. The slope is 0.536 and the units are thousands of dollars per year. 0.536 in thousands of dollars is $536. So, for every year an employee stays with the company, they will get a raise of approximately $536 per year. An alternate statement could be that the rate of change in an employee's salary is approximately 0.536 thousand dollars per year. Next, interpret the intercept. The value of the intercept is 40, and it is measured in thousands of dollars; so, in the scenario, it represents $40,000. This relates to when the value of x, or the number of years the person has been with the company, is 0. So, a person could expect their starting salary to be about $40,000 if they start working at this company. Lesson Summary -------------- A linear model is a comparison of two variables and their constant rate of change. Linear models are used to approximate data that appear to be linear, which means that the points appear to form a straight line and/or appear to have a constant rate of change. A regression line is calculated to predict the relationship between points. A regression line approximates the data as closely as possible with a straight line in slope-intercept form, y = mx + b. The slope is represented by m and the intercept is represented by b. The slope and intercept give a lot of information about sets of data. The slope represents the change in y for any 1 unit change in x. The intercept, also known as the y-intercept, is where the line of best fit intersects the y-axis. It represents the initial condition or starting point of the data. Video Transcript Interpreting Slope and Intercept Lauren is collecting information for her auto mechanics class. She surveys six different auto mechanic shops in her town and collects information to see if there is a relationship between the number of times the oil is changed in a vehicle and the longevity in the engine of the vehicle. Once she gathers her information, Lauren puts all of it into a scatterplot with a regression line. Now that she has collected all of her data, how can she interpret this data into usable information? In this lesson, you will learn how to interpret the meaning of slope and y-intercept in different examples of linear models. Identifying Slope and Intercept A linear model is a comparison of two values, usually x and y, and the consistent change between those values. The easiest way to understand and interpret slope and intercept in linear models is to first understand the slope-intercept formula: y = mx + b. M is the slope or the consistent change between x and y, and b is the y-intercept. Often, the y-intercept represents the starting point of the equation. Take a look at this graph: Regression Line The line in the center is known as a regression line, a straight line that attempts to predict the relationship between two points. This relationship is the same thing as the slope, and you may also hear the terms consistent change or interval. These three words are used interchangeably and mean the same thing in this case. The points around this line represent the data that is collected in this scenario. The equation for this line is y = .3136 x + .2644. Interpreting Slope Let's take a look at our regression equation. For this scenario we have .3136 and .2644. .3136 is the slope in this equation, and .2644 is the intercept in this equation. First, let's talk about slope and how we can interpret slope in this equation. Remember that the slope is the consistent change, or the relationship between two variables, in a linear model. For example, let's say you were getting paid eight dollars an hour at your job. The rate, eight dollars, would be multiplied by the number of hours that you worked to get how much you should be paid for the week. In this case, the two variables are the number of hours you worked and how much you get paid for the week. The relationship between the number of hours you work and how much you get paid is the amount you get paid per hour. In this case, you know the relationship between the two variables ahead of time, but sometimes you know the variables and not the relationship, also known as the slope. Notice on our equation that slope is .3136. So what does this mean? Remember, our two variables are the number of times the oil is changed in the vehicle and the longevity of the engine. The slope is a positive number, which means that when the one variable increases, the other also increases. Just like the amount you get paid at the end of the week increases when the number of hours you work increases, so does the longevity of your engine increase as the number of times you change out the oil increases. Since a positive slope tells us there is a positive relationship between the two variables, what does the number .3136 tell us? Remember, in the previous scenario, the eight told us how much you were being paid per hour. In this example, .3136 shows us how much the longevity of your engine increases. Let's look at it like this. You have your vehicle sitting in your garage. Maybe you've had it for a couple of years. Each time you change the oil out in your vehicle increases the likelihood that the engine will last by .3136 years. That's right! In this case, the slope represents the number of years that you increase your engine's lifespan every time you change the oil. Remember, this is just an example, and statistics doesn't always show us the full picture. Obviously, if your vehicle's engine is broken, changing the oil in it several times won't fix the problem! Now that you understand slope in this scenario, let's move on to the intercept. Interpreting Intercept We know that the intercept is .2644 in our equation, but what does that mean? First, the intercept is also called the y-intercept. This is because it is the place in the equation where the line intercepts the y-axis. The ordered pair for the intercept is (0, .2644). This means that x = 0 and y = .2644. To understand the intercept, you need to understand the ordered pair. The x variable represents the number of oil changes in the vehicle. Therefore, in this case we are saying that there have been zero oil changes in the vehicle. The y variable represents the number of years of longevity in the engine. Therefore, this ordered pair shows that the vehicle's engine longevity is .2644 years. If we put this together, the intercept tells us that if there are zero oil changes in the vehicle the engine will last .2644 years. Obviously, in reality, this will differ depending on the maintenance on the vehicle, the age of the vehicle and the freshness of the oil that is currently in the vehicle. You will find that many data sets will have more variables that influence the slope and the intercept of the equation. Practice Problem Lauren is now collecting data on the amount of gasoline used and how often a person sits idle when using their vehicle. After collecting the data and plotting the points, she has developed the following equation: y = 3.52 x + 17.32, where y equals the number of gallons used in a month, and x equals the number of hours spent idle in a month. Can you interpret the slope and intercept of this equation? First, let's talk about the slope. In this example, the slope represents the gallons of gas that are used. Therefore, for every hour the vehicle spends idle, 3.52 extra gallons of gas are used per month according to this linear model. While this may or may not be true in real life, this is the interpretation of this particular model. Second, it's time to interpret the intercept. Remember that the ordered pair for the intercept is (0, 17.32). If x equals the number of hours spent idle in a month, and y equals the number of gallons of gas used in a month, then this means that a person will still use at least 17.32 gallons even if he or she never has the vehicle idle. Lesson Summary Every linear model is different. However, you can identify some similarities that will help you interpret the slope and intercept of the model. Remember, a linear model is a comparison of two values, usually x and y, and the consistent change between the values. Linear models will have a regression line, a straight line that attempts to predict the relationship between two points. You can use the slope-intercept formula, y = mx + b, to identify the slope and intercept of the regression line. In this equation, m is the slope, or the consistent change between x and y, and b is the y-intercept. Often, the y-intercept represents the starting point of the equation; in our examples, the y-intercept represented what would happen if the x-value did not exist, and this is true for all linear models. You can use these interpretations to predict information about a set of data. Learn more in our other lessons on regression and correlation. Learning Outcomes After completing this lesson, take the next step and test your ability to: Recognize a linear model and regression line Use the slope-intercept formula to interpret real-world data Register to view this lesson Are you a student or a teacher? I am a student I am a teacher Unlock your education See for yourself why 30 million people use Study.com Become a Study.com member and start learning now. 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Try it now Statistics 101: Principles of Statistics 11 chapters 144 lessons 9 flashcard sets Chapter 1 Overview of Statistics Descriptive vs. Inferential Statistics \| Definition & Examples 5:11 min Population vs Sample in Statistics \| Differences & Examples 3:24 min Parameter vs. Statistic \| Definition, Differences & Example 5:18 min Parameter Estimation \| Definition, Methods & Examples 7:46 min Quantitative Data Overview, Types & Examples 4:11 min What is Categorical Data? - Definition & Examples 5:25 min Discrete & Continuous Data: Definition & Examples 3:32 min Nominal & Ordinal Data \| Definition, Measurements & Analysis 8:29 min Statistical Modeling Purpose, Types & Examples 10:20 min Experiment vs. Observational Study \| Definition & Examples 6:21 min Random Allocation & Random Selection \| Definition & Examples 6:13 min Convenience Sampling in Statistics \| Definition & Examples 6:27 min Randomized Controlled Trial \| Overview, Design & Examples 8:21 min Analyzing & Interpreting the Results of Randomized Experiments 4:46 min Confounding & Bias in Statistics: Definition & Examples 3:59 min Confounding Variables in Statistics \| Definition, Types & Tips 5:20 min Bias in Statistics: Definition & Examples 7:24 min Bias in Polls & Surveys: Definition, Common Sources & Examples 4:36 min Misleading Uses of Statistics 8:14 min Chapter 2 Summarizing Data What is the Center in a Data Set? - Definition & Options 5:08 min Overview on the Measures of Central Tendency in Psychology 6:00 min Measures of Central Tendency \| Definition, Formula & Examples 8:30 min Calculating the Mean, Median, Mode & Range: Practice Problems 7:13 min Skewed & Symmetric Distribution \| Definition & Graphs 5:22 min Unimodal & Bimodal Histogram \| Definition & Examples 5:29 min Mean vs. Median \| Usage, Calculation & Examples 6:30 min Spread of Data Overview & Examples 7:51 min Outliers in a Data Set \| Minimums & Maximums 4:40 min Quartiles & Interquartile Range \| Calculation & Examples 8:00 min Finding the Percentile of a Data Set \| Formula & Example 8:25 min Standard Deviation Equation, Formula & Examples 13:05 min The Effect of Linear Transformations on Measures of Center & Spread 6:16 min Population Variance \| Definition, Formula & Calculation 9:34 min Ranked Data Definition, Types & Analysis 6:54 min Chapter 3 Tables and Plots Relative Frequency \| Definition & Examples 4:48 min Cumulative Frequency \| Definition, Table & Example 5:17 min Cumulative & Relative Frequency \| Formula, Table & Calculations 5:47 min Creating & Reading Stem & Leaf Displays 4:27 min Creating & Interpreting Histograms: Process & Examples 5:43 min Frequency Polygon \| Graphs & Maker 5:48 min Creating & Interpreting Dot Plots: Process & Examples 7:35 min Box Plot \| Definition, Uses & Examples 6:29 min Pie Chart vs. Bar Graph \| Overview, Uses & Examples 9:36 min Making Arguments & Predictions from Univariate Data 8:35 min Bivariate Data Definition, Analysis & Examples 8:12 min Two-Way Table \| Definition, Examples & Usage 3:40 min Joint, Marginal & Conditional Frequencies \| Definition & Overview 9:57 min Chapter 4 Probability Mathematical Sets: Elements, Intersections & Unions 3:02 min Events as Subsets of a Sample Space: Definition & Example 4:51 min Probability of an Event \| Simple, Compound & Complementary 6:55 min Independent & Dependent Events \| Overview, Probability & Examples 12:06 min Probability of at Least One Event \| Overview & Calculation 5:27 min Conditional Probability \| Overview, Calculation & Examples 5:10 min Conditional Probability & Independence \| Rules & Examples 7:52 min Using Two-Way Tables to Evaluate Independence 8:09 min Applying Conditional Probability & Independence to Real Life Situations 12:32 min Addition Rule of Probability \| Formulas & Examples 10:57 min Multiplication Rule of Probability \| Definition & Examples 8:37 min Combination in Mathematics \| Definition, Formula & Examples 7:14 min Permutation Definition, Formula & Examples 6:58 min How to Calculate the Probability of Permutations 10:06 min Relative Frequency \| Formula & Examples 5:56 min Chapter 5 Discrete Probability Distributions Random Variable \| Overview, Types & Examples 9:53 min Expected Value & Discrete Random Variables \| Overview & Examples 5:25 min Developing Discrete Probability Distributions Theoretically & Finding Expected Values 9:21 min Developing Discrete Probability Distributions Empirically & Finding Expected Values 10:09 min Dice: Finding Expected Values of Games of Chance 13:36 min Blackjack: Finding Expected Values of Games of Chance with Cards 8:41 min Poker: Finding Expected Values of High Hands 9:38 min Poker: Finding Expected Values of Low Hands 8:38 min Lotteries: Finding Expected Values of Games of Chance 11:58 min Comparing Game Strategies Using Expected Values: Process & Examples 4:31 min How to Apply Discrete Probability Concepts to Problem Solving 7:35 min Binomial Experiment Definition, Requirements & Examples 4:46 min Finding Binomial Probabilities Using Formulas: Process & Examples 6:10 min Practice Problems for Finding Binomial Probabilities Using Formulas 7:15 min Binomial Distribution Table \| Definition, Purpose & Example 8:26 min Binomial Random Variable \| Definition, Formula & Examples 6:34 min Solving Problems with Binomial Experiments: Steps & Example 5:03 min Chapter 6 Continuous Probability Distributions Probability Distribution Graphs \| Discrete & Continuous 6:33 min Finding & Interpreting the Expected Value of a Continuous Random Variable 5:29 min Developing Continuous Probability Distributions Theoretically & Finding Expected Values 6:12 min Probabilities as Areas of Geometric Regions: Definition & Examples 7:06 min Normal Distribution \| Curve, Table & Examples 11:40 min Z Score \| Definition, Equation & Example 6:30 min Estimating Areas Under the Normal Curve Using Z-Scores 5:54 min Normal Distribution's Empirical Rule \| Overview & Percentages 4:41 min Using the Normal Distribution: Practice Problems 10:32 min Using Normal Distribution to Approximate Binomial Probabilities 6:34 min How to Apply Continuous Probability Concepts to Problem Solving 5:05 min Chapter 7 Sampling Simple Random Sampling in Statistics \| Definition & Examples 5:10 min Random Sampling Definition, Types & Examples 5:55 min Stratified Random Sampling \| Definition, Method & Characteristics 6:25 min Cluster Sampling \| Definition, Types & Examples 6:44 min Systematic Random Sampling \| Definition, Formula & Examples 8:37 min Law of Large Numbers \| Definition & Significance 5:14 min Central Limit Theorem \| Definition, Formula & Examples 5:06 min Sampling Distribution \| Definition & Formula 5:03 min Finding Probabilities About Means Using the Central Limit Theorem 4:24 min Chapter 8 Regression & Correlation Creating & Interpreting Scatterplots: Process & Examples 6:14 min Problem Solving Using Linear Regression: Steps & Examples 8:38 min Analyzing Residuals: Process & Examples 5:30 min Viewing now Interpreting the Slope & Intercept \| Definition, Method & Example 8:05 min Up next The Correlation Coefficient \| Definition, Formula & Calculation 9:57 min Watch next lesson The Correlation Coefficient: Practice Problems 8:14 min Correlation \| Overview, Interpretation & Limitation 14:31 min Causal Relationship Definition, Theories & Application 7:27 min Interpreting Linear Relationships Using Data: Practice Problems 6:15 min Transforming Nonlinear Data: Steps & Examples 9:25 min Coefficient of Determination \| Definition, Purpose & Formula 5:21 min Pearson Correlation Coefficient \| Formula & Examples 6:31 min Chapter 9 Statistical Estimation Point & Interval Estimations: Definition & Differences 6:17 min Calculating Confidence Intervals, Levels & Coefficients 5:14 min Finding Confidence Intervals with the Normal Distribution 5:01 min Sample Size & Confidence Interval \| Definition & Examples 4:13 min Student t-Distribution \| Definition, Properties & Examples 5:07 min Using the t Distribution to Find Confidence Intervals 6:57 min Biased vs. Unbiased Estimator \| Definition, Examples & Statistics 4:30 min Finding Confidence Intervals for Proportions: Formula & Example 6:50 min Chapter 10 Hypothesis Testing Hypothesis Testing Definition, Steps & Examples 8:32 min Conducting Hypothesis Testing for a Mean: Process & Examples 8:43 min Effect Size in Hypothesis Testing: Definition & Interpretation 3:11 min Type I & Type II Errors \| Definition & Examples 5:44 min The Relationship Between Confidence Intervals & Hypothesis Tests 9:24 min Hypothesis Testing Large Independent Samples 6:01 min What Are t-Tests? - Assessing Statistical Differences Between Groups 7:49 min Hypothesis Testing Matched Pairs 5:26 min Hypothesis Testing for a Proportion 6:12 min Hypothesis Testing for a Difference Between Two Proportions 10:09 min Chi-Square Test \| Definition, Purpose & Examples 6:37 min ANOVA Test Definition, Purpose & Examples 6:47 min Using ANOVA to Analyze Variances Between Multiple Groups 9:14 min Chapter 11 Studying for Statistics 101 Overview of Statistics Flashcards Summarizing Data Flashcards Tables and Plots Flashcards Probability Flashcards Discrete Probability Distributions Flashcards Continuous Probability Distributions Flashcards Regression & Correlation Flashcards Statistical Estimation Flashcards Hypothesis Testing in Statistics Flashcards Z-Scores & Standard Normal Curve Areas Statistical Table Critical Values of the t-Distribution Statistical Table Binomial Probabilities Statistical Tables Related Study Materials Interpreting the Slope & Intercept \| Definition, Method & Example LessonsCoursesTopics ##### Line of Best Fit \| Definition, Formula & Examples 6:00 ##### Line of Best Fit \| Definition, Formula & Equation 6:05 ##### Least-Squares Regression \| Line Formula, Method & Examples 8:26 ##### Residual Plot \| Definition, Interpretation & Examples 5:19 ##### How to Draw a Trend Line 3:43 ##### How to 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15706	https://brainly.com/question/50524396	[FREE] Evaluate the definite integral \int_{0}^{2} \sqrt{2x - x^2} \, dx. - brainly.com 2 Search Learning Mode Cancel Log in / Join for free Browser ExtensionTest PrepBrainly App Brainly TutorFor StudentsFor TeachersFor ParentsHonor CodeTextbook Solutions Log in Join for free Tutoring Session +35,5k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +25,8k Ace exams faster, with practice that adapts to you Practice Worksheets +7,9k Guided help for every grade, topic or textbook Complete See more / Mathematics Expert-Verified Expert-Verified Evaluate the definite integral ∫0 22 x−x 2d x. 1 See answer Explain with Learning Companion NEW Asked by ajwalker6221 • 04/23/2024 0:00 / -- Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 3150453 people 3M 0.0 0 Upload your school material for a more relevant answer The value of the definite integral ∫_0² √(2x-x²) dx is π. To evaluate this definite integral, we can start by substituting u = 2x - x², which gives du/dx = 2 - 2x. Then, we can rewrite the integral as: ∫_0² √(2x-x²) dx = ∫_0¹ √u du Now, we can use the substitution u = sin²(θ), which gives du/dθ = 2sin(θ)cos(θ). Then, we can rewrite the integral as: ∫_0¹ √u du = ∫_0^(π/2) sin²(θ) dθ This integral can be evaluated using the trigonometric identity sin²(θ) = (1 - cos(2θ))/2, which gives: ∫_0^(π/2) sin²(θ) dθ = (1/2) ∫_0^(π/2) (1 - cos(2θ)) dθ = (1/2) [θ - (1/2)sin(2θ)]_0^(π/2) = (1/2) [(π/2) - 0] = π/2 Therefore, the value of the original definite integral is π. Note: The substitutions used in this solution are common techniques for evaluating definite integrals involving square roots and trigonometric functions. The first substitution, u = 2x - x², helps to simplify the original integral, while the second substitution, u = sin²(θ), allows us to use the trigonometric identity sin²(θ) = (1 - cos(2θ))/2 to evaluate the integral. Answered by nidakhalid566 •11.3K answers•3.2M people helped Thanks 0 0.0 (0 votes) Expert-Verified⬈(opens in a new tab) This answer helped 3150453 people 3M 0.0 0 Upload your school material for a more relevant answer The value of the definite integral ∫0 22 x−x 2d x is 2 π. This integral represents the area under the upper half of a circle with radius 1 centered at (1,0). Therefore, by calculating this area, we arrive at our answer. Explanation To evaluate the definite integral ∫0 22 x−x 2d x, we first notice that 2 x−x 2 can be rewritten to make integration easier. We rewrite it as: 2 x−x 2=−(x 2−2 x)=−(x 2−2 x+1−1)=−((x−1)2−1). Now we can express the integral as: ∫0 22 x−x 2d x=∫0 21−(x−1)2d x This resembles the integral of a semicircle. The expression 1−(x−1)2 represents the upper half of a circle with a radius of 1 centered at (1, 0). Next, we find the area under this semicircle between 0 and 2. The area of a semicircle is given by 2 1π r 2. For our semicircle, r=1, hence: Area=2 1π(1)2=2 π Thus, the value of the definite integral is: ∫0 22 x−x 2d x=2 π Examples & Evidence An example that illustrates this is the area calculation of a semicircle which is a common problem in geometry. Similarly, the use of substitution can often simplify integrals that involve square roots, especially when recognizing geometric interpretations. The provided calculations and geometric interpretations are standard methods in calculus for evaluating definite integrals involving square roots. You can find similar problems in calculus textbooks that discuss areas under curves and their relationship with integrals. Thanks 0 0.0 (0 votes) Advertisement ajwalker6221 has a question! Can you help? Add your answer See Expert-Verified Answer ### Free Mathematics solutions and answers Community Answer Evaluate the integral ‚à´_0^a 9x² ‚àö(a² - x²) dx Community Answer 4.6 12 Jonathan and his sister Jennifer have a combined age of 48. If Jonathan is twice as old as his sister, how old is Jennifer Community Answer 11 What is the present value of a cash inflow of 1250 four years from now if the required rate of return is 8% (Rounded to 2 decimal places)? Community Answer 13 Where can you find your state-specific Lottery information to sell Lottery tickets and redeem winning Lottery tickets? (Select all that apply.) 1. Barcode and Quick Reference Guide 2. Lottery Terminal Handbook 3. Lottery vending machine 4. OneWalmart using Handheld/BYOD Community Answer 4.1 17 How many positive integers between 100 and 999 inclusive are divisible by three or four? Community Answer 4.0 9 N a bike race: julie came in ahead of roger. julie finished after james. david beat james but finished after sarah. in what place did david finish? Community Answer 4.1 8 Carly, sandi, cyrus and pedro have multiple pets. carly and sandi have dogs, while the other two have cats. sandi and pedro have chickens. everyone except carly has a rabbit. who only has a cat and a rabbit? Community Answer 4.1 14 richard bought 3 slices of cheese pizza and 2 sodas for $8.75. Jordan bought 2 slices of cheese pizza and 4 sodas for $8.50. How much would an order of 1 slice of cheese pizza and 3 sodas cost? A. $3.25 B. $5.25 C. $7.75 D. $7.25 Community Answer 4.3 192 Which statements are true regarding undefinable terms in geometry? Select two options. A point's location on the coordinate plane is indicated by an ordered pair, (x, y). A point has one dimension, length. A line has length and width. A distance along a line must have no beginning or end. A plane consists of an infinite set of points. New questions in Mathematics x-5 0 7 \ The table of ordered pairs shows the coordinates of two points on the graph of a line. Which equation describes the line?\na. y=x+7\nc. y=−5 x+2\nb. y=x−7\nd. y=5 x+2 Solve the equation (0.3)x+1=(0.7)x One number is 8 times another number. The numbers are both positive and have a difference of 70. Which of the following equations could be used to solve the problem? A. x−8 x=70 B. x−8=70 C. 8 x−x=70 Find the sum: $\begin{aligned} -0.25-1.5 & =-0.25+(-1.5) \ & =\square \end{aligned}$ Write an equation of the line that passes through a pair of points: (−3,1),(−2,−5) Previous questionNext question Learn Practice Test Open in Learning Companion Company Copyright Policy Privacy Policy Cookie Preferences Insights: The Brainly Blog Advertise with us Careers Homework Questions & Answers Help Terms of Use Help Center Safety Center Responsible Disclosure Agreement Connect with us (opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab) Brainly.com Dismiss Materials from your teacher, like lecture notes or study guides, help Brainly adjust this answer to fit your needs. Dismiss
15707	https://www.teachoo.com/3394/745/Ex-10.2--11---Show-that-2i---3j---4k--4i---6j---8k-are-collinear/category/Ex-10.2/	Show that the vectors 2i - 3j + 4k and -4i + 6j - 8k are collinear Everything Class 6 Class 6 Maths (Ganita Prakash) Class 6 Maths (old NCERT) Class 6 Science (Curiosity) Class 6 Science (old NCERT) Class 6 Social Science Class 6 English Class 7 Class 7 Maths (Ganita Prakash) Class 7 Maths (old NCERT) Class 7 Science (Curiosity) Class 7 Science (old NCERT) Class 7 Social Science Class 7 English Class 8 Class 8 Maths (Ganita Prakash) Class 8 Maths (old NCERT) Class 8 Science (Curiosity) Class 8 Science (old NCERT) Class 8 Social Science Class 8 English Class 9 Class 9 Maths Class 9 Science Class 9 Social Science Class 9 English Class 10 Class 10 Maths Class 10 Science Class 10 Social Science Class 10 English Class 11 Class 11 Maths Class 11 English Class 11 Computer Science (Python) Class 12 Class 12 Maths Class 12 English Class 12 Economics Class 12 Accountancy Class 12 Physics Class 12 Chemistry Class 12 Biology Class 12 Computer Science (Python) Class 12 Physical Education Courses GST and Accounting Course Excel Course Tally Course Finance and CMA Data Course Payroll Course Interesting Learn English Learn Excel Learn Tally Learn GST (Goods and Services Tax) Learn Accounting and Finance Popular GST Demo GST Tax Invoice Format Accounts Tax Practical Tally Ledger List Maths Updated 2025-26 Class 6 Maths (Ganita Prakash) Class 6 Maths (old NCERT) Class 7 Maths (Ganita Prakash) Class 7 Maths (old NCERT) Class 8 Maths (Ganita Prakash) Class 8 Maths (old NCERT) Class 9 Maths Class 10 Maths Class 11 Maths Class 12 Maths Sample Paper Class 10 Maths Sample Paper Class 12 Maths Accounts & Finance Learn Excel free Learn Tally free Learn GST (Goods and Services Tax) Learn Accounting and Finance GST and Income Tax Return Filing Course Tally Ledger List Popular Class 6 Science (Curiosity) Class 6 Science (old NCERT) Class 7 Science (Curiosity) Class 7 Science (old NCERT) Class 8 Science (Curiosity) Class 8 Science (old NCERT) Class 9 Science Class 10 Science Remove Ads Login Maths Remove ads Science GST Accounts Tax Englishtan Excel Social Science Class 6 Class 6 Maths (Ganita Prakash) Class 6 Maths (old NCERT) Class 6 Science (Curiosity) Class 6 Science (old NCERT) Class 6 Social Science Class 6 English Class 7 Class 7 Maths (Ganita Prakash) Class 7 Maths (old NCERT) Class 7 Science (Curiosity) Class 7 Science (old NCERT) Class 7 Social Science Class 7 English Class 8 Class 8 Maths (Ganita Prakash) Class 8 Maths (old NCERT) Class 8 Science (Curiosity) Class 8 Science (old NCERT) Class 8 Social Science Class 8 English Class 9 Class 9 Maths Class 9 Science Class 9 Social Science Class 9 English Class 10 Class 10 Maths Class 10 Science Class 10 Social Science Class 10 English Class 11 Class 11 Maths Class 11 Computer Science (Python) Class 11 English Class 12 Class 12 Maths Class 12 English Class 12 Economics Class 12 Accountancy Class 12 Physics Class 12 Chemistry Class 12 Biology Class 12 Computer Science (Python) Class 12 Physical Education Courses GST and Accounting Course Excel Course Tally Course Finance and CMA Data Course Payroll Course Interesting Learn English Learn Excel Learn Tally Learn GST (Goods and Services Tax) Learn Accounting and Finance Popular GST Demo GST Tax Invoice Format Accounts Tax Practical Tally Ledger List Remove Ads Login Chapter 10 Class 12 Vector Algebra Serial order wise Ex 10.2 Ex 10.2 Ex 10.2, 1 Ex 10.2, 2 Ex 10.2, 3 Important Ex 10.2, 4 Ex 10.2, 5 Important Ex 10.2, 6 Ex 10.2, 7 Important Ex 10.2, 8 Ex 10.2, 9 Ex 10.2, 10 Important Ex 10.2, 11 Important You are here Ex 10.2, 12 Ex 10.2, 13 Important Ex 10.2, 14 Ex 10.2, 15 Important Ex 10.2, 16 Ex 10.2, 17 Important Ex 10.2, 18 (MCQ) Important Ex 10.2, 19 (MCQ) Important Ex 10.3→ Chapter 10 Class 12 Vector Algebra Serial order wise Ex 10.1 Ex 10.2 Ex 10.3 Ex 10.4 Ex 10.5 (Supplementary NCERT) Examples Miscellaneous Supplementary examples and questions from CBSE Case Based Questions (MCQ) Ex 10.2, 11 - Chapter 10 Class 12 Vector Algebra Last updated at December 16, 2024 by Teachoo ADVERTISEMENT Powered by VidCrunch Next Stay Playback speed 1x Normal Quality Auto Back 720p 360p 240p 144p Auto Back 0.25x 0.5x 1x Normal 1.5x 2x / Skip Ads by Next: Ex 10.2, 12 →Remove AdsShare on WhatsApp Transcript Ex 10.2, 11 (Method 1) Show that the vectors 2𝑖 ̂ − 3𝑗 ̂ + 4𝑘 ̂ and − 4𝑖 ̂ + 6 𝑗 ̂ − 8𝑘 ̂ are collinear.Two vectors are collinear if they are parallel to the same line. Let 𝑎 ⃗ = 2𝑖 ̂ − 3𝑗 ̂ + 4𝑘 ̂ and 𝑏 ⃗ = –4𝑖 ̂ + 6𝑗 ̂ – 8𝑘 ̂ Magnitude of 𝑎 ⃗ = √(22+(−3)2+42) \|𝑎 ⃗ \| = √(4+9+16) = √29 Directions cosines of 𝑎 ⃗ = (2/√29,(−3)/√29,4/√29) Magnitude of 𝑏 ⃗ =√((−4)2+62+(−8)2) \|𝑏 ⃗ \| = √(16+36+64) = √116 = 2√29 Directions cosines of 𝑏 ⃗ = ((−4)/(2√29),6/(2√29),(−8)/(2√29)) = ((−2)/√29,3/√29,(−4)/√29) = −1(2/√29,(−3)/√29,4/√29) Hence, Direction cosines of 𝒂 ⃗ = (−1) × Direction cosines of 𝒃 ⃗ ∴ They have opposite directions Since 𝑎 ⃗ and 𝑏 ⃗ are parallel to the same line 𝑚 ⃗, they are collinear. Hence proved Ex 10.2, 11 (Method 2) Show that the vectors 2𝑖 ̂ − 3𝑗 ̂ + 4𝑘 ̂ and − 4𝑖 ̂ + 6 𝑗 ̂ − 8𝑘 ̂ are collinear.𝑎 ⃗ = 2𝑖 ̂ − 3𝑗 ̂ + 4𝑘 ̂ 𝑏 ⃗ = –4𝑖 ̂ + 6𝑗 ̂ – 8𝑘 ̂ Two vectors are collinear if their directions ratios are proportional 𝑎_1/𝑏_1 = 𝑎_2/𝑏_2 = 𝑏_3/𝑏_3 = 𝜆 2/(−4) = (−3)/6 = 4/(−8) = (−1)/2 Since, directions ratios are proportional Hence, 𝑎 ⃗ & 𝑏 ⃗ are collinear Show More Chapter 10 Class 12 Vector Algebra Serial order wise Ex 10.2 Ex 10.2, 1 Ex 10.2, 2 Ex 10.2, 3 Important Ex 10.2, 4 Ex 10.2, 5 Important Ex 10.2, 6 Ex 10.2, 7 Important Ex 10.2, 8 Ex 10.2, 9 Ex 10.2, 10 Important Ex 10.2, 11 Important You are here Ex 10.2, 12 Ex 10.2, 13 Important Ex 10.2, 14 Ex 10.2, 15 Important Ex 10.2, 16 Ex 10.2, 17 Important Ex 10.2, 18 (MCQ) Important Ex 10.2, 19 (MCQ) Important Ex 10.3→ Made by Davneet Singh Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. 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15708	https://www.math.chalmers.se/Math/Grundutb/GU/MMA200/A15/pidmod.pdf	Modules over a PID Hjalmar Rosengren, 17 September 2015 We will classify all ﬁnitely generated modules over a PID R. One very important case is R = Z, when we obtain a classiﬁcation of ﬁnitely generated abelian groups. We will show that any such group is isomorphic to Zs × Z/pk1 1 Z × · · · × Z/pkm m Z, (1) where pj are primes. This is a good example to keep in mind throughout the discussion. Another important case gives the “Jordan normal form” of a complex matrix, which generalizes diagonalization to non-diagonalizable matrices. This is useful for many applications of linear algebra. To formulate the main theorem, consider the equivalence relation on prime elements of R, deﬁned by p ∼q if p = eq with e a unit. Let P be a set of representatives for these equivalence classes. (For instance, in Z the equivalence classes are {±p} with p a prime number. Choosing always the positive representative, we can take P as the set of prime numbers.) Theorem 1. Any ﬁnitely generated module over a PID R is isomorphic to Rs × R/pk1 1 R × · · · × R/pkm m R (2) with pj ∈P. Moreover, this decomposition is unique up to rearranging the factors. The decomposition (2) is called the primary decomposition of M (an ideal of the form pkR with p prime is called a primary ideal). Note that some of the primes pj in Theorem 1 may be equal. Each p ∈P corresponds to a subfactor R/pl1R × · · · × R/pljR, (3) where we may order the factors so that l1 ≤· · · ≤lj. The uniqueness statement means that M uniquely determines the non-negative integer s, a ﬁnite subset of P and a ﬁnite sequence l1 ≤· · · ≤lj of positive integers for each prime in this ﬁnite subset. To prove Theorem 1, we follow Lang rather then Brzezinski. However, we give more details and split the proof into several lemmas for increased clarity. Note that the individual factors in (1) are cyclic groups, which can ei-ther be deﬁned as groups with one generators or as quotients of Z. The corresponding concept for modules is cyclic modules. 1 Lemma 2. A module over a commutative ring R is generated by one element if and only if it is isomorphic to a quotient R/I for some ideal I. Moreover, Rx ≃R/I with I = Ann(x) = {r ∈R; rx = 0} (the annihilator of x). This also holds over non-commutative rings, but for left ideals rather than ideals. Proof. If M = Rx, then φ(r) = rx gives a surjective homomorphism R →M with kernel I = Ann(x), so Rx ≃R/I. Conversely, if we let x = 1+I ∈R/I, then x generates R/I and Ann(x) = I. The following elementary result gives a “cancellation law” for free modules over an integral domain. Lemma 3. If M is a free module over an integral domain R and rx = 0 for r ∈R and x ∈M, then either r = 0 or x = 0. Proof. Let (ej)j=∈Λ be a basis for M and write x = P j∈Λ xjej. Then, 0 = P j∈Λ rxjej so by the deﬁnition of a basis rxj = 0 for each j. Since R is a domain we have either r = 0 or xj = 0 for all j, that is, x = 0. We will write dim(M) for the number of elements in a basis for a mod-ule. Recall that this makes sense for ﬁnitely generated free modules over commutative rings (but not over rings in general). Proposition 4. If F is a ﬁnitely generated free module over a PID R and M is a submodule, then M is again free and dim(M) ≤dim(F). The condition of R being a PID is vital. Indeed, let I be a non-principal ideal in a commutative ring R, considered as an R-submodule. Then dim(R) = 1 but I is not generated by a single element, so either I is not free or I is free with dim(I) ≥2. Proof. Let e1, . . . , en be a basis for F. Let Mk = M ∩(Re1 + · · · + Rek). We will prove by induction on k that Mk is free and dim Mk ≤k. Our starting point is the trivial case M0 = {0} and the endpoint case k = n is the statement of the theorem. Consider the map π : Mk →R deﬁned by π(x1e1 + · · · + xkek) = xk. 2 Then π is an R-module homomorphism so Im(π) is an ideal. By the PID property, Im(π) = aR for some a ∈R. If a = 0, clearly Mk = Mk−1 and we are done. Assuming from now on that a ̸= 0, pick x ∈Mk with π(x) = a. We claim that Mk = Mk−1 ⊕Rx or, equivalently, Mk = Mk−1 + Rx and Mk−1 ∩Rx = {0}. For the ﬁrst identity, pick any y ∈Mk. We have π(y) = ab for some b ∈R. Then, π(y −bx) = 0, so y −bx ∈Mk−1, which gives y ∈Mk−1 + Rx. For the second identity, if rx ∈Mk−1 then 0 = π(rx) = rπ(x) = ra. Since a ̸= 0 we get r = 0 (R is a domain) and thus rx = 0. Finally, we note that Rx is a free module with basis x. Otherwise, we would have rx = 0 for some r ̸= 0, which contradicts Lemma 3. Thus, it follows from the induction hypothesis that Mk is free with dim(Mk) = dim(Mk−1) + 1 ≤k. Corollary 5. Let M be a ﬁnitely generated module over a PID. Then any submodule of M is ﬁnitely generated. Proof. Suppose M is generated by v1, . . . , vn. Deﬁne f : Rn →M by f(x1, . . . , xn) = x1v1 + · · · + xnvn. Then, f is a homomorphism. If N is a submodule of M then f −1(N) = {x ∈Rn; f(x) ∈N} is a submodule of Rn. By Prop. 4, f −1(N) is ﬁnitely generated. Acting by f on a set of generators we obtain a ﬁnite set of generators for N. A torsion element of a module M is an element x such that ax = 0 for some non-zero a ∈R. The set of all torsion elements form a submodule, which we denote Mtor. A module with Mtor = {0} is called torsion free and a module with Mtor = M is called a torsion module. For instance, in (1) the ﬁrst factor Zs is torsion free and the remaining factors form a torsion module. It is easy to see that M/Mtor is always torsion free. By Lemma 3, a free module over an integral domain is torsion free. The converse holds for ﬁnitely generated modules over a PID. Lemma 6. If M is a ﬁnitely generated module over a PID and M is torsion free then M is free. Proof. Let v1, . . . , vn be generators for M and let e1, . . . , ek be a maximal set of linearly independent elements among these generators. Then, e1, . . . , ek generate a free module F ⊆M. We claim that we can ﬁnd non-zero elements aj ∈R such that ajvj ∈F. If vj is one of the generators ek this is true with aj = 1. Else, {vj, e1, . . . , ek} are linearly dependent, so we can write ajvj + x1e1 + · · · + xkek = 0, aj, x1, . . . , xk ∈R, 3 where not all the coeﬃcients are zero. Since e1, . . . , ek are linearly indepen-dent, aj ̸= 0. We now let a = a1 · · · an. Since ajvj ∈F we have that avj ∈F for each j. It follows that f(x) = ax is a homomorphism from M to F. Since M is torsion-free, f is injective. Thus, M is isomorphic to a submodule of a ﬁnitely generated free module. The conclusion now follows from Prop. 4. We are now ready for the following weak version of Theorem 1. Lemma 7. If M is a ﬁnitely generated module over a PID, then M ≃ Rs × Mtor for some s. Proof. As we remarked above, M/Mtor is torsion free. Thus, by Lemma 6, it is free. By Lemma 4 from the notes last time, Mtor is a direct summand of M, so M = F ⊕Mtor for some submodule F. It follows that F ≃M/Mtor, so F is free. Finally, by Cor. 5, F is ﬁnitely generated, so F ≃Rs for some s. We will now study torsion modules. We ﬁrst say a few words about greatest common divisors. When a and b are elements of a PID, we write gcd(a, b) = c for any element c such that (a, b) = (c) (it is determined up to multiplication by a unit). We can obtain c as the product of all prime powers that divide both a and b. Since c ∈(a, b) we can write c = as + bt for some s, t ∈R. When M is a module over R and a ∈R, let Ma = {x ∈M; ax = 0}. Clearly, Ma is a sub-module. Lemma 8. If M is a module over a PID R and a, b, c ∈R with a = bc and gcd(b, c) = 1, then Ma = Mb ⊕Mc. Proof. We can write 1 = sc + tb for some s, t ∈R. We need to show that Ma = Mb+Mc and that Mb∩Mc = {0}. Both statements follow from writing x = scx + tbx. Indeed, if x ∈Ma, then scx ∈Mb and tbx ∈Mc. Moreover, if x ∈Mb ∩Mc then scx = tbx = 0. We can now obtain a preliminary decomposition of a ﬁnitely generated torsion module. Lemma 9. If M is a ﬁnitely generated torsion module over a PID, then M = Mpk1 1 ⊕· · · ⊕Mpkm m (4) for certain distinct primes pj ∈P and positive integers kj. 4 Note that, in contrast to the decomposition (2), the pj appearing in (4) are distinct. We should think of each summand in (4) as gathering all factors in (2) that involve a ﬁxed prime. Later, we will show that each of these summands can be decomposed further as in (3). Proof of Lemma 9. The annihilator of M is deﬁned as Ann(M) = {a ∈R; ax = 0 for all x ∈M}. It is an ideal in R. We prove that Ann(M) ̸= {0}. Indeed, if v1, . . . , vn are generators for M, then ajvj = 0 for some non-zero aj ∈R and it follows that a1 · · · an ∈Ann(M). By the PID property, we can write Ann(M) = aR for some non-zero a ∈R. Since a PID is a UFD, we may factor a as a product of primes. Since multiplying a by a unit does not change the ideal aR, we may assume a = pk1 1 · · · pkm m , where pj ∈P. The desired result now follows by iterating Lemma 8. Note that, for any module M, Ann(Mpk) is an ideal in R containing pkR and thus has the form plR for some l ≤k. Thus, the following result shows that each summand in (4) can be decomposed as in (3). Lemma 10. Let M be a ﬁnitely generated module over a PID R, such that Ann(M) = pkR, where p ∈P and k ≥0. Then, M is isomorphic to a product R/(pl1R) × · · · × R/(plmR) (5) for some positive integers lj. To prove Lemma 10, we will argue by induction on the “size” of M. It’s not immediately obvious how the size should be measured, but it turns out that the following idea works. It’s easy to see that, for any module M, M/pM is a module over R/pR (cf. Prop. (3.14) in Brzezinski). Moreover, if M is ﬁnitely generated, then the cosets containing the generators generate M/pM as an R/pR-module. Thus, \|M\|p = dimR/pR(M/pM) is a non-negative in-teger. We will prove Lemma 10 by induction on \|M\|p. It is easy to check that if M is given by (5), then \|M\|p = m (cf. Lemma 9 below). Thus, we are actually performing induction over the number of factors in (5), but of course that does not make sense until we have proved the lemma. 5 Proof of Lemma 10. If M is the trivial module, we interpret (5) as an empty product. Otherwise, k ≥1 and there is an element x ∈M such that pk−1x ̸= 0. Note that x / ∈pM since x = py would give pky ̸= 0. This means in particular that pM ⊊M so that \|M\|p ≥1. Thus, \|M\|p = 0 only for the trivial module, which we can use as the starting case for the induction. Assume that the statement of the lemma holds for all modules N with \|N\|p < \|M\|p. Choose x as above and let N = M/Rx. We want to show that \|N\|p < \|M\|p. To see this, let ¯ y1, . . . , ¯ yn be R/pR-linearly indepen-dent elements in N/pN. We pick arbitrary representatives yi ∈M, so ¯ yi = yi+pM +Rx. (Note that pN = (pM +Rx)/Rx and hence by a standard isomorphism theorem, N/pN = (M/Rx)/((pM+Rx)/Rx) ≃M/(pM+Rx).) We claim that y1 + pM, . . . , yn + pM, x + pM are R/pR-linearly independent elements in M/pM. This will clearly imply \|N\|p < \|M\|p. To be completely explicit, we are assuming that a1y1 + · · · + anyn ∈pM + Rx = ⇒ ai ∈pR and want to prove that a1y1 + · · · + anyn + bx ∈pM = ⇒ ai, b ∈pR. This reduces to bx ∈pM ⇒b ∈pR. Equivalently, in the R/pR-module M/pM, ¯ b¯ x = 0 ⇒¯ b = 0. But since R/pR is a ﬁeld, it is enough to show that ¯ x ̸= 0, that is, that x / ∈pM. This was already observed above. Having proved that \|N\|p < \|M\|p, it follows from our induction hypothesis that N is isomorphic to a product like (5). If we can show that the sequence 0 →Rx →M →N →0 (6) splits, then M ≃N × Rx. Since, by Lemma 2, Rx ≃R/pkR, this would complete the proof. Using again Lemma 2, if N has the form (5) then we can write N = R¯ x1 ⊕· · · ⊕R¯ xn, (7) where ¯ xj ∈M/Rx are such that Ann(¯ xj) = pljR. To split the sequence, we need to ﬁnd representatives yj for the coset ¯ xj so that ψ(a1¯ x1 + · · · + am¯ xm) = a1y1 + · · · amym (8) is a well-deﬁned homomorphism from N to M. The only potential problem with this deﬁnition is that the coeﬃcients aj ∈R are not uniquely determined 6 by P aj¯ xj. If P aj¯ xj = P j bj¯ xj then, since the sum (7) is direct, (bj −aj) ∈ Ann(¯ xj) = pljR for each j and thus bj = aj + pljrj for some rj ∈R. The right-hand side of (8) is invariant under aj 7→bj provided that pljyj = 0. If we can ﬁnd such representatives yj for ¯ xj, then ψ splits the sequence and the proof is complete. To ﬁnd appropriate representatives yj, let xj be an arbitrary representa-tive of ¯ xj. Then pljxj ∈Rx, so we can write pljxj = pscx, where (c, p) = 1. Since pkx = 0, we may assume s ≤k. If s = k we can choose yj = xj. If s ≤ k −1, then pk−1−s+ljxj = pk−1cx. We want to prove that pk−1cx ̸= 0. To this end, we write 1 = tp+uc, which gives 0 ̸= pk−1x = tpkx+upk−1cx = upk−1cx. It follows that k −1 −s + lj ≤k −1, that is, s ≥lj. With yj = xj −ps−ljcx, we then have yj ∈xj + Rx and pljyj = 0. As we have seen, with this choice of yj the map (8) splits the sequence (6) and the proof is complete. Together, Lemma 7, Lemma 9 and Lemma 10 prove the existence part of Theorem 1. For the uniqueness, the following simple result will be useful. Lemma 11. If R is a PID, p and q are prime elements and M = R/qkR, then plM/pl+1M ≃ ( R/pR, p ∼q and l ≤k −1, 0, else as R-modules (and hence also as R/pR-modules). Proof. By Lemma 3, we can write M = Rx with Ann(x) = qkR. Then, plM is generated by plx and N = plM/pl+1M is generated by the coset y = plx + pl+1M. Clearly py = 0 so, again by Lemma 3, we have either N ≃ R/pR (when plx / ∈pl+1M so that y ̸= 0) or N = {0} (when plx ∈pl+1M). If p = q and l ≥k, then plx = 0 so we are in the second case. If p = q and l ≤k −1, then plx = pl+1z would give pk−1x = pk−1−lplx = pkz = 0, which contradicts Ann(x) = pkR. Then we are in the ﬁrst case. Finally, if p ̸∼q then 1 = tqk + upl+1 for some t and u. Then, z = upl+1z ∈pl+1M for any z ∈M, so we are in the second case. We can now prove the uniqueness part of Theorem 1. We need to show that if M denotes the module (2), then s, pj and kj can be con-structed uniquely from M (up to trivial reordering). It is clear that s = 7 dimR(M/M tor) so it’s enough to consider the torsion part. If M is a torsion module (that is, there is no factor Rs in (2)) let us compute the numbers dl(p) = dimR/pR(plM/pl+1M) (9) for p ∈P and l ≥0. By generalities on direct products, we can compute dl(p) by adding up the contribution from each factor. Thus, by Lemma 9, dl(p) is the number of indices j such that pj = p and l ≤kj−1. Consequently, dl−1(p) −dl(p) (10) is the number of indices j such that pj = p and kj = l. This shows that pj and kj can be reconstructed from the module M. It is sometimes useful to rewrite the decomposition (2) in the following alternative form. Theorem 12. Any ﬁnitely generated module M over a PID R is isomorphic to R/q1R × R/q2R × · · · × R/qmR, (11) where qj are non-zero elements of R and q1\|q2\| . . . \|qm. Moreover, this decom-position is unique (up to multiplying qj by units). The elements qj are called invariant factors and (11) the invariant factor decomposition. Note that, for some s, q1 = · · · = qs = 1 and the other qj are non-units. Then, M ≃Rs × M tor. In particular, a ﬁnitely generated torsion module is isomorphic to (11), where qj are non-zero non-units. To prove Theorem 12, we write M as in (2). As we have just explained, we may assume s = 0. We then rearrange the product in a rectangular array so that all factors involving a ﬁxed prime are placed in the same row, and the entries in each row are ordered as in (3). Moreover, the rows are aligned to the right so that the largest entry of each row end up in the right-most column. We then deﬁne qk as the product of all prime powers in the k-th column. An example should clarify the deﬁnition of qk. If R = Z/2Z × Z/2Z × Z/3Z × Z/5Z × Z/16Z × Z/25Z, we write R = Z/2Z × Z/2Z ×Z/16Z ×Z/3Z × Z/5Z ×Z/25Z. 8 Then, q1 = 2, q2 = 2 · 5 and q3 = 16 · 3 · 25. Having deﬁned qk in this way, it is obvious that qk \| qk+1. Moreover, by (12) below, each R/qkR is isomorphic to the direct product of those factors in (2) that were put in the k-th column (in the example above, we have for instance that Z/10Z ≃Z/2Z × Z/5Z). This shows the existence part. To prove uniqueness, we reverse the argument. Given a factorization (11) with qj \| qj+1 we can factorize the non-units qj into primes and construct a rectangular array of factors R/pkR as above. This leads to a decomposition of M as in (2). Since that factorization is unique, so is the factorization (11). To prove (12), we use the module case of the following fact. Lemma 13. If R is a PID and a, b ∈R with gcd(a, b) = 1, then R/abR ≃R/aR × R/bR as rings, and also as R-modules. Proof. Consider the map f : R →R/aR × R/bR given by f(x) = (¯ x, ¯ x). Clearly, Ker(f) = aR ∩bR = abR. To show that f is surjective, take (¯ x1, ¯ x2) arbitrary and write 1 = sa + tb. Then, if we let x = x1 + sa(x2 −x1) = x2 + tb(x1 −x2) we have f(x) = (¯ x1, ¯ x2). Note that f is both a ring homomorphism and an R-module homomorphism, since f(xy) = (xy, xy) = (¯ x, ¯ x)(¯ y, ¯ y) = f(x)f(y) in the ring and f(x, y) = (xy, xy) = x(¯ y, ¯ y) = xf(y) in the module. Iterating this lemma, we have that if a = pk1 1 · · · pkm m , with pj distinct prime elements, then R/aR ≃R/pk1 1 R × · · · × R/pkm m R, (12) both as rings and as R-modules. 9 Canonical forms We now come to an important application of the theory described above, canonical forms for matrices. Any linear operator on a ﬁnite-dimensional vector space can be represented by a matrix, which expresses how it acts on a basis. However, changing the basis leads to a diﬀerent matrix. We would like to ﬁnd a basis such that the matrix has an especially simple form. Moreover, the form should be canonical in the sense that this basis is unique (possibly up to reordering the basis vectors). As a conseqence, if we want to know whether two matrices are similar in the sense that they express the same linear map in diﬀerent bases, we can check whether their canonical forms agree or not. One familiar example is diagonalization, but not any matrix can be diagonalized. The Jordan canonical form explained below can be viewed as an analogue of diagonalization for arbitrary complex matrices. Let K be a ﬁeld, V an n-dimensional vector space over K and A ∈ EndK(V ), that is, A is a linear map from V to itself. We want to study A by viewing V as a module over R = K[x]. The module structure is obtained from the ring homomorphism Φ : K[x] →EndK(V ) given by Φ(p) = p(A). More explicitly, (k0+k1x+· · ·+kmxm)v = k0v+k1Av+k2A2v+· · ·+kmAmv, ki ∈K, v ∈V. We know that R is a PID and V is ﬁnitely generated (any basis for V generates V ). Note also that, as vector spaces over K, K[x] is inﬁnite-dimensional and EndK(V ) is n2-dimensional. Since Φ is K-linear, it follows that Ker(Φ) = Ann(M) is non-zero. In particular, V is a torsion module. We can now apply Thm. 1 and Thm. 12. We have V ≃R/q1R × · · · × R/qmR (13) for some polynomials qj. We may either choose qj = p kj j with pj irreducible polynomials or qj as non-constant polynomials with q1 \| q2 · · · \| qm. Note that the units in R are the non-zero constant polynomials, so if we normalize qj to be monic (leading coeﬃcient 1) then they are unique (in the ﬁrst case up to reordering). If φ is an R-module isomorphism from V to the right-hand side of (13), then φ(Av) = xφ(v). So to understand how A acts on the left we must understand how multiplication by x acts on the right. Consider ﬁrst the 10 action on R/qR, where q(x) = a0 + a1x + · · · + ak−1xk−1 + xk is a monic polynomial of degree k. We choose 1, x, x2, . . . , xk−1 as a basis for R/qR. Then, multiplication by x maps each basis element to the next, except that xk−1 is mapped to xk = −a0 −a1x −· · · −ak−1xk−1. Thus, the matrix for multiplication by x is given by Mq =        0 0 · · · −a0 1 0 −a1 0 1 −a2 . . . . . . 0 0 · · · −ak−1        , the so called companion matrix to q. We can then interpret (13) as saying that there is a basis for V where A is expressed by the block matrix      Mq1 0 · · · 0 0 Mq2 0 . . . . . . 0 0 · · · Mqm     . (14) In the ﬁrst case (qj = p kj j ), we call (14) the primary canonical form and in the second case the rational canonical form. (Brzezinski calls the ﬁrst of these ”rational canonical form”, which seems quite unorthodox.) The primary canonical form is unique up to reordering the blocks whereas the rational canonical form is unique. We remark that, by expanding along the right column, it is easy to see that det(xI −Mq) = q(x). Thus, the characteristic polynomial for A is in both cases given by det(xI −A) = q1 · · · qm. One important diﬀerence between these two forms is that the primary canonical form depends heavily on the base ﬁeld. Given, say, a matrix with integer entries, the primary canonical form will typically look diﬀerent if we work over Q, R or C. This is because these ﬁelds have diﬀerent irreducible polyomials: x2+1 is irreducible over R but not over C and x2−2 is irreducible over Q but not over R. By contrast, the rational canonical form does not change if we extend the ﬁeld. This follows from uniqueness; if K ⊆K′ the canonical form over K is also a canonical form over K′. One non-trivial consequence is that the notion of similarity does not depend on the ﬁeld. For instance, if A and B are integer matrices such that A = TBT −1 for a complex matrix T, then there is such a matrix T with rational (and hence integer) entries. 11 Let us look more closely at the primary decomposition in the case when K is algebraically closed. (In fact, it is enough to assume that K contains all eigenvalues of A.) For instance, we can take K = C. Then, any monic irreducible polynomial has the form x −λ for some λ ∈K. Consider the matrix for multiplication by x in the module R/qR, with q = (x −λ)k. Instead of choosing the basis vectors xj as above, it’s nicer to work with (x −λ)j. Then, x(x −λ)j = λ(x −λ)j + (x −λ)j+1, so we obtain instead of Mq the matrix Jq =        λ 0 0 · · · 0 1 λ 0 0 1 λ . . . . . . 0 0 0 · · · λ        . (15) In the corresponding basis, A takes the form      Jq1 0 · · · 0 0 Jq2 0 . . . . . . 0 0 · · · Jqm     . (16) This is called the Jordan canonical form. It is unique up to reordering the blocks. In all but the very simplest cases, it makes little sense to compute canon-ical forms by hand. Therefore, we will not focus on how that can be done, except for a brief discussion on the case K = C. Note that C[x]/(x−λ) ≃C, so with p(x) = x −λ the numbers (9) are given by dl = rank((A −λI)l) −rank((A −λI)l+1) = dim(Ker((A −λI)l+1))) −dim(Ker((A −λI)l))). It is easy to understand directly why the primary canonical form is deter-mined by the solutions to (A −λI)lv = 0. If J is the block (15), then we have e.g. (J −λ)2 =        0 0 0 · · · 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 . . . . . . 0 0 0 · · · 1 0 0        . 12 In general, (J −λ)k has only zeroes except for a diagonal of 1:s, which is pushed to the southwest as k increases. Thus, the solutions of the equation (J −λ)kv = 0 is the linear span of the k right-most columns. For the block matrix (16), the solution space is spanned by the k right-most columns in all blocks with λ on the diagonal. To give a simple example, suppose we are given a complex 6 × 6-matrix A. We compute the characteristic polynomial det(xI −A) = (x−1)4(x−2)2. We start with the eigenvalue x = 1 and look for eigenvectors. It turns out that the equation (A −I)v = 0 has a two-dimensional space of solutions. This means that there are exactly two Jordan blocks with diagonal entries 1. The sum of their dimension must be equal to 4. So the blocks have either size (3, 1) or (2, 2). Next, we look at the equation (A −I)2v = 0. In the ﬁrst case, the solution space would be 3-dimensional and in the second case 4-dimensional. Let’s say that we are in the ﬁrst case. We then turn to the eigenvalue 2 and ﬁnd that the space of eigenvectors is two-dimensional. There are then two blocks with diagonal entries 2, which necessarily have size 1. We conclude that the Jordan canonical form for A is         1 1 1 1 1 1 2 2         , where we didn’t write the zero entries. The prime powers corresponding to the Jordan blocks are (x −1)3 = x3 −3x2 + 3x −1, (x −1), (x −2) and (x −2), so the primary canonical form is         −1 1 −3 1 −3 1 2 2         . To get the rational canonical form we arrange the prime powers in an array as explained in the proof of Theorem 2. We get (x −1) (x −1)3 (x −2) (x −2) 13 and can read of the invariant factors as the columns: c1 = (x −1)(x −2) and c2 = (x −1)3(x −2). Note that c1c2 is indeed the characteristic polynomial. Since (x −1)3(x −2) = x4 −5x3 + 9x2 −7x + 2, the rational canonical form is         −2 1 3 −2 1 7 1 −9 1 5         . As a ﬁnal remark, note that since Ann(M) is a non-zero ideal in R, it is equal to qR for a unique monic polynomial q, called the minimal polynomial of A. Equivalently, q is the smallest degree monic polynomial such that q(A) = 0. In terms of the rational canonical form, it is clear that q = qm, the largest invariant factor of M. As was remarked above, the characteristic polynomial for A is p = q1 · · · qm. This implies the following useful result. Theorem 14 (Cayley–Hamilton). If p is the characteristic polynomial of a square matrix A, then p(A) = 0. 14
15709	https://math.jhu.edu/~brown/courses/s15/Lectures/421Week2S15.pdf	2.2. THE CONTRACTION PRINCIPLE 23 Now, applying the Lipschitz condition to the nth iterate of f leads directly to the condition that ∣f n(x) −f n(y)∣≤λn ∣x −y∣, so that every two orbits also converge to each other. Hence every orbit converges to this number x0. And ﬁnally, we can show that x0 is actually a ﬁxed point solution for f. To see this, again ∀x ∈I, and ∀n ∈N, ∣x0 −f(x0)∣ = ∣x0 −f n(x) + f n(x) −f n+1(x) + f n+1(x) −f(x0)∣ ≤ ∣x0 −f n(x)∣+ ∣f n(x) −f n+1(x)∣+ ∣f n+1(x) −f(x0)∣ ≤ ∣x0 −f n(x)∣+ λn ∣x −f(x)∣+ λ∣f n(x) −x0∣ = (1 + λ)∣x0 −f n(x)∣+ λn ∣x −f(x)∣. Again, the steps are straightforward. The ﬁrst step is another clever addition of zero, and the second is the triangle inequality. The third involves using the Lipschitz condition on two of the terms, and the last is a clean up of the leftovers. However, this last inequality is the most important. It must be valid for every choice of n ∈N. Hence choosing an increasing sequence of values for n, we see that as n →∞, both ∣x0 −f n(x)∣Ð →0 and λn Ð →0. Hence, it must be the case that ∣x0 −f(x0)∣= 0 or x0 = f(x0). Thus, x0 is a ﬁxed point solution for f on I and the theorem is established. □ Example 2.23. f(x) = √x on the interval [1,∞) is a 1 2-contraction. Why? f ∈C1 and 0 < f ′(x) = 1 2√x ≤1 2 on [1,∞), and strictly less than 1 2 on (1,∞). Thus, by Proposition 2.13, f is 1 2-Lipschitz. So by the Contraction Principle, then, there is a unique ﬁxed point for this discrete dynamical system. Can you ﬁnd it? Exercise 32. Without using any derivative information (that is, without using Propositions 2.13 or 2.18 above), show that f(x) = √x is a 1 2-contraction on [1,∞). Exercise 33. Find all periodic points (to an accuracy of 1 1000) of the discrete dynamical system given by the map f(x) = ln(x−1)+5 on the interval I = [2,100]. 2.2.2. Contractions in several variables. The Contraction Principle in several variables is basically the same as that of one variable. The only diﬀer-ence, really, is that the absolute signs are replaced by the more general metric in Rn. To start, recall in any metric space X, we can deﬁne a small open set via the strict inequality: Bϵ(x) = {y ∈X∣d(x,y) < ϵ}. Definition 2.24. A subset U ∈X is called open if ∀x ∈U, ∃ϵ > 0 such that Bϵ(x) ∈U. And U ∈X is called closed if its complement in X is open. Note that for any x ∈U, where U is open in X, we say U is a neighborhood of x in X, and write U(x) ∈X. Definition 2.25. a point x ∈X is called a boundary point of a subset U ∈X if every neighborhood of x contains at least one point in U and one point not in U. Definition 2.26. a subset U ∈X is called closed in X if it contains all of its boundary points in X. 24 2. SIMPLE DYNAMICS In a loose sense, one can think of a closed subset of real space as a set of solutions to either equations or inequalities of the form ≤or ≥. In this fashion, curves in the plane and surfaces in R3 are closed sets, although ones without interior points (every point is a boundary point). Often, in vector calculus, though, the closed sets constructed as domains for functions are open sets together with their closure, formed by adding to the open set the set of all boundary points. Example 2.27. For U(x) = Bϵ(x) the open ϵ-ball centered at x ∈Rn, its closure is U(x) = Bϵ(x) = {y ∈X∣d(x,y) ≤ϵ}. The boundary of U(x), sometimes written ∂U, is the set of all points y ∈Rn where d(x,y) = ϵ. From vector calculus, recall that this is the (n-1)-dimensional sphere in Rn of radius ϵ centered at x. Keep this in mind as we generalize the Contraction Principle from R to Rn, n > 1. Theorem 2.28 (The Contraction Principle). Let X ⊂Rn be closed and f ∶X → X a λ-contraction. Then f has a unique ﬁxed point x0 ∈X and ∀x ∈X, d(f n(x),x0) ≤λnd(x,x0). Notes: ●Again, we say here that the “dynamics are simple”. The orbit of every point in x does exactly the same thing: Converge exponentially to the ﬁxed point solution x0. ●This also means that every orbit converges to every other orbit also! ●What about periodic points? Can contractions have Periodic points other than ﬁxed points. The answer is no: Exercise 34. Show that a contraction cannot have a non-trivial pe-riodic point (periodic point of prime period greater than 1.) ●The Contraction Principle is also called the Contraction Mapping Theo-rem, or sometimes the Banach Fixed Point Theorem. Recall in dimension-1, the derivative of f (if it existed) can help to deﬁne the Lipschitz constant (Recall Propositions 2.13 and 2.18 above.) How about in several dimensions? In this context, recall that for a C1 function f ∶X ⊂Rn →Rn, the derivative at x ∈X, d fx ∶TxRn →Tf(x)Rn is a linear map from TxRn to Tf(x)Rn. The points in the domain X where the n × n-matrix d fx is of maximal rank are called regular. At a domain point, we can use the Euclidean norm for vectors to deﬁne a matrix norm for d fx as the maximal stretching ability of the unit vectors in the tangent space: ∣∣d fx∣∣= max ∣∣v∣∣=1∣∣d fx(v)∣∣. This non-negative number is not diﬃcult to ﬁnd for an n × n matrix A: If A is symmetric, then ∣∣A∣∣is just the spectral radius ρ(A), the absolute value of the largest (in magnitude) eigenvalue of A. For general n × n A, it is √ ρ(AT A). With this, there are two “derivative”-versions of the Contraction Principle of note. The ﬁrst is a sort of “global version” since the result holds over the entire domain. We will require that the space be strictly convex, however (See Figure 5). 2.2. THE CONTRACTION PRINCIPLE 25 Definition 2.29. A subset X ∈Rn is convex if for any two points x,y ∈X, the straight line segment joining x and y lies entirely in X. X is called strictly convex, if for any two boundary points x,y ∈X, the line segment joining x to y intersects the boundary of X only at x and y. Figure 5. Convexity: Non-convex, convex, but not strictly, and strictly convex sets, respectively. Theorem 2.30 (Global version). If X is the closure of a strictly convex set in Rn, and f ∶X →Rn a C1-map with ∣∣d fx∣∣≤λ < 1, ∀x ∈X, then f has a unique ﬁxed point x0, and ∀x ∈X, d(f n(x),x0) ≤λnd(x,x0). Note here that, technically, we really want that f is diﬀerentiable on the interior of X and continuous on the boundary. Exercise 35. Let f ∶R2 →R2 be the aﬃne map f(x,y) = ( 1 2x + 1 3y,−1 2x + 1 2y + 1). Find the ﬁxed point of f, and show that f is a contraction on R2. This next version is a local version, and presents a very useful tool for the analysis of what happens near (in a neighborhood of) a ﬁxed point: Theorem 2.31 (Local version). Let f be diﬀerentiable with a ﬁxed point x0 such that all of the eigenvalues of the derivative matrix d fx0 have absolute values less than 1. Then there exists ϵ > 0 and an open neighborhood U(x0) = Bϵ(x0) such that on the closure U of U, f(U) ⊂U and f is a contraction on U. Thus, on U, which will be a strictly convex set, the global version above applies. To see these two versions in action, here are two applications: 2.2.3. Application: Newton-Raphson Method. An iterative procedure for the location of a root of a C2-function f ∶R →R, called the Newton-Raphson Method, is an application of the local version of the contraction principle. This method is part of a series of approximation methods that utilize the Taylor Poly-nomials of a function to help identify important features of the function, and is typically found in most standard calculus texts as an application using the tangent line approximation to a function. Here, let f be C1 near an unknown root x∗. Then, for a point x0 near the root, the tangent line approximation to f at x0 is f(x) −f(x0) = f ′(x0)(x −x0). Under relatively mild conditions for f and for x0 “close” enough to x∗, the tangent line function f(x) = f(x0) + f ′(x0)(x −x0) will also have a root. Call this point 26 2. SIMPLE DYNAMICS x1. Solving the tangent line function for x1, we get x1 = x0 −f(x0) f ′(x0). Again, under mild conditions on f, the number x1 lies closer to x∗than x0, and can serve as a new approximation to x∗. Repeating this procedure yields a discrete dynamical system given by xn+1 = g(xn), where g(x) = x −f(x) f ′(x). Some things to note: ●If f is C2 near the root x∗, then as long as f ′(x) does not vanish in a neighborhood of x∗, then g is C1 there. And g′(x) = 1 −(f ′(x))2 −f(x)f ′′(x) (f ′(x))2 = f(x)f ′′(x) (f ′(x))2 . ●If there exists an open interval containing x∗, and if on this interval, there are positive constants δ,M, where ∣f ′(x)∣> δ (f ′ doesn’t get too small), and ∣f ′′(x)∣< M (f ′′ doesn’t get too large), then g′(x) is bounded. ●Notice that if x∗is considered a root of f, then g′(x∗) = f(x∗)f ′′(x∗) (f ′(x∗))2 = 0 and g(x∗) = x∗−f(x∗) f ′(x∗) = x∗. Hence g ﬁxes the root of f and, by continuity, “near x∗” g′(x) will remain small in magnitude. All this points to the contention that there will exist a small (closed) ϵ-neighborhood Bϵ(x∗) about x∗, where ∣g′(x)∣< 1 for all x ∈Bϵ(x∗). (Remember that the deriv-ative of g is 0 at x∗, is continuous in a neighborhood of x∗and cannot grow too quickly around x∗due to the two constraints on the derivatives of f. Once we have this, then by Proposition 2.18, g will form a contraction on B. Thus we have: Proposition 2.32. Let f ∶R →R be C2 with a root x∗. If ∃δ,M > 0, such that ∣f ′(x)∣> δ and ∣f ′′(x)∣< M on an open interval containing x∗, then g(x) = x −f(x) f ′(x). is a contraction near x∗with ﬁxed point x∗. 2.2.4. Application: Existence and Uniqueness of ODE solutions. A global version example of the contraction principle above involves the standard proof of the existence and uniqueness of solutions to the ﬁrst-order IVP (2.2.1) ˙ y = f(t,y), y(t0) = y0 in a neighborhood of (t0,y0) in the t,y-plane. The proof uses a dynamical approach, again with a ﬁrst approximation and then successive iterations using a technique attributed to Charles Picard and known as Picard Iterations. Theorem 2.33 (Picard-Lindel¨ of Theorem). Suppose f(t,y) is continuous in some rectangle R = {(t,y) ∈R2 ∣α < t < β,γ < y < δ}, containing the initial point (t0,y0), and f is Lipschitz continuous in y on R. Then, in some interval t0 −ϵ < t < t0 + ϵ contained in α < t < β, there is a unique solution y = φ(t) of Equation 2.2.1. 2.2. THE CONTRACTION PRINCIPLE 27 To prove this theorem, we will need to understand a bit about how spaces of functions behave. To start, recall from linear algebra that an (real) operator is simply a function f ∶U →V whose domain and codomain are (real) vector spaces. An (real) operator is called linear if ∀x,y ∈U and c1,c2 ∈R, we have f(c1x + c2y) = c1f(x) + c2f(y). Linear operators where both dim(U) = n and dim(V ) = m are ﬁnite-dimensional can be represented by matrices, so that f(x) = Ax, A an m×n matrix. Real-valued continuous functions on R also form a vector space using addition of functions and scalar multiplication as the operations. One can form linear operators on spaces of functions like this one also, but the operator is not represented a as matrix. A good example is the derivative operator d dx which acts on the vector space of all diﬀerentiable, real-valued functions of one independent variable, and takes them to other (in this case, at least) continuous functions. Think d dx(x2 + sinx) = 2x + cosx. This operator is linear due to the Sum and Constant Multiple Rules found in any standard single variable calculus text. There are numerous technical diﬃculties in discussing linear operators in general, but for now, simply accept this general description. Back to the case at hand, any possible solution y = φ(t) (if it exists) to Equa-tion 2.2.1 must be a diﬀerentiable function that satisﬁes (2.2.2) φ(t) = y0 + ∫ t t0 f(s,φ(s))ds for all t in some interval containing t0. Exercise 36. Show that this is true (Hint: Simply diﬀerentiate to recover the ODE.) At this point, existence of a solution to the ODE is assured in the case that f(t,y) is continuous on R, as the integral will then exist at least on some smaller interval t0 −ϵ < t < t0 + ϵ contained inside α < t < β. Note the following: ●One reason a solution may not exist all the way out to the edge of R? What if the edge of R is an asymptote in the t variable? ●A function does not have to be continuous to be integrable (step functions are one example of integrable functions that are not continuous. However, the integral of a step function IS continuous. And if f(t,y) included a step-like function in Equation 2.2.1, solutions may still exist and be continuous. As for uniqueness, suppose f(t,y) is continuous as above, and consider the following operator T, whose domain is the space of all diﬀerentiable functions on R, which takes a function ψ(t) to its image T(ψ(t)) (which we will denote Tψ to help remove some of the parentheses) deﬁned by Tψ = y0 + ∫ t t0 f(s,ψ(s))ds. We can apply T to many functions ψ(t) and the image will be a diﬀerent function Tψ (but still a function of t; see Example 2.34 below). However, looking back at Equation 2.2.2, if we apply T to an actual solution φ(t) to the IVP, the image Tφ should be the same as φ. A solution will be a ﬁxed point of the discrete dynamical 28 2. SIMPLE DYNAMICS system formed by T on the space of functions deﬁned and continuous on R, since Tφ = φ. Exercise 37. Find all ﬁxed points for the derivative operator d dx whose domain is all diﬀerentiable functions on R. Hence, instead of looking for solutions to the IVP, we can instead look for ﬁxed points of the operator T. How do we do this? Fortunately, this operator T has the nice property that it is a contraction. proof of Theorem. By assumption, f(t,y) is Lipschitz continuous in y on R. Hence there is a constant M > 0 where ∣f(t,y) −f(t,y1)∣≤M∣y −y1∣, ∀y,y1 ∈R. Choose a small number ϵ = C M , where C < 1. And deﬁne a distance within the set of continuous functions on the closed interval I = [t0 −ϵ,t0 + ϵ] by d(g,h) = max t∈I ∣g(t) −h(t)∣. Then we have d(Tg,Th) = max t∈I ∣Tg(t) −Th(t)∣ (2.2.3) = max t∈I ∣y0 + ∫ t t0 f(s,g(s))ds −y0 −∫ t t0 f(s,h(s))ds∣ (2.2.4) = max t∈I ∣∫ t t0 f(s,g(s)) −f(s,h(s))ds∣ (2.2.5) ≤ max t∈I ∫ t t0 ∣f(s,g(s)) −f(s,h(s))∣ds (2.2.6) ≤ max t∈I ∫ t t0 M ∣g(s) −h(s)∣ds (2.2.7) ≤ max t∈I ∫ t t0 M ⋅d(g,h)ds (2.2.8) ≤ max t∈I {M ⋅d(g,h) ⋅∣t −t0∣} (2.2.9) Exercise 38. The justiﬁcations of going from Step 2.2.5 to 2.2.6, Step 2.2.6 to 2.2.7, and Step 2.2.8 to 2.2.9 are adaptations of major concepts and/or theorems from Calculus I-II to functions of more than one independent variable. Find what theorems these are and show that these are valid justiﬁcations. Can you see now why the Lipschitz continuity of f is a necessary hypothesis to the theorem? Exercise 39. Justify why the remaining steps are true. Now notice in the last inequality that since I = [t0 −ϵ,t0 + ϵ], we have that ∣t −t0∣≤ϵ = C M . Hence d(Tg,Th) ≤ max t∈I {M ⋅d(g,h) ⋅∣t −t0∣} ≤ M ⋅d(g,h) ⋅C M = C ⋅d(g,h). 2.2. THE CONTRACTION PRINCIPLE 29 Hence T is a C-contraction and there is a unique ﬁxed point φ (which is a solution to the original IVP) on the interval I. Here φ(t) = Tφ(t) = y0 + ∫ t t0 f(s,φ(s))ds. □ We can actually use this construction to construct a solution to an ODE: Example 2.34. Solve the IVP y′ = 2t(1 + y), y(0) = 0 using the above Picard iterations construction. Here, f(t,y) = 2t(1 + y) is a polynomial in both t and y, so that f is obviously continuous in both variables, as well as Lipschitz continuous in y, on the whole plane R2. Hence unique solutions exist everywhere. To actually ﬁnd a solution, start with an initial guess. An obvious one is φ0(t) = 0. Notice that this choice of φ0(t) does not solve the ODE. But since the operator T is a contraction, iterating will lead us to a solution: Here φn+1(t) = Tφn(t). We get φ1(t) = Tφ0(t) = y0 + ∫ t 0 2s(1 + φn(s))ds = ∫ t 0 2s(1 + 0)ds = t2 φ2(t) = Tφ1(t) = y0 + ∫ t 0 2s(1 + φ1(s))ds = ∫ t 0 2s(1 + s2)ds = t2 + 1 2t4, φ3(t) = Tφ2(t) = y0 + ∫ t 0 2s(1 + φ2(s))ds = ∫ t 0 2s(1 + s2 + 1 2s4) ds = t2 + 1 2t4 + 1 6t6, φ4(t) = Tφ3(t) = y0 + ∫ t 0 2s(1 + φ3(s))ds = ∫ t 0 2s(1 + s2 + 1 2s4 + 1 6s6) ds = t2 + 1 2t4 + 1 6t6 + 1 24t8. Exercise 40. Find the pattern and write out a ﬁnite series expression for φn(t). Hint: Use induction. Exercise 41. Find a closed form expression for lim n→∞φn(t) and show that it is a solution of the IVP. Exercise 42. Now rewrite the original ODE in a standard form as a ﬁrst-order linear equation, and solve. To understand why Lipschitz continuity in the dependent variable y is a nec-essary condition for uniqueness of solutions, consider the following example: Example 2.35. Let ˙ y = y 2 3 , y(0) = 0 a ﬁrst-order, autonomous IVP. It should be clear that y(t) ≡0 is a solution. But so is yc(t) = { 1 27 (t + c)3 t < −c 0 t ≥−c ∀c ≥0. There are lots of solutions passing through the origin in ty-trajectory space. So-lutions exist but are deﬁnitely not unique here. What has failed in establishing uniqueness of solutions to this IVP in the Picard-Lindel¨ of Theorem? Here f(y) = y 2 3 30 2. SIMPLE DYNAMICS is certainly continuous at y = 0, but it is NOT Lipshitz continuous there. In fact, f ′(y) = 2 3y−1 3 is not diﬀerentiable at y = 0, and lim y→0f ′(y) = ∞. Exercise 43. Verify that the family of curves yc(t) above solve the IVP, and derive this family by solving the IVP as a separable ODE. Exercise 44. Verify that f(y) = y 2 3 is not Lipschitz continuous at y = 0. 2.3. Interval Maps The Contraction Principle above is a facet of some dynamical systems which display what is called “simple dynamics”: With very little information about the system (map or ODE), one can say just about everything there is to say about the system. Another way to put this is to say that, in a contraction, all orbits do exactly the same thing. Which is, they all converge to the same ﬁxed point (equilibrium solution in the case of a continuous dynamical system.) We can build on this idea by now beginning a study of a relatively simple family of discrete dynamical systems that display slightly more complicated behavior. Let f ∶I →I be a continuous map, where I = [0,1] (we will say f is a C0-map on I, or f ∈C0(I,I)). The graph of f sits inside the unit square [0,1]2 = [0,1] × [0,1] ⊂R2. 2.3.1. Cobwebbing. This graph intersects the line y = x at precisely the points where y = f(x) = x, or the ﬁxed points of the dynamical system given by f on I. Recall that the dynamical sys-tem is formed by iterating f on I, and all of the forward iterates of x0 under f comprise the orbit of x0, Ox0, where Ox0 = {x0, x1 = f(x0), x2 = f(x1) = f 2(x0), ...}. One can track Ox0 in I (and in [0,1]2) visually via the notion of cobwebbing. We use the example of f(x) = x2 on I to illustrate: Figure 6. A cobweb of f(x) = x2 on [0,1]. Choose a starting value x0 ∈I. Un-der f, the next term in the orbit is x1 = f(x0). Vertically, it is the height of the graph of f over x0. Making it the new input value to f means ﬁnding its corresponding place on the horizontal axis. This is easy to see visually. The vertical line x = x0 crosses the graph of f at the height x1 = f(x0). The hor-izontal line y = x1 = f(x0) crosses the diagonal y = x precisely at the point (x1,x1). Taking the output x1 and making it the new input to f (iterat-ing the function) means ﬁnding where 2.3. INTERVAL MAPS 31 the vertical line through this point will again intersect the graph of f (at one point). That point will be at the height x2 = f(x1) and constitutes the second value of the sequence Ox0. By only zig-zagging this way – moving vertically from the y = x line to the graph of f and then horizontally back to the y = x line – we can document the orbit of the point x0 without actually calculating the explicit function values. While this technique is only as accurate as the drawing of the graph, it is an excellent way to “see” an orbit without calculating it. Speciﬁc to this example f(x) = x2 on I, we have two ﬁxed points: x = 0 and x = 1 (the graph crosses the y = x line at these points). And if x0 is chosen to be strictly less than 1, then we can easily conclude via the cobweb that Ox0 Ð →0. Visually, it makes sense. Analytically, it is also intuitive; squaring a number between 0 and 1 always makes it smaller but keeping it positive. However, can you prove that every orbit goes to 0 except for the orbit O1? We will do something like this shortly. Exercise 45. Cobweb the following functions and describe the dynamics as completely as you are able: a. f(x) = x3 on [−1, 1] b. g(x) = ln(x + 1) on [0, ∞) c. h(x) = x2−3 x−2 on R d. k(x) = −5 3(x2 −x) on [0, 1] e. ℓ(x) = −10 3 (x2 −x) on [0, 1] . 2.3.2. Fixed point stability. What happens to orbits near a ﬁxed point of a discrete dynamical system (equilibrium solutions to a continuous one) are of pro-found importance in an analysis of a mathematical model. Often, the ﬁxed points are the only easily discoverable orbits of a hard-to-solve system. They play the role of a “steady-state” of the system. And like for functions in general, knowledge of a function’s derivatives at a point say important things about how a function behaves near a point. To begin this analysis, we will need some deﬁnitions which will allow us to talk about the nature of ﬁxed points in terms of what happens around them. This language is a lot like the way we classiﬁed equilibrium solutions in the ODEs class. For the moment, think of X as simply an interval in R with the metric just the absolute value of the diﬀerence between two points. These deﬁnitions are for all metric spaces X, though. The only caveat here is that in higher dimensions, there are some more subtle things that can happen near a ﬁxed point, as we will see. Definition 2.36. Let x0 be a ﬁxed point of the C0-map f ∶X →X. Then x0 is said to be ●Poisson stable if ∀ϵ > 0, ∃δ > 0 such that ∀x ∈X, if d(x,x0) < δ, then ∀n ∈N d(f n(x),x0) < ϵ. ●asymptotically stable, an attractor, or a sink if ∃ϵ > 0 such that ∀x ∈X, if d(x,x0) < ϵ, then Ox Ð →x0. ●a repeller or a source if ∃ϵ > 0 such that ∀x ∈X, if 0 < d(x,x0) < ϵ, then ∃N ∈N such that ∀n > N, d(f n(x),x0) > ϵ. Do we need semi-stable here for one-dimensional ﬁxed points? Remark 2.37. Asymptotically stable basically means that there is a neighbor-hood of the ﬁxed point where f restricted to that neighborhood is a contraction with x0 as the sole ﬁxed point. Poisson stable means that given any small neigh-borhood of the ﬁxed point, I can choose a smaller neighborhood where if I start 32 2. SIMPLE DYNAMICS in the smaller neighborhood, the forward orbit never leaves the larger neighbor-hood. Asymptotically stable points are always Poisson stable, but not necessarily vice versa. And a ﬁxed point is a repeller if in a small neighborhood of the ﬁxed point, all points that are not the ﬁxed point itself have forward orbit that leave the neighborhood and never return. Example 2.38. f(x) = 1 −x on [0,1] has a unique ﬁxed point x∗= 1 2. This ﬁxed point is Poisson stable, but not asymptotically stable. To see this, simply let δ = ϵ and write out the deﬁnition. Place a drawing here of this map. Remark 2.39. In the classiﬁcation of 2 × 2, ﬁrst-order, homogeneous, linear ODE systems, one can classify the type of the equilibrium solution at the origin via a knowledge of the eigenvalues of the coeﬃcient matrix. In this classiﬁcation, the sink (negative eigenvalues) was the asymptotically stable equilibrium, the source was the repeller, and the center (recall where the two eigenvalues were purely imaginary complex conjugates) was the Poisson stable equilibrium, Since we will review all of the 2 × 2 linear systems theory in a later Chapter, perhaps we can kill this remark, or rewrite it to reﬂect the future result. Example 2.40. Back to Figure 6 the graph of f(x) = x2 on [0,1]. This dynamical system has two ﬁxed points. One can see visually via the cobweb that x = 0 is asymptotically stable. Also, x = 1 is unstable and a repeller. Exercise 46. Show analytically that x = 0 is an attractor while x = 1 is a repeller. That is, show that the ﬁxed points satisfy the respective deﬁnitions. In this class, we will spend a fair amount of time on the maps f ∶[0,1] →[0,1]. There are basically two reasons for this: 1) They have applications beyond simple interval maps, and 2) maps of the unit in-terval are really all one need study when studying intervals. To see the second point, let f ∶R →R, but suppose that there exists a closed interval [a,b], b > a (a single point is considered a closed interval, so the condition that b > a means something with an inte-rior), where f∣[a,b] ∶[a,b] →[a,b]. ●Dynamically speaking, what happens to f(x) under iteration on [a,b] is no diﬀerent from what happens to g(y) on [0,1] under the linear transfor-mation of coordinates y = x−a b−a , where one must transform both the input and output variables appropriately.) 2.3. INTERVAL MAPS 33 Exercise 47. Find the map g ∶[0,1] →[0,1] on the unit interval in the plane that is equivalent dynamically to the map f(x) = x3 on [−1,1]. ●Let f be continuous on an unbounded interval I, either on one side or both. Then in the case that f has bounded image (possibly f has horizontal asymptotes, but this is not necessary), then one can simply study the new dynamical system formed by f, where the domain is the interval f(I), the set of ﬁrst image points f(x), for x ∈I. ●Combining both of these items into one may also be useful: For example, a coordinate transformation can map an unbounded interval to a bounded one (e.g., g(x) = 1 x, mapping [1,∞) onto (0,1], or h(x) = 2 π tan−1(x), taking R to the interval (−1,1)). Under proper care with regard to the orbits, one can transform the dynamical system to one on the bounded interval. We shall elaborate on this later. It is usually an f(x) from above that appears in applications, and mathemat-ically we usually only study maps like g(y). For an example, let’s go back to the Newton Raphson Method for root location. First, a quick deﬁnition: Definition 2.41. A ﬁxed point x0 for f ∶I →I where f ∈C1 is called super-attracting if f ′(x0) = 0. (Why? See the picture.) Figure 7. Follow the cobwebs to see how quickly nearby orbits converge to a superattractor vis a vis an attractor. Proposition 2.42. Let f ∶R →R be C2 with a root r. If ∃δ > 0, M > 0 such that ∣f ′(x)∣> δ and ∣f ′′(x)∣< M on a neighborhood of r, then r is a superattracting ﬁxed point of F(x) = x −f(x) f ′(x). Proof. As we have already calculated, F ′(r) = f(r)f ′′(r) [f ′(r)]2 = 0. □ We can go further, and I will state this part without proof: Since f is C2, then F is C1. Calculating F ′(x) and knowing that it is both continuous and 0 at x = r, there will be a small, closed interval [a,b], b > a with r in the interior, where ∣F ′(x)∣< 1. One can show that, restricted to this interval, F∣[a,b] ∶[a,b] →[a,b] 34 2. SIMPLE DYNAMICS is a λ-contraction, with a superattracting ﬁxed point at r. In this case, all orbits of F converge exponentially to r by λ2, even thought F is simply a λ-contraction. Perhaps we should prove this? 2.3.3. Monotonic maps. Interval maps are quite general, and display tons of diverse and interesting behavior. To begin exploring this behavior, we will need to specify some types of interval maps. The ﬁrst type designation is as follows: Definition 2.43. A map f ∶[a,b] →[a,b] be C0. We say f is ●increasing if for x > y, we have f(x) > f(y), ●nondecreasing if for x > y, we have f(x) ≥f(y), ●non-increasing if for x > y, we have f(x) ≤f(y), ●decreasing if for x > y, we have f(x) < f(y). It is easy and intuitive to see how these deﬁnitions work. You should draw some examples to diﬀerentiate these types. You should also work to understand how these diﬀerent types aﬀect the dynamics a lot. For example, increasing and non-decreasing maps can have many ﬁxed points (actually, the map f(x) = x has ALL points ﬁxed!). While all non-increasing maps (hence all decreasing maps also) can have only one ﬁxed point each. Further, increasing maps cannot have points of period two (why not?), while there does exist a decreasing map with ALL points of period two (can you ﬁnd it?). We will explore these in time. For now, we will start with a fact shared by ALL interval maps: Proposition 2.44. For the C0 map f ∶[a,b] →[a,b], where a,b ∈R, f must have a ﬁxed point. Figure 8. An interval map on [a,b] Visually, this should make sense. Imagine trying to draw the graph of a continuous function in the unit square in a way that it does NOT intersect the diagonal, “ﬁxed point” line. When you get tired of trying, read on. Proof. Suppose for now that f has no ﬁxed point on (a,b) (this seems plausible, since our example above f(x) = x2 on [0,1] satisﬁes this crite-rion). Then, it must be the case that for all x ∈(a,b), either (1) f(x) > x, or (2) f(x) < x. This means that the en-tire graph of f lies above the diagonal, or below it, respectively (See Figure 8). If we are in situation (1), then for any choice of x ∈(a,b), Ox will be an increasing sequence in [a,b]. It is also bounded above by b since the entire se-quence lives in [a,b]. Recall from Calculus the Monotone Sequence Theorem: Every bounded, monotone inﬁnite sequence converges. By deﬁnition, monotone means ei-ther (strictly) increasing or decreasing. For case (2), the orbit is strictly decreasing, and will be bounded below since the entire sequence live in the closed interval [a,b]. 2.3. INTERVAL MAPS 35 Thus we can say in each instance that Ox Ð →x0, for some x0 ∈[a,b] (we can also say lim n→∞f n(x) = x0). What can this point x0 look like? Well, for starters, it must be a ﬁxed point! To see this, f(x0) = f ( lim n→∞f n(x)) = lim n→∞f n+1(x) = lim n→∞f n(x) = x0. Since there are no ﬁxed points in (a,b) by assumption, it must be the case that either x0 = b (case (1)), or x0 = a (case (2)). □ Exercise 48. Reprove Proposition 2.44 using the Intermediate Value Theorem on the function g(x) = f(x) −x. This last proof immediately tells us the following: Proposition 2.45. Let the C0-map f ∶[a,b] →[a,b] be non-decreasing, and suppose there are no ﬁxed points on (a,b). Then either ●exactly one end point is ﬁxed and ∀x ∈[a,b], Ox converges to the ﬁxed end point, or ●Both end points are ﬁxed, one is an attractor and the other is a repeller. And if in the second case above, f is also increasing, then ∀x ∈(a,b), Ox is forward asymptotic to one end point, and backward asymptotic to the other. Example 2.46. Let f(x) = √x on the closed interval [1,b] for any b > 1. This is an example of the ﬁrst situation in the proposition. Here, actually, we can let f have the closed interval [1,∞) as its domain, and we gain the same result. One does have to be careful here, though, as the related function g(x) = √x 2 on [1,∞) has no ﬁxed points at all. Example 2.47. For the second case, think f(x) = x2 on [0,1]. Place some exercises here. We can make the above proposition a bit clearer by adding to our deﬁnitions of orbits: Definition 2.48. Let f ∶[a,b] →[a,b] be C0 and invertible (it has continuous inverse f −1(x)). Then ●Ox ∶= {y ∈[a,b]∣y = f n(x), n ∈Z} ●O+ x ∶= {y ∈[a,b]∣y = f n(x), n ∈N} ●O− x ∶= {y ∈[a,b]∣y = f −n(x), n ∈N}. 2.3.4. Homoclinic/heteroclinic points. Using the notation for the forward and backward orbits introduced above, we can close in on a special property of nondecreasing interval maps. First, we can now say deﬁnitively that ●In the ﬁrst case of Proposition 2.45, ∀x ∈[a,b], either O+ x Ð →a or O+ x Ð → b. Here, f will certainly look like a contraction. Must it be? Keep in mind that the deﬁnition of a contraction is very precise, and maps that behave like contractions may not actually be contractions. Think f(x) = x2 on the closed interval [0,.6]. What is Lip(f) here?
15710	https://geogebra.github.io/docs/manual/en/Lines_and_Axes/	Lines and Axes :: GeoGebra Manual Manual × Custom Search Sort by: Relevance Relevance Date HomeResourcesDownload Manual Commands 3D Commands Algebra Commands Chart Commands Conic Commands Discrete Math Commands Function Commands Geometry Commands GeoGebra Commands List Commands Logic Commands Optimization Commands Probability Commands Scripting Commands Spreadsheet Commands Statistics Commands Financial Commands Text Commands Transformation Commands Vector and Matrix Commands CAS Specific Commands Tools Movement Tools Point Tools Line Tools Special Line Tools Polygon Tools Circle and Arc Tools Conic Section Tools Measurement Tools Transformation Tools Special Object Tools Action Object Tools General Tools CAS Tools Spreadsheet Tools 3D Graphics Tools Custom Tools Manual English Manual Reference Manual Lines and Axes English ČeskyDeutschEnglishEspañolFrançaisMagyarItaliano日本語NederlandsJęzyk polskiTürkçe Edit this Page Contents Lines Axes Lines and Axes Contents Lines Axes Lines You can enter a line as a linear equation in x and y or in parametric form into the Input Bar. In both cases previously defined variables (e.g. numbers, points, vectors) can be used within the equation. You can enter a line’s name at the beginning of the input followed by a colon. 2D Type in g: 3x + 4y = 2 to enter line g as a linear equation. You can enter a line in parametric form thus: g: X = (-5, 5) + t (4, -3) Define the parameters m = 2 and b = -1. Then, you can enter the equation h: y = mx + b to get a line h in y-intercept-form. 3D You can enter a line in parametric form thus: g: X = (1, 6, 3) + λ (7, -4, 4); or via g: Line[(1, 6, 3), Vector[(7, -4, 4)]] You can enter a line as an intersection of 2 planes, by one of the following 3 equivalent input: IntersectPath[4x+7y=46,y+z=9] (4x + 7y = 46, y + z = 9) 7y = 46 - 4x = 7(9 - z) Axes The two coordinate axes are available in commands using the names xAxis and yAxis. The command PerpendicularLine[A, xAxis] constructs the perpendicular line to the x-axis through a given point A. © GeoGebra® \| License \| Help Center \| Built using Antora
15711	https://hjog.org/?p=1440	Ovarian cysts in pregnancy. When surgical treatment required and when monitoring preferred? – hjog.org Skip to content Hellenic Journal of Obstetrics & Gynecology ISSN: 2241-9276 \| e-ISSN: 2241-9551 hjog.org MENU MENU HOME FOR AUTHORS ABOUT Aims and Scope Journal Information Editorial Board CURRENT ISSUE ARCHIVE 2025 2025 Volume 24 – Issue 1 2025 Volume 24 – Issue 2 2025 Volume 24 – Issue 3 2024 2024 Volume 23 – Issue 1 2024 Volume 23 – Issue 2 2024 Volume 23 – Issue 3 2024 Volume 23 – Issue 4 2023 2023 Volume 22 – Issue 1 2023 Volume 22 – Issue 2 2023 Volume 22 – Issue 3 2023 Volume 22 – Issue 4 2022 2022 Volume 21 – Issue 1 2022 Volume 21 – Issue 2 2022 Volume 21 – Issue 3 2022 Volume 21 – Issue 4 2021 2021 Volume 20 – Issue 1 2021 Volume 20 – Issue 2 2021 Volume 20 – Issue 3 2021 Volume 20 – Issue 4 2020 2020 Volume 19 – Issue 1 2020 Volume 19 – Issue 2 2020 Volume 19 – Issue 3 2020 Volume 19 – Issue 4 2019 2019 Volume 18 – Issue 1 2019 Volume 18 – Issue 2 2019 Volume 18 – Issue 3 2019 Volume 18 – Issue 4 2018 2018 Volume 17 – Issue 1 2018 Volume 17 – Issue 2 2018 Volume 17 – Issue 3 2018 Volume 17 – Issue 4 2017 2017 Volume 16 – Issue 1 2017 Volume 16 – Issue 2 2017 Volume 16 – Issue 3 2017 Volume 16 – Issue 4 2016 2016 Volume 15 – Issue 1 2016 Volume 15 – Issue 2 2016 Volume 15 – Issue 3 2016 Volume 15 – Issue 4 2015 2015 Volume 14 – Issue 1 2015 Volume 14 – Issue 2 2015 Volume 14 – Issue 3 2015 Volume 14 – Issue 4 2014 2014 Volume 13 – Issue 1 2014 Volume 13 – Issue 2 2014 Volume 13 – Issue 3 2014 Volume 13 – Issue 4 2013 2013 Volume 12 – Issue 1 2013 Volume 12 – Issue 2 2012 2012 Volume 11 – Issue 1 2012 Volume 11 – Issue 2 2012 Volume 11 – Issue 3 2012 Volume 11 – Issue 4 2011 2011 Volume 10 – Issue 1 2011 Volume 10 – Issue 2 2011 Volume 10 – Issue 3 2011 Volume 10 – Issue 4 2010 2010 Volume 9 – Issue 1 2010 Volume 9 – Issue 2 2010 Volume 9 – Issue 3 2010 Volume 9 – Issue 4 2009 2009 Volume 8 – Issue 1 2009 Volume 8 – Issue 2 2009 Volume 8 – Issue 3 2009 Volume 8 – Issue 4 2008 2008 Volume 7 – Issue 1 2008 Volume 7 – Issue 2 2008 Volume 7 – Issue 3 2008 Volume 7 – Issue 4 2007 2007 Volume 6 – Issue 1 2007 Volume 6 – Issue 2 2007 Volume 6 – Issue 3 2007 Volume 6 – Issue 4 2006 2006 Volume 5 – Issue 1 2006 Volume 5 – Issue 2 2005 2005 Volume 4 – Issue 1 2005 Volume 4 – Issue 2 2005 Volume 4 – Issue 3 2005 Volume 4 – Issue 4 2004 2004 Volume 3 – Issue 1 2004 Volume 3 – Issue 2 2004 Volume 3 – Issue 3 2004 Volume 3 – Issue 4 SUBMIT YOUR MANUSCRIPT CONTACT US Ovarian cysts in pregnancy. When surgical treatment required and when monitoring preferred? Posted on December 28, 2017 February 4, 2022 by admin Case Report Kalmantis K, Petsa A, Alexopoulos E, Daskalakis G, Rodolakis A First Department of Obstetrics & Gynecology, Alexandra Maternity Hospital, Athens, Greece Correspondence: Kalmantis Konstantinos, 15 Ifigenias str P. Faliro, Athens Greece, Tel: 0030 6972839327, Email: kkalmantis@hotmail.com Abstract Introduction: Ovarian cysts occur in 3% of pregnancies. Usually, they do not have clinical symptoms and are found accidentally during the ultrasound screening performed in the 1st trimester of pregnancy. Materials and Methods: Articles were identified through electronic databases; no date or language restrictions were placed; relevant citations were hand searched. The search was conducted using the following terms: ovarian tumors, ovarian cyst, pregnancy, management, and outcome. Discussion: Vast majority is benign; these are functional cysts which resolve in the 2nd trimester. Discrimination of these cysts can be made by color ultrasonography. Ultrasound examination will provide useful information on the size, composition and nature of the tumor, in order to avoid unnecessary surgical interventions during pregnancy.Conclusions:Management of cysts in pregnancy should mainly include a detailed evaluation, through color sonography, based on specific I.O.T.A. criteria, and appropriate treatment should be decided, based on the main purpose of providing healthcare to the pregnant woman and ensuring successful outcome of the pregnancy. Keywords: ovarian cyst, pregnancy, ultrasound evaluation, sonographic features, management, outcome. Introduction Finding an ovarian tumor during pregnancy accounts for about 3%1. Usually, these cysts do not have clinical symptoms and are found accidentally during the ultrasound screening performed in the 1st trimester of pregnancy 2. The most common cyst in pregnancy is the persistent corpus luteum, which in 85% of cases resolves spontaneously in the early of the 2nd trimester 3-5 Cysts that persist beyond the 16th week of pregnancy are not functional and require further investigation. Rarely these are complicated by pain, torsion, and rupture and create noisy symptoms and bleeding 6. Finally, the incidence of ovarian cancer in pregnant women aged between 18 and 39 years old, is 1-8 per 100,000 women 7-8. Before the application of ultrasound in clinical practice, diagnosis of ovarian cysts in pregnancy was made by physical examination, especially in cases of pregnant women with symptoms such as abdominal pain or palpable mass. In these cases, immediate surgical intervention was preferred in order to avoid complications. In recent decades, the use of ultrasound changed the management and more information was available to predict the nature of a cyst in pregnancy. The main purpose was to identify and distinguish the cases of pregnant women that require surgical intervention 4,9,10. Materials and Methods The purpose of this paper is to provide a complete review for ovarian cysts in pregnancy as a guide for the appropriate management. The authors researched through electronic databases articles concerning this specific subject. No date or language restrictions were placed; relevant citations were hand searched. The search was conducted using the following terms: ovarian tumors, ovarian cyst, pregnancy, management, and outcome. Discussion Cysts in pregnancy are divided into benign and malignant. The most common benign ovarian tumors are functional cysts, followed by dermoid cysts, cystadenomas, endometriomas, and finally peritoneal and paraovarian cysts. Malignant ovarian tumors include non-epithelial (germ cell-sex cord) and epithelial tumors (borderline ovarian tumors, invasive ovarian tumors). Cysts are diagnosed by ultrasound, where high diagnostic accuracy allows discrimination of ovarian tumors in benign and malignant 7,9,10,11. Various studies and algorithms have been reported in literature, in order to assess the possibility of malignancy of an ovarian tumor during pregnancy 9,10-17. International Ovarian Tumor Analysis group (I.O.T.A.) described the morphological characteristics and angiogenesis of ovarian tumors (“Simple rules”), to determine their nature 18, 19. According to I.O.T.A., the characteristics of benign tumors are the following: 1) unilocular cyst of low echogenicity, 2) presence of solid components (diameter <7mm), 3) acoustic shadowing, 4) multilocular cyst, with smooth surface and a major diameter of <100mm, 5) no blood flow, while the malignant characteristics are: 1) solid tumor with irregular surface, 2) ascites, 3) more than 4 pappilary projections, 4) solid multilocular tumor with major diameter> 100mm and 5) strong blood flow. If there are one or more benign characteristics, the tumor is defined as benign. If there are one or more malignant characteristics, then the tumor is classified as malignant. If none of the above features exist or if malignant and benign characteristics coexist, then the tumor cannot be classified. In pregnancy, frequently encountered unilocular cysts of low echogenicity, without vascularization, with a size smaller than 50mm, with smooth surface. They remain asymptomatic and tend to regress during pregnancy. Transvaginal ultrasound color flow imaging, based on specific rules of I.O.T.A. (Simple rules) can help in the diagnosis of a cyst in pregnancy 18, 19. The sonographic features of cysts occurring in pregnancy are listed in Table 1. Recent studies have reported the usefulness of color ultrasound (Doppler), in the evaluation of cysts in pregnancy. The pulsatility index (PI), and the vascular resistance index (RI) have low values in malignancies, as the vessels of the tumors lack muscular layer. Thereby they present low resistances 20, 21. Shah et al, have reported high sensitivity and specificity in the detection of malignant tumors with PI <1 (sensitivity 93%, specificity 93%) and RI <0.6 (sensitivity 83%, specificity 93%)22. Nevertheless, there is a significant overlap rate between RI and RI in benign and malignant ovarian tumors. Especially in pregnancy, where increased blood flow of the uterus and hemodynamic changes from all three trimesters of the pregnancy affect even more RI and PI values 21. Also, three-dimensional color ultrasound provides additional information to distinguish cysts, from the first trimester of the pregnancy, without harming the fetus and hassling the pregnant woman 23, 24. MRI is safely used in the 2nd – 3rd trimester of pregnancy. Its role is mostly assistive (when the ultrasound diagnosis is uncertain or there are suspicious sonographic findings)25. It can accurately distinguish the content of the tumor (liquid or solid), displays the exact origin of pelvic formations and controls in detail the existence of vascularization 26-29. Correlation between tumor markers and pregnancy is unclear. Ca–125 is most frequently used. It is a glycoprotein that increases in primary ovarian carcinoma, as well as in benign conditions such as endometriosis, uterine fibroids, pelvic inflammation, menstruation and pregnancy 30-32. Particularly during the 1st and 3rd trimester, increased amounts of Ca–125 are produced from amnion and decidua. This marker is estimated mainly when its value is greater than 100IU/ml, while its main use is disease progression or response to therapy 33, 34. Management of cysts in pregnancy The decision whether a cyst in pregnancy will be managed conservatively or surgically depends on the sonographic features and the size of the cyst 35. Initially, conservative management of cysts is recommended, with regular monitoring, as 71% of them will be absorbed 36, 37. According to the Royal College of Obstetricians and Gynecologists, simple, unilateral, unilocular ovarian cysts <5cm have a low risk of malignancy and can be treated conservatively 38. Also, regular ultrasound examination assists early detection of progressive changes in the size and morphology of tumors 39. In asymptomatic tumors, without signs of malignancy, which are detected in the 1st trimester, re-examination is recommended at 16 weeks of gestation. If this does not persist and remains stable, ultrasound re-examination is recommended at 6-8 weeks postpartum 13. In case that the cyst persists or increases in size, after 16 weeks of gestation, and there are equivocal or suspicious sonographic findings for malignancy, MRI in the lower abdomen and surgical intervention are recommended. Moreover, surgical intervention is required when a complication occurs, such as rupture, torsion, and bleeding (Figure 1) 13, 40-42. Second trimester, between 16 and 23 weeks of gestation, is considered ideal for surgical intervention, since the risk for spontaneous abortion of 1st trimester and preterm delivery of 3rd trimester is reduced 43, 44. Figure 1. Management of cyst in pregnancy Conclusions Cysts occur in 3% of pregnancies. Vast majority is benign; these are functional cysts which resolve in the 2nd trimester. Discrimination of these cysts can be made by color ultrasonography. Ultrasound examination will provide useful information on the size, composition and nature of the tumor, in order to avoid unnecessary surgical interventions during pregnancy. Features, such as tumor size> 5cm, heterogeneous echostructure, irregular surface, presence of wall abnormalities, presence of solid components with vascularization, raise strong suspicion of malignancy. Color three-dimensional ultrasonography significantly increases diagnostic accuracy, while MRI has a complementary-supportive role, especially when there are suspicious sonographic findings. Tumor markers are not reliable in investigating cysts in pregnancy, due to their low specificity. Cysts in pregnancy should be evaluated in detail, through color sonography, based on specific I.O.T.A. criteria, and appropriate treatment should be decided, based on the main purpose of providing healthcare to the pregnant woman and ensuring successful outcome of the pregnancy. References Goffinet F.[Ovarian cysts and pregnancy].J Gynecol Obstet Biol Reprod (Paris). 2001 Nov;30(1 Suppl):S100-8. Sherard GB 3rd, Hodson CA, Williams HJ, Semer DA, Hadi HA, Tait DL. Adnexal masses and pregnancy: a 12-year experience. Am J Obstet Gynecol. 2003 Aug;189(2):358-62; discussion 362-3. Nelson MJ, Cavalieri R, Graham D, Sanders RC. Cysts in pregnancy discovered by sonography. J Clin Ultrasound. 1986 Sep;14(7):509-12. Hogston P, Lilford RJ. Ultrasound study of ovarian cysts in pregnancy: prevalence and significance.Br J Obstet Gynaecol. 1986 Jun;93(6):625-8. Platek DN, Henderson CE, Goldberg GL. The management of a persistent adnexal mass in pregnancy. Am J Obstet Gynecol. 1995 Oct;173(4): 1236-40. Whitecar MP, Turner S, Higby MK. Adnexal masses in pregnancy: a review of 130 cases undergoing surgical management. Am J Obstet Gynecol. 1999 Jul;181(1):19-24. Hoover K, Jenkins TR. Evaluation and management of adnexal mass in pregnancy. Am J Obstet Gynecol. 2011 Aug;205(2):97-102. Runowicz CD, Brewer M. Goff B, Section Editor. Adnexal mass in pregnancy, 2016. Glanc P, Salem S, Farine D. Adnexal masses in the pregnant patient: a diagnostic and management challenge. Ultrasound Q. 2008 Dec;24(4):225-40. Leiserowitz GS. Managing ovarian masses during pregnancy. Obstet Gynecol Surv. 2006 Jul;61(7):463-70. Usui R, Minakami H, Kosuge S, Iwasaki R, Ohwada M, Sato I. A retrospective survey of clinical, pathologic, and prognostic features of adnexal masses operated on during pregnancy. J Obstet Gynaecol Res. 2000 Apr;26(2):89-93. Bignardi T, Condous G. The management of ovarian pathology in pregnancy. Best Pract Res Clin Obstet Gynaecol. 2009 Aug;23(4):539-48. Aggarwal P, Kehoe S. Ovarian tumours in pregnancy: a literature review. Eur J Obstet Gynecol Reprod Biol. 2011 Apr;155(2):119-24. Bromley B, Benacerraf B. Adnexal masses during pregnancy: accuracy of sonographic diagnosis and outcome. J Ultrasound Med. 1997 Jul;16(7):447-52; quiz 453-4. Knudsen UB, Tabor A, Mosgaard B, Andersen ES, Kjer JJ, Hahn-Pedersen S, Toftager-Larsen K, Mogensen O. Management of ovarian cysts. Acta Obstet Gynecol Scand. 2004 Nov;83(11):1012-21. Levine D, Barbieir RL (ed). Up to date 2015. Sonographic differentiation of benign versus malignant adnexal masses. Telischak NA, Yeh BM, Joe BN, Westphalen AC, Poder L, Coakley FV. MRI of adnexal masses in pregnancy.AJR Am J Roentgenol. 2008 Aug;191(2):364-70. Timmerman D, Valentin L, Bourne TH, Collins WP, Verrelst H, Vergote I; International Ovarian Tumor Analysis (IOTA) Group. Terms, definitions and measurements to describe the sonographic features of adnexal tumors: a consensus opinion from the International Ovarian Tumor Analysis (IOTA) Group. Ultrasound Obstet Gynecol. 2000 Oct;16(5):500-5. Timmerman D, Ameye L, Fischerova D, Epstein E, Melis GB, Guerriero S, Van Holsbeke C, Savelli L, Fruscio R, Lissoni AA, Testa AC, Veldman J, Vergote I, Van Huffel S, Bourne T, Valentin L. Simple ultrasound rules to distinguish between benign and malignant adnexal masses before surgery: prospective validation by IOTA group.BMJ. 2010 Dec 14;341:c6839. Fleischer AC, Cullinan JA, Jones HW 3rd, Peery CV, Bluth RF, Kepple DM. Serial assessment of adnexal masses with transvaginal color Doppler sonography. Ultrasound Med Biol. 1995;21(4):435-41. Wheeler TC, Fleischer AC. Complex adnexal mass in pregnancy: predictive value of color Doppler sonography. J Ultrasound Med. 1997 Jun;16(6): 425-8. Shah D, Shah S, Parikh J, Bhatt C J, Vaishnav K, Bala D V. Doppler ultrasound: a good and reliable predictor of ovarian malignancy. J Obstet Gynecol India 2013;63(3):186-189 Testa AC, Ajossa S, Ferrandina G, Fruscella E, Ludovisi M, Malaggese M, Scambia G, Melis GB, Guerriero S. Does quantitative analysis of three- dimensional power Doppler angiography have a role in the diagnosis of malignant pelvic solid tumors? A preliminary study. Ultrasound Obstet Gynecol. 2005 Jul;26(1):67-72. Kalmantis K, Iavazzo C, Anastasiadou V, Antsaklis A. The role of 3-dimensional power Doppler imaging in the assessment of ovarian teratoma in pregnancy: a case report. Case Rep Med. 2011; 2011:896396. Patenaude Y, Pugash D, Lim K, Morin L; Diagnostic Imaging Committee., Lim K, Bly S, Butt K, Cargill Y, Davies G, Denis N, Hazlitt G, Morin L, Naud K, Ouellet A, Salem S; Society of Obstetricians and Gynaecologists of Canada. The use of magnetic resonance imaging in the obstetric patient. J Obstet Gynaecol Can. 2014 Apr;36(4):349-63. Curtis M, Hopkins MP, Zarlingo T, Martino C, Graciansky-Lengyl M, Jenison EL. Magnetic resonance imaging to avoid laparotomy in pregnancy. Obstet Gynecol. 1993 Nov;82(5):833-6. Buist MR, Golding RP, Burger CW, Vermorken JB, Kenemans P, Schutter EM, Baak JP, Heitbrink MA, Falke TH. Comparative evaluation of diagnostic methods in ovarian carcinoma with emphasis on CT and MRI. Gynecol Oncol. 1994 Feb;52(2):191-8. Yakasai IA, Bappa LA. Diagnosis and management of adnexal masses in pregnancy. J Surg Tech Case Rep. 2012 Jul;4(2):79-85. Araujo Júnior E, Sun SY, Campanharo FF, Nacaratto DC, Nardozza LM, Mattar R, Habib VV, Moron AF. Diagnosis of ovarian metastasis from gestational trophoblastic neoplasia by 3D power Doppler ultrasound and dynamic contrast-enhanced magnetic resonance imaging: case report. Case Rep Oncol. 2012 May;5(2):359-66. Bast RC Jr, Feeney M, Lazarus H, Nadler LM, Colvin RB, Knapp RC. Reactivity of a monoclonal antibody with human ovarian carcinoma. J Clin Invest. 1981 Nov;68(5):1331-7. Bast RC Jr, Klug TL, St John E, Jenison E, Niloff JM, Lazarus H, Berkowitz RS, Leavitt T, Griffiths CT, Parker L, Zurawski VR Jr, Knapp RC. A radioimmunoassay using a monoclonal antibody to monitor the course of epithelial ovarian cancer. N Engl J Med. 1983 Oct 13;309(15):883-7. Jacobs IJ, Fay TN, Yovich J, Stabile I, Frost C, Turner J, Oram DH, Grudzinskas JG. Serum levels of CA 125 during the first trimester of normal outcome, ectopic and anembryonic pregnancies. Hum Reprod. 1990 Jan;5(1):116-22. Mc Carthy A. 7th ed. United States: Blackwell publishing: 2007. Miscellaneous Medical Disorders Dewhurst’s textbook of Obstetrics and Gynaecology: 283-8. Aslam N, Ong C, Woelfer B, Nicolaides K, Jurkovic D. Serum CA125 at 11-14 weeks of gestation in women with morphologically normal ovaries. BJOG. 2000 May;107(5):689-90. Caspi B, Levi R, Appelman Z, Rabinerson D, Goldman G, Hagay Z. Conservative management of ovarian cystic teratoma during pregnancy and labor. Am J Obstet Gynecol. 2000 Mar;182(3):503-5. Schmeler KM, Mayo-Smith WW, Peipert JF, Weitzen S, Manuel MD, Gordinier ME. Adnexal masses in pregnancy: surgery compared with observation. Obstet Gynecol. 2005 May;105(5 Pt 1):1098-103. Leiserowitz GS. Managing ovarian masses during pregnancy. Obstet Gynecol Surv. 2006 Jul;61(7):463-70. RCOG. Ovarian cyst in post-menopausal women (Green-top 34). In: Green top Guidelines: RCOG: 2003. Zanetta G, Mariani E, Lissoni A, Ceruti P, Trio D, Strobelt N, Mariani S. A prospective study of the role of ultrasound in the management of adnexal masses in pregnancy. BJOG. 2003 Jun;110(6):578-83. Mavromatidis G, Sotiriadis A, Dinas K, Mamopoulos A, Rousso D. Large luteinized follicular cyst of pregnancy. Ultrasound Obstet Gynecol. 2010 Oct;36(4):517-20. Shah K, Anjurani S, Ramkumar V, Bhat P, Urala M. Ovarian mass in pregnancy: a review of six cases treated with surgery. Internet J Gynecol Obstet, 2013;14 (2) Fujishita A, Masuzaki H, Nakajima H, Ishimaru T, Yamabe T. The evaluation of ovarian tumor associated with pregnancy by the ultrasonographical method and serum CA125 levels. Nihon Sanka Fujinka Gakkai Zasshi. 1994 Sep;46(9):875-82. Agarwal N, Parul, Kriplani A, Bhatla N, Gupta A. Management and outcome of pregnancies complicated with adnexal masses. Arch Gynecol Obstet. 2003 Jan;267(3):148-52. Mazze RI, Källén B. Reproductive outcome after anesthesia and operation during pregnancy: a registry study of 5405 cases. Am J Obstet Gynecol. 1989 Nov;161(5):1178-85. Received 23-8-2017 Revised 20-9-2017 Accepted 6-10-2017 Posted in 2017 Volume 16 – Issue 3Tagged management, outcome., ovarian cyst, pregnancy, sonographic features, ultrasound evaluationPost navigation In utero predisposition→ ←Can the measurement of endometrial thickness by transvaginal ultrasound be used for the prediction of endometrial cancer in asymptomatic post-menopausal women? 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15712	https://www.cs.helsinki.fi/juha.karkkainen/teach/14-15/SPA/all-lectures.pdf	58093 String Processing Algorithms Lectures, Autumn 2014, period II Juha K¨ arkk¨ ainen Contents 0. Introduction 1. Sets of strings • Search trees, string sorting, binary search 2. Exact string matching • Finding a pattern (string) in a text (string) 3. Approximate string matching • Finding in the text something that is similar to the pattern 4. Suﬃx tree and suﬃx array • Preprocess a long text for fast string matching and all kinds of other tasks 2 0. Introduction Strings and sequences are one of the simplest, most natural and most used forms of storing information. • natural language, biosequences, programming source code, XML, music, any data stored in a ﬁle • Many algorithmic techniques are ﬁrst developed for strings and later generalized for more complex types of data such as graphs. The area of algorithm research focusing on strings is sometimes known as stringology. Characteristic features include • Huge data sets (document databases, biosequence databases, web crawls, etc.) require eﬃciency. Linear time and space complexity is the norm. • Strings come with no explicit structure, but many algorithms discover implicit structures that they can utilize. 3 About this course On this course we will cover a few cornerstone problems in stringology. We will describe several algorithms for the same problems: • the best algorithms in theory and/or in practice • algorithms using a variety of diﬀerent techniques The goal is to learn a toolbox of basic algorithms and techniques. On the lectures, we will focus on the clean, basic problem. Exercises may include some variations and extensions. We will mostly ignore any application speciﬁc issues. 4 Strings An alphabet is the set of symbols or characters that may occur in a string. We will usually denote an alphabet with the symbol Σ and its size with σ. We consider three types of alphabets: • Ordered alphabet: Σ = {c1, c2, . . . , cσ}, where c1 < c2 < · · · < cσ. • Integer alphabet: Σ = {0, 1, 2, . . . , σ −1}. • Constant alphabet: An ordered alphabet for a (small) constant σ. 5 The alphabet types are really used for classifying and analysing algorithms rather than alphabets: • Algorithms for ordered alphabet use only character comparisons. • Algorithms for integer alphabet can use more powerful operations such as using a symbol as an address to a table or arithmetic operations to compute a hash function. • Algorithms for constant alphabet can perform almost any operation on characters and even sets of characters in constant time. The assumption of a constant alphabet in the analysis of an algorithm often indicates one of two things: • The eﬀect of the alphabet on the complexity is complicated and the constant alphabet assumption is used to simplify the analysis. • The time or space complexity of the algorithm is heavily (e.g., linearly) dependent on the alphabet size and the algorithm is eﬀectively unusable for large alphabets. An algorithm is called alphabet-independent if its complexity has no dependence on the alphabet size. 6 A string is a sequence of symbols. The set of all strings over an alphabet Σ is Σ∗= ∞ [ k=0 Σk = Σ0 ∪Σ1 ∪Σ2 ∪. . . where Σk = k z }\| { Σ × Σ × · · · × Σ = {a1a2 . . . ak \| ai ∈Σ for 1 ≤i ≤k} = {(a1, a2, . . . , ak) \| ai ∈Σ for 1 ≤i ≤k} is the set of strings of length k. In particular, Σ0 = {ε}, where ε is the empty string. We will usually write a string using the notation a1a2 . . . ak, but sometimes using (a1, a2, . . . , ak) may avoid confusion. 7 There are many notations for strings. When describing algorithms, we will typically use the array notation to emphasize that the string is stored in an array: S = S[1..n] = SS . . . S[n] T = T[0..n) = TT . . . T[n −1] Note the half-open range notation [0..n) which is often convenient. In an abstract context, we often use other notations, for example: • α, β ∈Σ∗ • x = a1a2 . . . ak where ai ∈Σ for all i • w = uv, u, v ∈Σ∗(w is the concatenation of u and v) We will use \|w\| to denote the length of a string w. 8 Individual characters or their positions usually do not matter. The signiﬁcant entities are the substrings or factors. Deﬁnition 0.1: Let w = xyz for any x, y, z ∈Σ∗. Then x is a preﬁx, y is a factor (substring), and z is a suﬃx of w. If x is both a preﬁx and a suﬃx of w, then x is a border of w. Example 0.2: Let w = bonobo. Then • ε, b, bo, bon, bono, bonob, bonobo are the preﬁxes of w • ε, o, bo, obo, nobo, onobo, bonobo are the suﬃxes of w • ε, bo, bonobo are the borders of w • ε, b, o, n, bo, on, no, ob, bon, ono, nob, obo, bono, onob, nobo, bonob, onobo, bonobo are the factors of w. Note that ε and w are always suﬃxes, preﬁxes, and borders of w. A suﬃx/preﬁx/border of w is proper if it is not w, and nontrivial if it is not ε or w. 9 Some Interesting Strings The Fibonacci strings are deﬁned by the recurrence: F0 = ε F1 = b F2 = a Fi = Fi−1Fi−2 for i > 2 The inﬁnite Fibonacci string is the limit F∞. For all i > 1, Fi is a preﬁx of F∞. Example 0.3: F3 = ab F4 = aba F5 = abaab F6 = abaababa F7 = abaababaabaab F8 = abaababaabaababaababa Fibonacci strings have many interesting properties: • \|Fi\| = fi, where fi is the ith Fibonacci number. • F∞has exactly k + 1 distinct factors of length k. • For all i > 1, we can obtain Fi from Fi−1 by applying the substitutions a 7→ab and b 7→a to every character. 10 A De Bruijn sequence Bk of order k for an alphabet Σ of size σ is a cyclic string of length σk that contains every string of length k over the alphabet Σ as a factor exactly once. The cycle can be opened into a string of length σk + k −1 with the same property. Example 0.4: De Bruijn sequences for the alphabet {a, b}: B2 = aabb(a) B3 = aaababbb(aa) B4 = aaaabaabbababbbb(aaa) De Bruijn sequences are not unique. They can be constructed by ﬁnding Eulerian or Hamiltonian cycles in a De Bruijn graph. Example 0.5: De Bruijn graph for the alphabet {a, b} that can be used for constructing B2 (Hamiltonian cycle) or B3 (Eulerian cycle). aa bb ab ba b a b b a a a b 11 1. Sets of Strings Basic operations on a set of objects include: Insert: Add an object to the set Delete: Remove an object from the set. Lookup: Find if a given object is in the set, and if it is, possibly return some data associated with the object. There can also be more complex queries: Range query: Find all objects in a given range of values. There are many other operations too but we will concentrate on these here. 12 An eﬃcient execution of the operations requires that the set is stored as a suitable data structure. • A (balanced) binary search tree supports the basic operations in O(log n) time and range searching in O(log n + r) time, where n is the size of the set and r is the size of the result. • An ordered array supports lookup and range searching in the same time as binary search trees. It is simpler, faster and more space eﬃcient in practice, but does not support insertions and deletions. • A hash table supports the basic operations in constant time (usually randomized) but does not support range queries. A data structure is called dynamic if it supports insertions and deletions (tree, hash table) and static if not (array). Static data structures are constructed once for the whole set of objects. In the case of an ordered array, this involves another important operation, sorting. Sorting can be done in O(n log n) time using comparisons and even faster for integers. 13 The above time complexities assume that basic operations on the objects including comparisons can be performed in constant time. When the objects are strings, this is no more true: • The worst case time for a string comparison is the length of the shorter string. Even the average case time for a random set of n strings is O(logσ n) in many cases, including for basic operations in a balanced binary search tree. We will show an even stronger result for sorting later. And sets of strings are rarely fully random. • Computing a hash function is slower too. A good hash function depends on all characters and cannot be computed faster than the length of the string. For a string set R, there are also new types of queries: Preﬁx query: Find all strings in R that have the query string S as a preﬁx. This is a special type of range query. Lcp (longest common preﬁx) query: What is the length of the longest preﬁx of the query string S that is also a preﬁx of some string in R. Thus we need special set data structures and algorithms for strings. 14 Trie A simple but powerful data structure for a set of strings is the trie. It is a rooted tree with the following properties: • Edges are labelled with symbols from an alphabet Σ. • For every node v, the edges from v to its children have diﬀerent labels. Each node represents the string obtained by concatenating the symbols on the path from the root to that node. The trie for a string set R, denoted by trie(R), is the smallest trie that has nodes representing all the strings in R. The nodes representing strings in R may be marked. Example 1.1: trie(R) for R = {ali, alice, anna, elias, eliza}. a l i c e n n a e l i a s z a 15 The trie is conceptually simple but it is not simple to implement eﬃciently. The time and space complexity of a trie depends on the implementation of the child function: For a node v and a symbol c ∈Σ, child(v, c) is u if u is a child of v and the edge (v, u) is labelled with c, and child(v, c) = ⊥(null) if v has no such child. As an example, here is the insertion algorithm: Algorithm 1.2: Insertion into trie Input: trie(R) and a string S[0..m) ̸∈R Output: trie(R ∪{S}) (1) v ←root; j ←0 (2) while child(v, S[j]) ̸= ⊥do (3) v ←child(v, S[j]); j ←j + 1 (4) while j < m do (5) Create new node u (initializes child(u, c) to ⊥for all c ∈Σ) (6) child(v, S[j]) ←u (7) v ←u; j ←j + 1 (8) Mark v as representative of S 16 There are many implementation options for the child function including: Array: Each node stores an array of size σ. The space complexity is O(σN), where N is the number of nodes in trie(R). The time complexity of the child operation is O(1). Requires an integer alphabet. Binary tree: Replace the array with a binary tree. The space complexity is O(N) and the time complexity O(log σ). Works for an ordered alphabet. Hash table: One hash table for the whole trie, storing the values child(v, c) ̸= ⊥. Space complexity O(N), time complexity O(1). Requires an integer alphabet. A common simpliﬁcation in the analysis of tries is to assume a constant alphabet. Then the implementation does not matter: Insertion, deletion, lookup and lcp query for a string S take O(\|S\|) time. Note that a trie is a complete representation of the strings. There is no need to store the strings separately. 17 Preﬁx free string sets Many data structures and algorithms for a string set R become simpler if R is preﬁx free. Deﬁnition 1.3: A string set R is preﬁx free if no string in R is a preﬁx of another string in R. There is a simple way to make any string set preﬁx free: • Let $ ̸∈Σ be an extra symbol satisfying $ < c for all c ∈Σ. • Append $ to the end of every string in R. This has little or no eﬀect on most operations on the set. The length of each string increases by one only, and the additional symbol could be there only virtually. Example 1.4: The set {ali, alice, anna, elias, eliza} is not preﬁx free because ali is a preﬁx of alice, but {ali$, alice$, anna$, elias$, eliza$} is preﬁx free. 18 If R is preﬁx free, the leaves of trie(R) represent exactly R. This simpliﬁes the implementation of the trie. Example 1.5: The trie for {ali$, alice$, anna$, elias$, eliza$}. a l i $ c e $ n n a $ e l i a s $ z a $ 19 Compact Trie Tries suﬀer from a large number of nodes, close to \|\|R\|\| in the worst case. • For a string set R, we use \|R\| to denote the number of strings in R and \|\|R\|\| to denote the total length of the strings in R. The space requirement can be problematic, since typically each node needs much more space than a single symbol. Compact tries reduce the number of nodes by replacing branchless path segments with a single edge. Example 1.6: Compact trie for {ali$, alice$, anna$, elias$, eliza$}. a li $ ce$ nna$ eli as$ za$ 20 The space complexity of a compact trie is O(\|R\|) (in addition to the strings): • In a compact trie, every internal node (except possibly the root) has at least two children. In such a tree, there is always at least as many leaves as internal nodes. Thus the number of nodes is at most 2\|R\|. • The egde labels are factors of the input strings. If the input strings are stored separately, the edge labels can be represented in constant space using pointers to the strings. The time complexities are the same or better than for tries: • An insertion adds and a deletion removes at most two nodes. • Lookups may execute fewer calls to the child operation, though the worst case complexity is the same. • Preﬁx and range queries are faster even on a constant alphabet (exercise). 21 There is also an intermediate form of trie called leaf-path-compacted trie, where branchless path segments are compacted only when they end in a leaf. • Typically (though not in the worst case) this achieves most of the advantages of a compact trie. • For trie algorithms, this means stopping the normal search, when only one string is remaining in the subtree. Example 1.7: Leaf-path-compacted trie for {ali$, alice$, anna$, elias$, eliza$}. a l i $ ce$ nna$ e l i as$ za$ 22 Ternary Trie Tries can be implemented for ordered alphabets but a bit awkwardly using a comparison-based child function. Ternary trie is a simpler data structure based on symbol comparisons. Ternary trie is like a binary search tree except: • Each internal node has three children: smaller, equal and larger. • The branching is based on a single symbol at a given position as in a trie. The position is zero (ﬁrst symbol) at the root and increases along the middle branches but not along side branches. There are also compact ternary tries and leaf-path-compated ternary tries based on compacting branchless path segments. 23 Example 1.8: Ternary tries for {ali$, alice$, anna$, elias$, eliza$}. a l i c $ e $ n n a $ e l i a s $ z a $ a l i c e l i a $ e$ nna$ s$ za$ a l c a i $ e$ nna$ eli s$ za$ Ternary tries have the same asymptotic size as the corresponding (σ-ary) tries. 24 A ternary trie is balanced if each left and right subtree contains at most half of the strings in its parent tree. • The balance can be maintained by rotations similarly to binary search trees. C rotation C d D E b A B b A B d D E • We can also get reasonably close to a balance by inserting the strings in the tree in a random order. Note that there is no restriction on the size of the middle subtree. 25 In a balanced ternary trie each step down either • moves the position forward (middle branch), or • halves the number of strings remaining in the subtree (side branch). Thus, in a balanced ternary trie storing n strings, any downward traversal following a string S passes at most \|S\| middle edges and at most log n side edges. Thus the time complexity of insertion, deletion, lookup and lcp query is O(\|S\| + log n). In comparison based tries, where the child function is implemented using binary search trees, the time complexities could be O(\|S\| log σ), a multiplicative factor O(log σ) instead of an additive factor O(log n). Preﬁx and range queries behave similarly (exercise). 26 Longest Common Preﬁxes The standard ordering for strings is the lexicographical order. It is induced by an order over the alphabet. We will use the same symbols (≤, <, ≥, ̸≤, etc.) for both the alphabet order and the induced lexicographical order. We can deﬁne the lexicographical order using the concept of the longest common preﬁx. Deﬁnition 1.9: The length of the longest common preﬁx of two strings A[0..m) and B[0..n), denoted by lcp(A, B), is the largest integer ℓ≤min{m, n} such that A[0..ℓ) = B[0..ℓ). Deﬁnition 1.10: Let A and B be two strings over an alphabet with a total order ≤, and let ℓ= lcp(A, B). Then A is lexicographically smaller than or equal to B, denoted by A ≤B, if and only if 1. either \|A\| = ℓ 2. or \|A\| > ℓ, \|B\| > ℓand A[ℓ] < B[ℓ]. 27 An important concept for sets of strings is the LCP (longest common preﬁx) array and its sum. Deﬁnition 1.11: Let R = {S1, S2, . . . , Sn} be a set of strings and assume S1 < S2 < · · · < Sn. Then the LCP array LCPR[1..n] is deﬁned so that LCPR = 0 and for i ∈[2..n] LCPR[i] = lcp(Si, Si−1) Furthermore, the LCP array sum is ΣLCP(R) = X i∈[1..n] LCPR[i] . Example 1.12: For R = {ali$, alice$, anna$, elias$, eliza$}, ΣLCP(R) = 7 and the LCP array is: LCPR 0 ali$ 3 alice$ 1 anna$ 0 elias$ 3 eliza$ 28 A variant of the LCP array sum is sometimes useful: Deﬁnition 1.13: For a string S and a string set R, deﬁne lcp(S, R) = max{lcp(S, T) \| T ∈R} Σlcp(R) = X S∈R lcp(S, R \ {S}) The relationship of the two measures is shown by the following two results: Lemma 1.14: For i ∈[2..n], LCPR[i] = lcp(Si, {S1, . . . , Si−1}). Lemma 1.15: ΣLCP(R) ≤Σlcp(R) ≤2 · ΣLCP(R). The proofs are left as an exercise. The concept of distinguishing preﬁx is closely related and often used in place of the longest common preﬁx for sets. The distinguishing preﬁx of a string is the shortest preﬁx that separates it from other strings in the set. It is easy to see that dp(S, R \ S) = lcp(S, R \ S) + 1 (at least for a preﬁx free R). Example 1.16: For R = {ali$, alice$, anna$, elias$, eliza$}, Σlcp(R) = 13 and Σdp(R) = 18. 29 Theorem 1.17: The number of nodes in trie(R) is exactly \|\|R\|\| −ΣLCP(R) + 1, where \|\|R\|\| is the total length of the strings in R. Proof. Consider the construction of trie(R) by inserting the strings one by one in the lexicographical order using Algorithm 1.2. Initially, the trie has just one node, the root. When inserting a string Si, the algorithm executes exactly \|Si\| rounds of the two while loops, because each round moves one step forward in Si. The ﬁrst loop follows existing edges as long as possible and thus the number of rounds is LCPR[i] = lcp(Si, {S1, . . . , Si−1}). This leaves \|Si\| −LCPR[i] rounds for the second loop, each of which adds one new node to the trie. Thus the total number of nodes in the trie at the end is: 1 + X i∈[1..n] \|Si\| −LCPR[i] = \|\|R\|\| −ΣLCP(R) + 1 . □ The proof reveals a close connection between LCPR and the structure of the trie. We will later see that LCPR is useful as an actual data structure in its own right. 30 String Sorting Ω(n log n) is a well known lower bound for the number of comparisons needed for sorting a set of n objects by any comparison based algorithm. This lower bound holds both in the worst case and in the average case. There are many algorithms that match the lower bound, i.e., sort using O(n log n) comparisons (worst or average case). Examples include quicksort, heapsort and mergesort. If we use one of these algorithms for sorting a set of n strings, it is clear that the number of symbol comparisons can be more than O(n log n) in the worst case. Determining the order of A and B needs at least lcp(A, B) symbol comparisons and lcp(A, B) can be arbitrarily large in general. On the other hand, the average number of symbol comparisons for two random strings is O(1). Does this mean that we can sort a set of random strings in O(n log n) time using a standard sorting algorithm? 31 The following theorem shows that we cannot achieve O(n log n) symbol comparisons for any set of strings (when σ = no(1)). Theorem 1.18: Let A be an algorithm that sorts a set of objects using only comparisons between the objects. Let R = {S1, S2, . . . , Sn} be a set of n strings over an ordered alphabet Σ of size σ. Sorting R using A requires Ω(n log n logσ n) symbol comparisons on average, where the average is taken over the initial orders of R. • If σ is considered to be a constant, the lower bound is Ω(n(log n)2). • Note that the theorem holds for any comparison based sorting algorithm A and any string set R. In other words, we can choose A and R to minimize the number of comparisons and still not get below the bound. • Only the initial order is random rather than “any”. Otherwise, we could pick the correct order and use an algorithm that ﬁrst checks if the order is correct, needing only O(n + ΣLCP(R)) symbol comparisons. An intuitive explanation for this result is that the comparisons made by a sorting algorithm are not random. In the later stages, the algorithm tends to compare strings that are close to each other in lexicographical order and thus are likely to have long common preﬁxes. 32 Proof of Theorem 1.18. Let k = ⌊(logσ n)/2⌋. For any string α ∈Σk, let Rα be the set of strings in R having α as a preﬁx. Let nα = \|Rα\|. Let us analyze the number of symbol comparisons when comparing strings in Rα against each other. • Each string comparison needs at least k symbol comparisons. • No comparison between a string in Rα and a string outside Rα gives any information about the relative order of the strings in Rα. • Thus A needs to do Ω(nα log nα) string comparisons and Ω(knα log nα) symbol comparisons to determine the relative order of the strings in Rα. Thus the total number of symbol comparisons is Ω P α∈Σk knα log nα and X α∈Σk knα log nα ≥k(n −√n) log n −√n σk ≥k(n −√n) log(√n −1) = Ω(kn log n) = Ω(n log n logσ n) . Here we have used the facts that σk ≤√n, that P α∈Σk nα > n −σk ≥n −√n, and that P α∈Σk nα log nα > (n −√n) log((n −√n)/σk) (see exercises). □ 33 The preceding lower bound does not hold for algorithms specialized for sorting strings. Theorem 1.19: Let R = {S1, S2, . . . , Sn} be a set of n strings. Sorting R into the lexicographical order by any algorithm based on symbol comparisons requires Ω(ΣLCP(R) + n log n) symbol comparisons. Proof. If we are given the strings in the correct order and the job is to verify that this is indeed so, we need at least ΣLCP(R) symbol comparisons. No sorting algorithm could possibly do its job with less symbol comparisons. This gives a lower bound Ω(ΣLCP(R)). On the other hand, the general sorting lower bound Ω(n log n) must hold here too. The result follows from combining the two lower bounds. □ • Note that the expected value of ΣLCP(R) for a random set of n strings is O(n logσ n). The lower bound then becomes Ω(n log n). We will next see that there are algorithms that match this lower bound. Such algorithms can sort a random set of strings in O(n log n) time. 34 String Quicksort (Multikey Quicksort) Quicksort is one of the fastest general purpose sorting algorithms in practice. Here is a variant of quicksort that partitions the input into three parts instead of the usual two parts. Algorithm 1.20: TernaryQuicksort(R) Input: (Multi)set R in arbitrary order. Output: R in ascending order. (1) if \|R\| ≤1 then return R (2) select a pivot x ∈R (3) R< ←{s ∈R \| s < x} (4) R= ←{s ∈R \| s = x} (5) R> ←{s ∈R \| s > x} (6) R< ←TernaryQuicksort(R<) (7) R> ←TernaryQuicksort(R>) (8) return R< · R= · R> 35 In the normal, binary quicksort, we would have two subsets R≤and R≥, both of which may contain elements that are equal to the pivot. • Binary quicksort is slightly faster in practice for sorting sets. • Ternary quicksort can be faster for sorting multisets with many duplicate keys. Sorting a multiset of size n with σ distinct elements takes O(n log σ) comparisons (exercise). The time complexity of both the binary and the ternary quicksort depends on the selection of the pivot (exercise). In the following, we assume an optimal pivot selection giving O(n log n) worst case time complexity. 36 String quicksort is similar to ternary quicksort, but it partitions using a single character position. String quicksort is also known as multikey quicksort. Algorithm 1.21: StringQuicksort(R, ℓ) Input: (Multi)set R of strings and the length ℓof their common preﬁx. Output: R in ascending lexicographical order. (1) if \|R\| ≤1 then return R (2) R⊥←{S ∈R \| \|S\| = ℓ}; R ←R \ R⊥ (3) select pivot X ∈R (4) R< ←{S ∈R \| S[ℓ] < X[ℓ]} (5) R= ←{S ∈R \| S[ℓ] = X[ℓ]} (6) R> ←{S ∈R \| S[ℓ] > X[ℓ]} (7) R< ←StringQuicksort(R<, ℓ) (8) R= ←StringQuicksort(R=, ℓ+ 1) (9) R> ←StringQuicksort(R>, ℓ) (10) return R⊥· R< · R= · R> In the initial call, ℓ= 0. 37 Example 1.22: A possible partitioning, when ℓ= 2. al p habet al i gnment al l ocate al g orithm al t ernative al i as al t ernate al l = ⇒ al i gnment al g orithm al i as al l ocate al l al p habet al t ernative al t ernate Theorem 1.23: String quicksort sorts a set R of n strings in O(ΣLCP(R) + n log n) time. • Thus string quicksort is an optimal symbol comparison based algorithm. • String quicksort is also fast in practice. 38 Proof of Theorem 1.23. The time complexity is dominated by the symbol comparisons on lines (4)–(6). We charge the cost of each comparison either on a single symbol or on a string depending on the result of the comparison: S[ℓ] = X[ℓ]: Charge the comparison on the symbol S[ℓ]. • Now the string S is placed in the set R=. The recursive call on R= increases the common preﬁx length to ℓ+ 1. Thus S[ℓ] cannot be involved in any future comparison and the total charge on S[ℓ] is 1. • Only lcp(S, R \ {S}) symbols in S can be involved in these comparisons. Thus the total number of symbol comparisons resulting equality is at most Σlcp(R) = Θ(ΣLCP(R)). (Exercise: Show that the number is exactly ΣLCP(R).) S[ℓ] ̸= X[ℓ]: Charge the comparison on the string S. • Now the string S is placed in the set R< or R>. The size of either set is at most \|R\|/2 assuming an optimal choice of the pivot X. • Every comparison charged on S halves the size of the set containing S, and hence the total charge accumulated by S is at most log n. Thus the total number of symbol comparisons resulting inequality is at most O(n log n). □ 39 Radix Sort The Ω(n log n) sorting lower bound does not apply to algorithms that use stronger operations than comparisons. A basic example is counting sort for sorting integers. Algorithm 1.24: CountingSort(R) Input: (Multi)set R = {k1, k2, . . . kn} of integers from the range [0..σ). Output: R in nondecreasing order in array J[0..n). (1) for i ←0 to σ −1 do C[i] ←0 (2) for i ←1 to n do C[ki] ←C[ki] + 1 (3) sum ←0 (4) for i ←0 to σ −1 do // cumulative sums (5) tmp ←C[i]; C[i] ←sum; sum ←sum + tmp (6) for i ←1 to n do // distribute (7) J[C[ki]] ←ki; C[ki] ←C[ki] + 1 (8) return J • The time complexity is O(n + σ). • Counting sort is a stable sorting algorithm, i.e., the relative order of equal elements stays the same. 40 Similarly, the Ω(ΣLCP(R) + n log n) lower bound does not apply to string sorting algorithms that use stronger operations than symbol comparisons. Radix sort is such an algorithm for integer alphabets. Radix sort was developed for sorting large integers, but it treats an integer as a string of digits, so it is really a string sorting algorithm. There are two types of radix sorting: MSD radix sort starts sorting from the beginning of strings (most signiﬁcant digit). LSD radix sort starts sorting from the end of strings (least signiﬁcant digit). 41 The LSD radix sort algorithm is very simple. Algorithm 1.25: LSDRadixSort(R) Input: (Multi)set R = {S1, S2, . . . , Sn} of strings of length m over alphabet [0..σ). Output: R in ascending lexicographical order. (1) for ℓ←m −1 to 0 do CountingSort(R,ℓ) (2) return R • CountingSort(R,ℓ) sorts the strings in R by the symbols at position ℓ using counting sort (with ki replaced by Si[ℓ]). The time complexity is O(\|R\| + σ). • The stability of counting sort is essential. Example 1.26: R = {cat, him, ham, bat}. cat him ham bat = ⇒ hi m ha m ca t ba t = ⇒ h a m c a t b a t h i m = ⇒ b at c at h am h im It is easy to show that after i rounds, the strings are sorted by suﬃx of length i. Thus, they are fully sorted at the end. 42 The algorithm assumes that all strings have the same length m, but it can be modiﬁed to handle strings of diﬀerent lengths (exercise). Theorem 1.27: LSD radix sort sorts a set R of strings over the alphabet [0..σ) in O(\|\|R\|\| + mσ) time, where \|\|R\|\| is the total length of the strings in R and m is the length of the longest string in R. Proof. Assume all strings have length m. The LSD radix sort performs m rounds with each round taking O(n + σ) time. The total time is O(mn + mσ) = O(\|\|R\|\| + mσ). The case of variable lengths is left as an exercise. □ • The weakness of LSD radix sort is that it uses Ω(\|\|R\|\|) time even when ΣLCP(R) is much smaller than \|\|R\|\|. • It is best suited for sorting short strings and integers. 43 MSD radix sort resembles string quicksort but partitions the strings into σ parts instead of three parts. Example 1.28: MSD radix sort partitioning. al p habet al i gnment al l ocate al g orithm al t ernative al i as al t ernate al l = ⇒ al g orithm al i gnment al i as al l ocate al l al p habet al t ernative al t ernate 44 Algorithm 1.29: MSDRadixSort(R, ℓ) Input: (Multi)set R = {S1, S2, . . . , Sn} of strings over the alphabet [0..σ) and the length ℓof their common preﬁx. Output: R in ascending lexicographical order. (1) if \|R\| < σ then return StringQuicksort(R, ℓ) (2) R⊥←{S ∈R \| \|S\| = ℓ}; R ←R \ R⊥ (3) (R0, R1, . . . , Rσ−1) ←CountingSort(R, ℓ) (4) for i ←0 to σ −1 do Ri ←MSDRadixSort(Ri, ℓ+ 1) (5) return R⊥· R0 · R1 · · · Rσ−1 • Here CountingSort(R,ℓ) not only sorts but also returns the partitioning based on symbols at position ℓ. The time complexity is still O(\|R\| + σ). • The recursive calls eventually lead to a large number of very small sets, but counting sort needs Ω(σ) time no matter how small the set is. To avoid the potentially high cost, the algorithm switches to string quicksort for small sets. 45 Theorem 1.30: MSD radix sort sorts a set R of n strings over the alphabet [0..σ) in O(ΣLCP(R) + n log σ) time. Proof. Consider a call processing a subset of size k ≥σ: • The time excluding the recursive calls but including the call to counting sort is O(k + σ) = O(k). The k symbols accessed here will not be accessed again. • At most dp(S, R \ {S}) ≤lcp(S, R \ {S}) + 1 symbols in S will be accessed by the algorithm. Thus the total time spent in this kind of calls is O(Σdp(R)) = O(Σlcp(R) + n) = O(ΣLCP(R) + n). The calls for a subsets of size k < σ are handled by string quicksort. Each string is involved in at most one such call. Therefore, the total time over all calls to string quicksort is O(ΣLCP(R) + n log σ). □ • There exists a more complicated variant of MSD radix sort with time complexity O(ΣLCP(R) + n + σ). • Ω(ΣLCP(R) + n) is a lower bound for any algorithm that must access symbols one at a time. • In practice, MSD radix sort is very fast, but it is sensitive to implementation details. 46 Lcp-Comparisons General (non-string) comparison-based sorting algorithms are not optimal for sorting strings because of an imbalance between eﬀort and result in a string comparison: it can take a lot of time but the result is only a bit or a trit of useful information. String quicksort solves this problem by processing the obtained information immediately after each symbol comparison. An opposite approach is to replace a standard string comparison with an lcp-comparison, which is the operation LcpCompare(A, B, k): • The return value is the pair (x, ℓ), where x ∈{<, =, >} indicates the order, and ℓ= lcp(A, B), the length of the longest common preﬁx of strings A and B. • The input value k is the length of a known common preﬁx, i.e., a lower bound on lcp(A, B). The comparison can skip the ﬁrst k characters. Extra time spent in the comparison is balanced by the extra information obtained in the form of the lcp value. 47 The following result shows how we can use the information from earlier comparisons to obtain a lower bound or even the exact value for an lcp. Lemma 1.31: Let A, B and C be strings. (a) lcp(A, C) ≥min{lcp(A, B), lcp(B, C)}. (b) If A ≤B ≤C, then lcp(A, C) = min{lcp(A, B), lcp(B, C)}. (c) If lcp(A, B) ̸= lcp(B, C), then lcp(A, C) = min{lcp(A, B), lcp(B, C)}. Proof. Assume ℓ= lcp(A, B) ≤lcp(B, C). The opposite case lcp(A, B) ≥lcp(B, C) is symmetric. (a) Now A[0..ℓ) = B[0..ℓ) = C[0..ℓ) and thus lcp(A, C) ≥ℓ. (b) Either \|A\| = ℓor A[ℓ] < B[ℓ] ≤C[ℓ]. In either case, lcp(A, C) = ℓ. (c) Now lcp(A, B) < lcp(B, C). If lcp(A, C) > min{lcp(A, B), lcp(B, C)}, then lcp(A, B) < min{lcp(A, C), lcp(B, C)}, which violates (a). □ The above means that the three lcp values between three strings can never be three diﬀerent values. At least two of them are the same and the third one is the same or bigger. 48 It can also be possible to determine the order of two strings without comparing them directly. Lemma 1.32: Let A, B, B′ and C be strings such that A ≤B ≤C and A ≤B′ ≤C. (a) If lcp(A, B) > lcp(A, B′), then B < B′. (b) If lcp(B, C) > lcp(B′, C), then B > B′. Proof. We show (a); (b) is symmetric. Assume to the contrary that B ≥B′. Then by Lemma 1.31, lcp(A, B) = min{lcp(A, B′), lcp(B′, B)} ≤lcp(A, B′), which is a contradiction. □ Intuitively, the above result makes sense if you think of lcp(·, ·) as a measure of similarity between two strings. The higher the lcp, the closer the two strings are lexicographically. 49 String Mergesort String mergesort is a string sorting algorithm that uses lcp-comparisons. It has the same structure as the standard mergesort: sort the ﬁrst half and the second half separately, and then merge the results. Algorithm 1.33: StringMergesort(R) Input: Set R = {S1, S2, . . . , Sn} of strings. Output: R sorted and augmented with LCPR values. (1) if \|R\| = 1 then return ((S1, 0)) (2) m ←⌊n/2⌋ (3) P ←StringMergesort({S1, S2, . . . , Sm}) (4) Q ←StringMergesort({Sm+1, Sm+2, . . . , Sn}) (5) return StringMerge(P, Q) The output is of the form ((T1, ℓ1), (T2, ℓ2), . . . , (Tn, ℓn)) where ℓi = lcp(Ti, Ti−1) for i > 1 and ℓ1 = 0. In other words, ℓi = LCPR[i]. Thus we get not only the order of the strings but also a lot of information about their common preﬁxes. The procedure StringMerge uses this information eﬀectively. 50 Algorithm 1.34: StringMerge(P,Q) Input: Sequences P = (S1, k1), . . . , (Sm, km) and Q = (T1, ℓ1), . . . , (Tn, ℓn) Output: Merged sequence R (1) R ←∅; i ←1; j ←1 (2) while i ≤m and j ≤n do (3) if ki > ℓj then append (Si, ki) to R; i ←i + 1 (4) else if ℓj > ki then append (Tj, ℓj) to R; j ←j + 1 (5) else // ki = ℓj (6) (x, h) ←LcpCompare(Si, Tj, ki) (7) if x = ”<” then (8) append (Si, ki) to R; i ←i + 1 (9) ℓj ←h (10) else (11) append (Tj, ℓj) to R; j ←j + 1 (12) ki ←h (13) while i ≤m do append (Si, ki) to R; i ←i + 1 (14) while j ≤n do append (Tj, ℓj) to R; j ←j + 1 (15) return R 51 Lemma 1.35: StringMerge performs the merging correctly. Proof. We will show that the following invariant holds at the beginning of each round in the loop on lines (2)–(12): Let X be the last string appended to R (or ε if R = ∅). Then ki = lcp(X, Si) and ℓj = lcp(X, Tj). The invariant is clearly true in the beginning. We will show that the invariant is maintained and the smaller string is chosen in each round of the loop. • If ki > ℓj, then lcp(X, Si) > lcp(X, Tj) and thus – Si < Tj by Lemma 1.32. – lcp(Si, Tj) = lcp(X, Tj) because, by Lemma 1.31, lcp(X, Tj) = min{lcp(X, Si), lcp(Si, Tj)}. Hence, the algorithm chooses the smaller string and maintains the invariant. The case ℓj > ki is symmetric. • If ki = ℓj, then clearly lcp(Si, Tj) ≥ki and the call to LcpCompare is safe, and the smaller string is chosen. The update ℓj ←h or ki ←h maintains the invariant. □ 52 Theorem 1.36: String mergesort sorts a set R of n strings in O(ΣLCP(R) + n log n) time. Proof. If the calls to LcpCompare took constant time, the time complexity would be O(n log n) by the same argument as with the standard mergesort. Whenever LcpCompare makes more than one, say t + 1 symbol comparisons, one of the lcp values stored with the strings increases by t. Since the sum of the ﬁnal lcp values is exactly ΣLCP(R), the extra time spent in LcpCompare is bounded by O(ΣLCP(R)). □ • Other comparison based sorting algorithms, for example heapsort and insertion sort, can be adapted for strings using the lcp-comparison technique. 53 String Binary Search An ordered array is a simple static data structure supporting queries in O(log n) time using binary search. Algorithm 1.37: Binary search Input: Ordered set R = {k1, k2, . . . , kn}, query value x. Output: The number of elements in R that are smaller than x. (1) left ←0; right ←n + 1 // output value is in the range [left..right) (2) while right −left > 1 do (3) mid ←⌊(left + right)/2⌋ (4) if kmid < x then left ←mid (5) else right ←mid (6) return left With strings as elements, however, the query time is • O(m log n) in the worst case for a query string of length m • O(log n logσ n) on average for a random set of strings. 54 We can use the lcp-comparison technique to improve binary search for strings. The following is a key result. Lemma 1.38: Let A, B, B′ and C be strings such that A ≤B ≤C and A ≤B′ ≤C. Then lcp(B, B′) ≥lcp(A, C). Proof. Let Bmin = min{B, B′} and Bmax = max{B, B′}. By Lemma 1.31, lcp(A, C) = min(lcp(A, Bmax), lcp(Bmax, C)) ≤lcp(A, Bmax) = min(lcp(A, Bmin), lcp(Bmin, Bmax)) ≤lcp(Bmin, Bmax) = lcp(B, B′) □ 55 During the binary search of P in {S1, S2, . . . , Sn}, the basic situation is the following: • We want to compare P and Smid. • We have already compared P against Sleft and Sright, and we know that Sleft ≤P, Smid ≤Sright. • By using lcp-comparisons, we know lcp(Sleft, P) and lcp(P, Sright). By Lemmas 1.31 and 1.38, lcp(P, Smid) ≥lcp(Sleft, Sright) = min{lcp(Sleft, P), lcp(P, Sright)} Thus we can skip min{lcp(Sleft, P), lcp(P, Sright)} ﬁrst characters when comparing P and Smid. 56 Algorithm 1.39: String binary search (without precomputed lcps) Input: Ordered string set R = {S1, S2, . . . , Sn}, query string P. Output: The number of strings in R that are smaller than P. (1) left ←0; right ←n + 1 (2) llcp ←0 // llcp = lcp(Sleft, P) (3) rlcp ←0 // rlcp = lcp(P, Sright) (4) while right −left > 1 do (5) mid ←⌊(left + right)/2⌋ (6) mlcp ←min{llcp, rlcp} (7) (x, mlcp) ←LcpCompare(P, Smid, mlcp) (8) if x = “ < ” then right ←mid; rlcp ←mclp (9) else left ←mid; llcp ←mclp (10) return left • The average case query time is now O(log n). • The worst case query time is still O(m log n) (exercise). 57 We can further improve string binary search using precomputed information about the lcp’s between the strings in R. Consider again the basic situation during string binary search: • We want to compare P and Smid. • We have already compared P against Sleft and Sright, and we know lcp(Sleft, P) and lcp(P, Sright). The values left and right are fully determined by mid independently of P. That is, P only determines whether the search ends up at position mid at all, but if it does, left and right are always the same. Thus, we can precompute and store the values LLCP[mid] = lcp(Sleft, Smid) RLCP[mid] = lcp(Smid, Sright) Now we know all lcp values between P, Sleft, Smid, Sright except lcp(P, Smid). The following lemma shows how to utilize this. 58 Lemma 1.40: Let A, B, B′ and C be strings such that A ≤B ≤C and A ≤B′ ≤C. (a) If lcp(A, B) > lcp(A, B′), then B < B′ and lcp(B, B′) = lcp(A, B′). (b) If lcp(A, B) < lcp(A, B′), then B > B′ and lcp(B, B′) = lcp(A, B). (c) If lcp(B, C) > lcp(B′, C), then B > B′ and lcp(B, B′) = lcp(B′, C). (d) If lcp(B, C) < lcp(B′, C), then B < B′ and lcp(B, B′) = lcp(B, C). (e) If lcp(A, B) = lcp(A, B′) and lcp(B, C) = lcp(B′, C), then lcp(B, B′) ≥max{lcp(A, B), lcp(B, C)}. Proof. Cases (a)–(d) are symmetrical, we show (a). B < B′ follows from Lemma 1.32. Then by Lemma 1.31, lcp(A, B′) = min{lcp(A, B), lcp(B, B′)}. Since lcp(A, B′) < lcp(A, B), we must have lcp(A, B′) = lcp(B, B′). In case (e), we use Lemma 1.31: lcp(B, B′) ≥min{lcp(A, B), lcp(A, B′)} = lcp(A, B) lcp(B, B′) ≥min{lcp(B, C), lcp(B′, C)} = lcp(B, C) Thus lcp(B, B′) ≥max{lcp(A, B), lcp(B, C)}. □ 59 Algorithm 1.41: String binary search (with precomputed lcps) Input: Ordered string set R = {S1, S2, . . . , Sn}, arrays LLCP and RLCP, query string P. Output: The number of strings in R that are smaller than P. (1) left ←0; right ←n + 1 (2) llcp ←0; rlcp ←0 (3) while right −left > 1 do (4) mid ←⌊(left + right)/2⌋ (5) if LLCP[mid] > llcp then left ←mid (6) else if LLCP[mid] < llcp then right ←mid; rlcp ←LLCP[mid] (7) else if RLCP[mid] > rlcp then right ←mid (8) else if RLCP[mid] < rlcp then left ←mid; llcp ←RLCP[mid] (9) else (10) mlcp ←max{llcp, rlcp} (11) (x, mlcp) ←LcpCompare(P, Smid, mlcp) (12) if x = “ < ” then right ←mid; rlcp ←mclp (13) else left ←mid; llcp ←mclp (14) return left 60 Theorem 1.42: An ordered string set R = {S1, S2, . . . , Sn} can be preprocessed in O(ΣLCP(R) + n) time and O(n) space so that a binary search with a query string P can be executed in O(\|P\| + log n) time. Proof. The values LLCP[mid] and RLCP[mid] can be computed in O(lcp(Smid, R \ {Smid}) + 1) time. Thus the arrays LLCP and RLCP can be computed in O(Σlcp(R) + n) = O(ΣLCP(R) + n) time and stored in O(n) space. The main while loop in Algorithm 1.41 is executed O(log n) times and everything except LcpCompare on line (11) needs constant time. If a given LcpCompare call performs t + 1 symbol comparisons, mclp increases by t on line (11). Then on lines (12)–(13), either llcp or rlcp increases by at least t, since mlcp was max{llcp, rlcp} before LcpCompare. Since llcp and rlcp never decrease and never grow larger than \|P\|, the total number of extra symbol comparisons in LcpCompare during the binary search is O(\|P\|). □ Other comparison-based data structures such as binary search trees can be augmented with lcp information in the same way (study groups). 61 Hashing and Fingerprints Hashing is a powerful technique for dealing with strings based on mapping each string to an integer using a hash function: H : Σ∗→[0..q) ⊂N The most common use of hashing is with hash tables. Hash tables come in many ﬂavors that can be used with strings as well as with any other type of object with an appropriate hash function. A drawback of using a hash table to store a set of strings is that they do not support lcp and preﬁx queries. Hashing is also used in other situations, where one needs to check whether two strings S and T are the same or not: • If H(S) ̸= H(T), then we must have S ̸= T. • If H(S) = H(T), then S = T and S ̸= T are both possible. If S ̸= T, this is called a collision. When used this way, the hash value is often called a ﬁngerprint, and its range [0..q) is typically large as it is not restricted by a hash table size. 62 Any good hash function must depend on all characters. Thus computing H(S) needs Ω(\|S\|) time, which can defeat the advantages of hashing: • A plain comparison of two strings is faster than computing the hashes. • The main strength of hash tables is the support for constant time insertions and deletions, but inserting a string S into a hash table needs Ω(\|S\|) time when the hash computation time is included. Compare this to the O(\|S\|) time for a trie under a constant alphabet and the O(\|S\| + log n) time for a ternary trie. However, a hash table can still be competitive in practice. Furthermore, there are situations, where a full computation of the hash function can be avoided: • A hash value can be computed once, stored, and used many times. • Some hash functions can be computed more eﬃciently for a related set of strings. An example is the Karp–Rabin hash function. 63 Deﬁnition 1.43: The Karp–Rabin hash function for a string S = s0s1 . . . sm−1 over an integer alphabet is H(S) = (s0rm−1 + s1rm−2 + · · · + sm−2r + sm−1) mod q for some ﬁxed positive integers q and r. Lemma 1.44: For any two strings A and B, H(AB) = (H(A) · r\|B\| + H(B)) mod q H(B) = (H(AB) −H(A) · r\|B\|) mod q Proof. Without the modulo operation, the result would be obvious. The modulo does not interfere because of the rules of modular arithmetic: (x + y) mod q = ((x mod q) + (y mod q)) mod q (xy) mod q = ((x mod q)(y mod q)) mod q □ Thus we can quickly compute H(AB) from H(A) and H(B), and H(B) from H(AB) and H(A). We will see applications of this later. If q and r are coprime, then r has a multiplicative inverse r−1 modulo q, and we can also compute H(A) = ((H(AB) −H(B)) · (r−1)\|B\|) mod q. 64 The parameters q and r have to be chosen with some care to ensure that collisions are rare for any reasonable set of strings. • The original choice is r = σ and q is a large prime. • Another possibility is that q is a power of two and r is a small prime (r = 37 has been suggested). This is faster in practice, because the slow modulo operations can be replaced by bitwise shift operations. If q = 2w, where w is the machine word size, the modulo operations can be omitted completely. • If q and r were both powers of two, then only the last ⌈(log q)/ log r⌉ characters of the string would aﬀect the hash value. More generally, q and r should be coprime, i.e, have no common divisors other than 1. • The hash function can be randomized by choosing q or r randomly. For example, if q is a prime and r is chosen uniformly at random from [0..q), the probability that two strings of length m collide is at most m/q. • A random choice over a set of possibilities has the additional advantage that we can change the choice if the ﬁrst choice leads to too many collisions. 65 Automata Finite automata are a well known way of representing sets of strings. In this case, the set is often called a (regular) language. A trie is a special type of an automaton. • The root is the initial state, the leaves are accept states, ... • Trie is generally not a minimal automaton. • Trie techniques including path compaction can be applied to automata. Automata are much more powerful than tries in representing languages: • Inﬁnite languages • Nondeterministic automata • Even an acyclic, deterministic automaton can represent a language of exponential size. Automata support set inclusion testing but not other trie operations: • No insertions and deletions • No satellite data, i.e., data associated to each string 66 Sets of Strings: Summary Eﬃcient algorithms and data structures for sets of strings: • Storing and searching: trie and ternary trie and their compact versions, string binary search, Karp–Rabin hashing. • Sorting: string quicksort and mergesort, LSD and MSD radix sort. Lower bounds: • Many of the algorithms are optimal. • General purpose algorithms are asymptotically slower. The central role of longest common preﬁxes: • LCP array LCPR and its sum ΣLCP(R). • Lcp-comparison technique. 67 2. Exact String Matching Let T = T[0..n) be the text and P = P[0..m) the pattern. We say that P occurs in T at position j if T[j..j + m) = P. Example: P = aine occurs at position 6 in T = karjalainen. In this part, we will describe algorithms that solve the following problem. Problem 2.1: Given text T[0..n) and pattern P[0..m), report the ﬁrst position in T where P occurs, or n if P does not occur in T. The algorithms can be easily modiﬁed to solve the following problems too. • Existence: Is P a factor of T? • Counting: Count the number of occurrences of P in T. • Listing: Report all occurrences of P in T. 68 The naive, brute force algorithm compares P against T[0..m), then against T[1..1 + m), then against T[2..2 + m) etc. until an occurrence is found or the end of the text is reached. The text factor T[j..j + m) that is currently being compared against the pattern is called the text window. Algorithm 2.2: Brute force Input: text T = T[0 . . . n), pattern P = P[0 . . . m) Output: position of the ﬁrst occurrence of P in T (1) i ←0; j ←0 (2) while i < m and j < n do (3) if P[i] = T[j] then i ←i + 1; j ←j + 1 (4) else j ←j −i + 1; i ←0 (5) if i = m then return j −m else return n The worst case time complexity is O(mn). This happens, for example, when P = am−1b = aaa..ab and T = an = aaaaaa..aa. 69 (Knuth–)Morris–Pratt The Brute force algorithm forgets everything when it shifts the window. The Morris–Pratt (MP) algorithm remembers matches. It never goes back to a text character that already matched. The Knuth–Morris–Pratt (KMP) algorithm remembers mismatches too. Example 2.3: Brute force ainaisesti-ainainen ainai/ / nen (6 comp.) / / ainainen (1) / ainainen (1) ai/ / nainen (3) / ainainen (1) / / ainainen (1) Morris–Pratt ainaisesti-ainainen ainai/ / nen (6) ai/ / nainen (1) / / ainainen (1) Knuth–Morris–Pratt ainaisesti-ainainen ainai/ / nen (6) / / ainainen (1) 70 MP and KMP algorithms never go backwards in the text. When they encounter a mismatch, they ﬁnd another pattern position to compare against the same text position. If the mismatch occurs at pattern position i, then fail[i] is the next pattern position to compare. The only diﬀerence between MP and KMP is how they compute the failure function fail. Algorithm 2.4: Knuth–Morris–Pratt / Morris–Pratt Input: text T = T[0 . . . n), pattern P = P[0 . . . m) Output: position of the ﬁrst occurrence of P in T (1) compute fail[0..m] (2) i ←0; j ←0 (3) while i < m and j < n do (4) if i = −1 or P[i] = T[j] then i ←i + 1; j ←j + 1 (5) else i ←fail[i] (6) if i = m then return j −m else return n • fail[i] = −1 means that there is no more pattern positions to compare against this text positions and we should move to the next text position. • fail[m] is never needed here, but if we wanted to ﬁnd all occurrences, it would tell how to continue after a full match. 71 We will describe the MP failure function here. The KMP failure function is left for the exercises. • When the algorithm ﬁnds a mismatch between P[i] and T[j], we know that P[0..i) = T[j −i..j). • Now we want to ﬁnd a new i′ < i such that P[0..i′) = T[j −i′..j). Speciﬁcally, we want the largest such i′. • This means that P[0..i′) = T[j −i′..j) = P[i −i′..i). In other words, P[0..i′) is the longest proper border of P[0..i). Example: ai is the longest proper border of ainai. • Thus fail[i] is the length of the longest proper border of P[0..i). • P[0..0) = ε has no proper border. We set fail = −1. 72 Example 2.5: Let P = ainainen. i P[0..i) border fail[i] 0 ε – -1 1 a ε 0 2 ai ε 0 3 ain ε 0 4 aina a 1 5 ainai ai 2 6 ainain ain 3 7 ainaine ε 0 8 ainainen ε 0 The (K)MP algorithm operates like an automaton, since it never moves backwards in the text. Indeed, it can be described by an automaton that has a special failure transition, which is an ε-transition that can be taken only when there is no other transition to take. -1 1 2 3 4 5 6 7 8 0 a n a i n i e n Σ 73 An eﬃcient algorithm for computing the failure function is very similar to the search algorithm itself! • In the MP algorithm, when we ﬁnd a match P[i] = T[j], we know that P[0..i] = T[j −i..j]. More speciﬁcally, P[0..i] is the longest preﬁx of P that matches a suﬃx of T[0..j]. • Suppose T = #P[1..m), where # is a symbol that does not occur in P. Finding a match P[i] = T[j], we know that P[0..i] is the longest preﬁx of P that is a proper suﬃx of P[0..j]. Thus fail[j + 1] = i + 1. Algorithm 2.6: Morris–Pratt failure function computation Input: pattern P = P[0 . . . m) Output: array fail[0..m] for P (1) i ←−1; j ←0; fail[j] ←i (2) while j < m do (3) if i = −1 or P[i] = P[j] then i ←i + 1; j ←j + 1; fail[j] ←i (4) else i ←fail[i] (5) return fail • When the algorithm reads fail[i] on line 4, fail[i] has already been computed. 74 Theorem 2.7: Algorithms MP and KMP preprocess a pattern in time O(m) and then search the text in time O(n) for ordered alphabet. Proof. We show that the text search requires O(n) time. Exactly the same argument shows that pattern preprocessing needs O(m) time. It is suﬃcient to count the number of comparisons that the algorithms make. After each comparison P[i] vs. T[j], one of the two conditional branches is executed: then Here j is incremented. Since j never decreases, this branch can be taken at most n + 1 times. else Here i decreases since fail[i] < i. Since i only increases in the then-branch, this branch cannot be taken more often than the then-branch. □ 75 Shift-And (Shift-Or) When the MP algorithm is at position j in the text T, it computes the longest preﬁx of the pattern P[0..m) that is a suﬃx of T[0..j]. The Shift-And algorithm computes all preﬁxes of P that are suﬃxes of T[0..j]. • The information is stored in a bitvector D of length m, where D.i = 1 if P[0..i] = T[j −i..j] and D.i = 0 otherwise. (D.0 is the least signiﬁcant bit.) • When D.(m −1) = 1, we have found an occurrence. The bitvector D is updated at each text position j: • There are precomputed bitvectors B[c], for all c ∈Σ, where B[c].i = 1 if P[i] = c and B[c].i = 0 otherwise. • D is updated in two steps: 1. D ←(D << 1) \| 1 (the bitwise shift and the bitwise or). Now D tells, which preﬁxes would match if T[j] would match every character. 2. D ←D & B[T[j]] (the bitwise and). Remove the preﬁxes where T[j] does not match. 76 Let w be the wordsize of the computer, typically 64. Assume ﬁrst that m ≤w. Then each bitvector can be stored in a single integer and the bitwise operations can be executed in constant time. Algorithm 2.8: Shift-And Input: text T = T[0 . . . n), pattern P = P[0 . . . m) Output: position of the ﬁrst occurrence of P in T Preprocess: (1) for c ∈Σ do B[c] ←0 (2) for i ←0 to m −1 do B[P[i]] ←B[P[i]] + 2i // B[P[i]].i ←1 Search: (3) D ←0 (4) for j ←0 to n −1 do (5) D ←((D << 1) \| 1) & B[T[j]] (6) if D & 2m−1 ̸= 0 then return j −m + 1 // D.(m −1) = 1 (7) return n Shift-Or is a minor optimization of Shift-And. It is the same algorithm except the roles of 0’s and 1’s in the bitvectors have been swapped. Then line 5 becomes D ←(D << 1) \| B[T[j]]. Note that the “\| 1” was removed, because the shift already brings the correct bit to the least signiﬁcant bit position. 77 Example 2.9: P = assi, T = apassi, bitvectors are columns. B[c], c ∈{a,i,p,s} a i p s a 1 0 0 0 s 0 0 0 1 s 0 0 0 1 i 0 1 0 0 D at each step a p a s s i a 0 1 0 1 0 0 0 s 0 0 0 0 1 0 0 s 0 0 0 0 0 1 0 i 0 0 0 0 0 0 1 The Shift-And algorithm can also be seen as a bitparallel simulation of the nondeterministic automaton that accepts a string ending with P. 0 1 2 3 -1 a s s i Σ After processing T[j], D.i = 1 if and only if there is a path from the initial state (state -1) to state i with the string T[0..j]. 78 On an integer alphabet when m ≤w: • Preprocessing time is O(σ + m). • Search time is O(n). If m > w, we can store the bitvectors in ⌈m/w⌉machine words and perform each bitvector operation in O(⌈m/w⌉) time. • Preprocessing time is O(σ⌈m/w⌉+ m). • Search time is O(n⌈m/w⌉). If no pattern preﬁx longer than w matches a current text suﬃx, then only the least signiﬁcant machine word contains 1’s. There is no need to update the other words; they will stay 0. • Then the search time is O(n) on average. Algorithms like Shift-And that take advantage of the implicit parallelism in bitvector operations are called bitparallel. 79 Karp–Rabin The Karp–Rabin hash function (Deﬁnition 1.43) was originally developed for solving the exact string matching problem. The idea is to compute the hash values or ﬁngerprints H(P) and H(T[j..j + m)) for all j ∈[0..n −m]. • If H(P) ̸= H(T[j..j + m)), then we must have P ̸= T[j..j + m). • If H(P) = H(T[j..j + m), the algorithm compares P and T[j..j + m) in brute force manner. If P ̸= T[j..j + m), this is a false positive. The text factor ﬁngerprints are computed in a sliding window fashion. The ﬁngerprint for T[j + 1..j + 1 + m) = αT[j + m] is computed from the ﬁngerprint for T[j..j + m) = T[j]α in constant time using Lemma 1.44: H(T[j + 1..j + 1 + m)) = (H(T[j]α) −H(T[j]) · rm−1) · r + H(T[j + m])) mod q = (H(T[j..j + m)) −T[j] · rm−1) · r + T[j + m]) mod q . A hash function that supports this kind of sliding window computation is known as a rolling hash function. 80 Algorithm 2.10: Karp-Rabin Input: text T = T[0 . . . n), pattern P = P[0 . . . m) Output: position of the ﬁrst occurrence of P in T (1) Choose q and r; s ←rm−1 mod q (2) hp ←0; ht ←0 (3) for i ←0 to m −1 do hp ←(hp · r + P[i]) mod q // hp = H(P) (4) for j ←0 to m −1 do ht ←(ht · r + T[j]) mod q (5) for j ←0 to n −m −1 do (6) if hp = ht then if P = T[j . . . j + m) then return j (7) ht ←((ht −T[j] · s) · r + T[j + m]) mod q (8) if hp = ht then if P = T[j . . . j + m) then return j (9) return n On an integer alphabet: • The worst case time complexity is O(mn). • The average case time complexity is O(m + n). Karp–Rabin is not competitive in practice for a single pattern, but can be for multiple patterns (exercise). 81 Horspool The algorithms we have seen so far access every character of the text. If we start the comparison between the pattern and the current text position from the end, we can often skip some text characters completely. There are many algorithms that start from the end. The simplest are the Horspool-type algorithms. The Horspool algorithm checks ﬁrst the last character of the text window, i.e., the character aligned with the last pattern character. If that doesn’t match, it moves (shifts) the pattern forward until there is a match. Example 2.11: Horspool ainaisesti-ainainen ainaine/ n ainaine / / n ainainen 82 More precisely, suppose we are currently comparing P against T[j..j + m). Start by comparing P[m −1] to T[k], where k = j + m −1. • If P[m −1] ̸= T[k], shift the pattern until the pattern character aligned with T[k] matches, or until the full pattern is past T[k]. • If P[m −1] = T[k], compare the rest in a brute force manner. Then shift to the next position, where T[k] matches. The length of the shift is determined by the shift table that is precomputed for the pattern. shift[c] is deﬁned for all c ∈Σ: • If c does not occur in P, shift[c] = m. • Otherwise, shift[c] = m −1 −i, where P[i] = c is the last occurrence of c in P[0..m −2]. Example 2.12: P = ainainen. c last occ. shift a ainainen 4 e ainainen 1 i ainainen 3 n ainainen 2 Σ \ {a,e,i,n} — 8 83 Algorithm 2.13: Horspool Input: text T = T[0 . . . n), pattern P = P[0 . . . m) Output: position of the ﬁrst occurrence of P in T Preprocess: (1) for c ∈Σ do shift[c] ←m (2) for i ←0 to m −2 do shift[P[i]] ←m −1 −i Search: (3) j ←0 (4) while j + m ≤n do (5) if P[m −1] = T[j + m −1] then (6) i ←m −2 (7) while i ≥0 and P[i] = T[j + i] do i ←i −1 (8) if i = −1 then return j (9) j ←j + shift[T[j + m −1]] (10) return n 84 On an integer alphabet: • Preprocessing time is O(σ + m). • In the worst case, the search time is O(mn). For example, P = bam−1 and T = an. • In the best case, the search time is O(n/m). For example, P = bm and T = an. • In the average case, the search time is O(n/ min(m, σ)). This assumes that each pattern and text character is picked independently by uniform distribution. In practice, a tuned implementation of Horspool is very fast when the alphabet is not too small. 85 BNDM Starting the matching from the end enables long shifts. • The Horspool algorithm bases the shift on a single character. • The Boyer–Moore algorithm uses the matching suﬃx and the mismatching character. • Factor based algorithms continue matching until no pattern factor matches. This may require more comparisons but it enables longer shifts. Example 2.14: Horspool shift varmasti-aikai/ sen-ainainen ainaisen-ainainen ainaisen-ainainen Boyer–Moore shift varmasti-aikai/ sen-ainainen ainaisen-ainainen ainaisen-ainainen Factor shift varmasti-ai/ kaisen-ainainen ainaisen-ainainen ainaisen-ainainen 86 Factor based algorithms use an automaton that accepts suﬃxes of the reverse pattern P R (or equivalently reverse preﬁxes of the pattern P). • BDM (Backward DAWG Matching) uses a deterministic automaton that accepts exactly the suﬃxes of P R. DAWG (Directed Acyclic Word Graph) is also known as suﬃx automaton. • BNDM (Backward Nondeterministic DAWG Matching) simulates a nondeterministic automaton. Example 2.15: P = assi. s s -1 0 1 2 3 ε a i • BOM (Backward Oracle Matching) uses a much simpler deterministic automaton that accepts all suﬃxes of P R but may also accept some other strings. This can cause shorter shifts but not incorrect behaviour. 87 Suppose we are currently comparing P against T[j..j + m). We use the automaton to scan the text backwards from T[j + m −1]. When the automaton has scanned T[j + i..j + m): • If the automaton is in an accept state, then T[j + i..j + m) is a preﬁx of P. ⇒If i = 0, we found an occurrence. ⇒Otherwise, mark the preﬁx match by setting shift = i. This is the length of the shift that would achieve a matching alignment. • If the automaton can still reach an accept state, then T[j + i..j + m) is a factor of P. ⇒Continue scanning. • When the automaton can no more reach an accept state: ⇒Stop scanning and shift: j ←j + shift. 88 BNDM does a bitparallel simulation of the nondeterministic automaton, which is quite similar to Shift-And. The state of the automaton is stored in a bitvector D. When the automaton has scanned T[j + i..j + m): • D.k = 1 if and only if there is a path from the initial state to state k with the string (T[j + i..j + m))R, and thus T[j + i..j + m) = P[m −k −1..2m −k −i −1). • If D.(m −1) = 1, then T[j + i..j + m) is a preﬁx of the pattern. • If D = 0, then the automaton can no more reach an accept state. Updating D uses precomputed bitvectors B[c], for all c ∈Σ: • B[c].i = 1 if and only if P[m −1 −i] = P R[i] = c. The update when reading T[j + i] is familiar: D = (D << 1) & B[T[j + i]] • Note that there is no “\| 1”. This is because D.(−1) = 0 always after reading at least one character, so the shift brings the right bit to D.0. • Before reading anything D.(−1) = 1. This exception is handled by starting the computation with the ﬁrst shift already performed. Because of this, the shift is done at the end of the loop. 89 Algorithm 2.16: BNDM Input: text T = T[0 . . . n), pattern P = P[0 . . . m) Output: position of the ﬁrst occurrence of P in T Preprocess: (1) for c ∈Σ do B[c] ←0 (2) for i ←0 to m −1 do B[P[m −1 −i]] ←B[P[m −1 −i]] + 2i Search: (3) j ←0 (4) while j + m ≤n do (5) i ←m; shift ←m (6) D ←2m −1 // D ←1m (7) while D ̸= 0 do // Now T[j + i..j + m) is a pattern factor (8) i ←i −1 (9) D ←D & B[T[j + i]] (10) if D & 2m−1 ̸= 0 then // Now T[j + i..j + m) is a pattern preﬁx (11) if i = 0 then return j (12) else shift ←i (13) D ←D << 1 (14) j ←j + shift (15) return n 90 Example 2.17: P = assi, T = apassi. B[c], c ∈{a,i,p,s} a i p s i 0 1 0 0 s 0 0 0 1 s 0 0 0 1 a 1 0 0 0 D when scanning apas backwards s a p a i 1 0 0 0 s 1 1 0 0 s 1 1 0 0 a 1 0 1 0 ⇒shift = 2 D when scanning apassi backwards i s s a p a i 1 1 0 0 0 s 1 0 1 0 0 s 1 0 0 1 0 a 1 0 0 0 1 ⇒ occurrence 91 On an integer alphabet when m ≤w: • Preprocessing time is O(σ + m). • In the worst case, the search time is O(mn). For example, P = am−1b and T = an. • In the best case, the search time is O(n/m). For example, P = bm and T = an. • In the average case, the search time is O(n(logσ m)/m). This is optimal! It has been proven that any algorithm needs to inspect Ω(n(logσ m)/m) text characters on average. When m > w, there are several options: • Use multi-word bitvectors. • Search for a pattern preﬁx of length w and check the rest when the preﬁx is found. • Use BDM or BOM. 92 • The search time of BDM and BOM is O(n(logσ m)/m), which is optimal on average. (BNDM is optimal only when m ≤w.) • MP and KMP are optimal in the worst case but not in the average case. • There are also algorithms that are optimal in both cases. They are based on similar techniques, but we will not describe them here. 93 Crochemore The Crochemore algorithm resembles the Morris–Pratt algorithm at a high level: • When the pattern P is aligned against a text factor T[j..j + m), they compute the longest common preﬁx ℓ= lcp(P, T[j..j + m)) and report an occurrence if ℓ= m. Otherwise, they shift the pattern forward. • MP shifts the pattern forward by ℓ−fail[ℓ] positions. In the next lcp computation, MP skips the ﬁrst fail[ℓ] characters (cf. lcp-comparison). • Crochemore either does the same shift and skip as MP, or a shorter shift than MP and starts the lcp comparison from scratch. Note that the latter case is inoptimal but always safe: no occurrence is missed. Despite sometimes shorter shifts and less eﬃcient lcp computation, Crochemore runs in linear time. More remarkably, it does so without any preprocessing and using only constant extra space in addition to P and T. We will only outline the main ideas of the algorithm without detailed proofs. Even then we will need some concepts from combinatorics on words, a branch of mathematics that studies combinatorial properties of strings. 94 Deﬁnition 2.18: Let S[0..m) be a string. An integer p ∈[1..m] is a period of S, if S[i] = S[i + p] for all i ∈[0..m −p). The smallest period of S is denoted per(S). S is k-periodic if m ≥k · per(S). Example 2.19: The periods of S1 = aabaaabaa are 4,7,8 and 9. The periods of S2 = abcabcabcabca are 3, 6, 9, 12 and 13. S2 is 3-periodic but S1 is not. There is a strong connection between periods and borders. Lemma 2.20: p is a period of S[0..m) if and only if S has a proper border of length m −p. Proof. Both conditions hold if and only if S[0..m −p) = S[p..m). □ Corollary 2.21: The length of the longest proper border of S is m −per(S). 95 Recall that fail[ℓ] in MP is the length of the longest proper border of P[0..ℓ). Thus the pattern shift by MP is ℓ−fail[ℓ] = per(P[0..ℓ)) and the lcp skip is fail[ℓ] = ℓ−per(P[0..ℓ)). Thus knowing per(P[0..ℓ)) is suﬃcient to emulate MP shift and skip. The Crochemore algorithm has two cases: • If P[0..ℓ) is 3-periodic, then compute per(P[0..ℓ)) and do the MP shift and skip. • If P[0..ℓ) is not 3-periodic, then shift by ⌊ℓ/3⌋+ 1 ≤per(P[0..ℓ)) and start the lcp comparison from scratch. To ﬁnd out if P[0..ℓ) is 3-periodic and to compute per(P[0..ℓ)) if it is, Crochemore uses another combinatorial concept. 96 Deﬁnition 2.22: Let MS(S) denote the lexicographically maximal suﬃx of a string S. If S = MS(S), S is called self-maximal. Period computation is easier for maximal suﬃxes and self-maximal strings than for arbitrary strings. Lemma 2.23: Let S[0..m) be a self-maximal string and let p = per(S). For any c ∈Σ, MS(Sc) = Sc and per(Sc) = p if c = S[m −p] MS(Sc) = Sc and per(Sc) = m + 1 if c < S[m −p] MS(Sc) ̸= Sc if c > S[m −p] Furthermore, let r = m mod p and R = S[m −r..m). Then R is self-maximal and MS(Sc) = MS(Rc) if c > S[m −p] 97 Crochemore’s algorithm computes the maximal suﬃx and its period for P[0..ℓ) incrementally using Lemma 2.23. The following algorithm updates the maximal suﬃx information when the match is extended by one character. Algorithm 2.24: Update-MS(P, ℓ, s, p) Input: a string P and integers ℓ, s, p such that MS(P[0..ℓ)) = P[s..ℓ) and p = per(P[s..ℓ)). Output: a triple (ℓ+ 1, s′, p′) such that MS(P[0..ℓ+ 1)) = P[s′..ℓ+ 1) and p′ = per(P[s′..ℓ+ 1)). (1) if ℓ= 0 then return (1, 0, 1) (2) i ←ℓ (3) while i < ℓ+ 1 do // P[s..i) is self-maximal and p = per(P[s..i)) (4) if P[i] > P[i −p] then (5) i ←i −((i −s) mod p) (6) s ←i (7) p ←1 (8) else if P[i] < P[i −p] then (9) p ←i −s + 1 (10) i ←i + 1 (11) return (ℓ+ 1, s, p) 98 As the ﬁnal piece of the Crochemore algorithm, the following result shows how to use the maximal suﬃx information to obtain information about the periodicity of the full string. Lemma 2.25: Let S[0..m) be a string and let S[s..m) = MS(S) and p = per(MS(S)). • S is 3-periodic if and only if p ≤m/3 and S[0..s) = S[p..p + s). • If S is 3-periodic, then per(S) = p. The algorithm is given on the next slide. 99 Algorithm 2.26: Crochemore Input: strings T[0..n) (text) and P[0..m) (pattern). Output: position of the ﬁrst occurrence of P in T (1) j ←ℓ←p ←s ←0 (2) while j + m ≤n do (3) while j + ℓ< n and ℓ< m and T[j + ℓ] = P[ℓ] do (4) (ℓ, s, p) ←Update-MS(P, ℓ, s, p) // ℓ= lcp(P, T[j..j + m)) (5) if ℓ= m then return j // MS(P[0..ℓ)) = P[s..ℓ) and p = per(P[s..ℓ)) (6) if p ≤ℓ/3 and P[0..s) = P[p..p + s) then // per(P[0..ℓ)) = p (7) j ←j + p (8) ℓ←ℓ−p (9) else // per(P[0..ℓ)) > ℓ/3 (10) j ←j + ⌊ℓ/3⌋+ 1 (11) (ℓ, s, p) ←(0, 0, 0) (12) return n 100 For ordered alphabet: • The time complexity is O(n). • The algorithm uses only a constant number of integer variables in addition to the strings P and T. Crochemore is not competitive in practice. However, there are situations, where the pattern can be very long and the space complexity is more important than speed. There are also other linear time, constant extra space algorithms. All of them are based on string periodicity in some way. 101 Aho–Corasick Given a text T and a set P = {P1.P2, . . . , Pk} of patterns, the multiple exact string matching problem asks for the occurrences of all the patterns in the text. The Aho–Corasick algorithm is an extension of the Morris–Pratt algorithm for multiple exact string matching. Aho–Corasick uses the trie trie(P) as an automaton and augments it with a failure function similar to the Morris-Pratt failure function. Example 2.27: Aho–Corasick automaton for P = {he, she, his, hers}. 1 2 0 h s e e s r i s 3 4 5 6 7 8 9 Σ -1 h 102 Let Sv denote the string represented by a node v in the trie. The components of the AC automaton are: • root is the root and child() the child function of the trie. • fail(v) = u such that Su is the longest proper suﬃx of Sv represented by any trie node u. • patterns(v) is the set of pattern indices i such that Pi is a suﬃx of Sv. Example 2.28: For the automaton in Example 2.27, patterns(2) = {1} ({he}), patterns(5) = {1, 2} ({he, she}), patterns(7) = {3} ({his}), patterns(9) = {4} ({hers}), and patterns(v) = ∅for all other nodes v. 103 At each stage of the matching, the algorithm computes the node v such that Sv is the longest suﬃx of T[0..j] represented by any node. Algorithm 2.29: Aho–Corasick Input: text T, pattern set P = {P1, P2, . . . , Pk}. Output: all pairs (i, j) such that Pi occurs in T ending at j. (1) (root, child(), fail(), patterns()) ←Construct-AC-Automaton(P) (2) v ←root (3) for j ←0 to n −1 do (4) while child(v, T[j]) = ⊥do v ←fail(v) (5) v ←child(v, T[j]) (6) for i ∈patterns(v) do output (i, j) The construction of the automaton is done in two phases: the trie construction and the failure links computation. Algorithm 2.30: Construct-AC-Automaton Input: pattern set P = {P1, P2, . . . , Pk}. Output: AC automaton: root, child(), fail() and patterns(). (1) (root, child(), patterns()) ←Construct-AC-Trie(P) (2) (fail(), patterns()) ←Compute-AC-Fail(root, child(), patterns()) (3) return (root, child(), fail(), patterns()) 104 Algorithm 2.31: Construct-AC-Trie Input: pattern set P = {P1, P2, . . . , Pk}. Output: AC trie: root, child() and patterns(). (1) Create new node root (2) for i ←1 to k do (3) v ←root; j ←0 (4) while child(v, Pi[j]) ̸= ⊥do (5) v ←child(v, Pi[j]); j ←j + 1 (6) while j < \|Pi\| do (7) Create new node u (8) child(v, Pi[j]) ←u (9) v ←u; j ←j + 1 (10) patterns(v) ←{i} (11) return (root, child(), patterns()) Lines (3)–(10) perform the standard trie insertion (Algorithm 1.2). • Line (10) marks v as a representative of Pi. • The creation of a new node v initializes patterns(v) to ∅ (in addition to initializing child(v, c) to ⊥for all c ∈Σ). 105 Algorithm 2.32: Compute-AC-Fail Input: AC trie: root, child() and patterns() Output: AC failure function fail() and updated patterns() (1) Create new node fallback (2) for c ∈Σ do child(fallback, c) ←root (3) fail(root) ←fallback (4) queue ←{root} (5) while queue ̸= ∅do (6) u ←popfront(queue) (7) for c ∈Σ such that child(u, c) ̸= ⊥do (8) v ←child(u, c) (9) w ←fail(u) (10) while child(w, c) = ⊥do w ←fail(w) (11) fail(v) ←child(w, c) (12) patterns(v) ←patterns(v) ∪patterns(fail(v)) (13) pushback(queue, v) (14) return (fail(), patterns()) The algorithm does a breath ﬁrst traversal of the trie. This ensures that correct values of fail() and patterns() are already computed when needed. 106 fail(v) is correctly computed on lines (8)–(11): • Let fail∗(v) = {v, fail(v), fail(fail(v)), . . . , root}. These nodes are exactly the trie nodes that represent suﬃxes of Sv. • Let u = parent(v) and child(u, c) = v. Then Sv = Suc and a string S is a suﬃx of Su iﬀSc is suﬃx of Sv. Thus for any node w – If w ∈fail∗(v) \ {root}, then parent(w) ∈fail∗(u). – If w ∈fail∗(u) and child(w, c) ̸= ⊥, then child(w, c) ∈fail∗(v). • Therefore, fail(v) = child(w, c), where w is the ﬁrst node in fail∗(u) other than u such that child(w, c) ̸= ⊥, or fail(v) = root if no such node w exists. patterns(v) is correctly computed on line (12): patterns(v) = {i \| Pi is a suﬃx of Sv} = {i \| Pi = Sw and w ∈fail∗(v)} = {i \| Pi = Sv} ∪patterns(fail(v)) 107 Assuming σ is constant: • The search time is O(n). • The space complexity is O(m), where m = \|\|P\|\|. – The implementation of patterns() requires care (exercise). • The preprocessing time is O(m), where m = \|\|P\|\|. – The only non-trivial issue is the while-loop on line (10). – Let root, v1, v2, . . . , vℓbe the nodes on the path from root to a node representing a pattern Pi. Let wj = fail(vj) for all j. Let depth(v) be the depth of a node v (depth(root) = 0). – When processing vj and computing wj = fail(vj), we have depth(wj) = depth(wj−1) + 1 before line (10) and depth(wj) ≤depth(wj−1) + 1 −tj after line (10), where tj is the number of rounds in the while-loop. – Thus, the total number of rounds in the while-loop when processing the nodes v1, v2, . . . , vℓis at most ℓ= \|Pi\|, and thus over the whole algorithm at most \|\|P\|\|. The analysis when σ is not constant is left as an exercise. 108 Summary: Exact String Matching Exact string matching is a fundamental problem in stringology. We have seen several diﬀerent algorithms for solving the problem. The properties of the algorithms vary with respect to worst case time complexity, average case time complexity, type of alphabet (ordered/integer) and even space complexity. The algorithms use a wide range of completely diﬀerent techniques: • There exists numerous algorithms for exact string matching but most of them use variations or combinations of the techniques we have seen (study groups). • Many of the techniques can be adapted to other problems. All of the techniques have some uses in practice. 109 3. Approximate String Matching Often in applications we want to search a text for something that is similar to the pattern but not necessarily exactly the same. To formalize this problem, we have to specify what does “similar” mean. This can be done by deﬁning a similarity or a distance measure. A natural and popular distance measure for strings is the edit distance, also known as the Levenshtein distance. 110 Edit distance The edit distance ed(A, B) of two strings A and B is the minimum number of edit operations needed to change A into B. The allowed edit operations are: S Substitution of a single character with another character. I Insertion of a single character. D Deletion of a single character. Example 3.1: Let A = Lewensteinn and B = Levenshtein. Then ed(A, B) = 3. The set of edit operations can be described with an edit sequence: NNSNNNINNNND or with an alignment: Lewens-teinn Levenshtein-In the edit sequence, N means No edit. 111 There are many variations and extension of the edit distance, for example: • Hamming distance allows only the subtitution operation. • Damerau–Levenshtein distance adds an edit operation: T Transposition swaps two adjacent characters. • With weighted edit distance, each operation has a cost or weight, which can be other than one. • Allow insertions and deletions (indels) of factors at a cost that is lower than the sum of character indels. We will focus on the basic Levenshtein distance. Levenshtein distance has the following two useful properties, which are not shared by all variations (exercise): • Levenshtein distance is a metric. • If ed(A, B) = k, there exists an edit sequence and an alignment with k edit operations, but no edit sequence or alignment with less than k edit operations. An edit sequence and an alignment with ed(A, B) edit operations is called optimal. 112 Computing Edit Distance Given two strings A[1..m] and B[1..n], deﬁne the values dij with the recurrence: d00 = 0, di0 = i, 1 ≤i ≤m, d0j = j, 1 ≤j ≤n, and dij = min      di−1,j−1 + δ(A[i], B[j]) di−1,j + 1 di,j−1 + 1 1 ≤i ≤m, 1 ≤j ≤n, where δ(A[i], B[j]) = 1 if A[i] ̸= B[j] 0 if A[i] = B[j] Theorem 3.2: dij = ed(A[1..i], B[1..j]) for all 0 ≤i ≤m, 0 ≤j ≤n. In particular, dmn = ed(A, B). 113 Example 3.3: A = ballad, B = handball d h a n d b a l l 0 1 2 3 4 5 6 7 8 b 1 1 2 3 4 4 5 6 7 a 2 2 1 2 3 4 4 5 6 l 3 3 2 2 3 4 5 4 5 l 4 4 3 3 3 4 5 5 4 a 5 5 4 4 4 4 4 5 5 d 6 6 5 5 4 5 5 5 6 ed(A, B) = dmn = d6,8 = 6. 114 Proof of Theorem 3.2. We use induction with respect to i + j. For brevity, write Ai = A[1..i] and Bj = B[1..j]. Basis: d00 = 0 = ed(ϵ, ϵ) di0 = i = ed(Ai, ϵ) (i deletions) d0j = j = ed(ϵ, Bj) (j insertions) Induction step: We show that the claim holds for dij, 1 ≤i ≤m, 1 ≤j ≤n. By induction assumption, dpq = ed(Ap, Bq) when p + q < i + j. Let Eij be an optimal edit sequence with the cost ed(Ai, Bj). We have three cases depending on what the last operation symbol in Eij is: N or S: Eij = Ei−1,j−1N or Eij = Ei−1,j−1S and ed(Ai, Bj) = ed(Ai−1, Bj−1)+δ(A[i], B[j]) = di−1,j−1 +δ(A[i], B[j]). I: Eij = Ei,j−1I and ed(Ai, Bj) = ed(Ai, Bj−1) + 1 = di,j−1 + 1. D: Eij = Ei−1,jD and ed(Ai, Bj) = ed(Ai−1, Bj) + 1 = di−1,j + 1. One of the cases above is always true, and since the edit sequence is optimal, it must be one with the minimum cost, which agrees with the deﬁnition of dij. □ 115 The recurrence gives directly a dynamic programming algorithm for computing the edit distance. Algorithm 3.4: Edit distance Input: strings A[1..m] and B[1..n] Output: ed(A, B) (1) for i ←0 to m do di0 ←i (2) for j ←1 to n do d0j ←j (3) for j ←1 to n do (4) for i ←1 to m do (5) dij ←min{di−1,j−1 + δ(A[i], B[j]), di−1,j + 1, di,j−1 + 1} (6) return dmn The time and space complexity is O(mn). 116 The space complexity can be reduced by noticing that each column of the matrix (dij) depends only on the previous column. We do not need to store older columns. A more careful look reveals that, when computing dij, we only need to store the bottom part of column j −1 and the already computed top part of column j. We store these in an array C[0..m] and variables c and d as shown below: d0,j−1 dm,j−1 di−1,j d0,j dm,j di−1,j d0,j di−1,j−1 c di,j di,j−1 di,j−1 dm,j−1 di,j d di−1,j−1 C[0..m] 117 Algorithm 3.5: Edit distance in O(m) space Input: strings A[1..m] and B[1..n] Output: ed(A, B) (1) for i ←0 to m do C[i] ←i (2) for j ←1 to n do (3) c ←C; C ←j (4) for i ←1 to m do (5) d ←min{c + δ(A[i], B[j]), C[i −1] + 1, C[i] + 1} (6) c ←C[i] (7) C[i] ←d (8) return C[m] Note that because ed(A, B) = ed(B, A) (exercise), we can always choose A to be the shorter string so that m ≤n. 118 It is also possible to ﬁnd optimal edit sequences and alignments from the matrix dij. An edit graph is a directed graph, where the nodes are the cells of the edit distance matrix, and the edges are as follows: • If A[i] = B[j] and dij = di−1,j−1, there is an edge (i −1, j −1) →(i, j) labelled with N. • If A[i] ̸= B[j] and dij = di−1,j−1 + 1, there is an edge (i −1, j −1) →(i, j) labelled with S. • If dij = di,j−1 + 1, there is an edge (i, j −1) →(i, j) labelled with I. • If dij = di−1,j + 1, there is an edge (i −1, j) →(i, j) labelled with D. Any path from (0, 0) to (m, n) is labelled with an optimal edit sequence. 119 Example 3.6: A = ballad, B = handball d h a n d b a l l 0 ⇒1 ⇒2 ⇒3 ⇒4 →5 →6 →7 →8 b → ⇒ → → → ⇒ 1 1 →2 →3 →4 4 →5 →6 →7 a → → → ⇒ ⇒ 2 2 1 ⇒2 →3 →4 4 →5 →6 l → → → → ⇒ ⇒ → → → ⇒ → 3 3 2 2 ⇒3 →4 →5 4 →5 l → → → → → → ⇒ ⇒ → → → ⇒ 4 4 3 3 3 ⇒4 →5 5 4 a → → → → → → → → → → ⇒ ⇒ 5 5 4 4 4 4 4 ⇒5 5 d → → → → → → → → → → → ⇒ ⇒ ⇒ 6 6 5 5 4 →5 5 5 ⇒6 There are 7 paths from (0, 0) to (6, 8) corresponding to 7 diﬀerent optimal edit sequences and alignments, including the following three: IIIINNNNDD SNISSNIS SNSSINSI ----ballad ba-lla-d ball-ad-handball--handball handball 120 Approximate String Matching Now we are ready to tackle the main problem of this part: approximate string matching. Problem 3.7: Given a text T[1..n], a pattern P[1..m] and an integer k ≥0, report all positions j ∈[1..m] such that ed(P, T(j −ℓ...j]) ≤k for some ℓ≥0. The factor T(j −ℓ...j] is called an approximate occurrence of P. There can be multiple occurrences of diﬀerent lengths ending at the same position j, but usually it is enough to report just the end positions. We ask for the end position rather than the start position because that is more natural for the algorithms. 121 Deﬁne the values gij with the recurrence: g0j = 0, 0 ≤j ≤n, gi0 = i, 1 ≤i ≤m, and gij = min      gi−1,j−1 + δ(P[i], T[j]) gi−1,j + 1 gi,j−1 + 1 1 ≤i ≤m, 1 ≤j ≤n. Theorem 3.8: For all 0 ≤i ≤m, 0 ≤j ≤n: gij = min{ed(P[1..i], T(j −ℓ...j]) \| 0 ≤ℓ≤j} . In particular, j is an ending position of an approximate occurrence if and only if gmj ≤k. 122 Proof. We use induction with respect to i + j. Basis: g00 = 0 = ed(ϵ, ϵ) g0j = 0 = ed(ϵ, ϵ) = ed(ϵ, T(j −0..j]) (min at ℓ= 0) gi0 = i = ed(P[1..i], ϵ) = ed(P[1..i], T(0 −0..0]) (0 ≤ℓ≤j = 0) Induction step: Essentially the same as in the proof of Theorem 3.2. 123 Example 3.9: P = match, T = remachine, k = 1 g r e m a c h i n e 0 0 0 0 0 0 0 0 0 0 m ⇒ 1 1 1 0 1 1 1 1 1 1 a ⇒ 2 2 2 1 0 1 2 2 2 2 t ⇒ 3 3 3 2 1 1 2 3 3 3 c ⇒ 4 4 4 3 2 1 2 3 4 4 h ⇒ 5 5 5 4 3 2 1 2 3 4 One occurrence ending at position 6. 124 Algorithm 3.10: Approximate string matching Input: text T[1..n], pattern P[1..m], and integer k Output: end positions of all approximate occurrences of P (1) for i ←0 to m do gi0 ←i (2) for j ←1 to n do g0j ←0 (3) for j ←1 to n do (4) for i ←1 to m do (5) gij ←min{gi−1,j−1 + δ(A[i], B[j]), gi−1,j + 1, gi,j−1 + 1} (6) if qmj ≤k then output j • Time and space complexity is O(mn) on ordered alphabet. • The space complexity can be reduced to O(m) by storing only one column as in Algorithm 3.5. 125 Ukkonen’s Cut-oﬀHeuristic We can speed up the algorithm using the diagonal monotonicity of the matrix (gij): A diagonal d, −m ≤d ≤n, consists of the cells gij with j −i = d. Every diagonal in (gij) is monotonically non-decreasing. Example 3.11: Diagonals -3 and 2. g r e m a c h i n e 0 0 0 0 0 0 0 0 0 0 m – 1 1 1 0 1 1 1 1 1 1 a – 2 2 2 1 0 1 2 2 2 2 t – 3 3 3 2 1 1 2 3 3 3 c – – 4 4 4 3 2 1 2 3 4 4 h – – 5 5 5 4 3 2 1 2 3 4 126 Lemma 3.12: For every i ∈[1..m] and every j ∈[1..n], gij = gi−1,j−1 or gij = gi−1,j−1 + 1. Proof. By deﬁnition, gij ≤gi−1,j−1 + δ(P[i], T[j]) ≤gi−1,j−1 + 1. We show that gij ≥gi−1,j−1 by induction on i + j. The induction assumption is that gpq ≥gp−1,q−1 when p ∈[1..m], q ∈[1..n] and p + q < i + j. At least one of the following holds: 1. gij = gi−1,j−1 + δ(P[i], T[j]). Then gij ≥gi−1,j−1. 2. gij = gi−1,j + 1 and i > 1. Then gij = gi−1,j + 1 ind. assump. ≥ gi−2,j−1 + 1 deﬁnition ≥ gi−1,j−1 3. gij = gi,j−1 + 1 and j > 1. Then gij = gi,j−1 + 1 ind. assump. ≥ gi−1,j−2 + 1 deﬁnition ≥ gi−1,j−1 4. gij = gi−1,j + 1 and i = 1. Then gij = 0 + 1 > 0 = gi−1,j−1. 5. gij = gi,j−1 + 1 and j = 1. Then gij = i + 1 = (i −1) + 2 = gi−1,j−1 + 2, which cannot be true. Thus this case can never happen. □ 127 We can reduce computation using diagonal monotonicity: • Whenever the value on a diagonal d grows larger than k, we can discard d from consideration, because we are only interested in values at most k on the row m. • We keep track of the smallest undiscarded diagonal d. Each column is computed only up to diagonal d + 1. Example 3.13: P = strict, T = datastructure, k = 1 g d a t a s t r u c t u r e 0 0 0 0 0 0 0 0 0 0 0 0 0 0 s 1 1 1 1 1 0 1 1 1 1 1 1 1 1 t 2 2 2 1 2 1 0 1 2 2 1 2 2 2 r 2 2 2 1 0 1 2 2 2 i 2 1 1 2 3 3 c 2 2 1 2 3 t 2 1 2 128 The position of the smallest undiscarded diagonal on the current column is kept in a variable top. Algorithm 3.14: Ukkonen’s cut-oﬀalgorithm Input: text T[1..n], pattern P[1..m], and integer k Output: end positions of all approximate occurrences of P (1) top ←min(k + 1, m) (2) for i ←0 to top do gi0 ←i (3) for j ←1 to n do g0j ←0 (4) for j ←1 to n do (5) for i ←1 to top do (6) gij ←min{gi−1,j−1 + δ(A[i], B[j]), gi−1,j + 1, gi,j−1 + 1} (7) while gtop,j > k do top ←top −1 (8) if top = m then output j (9) else top ←top + 1; gtop,j ←k + 1 129 The time complexity is proportional to the computed area in the matrix (gij). • The worst case time complexity is still O(mn) on ordered alphabet. • The average case time complexity is O(kn). The proof is not trivial. There are many other algorithms based on diagonal monotonicity. Some of them achieve O(kn) worst case time complexity. 130 Myers’ Bitparallel Algorithm Another way to speed up the computation is bitparallelism. Instead of the matrix (gij), we store diﬀerences between adjacent cells: Vertical delta: ∆vij = gij −gi−1,j Horizontal delta: ∆hij = gij −gi,j−1 Diagonal delta: ∆dij = gij −gi−1,j−1 Because gi0 = i ja g0j = 0, gij = ∆v1j + ∆v2j + · · · + ∆vij = i + ∆hi1 + ∆hi2 + · · · + ∆hij Because of diagonal monotonicity, ∆dij ∈{0, 1} and it can be stored in one bit. By the following result, ∆hij and ∆vij can be stored in two bits. Lemma 3.15: ∆hij, ∆vij ∈{−1, 0, 1} for every i, j that they are deﬁned for. The proof is left as an exercise. 131 Example 3.16: ‘–’ means −1, ‘=’ means 0 and ‘+’ means +1 r e m a c h i n e 0 = 0 = 0 = 0 = 0 = 0 = 0 = 0 = 0 = 0 m + + + + + = = + + + + + + + + + + + + 1 = 1 = 1 – 0 + 1 = 1 = 1 = 1 = 1 = 1 a + + + + + = + = – = = + + + + + + + + 2 = 2 = 2 – 1 – 0 + 1 + 2 = 2 = 2 = 2 t + + + + + = + = + + = + = + + + + + + 3 = 3 = 3 – 2 – 1 = 1 + 2 + 3 = 3 = 3 c + + + + + = + = + = = + = + = + + + + 4 = 4 = 4 – 3 – 2 – 1 + 2 + 3 + 4 = 4 h + + + + + = + = + = + = – = – = – = = 5 = 5 = 5 – 4 – 3 – 2 – 1 + 2 + 3 + 4 132 In the standard computation of a cell: • Input is gi−1,j, gi−1,j−1, gi,j−1 and δ(P[i], T[j]). • Output is gij. In the corresponding bitparallel computation: • Input is ∆vin = ∆vi,j−1, ∆hin = ∆hi−1,j and Eqij = 1 −δ(P[i], T[j]). • Output is ∆vout = ∆vi,j and ∆hout = ∆hi,j. gi−1,j−1 ∆hin − − − − − − − − → gi−1,j ∆vin     y     y∆vout gi,j−1 − − − − − − − − − → ∆hout gij The algorithm does not compute the ∆d values but they are useful in the proofs. 133 The computation rule is deﬁned by the following result. Lemma 3.17: If Eq = 1 or ∆vin = −1 or ∆hin = −1, then ∆d = 0, ∆vout = −∆hin and ∆hout = −∆vin. Otherwise ∆d = 1, ∆vout = 1 −∆hin and ∆hout = 1 −∆vin. Proof. We can write the recurrence for gij as gij = min{gi−1,j−1 + δ(P[i], T[j]), gi,j−1 + 1, gi−1,j + 1} = gi−1,j−1 + min{1 −Eq, ∆vin + 1, ∆hin + 1}. Then ∆d = gij −gi−1,j−1 = min{1 −Eq, ∆vin + 1, ∆hin + 1} which is 0 if Eq = 1 or ∆vin = −1 or ∆hin = −1 and 1 otherwise. Clearly ∆d = ∆vin + ∆hout = ∆hin + ∆vout. Thus ∆vout = ∆d −∆hin and ∆hout = ∆d −∆vin. □ 134 To enable bitparallel operation, we need two changes: • The ∆v and ∆h values are “trits” not bits. We encode each of them with two bits as follows: Pv = 1 if ∆v = +1 0 otherwise Mv = 1 if ∆v = −1 0 otherwise Ph = 1 if ∆h = +1 0 otherwise Mh = 1 if ∆h = −1 0 otherwise Then ∆v = Pv −Mv ∆h = Ph −Mh • We replace arithmetic operations (+, −, min) with Boolean (logical) operations (∧, ∨, ¬). 135 Now the computation rules can be expressed as follows. Lemma 3.18: Pvout = Mhin ∨¬(Xv ∨Phin) Mvout = Phin ∧Xv Phout = Mvin ∨¬(Xh ∨Pvin) Mhout = Pvin ∧Xh where Xv = Eq ∨Mvin and Xh = Eq ∨Mhin. Proof. We show the claim for Pv and Mv only. Ph and Mh are symmetrical. By Lemma 3.17, Pvout = (¬∆d ∧Mhin) ∨(∆d ∧¬Phin) Mvout = (¬∆d ∧Phin) ∨(∆d ∧0) = ¬∆d ∧Phin Because ∆d = ¬(Eq ∨Mvin ∨Mhin) = ¬(Xv ∨Mhin) = ¬Xv ∧¬Mhin, Pvout = ((Xv ∨Mhin) ∧Mhin) ∨(¬Xv ∧¬Mhin ∧¬Phin) = Mhin ∨¬(Xv ∨Mhin ∨Phin) = Mhin ∨¬(Xv ∨Phin) Mvout = (Xv ∨Mhin) ∧Phin = (Xv ∧Phin) ∨(Mhin ∧Phin) = Xv ∧Phin All the steps above use just basic laws of Boolean algebra except the last step, where we use the fact that Mhin and Phin cannot be 1 simultaneously. □ 136 According to Lemma 3.18, the bit representation of the matrix can be computed as follows. for i ←1 to m do Pvi0 ←1; Mvi0 ←0 for j ←1 to n do Ph0j ←0; Mh0j ←0 for i ←1 to m do Xhij ←Eqij ∨Mhi−1,j Phij ←Mvi,j−1 ∨¬(Xhij ∨Pvi,j−1) Mhij ←Pvi,j−1 ∧Xhij for i ←1 to m do Xvij ←Eqij ∨Mvi,j−1 Pvij ←Mhi−1,j ∨¬(Xvij ∨Phi−1,j) Mvij ←Phi−1,j ∧Xvij This is not yet bitparallel though. 137 To obtain a bitparallel algorithm, the columns Pv∗j, Mv∗j, Xv∗j, Ph∗j, Mh∗j, Xh∗j and Eq∗j are stored in bitvectors. Now the second inner loop can be replaced with the code Xv∗j ←Eq∗j ∨Mv∗,j−1 Pv∗j ←(Mh∗,j << 1) ∨¬(Xv∗j ∨(Ph∗j << 1)) Mv∗j ←(Ph∗j << 1) ∧Xv∗j A similar attempt with the for ﬁrst inner loop leads to a problem: Xh∗j ←Eq∗j ∨(Mh∗j << 1) Ph∗j ←Mv∗,j−1 ∨¬(Xh∗j ∨Pv∗,j−1) Mh∗j ←Pv∗,j−1 ∧Xh∗j Now the vector Mh∗j is used in computing Xh∗j before Mh∗j itself is computed! Changing the order does not help, because Xh∗j is needed to compute Mh∗j. To get out of this dependency loop, we compute Xh∗j without Mh∗j using only Eq∗j and Pv∗,j−1 which are already available when we compute Xh∗j. 138 Lemma 3.19: Xhij = ∃ℓ∈[1, i] : Eqℓj ∧(∀x ∈[ℓ, i −1] : Pvx,j−1). Proof. We use induction on i. Basis i = 1: The right-hand side reduces to Eq1j, because ℓ= 1. By Lemma 3.18, Xh1j = Eq1j ∨Mh0j, which is Eq1j because Mh0j = 0 for all j. Induction step: The induction assumption is that Xhi−1,j is as claimed. Now we have ∃ℓ∈[1, i] : Eqℓj ∧(∀x ∈[ℓ, i −1] : Pvx,j−1) = Eqij ∨∃ℓ∈[1, i −1] : Eqℓj ∧(∀x ∈[ℓ, i −1] : Pvx,j−1) = Eqij ∨(Pvi−1,j−1 ∧∃ℓ∈[1, i −1] : Eqℓj ∧(∀x ∈[ℓ, i −2] : Pvx,j−1)) = Eqij ∨(Pvi−1,j−1 ∧Xhi−1,j) (ind. assump.) = Eqij ∨Mhi−1,j (Lemma 3.18) = Xhij (Lemma 3.18) □ 139 At ﬁrst sight, we cannot use Lemma 3.19 to compute even a single bit in constant time, let alone a whole vector Xh∗j. However, it can be done, but we need more bit operations: • Let ⊻denote the xor-operation: 0 ⊻1 = 1 ⊻0 = 1 and 0 ⊻0 = 1 ⊻1 = 0. • A bitvector is interpreted as an integer and we use addition as a bit operation. The carry mechanism in addition plays a key role. For example 0001 + 0111 = 1000. In the following, for a bitvector B, we will write B = B[1..m] = B[m]B[m −1] . . . B The reverse order of the bits reﬂects the interpretation as an integer. 140 Lemma 3.20: Denote X = Xh∗j, E = Eq∗j, P = Pv∗,j−1 and let Y = (((E ∧P) + P) ⊻P) ∨E. Then X = Y . Proof. By Lemma 3.19, X[i] = 1 iﬀand only if a) E[i] = 1 or b) ∃ℓ∈[1, i] : E[ℓ. . . i] = 00 · · · 01 ∧P[ℓ. . . i −1] = 11 · · · 1. and X[i] = 0 iﬀand only if c) E1...i = 00 · · · 0 or d) ∃ℓ∈[1, i] : E[ℓ. . . i] = 00 · · · 01 ∧P[ℓ. . . i −1] ̸= 11 · · · 1. We prove that Y [i] = X[i] in all of these cases: a) The deﬁnition of Y ends with “∨E” which ensures that Y [i] = 1 in this case. 141 b) The following calculation shows that Y [i] = 1 in this case: i ℓ E[ℓ. . . i] =00...01 P[ℓ. . . i] =b1...11 (E ∧P)[ℓ. . . i] =00...01 ((E ∧P) + P)[ℓ. . . i] =¯ b0...0c (((E ∧P) + P) ⊻P)[ℓ. . . i] =11...1¯ c Y = ((((E ∧P) + P) ⊻P) ∨E)[ℓ. . . i] =11...11 where b is the unknown bit P[i], c is the possible carry bit coming from the summation of bits 1 . . . , ℓ−1, and ¯ b and ¯ c are their negations. c) Because for all bitvectors B, 0 ∧B = 0 ja 0 + B = B, we get Y = (((0 ∧P) + P) ⊻P) ∨0 = (P ⊻P) ∨0 = 0. d) Consider the calculation in case b). A key point there is that the carry bit in the summation travels from position ℓto i and produces ¯ b to position i. The diﬀerence in this case is that at least one bit P[k], ℓ≤k < i, is zero, which stops the carry at position k. Thus ((E ∧P) + P)[i] = b and Y [i] = (b ⊻b) ∨0 = 0. □ 142 As a ﬁnal detail, we compute the bottom row values gmj using the equalities gm0 = m ja gmj = gm,j−1 + ∆hmj. Algorithm 3.21: Myers’ bitparallel algorithm Input: text T[1..n], pattern P[1..m], and integer k Output: end positions of all approximate occurrences of P (1) for c ∈Σ do B[c] ←0m (2) for i ←1 to m do B[P[i]][i] = 1 (3) Pv ←1m; Mv ←0; g ←m (4) for j ←1 to n do (5) Eq ←B[T[j]] (6) Xh ←(((Eq ∧Pv) + Pv) ⊻Pv) ∨Eq (7) Ph ←Mv ∨¬(Xh ∨Pv) (8) Mh ←Pv ∧Xh (9) Xv ←Eq ∨Mv (10) Pv ←(Mh << 1) ∨¬(Xv ∨(Ph << 1)) (11) Mv ←(Ph << 1) ∧Xv (12) g ←g + Ph[m] −Mh[m] (13) if g ≤k then output j 143 On an integer alphabet, when m ≤w: • Pattern preprocessing time is O(m + σ). • Search time is O(n). When m > w, we can store each bit vector in ⌈m/w⌉machine words: • The worst case search time is O(n⌈m/w⌉). • Using Ukkonen’s cut-oﬀheuristic, it is possible reduce the average case search time to O(n⌈k/w⌉). There are also algorithms for bitparallel simulation of a nondeterministic automaton that recognizes the aprroximate occurrences of the pattern. Example 3.22: P = pattern, k = 3 a t t e r n p a t t e r n p Σ ε Σ ε Σ ε Σ ε Σ ε Σ ε Σ ε Σ Σ Σ Σ Σ Σ Σ Σ a t t e r n p Σ ε Σ ε Σ ε Σ ε Σ ε Σ ε Σ ε Σ Σ Σ Σ Σ Σ Σ Σ a t t e r n p Σ ε Σ ε Σ ε Σ ε Σ ε Σ ε Σ ε Σ Σ Σ Σ Σ Σ Σ Σ no errors 1 error 2 errors 3 errors 144 Another way to utilize Lemma 3.15 (∆hij, ∆vij ∈{−1, 0, 1}) is to use precomputed tables to process multiple matrix cells at a time. • There are at most 3m diﬀerent columns. Thus there exists a deterministic automaton with 3m states and σ3m transitions that can ﬁnd all approximate occurrences in O(n) time. However, the space and constructions time of the automaton can be too big to be practical. • There is a super-alphabet algorithm that processes O(logσ n) characters at a time and O(log2 σ n) matrix cells at a time using lookup tables of size O(n). This gives time complexity O(mn/ log2 σ n). • A practical variant uses smaller lookup tables to compute multiple entries of a column at a time. 145 Baeza-Yates–Perleberg Filtering Algorithm A ﬁltering algorithm for approximate string matching searches the text for factors having some property that satisﬁes the following conditions: 1. Every approximate occurrence of the pattern has this property. 2. Strings having this property are reasonably rare. 3. Text factors having this property can be found quickly. Each text factor with the property is a potential occurrence, which is then veriﬁed for whether it is an actual approximate occurrence. Filtering algorithms can achieve linear or even sublinear average case time complexity. 146 The following lemma shows the property used by the Baeza-Yates–Perleberg algorithm and proves that it satisﬁes the ﬁrst condition. Lemma 3.23: Let P1P2 . . . Pk+1 = P be a partitioning of the pattern P into k + 1 nonempty factors. Any string S with ed(P, S) ≤k contains Pi as a factor for some i ∈[1..k + 1]. Proof. Each single symbol edit operation can change at most one of the pattern factors Pi. Thus any set of at most k edit operations leaves at least one of the factors untouched. □ 147 The algorithm has two phases: Filtration: Search the text T for exact occurrences of the pattern factors Pi. Using the Aho–Corasick algorithm this takes O(n) time for a constant alphabet. Veriﬁcation: An area of length O(m) surrounding each potential occurrence found in the ﬁltration phase is searched using the standard dynamic programming algorithm in O(m2) time. The worst case time complexity is O(m2n), which can be reduced to O(mn) by combining any overlapping areas to be searched. 148 Let us analyze the average case time complexity of the veriﬁcation phase. • The best pattern partitioning is as even as possible. Then each pattern factor has length at least r = ⌊m/(k + 1)⌋. • The expected number of exact occurrences of a random string of length r in a random text of length n is at most n/σr. • The expected total veriﬁcation time is at most O m2(k + 1)n σr ≤O m3n σr . This is O(n) if r ≥3 logσ m. • The condition r ≥3 logσ m is satisﬁed when (k + 1) ≤m/(3 logσ m + 1). Theorem 3.24: The average case time complexity of the Baeza-Yates–Perleberg algorithm is O(n) when k ≤m/(3 logσ m + 1) −1. 149 Many variations of the algorithm have been suggested: • The ﬁltration can be done with a diﬀerent multiple exact string matching algorithm. • The veriﬁcation time can be reduced using a technique called hierarchical veriﬁcation. • The pattern can be partitioned into fewer than k + 1 pieces, which are searched allowing a small number of errors. A lower bound on the average case time complexity is Ω(n(k + logσ m)/m), and there exists a ﬁltering algorithm matching this bound. 150 Summary: Approximate String Matching We have seen two main types of algorithms for approximate string matching: • Basic dynamic programming time complexity is O(mn). The time complexity can be improved to O(kn) using diagonal monotonicity, and to O(n⌈m/w⌉) using bitparallelism. • Filtering algorithms can improve average case time complexity and are the fastest in practice when k is not too large. Similar techniques can be useful for other variants of edit distance but not always straightforwardly. 151 4. Suﬃx Trees and Arrays Let T = T[0..n) be the text. For i ∈[0..n], let Ti denote the suﬃx T[i..n). Furthermore, for any subset C ∈[0..n], we write TC = {Ti \| i ∈C}. In particular, T[0..n] is the set of all suﬃxes of T. Suﬃx tree and suﬃx array are search data structures for the set T[0..n]. • Suﬃx tree is a compact trie for T[0..n]. • Suﬃx array is an ordered array for T[0..n]. They support fast exact string matching on T: • A pattern P has an occurrence starting at position i if and only if P is a preﬁx of Ti. • Thus we can ﬁnd all occurrences of P by a preﬁx search in T[0..n]. A data structure supporting fast string matching is called a text index. There are numerous other applications too, as we will see later. 152 The set T[0..n] contains \|T[0..n]\| = n + 1 strings of total length \|\|T[0..n]\|\| = Θ(n2). It is also possible that ΣLCP(T[0..n]) = Θ(n2), for example, when T = an or T = XX for any string X. • A basic trie has Θ(n2) nodes for most texts, which is too much. • A compact trie with O(n) nodes and an ordered array with n + 1 entries have linear size. • A compact ternary trie has O(n) nodes too. However, the construction algorithms and some other algorithms we will see are not straightforward to adapt for it. Even for a compact trie or an ordered array, we need a specialized construction algorithm, because any general construction algorithm would need Ω(ΣLCP(T[0..n])) time. 153 Suﬃx Tree The suﬃx tree of a text T is the compact trie of the set T[0..n] of all suﬃxes of T. We assume that there is an extra character $ ̸∈Σ at the end of the text. That is, T[n] = $ and Ti = T[i..n] for all i ∈[0..n]. Then: • No suﬃx is a preﬁx of another suﬃx, i.e., the set T[0..n] is preﬁx free. • All nodes in the suﬃx tree representing a suﬃx are leaves. This simpliﬁes algorithms. Example 4.1: T = banana$. 1 3 5 6 2 4 0 $ $ $ na$ na $ na na$ banana$ a 154 As with tries, there are many possibilities for implementing the child operation. We again avoid this complication by assuming that σ is constant. Then the size of the suﬃx tree is O(n): • There are exactly n + 1 leaves and at most n internal nodes. • There are at most 2n edges. The edge labels are factors of the text and can be represented by pointers to the text. Given the suﬃx tree of T, all occurrences of P in T can be found in time O(\|P\| + occ), where occ is the number of occurrences. 155 Brute Force Construction Let us now look at algorithms for constructing the suﬃx tree. We start with a brute force algorithm with time complexity Θ(ΣLCP(T[0..n])). Later we will modify this algorithm to obtain a linear time complexity. The idea is to add suﬃxes to the trie one at a time starting from the longest suﬃx. The insertion procedure is essentially the same as we saw in Algorithm 1.2 (insertion into trie) except it has been modiﬁed to work on a compact trie instead of a trie. 156 Let Su denote the string represented by a node u. The suﬃx tree representation uses four functions: child(u, c) is the child v of node u such that the label of the edge (u, v) starts with the symbol c, and ⊥if u has no such child. parent(u) is the parent of u. depth(u) is the length of Su. start(u) is the starting position of some occurrence of Su in T. Then • Su = T[start(u) . . . start(u) + depth(u)). • T[start(u) + depth(parent(u)) . . . start(u) + depth(u)) is the label of the edge (parent(u), u). 157 A locus in the suﬃx tree is a pair (u, d) where depth(parent(u)) < d ≤depth(u). It represents • the uncompact trie node that would be at depth d along the edge (parent(u), u), and • the corresponding string S(u,d) = T[start(u) . . . start(u) + d). Every factor of T is a preﬁx of a suﬃx and thus has a locus along the path from the root to the leaf representing that suﬃx. During the construction, we need to create nodes at an existing locus in the middle of an edge, splitting the edge into two edges: CreateNode(u, d) // d < depth(u) (1) i ←start(u); p ←parent(u) (2) create new node v (3) start(v) ←i; depth(v) ←d (4) child(v, T[i + d]) ←u; parent(u) ←v (5) child(p, T[i + depth(p)]) ←v; parent(v) ←p (6) return v 158 Now we are ready to describe the construction algorithm. Algorithm 4.2: Brute force suﬃx tree construction Input: text T[0..n] (T[n] = $) Output: suﬃx tree of T: root, child, parent, depth, start (1) create new node root; depth(root) ←0 (2) u ←root; d ←0 // (u, d) is the active locus (3) for i ←0 to n do // insert suﬃx Ti (4) while d = depth(u) and child(u, T[i + d]) ̸= ⊥do (5) u ←child(u, T[i + d]); d ←d + 1 (6) while d < depth(u) and T[start(u) + d] = T[i + d] do d ←d + 1 (7) if d < depth(u) then // (u, d) is in the middle of an edge (8) u ←CreateNode(u, d) (9) CreateLeaf(i, u) (10) u ←root; d ←0 CreateLeaf(i, u) // Create leaf representing suﬃx Ti (1) create new leaf w (2) start(w) ←i; depth(w) ←n −i + 1 (3) child(u, T[i + d]) ←w; parent(w) ←u // Set u as parent (4) return w 159 Suﬃx Links The key to eﬃcient suﬃx tree construction are suﬃx links: slink(u) is the node v such that Sv is the longest proper suﬃx of Su, i.e., if Su = T[i..j) then Sv = T[i + 1..j). Example 4.3: The suﬃx tree of T = banana$ with internal node suﬃx links. 1 3 5 6 2 4 0 $ $ $ na$ na $ na na$ banana$ a 160 Suﬃx links are well deﬁned for all nodes except the root. Lemma 4.4: If the suﬃx tree of T has a node u representing T[i..j) for any 0 ≤i < j ≤n, then it has a node v representing T[i + 1..j). Proof. If u is the leaf representing the suﬃx Ti, then v is the leaf representing the suﬃx Ti+1. If u is an internal node, then it has two child edges with labels starting with diﬀerent symbols, say a and b, which means that T[i..j)a and T[i..j)b are both factors of T. Then, T[i + 1..j)a and T[i + 1..j)b are factors of T too, and thus there must be a branching node v representing T[i + 1..j). □ Usually, suﬃx links are needed only for internal nodes. For root, we deﬁne slink(root) = root. 161 Suﬃx links are the same as Aho–Corasick failure links but Lemma 4.4 ensures that depth(slink(u)) = depth(u) −1. This is not the case for an arbitrary trie or a compact trie. Suﬃx links are stored for compact trie nodes only, but we can deﬁne and compute them for any locus (u, d): slink(u, d) (1) v ←slink(parent(u)) (2) while depth(v) < d −1 do (3) v ←child(v, T[start(u) + depth(v) + 1]) (4) return (v, d −1) parent(u) (u, d) u slink(u) slink(u, d) slink(parent(u)) 162 The same idea can be used for computing the suﬃx links during or after the brute force construction. ComputeSlink(u) (1) d ←depth(u) (2) v ←slink(parent(u)) (3) while depth(v) < d −1 do (4) v ←child(v, T[start(u) + depth(v) + 1]) (5) if depth(v) > d −1 then // no node at (v, d −1) (6) v ←CreateNode(v, d −1) (7) slink(u) ←v The procedure CreateNode(v, d −1) sets slink(v) = ⊥. The algorithm uses the suﬃx link of the parent, which must have been computed before. Otherwise the order of computation does not matter. 163 The creation of a new node on line (6) is never needed in a fully constructed suﬃx tree, but during the brute force algorithm the necessary node may not exist yet: • If a new internal node ui was created during the insertion of the suﬃx Ti, there exists an earlier suﬃx Tj, j < i that branches at ui into a diﬀerent direction than Ti. • Then slink(ui) represents a preﬁx of Tj+1 and thus exists at least as a locus on the path labelled Tj+1. However, it might not become a branching node until the insertion of Ti+1. • In such a case, ComputeSlink(ui) creates slink(ui) a moment before it would otherwise be created by the brute force construction. 164 McCreight’s Algorithm McCreight’s suﬃx tree construction is a simple modiﬁcation of the brute force algorithm that computes the suﬃx links during the construction and uses them as short cuts: • Consider the situation, where we have just added a leaf wi representing the suﬃx Ti as a child to a node ui. The next step is to add wi+1 as a child to a node ui+1. • The brute force algorithm ﬁnds ui+1 by traversing from the root. McCreight’s algorithm takes a short cut to slink(ui). slink(ui) ui wi wi+1 ui+1 • This is safe because slink(ui) represents a preﬁx of Ti+1. 165 Algorithm 4.5: McCreight Input: text T[0..n] (T[n] = $) Output: suﬃx tree of T: root, child, parent, depth, start, slink (1) create new node root; depth(root) ←0; slink(root) ←root (2) u ←root; d ←0 // (u, d) is the active locus (3) for i ←0 to n do // insert suﬃx Ti (4) while d = depth(u) and child(u, T[i + d]) ̸= ⊥do (5) u ←child(u, T[i + d]); d ←d + 1 (6) while d < depth(u) and T[start(u) + d] = T[i + d] do d ←d + 1 (7) if d < depth(u) then // (u, d) is in the middle of an edge (8) u ←CreateNode(u, d) (9) CreateLeaf(i, u) (10) if slink(u) = ⊥then ComputeSlink(u) (11) u ←slink(u); d ←d −1 166 Theorem 4.6: Let T be a string of length n over an alphabet of constant size. McCreight’s algorithm computes the suﬃx tree of T in O(n) time. Proof. Insertion of a suﬃx Ti takes constant time except in two points: • The while loops on lines (4)–(6) traverse from the node slink(ui) to ui+1. Every round in these loops increments d. The only place where d decreases is on line (11) and even then by one. Since d can never exceed n, the total time on lines (4)–(6) is O(n). • The while loop on lines (3)–(4) during a call to ComputeSlink(ui) traverses from the node slink(parent(ui)) to slink(ui). Let d′ i be the depth of parent(ui). Clearly, d′ i+1 ≥d′ i −1, and every round in the while loop during ComputeSlink(ui) increases d′ i+1. Since d′ i can never be larger than n, the total time in the loop on lines (3)–(4) in ComputeSlink is O(n). □ 167 There are other linear time algorithms for suﬃx tree construction: • Weiner’s algorithm was the ﬁrst. It inserts the suﬃxes into the tree in the opposite order: Tn, Tn−1, . . . , T0. • Ukkonen’s algorithm constructs suﬃx tree ﬁrst for T[0..1) then for T[0..2), etc.. The algorithm is structured diﬀerently, but performs essentially the same tree traversal as McCreight’s algorithm. • All of the above are linear time only for constant alphabet size. Farach’s algorithm achieves linear time for an integer alphabet of polynomial size. The algorithm is complicated and unpractical. • Practical linear time construction for an integer alphabet is possible via suﬃx array. 168 Applications of Suﬃx Tree Let us have a glimpse of the numerous applications of suﬃx trees. Exact String Matching As already mentioned earlier, given the suﬃx tree of the text, all occ occurrences of a pattern P can be found in time O(\|P\| + occ). Even if we take into account the time for constructing the suﬃx tree, this is asymptotically as fast as Knuth–Morris–Pratt for a single pattern and Aho–Corasick for multiple patterns. However, the primary use of suﬃx trees is in indexed string matching, where we can aﬀord to spend a lot of time in preprocessing the text, but must then answer queries very quickly. 169 Approximate String Matching Several approximate string matching algorithms achieving O(kn) worst case time complexity are based on suﬃx trees. Filtering algorithms that reduce approximate string matching to exact string matching such as partitioning the pattern into k + 1 factors, can use suﬃx trees in the ﬁltering phase. Another approach is to generate all strings in the k-neighborhood of the pattern, i.e., all strings within edit distance k from the pattern and search for them in the suﬃx tree. The best practical algorithms for indexed approximate string matching are hybrids of the last two approaches. For example, partition the pattern into ℓ≤k + 1 factors and ﬁnd approximate occurrences of the factors with edit distance ⌊k/ℓ⌋using the neighborhood method in the ﬁltering phase. 170 Text Statistics Suﬃx tree is useful for computing all kinds of statistics on the text. For example: • Every locus in the suﬃx tree represents a factor of the text and, vice versa, every factor is represented by some locus. Thus the number of distinct factors in the text is exactly the number of distinct locuses, which can be computed by a traversal of the suﬃx tree in O(n) time even though the resulting value is typically Θ(n2). • The longest repeating factor of the text is the longest string that occurs at least twice in the text. It is represented by the deepest internal node in the suﬃx tree. 171 Generalized Suﬃx Tree A generalized suﬃx tree of two strings S and T is the suﬃx tree of the string S£T$, where £ and $ are symbols that do not occur elsewhere in S and T. Each leaf is marked as an S-leaf or a T-leaf according to the starting position of the suﬃx it represents. Using a depth ﬁrst traversal, we determine for each internal node if its subtree contains only S-leafs, only T-leafs, or both. The deepest node that contains both represents the longest common factor of S and T. It can be computed in linear time. The generalized suﬃx tree can also be deﬁned for more than two strings. 172 AC Automaton for the Set of Suﬃxes As already mentioned, a suﬃx tree with suﬃx links is essentially an Aho–Corasick automaton for the set of all suﬃxes. • We saw that it is possible to follow suﬃx link / failure transition from any locus, not just from suﬃx tree nodes. • Following such an implicit suﬃx link may take more than a constant time, but the total time during the scanning of a string with the automaton is linear in the length of the string. This can be shown with a similar argument as in the construction algorithm. Thus suﬃx tree is asymptotically as fast to operate as the AC automaton, but needs much less space. 173 Matching Statistics The matching statistics of a string S[0..n) with respect to a string T is an array MS[0..n), where MS[i] is a pair (ℓi, pi) such that 1. S[i..i + ℓi) is the longest preﬁx of Si that is a factor of T, and 2. T[pi..pi + ℓi) = S[i..i + ℓi). Matching statistics can be computed by using the suﬃx tree of T as an AC-automaton and scanning S with it. • If before reading S[i] we are at the locus (v, d) in the automaton, then S[i −d..i) = T[j..j + d), where j = start(v). If reading S[i] causes a failure transition, then MS[i −d] = (d, j). • Following the failure transition decrements d and thus increments i −d by one. Following a normal transition/edge, increments both i and d by one, and thus i −d stays the same. Thus all entries are computed. From the matching statistics, we can easily compute the longest common factor of S and T. Because we need the suﬃx tree only for T, this saves space compared to a generalized suﬃx tree. Matching statistics are also used in some approximate string matching algorithms. 174 LCA Preprocessing The lowest common ancestor (LCA) of two nodes u and v is the deepest node that is an ancestor of both u and v. Any tree can be preprocessed in linear time so that the LCA of any two nodes can be computed in constant time. The details are omitted here. • Let wi and wj be the leaves of the suﬃx tree of T that represent the suﬃxes Ti and Tj. The lowest common ancestor of wi and wj represents the longest common preﬁx of Ti and Tj. Thus lcp(Ti, Tj) = depth(LCA(wi, wj)) , which can be computed in constant time using the suﬃx tree with LCA preprocessing. • The longest common preﬁx of two suﬃxes Si and Tj from two diﬀerent strings S and T is called the longest common extension. Using the generalized suﬃx tree with LCA preprocessing, the longest common extension for any pair of suﬃxes can be computed in constant time. Some O(kn) worst case time approximate string matching algorithms use longest common extension data structures. 175 Longest Palindrome A palindrome is a string that is its own reverse. For example, saippuakauppias is a palindrome. We can use the LCA preprocessed generalized suﬃx tree of a string T and its reverse T R to ﬁnd the longest palindrome in T in linear time. • Let ki be the length of the longest common extension of Ti+1 and T R n−i, which can be computed in constant time. Then T[i −ki..i + ki] is the longest odd length palindrome with the middle at i. • We can ﬁnd the longest odd length palindrome by computing ki for all i ∈[0..n) in O(n) time. • The longest even length palindrome can be found similarly in O(n) time. The longest palindrome overall is the longer of the two. 176 Suﬃx Array The suﬃx array of a text T is a lexicographically ordered array of the set T[0..n] of all suﬃxes of T. More precisely, the suﬃx array is an array SA[0..n] of integers containing a permutation of the set [0..n] such that TSA < TSA < · · · < TSA[n]. A related array is the inverse suﬃx array SA−1 which is the inverse permutation, i.e., SA−1[SA[i]] = i for all i ∈[0..n]. The value SA−1[j] is the lexicographical rank of the suﬃx Tj As with suﬃx trees, it is common to add the end symbol T[n] = $. It has no eﬀect on the suﬃx array assuming $ is smaller than any other symbol. Example 4.7: The suﬃx array and the inverse suﬃx array of the text T = banana$. i SA[i] TSA[i] 0 6 $ 1 5 a$ 2 3 ana$ 3 1 anana$ 4 0 banana$ 5 4 na$ 6 2 nana$ j SA−1[j] 0 4 banana$ 1 3 anana$ 2 6 nana$ 3 2 ana$ 4 5 na$ 5 1 a$ 6 0 $ 177 Suﬃx array is much simpler data structure than suﬃx tree. In particular, the type and the size of the alphabet are usually not a concern. • The size on the suﬃx array is O(n) on any alphabet. • We will later see that the suﬃx array can be constructed in the same asymptotic time it takes to sort the characters of the text. Suﬃx array construction algorithms are quite fast in practice too. Probably the fastest way to construct a suﬃx tree is to construct a suﬃx array ﬁrst and then use it to construct the suﬃx tree. (We will see how in a moment.) Suﬃx arrays are rarely used alone but are augmented with other arrays and data structures depending on the application. We will see some of them in the next slides. 178 Exact String Matching As with suﬃx trees, exact string matching in T can be performed by a preﬁx search on the suﬃx array. The answer can be conveniently given as a contiguous interval SA[b..e) that contains the suﬃxes with the given preﬁx. The interval can be found using string binary search. • If we have the additional arrays LLCP and RLCP, the result interval can be computed in O(\|P\| + log n) time. • Without the additional arrays, we have the same time complexity on average but the worst case time complexity is O(\|P\| log n). • We can then count the number of occurrences in O(1) time, list all occ occurrences in O(occ) time, or list a sample of k occurrences in O(k) time. An alternative algorithm for computing the interval SA[b..e) is called backward search. It is commonly used with compressed representations of suﬃx arrays and will be covered in the course Data Compression Techniques. 179 LCP Array Eﬃcient string binary search uses the arrays LLCP and RLCP. However, for many applications, the suﬃx array is augmented with the lcp array of Deﬁnition 1.11 (Lecture 2). For all i ∈[1..n], we store LCP[i] = lcp(TSA[i], TSA[i−1]) Example 4.8: The LCP array for T = banana$. i SA[i] LCP[i] TSA[i] 0 6 $ 1 5 0 a$ 2 3 1 ana$ 3 1 3 anana$ 4 0 0 banana$ 5 4 0 na$ 6 2 2 nana$ 180 Using the solution of Exercise 2.4 (construction of compact trie from sorted array and LCP array), the suﬃx tree can be constructed from the suﬃx and LCP arrays in linear time. However, many suﬃx tree applications can be solved using the suﬃx and LCP arrays directly. For example: • The longest repeating factor is marked by the maximum value in the LCP array. • The number of distinct factors can be compute by the formula n(n + 1) 2 + 1 − n X i=1 LCP[i] since it equals the number of nodes in the uncompact suﬃx trie, for which we can use Theorem 1.17. • Matching statistics of S with respect to T can be computed in linear time using the generalized suﬃx array of S and T (i.e., the suﬃx array of S£T$) and its LCP array (exercise). 181 LCP Array Construction The LCP array is easy to compute in linear time using the suﬃx array SA and its inverse SA−1. The idea is to compute the lcp values by comparing the suﬃxes, but skip a preﬁx based on a known lower bound for the lcp value obtained using the following result. Lemma 4.9: For any i ∈[0..n), LCP[SA−1[i]] ≥LCP[SA−1[i −1]] −1 Proof. For each j ∈[0..n), let Φ(j) = SA[SA−1[j] −1]. Then TΦ(j) is the immediate lexicographical predecessor of Tj and LCP[SA−1[j]] = lcp(Tj, TΦ(j)). • Let ℓ= LCP[SA−1[i −1]] and ℓ′ = LCP[SA−1[i]]. We want to show that ℓ′ ≥ℓ−1. If ℓ= 0, the claim is trivially true. • If ℓ> 0, then for some symbol c, Ti−1 = cTi and TΦ(i−1) = cTΦ(i−1)+1. Thus TΦ(i−1)+1 < Ti and lcp(Ti, Tφ(i−1)+1) = lcp(Ti−1, TΦ(i−1)) −1 = ℓ−1. • If Φ(i) = Φ(i −1) + 1, then ℓ′ = lcp(Ti, TΦ(i)) = lcp(Ti, TΦ(i−1)+1) = ℓ−1. • If Φ(i) ̸= Φ(i −1) + 1, then TΦ(i−1)+1 < TΦ(i) < Ti and ℓ′ = lcp(Ti, TΦ(i)) ≥lcp(Ti, TΦ(i−1)+1) = ℓ−1. □ 182 The algorithm computes the lcp values in the order that makes it easy to use the above lower bound. Algorithm 4.10: LCP array construction Input: text T[0..n], suﬃx array SA[0..n], inverse suﬃx array SA−1[0..n] Output: LCP array LCP[1..n] (1) ℓ←0 (2) for i ←0 to n −1 do (3) k ←SA−1[i] (4) j ←SA[k −1] // j = Φ(i) (5) while T[i + ℓ] = T[j + ℓ] do ℓ←ℓ+ 1 (6) LCP[k] ←ℓ (7) if ℓ> 0 then ℓ←ℓ−1 (8) return LCP The time complexity is O(n): • Everything except the while loop on line (5) takes clearly linear time. • Each round in the loop increments ℓ. Since ℓis decremented at most n times on line (7) and cannot grow larger than n, the loop is executed O(n) times in total. 183 RMQ Preprocessing The range minimum query (RMQ) asks for the smallest value in a given range in an array. Any array can be preprocessed in linear time so that RMQ for any range can be answered in constant time. In the LCP array, RMQ can be used for computing the lcp of any two suﬃxes. Lemma 4.11: The length of the longest common preﬁx of two suﬃxes Ti < Tj is lcp(Ti, Tj) = min{LCP[k] \| k ∈[SA−1[i] + 1..SA−1[j]]}. The lemma can be seen as a generalization of Lemma 1.31 (Lecture 3) and holds for any sorted array of strings. The proof is left as an exercise. • The RMQ preprocessing of the LCP array supports the same kind of applications as the LCA preprocessing of the suﬃx tree, but RMQ preprocessing is simpler than LCA preprocessing. • The RMQ preprocessed LCP array can also replace the LLCP and RLCP arrays in binary searching. 184 We will next describe the RMQ data structure for an arbitrary array L[1..n] of integers. • We precompute and store the minimum values for the following collection of ranges: – Divide L[1..n] into blocks of size log n. – For all 0 ≤ℓ≤log(n/ log n)), include all ranges that consist of 2ℓ blocks. There are O(log n · n log n) = O(n) such ranges. – Include all preﬁxes and suﬃxes of blocks. There are a total of O(n) of them. • Now any range L[i..j] that overlaps or touches a block boundary can be exactly covered by at most four ranges in the collection. The minimum value in L[i..j] is the minimum of the minimums of the covering ranges and can be computed in constant time. 185 Ranges L[i..j] that are completely inside one block are handled diﬀerently. • Let NSV (i) = min{k > i \| L[k] < L[i]} (NSV=Next Smaller Value). Then the position of the minimum value in the range L[i..j] is the last position in the sequence i, NSV (i), NSV (NSV (i)), . . . that is in the range. We call these the NSV positions for i. • For each i, store the NSV positions for i up to the end of the block containing i as a bit vector B(i). Each bit corresponds to a position within the block and is one if it is an NSV position. The size of B(i) is log n bits and we can assume that it ﬁts in a single machine word. Thus we need O(n) words to store B(i) for all i. • The position of the minimum in L[i..j] is found as follows: – Turn all bits in B(i) after position j into zeros. This can be done in constant time using bitwise shift -operations. – The right-most 1-bit indicates the position of the minimum. It can be found in constant time using a lookup table of size O(n). All the data structures can be constructed in O(n) time (exercise). 186 Enhanced Suﬃx Array The enhanced suﬃx array adds two more arrays to the suﬃx and LCP arrays to make the data structure fully equivalent to suﬃx tree. • The idea is to represent a suﬃx tree node v representing a factor Sv by the suﬃx array interval of the suﬃxes that begin with Sv. That interval contains exactly the suﬃxes that are in the subtree rooted at v. • The additional arrays support navigation in the suﬃx tree using this representation: one array along the regular edges, the other along suﬃx links. With all the additional arrays the suﬃx array is not very space eﬃcient data structure any more. Nowadays suﬃx arrays and trees are often replaced with compressed text indexes that provide the same functionality in much smaller space. These will be covered in the course Data Compression Techniques. 187 Burrows–Wheeler Transform The Burrows–Wheeler transform (BWT) is an important technique for text compression, text indexing, and their combination compressed text indexing. Let T[0..n] be the text with T[n] = $. For any i ∈[0..n], T[i..n]T[0..i) is a rotation of T. Let M be the matrix, where the rows are all the rotations of T in lexicographical order. All columns of M are permutations of T. In particular: • The ﬁrst column F contains the text characters in order. • The last column L is the BWT of T. Example 4.12: The BWT of T = banana$ is L = annb$aa. F L $ b a n a n a a $ b a n a n a n a $ b a n a n a n a $ b b a n a n a $ n a $ b a n a n a n a $ b a 188 Here are some of the key properties of the BWT. • The BWT is easy to compute using the suﬃx array: L[i] = $ if SA[i] = 0 T[SA[i] −1] otherwise • The BWT is invertible, i.e., T can be reconstructed from the BWT L alone. The inverse BWT can be computed in the same time it takes to sort the characters. • The BWT L is typically easier to compress than the text T. Many text compression algorithms are based on compressing the BWT. • The BWT supports backward searching, a diﬀerent technique for indexed exact string matching. This is used in many compressed text indexes. BWT will be covered in more detail in the course Data Compression Techniques. 189 Suﬃx Array Construction Suﬃx array construction means simply sorting the set of all suﬃxes. • Using standard sorting or string sorting the time complexity is Ω(ΣLCP(T[0..n])). • Another possibility is to ﬁrst construct the suﬃx tree and then traverse it from left to right to collect the suﬃxes in lexicographical order. The time complexity is O(n) on a constant alphabet. Specialized suﬃx array construction algorithms are a better option, though. 190 Preﬁx Doubling Our ﬁrst specialized suﬃx array construction algorithm is a conceptually simple algorithm achieving O(n log n) time. Let T ℓ i denote the text factor T[i.. min{i + ℓ, n + 1}) and call it an ℓ-factor. In other words: • T ℓ i is the factor starting at i and of length ℓexcept when the factor is cut short by the end of the text. • T ℓ i is the preﬁx of the suﬃx Ti of length ℓ, or Ti when \|Ti\| < ℓ. The idea is to sort the sets T ℓ [0..n] for ever increasing values of ℓ. • First sort T 1 [0..n], which is equivalent to sorting individual characters. This can be done in O(n log n) time. • Then, for ℓ= 1, 2, 4, 8, . . . , use the sorted set T ℓ [0..n] to sort the set T 2ℓ [0..n] in O(n) time. • After O(log n) rounds, ℓ> n and T ℓ [0..n] = T[0..n], so we have sorted the set of all suﬃxes. 191 We still need to specify, how to use the order for the set T ℓ [0..n] to sort the set T 2ℓ [0..n]. The key idea is assigning order preserving names (lexicographical names) for the factors in T ℓ [0..n]. For i ∈[0..n], let Nℓ i be an integer in the range [0..n] such that, for all i, j ∈[0..n]: Nℓ i ≤Nℓ j if and only if T ℓ i ≤T ℓ j . Then, for ℓ> n, Nℓ i = SA−1[i]. For smaller values of ℓ, there can be many ways of satisfying the conditions and any one of them will do. A simple choice is Nℓ i = \|{j ∈[0, n] \| T ℓ j < T ℓ i }\| . Example 4.13: Preﬁx doubling for T = banana$. N1 4 b 1 a 5 n 1 a 5 n 1 a 0 $ N2 4 ba 2 an 5 na 2 an 5 na 1 a$ 0 $ N4 4 bana 3 anan 6 nana 2 ana$ 5 na$ 1 a$ 0 $ N8 = SA−1 4 banana$ 3 anana$ 6 nana$ 2 ana$ 5 na$ 1 a$ 0 $ 192 Now, given Nℓ, for the purpose of sorting, we can use • Nℓ i to represent T ℓ i • the pair (Nℓ i , Nℓ i+ℓ) to represent T 2ℓ i = T ℓ i T ℓ i+ℓ. Thus we can sort T 2ℓ [0..n] by sorting pairs of integers, which can be done in O(n) time using LSD radix sort. Theorem 4.14: The suﬃx array of a string T[0..n] can be constructed in O(n log n) time using preﬁx doubling. • The technique of assigning order preserving names to factors whose lengths are powers of two is called the Karp–Miller–Rosenberg naming technique. It was developed for other purposes in the early seventies when suﬃx arrays did not exist yet. • The best practical variant is the Larsson–Sadakane algorithm, which uses ternary quicksort instead of LSD radix sort for sorting the pairs, but still achieves O(n log n) total time. 193 Let us return to the ﬁrst phase of the preﬁx doubling algorithm: assigning names N1 i to individual characters. This is done by sorting the characters, which is easily within the time bound O(n log n), but sometimes we can do it faster: • On an ordered alphabet, we can use ternary quicksort for time complexity O(n log σT) where σT is the number of distinct symbols in T. • On an integer alphabet of size nc for any constant c, we can use LSD radix sort with radix n for time complexity O(n). After this, we can replace each character T[i] with N1 i to obtain a new string T ′: • The characters of T ′ are integers in the range [0..n]. • The character T ′[n] = 0 is the unique, smallest symbol, i.e., $. • The suﬃx arrays of T and T ′ are exactly the same. Thus we can construct the suﬃx array using T ′ as the text instead of T. As we will see next, the suﬃx array of T ′ can be constructed in linear time. Then sorting the characters of T to obtain T ′ is the asymptotically most expensive operation in the suﬃx array construction of T for any alphabet. 194 Recursive Suﬃx Array Construction Let us now describe linear time algorithms for suﬃx array construction. We assume that the alphabet of the text T[0..n) is [1..n] and that T[n] = 0 (=$ in the examples). The outline of the algorithms is: 0. Choose a subset C ⊂[0..n]. 1. Sort the set TC. This is done as follows: (a) Construct a reduced string R of length \|C\|, whose characters are order preserving names of text factors starting at the positions in C. (b) Construct the suﬃx array of R recursively. 2. Sort the set T[0..n] using the order of TC. 195 Assume that • \|C\| ≤αn for a constant α < 1, and • excluding the recursive call, all steps in the algorithm take linear time. Then the total time complexity can be expressed as the recurrence t(n) = O(n) + t(αn), whose solution is t(n) = O(n). To make the scheme work, the set C must satisfy two nontrivial conditions: 1. There exists an appropriate reduced string R. 2. Given sorted TC the suﬃx array of T is easy to construct. Finding sets C that satisfy both conditions is diﬃcult, but there are two diﬀerent methods leading to two diﬀerent algorithms: • DC3 uses diﬀerence cover sampling • SAIS uses induced sorting 196 Diﬀerence Cover Sampling A diﬀerence cover Dq modulo q is a subset of [0..q) such that all values in [0..q) can be expressed as a diﬀerence of two elements in Dq modulo q. In other words: [0..q) = {i −j mod q \| i, j ∈Dq} . Example 4.15: D7 = {1, 2, 4} 1 −1 = 0 1 −4 = −3 ≡4 (mod q) 2 −1 = 1 2 −4 = −2 ≡5 (mod q) 4 −2 = 2 1 −2 = −1 ≡6 (mod q) 4 −1 = 3 In general, we want the smallest possible diﬀerence cover for a given q. • For any q, there exist a diﬀerence cover Dq of size O(√q). • The DC3 algorithm uses the simplest non-trivial diﬀerence cover D3 = {1, 2}. 197 A diﬀerence cover sample is a set TC of suﬃxes, where C = {i ∈[0..n] \| (i mod q) ∈Dq} . Example 4.16: If T = banana$ and D3 = {1, 2}, then C = {1, 2, 4, 5} and TC = {anana$, nana$, na$, a$}. Once we have sorted the diﬀerence cover sample TC, we can compare any two suﬃxes in O(q) time. To compare suﬃxes Ti and Tj: • If i ∈C and j ∈C, then we already know their order from TC. • Otherwise, ﬁnd ℓsuch that i + ℓ∈C and j + ℓ∈C. There always exists such ℓ∈[0..q). Then compare: Ti = T[i..i + ℓ)Ti+ℓ Tj = T[j..j + ℓ)Tj+ℓ That is, compare ﬁrst T[i..i + ℓ) to T[j..j + ℓ), and if they are the same, then Ti+ℓto Tj+ℓusing the sorted TC. Example 4.17: D3 = {1, 2} and C = {1, 2, 4, 5, . . . } T0 = TT1 T1 = TT2 T0 = TTT2 T2 = TTT4 T0 = TT1 T3 = TT4 198 Algorithm 4.18: DC3 Step 0: Choose C. • Use diﬀerence cover D3 = {1, 2}. • For k ∈{0, 1, 2}, deﬁne Ck = {i ∈[0..n] \| i mod 3 = k}. • Let C = C1 ∪C2 and ¯ C = C0. Example 4.19: i 0 1 2 3 4 5 6 7 8 9 10 11 12 T[i] y a b b a d a b b a d o $ ¯ C = C0 = {0, 3, 6, 9, 12}, C1 = {1, 4, 7, 10}, C2 = {2, 5, 8, 11} and C = {1, 2, 4, 5, 7, 8, 10, 11}. 199 Step 1: Sort TC. • For k ∈{1, 2}, Construct the strings Rk = (T 3 k , T 3 k+3, T 3 k+6, . . . , T 3 max Ck) whose characters are 3-factors of the text, and let R = R1R2. • Replace each factor T 3 i in R with an order preserving name N3 i ∈[1..\|R\|]. The names can be computed by sorting the factors with LSD radix sort in O(n) time. Let R′ be the result appended with 0. • Construct the inverse suﬃx array SA−1 R′ of R′. This is done recursively using DC3 unless all symbols in R′ are unique, in which case SA−1 R′ = R′. • From SA−1 R′ , we get order preserving names for suﬃxes in TC. For i ∈C, let Ni = SA−1 R′ [j], where j is the position of T 3 i in R. For i ∈¯ C, let Ni = ⊥. Also let Nn+1 = Nn+2 = 0. Example 4.20: R abb ada bba do$ bba dab bad o$ R′ 1 2 4 7 4 6 3 8 0 SA−1 R′ 1 2 5 7 4 6 3 8 0 i 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 T[i] y a b b a d a b b a d o $ Ni ⊥ 1 4 ⊥ 2 6 ⊥ 5 3 ⊥ 7 8 ⊥ 0 0 200 Step 2(a): Sort T ¯ C. • For each i ∈¯ C, we represent Ti with the pair (T[i], Ni+1). Then Ti ≤Tj ⇐ ⇒(T[i], Ni+1) ≤(T[j], Nj+1) . Note that Ni+1 ̸= ⊥for all i ∈¯ C. • The pairs (T[i], Ni+1) are sorted by LSD radix sort in O(n) time. Example 4.21: i 0 1 2 3 4 5 6 7 8 9 10 11 12 T[i] y a b b a d a b b a d o $ Ni ⊥ 1 4 ⊥ 2 6 ⊥ 5 3 ⊥ 7 8 ⊥ T12 < T6 < T9 < T3 < T0 because ($, 0) < (a, 5) < (a, 7) < (b, 2) < (y, 1). 201 Step 2(b): Merge TC and T ¯ C. • Use comparison based merging algorithm needing O(n) comparisons. • To compare Ti ∈TC and Tj ∈T ¯ C, we have two cases: i ∈C1 : Ti ≤Tj ⇐ ⇒(T[i], Ni+1) ≤(T[j], Nj+1) i ∈C2 : Ti ≤Tj ⇐ ⇒(T[i], T[i + 1], Ni+2) ≤(T[j], T[j + 1], Nj+2) Note that none of the N-values is ⊥. Example 4.22: i 0 1 2 3 4 5 6 7 8 9 10 11 12 T[i] y a b b a d a b b a d o $ Ni ⊥ 1 4 ⊥ 2 6 ⊥ 5 3 ⊥ 7 8 ⊥ T1 < T6 because (a, 4) < (a, 5). T3 < T8 because (b, a, 6) < (b, a, 7). 202 Theorem 4.23: Algorithm DC3 constructs the suﬃx array of a string T[0..n) in O(n) time plus the time needed to sort the characters of T. There are many variants: • DC3 is an optimal algorithm under several parallel and external memory computation models, too. There exists both parallel and external memory implementations of DC3. • Using a larger value of q, we obtain more space eﬃcient algorithms. For example, using q = log n, the time complexity is O(n log n) and the space needed in addition to the text and the suﬃx array is O(n/√log n). 203 Induced Sorting Deﬁne three type of suﬃxes −, + and ∗as follows: C−= {i ∈[0..n) \| Ti > Ti+1} C+ = {i ∈[0..n) \| Ti < Ti+1} C∗= {i ∈C+ \| i −1 ∈C−} Example 4.24: i 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 T[i] m m i s s i s s i i p p i i $ type of Ti − − ∗ − − ∗ − − ∗ + − − − − For every a ∈Σ and x ∈{−, +.∗} deﬁne Ca = {i ∈[0..n] \| T[i] = a} Cx a = Ca ∩Cx Then C− a = {i ∈Ca \| Ti < a∞} C+ a = {i ∈Ca \| Ti > a∞} and thus, if i ∈C− a and j ∈C+ a , then Ti < Tj. Hence the suﬃx array is nC1C2 . . . Cσ−1 = nC− 1 C+ 1 C− 2 C+ 2 . . . C− σ−1C+ σ−1. 204 The basic idea of induced sorting is to use information about the order of Ti to induce the order of the suﬃx Ti−1 = T[i −1]Ti. The main steps are: 1. Sort the sets C∗ a, a ∈[1..σ). 2. Use C∗ a, a ∈[1..σ), to induce the order of the sets C− a , a ∈[1..σ). 3. Use C− a , a ∈[1..σ), to induce the order of the sets C+ a , a ∈[1..σ). The suﬃxes involved in the induction steps can be indentiﬁed using the following rules (proof is left as an exercise). Lemma 4.25: For all a ∈[1..σ) (a) i −1 ∈C− a iﬀi > 0 and T[i −1] = a and one of the following holds 1. i = n 2. i ∈C∗ 3. i ∈C−and T[i −1] ≥T[i]. (b) i −1 ∈C+ a iﬀi > 0 and T[i −1] = a and one of the following holds 1. i ∈C−and T[i −1] < T[i] 2. i ∈C+ and T[i −1] ≤T[i]. 205 To induce C−suﬃxes: 1. Set C− a empty for all a ∈[1..σ). 2. For all suﬃxes Ti such that i −1 ∈C−in lexicographical order, append i −1 into C− T[i−1]. By Lemma 4.25(a), Step 2 can be done by checking the relevant conditions for all i ∈nC− 1 C∗ 1C− 2 C∗ 2 . . . . Algorithm 4.26: InduceMinusSuﬃxes Input: Lexicographically sorted lists C∗ a, a ∈Σ Output: Lexicographically sorted lists C− a , a ∈Σ (1) for a ∈Σ do C− a ←∅ (2) pushback(n −1, C− T[n−1]) (3) for a ←1 to σ −1 do (4) for i ∈C− a do // include elements added during the loop (5) if i > 0 and T[i −1] ≥a then pushback(i −1, C− T[i−1]) (6) for i ∈C∗ a do pushback(i −1, C− T[i−1]) Note that since Ti−1 > Ti by deﬁnition of C−, we always have i inserted before i −1. 206 Inducing +-type suﬃxes goes similarly but in reverse order so that again i is always inserted before i −1: 1. Set C+ a empty for all a ∈[1..σ). 2. For all suﬃxes Ti such that i −1 ∈C+ in descending lexicographical order, append i −1 into C+ T[i−1]. Algorithm 4.27: InducePlusSuﬃxes Input: Lexicographically sorted lists C− a , a ∈Σ Output: Lexicographically sorted lists C+ a , a ∈Σ (1) for a ∈Σ do C+ a ←∅ (2) for a ←σ −1 downto 1 do (3) for i ∈C+ a in reverse order do // include elements added during loop (4) if i > 0 and T[i −1] ≤a then pushfront(i −1, C+ T[i−1]) (5) for i ∈C− a in reverse order do (6) if i > 0 and T[i −1] < a then pushfront(i −1, C+ T[i−1]) 207 We still need to explain how to sort the ∗-type suﬃxes. Deﬁne F[i] = min{k ∈[i + 1..n] \| k ∈C∗or k = n} Si = T[i..F[i]] S′ i = Siσ where σ is a special symbol larger than any other symbol. Lemma 4.28: For any i, j ∈[0..n), Ti < Tj iﬀS′ i < S′ j or S′ i = S′ j and TF[i] < TF[j]. Proof. The claim is trivially true except in the case that Sj is a proper preﬁx of Si (or vice versa). In that case, Si > Sj but S′ i < S′ j and thus Ti < Tj by the claim. We will show that this is correct. Let ℓ= F[j] and k = i + ℓ−j. Then • ℓ∈C∗and thus ℓ−1 ∈C−. By Lemma 4.25(b), T[ℓ−1] > T[ℓ]. • T[k −1..k] = T[ℓ−1..ℓ] and thus T[k −1] > T[k]. If we had k ∈C+, we would have k ∈C∗. Since this is not the case, we must have k ∈C−. • Let a = T[ℓ]. Since ℓ∈C+ a and k ∈C− a , we must have Tk < a∞< Tℓ. • Since T[i..k) = T[j..ℓ) and Tk < Tℓ, we have Ti < Tj. □ 208 Algorithm 4.29: SAIS Step 0: Choose C. • Compute the types of suﬃxes. This can be done in O(n) time based on Lemma 4.25. • Set C = ∪a∈[1..σ)C∗ a ∪{n}. Note that \|C\| ≤n/2, since for all i ∈C, i −1 ∈C−⊆¯ C. Example 4.30: i 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 T[i] m m i s s i s s i i p p i i $ type of Ti − − ∗ − − ∗ − − ∗ + − − − − C∗ i = {2, 5, 8}, C∗ m = C∗ p = C∗ s = ∅, C = {2, 5, 8, 14}. 209 Step 1: Sort TC. • Sort the strings S′ i, i ∈C∗. Since the total length of the strings S′ i is O(n), the sorting can be done in O(n) time using LSD radix sort. • Assign order preserving names Ni ∈[1..\|C\| −1] to the string S′ i so that Ni ≤Nj iﬀS′ i ≤S′ j. • Construct the sequence R = Ni1Ni2 . . . Nk0, where i1 < i3 < · · · < ik are the -type positions. • Construct the suﬃx array SAR of R. This is done recursively unless all symbols in R are unique, in which case a simple counting sort is suﬃcient. • The order of the suﬃxes of R corresponds to the order of ∗-type suﬃxes of T. Thus we can construct the lexicographically ordered lists C∗ a, a ∈[1..σ). Example 4.31: i 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 T[i] m m i s s i s s i i p p i i $ Ni 2 2 1 0 R = [issiσ][issiσ][iippii$σ]$ = 2210, SAR = (3, 2, 1, 0), C∗ i = (8, 5, 2) 210 Step 2: Sort T[0..n]. • Run InduceMinusSuﬃxes to construct the sorted lists C− a , a ∈[1..σ). • Run InducePlusSuﬃxes to construct the sorted lists C+ a , a ∈[1..σ). • The suﬃx array is SA = nC− 1 C+ 1 C− 2 C+ 2 . . . C− σ−1C+ σ−1. Example 4.32: i 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 T[i] m m i s s i s s i i p p i i $ type of Ti − − ∗ − − ∗ − − ∗ + − − − − n = 14 ⇒ C− i = (13, 12) C− i C∗ i = (13, 12, 8, 5, 2) ⇒ C− m = (1, 0), C− p = (11, 10), C− s = (7, 4, 6, 3) ⇒ C+ i = (8, 9, 5, 2) ⇒ SA = C$C− i C+ i C− m C− p C− s = (14, 13, 12, 8, 9, 5, 2, 1, 0, 11, 10, 7, 4, 6, 3) 211 Theorem 4.33: Algorithm SAIS constructs the suﬃx array of a string T[0..n) in O(n) time plus the time needed to sort the characters of T. • In Step 1, to sort the strings S′ i, i ∈C∗, SAIS does not actually use LSD radix sort but the following procedure: 1. Construct the sets C∗ a, a ∈[1..σ) in arbitrary order. 2. Run InduceMinusSuﬃxes to construct the lists C− a , a ∈[1..σ). 3. Run InducePlusSuﬃxes to construct the lists C− a , a ∈[1..σ). 4. Remove non--type positions from C+ 1 C+ 2 . . . C+ σ−1. With this change, most of the work is done in the induction procedures. This is very fast in practice, because all the lists Cx a are accessed sequentially during the procedures. • The currently fastest suﬃx sorting implementation in practice is probably divsufsort by Yuta Mori. It sorts the -type suﬃxes non-recursively in O(n log n) time and then continues as SAIS. 212 Summary: Suﬃx Trees and Arrays The most important data structures for string processing: • Designed for indexed exact string matching. • Used in eﬃcient solutions to a huge variety of diﬀerent problems. Construction algorithms are among the most important algorithms for string processing: • Linear time for constant and integer alphabet. Often augmented with additional data structures: • suﬃx links, LCA preprocessing • LCP array, RMQ preprocessing, BWT, ... 213
15713	https://my.clevelandclinic.org/health/diseases/12100-sickle-cell-disease	Sickle Cell Disease (SCD): Types, Symptoms & Causes Locations: Abu Dhabi\|Canada\|Florida\|London\|Nevada\|Ohio\| 800.223.2273 100 Years of Cleveland Clinic MyChart Need Help? Giving Careers Search ClevelandClinic.org Find A Doctor Locations & Directions Patients & Visitors Health Library Institutes & Departments Appointments Home/ Health Library/ Diseases & Conditions/ Sickle Cell Disease Advertisement Advertisement Sickle Cell Disease Sickle cell disease affects the hemoglobin within your red blood cells. A genetic mutation causes abnormal hemoglobin to clump together, causing the red blood cells to turn sickle shaped. These sickle-shaped cells cause blockages in your blood flow, which can lead to anemia, pain, infections and severe complications. Advertisement Cleveland Clinic is a non-profit academic medical center. Advertising on our site helps support our mission. We do not endorse non-Cleveland Clinic products or services.Policy Care at Cleveland Clinic Sickle Cell Disease Treatment Find a Doctor and Specialists Make an Appointment ContentsOverviewSymptoms and CausesDiagnosis and TestsManagement and TreatmentPreventionOutlook / PrognosisLiving WithAdditional Common Questions ContentsOverviewSymptoms and CausesDiagnosis and TestsManagement and TreatmentPreventionOutlook / PrognosisLiving WithAdditional Common Questions Overview Sickle cell disease causes red blood cells to stiffen and change shape, which can cause pain and inflammation. What is sickle cell disease? Sickle cell disease (SCD) is the most common inherited blood disorder that affects your red blood cells. In SCD, a protein called hemoglobin, located within red blood cells, is abnormal. Hemoglobin is important. It’s the molecule that carries oxygen in your blood and throughout your body. Advertisement Cleveland Clinic is a non-profit academic medical center. Advertising on our site helps support our mission. We do not endorse non-Cleveland Clinic products or services.Policy Normal red blood cells are round and flexible. This allows them to move easily through small blood vessels (capillaries) in your body to deliver oxygen to your organs and tissues. In SCD, there’s an abnormal form of hemoglobin called hemoglobin S. This changes the shape of the red blood cell to a crescent shape and causes red blood cells to become rigid, lack flexibility and stick together. This can block blood flow, preventing oxygen from getting to the vital organs and tissues throughout your body. It can lead to serious complications including pain, infections, and organ damage and failure. Additionally, sickle-shaped cells don’t last as long as normal-shaped red blood cells, causing a constant shortage of red blood cells and leading to anemia. Sickle cell disease is a lifelong condition. But there are treatment options that can reduce your symptoms and prolong your life. Types of sickle cell disease There are several types of sickle cell disease. The different types depend on the genes a person inherits from their parents. Hemoglobin SS (HbSS) HbSS is a severe form, affecting 65% of people who have SCD. People with this form inherited one gene encoded with hemoglobin S from each parent. Most or all of your hemoglobin is abnormal, causing chronic anemia. Advertisement Hemoglobin SC (HbSC) HbSC is a mild to moderate form that affects about 25% of people with the disease. People with this form inherited a hemoglobin S gene from one parent. They inherited another abnormal type — hemoglobin C — from their other parent. Hemoglobin (HbS) beta thalassemia People with this form inherited a hemoglobin S gene from one parent. They inherited an abnormal type called beta thalassemia from their other parent. There are two subtypes: “Plus” (HbS beta +): This subtype affects about 8% of people with SCD and tends to be more mild. “Zero” (HbS beta 0): This subtype affects about 2% of people with SCD and is more severe, similar to hemoglobin SS disease. There are other, more rare forms, including hemoglobin SD (HbSD), hemoglobin SE (HbSE) and hemoglobin SO (HbSO). People with one of these forms inherited one hemoglobin S gene and one gene that encodes with another abnormal gene (D, E or O). What’s the difference between sickle cell anemia and sickle cell disease? Copy debug info Play Play 00:00 Play Seek 10 seconds backwards Seek 10 seconds forward 00:00 / 00:00 Unmute Settings Picture in picture Fullscreen Rabi Hanna, MD, explains what sickle cell disease is. Sickle cell disease is an umbrella term for the many different types of sickle cell disorders. Healthcare providers reserve the term “sickle cell anemia” for the types of SCD that cause the most severe anemia. These types are hemoglobin SS and hemoglobin beta zero thalassemia. Sickle cell trait vs. disease People who have sickle cell trait inherited a hemoglobin S gene from only one parent. They inherited a normal gene from their other parent. People with sickle cell trait typically don’t have any symptoms of sickle cell disease. But ongoing research may show that these people may have symptoms. They can pass on the abnormal gene to their own children. Rarely, severe dehydration and intense physical activity can lead to serious health issues in people with sickle cell trait. How common is sickle cell disease? Researchers estimate sickle cell disease affects about 100,000 Americans. The disorder occurs in about 1 in every 365 Black births. It occurs in about 1 in every 16,300 Hispanic American births. About 1 in every 12 people of Black or African descent carries the sickle cell trait. Symptoms and Causes What causes sickle cell disease? A genetic mutation in the HBB gene causes sickle cell disease. The HBB gene is responsible for making a part of the hemoglobin. People with SCD received two mutated HBB genes coded for abnormal hemoglobin — one from each parent. People inherit SCD in an autosomal recessive manner. This means each parent of a child with SCD carries one copy of the mutated gene, but they typically don’t show signs and symptoms of the condition. Who’s at risk of developing sickle cell disease (SCD)? Certain groups of people are more likely to develop SCD, including: People of African descent, including African Americans. Hispanic Americans from South America and Central America. People of Mediterranean, Middle Eastern, Indian and Asian descent. What are the symptoms of sickle cell disease? Sickle cell disease symptoms begin to show when a child is about 5 to 6 months old. Signs and symptoms of SCD vary from person to person. Some people have mild symptoms, while others develop more serious complications. Sickle cell disease symptoms include: Advertisement Frequent pain episodes. Anemia, causing fatigue, paleness and weakness. Jaundice (yellowing of their skin and the whites of their eyes). Painful swelling of their hands and feet. What are the complications of this condition? Sickle cell disease can affect many parts of your body. Some of the effects are acute (they start suddenly) and some are chronic (they last for a long time). Sickle cell complications begin early and continue throughout your life. Pain Pain is the most common complication of sickle cell disease. Sickled cells passing through blood vessels can get stuck and block blood flow, which causes pain. You may have an acute pain crisis, which is also called a sickle cell crisis, vaso-occlusive crisis (VOC) or vaso-occlusive episode (VOE). These pain crises may be mild or severe and can start suddenly and last for any length of time. Pain crises most often affect your chest, back, legs and arms. You might also have chronic pain, which is when pain lasts longer than six months. Anemia Red blood cells die early with sickle cell disease, which can lead to anemia. Anemia occurs when you don’t have enough healthy red blood cells to carry oxygen throughout your body. Anemia can cause severe fatigue, along with jaundice, irritability, dizziness and lightheadedness. Advertisement Acute chest syndrome Acute chest syndrome is a life-threatening medical emergency. It can cause lung injury, difficulty breathing and low oxygen to the rest of your body. This complication of SCD occurs when sickled cells block blood and oxygen from reaching your lungs. Blood clots Sickled cells make it more likely for blood to clot. This increases your chances of developing a blood clot in a deep vein (deep vein thrombosis, or DVT). A DVT can break off and travel to your lungs (pulmonary embolism, or PE). Stroke When sickled cells get stuck in a blood vessel, it blocks blood flow to your brain. This makes it harder for your brain to get the oxygen it needs to work correctly. This can lead to a stroke. About 10% of people with SCD will have a symptomatic (clinical) stroke. Strokes are most common in people with sickle cell anemia. Vision problems Sickled cells can block blood flow in the blood vessels in your eyes. This blockage occurs most often in the retina of your eye. You may not have any symptoms and then suddenly have vision loss, which can lead to permanent blindness. Priapism Sickled cells in your penis can cause a persistent, painful erection (priapism). In addition to causing pain, priapism can cause permanent damage and erectile dysfunction. A prolonged priapism (lasting longer than four hours) is a medical emergency. Advertisement Organ damage and failure People with sickle cell disease are at risk for problems related to their heart, lung, kidneys and other organs. This is because blood and oxygen aren’t reaching them. SCD can lead to multi-organ failure. Does sickle cell trait or sickle cell disease affect pregnancy? Many people with sickle cell disease have healthy pregnancies, but the risks are higher. SCD can increase the risk of high blood pressure, blood clots, miscarriage, low birth weight and premature birth. Diagnosis and Tests How is sickle cell disease diagnosed? In the United States, hospitals test all babies for sickle cell disease as part of routine newborn screenings. This test pricks your baby’s heel to get a sample of their blood. It checks for various other conditions, as well. Your child’s healthcare provider will obtain a hemoglobin electrophoresis to confirm the diagnosis. Your healthcare provider can also diagnose sickle cell disease before your baby is born using prenatal testing. These tests include chorionic villus sampling and amniocentesis. Management and Treatment Is there a cure for sickle cell disease? A bone marrow transplant (stem cell transplant) can cure sickle cell disease. The transplant requires a donor who’s a healthy, genetic match, such as a sibling. In this procedure, you receive healthy marrow from the donor. However, only about 18% of people with SCD have a compatible donor. In addition, there are risks and complications involved with a transplant. Your healthcare provider will discuss these issues with you. What is the treatment for sickle cell disease? Sickle cell disease treatment includes medications, transfusions, blood and marrow transplant and gene therapy. Sickle cell disease treatment may begin with antibiotics. Newborns with severe SCD will receive antibiotics twice a day until they’re 5 years old to prevent infection. Other sickle cell disease medications Most people with SCD use medications to make their disease less severe and treat symptoms. These medications include: Voxelotor: Voxelotor can prevent red blood cells from sickling and binding together. It may reduce the destruction of some red blood cells, which improves blood flow to your organs and lowers your risk for anemia. Crizanlizumab: This medicine helps prevent sickled red blood cells from sticking to your blood vessel walls. This can improve blood flow and reduce inflammation and pain crises. Hydroxyurea: Hydroxyurea can reduce or prevent several complications of SCD. This includes frequent pain crises, acute chest syndrome and severe anemia. L-glutamine: This medication is a pain reliever that can help reduce the number of pain crises you have. Other pain medication options include nonsteroidal anti-inflammatory drugs (NSAIDs) and opiates. Transfusions Your healthcare provider may recommend certain transfusions to treat and prevent SCD complications. These transfusions may include: Acute transfusions: Acute blood transfusions can help treat complications that cause severe anemia. Your provider may also use an acute transfusion to treat crises. This includes strokes, acute chest syndrome and organ failure. Red blood cell transfusions: Red blood cell transfusions can help increase the number of red blood cells in your body and provide normal, non-sickled red blood cells. Stem cell transplant (also known as blood or marrow transplant) A stem cell transplant can cure SCD. Sometimes called blood or marrow transplant, SCT requires a donor who’s a good match, like a sibling, and ongoing studies are looking to optimize the transplant from alternative donors, such as birthing parents or siblings who only half-matched. Your healthcare provider will discuss the risks and benefits of this treatment in your specific case. Gene therapy Researchers are currently testing gene therapy to treat SCD. This calls for correcting an abnormal hemoglobin gene or putting a normal hemoglobin gene into a person’s stem cells. There’s promising early data and the hope is that gene therapy might one day be a routine treatment for SCD. Care at Cleveland Clinic Sickle Cell Disease Treatment Find a Doctor and Specialists Make an Appointment Prevention Can this be prevented? You can’t prevent sickle cell disease because it’s a genetic condition. If you’re pregnant, consider talking to your provider about genetic testing or genetic counseling. Outlook / Prognosis What can I expect if I have sickle cell disease? People with sickle cell disease have a reduced life expectancy. New treatments for SCD are improving life expectancy and quality of life. People with sickle cell disease can survive beyond their 50s with optimal management of the disease. Living With How do I take care of my child if they have sickle cell disease? If your child has sickle cell disease, there are many things you can do to help manage their condition: Take your child to see their healthcare provider regularly. Make sure your child gets all their recommended vaccines. Help your child get regular exercise and eat a heart-healthy diet. During a pain crisis, have your child drink lots of fluids and take a nonsteroidal anti-inflammatory drug (NSAID). If you can’t manage their pain at home, take them to the hospital for stronger pain medication. When should I go to the ER? Sickle cell disease can lead to many different life-threatening complications. If you or your child experiences any of the following symptoms of complications, call 911 or go to the nearest emergency room: Severe pain. Symptoms of severe anemia, including fatigue, dizziness and shortness of breath. Fever of 101.3 degrees Fahrenheit (38.5 degrees Celsius). Vision problems. Difficulty breathing. Erection lasting for four or more hours. Symptoms of acute chest syndrome, including chest pain, coughing and fever. Symptoms of stroke, including sudden weakness, numbness on one side of your body and confusion. Additional Common Questions Why does sickle cell disease cause pain? Sickled red blood cells are C-shaped and look kind of like crescent moons. When they travel through your blood vessels, they can get stuck and clog up the flow of blood, causing pain. Is sickle cell an autoimmune disease? No. Sickle cell disease shares some characteristics with autoimmune diseases. But healthcare providers don’t consider SCD an autoimmune disorder. They consider SCD a genetic condition. A note from Cleveland Clinic Sickle cell disease is a lifelong condition. While there’s a cure available, stem cell transplants aren’t always available and come with many risks. However, early diagnosis and treatment can help reduce your symptoms and chances of complications. With ongoing care, you can go on to lead a full, active life. Care at Cleveland Clinic Cleveland Clinic cares for people of all ages with sickle cell disease. We walk you through your diagnosis and treatment and offer education and support services. Sickle Cell Disease Treatment Find a Doctor and Specialists Make an Appointment Medically Reviewed Last reviewed on 08/03/2023. 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15714	https://kam.mff.cuni.cz/~chvatal/MorSol00.pdf	BULLETIN (New Series) OF THE AMERICAN MATHEMATICAL SOCIETY Volume 37, Number 4, Pages 437–458 S 0273-0979(00)00877-6 Article electronically published on June 26, 2000 THE ERD˝ OS-SZEKERES PROBLEM ON POINTS IN CONVEX POSITION – A SURVEY W. MORRIS AND V. SOLTAN Abstract. In 1935 Erd˝ os and Szekeres proved that for any integer n ≥3 there exists a smallest positive integer N(n) such that any set of at least N(n) points in general position in the plane contains n points that are the vertices of a convex n-gon. They also posed the problem to determine the value of N(n) and conjectured that N(n) = 2n−2 + 1 for all n ≥3. Despite the eﬀorts of many mathematicians, the Erd˝ os-Szekeres problem is still far from being solved. This paper surveys the known results and questions related to the Erd˝ os-Szekeres problem in the plane and higher dimensions, as well as its generalizations for the cases of families of convex bodies and the abstract convexity setting. 1. Introduction The following problem attracts the attention of many mathematicians by its beauty and elementary character. The Erd˝ os-Szekeres Problem 1.1. (, ) For any integer n ≥3, determine the smallest positive integer N(n) such that any set of at least N(n) points in general position in the plane (i.e., no three of the points are on a line) contains n points that are the vertices of a convex n-gon. The interest of Erd˝ os and Szekeres in this problem was initiated by Esther Klein (later Mrs. Szekeres), who observed that any set of ﬁve points in general position in the plane contains four points that are the vertices of a convex quadrilateral. Indeed, there are three distinct types of placement of ﬁve points in the plane, no three on a line, as shown on Figure 1.1. In any of these cases, one can pick out at least one convex quadrilateral determined by the points. Klein suggested the following more general problem, namely the problem on the existence of a ﬁnite number N(n) such that from any set containing at least N(n) points in general position in the plane, it is possible to select n points forming a convex polygon. As observed by Erd˝ os and Szekeres, there are two particular questions related to this problem: (1) Does the number N(n) exist? (2) If so, how is N(n) determined as a function of n? In their paper , Erd˝ os and Szekeres proved the existence of the number Received by the editors December 20, 1999, and in revised form April 4, 2000. 2000 Mathematics Subject Classiﬁcation. Primary 52C10. Key words and phrases. Erd˝ os-Szekeres problem, Ramsey theory, convex polygons and poly-hedra, generalized convexity. c ⃝2000 American Mathematical Society 437 438 W. MORRIS AND V. SOLTAN s s s s s A A A A ZZZ Z s s s s s XXX X J J J J J J s s s s s QQ Q J J J J J J Figure 1.1. Any ﬁve points in general position determine a con-vex quadrilateral. N(n) by two diﬀerent methods. The ﬁrst of them uses Ramsey’s theorem and, as a result, gives the inequality N(n) ≤R4(5, n), where R4(5, n) is a Ramsey number (see Section 2 for details). The second method is based on some geometric considerations, resulting in a better upper bound N(n) ≤ ¡2n−4 n−2 ¢ + 1. In the same paper Erd˝ os and Szekeres formulated the following conjecture. Conjecture 1.2 (). N(n) = 2n−2 + 1 for all n ≥3. Many years later (see , , and ) Erd˝ os stated that “Szekeres conjectured N(n) = 2n−2+1.” He amended this in to “Probably N(n) = 2n−2+1.” Szekeres was more forceful in , saying “Of course we ﬁrmly believe that N(n) = 2n−2 +1 is the correct value.” Another statement of faith in the conjectured value for N(n) may be found in . Shortly before he died, Erd˝ os wrote: “I would certainly pay $500 for a proof of Szekeres’ conjecture.” Klein and Szekeres married shortly after the publication of , prompting Erd˝ os to call Problem 1.1 the “Happy End Problem”. The books and contain picturesque descriptions of the Erd˝ os-Szekeres problem origins. Their second paper contains an example of a set of 2n−2, n ≥3, points in general position in the plane, no n of which determine a convex polygon. In other words, Erd˝ os and Szekeres have shown that N(n) ≥2n−2 + 1 for all n ≥3. Despite its elementary character and the eﬀort of many mathematicians, the Erd˝ os-Szekeres problem is solved for the values n = 3, 4, and 5 only. The case n = 3 is trivial, and n = 4 is due to Klein. The original paper by Erd˝ os and Szekeres notes that E. Makai proved the equality N(5) = 9, while the ﬁrst published proof of this result is due to Kalbﬂeisch et al. . The next step in solving Problem 1.1 is to answer the following question. Question 1.3. Does any set of at least 17 points in general position in the plane contain 6 points that are the vertices of a convex hexagon? For larger values of n, the best known upper bound N(n) ≤ ¡2n−5 n−3 ¢ + 2 was recently proved by T´ oth and Valtr . Later Erd˝ os posed a similar problem on empty convex polygons. Problem 1.4 (). For any positive integer n ≥3, determine the smallest posi-tive integer H(n), if it exists, such that any set X of at least H(n) points in general position in the plane contains n points which are the vertices of an empty convex polygon, i.e., a polygon whose interior does not contain any point of X. Trivially, H(3) = 3 and H(4) = 5, as easily follows from Figure 1.1. In 1978 Harborth proved that H(5) = 10, while Horton showed in 1983 that H(n) THE ERD ˝ OS-SZEKERES PROBLEM ON POINTS IN CONVEX POSITION 439 does not exist for all n ≥7. The only open case in Problem 1.4 is given by the following question of Erd˝ os. Question 1.5 (). Does the number H(6) exist? There were many attempts to prove or disprove the existence of H(6) (see Section 3 for details). The best known result in this direction belongs to Overmars et al. , who showed in 1989 that H(6) ≥27, if it exists. To give positive or negative answers to Questions 1.3 and 1.5, several algorithms for detecting a largest convex polygon or a largest empty convex polygon determined by a given set of points were elaborated. Detailed descriptions of these algorithms can be found in , , , . Due to a wide interest in Problem 1.1, the original papers by Erd˝ os and Szekeres were reprinted (see , , and ), and a long list of reviews and books discussed the problem in broad strokes (see , , , , , , , , , , , , , , , ). Nevertheless, none of them covers the whole variety of existing results and open problems. The purpose of this survey is to reﬂect the recent stage of the Erd˝ os-Szekeres problem, as well as its various generalizations and related questions. The content of the survey is indicated by section headings as follows. 1. Introduction. 2. Bounds on N(n). 3. The Erd˝ os problem on empty convex polygons. 4. Higher dimensional extensions. 5. Other generalizations and related results. In particular, we formulate a new conjecture for the higher dimensional version of the Erd˝ os-Szekeres problem (see Conjecture 4.2). In what follows, we use standard notation: Ed denotes Euclidean d-space; \|X\| and convX are the cardinality and the convex hull of a set X ⊂Ed, respectively; extP is the set of vertices (extreme points) of a convex polytope P ⊂Ed. We say that a set X ⊂Ed is in convex position if x ̸∈conv(X \ x) for any point x ∈X. In other words, X ⊂Ed is in convex position provided X is a set of vertices of a convex polytope in Ed. A convex polytope whose vertices belong to a set Y ⊂Ed is called empty provided the interior of the polytope does not contain any point of Y . Recall that a set Z ⊂Ed is in general position if no d + 1 of its points lie in a hyperplane. 2. Bounds on N(n) 2.1. Upper Bounds from Ramsey Theory. As was mentioned above, the ﬁrst proof on the existence of N(n) belongs to Erd˝ os and Szekeres and is based on the following fundamental result of Ramsey . Theorem 2.1 (). For any positive integers k, l1, l2, . . . , lr there exists a small-est positive integer m0 satisfying the following condition. For any integer m ≥m0, if the k-element subsets of {1, 2, . . . , m} are colored with colors 1, 2, . . . , r, then there exists an i, 1 ≤i ≤r, and an li-element subset T ⊂{1, 2, . . . , m} so that each of the k-element subsets of T is i-colored. The smallest number m0 for which the conclusion of Ramsey’s theorem holds is usually denoted by Rk(l1, l2, . . . , lr). A proof of Ramsey’s theorem, for the case 440 W. MORRIS AND V. SOLTAN r = 2, was independently discovered by Szekeres (see ) for the purpose of showing the ﬁniteness of N(n). The paper discusses the importance of in the development of Ramsey theory. In the following theorem we give three diﬀerent methods for getting upper bounds on N(n) based on Ramsey’s theorem. In each case, an easy argument shows that N(n) ≤Rk(l1, l2) for an appropriate choice of k, l1, l2. Theorem 2.2. For any positive integer n ≥3 the number N(n) exists and N(n) ≤min{R4(n, 5), R3(n, n)}. Proof. 1) Let X be any set of at least R4(n, 5) points in general position in the plane. The original proof of colors the 4-element subsets of X with color 1 if the points are in convex position and colors them with color 2 otherwise. Klein’s argument shows that it is impossible for all of the 4-element subsets of a 5-element subset of X to be of color 2. Hence it must be true that X contains an n-element subset for which all 4-element subsets are of color 1; i.e. all of them are in convex position. It easily follows that all the n points are in convex position. Hence N(n) ≤R4(n, 5). 2) Lewin reported that Tarsy, an undergraduate student, had come up with the following independent proof while taking an exam in a combinatorics course. Let X = {x1, x2, . . . , xm} be a set of points in general position in the plane, with m ≥R3(n, n). Color a 3-element subset {xi, xj, xk} ⊂X, i < j < k, with color 1 if one encounters the points in the order (xi, xj, xk) by passing clockwise around their convex hull. Color the subset with color 2 otherwise. It is easy to see that a 4-element subset of X is in convex position if and only if all of its 3-element subsets are colored with the same color. This implies that an n-element subset of X is in convex position if and only if all of its 3-element subsets are colored with the same color. Hence N(n) ≤R3(n, n). 3) The most recent proof involving a Ramsey-theoretic upper bound on N(n) was discovered by Johnson . Color a 3-element subset S of a planar set X in general position with color 1 if there is an even number of points of X in the interior of convS, and color S with color 2 otherwise. A one-line proof then shows that a 4-element subset of X is in convex position if and only if all of its 3-element subsets have the same color. This once again implies that an n-element subset of X is in convex position if and only if all of its 3-element subsets have the same color. Therefore, N(n) ≤R3(n, n). Lewin points out that R3(n, n) seems to be lower than R4(n, 5). The best known bounds on the Ramsey numbers R3(n, n) are 2bn2 ≤R3(n, n) ≤2cn, for some constants b and c (see ). The next section shows that these bounds are far from the true value of N(n). 2.2. Caps and Cups - Better Upper Bounds. We will assume in this section that a coordinate system (x, y) is introduced in the plane. Let X = {(x1, y1), (x2, y2), . . . , (xm, ym)} be a set of points in general position in the plane, with xi ̸= xj for all i ̸= j. A subset of points {(xi1, yi1), (xi2, yi2), . . . , (xir, yir)} is called in an r-cup if xi1 < xi2 < . . . < xir and yi1 −yi2 xi1 −xi2 < yi2 −yi3 xi2 −xi3 < . . . < yir−1 −yir xir−1 −xir . THE ERD ˝ OS-SZEKERES PROBLEM ON POINTS IN CONVEX POSITION 441 t t t ZZZZ t t t t ZZZZ Figure 2.1. Examples of a 3-cup and a 4-cap. Similarly, the subset is called an r-cap if xi1 < xi2 < . . . < xir and yi1 −yi2 xi1 −xi2 > yi2 −yi3 xi2 −xi3 > . . . > yir−1 −yir xir−1 −xir . In other words, the set of r points forms an r-cup (respectively, an r-cap) pro-vided the sequence of slopes of the segments [(xi1, yi1), (xi2, yi2)], [(xi2, yi2), (xi3, yi3)], . . . , [(xir−1, yir−1), (xir, yir)] is monotonically increasing (respectively, decreasing). See, e.g., Figure 2.1. Deﬁne f(k, l) to be the smallest positive integer for which X contains a k-cup or an l-cap whenever X has at least f(k, l) points. Theorem 2.3 (). f(k, l) ≤ ¡k+l−4 k−2 ¢ + 1. Proof. The inequality follows from the boundary conditions f(k, 3) = f(3, k) = k and the recurrence f(k, l) ≤f(k −1, l) + f(k, l −1) −1. We sketch a proof of the recurrence. Suppose that X contains f(k −1, l) + f(k, l −1) −1 points. Let Y be the set of left endpoints of (k −1)-cups of X. If X \ Y contains f(k −1, l) points, then it contains an l-cap. Otherwise, Y contains f(k, l−1) points. Suppose that Y contains an (l −1)-cap {(xi1, yi1), (xi2, yi2), . . . , (xil−1, yil−1)}. Let {(xj1, yj1), (xj2, yj2), . . . , (xjk−1, yjk−1)} be a (k −1)-cup with il−1 = j1. A quick sketch then shows that either (xil−2, yil−2) can be added to the (k −1)-cup to create a k-cup or (xj2, yj2) can be added to the (l −1)-cap to create an l-cap. Because N(n) ≤f(n, n), we get from Theorem 2.3 that N(n) ≤ ¡2n−4 n−2 ¢ + 1. This upper bound was not improved upon until 63 years later, when Chung and Graham managed to modify the above argument to show that N(n) ≤ ¡2n−4 n−2 ¢ . A further modiﬁcation by Kleitman and Pachter implied N(n) ≤ ¡2n−4 n−2 ¢ + 7 − 2n. Shortly thereafter, T´ oth and Valtr gave the following simple argument to roughly cut the Erd˝ os-Szekeres bound in half. Theorem 2.4 (). N(n) ≤ ¡2n−5 n−3 ¢ + 2. Proof. Let a be an extreme point of a planar set X. Denote by b a point outside of convX so that no line determined by the points of X \ {a} intersects the segment [a, b]. Let l be a line through b that does not intersect convX. It is easily seen that the projective transformation T that maps l to the line at inﬁnity and maps the segment [a, b] to the vertical ray emanating downward from T(a) has the following properties: (i) a subset Y of X is in convex position if and only if T(Y ) is, 442 W. MORRIS AND V. SOLTAN (ii) a subset Z of X containing a is in convex position if and only if T(Z \ a) is a cap. Suppose now that X contains ¡2n−5 n−3 ¢ +2 points. Then T(X \a) determines either a (n −1)-cap or a n-cup (see Theorem 2.3). This implies that X contains n points in convex position. Stirling’s formula shows that ¡2n−5 n−3 ¢ is smaller than 4n and is, asymptotically, larger than (4 −c)n for any constant c > 0. Chung and Graham have oﬀered $100 for the ﬁrst proof that N(n) = O((4 −c)n) for some constant c > 0. (They oﬀer no money for showing that no such c exists.) The simplicity of the argument of T´ oth and Valtr seems to indicate that further reductions in the upper bound are within reach. On the other hand, substantially diﬀerent techniques might be needed to claim the $100 prize. 2.3. Construction for the Lower Bound. We begin with a theorem that states that the inequality for f(k, l) in Theorem 2.3 is actually an equality. Theorem 2.5 (). f(k, l) = ¡k+l−4 k−2 ¢ + 1. Proof. Note that we have already observed this to be the case when k or l is 3. We proceed by induction. Suppose that we have a set A of ¡k+l−5 k−3 ¢ points with no (k −1)-cup and no l-cap, and a set B of ¡k+l−5 k−2 ¢ points with no k-cup and no (l −1)-cap. Translate these sets so that the following conditions are satisﬁed: (i) every point of B has greater ﬁrst coordinate than the ﬁrst coordinates of points of A, (ii) the slope of any line connecting a point of A to a point of B is greater than the slope of any line connecting two points of A or two points of B. Let X = A ∪B be the resulting conﬁguration. Any cup in X that contains elements of both A and B may have only one element of B. It follows that X contains no k-cup. We similarly see that X contains no l-cap. Thus f(k, l) ≥ µk + l −5 k −3 ¶ + µk + l −5 k −2 ¶ + 1 = µk + l −4 k −2 ¶ + 1. 2 Now we are ready to prove the inequality N(n) ≥2n−2 + 1. This lower bound on N(n) was essentially proved in . Some inaccuracies in the proof were later corrected by Kalbﬂeisch and Stanton . (Erd˝ os refers to their corrections in .) We sketch the main ideas of the construction as it is presented in . Theorem 2.6 (,)). N(n) ≥2n−2 + 1. Proof. To prove the inequality, we construct a set X of 2n−2 points with no subset of n points in convex position. For i = 0, 1, . . . , n−2, let Ti be a set of ¡n−2 i ¢ points containing no (i + 2)-cap and no (n −i)-cup and having the property that no two points in the set are connected by a line having slope of absolute value greater than 1. For i = 0, 1, . . . , n −2, place a small copy of Ti in a neighborhood of the point on the unit circle making an angle of π 4 − iπ 2(n−2) with the positive x-axis. Let X be the union of the Ti, i = 1, 2, . . . , n −2. Then \|X\| = n−2 X i=0 µn −2 i ¶ = 2n−2. THE ERD ˝ OS-SZEKERES PROBLEM ON POINTS IN CONVEX POSITION 443 Suppose that Y is a subset of X in convex position. Let k and l be the smallest and the largest values of i so that Y ∩Ti ̸= ∅. If k = l, then Y contains no (k+2)-cap and no (n −k)-cup. The construction guarantees that: (a) Y ∩Tk is a cap of at most k + 1 points, (b) Y ∩Tl is a cup of at most n −l −1 points, (c) \|Y ∩Ti\| ≤1 for all i = k + 1, k + 2, . . . , l −1. Thus \|Y \| ≤k + 1 + l −k −1 + n −l −1 = n −1. Hence no subset of X in convex position contains n points. An interesting conjecture was formulated by Erd˝ os et al. that connects the proof of the upper bound on N(n) given in to the conjectured lower bound on N(n). Let m(n, k, l) be the smallest number such that any set of m(n, k, l) points in general position in the plane contains either a set of n points in convex position, or a k-cup, or an l-cap. It is proved in that m(n, k, l) ≤Pk−2 i=n−l ¡k−2 i ¢ . The authors of conjecture that equality holds and prove its equivalence to the conjecture N(n) = 2n−2 + 1. 2.4. The case n = 5. Erd˝ os and Szekeres note already in that Makai had proved the equality N(5) = 9. Credit for this result is given in to Makai and Tur´ an. A proof did not appear in the literature until Kalbﬂeisch et al. . Since their proof was rather long, Bonnice and Lov´ asz (, pp. 88–89, 501–506) independently published much simpler proofs. In what follows we outline the proof of Bonnice . Theorem 2.7. N(5) = 9. The proof of Theorem 2.7 is based on Lemma 2.8 below. Given a ﬁnite set X of points in the plane, the statement that X is (k1, k2, . . . , kj) will mean that \|X\| = k1+k2+. . .+kj and the convex hull of X is a k1-gon; that, when the vertex set of convX is taken away from X, the convex hull of the remaining points is a k2-gon; etc. Also, if abcd is a convex quadrilateral with vertices ordered counterclockwise, beam ab:cd denotes the section of the plane obtained by deleting conv{a, b, c, d} from the convex section of the plane bounded by segment [a, b] and rays [a, d), [b, c). Similarly, if x, y, and z are not collinear, beam x:yz will denote the inﬁnite section of the plane obtained by deleting conv{x, y, z} from the convex cone which has vertex x and is bounded by rays [x, y) and [x, z). Lemma 2.8. If a planar set Y is (3, 3, 2), or (4, 3, 1), or (3, 4, 2), then Y deter-mines a convex pentagon. Proof. First, assume that Y is (3, 3, 2). Let y1, y2, y3 be the vertices of convY ; let triangle v1v2v3 be the second triangle, conv(Y \ {y1, y2, y3}); and let z1, z2 be the two points of Y interior to v1v2v3. We may assume that line (z1, z2) intersects sides [v1, v2] and [v1v3] of v1v2v3 such that [z1, z2) intersects [v1, v2). The vertices y1, y2, and y3 of the outside triangle are in the union of beams z1z2:v2v3, z1:v1v3, and z2:v1v2. If one of these vertices, say y1, is in beam z1z2:v2v3, then z1z2v2y1v3 is a convex pentagon. Thus we may assume that two of the points y1, y2, y3, say y1 and y2, are in beam z1:v1v3. Since conv{y1, y2, y3} con-tains all of v1, v2, v3, triangle v1v2v3 lies in one of the open half-planes determined 444 W. MORRIS AND V. SOLTAN t t t t t t t t Figure 2.2. A set of eight points determining no convex pentagon. by line (y1, y2). Thus line (y1, y2) does not intersect conv{v1, v2, v3}, and therefore z1v1v3y1y2 is a convex pentagon. Now assume that Y is (4, 3, 1). Let y1, y2, y3, and y4 be the vertices of convY ; let v1v2v3 be the inside triangle, conv(Y \ {y1, y2, y3, y4}); and let z denote the point of Y inside it. Partitioning the plane outside v1v2v3 into beams z:v1v3, z:v1v2, and z:v2v3, we may assume that two of the four points y1, y2, y3, y4, say y1 and y2, are in beam z:v1v3. Then, as above, zv1v3y1y2 is a convex pentagon. Finally, assume that Y is (3, 4, 2). The technique is the same: let y1y2y3, v1v2v3v4, and z1z2 be the triangle, quadrilateral, and line segment given by the fact that Y is (3, 4, 2). If line (z1, z2) cuts a vertex, say v1, of v1v2v3v4, then z1z2v2v3v4 is a convex pentagon. So assume that line (z1, z2) cuts sides [v1, v4] and [v2, v3] such that rays [z1, z2) and [v2, v3) intersect. As above, if there is a point of {y1, y2, y3} in one of the beams z1z2:v3v4, z2z1:v1v2, a convex pentagon is formed. Thus we may assume that {y1, y2, y3} lies in the union of beams z1:v1v4 and z2:v2v3. In particular, we may assume that both y1 and y2 are in z1:v1v4; whence, as before, z1v1v4y1y2 is a convex pentagon. Proof of Theorem 2.7. If a set of nine points in general position in the plane deter-mines no convex pentagon, it is one of the following: (4, 4, 1), (4, 3, 2), (3, 4, 2), or (3, 3, 3). Each of the ﬁrst two cases has a subset of 8 points which is (4, 3, 1), and each of the last two cases has a subset which is (3, 3, 2). Thus Lemma 2.8 applies to all cases to show that N(5) ≤9. The opposite inequality easily follows from Figure 2.2. As mentioned by Bonnice , the same approach can hardly be applied to the case n = 6. Indeed, assuming that N(6) = 17, one can see that a set X of 17 points in the plane can determine 70 distinct tuples (k1, k2, . . . , kj) representing the diﬀerent ways the successive convex hulls of X might nest if it determines no convex hexagon. 3. The Erd˝ os problem on empty polygons In 1978 Erd˝ os , , posed a new problem on convex polygons. Problem 3.1 (). For any positive integer n ≥3, determine the smallest posi-tive integer H(n), if it exists, such that any set X of at least H(n) points in general position in the plane contains n points which are the vertices of an empty convex polygon, i.e., a polygon whose interior does not contain any point of X. THE ERD ˝ OS-SZEKERES PROBLEM ON POINTS IN CONVEX POSITION 445 t t t t t t t t t Figure 3.1. A set of nine points with no empty convex pentagon. Trivially, H(3) = 3, and from Figure 1.1 we easily conclude that H(4) = 5. Using a direct geometric approach, Harborth showed in 1978 that H(5) = 10. The inequality H(5) ≥10 immediately follows from Figure 3.1, where a set of nine points in general position determines no empty convex pentagon (there are still two convex pentagons, neither being empty). In 1983 Horton showed that H(n) does not exist for all n ≥7. This statement is due to the following analytic construction of a planar set Sk of 2k (k ≥1) points in general position determining no empty convex 7-gon. Let a1a2 · · · ak be the binary representation of the integer i, 0 ≤i < 2k, where leading 0’s are omitted. Put c = 2k +1 and deﬁne d(i) = Pk j=0 ajcj−1. Now a simple analytical consideration shows that any convex polygon determined by the set Sk = {(i, d(i)) : i = 0, 1, . . . , 2k −1} has at most six vertices. Valtr deﬁnes a Horton set inductively as follows. The empty set and any one-point set are Horton sets. The points of a Horton set H are in general position in the plane, with distinct x-coordinates. Furthermore, H can be partitioned into two sets A and B such that: 1. Each of A and B is a Horton set. 2. The set A is below any line connecting two points of B, and the set B lies above any line connecting two points of A. 3. The x-coordinates of the points of A and B alternate. One can easily prove by induction on n = \|H\| that if {(x1, y1), (x2, y2), (x3, y3), (x4, y4)} is a 4-cup (respectively, 4-cap) in H, then there is a point (x, y) of H lying above (respectively, below) one of the segments [(xi, yi), (xi+1, yi+1)], i = 1, 2, 3. It immediately follows that H contains no empty 7-gon, because otherwise A would contain a 4-cup or B would contain a 4-cap. Note that the above sets Sk ﬁt the deﬁnition of Horton sets. Valtr , uses Horton sets in several generalizations of the empty polygon problem, as we will see in Sections 4 and 5. In this connection the following question of Erd˝ os (and later of Horton ) still remains open. Question 3.2. Does the number H(6) exist? 446 W. MORRIS AND V. SOLTAN Horton expresses the belief that H(6) exists. B´ ar´ any and Valtr present a conjecture which would imply the existence of H(6). Trying to determine the lower bound on H(6), Avis and Rappaport elaborated a method to determine whether a given set of points in the plane contains an empty convex 6-gon, and by using this approach they found a set of 20 points in general position containing no empty convex 6-gon. Overmars et al. , modifying considerations of Dobkin et al. , constructed an algorithm of time complexity O(n2) that solves the following problem: for a given set V in the plane, containing no empty convex 6-gon, and for a point z ̸∈V , determine whether the set {z} ∪V contains an empty convex 6-gon. Using this algorithm, they found a set of 26 points containing no empty convex 6-gon. Hence, H(6) ≥27, if it exists. As with the original Erd˝ os-Szekeres problem, the theory for the empty polygon problem is limited to that which can be proved using cups and caps. There remains once again a large gap that probably will require some new paradigms to be bridged. 4. Higher dimensional extensions 4.1. The Erd˝ os-Szekeres Problem in Higher Dimensions. An observation that the Erd˝ os-Szekeres problem can be generalized for higher dimensions was al-ready mentioned by its authors (see ) and later rediscovered by Danzer et al. . Recall that a set X of points in Euclidean space Ed is in general position if no d + 1 points of X lie in a hyperplane. (Clearly, a set X is in general position if and only if for any positive integer k, 1 ≤k ≤d, no k + 1 points of X lie in a k-dimensional plane.) Furthermore, X is said to be in convex position if no point of X lies in the convex hull of the remaining points. In other words, a set X in general position is in convex position if and only if it is the vertex set of a convex d-polytope in Ed. Following , we deﬁne Nd(n), d ≥2, n ≥1 to be the smallest positive integer such that any set of Nd(n) points in general position in Ed contains n points in convex position. Similarly to the planar case, one can pose the following two problems: 1) Do the numbers Nd(n) exist for all d ≥2 and n ≥1? 2) If yes, what are the values of Nd(n)? The existence of Nd(n) can be established analogously to the planar case (see Section 1) by implementing the following steps: (a) A set X of at least d+2 points in general position in Ed is in convex position if and only if any d + 2 of them are in convex position. (This fact is a direct consequence of Carath´ eodory’s theorem : a point z belongs to the convex hull of a set A ⊂Ed if and only if z belongs to the convex hull of at most d + 1 points of A.) (b) Any set of d + 3 points in general position in Ed contains d + 2 points in convex position. (A stronger version of this statement was proved by Motzkin ; see also , who showed that the number of nonconvex (d+2)-subsets of a general (d + 3)-set in Ed equals either 0, or 2, or 4 for all d ≥2.) (c) If Rd+2(n, d + 3) is a Ramsey number, then any set of Rd+2(n, d + 3) points in general position in Ed contains n points in convex position. (This statement is a direct generalization of the original proof by Erd˝ os and Szekeres for the planar case.) THE ERD ˝ OS-SZEKERES PROBLEM ON POINTS IN CONVEX POSITION 447 As a result, we conclude that the numbers Nd(n) exist for all positive integers d ≥2 and n ≥1, and Nd(n) ≤Rd+2(n, d + 3). Valtr gave another idea for proving the existence of numbers Nd(n). He considers any set X of at least N2(n) points in general position in Ed and its projection Y onto a two-dimensional subspace L ⊂Ed such that Y is in general position in L. Since \|Y \| ≥N2(n), one can select in Y a subset of n points in convex position. It is easily seen that the prototypes of these points in X are in convex position. This consideration implies the inequality Nd(n) ≤N2(n), d ≥2. A similar consideration is true for the case of an m-dimensional plane in Ed, 2 < m < d. Hence we obtain Nd(n) ≤Nd−1(n) ≤. . . ≤N2(n) ≤ µ2n −5 n −3 ¶ + 2. K´ arolyi has recently proved that Nd(n) ≤Nd−1(n −1) + 1, and this implies Nd(n) ≤ µ2n −2d −1 n −d ¶ + d. The paper also contains the intriguing result that for any n ≥1 and d ≥3 there is a smallest integer Md(n) so that if P is any set of Md(n) points in general position in Ed and if p ∈P, then there is a subset of P consisting of n points in convex position and containing p. Johnson showed that his proof of the existence of N(n) can be modiﬁed to get Nd(n) ≤Rd+1(n, n, . . . , n), where the last term has d −1 copies of n. We note here that no one has yet succeeded in generalizing the “caps and cups” arguments of Erd˝ os and Szekeres for the case d ≥3. The only known general lower bound for Nd(n) is due to K´ arolyi and Valtr . They prove that for each d ≥2 there exists a constant c = c(d) so that Nd(n) = Ω(c d−1 √n). Table 4.1 shows the known values of Nd(n), where the respective value is placed at the intersection of column d and row n. Indeed, the equalities Nd(n) = n if n ≤d + 1 are trivial, and Nd(d + 2) = d + 3, mentioned by Danzer et al. (see also Gr¨ unbaum ), was proved by Motzkin . We note here that the equality N2(4) = 5, which is due to Klein, is a particular case of Nd(d+2) = d+3. The next range of values of Nd(n) is given by the following new theorem. Theorem 4.1. Nd(n) = 2n −d −1 for d + 2 ≤n ≤⌊3d/2⌋+ 1. Proof. As a consequence of a stronger assertion by Bisztriczky and Soltan (see Section 4.2 below) one has Nd(n) ≤2n −d −1 for d ≥2 and d + 2 ≤n ≤ ⌊3d/2⌋+ 1. For any n ≥d + 2, Bisztriczky and Harborth constructed a set X = {x1, x2, . . . , x2n−d−3} in Ed so that X ∪{o} is in general position and every set of n−1 points of X contains the origin o of Ed in the interior of its convex hull. They showed that no set of n points of X ∪{o} could be the set of vertices of an empty convex polytope. If we now scale the points of X by positive scalars λi so that for each i ≥n, λixi is in the interior of the convex hull of every set of n −1 points of {λ1x1, λ2x2, . . . , λi−1xi−1}, one can similarly see that no set of n points of {λ1x1, λ2x2, . . . , λ2n−d−3x2n−d−3, o} is in convex position. Thus Nd(n) ≥2n−d−1 for n ≥d+2. 448 W. MORRIS AND V. SOLTAN n 14 18 13 17 16 12 15 14 11 14 13 12 10 13 12 11 10 9 11 10 9 9 8 10 9 8 8 8 7 9 8 7 7 7 7 6 9 7 6 6 6 6 6 5 9 6 5 5 5 5 5 5 4 5 4 4 4 4 4 4 4 3 3 3 3 3 3 3 3 3 2 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 2 3 4 5 6 7 8 9 . . . d Table 4.1. Known values of Nd(n) when d < 10. For n > ⌊3d/2⌋+ 1, there are known only two values of Nd(n): N2(5) = 9; see Section 2 and N3(6) = 9, due to Bisztriczky and Soltan . Their proof is similar to that of Bonnice and is based on selecting three subsets, P, Q, R, of a set X ⊂E3 of 9 points in general position such that P = ext convX, Q = ext conv(X \ P), R = ext conv(X \ (P ∪Q)). Considering separately the cases (i) \|P\| ≥6, (ii) \|P\| = 5 and \|Q\| = 4, (iii) \|P\| = 4 and \|Q\| = 5, (iv) \|P\| = 4, \|Q\| = 4, and \|R\| = 1, they show each time the existence of a subset of 6 points in X in convex position. It is interesting to mention that all the known values of Nd(n), with d ≥2 and n > ⌊(3d + 1)/2⌋, satisfy the equality Nd(n) = 4Nd(n −d) −3. This also agrees with the conjecture N2(n) = 2n−2 + 1 of Erd˝ os and Szekeres. We therefore oﬀer the following conjecture. Conjecture 4.2. Nd(n) = 4Nd(n −d) −3 for all d ≥2 and n > ⌊(3d + 1)/2⌋. Gr¨ unbaum (, pp. 22-23) discovered a variant of the Erd˝ os-Szekeres problem in higher dimensions. Namely, he established the existence of the minimum number Bd(n), d ≥2 and n ≥1, such that any set X ⊂Ed consisting of at least Bd(n) points in general position contains a subset of n points lying on the boundary of a convex body in Ed. His proof is based on Ramsey’s theorem and the following assertion: a ﬁnite set Y ⊂Ed lies on the boundary of a convex body in Ed if and only if each of its subsets of at most 2d + 1 points lies on the boundary of a convex body in Ed. The last statement is a direct consequence of the Steinitz theorem : a point z ∈Ed belongs to the interior of the convex hull of a set S ⊂Ed if and only if z belongs to the interior of the convex hull of at most 2d points of S. THE ERD ˝ OS-SZEKERES PROBLEM ON POINTS IN CONVEX POSITION 449 Later Bisztriczky and Soltan showed that in the deﬁnition of Bd(n) the set X ⊂Ed can be arbitrary (not necessarily in general position), and they also proved the equality Bd(n) = Nd(n) for all d ≥2 and n ≥1. Their proof is based on a simple idea that any ﬁnite set in Ed can be approximated by a set in general position. 4.2. Empty Convex Polytopes. Generalizing the Erd˝ os problem on empty con-vex polygons (see Section 2), Bisztriczky and Soltan deﬁned Hd(n) to be the smallest positive integer, if it exists, such that any set X of Hd(n) points in general position in Ed contains a subset of n points that are the vertices of an empty convex polytope, i.e., a polytope whose interior does not contain any point of X. Valtr proved the following deep results on the existence of Hd(n), generaliz-ing considerations of Horton . 1) Hd(n) exists for all n ≤2d + 1, d ≥2, and Hd(2d + 1) ≤Nd(4d + 1); 2) H3(n) does not exist for all n ≥22; 3) Hd(n) does not exist if n ≥2d−1(P(d −1) + 1) and d ≥4, where P(d −1) is the product of the ﬁrst d −1 prime numbers. By using simple geometric arguments, Bisztriczky and Soltan showed that Hd(n) ≤2n −d −1 for d + 2 ≤n ≤⌊3d/2⌋+ 1. Later Bisztriczky and Harborth proved the opposite inequality Hd(n) ≥2n −d −1 if Hd(n) exists. Their proof is based on the construction of a set X ⊂Ed of cardinality 2n −d −3 such that the intersection of all convex hulls of subsets Y ⊂X, \|Y \| = n −1, is nonempty. Combining the results of , and Theorem 4.1, one gets Hd(n) = Nd(n) = 2n −d −1 for d ≥2 and d + 2 ≤n ≤⌊3d/2⌋+ 1. (1) For n > ⌊3d/2⌋+ 1, there are known only two values of Hd(n): H2(5) = 10, proved by Harborth , and H3(6) = 9, proved by Bisztriczky and Soltan . Since H3(6) can be considered as a particular case of Hd(⌊(3d + 1)/2⌋+ 1), we pose the following problem. Problem 4.3. Determine the value of Hd(⌊(3d + 1)/2⌋+ 1). Due to equality (1), it is suﬃcient to consider in the problem above the case when d is odd: e.g. the numbers H5(9), H7(12), H9(15), etc. Some problems on the existence of empty convex polytopes in a two-colored set of points in Ed are considered by Borwein . 5. Other generalizations and related results 5.1. Many Convex n-gons. After the existence of convex n-gons has been proved, it is natural to ask how many there are. For a planar point set of r points in general position there are, of course, ¡r 3 ¢ triangles determined by the set. The number of convex quadrilaterals formed by such a set is positive, for r ≥5, as noted by Klein . To show that the number of such convex quadrilaterals is at least ¡r−2 3 ¢ was a problem in the Eleventh International Mathematical Olympiad, Bucharest, 1969 (see , , and ). More generally, we can pose the following problem. Problem 5.1. Determine the minimum number of convex n-gons in a planar set of r points in general position. 450 W. MORRIS AND V. SOLTAN Clearly, a similar problem can be posed for higher dimensions. B´ ar´ any and Valtr proved that suﬃciently large planar point sets have many collections of subsets in convex position of a given size n. Theorem 5.2 (). For every integer n ≥4 there is a constant cn > 0 with the property that every suﬃciently large ﬁnite planar set X in general position contains n subsets Y1, . . . , Yn with \|Yi\| ≥cn\|X\|, i = 1, . . . , n, such that any set {y1, . . . , yn} satisfying yi ∈Yi for all i = 1, . . . , n is in convex position. Special cases of this theorem were previously proved by Solymosi and Nielsen . The inﬁmum of the constants cn for which Theorem 5.2 is true is shown in to be at least ³ N(n)2( N(n)−1 2 )´−1 . A note at the end of states that Solymosi has proved the inequality cn ≥2−16n2. Also proved in is c4 ≥ 1 22. Erd˝ os notes in that a discussion with P. Hammer prompted him to study the function s(r), the minimum number of convex subsets (of any number n ≥3 of points) contained in a set of r points in general position in the plane. Erd˝ os proves that there exist constants a and b so that ralogr < s(r) < rblogr. He also speculates that limn→∞logs(r)/(logn)2 exists. The problem of counting the number of empty n-gons has received considerable interest. Let fn(r) denote the minimum number of empty n-gons in a set of r points in general position in the plane. Note that Horton proved fn(r) = 0 for n ≥7. Purdy announced the equality f3(r) = O(r2), while Harborth showed that f3(r) = r2 −5r + 7 for 3 ≤r ≤9 and f3(10) = 58. Katchalski and Meir continued the investigation of fn(r) for smaller n by proving that ¡n 2 ¢ ≤f3(r) ≤Kr2 for some constant K < 200. B´ ar´ any and F¨ uredi followed up on the work of Katchalski and Meir by proving some new bounds: r2 −O(rlogr) ≤f3(r) ≤2r2, 1 4r2 −O(r) ≤f4(r) ≤3r2, ⌊r 10⌋≤f5(r) ≤2r2, f6(r) ≤1 2r2. The upper bounds were improved by Valtr , and later by Dumitrescu , to f3(r) < 1.68r2, f4(r) < 2.132r2, f5(r) < 1.229r2, f6 < 0.298r2. Valtr mentioned personal correspondence from B´ ar´ any in which a lower bound of f4(r) ≥1 2r2 −O(r) is given. We close this section by mentioning the papers by Ambarcumjan , Karolyi , Hosono and Urabe , and Urabe , which deal with some combinatorial problems on clustering of ﬁnite planar sets, i.e. partitioning a set into subsets in convex position. 5.2. Replacing Points with Convex Bodies. Bisztriczky and Fejes T´ oth (see , , ) showed that the Erd˝ os-Szekeres problem has a generalization for the case of convex bodies in the plane. We say that a family of pairwise disjoint convex bodies is in convex position if none of its members is contained in the convex hull of the union of the others. THE ERD ˝ OS-SZEKERES PROBLEM ON POINTS IN CONVEX POSITION 451 Theorem 5.3 (). For any integer n ≥4, there is a smallest positive integer g(n) such that if F is a family of pairwise disjoint convex bodies in the plane, any three of which are in convex position and \|F\| > g(n), then some n convex bodies of F are in convex position. The authors prove the existence of g(n) using Ramsey’s theorem. They also made the bold conjecture that g(n) = N(n) −1. This conjecture is supported in , where it is shown that g(5) = 8. A family F of convex bodies in the plane is said to have property Hn k , for 3 ≤k < n, if every set of k elements of F is in convex position and no n of them are in convex position. The third paper of these authors is devoted to studying h(k, n), which is the maximum cardinality of a family of mutually disjoint convex bodies satisfying property Hn k . Clearly, h(k, n) ≤h(3, n) = g(n), so h(k, n) exists for all 3 ≤k < n. An upper bound h(4, n) ≤(n −4) ¡2n−4 n−2 ¢ −n + 7, similar to the bound of on N(n), is proved. This is shown to imply the (very large) upper bound g(n) ≤R4((n −4) µ2n −4 n −2 ¶ −n + 8, 5). Smaller upper bounds are derived in for h(k, n) if k ≥5. The bounds on h(k, n) from are considerably improved by Pach and T´ oth , . The best known bounds are 2n−2 ≤h(3, n) ≤ µ2n −4 n −2 ¶2 , 2⌊n + 1 4 ⌋2 ≤h(4, n) ≤n3, n −1 + ⌊n −1 3 ⌋≤h(5, n) ≤6n −12, n −1 + ⌊n −1 k −2⌋≤h(k, n) ≤k −4 k −5n, k > 5. Pach and Solymosi extend the results of B´ ar´ any and Valtr (see Section 5.1), replacing points by compact convex sets. Speciﬁcally, they prove that for every n ≥4 there is a positive constant cn = 2O(n2), so that the following is true: every family F of r pairwise disjoint compact convex sets in general position in the plane has n disjoint ⌊cnr⌋-membered subfamilies Fi, 1 ≤i ≤n, such that no matter how we pick one set from each Fi, they are always in convex position. Note that the exponent for cn here is better than that of . Recent research by Pach and T´ oth investigates Erd˝ os-Szekeres type problems in which the points are replaced by convex sets that are not necessarily disjoint. 5.3. Restricted Planar Point Sets. The size of the coordinates of the points in the conﬁgurations given by Kalbﬂeisch and Stanton that meet the conjectured upper bound on N(n) grows very quickly. A step toward showing that this is unavoidable was taken by Alon et al. . Suppose that X is a set of points in the plane, with no three on a line. Let q(A) be the ratio of the largest distance between two points of X to the smallest distance between two points of X. The authors of prove that if \|X\| = k and q(X) ≤α √ k, then there is a constant β, depending on α, so that X contains a subset of size at least βk 1 4 in convex position. In other words, the restriction of 452 W. MORRIS AND V. SOLTAN the Erd˝ os-Szekeres problem to point sets with relatively uniform distances between points yields a function N(n) that is at most a fourth-degree polynomial. The results of were improved by Valtr . Under the same conditions, \|X\| = k and q(X) ≤α √ k, Valtr shows that there is a constant β = β(α) so that X contains a subset of size at least βk 1 3 in convex position. Furthermore, he constructs, for any c > 5.96 and k large enough, a set Ak of k points with no subset in convex position larger than ck 1 3 . The sets Ak have the property that they contain no empty convex 7-gons. 5.4. Polygons That Are Empty Modulo q. Recall that Horton proved the existence of arbitrarily large sets of points in general position in the plane that contain no empty convex 7-gons. In view of this, the following conjecture of is surprising. Conjecture 5.4. For any two positive integers q and n, n ≥3, there is a smallest positive integer C(n, q) so that any set X of C(n, q) points in general position in the plane contains a subset n points in convex position for which the number of points of X in its interior is divisible by q. Bialostocki et al. prove in that the conjecture is true in the cases n ≡2(mod q) and n ≥q +3. Extremely large upper bounds on C(n, q) in both cases are obtained by the Ramsey theoretic argument. Caro found a better upper bound that also holds for a more general function. Let X be a set of points in general position in the plane, and let G be an abelian group. If w is a function from X to G and K is a subset of X in convex position, then K is said to have zero-sum interior modulo G if X x∈interiorK w(x) = 0 (in G). Theorem 5.5 (). For any two integers n and q, n ≥q + 2, there is an integer E(n, q) satisfying the following conditions: (1) Let X be a set of points in general position in the plane, and let G be an abelian group of order q. Assume w : X →G. Then \|X\| ≥E(n, q) implies that X contains a set of n points in convex position that has zero-sum interior. (2) For a given q, one has E(n, q) ≤2c(q)n, where c(q) depends only on q but not on n or the structure of G. Caro speculates that the bound for E(n, q) can be considerably improved when n ≥q + 2. A recent result of K´ arolyi et al. is that Conjecture 5.4 is true for n ≥5 6q + O(1). 5.5. Duality. The Erd˝ os-Szekeres problem has a dual one in terms of arrangements of lines in the plane. Attention to this equivalent problem was ﬁrst drawn by Goodman and Pollack . An arrangement of lines is called simple if no two of the lines are parallel and no three of them meet in a point. The dual problem is then to determine the smallest integer N(n) so that every simple arrangement of N(n) lines together with a point q not on any line contains a sub-arrangement of n lines for which the cell containing q is a convex n-gon. One generalization is to consider the smallest integer p(n) so that every simple arrangement of p(n) lines contains a sub-arrangement of n lines determining a convex n-gon. Harborth and M¨ oller show that this problem is only interesting THE ERD ˝ OS-SZEKERES PROBLEM ON POINTS IN CONVEX POSITION 453 if the arrangements are considered to be in the projective plane. To do this, one identiﬁes opposite unbounded cells of the arrangements. It is trivially true that p(n) ≤N(n). It is shown in that p(6) = 9 and p(n) ≥1 + 2 k+1 2 . A second generalization is to replace the lines by pseudolines. Goodman and Pollack conjectured that the inequality N(n) ≤2n−2 + 1 holds even if “lines” in the dual Erd˝ os-Szekeres problem are replaced by “pseudolines”. If we denote by Nps(n) the analogous function for pseudolines, one can see that the arguments of and remain valid and show that Nps(n) ≤ ¡2n−5 n−2 ¢ + 2. A non-stretchable arrangement of 16 lines for which no subarrangement of 6 lines forms a polygon containing a speciﬁed point is given by Morris . Harborth and M¨ oller also ask if the function p(n) of their problem is altered by substituting pseudolines for lines. 5.6. Generalized Convexity. Many of the known results on the Erd˝ os-Szekeres problem have been proved using only some very simple combinatorial properties of the plane. It is natural to ask what the most general framework is for studying this problem. One such framework is that of generalized convexity (see the books by Soltan and Van de Vel for an overview of this topic). We start with a ﬁnite set X and a collection F of subsets of X. The pair (X, F) is called a convexity on X if the following hold: 1) ∅and X are in F, 2) F is closed under intersection. For any subset A of X, deﬁne co(A) to be the smallest member of F containing A. A subset A of X is called convexly independent if a / ∈co(A \ a) for all a ∈A. A set X of points in Ed is said to realize a convexity (X, F) if A ∈F precisely when A = K ∩X for some convex subset K of Ed. If (X, F) is realizable and satisﬁes a nondegeneracy assumption, then we have seen that there is a function N(n) so that X contains a convexly independent set of size n whenever \|X\| ≥N(n). For nondegeneracy, one simply stipulates that, for some k, all subsets of X of size at most k are convexly independent. One would like to replace the condition of realizability by a simpler combinatorial condition. A convexity (X, F) is said to have the anti-exchange property if for any subset A of X and x, y / ∈co(A), x ∈co(A ∪y) implies y / ∈co(A ∪x). Several names have been given to convexities with the anti-exchange property, most notably convex geometry (see ) and antimatroid (see ). Note that Coppel uses the term convex geometry to refer to a diﬀerent set of axioms. A basis of a set A ⊆X is a minimal set B ⊆A such that co(B) = co(A). The anti-exchange property is equivalent to the property that every set A ⊆X has a unique basis. The anti-exchange property by itself is not a strong enough property to provide a structure in which one can carry through Szekeres’ Ramsey-theoretic proof of the Erd˝ os-Szekeres theorem. We will see that the addition of one more property is suﬃcient for this purpose. The Carath´ eodory number of a convexity F is the least positive integer c such that co(Y ) = ∪{co(Z) : Z ⊆Y, \|Z\| ≤c} for any Y ⊆X. A set Y ⊆X is said to be in nice position if any c points of Y are convexly independent. Let c be the Carath´ eodory number of a convexity (X, F). We say that (X, F) satisﬁes the simplex partition property if for any set {z1, z2, . . . , zc+2} of c + 2 points in nice position, with {zc+1, zc+2} ⊆co(z1, . . . , zc), the point zc+2 belongs to exactly one of the sets co(z1, . . . , zi−1, zc+1, zi+1, . . . , zc), i = 1, . . . , c. 454 W. MORRIS AND V. SOLTAN Lemma 5.6. Let convexity (X, F) satisfy the anti-exchange property and the sim-plex partition property. Any set of c + 2 points of X in nice position contains c + 1 convexly independent points. Proof. Let {z1, . . . , zc+2} be a set of points of X in nice position. We may as-sume that zc+2 ∈co(z2, z3, . . . , zc+1) and zc+1 ∈co(z1, z3, z4, . . . , zc, zc+2). We then claim that A = {z1, z2, z4, . . . , zc+2} is convexly independent. The simplex partition property implies that zi / ∈co(A\zi) for i = c + 1, c + 2. The anti-exchange property implies that zi / ∈co(A\zi) for i = 1, 2, 4, . . . , c. Theorem 5.7. Let (X, F) be a convexity with the anti-exchange property and the simplex partition property, and with Caratheodory number c ≥3. Then for any positive integer n there exists a positive integer N(n) such that any set Y ⊆X of N(n) points in nice position contains n convexly independent points. Proof. Put N(n) = Rc+1(n, c + 2). Another variant of the Erd˝ os-Szekeres theorem for convexities satisfying almost the same properties as the above theorem is given by Korte and Lov´ asz . If (X, F) is a convexity, then a set A ⊆X is called free if it is both convexly independent and a member of F. For realizable (X, F), a free set is the set of vertices of an empty convex polytope. The Helly number of (X, F) is the smallest integer h such that for any subfamily B of F, if each h or fewer members of B have nonempty intersection, then the intersection of all members of B is nonempty. 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Department of Mathematical Sciences, George Mason University, 4400 University Drive, Fairfax, VA 22030 Department of Mathematical Sciences, George Mason University, 4400 University Drive, Fairfax, VA 22030
15715	https://epubs.siam.org/doi/10.1137/050626363	Cycle Decompositions of $K_{\lowercase{n,n}}-I$ \| SIAM Journal on Discrete Mathematics Skip to main content Search Search This Journal This Journal Anywhere Books Journals Proceedings Quick Search in Journals Enter Search Terms Search Quick Search anywhere Enter Search Terms Search Quick Search anywhere Enter Search Terms Search Quick Search anywhere Enter Search Terms Search Quick Search anywhere Enter Search Terms Search Advanced Search 0 Register / Sign In Access via your Institution Skip main navigationClose Drawer Menu Open Drawer Menu Menu Journals SIAM Review Multiscale Modeling & Simulation SIAM Journal on Applied Algebra and Geometry SIAM Journal on Applied Dynamical Systems SIAM Journal on Applied Mathematics SIAM Journal on Computing SIAM Journal on Control and Optimization SIAM Journal on Discrete Mathematics SIAM Journal on Financial Mathematics SIAM Journal on Imaging Sciences SIAM Journal on Life Sciences SIAM Journal on Mathematical Analysis SIAM Journal on Mathematics of Data Science SIAM Journal on Matrix Analysis and Applications SIAM Journal on Numerical Analysis SIAM Journal on Optimization SIAM Journal on Scientific Computing SIAM/ASA Journal on Uncertainty Quantification Theory of Probability & Its Applications Locus E-books Bookstore Proceedings For Authors Journal Authors Book Authors ICM Authors For Librarians Collections Epidemiology Collection High Impact Article Collection JOIN SIAM HELP/CONTACT US Journal Home Current Issue All Issues About About this Journal Editorial Policy Editorial Board Instructions for Authors Instructions for Referees Submit Subscribe Share Share on Facebook X LinkedIn Email HomeSIAM Journal on Discrete MathematicsVol. 20, Iss. 3 (2006)10.1137/050626363 Previous articleNext article Cycle Decompositions of 𝐾\l o w e r c a s e 𝑛,𝑛−𝐼 Authors: Jun Ma, Liqun Pu, and Hao ShenAuthors Info & Affiliations Get Access BibTeX Tools Add to favorites Download Citations Track Citations Contents ###### PREVIOUS ARTICLE A Note on Quasi‐triangulated Graphs Previous###### NEXT ARTICLE Shortest Paths in the Tower of Hanoi Graph and Finite Automata Next Abstract References Information & Authors Metrics & Citations Get Access References Figures Tables Media Share Abstract Let 𝐾 𝑛,𝑛 denote the complete bipartite graph with n vertices in each bipartition set and 𝐾 𝑛,𝑛−𝐼 denote 𝐾 𝑛,𝑛 with a 1‐factor removed. An m‐cycle system of 𝐾 𝑛,𝑛−𝐼 is a collection T of m‐cycles such that each edge of 𝐾 𝑛,𝑛−𝐼 is contained in a unique m‐cycle of T. In this paper, it is proved that the necessary and sufficient conditions for the existence of an m‐cycle system of 𝐾 𝑛,𝑛−𝐼 are 𝑛≡1(m⁢o⁢d 2), 𝑚≡0(m⁢o⁢d 2), 4≤𝑚≤2⁢𝑛, and 𝑛⁢(𝑛−1)≡0(m⁢o⁢d 𝑚). MSC codes 05C38 Keywords decomposition cycle complete bipartite graph 1‐factor Get full access to this article View all available purchase options and get full access to this article. Get Access Sign in as an individual or via your institution References 1. D. Archdeacon, M. Debowsky, J. Dinitz, and H. Gavlas, Cycle systems in the complete bipartite graph minus a one‐factor, Discrete Math., 284 (2004), pp. 37–43. Crossref Web of Science Google Scholar 2. D. Sotteau, Decompositions of 𝐾 𝑚,𝑛⁡(𝐾∗𝑚,𝑛) into cycles (circuits) of length 2⁢𝑘, J. Combin. Theory Ser. B, 29 (1981), pp. 75–81. Crossref Web of Science Google Scholar 3. B. Alspach and H. Gavlas, Cycle decompositions of 𝐾 𝑛 and 𝐾 𝑛−𝐼, J. Combin. Theory Ser. B, 81 (2001), pp. 77–99. Crossref Web of Science Google Scholar 4. M. Šajna, Cycle decompositions, III: Complete graphs and fixed length cycles, J. Combin. Des., 10 (2002), pp. 27–78. Crossref Web of Science Google Scholar 5. R. Laskar and B. Auerbach, On decomposition of r‐partite graphs into edge‐disjoint Hamilton circuits, Discrete Math., 14 (1976), pp. 265–268. Crossref Web of Science Google Scholar Show all references Information & Authors Information Authors Information Published In SIAM Journal on Discrete Mathematics Volume 20 • Issue 3 • January 2006 Pages: 603 - 609 DOI: 10.1137/050626363 ISSN (online): 1095-7146 Copyright Copyright © 2006 Society for Industrial and Applied Mathematics. History Submitted: 9 March 2005 Accepted: 23 February 2006 Published online: 25 August 2006 MSC codes 05C38 Keywords decomposition cycle complete bipartite graph 1‐factor Authors Affiliations Expand All Jun Ma View all articles by this author Liqun Pu View all articles by this author Hao Shen View all articles by this author Metrics & Citations Metrics Citations Metrics Metrics Downloads Citations No data available. 48 15 Total 6 Months 12 Months Total number of downloads Citations If you have the appropriate software installed, you can download article citation data to the citation manager of your choice. Simply select your manager software from the list below and click Download. Format - [x] Direct import Cited By Decomposition of the complete bipartite graphs plus a 1-factor Km,m⊕I into paths and cycles of length four 1 Jan 2024 Cross Ref Decomposing the tensor product of complete graphs into cycles of length eight Journal of Information and Optimization Sciences, Vol. 43, No. 4 \| 14 June 2022 Cross Ref C10 -decompositions of the tensor product of complete graphs Journal of Information and Optimization Sciences, Vol. 42, No. 6 \| 17 June 2021 Cross Ref Six-cycle systems Mathematica Slovaca, Vol. 71, No. 3 \| 8 June 2021 Cross Ref 2p -cycle decompositions of some regular graphs and digraphs Discrete Mathematics, Vol. 341, No. 8 \| 1 Aug 2018 Cross Ref Switching techniques for edge decompositions of graphs Surveys in Combinatorics 2017 \| 30 Jun 2017 Cross Ref Decomposition of Complete Bipartite Graphs into Cycles of Distinct Even Lengths Graphs and Combinatorics, Vol. 32, No. 4 \| 1 December 2015 Cross Ref Decomposing Complete Equipartite Multigraphs into Cycles of Variable Lengths: The Amalgamation‐detachment Approach Journal of Combinatorial Designs, Vol. 24, No. 4 \| 9 December 2014 Cross Ref Decomposition of the complete bipartite multigraph into cycles and stars Discrete Mathematics, Vol. 338, No. 8 \| 1 Aug 2015 Cross Ref DECOMPOSITIONS OF MULTICROWNS INTO CYCLES AND STARS Taiwanese Journal of Mathematics, Vol. 19, No. 4 \| 1 Aug 2015 Cross Ref Decomposition of the complete bipartite graph with a 1-factor removed into cycles and stars Discrete Mathematics, Vol. 313, No. 20 \| 1 Oct 2013 Cross Ref Multidecompositions of Several Graph Products Graphs and Combinatorics, Vol. 29, No. 3 \| 4 January 2012 Cross Ref Decomposing Various Graphs into Short Even-Length Cycles Annals of Combinatorics, Vol. 16, No. 3 \| 1 May 2012 Cross Ref On DRC-covering of K(n) t λ by cycles Computer Science and Information Systems, Vol. 6, No. 2 \| 1 Jan 2009 Cross Ref Cycle Systems in the Complete Bipartite Graph Plus a One-Factor Liqun Pu, Hao Shen, Jun Ma,and San Ling SIAM Journal on Discrete Mathematics, Vol. 21, No. 4 \| 25 January 2008 AbstractPDF (161 KB) Figures Tables Media Share Share Copy the content Link Copy Link Copied! Copying failed. Share with email Email a colleague Share on social media FacebookX (formerly Twitter)LinkedInemail Get Access Get Access PurchaseSave for laterItem saved, go to cart Article Pay-Per-View $40.00 Add to cart Article Pay-Per-View Checkout Access via your Institution Questions about how to access this content? Contact SIAM at service@siam.org. References References 1. D. Archdeacon, M. Debowsky, J. Dinitz, and H. Gavlas, Cycle systems in the complete bipartite graph minus a one‐factor, Discrete Math., 284 (2004), pp. 37–43. Crossref Web of Science Google Scholar 2. D. Sotteau, Decompositions of 𝐾 𝑚,𝑛⁡(𝐾∗𝑚,𝑛) into cycles (circuits) of length 2⁢𝑘, J. Combin. Theory Ser. B, 29 (1981), pp. 75–81. Crossref Web of Science Google Scholar 3. B. Alspach and H. Gavlas, Cycle decompositions of 𝐾 𝑛 and 𝐾 𝑛−𝐼, J. Combin. Theory Ser. B, 81 (2001), pp. 77–99. Crossref Web of Science Google Scholar 4. M. Šajna, Cycle decompositions, III: Complete graphs and fixed length cycles, J. Combin. Des., 10 (2002), pp. 27–78. Crossref Web of Science Google Scholar 5. R. Laskar and B. Auerbach, On decomposition of r‐partite graphs into edge‐disjoint Hamilton circuits, Discrete Math., 14 (1976), pp. 265–268. Crossref Web of Science Google Scholar Recommended Content Cycle Systems in the Complete Bipartite Graph Plus a One-Factor Liqun Pu , Hao Shen , Jun Ma , San Ling Abstract Let $K_{n,n}$ denote the complete bipartite graph with n vertices in each partite set and $K_{n,n}+I$ denote $K_{n,n}$ with a one-factor added. It is proved in this paper that there exists an m-cycle system of $K_{n,n}+I$ if and only if $n \equiv 1 (\rm{mod} 2)$, $m \equiv 0 (\rm{mod} 2)$, $4 \leq m \leq 2n$, and $n(n+1) \equiv$ 0 (mod m). The Existence of $r\times4$ Grid-Block Designs with $r=3,4$ Rucong Zhang , Gennian Ge , Alan C. H. Ling , Hung-Lin Fu , Yukiyasu Mutoh Abstract For a v-set V, let $\mathcal{A}$ be a collection of $r\times c$ arrays with elements in V. A pair $(V,\mathcal{A})$ is called an $r\times c$ grid-block design if every two distinct elements i and j in V occur exactly once in the same row or in the same column of an array in $\mathcal{A}$. This design originated from the use of DNA library screening. In this paper, we show the existence of $r\times 4$ grid-block designs with $r=3,4$. We settle completely for the case of $r=4$ and almost completely for the case of $r=3$, leaving 15 orders undetermined. Structure of Expansion-Contraction Matrices in the Inclusion Principle for Dynamic Systems Lubomír Bakule , José Rodellar , Josep M. Rossell Abstract This paper characterizes a general structure of complementary matrices involved in the input-state-output inclusion principle for linear time invariant dynamic systems. Aggregations and restrictions are adopted as two practically important classes within this scheme. Further, contractibility conditions for feedback controllers are presented, including the decentralized control design. The identified structure enables the formulation of conditions on the complementary matrices which can be useful for analysis and synthesis of control problems. Mod (2p + 1)-Orientations and $K_{1,2p+1}$-Decompositions Hong-Jian Lai Abstract In this paper, we establish an equivalence between the contractible graphs with respect to the mod $(2p+1)$-orientability and the graphs with $K_{1, 2p+1}$-decompositions. This is applied to disprove a conjecture proposed by Barat and Thomassen that every 4-edge-connected simple planar graph G with $\|E(G)\|\equiv 0$ (mod 3) has a claw decomposition. Decycling Cartesian Products of Two Cycles David A. Pike , Yubo Zou Abstract The decycling number $\nabla(G)$ of a graph G is the smallest number of vertices which can be removed from G so that the resultant graph contains no cycles. In this paper, we study the decycling number for the family of graphs consisting of the Cartesian product of two cycles. We completely solve the problem of determining the decycling number of $C_m \square C_n$ for all m and n. Moreover, we find a vertex set T that yields a maximum induced tree in $C_m\square C_n$. 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15716	https://physics.stackexchange.com/questions/288172/why-is-electric-field-lines-away-from-and-toward	Skip to main content Why is electric field lines away from (+) and toward (-)? Ask Question Asked Modified 6 years ago Viewed 34k times This question shows research effort; it is useful and clear 4 Save this question. Show activity on this post. I have a questions about the electric field lines. Well in the basic learning, we know that: The electric field lines extend away from a positive charge They move forward a negative charge Let's take parallel plates, which make a uniform electric field.If we take the basic learning, which I mentioned, in accounts, it's very easy to understand that this is the direction of a positive charge object if we put the object between the plate (move away the positive charged plate and toward negatively charged plate). The problem is if we put a negative charged object between. Isn't the electric field reversed ? electrostatics electric-fields conventions Share CC BY-SA 3.0 Improve this question Follow this question to receive notifications edited Oct 22, 2016 at 15:56 Spirine 2,1131717 silver badges2727 bronze badges asked Oct 22, 2016 at 15:22 PandoraU.U.DPandoraU.U.D 29722 gold badges44 silver badges1111 bronze badges Add a comment \| 3 Answers 3 Reset to default This answer is useful 6 Save this answer. Show activity on this post. The direction of the field is defined to be the direction of the force on a positively charged test particle. Positive charges always move away from other +ve charges and towards -ve charges. As @Charlie says, it is a convention, like driving on the right (or left), or which pin on a plug is "live". So that everyone can agree on the result of a calculation, we all have to define it the same way. It could be defined the other way round, but it isn't. And we can't have both - that would be confusing. Share CC BY-SA 3.0 Improve this answer Follow this answer to receive notifications answered Oct 22, 2016 at 18:37 sammy gerbilsammy gerbil 27.6k66 gold badges3636 silver badges7272 bronze badges 2 2 Oh I get it. So (I just want to be sure that I misunderstood). The conclusion is the electric field lines are describe to go away from positive charges and stop at negative charges. This statement is the standard and complete one despite being reversed. You say we can't have both because of the fact that things would go confusing. Am I correct !? PandoraU.U.D – PandoraU.U.D 2016-10-24 03:36:39 +00:00 Commented Oct 24, 2016 at 3:36 @PandoraU.U.D that is correct. It is convention. dimes – dimes 2021-02-07 22:21:17 +00:00 Commented Feb 7, 2021 at 22:21 Add a comment \| This answer is useful 2 Save this answer. Show activity on this post. The electrostatic field is a smart way to work with the Coulomb's law. We know that a charge Q located in P will produce a force Fe=Qq4πϵ0∥PM∥3PM on a charge q located in M. However, this expression of Fe is not very convenient when you work with a continuous distribution of charges, that's why we prefer to work with what we call the electrostatic field, no more than a mathematical object which facilitate the maths behind physics. We define the electrostatic field created by the charge Q(P) as the field E(M) that, when multiplied by q, gives the force of the charge Q on a charge q located in M, ie. E(M)=Q4πϵ0∥PM∥3PM Using this definition of the electrostatic field, we can notice that it always goes away from positive charges, and toward negative charges. A field line is defined as a line that is always tangent to the field, and is oriented by the field. Since the electrostatic field is always directed away from positive charges and toward negative charges, field lines must go away from positive charges and toward negative ones. However, we can't say that field lines are oriented by the motion of any charged object: indeed, as you pointed out, the electromagnetic force acting on a charge q depends on the signe of q since this force is F=qE. The electrostatic field doesn't change, it is only the electrostatic force that is reversed by replacing a positive charge by a negative one. Share CC BY-SA 3.0 Improve this answer Follow this answer to receive notifications edited Oct 22, 2016 at 16:13 answered Oct 22, 2016 at 15:40 SpirineSpirine 2,1131717 silver badges2727 bronze badges 7 Oh I understand your explanation. Thank you :). But I mean how do we know the electric field always directs always from a positive charge and toward a negative charge. I understand how the theory work. But still don't know how this matter is figured out. PandoraU.U.D – PandoraU.U.D 2016-10-22 16:02:14 +00:00 Commented Oct 22, 2016 at 16:02 1 I guess that the answer is "convention": you get correct results as long as you chose a convention (electric field from positive to negative charge or viceversa) and you stick to the one you chose. One could have chosen the opposite one (redefining what needs to be redefined) and still gets right results. Charlie – Charlie 2016-10-22 16:21:00 +00:00 Commented Oct 22, 2016 at 16:21 @Charlie Even if the orientation is caused by conventions (why does a proton have a positive charge ?), it is not directly a convention. The convention was choosen when Coulomb stated his law, and the orientation of the electrostatic field can be understand as a consequence of Coulomb's law. Spirine – Spirine 2016-10-22 16:28:09 +00:00 Commented Oct 22, 2016 at 16:28 2 @Spirine The Coulomb's law, vectorially speaking, just tells you that two charges of the same sign repulse each other and two charges with opposite sign attract each other. When you define the electric field from the electrostatic force you have to choose a convention. Charlie – Charlie 2016-10-22 16:37:53 +00:00 Commented Oct 22, 2016 at 16:37 Oh thank you everyone for helping me a lot. So in conclusion (just that I understand well). The sum is that: PandoraU.U.D – PandoraU.U.D 2016-10-24 03:33:50 +00:00 Commented Oct 24, 2016 at 3:33 \| Show 2 more comments This answer is useful 0 Save this answer. Show activity on this post. It's arbitrary. The physical properties of the "positive" and "negative" charge types play no role. We have chosen to use positive test charges by convention. Why? Because, since the electric field at a point is described by a vector, we can describe things in terms of negative and positive. It turns out that when using the equation F=qE, where F and E are vectors, if q is allowed to be positive or negative, and we think of the test charges in the electric field as being positive, it is easier to understand and talk about the electric field model. Remember that fields are just used as a model, albeit a very helpful and relevant model. Share CC BY-SA 4.0 Improve this answer Follow this answer to receive notifications edited Aug 23, 2019 at 20:19 Thomas Fritsch 42.8k1313 gold badges7878 silver badges150150 bronze badges answered Aug 23, 2019 at 20:12 BooleanBuckeyeBooleanBuckeye 1 Add a comment \| Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions electrostatics electric-fields conventions See similar questions with these tags. 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15717	https://arxiv.org/pdf/2210.14301	arXiv:2210.14301v1 [math.CO] 25 Oct 2022 Some complementary Gray codes Adam Hoyt 18-7825 Avenue Mountain Sights Montr´ eal QC H4P 2B1, Canada ahoyt@connect.carleton.ca Brett Stevens ∗ School of Mathematics and Statistics Carleton University 1125 Colonel By Drive Ottawa ON K1S 5B6, Canada brett@math.carleton.ca Abstract A complementary Gray code for binary n-tuples is one that, when all the tuples are complemented, is identical to itself; this is equiv-alent to the complement of the first half of the code being identical to the second half. We generalize the notion of complementary to q-ary n-tuples, fixed size combinations of an n-set and permutations and, in each case, construct complementary Gray codes. We relax, as weakly as possible, the notions of complementary to cases where necessary conditions for existence are violated and construct Gray codes within the weakened definitions: these include binary n-tuples when n is odd and Lee metric q-ary n-tuples when n is odd and q is even. Finally a lemma used in the construction for permutations offers the first known cyclic Gray code for the permutations of a particular family of multisets. ∗ Supported by NSERC, CFI, OIT and the Ontario MRI. 11 Introduction A Gray code for binary n-tuples is a linear ordering of the tuples so that consecutive tuples differ in exactly one bit position. It is cyclic if the first and last tuples also differ in only one bit position. Binary n-tuples can be associated with subsets of an n-set by considering the tuple to be the incidence vector of the subset. In this context, the complement of a tuple, w, is the incidence vector of the complement of the subset corresponding to w; it can be directly constructed by swapping the positions of all 0s and 1s. We denote the complement of w by w.A “complementary” Gray code is a Gray code where the complement, w,of any word, w, occurs exactly half way through the code after w appears. The existence of complementary Gray codes for binary n-tuples for n even is well known, although we are unaware of its original publication . When n is even we construct such a complementary Gray code as follows: let P be any Hamilton path from 00 · · · 0 to 11 · · · 1 in the binary ( n−1)-dimensional hypercube. Then 0P, 1P gives a complementary Gray code. We use P to denote the complement of the words in P in the same order . Similarly, when we use P with any operation we mean to apply that operation to each element of P and leave the resulting elements in the same order as from P . For example. P + x indicates adding x to every element of P and keeping the resulting objects in the same order as P . 0 P means prefixing every element of P with a 0. There are several ways to make the necessary ingredient, the required Hamilton path, for this construction. For instance, Frank Ruskey suggests the following construction: E(n) : 0n → 1n O(n) : 0n → 01 n−1 E(n + 1) = O(n)0 , BRRC (n)1 O(n + 1) = E(n)0 , BRC (n)1 where BRC (n) is the complement of Gray’s original binary reflected Gray code and BRRC (n) is the reversal of the complement of the original bi-nary reflected Gray code . The monotone Gray codes constructed by Savage and Winkler are also suitable ingredients for the construction . For example, Table 1 shows a complementary binary Gray code with 4 bits. We call the parity of the number of 1s in a binary n-tuple, w, the parity of w. Since any binary Gray code changes precisely one bit between every 0000 0001 0011 0010 0110 0100 0101 0111 1111 1110 1100 1101 1001 1011 1010 1000 Table 1: A complementary binary Gray code with 4 bits. pair of consecutive words, the parity of the words alternates between even and odd number. When n ≥ 2, the pair of words at distance 2 n−1 in the code must have the same parity since the distance between them is even. However, when n is odd, complementary words have opposite parity. Thus a complementary n-bit binary Gray code cannot exist when n is odd. In this paper we will suitably define and construct complementary Gray codes for binary n-tuples for n odd, q-ary n-tuples, combinations of an n-set and permutations of n-objects. When the existence of a code with a strict definition of “complementary” is forbidden by necessary conditions, we will relax the definition as little as possible to construct a code. We start by introducing some definitions and notations that we will use in this article. For a word, w ∈ Znq , we denote the set of non-zero indices as the support of w and the cardinality of this set is the word’s weight . When q = 2 there is a bijective correspondence between words of length n, w ∈ Zn 2 and the subsets Sw ⊆ { 1, 2, · · · , n }. The bijection map is simply the support of the word. For example, 001101 corresponds to the set {3, 4, 6}. There are two natural metrics that can be placed on Znq . The Hamming distance between v, w ∈ Znq is the number of entries where v and w differ, or equivalently, the weight of v − w. Let v = v1, v 2, · · · , v n and w = w1, w 2, · · · , w n. The Lee distance in the ith coordinate is the size of the difference between vi and wi in Zq regarded as a cycle; in other words min {\| vi − wi\|, q − \| vi − wi\|} . The Lee distance between words is then sum of the Lee distances over all coordinates : n ∑ i=1 min {\| vi − wi\|, q − \| vi − wi\|} . In graph theory terminology, Hamming distances correspond to dis-tances in the graph Kq✷Kq✷ · · · ✷Kq and Lee distances correspond to distances in the multi-dimensional torus Cq ✷Cq ✷ · · · ✷Cq , where vertices are labelled by length n q -ary words. In either case, a (cyclic) Gray code with consecutive words at distance one is equivalent to a Hamil-ton path (cycle) in the corresponding graph. 2 Gray codes of q-ary n-tuples We first address the problem of defining and constructing complementary Gray codes for binary n-tuples when n is odd. 2.1 Binary n-tuples for n odd When n is odd, there are two possible ways to relax the notion of “com-plementary”: we can drop the requirement that all words appear in the complementary code or we can keep all words, but allow complementary words to be almost, rather than exactly, halfway through the code. We show that one can construct Gray codes for all odd n in either case and do so as optimally as possible: leaving out the fewest words or having complementary words be exactly one position off from halfway. We start with a result that will be used in the construction of these Gray codes. Lemma 2.1. Let G be a binary Gray code on n bits with G(i) denoting the ith word (taken modulo 2n). Then G′(i) = { G(⌊i/ 2⌋)0 if i ≡ 0, 3 mod 4 G(⌊i/ 2⌋)1 if i ≡ 1, 2 mod 4 defines a binary Gray code on n + 1 bits. G′ is cyclic if and only if G is cyclic. Applying this lemma repeatedly to the 1 bit binary cyclic Gray code, 0, 1, yields an alternative construction of Gray’s original binary reflected Gray code. We can use Lemma 2.1 to construct a complementary Gray code missing only two words. Theorem 2.2. If n is odd there exists a Gray ordering of Zn 2 \ { 00 · · · 0, 11 · · · 1} such that the distance between any pair of complementary words is 2n−1 −1.Proof. Let G be a binary Gray code on n − 2 bits that begins with 00 · · · 0and ends with 11 · · · 1, either using Ruskey’s construction or monotone bi-nary Gray codes, both discussed in Section 1. Applying Lemma 2.1 yields an ( n − 1)-bit Gray code, H, which begins with words 00 · · · 00, 00 · · · 01 and ends with 11 · · · 11, 11 · · · 10. Delete the first word, 00 · · · 0 to produce H′. Now construct Γ = 0 H′, 1H′ Table 2 gives an example of this Gray code for n = 5 bits. 00001 00011 00010 00110 00111 00101 00100 01100 01101 01001 01000 01010 01011 01111 01110 11110 11100 11101 11001 11000 11010 11011 10011 10010 10110 10111 10101 10100 10000 10001 Table 2: A complementary Gray ordering of Z52 \ { 00000 , 11111 }.We can also use Lemma 2.1 to construct a Gray code containing all words such that the distance between complementary words is only one off half the length of the code. Theorem 2.3. If n is odd then there exists a Gray ordering of all n-bit binary words such that the distance between any pair of complementary words is 2n−1 ± 1.Proof. Let G be any self complementary ( n − 1)-bit Gray code. Apply Lemma 2.1 to yield an n-bit Gray code, H. Consider the word w and its complement w. Their length n − 1 prefixes are also complements and so are at distance 2 n−2 in G. By the construction of Lemma 2.1 w and w are at a distance at most 2 n−1 + 1 and at least 2 n−1 − 1. The parity of the words’ weights establishes that complements can’t be at distance 2 n−1. Table 3 gives an example of this Gray code for n = 5 bits. 00000 00001 00011 00010 00110 00111 00101 00100 01100 01101 01001 01000 01010 01011 01111 01110 11110 11111 11101 11100 11000 11001 11011 11010 10010 10011 10111 10110 10100 10101 10001 10000 Table 3: A Gray ordering of Z52 with complementary words at distance 24 ± 1 = 15 or 17. 2.2 q-ary n-tuples for q ≥ 3. How do we generalize the idea of complementary to q-ary n-tuples? One often thinks of complements in Zn 2 as being the two words that add to 11 · · · 1. However since +1 = −1 in Z2 this can also be a subtraction. This is the more natural notion to generalize. In an ordering of Znq , we can ask that consecutive words be distance one in either the Hamming or the Lee metric. We will address both metrics. Definition 2.4. A quasi-complementary Hamming (Lee) metric Gray code of q-ary n-tuples is an ordering of the qn words such that consecutive words have Hamming (Lee) distance 1 and G(( i + qn−1) mod qn) = G(i) + 111 · · · 1where G(i) is the ith tuple. Addition of tuples is done in Znq and addition of indices is performed in Zqn .This definition coincides with the standard definition of complementary when q = 2. It ensures that all words that are separated by a multiple of qn−1 are as far apart in the Hamming metric as possible. Additionally, in the Lee metric, the separation in the code between words from the set is a linear function of their Lee distance. An example of a quasi-complementary Lee metric Gray code of ternary 3-tuples is given in Table 4. 000 002 001 021 022 020 010 012 011 111 110 112 102 100 101 121 120 122 222 221 220 210 211 212 202 201 200 Table 4: A quasi-complementary Lee metric Gray code for Z33.There is a necessary condition that follows if we insist that consecutive words have distance one in the Lee metric: Proposition 2.5. If a quasi-complementary Lee metric Gray code of q-ary n-tuples exists, then either n is even, q is odd or n = 1 .Proof. Let Ci denote the 2-regular connected graph on i nodes, an i-cycle. If n ≥ 3 is odd and q is even then the desired Gray code is a Hamiltonian cycle in ✷0≤i<n Cq = Cq ✷Cq ✷ · · · ✷Cq which is a bipartite graph since it is the Cartesian product of bipartite graphs. Since the code is quasi-complementary the words 00 · · · 0 and 11 · · · 1 are distance qn−1 apart, which is even but are at an odd Lee distance which is a contradiction. The code for n = 1 is trivially constructed by taking the elements of Zq in the order they are generated from 1. Our primary construction is the higher modulus analog of standard construction of binary n-tuples when n is even, given on Page 1. Lemma 2.6. If a Hamming (Lee) metric Gray code of q-ary (n − 1) -tuples exists that starts with 00 · · · 0 and ends with 11 · · · 1 then a quasi-complementary Hamming (Lee) metric Gray code of q-ary n-tuples exists. Proof. Let A be the Gray code of q-ary ( n − 1)-tuples satisfying the hy-pothesis. Then 0A + 00 · · · 0, 0A + 11 · · · 1, . . . , 0A + ( q − 1)( q − 1) · · · (q − 1) is the desired quasi-complementary Gray code. We now establish the existence of the necessary ingredient codes for the construction. First when n is even. Lemma 2.7. If n′ is odd then there exists a Lee metric Gray code of q-ary n′-tuples that starts with 00 · · · 0 and ends with 11 · · · 1 for any q ≥ 3.Proof. We extend a monotonic binary Gray code to q-ary words by replac-ing each binary word, w, with a ( q − 1)-ary Gray code on the support of w with symbols {1, 2, . . . , q − 1}. We will replace every binary word w by an instance of the ( q − 1)-ary reflected Gray code . The positions with 0 in w will have 0 fixed in these positions for the entire code that is replacing w. The alphabet of the code replacing w is {1, 2, . . . , q − 1} with sentinels 1 and q − 1. The ( q − 1)-ary reflected Gray code never changes one sentinel directly to the other so it will remain valid for the Lee metric on Zq . Ad-ditionally, the ( q − 1)-ary reflected Gray code has the property that if all the values start set to sentinels then the code will end with all positions set to sentinels. The only decision to make is with which sentinel values the positions start. If w′ is the word preceding w in the binary monotone Gray code, then any position where w and w′ have a common 1, the ( q − 1)-ary reflected Gray code replacing w will be initially set at the sentinel value where the code replacing w′ ended in this position. There is at most one position in w that is non-zero which does not inherit its initial sentinel values from w′ in this way (because the Hamming distance of w and w′ is one). For this position, if there is one, we can choose either sentinel value because both ‘1’ and ‘ q − 1’ are Lee distance one from ‘0’. The last word in the binary monotone Gray code is 11 · · · 1. This is replaced by an instance of the ( q − 1)-ary reflected Gray code which will end in a word that has either a 1 or q − 1 in each position. Which of these sentinels each position contains depends only and directly on the most recent choice of a sentinel value for this position to start an instance of a (q − 1)-ary reflected Gray code. In order to finish with the word 11 · · · 1 we simply make the right choice for each position. Table 5 gives the Gray code for q = 4 and n′ = 3 constructed by Lemma 2.7. The positions where the correct choice of sentinel start posi-tions must be made to guarantee ending in 111 are underlined. 000 001 002 003 013 012 011 021 022 023 013 012 011 010 020 030 130 120 110 210 220 230 330 320 310 300 200 100 101 102 103 203 202 201 301 302 303 33 3 332 331 321 322 323 313 312 311 211 212 213 223 222 221 231 232 233 133 132 131 121 122 123 113 112 111 Table 5: The Gray code for 4-ary 3-tuples constructed by Lemma 2.7. Corollary 2.8. There exists a quasi-complementary Lee metric Gray code of q-ary n-tuples whenever n ∈ N is even. When q is odd we construct an different ingredient code. Lemma 2.9. If q is odd then there exists a Lee metric Gray code of q-ary n-tuples that starts with 00 · · · 0 and ends with 11 · · · 1 for any n ≥ 1.Proof. By appealing to Lemma 2.7 we may assume that n is even and therefore that there exists a Gray code, G, of q-ary ( n − 1)-tuples that starts with 00 · · · 0 and ends with 11 · · · 1. Now 0G, (q − 1) GR, (q − 2) G, (q − 3) GR, . . . , 1G is the desired Gray code for q-ary n-tuples. The notation GR indicates the reversal of G: the tuples from G listed in the reverse order. Corollary 2.10. There exists a quasi-complementary Lee metric Gray code of q-ary n-tuples whenever q ≥ 3 is odd. Table 4 shows a quasi-complementary Lee metric Gray code for q = n = 3. In the case that n > 1 is odd and q is even we show that one can con-struct the best possible code in the first of the senses described in Subsec-tion 2.1, i.e. removing words from the code. Since the Lee metric distance between w and w + 11 · · · 1 is unavoidably odd, we must remove at least one word between them (in other words, miss the vertex in the cycle on the Lee metric graph) so that the separation between them in the code matches the parity of the distance between them. Thus, at least q words must be removed in total from the code and this lower bound on the number of missed words can be achieved. Theorem 2.11. If n is odd and q is even, then there exists a quasi-complementary Lee metric Gray code of q-ary n-tuples, missing w + ii · · · i, 0 ≤ i < q for any choice of w ∈ Znq .Proof. We use Lemma 2.6 and build the ingredient in a similar manner to Lemma 2.7. There exists a binary monotonic Gray code of n-tuples that starts with 100 · · · 0, 00 · · · 0 and ends at 11 · · · 1 . We will expand each word to a ( q − 1)-ary reflected Gray code on its support just as in Lemma 2.7. This will produce a q-ary Gray code, 100 · · · 0, 200 · · · 0, . . . , (q − 1)00 · · · 0, 00 · · · 0, . . . , 11 · · · 1. We want this code to begin with 00 · · · 0 so we remove the initial tuples in pairs, i00 · · · 0 and ( i + 1)00 · · · 0, and insert them between the consecutive tuples i00 · · · 0( q − 1)0 · · · 0 and ( i + 1)00 · · · 0( q − 1)0 · · · 0 (or consecutive tuples i00 · · · 010 · · · 0 and ( i + 1)00 · · · 010 · · · 0) which appear in the first (q − 1)-ary reflected Gray code placed on the first binary tuple of weight 2 that includes the first coordinate. To ensure that the target pair of consecutive words exist consecutively in the ( q − 1)-ary reflected Gray code, we must treat the non-zero coordinate which is not the first as the “low order” position in the code, the one which changes most often. Since q − 1 is odd the tuple ( q − 1)00 · · · 0 will be left over and this is discarded, as we knew one tuple must be. In the monotonic binary Gray code between the first weight 2 tuple including the first coordinate and the last, full weight, tuple, every coordinate will be zero at least once. This means that this slight alteration of the code does not interfere with the choice of sentinels necessary to ensure that we end with the q-ary tuple 11 · · · 1 in exactly the same manner at Lemma 2.7. We now have a code which satisfies the hypotheses of Lemma 2.6 and is missing ( q − 1)00 · · · 0. The Lee metric Gray code constructed will be missing the words 0( q − 1)00 · · · 0 + ii · · · i, 0 ≤ i < q . We can choose the missing words arbitrarily to w +ii · · · i by the addition of w −0( q −1)00 · · · 0to every word in the sequence. The second relaxation of “complementary” in the binary case with an odd number of bits asks for complementary words to be separated by one more or one less than half the length of the code: qn−1. We show that this is not possible when q ≥ 4 is even. Theorem 2.12. When q ≥ 4 is even and n ≥ 3 is odd, there cannot exist a Lee metric Gray code of q-ary n-tuples with the property that words w and w + 11 · · · 1 have separation qn−1 ± 1.Proof. Let such a code be G and the ith word be G(i). We will generate a contradiction. If G(i) + 11 · · · 1 = G(i + qn−1 + ǫ)then let e(i) = ǫ = ±1. And let p(i) be the position of the word that changes between G(i) and G(i + 1). If two consecutive ǫs are both positive, e(i) = e(i + 1) = 1, then it is easy to check that e(i + 2) = 1 since the word at separation qn−1 − 1 from G(i + 2) is already determined to be G(i) + 11 · · · 1 6 = G(i + 2) + 11 · · · 1. And since it is not possible that all ǫs are 1, this forbids consecutive ǫs from being positive. Similarly it can be shown that two consecutive negative ǫs are impossible. So the ǫs must alternate between 1 and -1. Now that we know the behaviour of e(), we examine the behaviour of p(). Consider a place in the code where p(i) = 1 and p(i + 1) 6 = 1 (one must exist because n > 1). Then G(i) = xw, G(i + 1) = ( x ± 1) w, G(i + 2) = ( x ± 1) w′, where x is a q-ary digit and w, w ′ ∈ Zn−1 q . If e(i) = 1, for j = i + qn−1, we have G(j) = ( x ± 1 + 1)( w + 11 · · · 1) , G(j + 1) = ( x + 1)( w + 11 · · · 1) ,G(j + 2) =? , G(j + 3) = ( x ± 1 + 1)( w′ + 11 · · · 1) . We consider what word is G(j + 2)? If p(j + 1) = 1 then G(j + 2) = (x ± − 1 + 1)( w + 11 · · · 1) which is to be distance one in the Lee metric from G(j + 3) = ( x ± 1 + 1)( w′ + 11 · · · 1) and thus x ± − 1 + 1 = x ± 1 + 1 from which we can conclude that 1 = −1 but this is not possible since q > 2. Thus we have that p(j + 1) 6 = 1 and so G(j + 2) = ( x + 1) w′′ . Since x + 1 6 = x ± 1 + 1 we must have that p(j + 2) = 1 = p(j) and w′′ = w′.This forces p(i + 2) = 1. The same argument can be applied to p(i + 1) and p(i + 2), which differ, to conclude that p(i + 3) = p(i + 1) and in general p(i) = p(i mod 2) for all i. Since n > 2 this gives a contradiction. The proof when e(i) = −1 is similar. Even though a Gray code with complementary words separated by qn−1 ± 1 does not exist, we can bound the change in separation of words that makes such a Lee metric Gray code possible. Theorem 2.13. When q ≥ 4 is even and n is odd, there exists a Lee metric Gray code of q-ary n-tuples with the property that words w and w + 11 · · · 1 have separation qn−1 + 1 or qn−1 − (q − 1) .Proof. By Corollary 2.8 there exists a quasi-complementary Lee metric Gray code, G, of q-ary ( n − 1)-tuples. Let G(i) denote the ith word in G. Define the code, G′ on n-tuples by G′(i) = { G(⌊i/q ⌋)( i mod q) if ⌊i/q ⌋ ≡ 0 mod 2 G(⌊i/q ⌋)(( q − 1 − i) mod q) if ⌊i/q ⌋ ≡ 1 mod 2 It is an unknown if the absolute value of the change (“error”) in the separation could be bounded by something smaller than q − 1 but greater than 1. We can also show that if we ask that consecutive words be of distance one in the Hamming metric then we can construct quasi-complementary Gray codes regardless of the parity of n and q. Because distance 1 in the Lee metric implies distance 1 in the Hamming metric, the only case left to prove is the case of n odd and q even. Again we construct an ingredient code for use with Lemma 2.6. Lemma 2.14. If n′ is even then there exists a Hamming metric Gray code of q-ary n′-tuples that starts with 00 · · · 0 and ends with 11 · · · 1 for any even q ≥ 4.Proof. By Lemma 2.7 there exists a Lee metric Gray code, G, of q-ary (n′ − 1)-tuples that starts with 00 · · · 0 and ends with 11 · · · 1. Let G(i) be the ith word in this code. Construct the desired code G′ in the following manner G′(i) = { G(⌊i/q ⌋)( i mod q) if ⌊i/q ⌋ ≡ 0 mod 2 G(⌊i/q ⌋)(( q − 1 − i) mod q) if ⌊i/q ⌋ ≡ 1 mod 2 Now swapping G′(qn − 2) and G′(qn − 1) produces the desired code. Corollary 2.15. There exists a quasi-complementary Hamming metric Gray code of q-ary n-tuples whenever n ≥ 1 and q ≥ 3. 3 n-subsets of a 2n-set The complement of an n-subset of a 2 n-set is itself an n-subset. This al-lows us to sensibly define a complementary Gray code for these objects. The weakest form of a Gray code for fixed size subsets of an n-set is that where the symmetric difference of consecutive sets has cardinality two; this cor-responds to consecutive incidence vectors having Hamming distance two. However, in the Gray code literature of k-subsets of an n-set, it is cus-tomary to ask for stronger “closeness” properties . The strong minimal change property asks that consecutive subsets, when written as sets in lex-icographic order, be Hamming distance one; this is equivalent to changing two bit positions in the incidence vectors which must have only ‘0’s between them. When there are more than 2 bit positions then this property is only preserved by complementation if the two elements changing are consecu-tive or, equivalently, that the pair of bits that change are adjacent in the incidence vector. This is known as the adjacent interchange property . Such Gray codes of k-subsets of an n-set exist if and only if k = 0 , 1, n or n − 1or n is even and k is odd . Additionally these codes cannot be cyclic because the incidence vectors 1 k0n−k and 0 n−k1k have degree one in the corresponding transition graph. Since complementary Gray codes must be cyclic we cannot hope for the adjacent interchange property .We will modify the strong minimal change property to be preserved under complementation and look for Gray codes with this new property. Definition 3.1. A Gray code of k-subsets of an n-set is said to have the complementary strong minimal change property if the elements of each con-secutive pair of subsets differ in the exchange of two elements such that all the elements between the two either appear in both sets or appear in neither set. In the incidence vector representation, this is equivalent to the two bit positions which change having either all ‘1’s or all ‘0’s between them. We note that two incidence vectors which differ by a complementary strong minimal change have Hamming distance two and also differ by the reversal of a subword, the word between and including the two positions which change. Subword reversal is one of the ways that Frank Ruskey presents the various stronger change properties of k-subsets of an n-set. The simplicity of describing the complementary strong minimal change by subword reversals and reversals’ prior use in the literature makes this new minimal change property natural. The strongest form of the complemen-tary strong minimal change would only permit two bits to change if they were adjacent or had a single bit position between them. The existence of Gray codes meeting this strong notion of a complementary is an open and interesting question. We will construct these complementary Gray codes in a manner very similar to the method of Lemma 2.6. That is, we will find a Gray code for the n-subsets of a (2 n − 1)-set such that the first and last sets are as complementary as possible. We will then append a 0 to each of the inci-dence vectors. The complement of the first will now be Hamming distance two from the last and so we complete the code by running through the complements of each incidence vector from the list we constructed. Theorem 3.2. There exists a complementary Gray code for the n-subsets of a 2n-set with the complementary strong minimal change property .Proof. The Eades and McKay strong minimal change property Gray code for the k-subsets of an m-set has the property that the first incidence vector is 0 m−k1k and the final vector is 1 k0m−k[1, 6]. We use this code with k = n and m = 2 n − 1. If we append a 0 to all vectors then the first is 0 n−11n0and its complement, 1 n−10n1 is a single strong minimal change from the final vector 1 n0n. So we simply finish the code with the complements of all the vectors (sets) from the first half. Table 6 shows a complementary Gray code for the 3-subsets of a 6-set with the complementary strong minimal change property represented as subsets. Table 7 shows the same code represented as incidence vectors. 234 235 245 345 346 246 236 256 356 456 156 146 136 126 125 135 145 134 124 123 Table 6: A complementary Gray code for the 3-subsets of a 6-set with the complementary strong minimal change property represented as subsets. 4 Permutations In 1974 Martin Gardner asked the question “True or false: If a1a2 . . . a n is initially 12 . . . n , Algorithm P [“Plain Changes”, or the Trotter-Johnson algorithm for an adjacent interchange Gray code of permutations [2, 10]] 011100 011010 010110 001110 001101 010101 011001 010011 001011 000111 100011 100101 101001 110001 110010 101010 100110 101100 110100 111000 Table 7: A complementary Gray code for the 3-subsets of a 6-set with the complementary strong minimal change property represented as incidence vectors. begins by visiting all n!/2 permutations in which 1 precedes 2; then the next permutation is n . . . 21.”. This statement is true when n ≤ 4 but it fails for all greater n because after n = 4, ( n − 1)! /2 is even and so the larger symbols, n, n − 1, . . . are on the right hand side of the permutation after n!/2 steps, not on the left. Seemingly implicit in this question and its false answer is the problem of determining if there exists an adjacent interchange Gray code for permutations which has the property that it begins by listing “all n!/2 permutations in which 1 precedes 2; then the next permutation is n . . . 21.” Such a Gray code is the most natural candidate for a complementary (really “reverse”) Gray code for permutations; by completing the second half of the code with the reversals of the elements from the first n!/2 permutations, it has the property that the ( i + n!/2)th permutation is the reversal of the ith. Reversal is a better extension of “complement” than the algebraic inverse because some permutations are their own inverses and this is an obstruction to the existence of an “inverse” Gray code. Considering the parity of permutations and the length of a Gray code of permutations, it is easy to generate simple necessary conditions. Proposition 4.1. If a reverse adjacent interchange Gray code for permu-tations of 12 . . . n exists for n ≥ 5 then n ≡ 0, 1 mod 4 . We show that these necessary conditions are sufficient: Theorem 4.2. A reverse adjacent interchange Gray code for permutations of 12 . . . n exists if and only if n ≤ 4 or n ≡ 0, 1 mod 4 . When n ≡ 0 mod 4, we will need our codes to have a particular prop-erty as they approach the permutation n . . . 21, the reversal of the first permutation, namely that the permutations just previous to this all have the symbols 1 , 2, . . . , n − 1 in the order ( n − 1)( n − 2) · · · 312 and the sym-bol n in the four left-most positions moving from right to left as shown in Table 8. We call this property P . We note that when n ≤ 7 the positions of ( n − 1) will include positions to the right of the 1 and 2. (n − 1)( n − 2)( n − 3) n(n − 4) . . . 312 (n − 1)( n − 2) n(n − 3)( n − 4) . . . 312 (n − 1) n(n − 2)( n − 3)( n − 4) . . . 312 n(n − 1)( n − 2)( n − 3)( n − 4) . . . 312; Table 8: Required sequence of permutations of order n, for Property P . Theorem 4.3. If n ≡ 0 mod 4 , n ≥ 8 and there exists a reverse Gray code for permutations of 12 . . . (n − 3) then there exists a reverse Gray code for permutations of 12 . . . n with property P .Proof. We construct the desired Gray code from three different codes • G1, the hypothesized Gray code of order n − 3, • G2, an adjacent interchange property Gray code of the 3-subsets of an n set (which exists because n is even and 3 is odd ), and • G3, the unique (up to reversal and starting position) cyclic Gray code for permutations of 123, shown in Table 9 123 213 231 321 312 132 Table 9: The cyclic Gray code of permutations of 123. We start with the permutation 123 . . . (n − 3)( n − 2)( n − 1) n and apply the various Gray codes to this word in the following manner. We first run G2 on the positions of the symbols ( n−2), ( n−1) and n. If these start in the right three positions they will end in the left three positions and vice versa. In either case they will remain in the same relative order to each other. We then apply the next transformation from G3, on the all-adjacent symbols (n − 2), ( n − 1) and n. We run G3 alternately forwards then backwards assuring that the relative positions of its symbols are either ( n − 2)( n − 1) n or ( n − 2) n(n − 1) depending on the number of iterations of G1 that have passed. At this point we run through G2 again. Once we have completed all transformations from G3 with a step of G2 after each one, we perform an iteration of G1 on the all-adjacent symbols 12 . . . (n − 3) and begin the iterations of G3 again each followed by a step of G2.Continue in this manner until we produce the permutation (n − 3)( n − 4) . . . 312( n − 2) n(n − 1) . Note that this is the final iterate from G1 and the odd parity of this in G1 gives both the position of the largest three symbols being at the right and their order. We now run the Plain changes algorithm, moving ( n − 1) most frequently, n next frequently and ( n − 2) least frequently, all initially towards the left. All adjacent interchanges from now on will involve at least one of these symbols. We run this process until we reach the permutation (n − 3)( n − 2)( n − 4)( n − 5) . . . 312 n(n − 1) . At this point the only remaining interchanges permitted are ( n − 3) with (n − 2) and interchanges involving one or both of n and ( n − 1). There are essentially five permitted moves: • swap ( n − 2) and ( n − 3) (they must be adjacent to do this); • move ( n − 1) to the left; • move ( n − 1) to the right; • move n to the left and • move n to the right. We now translate the problem of finishing this Gray code into finding a suitable Hamilton path on a specific family of graphs, Γ n. Their vertex sets are V (Γ n) = {(y, z, b )\| where b ∈ { 0, 1}, 1 ≤ y, z ≤ n and y 6 = z} There is an edge between ( y, z, 0) and ( y, z, 1) whenever y 6 = 2, z 6 = 2, (y, z ) 6 = (1 , 3) , (3 , 1) or ( y, z ) = (1 , 2) , (2 , 1). There is an edge between (y, z, b ) and ( y ± 1, z, b ) and ( y, z ± 1, b ) whenever those vertices are defined. Finally there is an edge between ( y, y + 1 , b ) and ( y + 1 , y, b ) whenever this pair of vertices is defined. The set of vertices is in bijection to the permutations (with 1 preceding 2) that remain to be added to the Gray code. The positions of ( n − 1) and n are encoded by y and z respectively and b = 0 if ( n − 2) precedes ( n − 3) and b = 1 otherwise. The edges connect any pair of vertices that differ precisely by an allowed adjacent interchange. The restrictions on y and z for edges between ( y, z, 0) and ( y, z, 1) correspond to the fact that ( n − 2) and ( n − 3) cannot be swapped if n or ( n − 1) are between them in the (4 , 3, 1) , (4 , 3, 0) , (4 , 2, 0) , (4 , 2, 1) , (3 , 2, 1) , (2 , 3, 1) , (2 , 3, 0) , (3 , 2, 0) , (3 , 1, 0) , (4 , 1, 0) , (4 , 1, 1) , (3 , 1, 1) , (2 , 1, 1) , (1 , 2, 1) , (1 , 3, 1) , (1 , 4, 1) , (2 , 4, 1) , (3 , 4, 1) , (3 , 4, 0) , (2 , 4, 0) , (1 , 4, 0) , (1 , 3, 0) , (1 , 2, 0) , (2 , 1, 0) Table 10: False Hamilton path on Γ 4.(n, n − 1, 1), (n − 1, n, 1), (n − 1, n, 0), (n, n − 1, 0), (n, n − 2, 0), (n − 1, n − 2, 0), ( n − 1, n − 3, 0), ( n, n − 3, 0), ( n, n − 3, 1), ( n, n − 2, 1), (n − 1, n − 2, 1), ( n − 1, n − 3, 1), ( n − 2, n − 3, 1), Follow Hamilton path in Γ′ n−4 with y′ = y + 2 and z′ = z + 2 . Continue: (4 , 3, 0), (4 , 2, 0), (5 , 2, 0), . . . , ( n, 2, 0), ( n, 3, 0), . . . , ( n, n −4, 0), ( n−1, n −4, 0), (n − 1, n − 5, 0), . . . , ( n − 1, 3, 0), (n − 1, 3, 1), (n − 1, 4, 1), . . . , (n − 1, n − 4, 1), (n, n − 4, 1), ( n, n − 5, 1), . . . , ( n, 2, 1), (n − 1, 2, 1), . . . , (3 , 2, 1), (2 , 3, 1), (2 , 4, 1), . . . , (2 , n, 1), (3 , n, 1), . . . , ( n − 2, n, 1), (n−2, n −1, 1), ( n−3, n −1, 1), . . . , (3 , n −1, 1), (3 , n −1, 0), (4 , n −1, 0), . . . , ( n − 2, n − 1, 0), (n − 2, n, 0), (n − 3, n, 0), (n − 4, n, 0), . . . , (2 , n, 0), (2 , n − 1, 0), . . . , (2 , 3, 0), (3 , 2, 0), (3 , 1, 0), (4 , 1, 0), . . . , (n, 1, 0), (n, 1, 1), (n − 1, 1, 1), . . . , (2 , 1, 1), (1 , 2, 1), (1 , 3, 1), . . . , (1 , n, 1), (1 , n, 0), (1 , n − 1, 0), . . . , (1 , 2, 0), (2 , 1, 0) Table 11: Hamilton path on Γ n (read left to right and top to bottom). permutation. The permutation we have reached corresponds to the vertex (n, n − 1, 1); for our reverse property we require to finish at (2 , 1, 0) and we must visit each vertex. We will show that the desired Hamilton path exists for all even n, except 4 (which we do not need as we already have a reverse Gray code for n = 4). For n = 2 the path is (2 , 1, 1) , (1 , 2, 1) , (1 , 2, 0) , (2 , 1, 0) . For n = 4 (this will fail to be a valid Hamilton path in Γ 4 but it will be useful in the recursive construction) the path is given in Table 10. To construct the Hamilton path in Γ n in general we will make use of the fact that an isomorphic copy of Γ n−4 exists centred inside Γ n. We will use the Hamilton path in this Γ n−4 in the recursion. The general form of the recursion is given in Table 11. The only exception to this pattern is when n = 6 for which we show the modified portion in Table 12. The code so far contains all permutations in which 1 precedes 2, begins (6 , 5, 1) , (5 , 6, 1) , (5 , 6, 0) , (6 , 5, 0) , (6 , 4, 0) , (6 , 4, 1) , (5 , 4, 1) , (5 , 4, 0) , (5 , 3, 0) , (5 , 3, 1) , (4 , 3, 1) , Follow Hamilton path in Γ′ 2 with y′ = y + 2 and z′ = z + 2 . Continue: (4 , 3, 0) , (4 , 2, 0) , (5 , 2, 0) , (6 , 2, 0) , (6 , 3, 0) , (6 , 3, 1) , (6 , 2, 1) , (5 , 2, 1) , (4 , 2, 1) , (3 , 2, 1) , (2 , 3, 1) , (2 , 4, 1) , . . . , (2 , n, 1) , (3 , n, 1) , . . . , The rest is the same as the general recursion for Γn. Table 12: Modified portion of Hamilton path on Γ 6 (read left to right and top to bottom). with 12 . . . n and ends with n . . . 312 which is adjacent to n . . . 321, the reversal of the first permutation. We now complete the code by appending the reversals of the first n!/2 permutations. We note that the structure of the recursion on the Γ n produces a code with property P . Theorem 4.3 uses the existence of codes for n ≡ 1 mod 4 to construct the code of order n+3, the next largest n ≡ 0 mod 4. We now give Theorem 4.4 which completes the induction; it is a construction for proceeding from n to n + 1 when n ≡ 0 mod 4. Theorem 4.4. If n ≡ 1 mod 4 , n ≥ 5 and there exists a reverse Gray code for permutations of 12 . . . (n − 1) with property P , then there exists a reverse Gray code for permutations of 12 . . . n .Proof. We run a modification of the “Plain changes” algorithm . We take the posited reverse Gray code, G, for permutations of 12 . . . (n − 1), replicate each permutation n times and insert the symbol n in all possible n positions starting with 12 . . . (n − 1) n and sweeping n from right to left, then left to right alternately, as occurs in the Trotter-Johnson Algorithm [2, 10]. Because G has property P we will eventually reach the word n(n − 2)( n − 3)( n − 4)( n − 1)( n − 5) . . . 312 At this point we finish the code as shown in Table 13. Then swap the 1 and the 2 to produce the reversal of the very first permutation and complete the code with the reversal of the first n!/2 permutations. Since the Trotter-Johnson algorithm produces reverse Gray codes for n ≤ 4, the existence of all the desired Gray codes is established and The-orem 4.2 is a corollary of Theorems 4.3 and 4.4. Table 14 displays a full reverse Gray code for permutations of order 5. We finish with a related but slightly off topic extension of the investi-gation of the graph, Γ n, which was introduced in the proof of Theorem 4.3. When n is odd and we only take the b = 0 layer, a Hamilton cycle in Γ n is a Gray code for all permutations of the multiset {1, 2, 3, 3, . . . , 3}. The only known Gray codes we could find for this combinatorial family were not cyclic [4, 5] so we feel it worth discussing. For n = 3 the Hamilton cycle is (1 , 3) , (2 , 3) , (3 , 2) , (3 , 1) , (2 , 1) , (1 , 2) , For larger n we use the recursion (1 , 3) , (2 , 3) , Follow the Hamilton cycle in Γ′ n−2 with y′ = y + 1 and z′ = z + 1 . Continue: (2 , 4) , (1 , 4) , (1 , 5) , (1 , 6) , . . . , (1 , n ), (2 , n ), . . . , (n − 1, n ), (n, n − 1) , (n, n − 2) , (n, n − 3) , . . . , (n, 1) , (n − 1, 1) , . . . , (2 , 1) , (1 , 2) An interesting open question in reverse Gray codes for permutations is whether weakened reverse codes can be found for n ≡ 2, 3 mod 4 in either of the two senses discussed in Section 2. References P. Eades and B. McKay, An algorithm for generating subsets of a fixed size with a strong minimal change property. Inform. Process. Lett. 19(1984), 131–133. Selmer M. Johnson, Generation of permutations by adjacent transpo-sition. Math. Comp. 17(1963), 282–285. Donald E. Knuth, The art of computer programming. Vol. 4, Fasc. 2: Generating all tuples and permutations (Addison-Wesley, Upper Saddle River, NJ, 2005). Chun Wa Ko and Frank Ruskey, Generating permutations of a bag by interchanges. Inform. Process. Lett. 41(1992), 263–269. D. H. Lehmer, Permutation by adjacent interchanges. Amer. Math. Monthly 72(1965), 36–46. Frank Ruskey, Simple combinatorial Gray codes constructed by revers-ing sublists. Algorithms and computation (Hong Kong, 1993), Lecture Notes in Comput. Sci. 762(1993), 201–208. Frank Ruskey, Personal communication (2008). C. Savage, A survey of combinatorial Gray codes. SIAM Rev. 39(1997), 605–629. Carla D. Savage and Peter Winkler, Monotone Gray codes and the middle levels problem. J. Combin. Theory Ser. A 70(1995), 230–248. H.F. Trotter, PERM (Algorithm 115). Communications of the ACM 5(1962), 434–435. J. H. van Lint, Introduction to coding theory (Springer-Verlag, Berlin, third edition, 1999). n (n − 2) (n − 3) (n − 4) (n − 1) (n − 5) . . . 3 1 2 n (n − 2) (n − 3) (n − 1) (n − 4) (n − 5) . . . 3 1 2 n (n − 2) (n − 1) (n − 3) (n − 4) (n − 5) . . . 3 1 2(n − 2) n (n − 1) (n − 3) (n − 4) (n − 5) . . . 3 1 2(n − 2) n (n − 3) (n − 1) (n − 4) (n − 5) . . . 3 1 2(n − 2) (n − 3) n (n − 1) (n − 4) (n − 5) . . . 3 1 2(n − 2) (n − 3) (n − 1) n (n − 4) (n − 5) . . . 3 1 2(n − 2) (n − 3) (n − 1) (n − 4) n (n − 5) . . . 3 1 2(n − 2) (n − 3) (n − 1) (n − 4) (n − 5) n . . . 3 1 2...(n − 2) (n − 3) (n − 1) (n − 4) (n − 5) . . . 3 1 2 n (n − 2) (n − 1) (n − 3) (n − 4) (n − 5) . . . 3 1 2 n (n − 1) (n − 2) (n − 3) (n − 4) (n − 5) . . . 3 1 2 n (n − 1) (n − 2) (n − 3) (n − 4) (n − 5) . . . 3 1 n 2(n − 2) (n − 1) (n − 3) (n − 4) (n − 5) . . . 3 1 n 2(n − 2) (n − 1) (n − 3) (n − 4) (n − 5) . . . 3 n 1 2(n − 1) (n − 2) (n − 3) (n − 4) (n − 5) . . . 3 n 1 2(n − 1) (n − 2) (n − 3) (n − 4) (n − 5) . . . n 3 1 2(n − 2) (n − 1) (n − 3) (n − 4) (n − 5) . . . n 3 1 2...(n − 1) (n − 2) (n − 3) n (n − 4) (n − 5) . . . 3 1 2(n − 2) (n − 1) (n − 3) n (n − 4) (n − 5) . . . 3 1 2(n − 2) (n − 1) n (n − 3) (n − 4) (n − 5) . . . 3 1 2(n − 1) (n − 2) n (n − 3) (n − 4) (n − 5) . . . 3 1 2(n − 1) n (n − 2) (n − 3) (n − 4) (n − 5) . . . 3 1 2 n (n − 1) (n − 2) (n − 3) (n − 4) (n − 5) . . . 3 1 2Table 13: The end of the first half of the code in the case n ≡ 1 mod 4. 12345 12354 12534 15234 51234 51243 15243 12543 12453 12435 14235 14253 14523 15423 51423 54123 45123 41523 41253 41235 41325 41352 41532 45132 54132 51432 15432 14532 14352 14325 13425 13452 13542 15342 51342 51324 15324 13524 13254 13245 31245 31254 31524 35124 53124 Now we stop running Plain Changes and start the Theorem 4.4 modified order: 53142 53412 35412 35142 31542 31452 31425 34125 43125 43152 34152 34512 43512 45312 54312 We are halfway. We finish with the permutations from the first half in order, each reversed: 54321 45321 43521 43251 43215 34215 34251 34521 35421 53421 53241 35241 32541 32451 32415 32145 32154 32514 35214 53214 52314 25314 23514 23154 23145 23415 23451 23541 25341 52341 52431 25431 24531 24351 24315 42315 42351 42531 45231 54231 54213 45213 42513 42153 42135 24135 21435 21453 24153 24513 25413 52413 52143 52134 25134 25143 21543 21534 21354 21345 Table 14: A reverse Gray code for the permutations of order 5.
15718	https://mathexp.eu/bostan/publications/BoMo21.pdf	A Simple and Fast Algorithm for Computing the N-th Term of a Linearly Recurrent Sequence Alin Bostan⋆and Ryuhei Mori† ⋆Inria, Palaiseau, France and †Tokyo Institute of Technology, Japan alin.bostan@inria.fr, mori@c.titech.ac.jp Abstract We present a simple and fast algorithm for computing the N-th term of a given linearly recurrent sequence. Our new algorithm uses O(M(d) log N) arithmetic op-erations, where d is the order of the recurrence, and M(d) denotes the number of arithmetic operations for computing the product of two polynomials of degree d. The state-of-the-art algorithm, due to Fiduccia (1985), has the same arithmetic complexity up to a constant factor. Our algorithm is simpler, faster and obtained by a totally diﬀerent method. We also discuss several algorithmic applications, notably to polynomial modu-lar exponentiation and powering of matrices. Keywords: Algebraic Algorithms; Computational Complexity; Linearly Recurrent Sequence; Rational Power Series; Fast Fourier Transform 1 Introduction 1.1 General context Computing eﬃciently selected terms in sequences is a basic and fundamental algo-rithmic problem, whose applications are ubiquitous, for instance in theoretical computer science [65, 49], algebraic complexity theory [59, 73], computer alge-bra [28, 71, 48], cryptography [31, 32, 29, 33], algorith-mic number theory [72, 1], eﬀective algebraic geome-try [12, 36], numerical analysis [52, 51] and computa-tional biology . In simple terms, the problem can be formulated as follows: Given a sequence (un)n≥0 in an eﬀective ring0 R, and given a positive integer N ∈N, compute the term uN as fast as possible. Here, the input (un)n≥0 ∈RN is assumed to be a recurrent sequence, speciﬁed by a data structure 0The ring R is assumed to be commutative with unity and eﬀective in the sense that its elements are represented using some data structure, and there exist algorithms for performing the basic ring operations (+, −, ×) and for testing equality of elements in R. consisting in a recurrence relation and suﬃciently many initial terms that uniquely determine its terms. Eﬃciency is measured in terms of ring operations (algebraic model), or of bit operations (Turing machine model). The cost of an algorithm is respectively esti-mated in terms of arithmetic complexity or of binary complexity. Both measures have their own usefulness: the algebraic model is relevant when ring operations have essentially unit cost (typically, if R is a ﬁnite ring such as the prime ﬁeld Fp := Z/pZ), while the bit com-plexity model is relevant when elements of R have a vari-able bitsize, and thus ring operations in R have variable cost (typically, when R is the ring Z of integer numbers). The recurrence relation satisﬁed by the input se-quence (un)n≥0 might be of several types: (C) linear with constant coeﬃcients, that is of the form, un+d = cd−1un+d−1 + · · · + c0un, n ≥0, for some given c0, . . . , cd−1 in R. In this case we simply say that (un)n≥0 is linearly recurrent (or, C-recursive). The most basic examples are the geometric sequence (qn)n≥0, for q ∈R, and the Fibonacci sequence (Fn)n with Fn+2 = Fn+1 + Fn for n ≥0 and F0 = 0, F1 = 1. (P) linear with polynomial coeﬃcients, of the form, un+d = cd−1(n)un+d−1 + · · · + c0(n)un, n ≥0, for some given rational functions c0(x), . . . , cd−1(x) in R(x). In this case the sequence is called holonomic (or, P-recursive). Among the most basic examples, other than the C-recursive ones, there is the factorial sequence (n!)n≥0 = (1, 1, 2, 6, 24, 120, . . .) and the Motzkin sequence (un)n≥0 = (1, 1, 2, 4, 9, 21, 51, . . .) speciﬁed by the recurrence un+1 = 2n+3 n+3 · un + 3n n+3 · un−1 and the initial conditions u0 = u1 = 1. (Q) linear with polynomial coeﬃcients in q and qn, that is of the form, un+d = cd−1(q, qn)un+d−1+· · ·+c0(q, qn)un, n ≥0, Copyright c ⃝2021 by SIAM Unauthorized reproduction of this article is prohibited for some q ∈ R and some rational functions c0(x, y), . . . , cd−1(x, y) in R(x, y). In this case, the sequence is called q-holonomic; such a sequence can be seen as a q-deformation of a holonomic sequence (in the sense that when q 7→1, the limit sequence tends to be holonomic). A typical example is the q-factorial [n]q! := (1 + q)· · ·(1 + q + · · · + qn−1). In all these classes of examples, the recurrence is linear, and the integer d that governs the length of the recurrence relation is called the order of the corresponding linear recurrence. Of course, some interesting sequences satisfy non-linear recurrences, as is the case for the so-called Somos-4 sequence (1, 1, 1, 1, 2, 3, 7, 23, . . .) deﬁned by: un+4 = un+3un+1+u2 n+2 /un together with u0 = · · · = u3 = 1, but we will not consider this larger class in what follows. To compute the N-th term uN of a sequence of type (P), resp. (Q), the best known algorithms are presented in [16, 12], resp. in . In the algebraic model, they rely on an algorithmic technique called baby-step/giant-step, allowing to compute uN in a number of operations in R that is almost linear in √ N, up to logarithmic factors. This should be contrasted with the direct iterative algorithm, of arithmetic complexity linear in N. In the bit model, the same references provide dif-ferent algorithms based on a diﬀerent technique, called binary splitting; these algorithms are quasi-optimal in the sense that they are able to compute uN in a num-ber of bit operations almost linear (up to logarithmic factors) in the bitsize of the output value uN. Once again, this should be contrasted with the direct itera-tive algorithm, whose binary complexity is larger by at least one order of magnitude (e.g., in case (P) the naive algorithm has bit complexity O(N 3)). 1.2 The case of C-recursive sequences In what follows, we will restrict our attention to the case (C) only. This case obviously is a subcase of both cases (P) and (Q). It presents an exceptional feature with respect to the algebraic model: contrary to the general cases (P) and (Q), in case (C) it is possible to compute the term uN using a number of arithmetic operations in R that is only logarithmic in N. For the geometric sequence un = qn, this is known since Pingala (∼200 BC) who seemingly is the inventor of the algorithmic method of binary powering, or square-and-multiply [46, §4.6.3]. The corresponding algorithm is recursive and based on the equalities qN = ( (qN/2)2, if N is even, q · (q N−1 2 )2, else. The arithmetic complexity of this algorithm is bounded by 2 log N multiplications1 in R, which represents a tremendous improvement compared to the naive iter-ative algorithm that computes the term qN in N −1 multiplications in R, by simply unrolling the recurrence un+1 = q · un with q0 = 1. In the general case (C), Miller and Spencer Brown showed in 1966 that a similar complexity can be obtained by converting2 the scalar recurrence of order d (1.1) un+d = cd−1un+d−1 + · · · + c0un, n ≥0, into a vector recurrence of order 1 (1.2)      un un+1 . . . un+d−1      \| {z } vn =      1 ... 1 c0 c1 · · · cd−1      \| {z } M ×      un−1 un . . . un+d−2      \| {z } vn−1 , n ≥1, and by using binary powering in the ring Md(R), of d×d matrices with coeﬃcients in R, in order to compute M N recursively by M N = ( (M N/2)2, if N is even, M · (M N−1 2 )2, else. From there, uN can be read oﬀthe matrix-vector product vN = M N · v0. The complexity of this method is O(dθ log N) operations in R, where θ ∈[2, 3] is any feasible exponent for matrix multiplication in Md(R). Strangely enough, the paper of Miller and Spencer Brown was largely overlooked in the subsequent literature, and their result has been rediscovered sev-eral times in the 1980s. For instance, Shortt pro-posed a O(log N) algorithm for computing the N-th Fi-bonacci number3,4 and extended it together with Wil-1In this article, notation log refers to the logarithm in base 2. 2[15, p. 74] calls this un truc bien connu (“a well-known trick”). 3Shortt’s algorithm had actually appeared before, in the 1969 edition of Knuth’s book [45, p. 552], as a solution of Ex. 26 (p. 421, §4.6.3). The algorithm is based on the so-called doubling formulas (F2n, F2n−1) = (F 2 n+2FnFn−1, F 2 n+F 2 n−1), actually due to Lucas (1876) and Catalan (1886), see e.g. [20, Ch. XVII]. The currently best implementation for computing FN over Z (mpz_ﬁb_ui from GMP) uses a variant of this method, requiring just two squares (and a few additions) per binary digit of N. 4Already in 1899, G. de Rocquigny asked “for an expeditious procedure to compute a very distant term of the Fibonacci sequence” . In response, several methods (including the one mentioned by Knuth in [45, p. 552]) have been published one year later by Rosace (alias), E.-B. Escott, E. Malo, C.-A. Laisant and G. Picou . This fact does not seem to have been noticed in the modern algorithmic literature before the current paper, although the reference is mentioned in Dickson’s formidable encyclopedic book [20, p. 404]. Copyright c ⃝2021 by SIAM Unauthorized reproduction of this article is prohibited son to the computation of order-d Fibonacci num-bers in O(d3 log N) arithmetic operations. The same cost has also been obtained by Dijkstra and Ur-banek . Pettorossi , and independently Gries and Levin , improved the algorithm and lowered the cost to O(d2 log N), essentially by taking into ac-count the sparse structure of the matrix M. See also [22, 50, 23, 24, 39, 63, 30, 44] for similar algorithms. 1.3 Fiduccia’s algorithm The currently best algo-rithm is due to Fiduccia5 . It is based on the fol-lowing observation: the matrix M in (1.2) is the trans-pose of the companion matrix C which represents the R-linear multiplication-by-x map from the quotient ring R[x]/(Γ) to itself, where Γ = xd −Pd−1 i=0 cixi. There-fore, denoting by e the row vector e = [1 0 · · · 0], the N-th term uN equals (1.3) uN = e·vN = e·M N·v0 = CN · eT T·v0 = ⟨xN mod Γ, v0⟩, where the inner product takes place between the vector v0 = u0 · · · ud−1 of initial terms of (un)n≥0, and the vector whose entries are the coeﬃcients of the remainder (xN mod Γ) of the Euclidean division of xN by Γ. Thus, computing uN is reduced to computing the coeﬃcients of (xN mod Γ), and this can be performed eﬃciently by using binary powering in the quotient ring A := R[x]/(Γ), at the cost of O(log N) multiplications in A. Each multiplication in A may be performed using O(M(d)) operations in R [27, Ch. 9, Corollary 9.7], where M(d) denotes the arithmetic cost of polynomial multiplication in R[x] in degree d. Using Fast Fourier Transform (FFT) methods, one may take M(d) = O(d log d) when R contains enough roots of unity, and M(d) = O(d log d log log d) in general [27, Ch. 8]. In conclusion, Fiduccia’s algorithm allows the com-putation of the N-th term uN of a linearly recurrent se-quence of order d using O(M(d) log N) operations in R. Since 1985, this is the state-of-the-art algorithm for this task in case (C). A closer inspection of the proof of [27, Corollary 9.7] shows that a more precise estimate for the arithmetic cost of Fiduccia’s algorithm is (1.4) F(N, d) = 3 M(d)⌊log N⌋+ O(d log N) operations in R. This comes from the fact that squar-ing6 in A = R[x]/(Γ) is based on a polynomial product 5The idea already appears in the 1982 conference paper . We have discovered that the same algorithm had been sketched by D. Knuth in the corrections/changes to published in 1981 in [47, p. 28], where he attributes the result to R. Brent. Almost surely, C. Fiduccia was not aware about this fact. 6Note that multiplying by x in A is much easier and has linear arithmetic cost O(d). in degree less than d followed by an Euclidean division by Γ of a polynomial of degree less than 2d. The Eu-clidean division is reduced to a power series division by the reversal Q(x) := xd·Γ(1/x) of Γ, followed by a poly-nomial product in degree less than d. The reciprocal of Q(x) is precomputed modulo xd once and for all (using a formal Newton iteration) in 3 M(d)+O(d) operations in R, and then each squaring in A also takes 3 M(d)+O(d) operations in R. The announced cost from (1.4) follows from the fact that binary powering uses ⌊log N⌋squar-ings and at most ⌊log N⌋multiplications by x. 1.4 Main results We propose in this paper a new and simpler algorithm, with a better cost. More precisely, our ﬁrst main complexity result is: Theorem 1. One can compute the N-th term of a linearly recurrent sequence of order d with coeﬃcients in a ring R using T(N, d) = 2 M(d)⌈log(N + 1)⌉+ M(d) arithmetic operations in R. The proof of this result is based on a very natural algorithm, which will be presented in Section 2.1. Let us remark that it improves by a factor of 1.5 the complexity of Fiduccia’s algorithm. This factor is even higher in the FFT setting, where polynomial multiplication is assumed to be performed using Fast Fourier Transform techniques. In this setting, we obtain the following complexity result, which will be proved in Section 4. Theorem 2. One can compute the N-th term of a linearly recurrent sequence of order d with coeﬃcients in a ﬁeld K supporting FFT using7 ∼2 3 M(d) log(N) arithmetic operations in K. Algorithms 1 and 2 (underlying Theorem 1) and Algorithm 11 (underlying Theorem 2) are both of LSB-ﬁrst (least signiﬁcant bit ﬁrst) type. This prevents them from computing simultaneously several consecu-tive terms of high indices, such as uN, . . . , uN+d−1. This makes a notable diﬀerence with Fiduccia’s algorithm from §1.3. For this reason, we will design a second al-gorithm, of MSB-ﬁrst (most signiﬁcant bit ﬁrst) type, by “transposing” Algorithm 1. This leads to the following complexity result. 7Here, and in what follows, the “equivalent” sign ∼is to be understood for ﬁxed N and d going to inﬁnity. Copyright c ⃝2021 by SIAM Unauthorized reproduction of this article is prohibited Theorem 3. One can compute the terms of indices N −d+1, . . . , N of a linearly recurrent sequence of order d with coeﬃcients in a ring R using (2 M(d) + d)⌈log(N + 1)⌉+ O(M(d)) arithmetic operations in R. The method underlying this complexity result is based on Algorithms 5, 6 and 8, which are presented in Section 3. Along the way, using the MSB-ﬁrst Algorithm 5, we improve the cost of polynomial modular exponentiation, which is a central algorithmic task in computer algebra, with many applications. Since this result has an interest per se, we isolate it here as our last complexity result. Theorem 4. Given N ∈N and a polynomial Γ(x) in R[x] of degree d, one can compute xN mod Γ(x) using (2 M(d) + d)⌈log(N + 1)⌉+ M(d) arithmetic operations in R. This complexity of ∼2 M(d) log N compares favor-ably with the currently best estimate of ∼3 M(d) log N obtained by square-and-multiply in the quotient ring R[x]/(Γ(x)), combined with fast modular multiplica-tions performed either classically [27, Corollary 9.7], or using Montgomery’s algorithm . In the FFT setting, the gain is even larger, and our results improve on the best estimates, due to Mihăilescu . 1.5 Structure of the paper In Section 2 we propose our LSB-ﬁrst (least signiﬁcant bit ﬁrst) algorithm for computing the N-th term of a C-recursive sequence. We design in Section 3 a second algorithm, which is an MSB-ﬁrst (most signiﬁcant bit ﬁrst) variant, and discuss several algorithmic applications, including polynomial modular exponentiation and powering of matrices. In Section 4 we specialize and analyze Algorithm 1 in the speciﬁc FFT setting, where polynomial multiplication is based on Discrete Fourier Transform techniques, and we compare it with the FFT-based Fiduccia’s algorithm. We conclude in Section 5 by a summary of results and plans of future work. 2 The LSB-ﬁrst algorithm and applications We will prove Theorem 1 in §2.1, where we propose the ﬁrst main algorithms (Algorithms 1 and 2), which are faster than Fiduccia’s algorithm. Then, in §2.2 we instantiate them in the particular case of the Fibonacci sequence. The resulting algorithm is competitive with state-of-the-art algorithms. 2.1 LSB-ﬁrst algorithm: Proof of Theorem 1 In this section, we give the proof of Theorem 1. Let us denote by F(x) ∈R the generating function of the sequence (un)n≥0, F(x) := X n≥0 unxn. We use the following classical characterization of gener-ating functions of linearly recurrent sequences . To be self-contained, we include a proof. Lemma 1. A sequence (un)n≥0 is linearly recurrent of order d if and only if its generating function is F(x) = P(x)/Q(x) for some polynomials P(x) of degree at most d −1 and Q(x) of degree d satisfying Q(0) = 1. Proof. Assume (un)n≥0 is a linearly recurrent sequence of order d. Let us deﬁne Q(x) := 1−cd−1x−· · ·−c0xd, that is the reversal of the characteristic polynomial Γ(x) = xd −Pd−1 i=0 cixi of recurrence (1.1). Then, P(x) := Q(x) · F(x) is a polynomial of degree less than d. This is immediately seen by checking that, for any n ≥0, the coeﬃcient of xn+d in the power series Q(x) · F(x) is equal to un+d −cd−1un+d−1 −· · · −c0un, hence it is zero by (1.1). Conversely, assume that the generating function of a sequence (un)n≥0 is F(x) = P(x)/Q(x) for some polynomials P(x) and Q(x) satisfying the conditions in the lemma. Then, Q(x)F(x) = P(x) implies that the coeﬃcient of xn+d in Q(x)F(x) is zero for all integers n ≥0. Hence, the linear recurrence equation (1.1) must be satisﬁed, where 1 −cd−1x −· · · −c0xd := Q(x). Proof of Theorem 1. From the values u0, . . . , ud−1 and c0, . . . cd−1, we obtain Q(x) := 1 −cd−1x −· · · −c0xd without any operations in R, and can compute P(x) := (u0 +· · ·+ud−1xd−1)·Q(x) mod xd using at most M(d) operations in R. According to Lemma 1, our goal is to compute uN = [xN] P(x) Q(x), where [xN] F(x) denotes the coeﬃcient of xN in F(x) ∈ R. From Lemma 1, we can assume that Q(0) = 1. But, for the sake of generality, in the following, we only assume that Q(0) is invertible in R. If N = 0, we have uN = P(0)/Q(0). If N ≥1, we multiply Q(−x) by the numerator P(x) and the denominator Q(x), and obtain uN = [xN] P(x)Q(−x) Q(x)Q(−x). Here, Q(x)Q(−x) is an even polynomial, which means that all coeﬃcients of x2n+1 for integers n ≥0 in this Copyright c ⃝2021 by SIAM Unauthorized reproduction of this article is prohibited Algorithm 1 (OneCoeﬀ) Input: P(x), Q(x), N Output: [xN] P (x) Q(x) Assumptions: Q(0) invertible and deg(P) < deg(Q) =: d 1: while N ≥1 do 2: U(x) ←P(x)Q(−x) ▷U = P2d−1 i=0 Uixi 3: if N is even then 4: P(x) ←Pd−1 i=0 U2ixi 5: else 6: P(x) ←Pd−1 i=0 U2i+1xi 7: A(x) ←Q(x)Q(−x) ▷A = P2d i=0 Aixi 8: Q(x) ←Pd i=0 A2ixi 9: N ←⌊N/2⌋ 10: return P(0)/Q(0) polynomial are equal to 0. Hence, there exists a unique polynomial V (x) satisfying V (x2) = Q(x)Q(−x). Here, the degree of V (x) is d and V (0) = Q(0)2 is invertible in R. Now, we obtain uN = [xN] U(x) V (x2) for U(x) := P(x)Q(−x). The numerator U(x) has degree at most 2d −1. Here, 1/V (x2) is an even formal power series. That implies that if N is even (or odd), we can ignore the odd (or even) part of U(x). Let Ue(x) and Uo(x) be the even and odd parts of U(x), respectively, i.e., Ue(x) := Pd−1 i=0 U2ixi and Uo(x) := Pd−1 i=0 U2i+1xi for U(x) = P2d−1 i=0 Uixi. Then, we obtain the decomposition U(x) = Ue(x2) + xUo(x2). Hence, uN = ( [xN] Ue(x2) V (x2) , if N is even [xN] xUo(x2) V (x2) , else. = ( [xN/2] Ue(x) V (x) , if N is even [x(N−1)/2] Uo(x) V (x) , else. Here, Ue(x) and Uo(x) are polynomials of degree at most d −1, while V (x) is a polynomial of degree d satisfying that V (0) is invertible in R. Hence, we can repeat this reduction until N ≥1. Our algorithm for computing [xN] P(x)/Q(x) is summarized in Algorithm 1, and its immediate con-sequence for computing the N-th term of the lin-early recurrent sequence (un)n≥0 deﬁned by eq. (1.1) is displayed in Algorithm 2. Algorithm 1 has com-plexity 2 M(d)⌈log(N + 1)⌉and Algorithm 2 has cost 2 M(d)⌈log(N +1)⌉+M(d), which proves Theorem 1. Note that Algorithms 1 and 2 use an idea similar to the ones in [11, 10] which were dedicated to the larger Algorithm 2 (OneTerm) Input: rec. (1.1), u0, . . . , ud−1, N Output: uN Assumptions: Γ(x) = xd −Pd−1 i=0 cixi with c0 ̸= 0 1: Q(x) ←xdΓ(1/x) 2: P(x) ←(u0 + · · · + ud−1xd−1) · Q(x) mod xd 3: return [xN]P(x)/Q(x) ▷using Algorithm 1 class of algebraic power series, but restricted to positive characteristic only. Algorithm 1 also shares common features with the technique of section operators [2, Lemma 4.1] used by Allouche and Shallit to compute the N-th term of k-regular sequences [2, Corollary 4.5] in O(log N) ring operations. Algorithm 1 can be interpreted at the level of re-currences as computing ∼log N new recurrences pro-duced by the Graeﬀe process, which is a classical tech-nique to compute the largest root of a real polyno-mial [40, 57, 58]. Interestingly, the Graeﬀe process has been used in a purely algebraic context by Schönhage in [67, §3] for computing the reciprocal of a power series, see also [14, §2]. However, our paper seems to be the ﬁrst reference where the Graeﬀe process and the section operators approach are combined together. 2.2 New algorithm for Fibonacci numbers To illustrate the mechanism of Algorithm 1, let us instan-tiate it in the particular case of the Fibonacci sequence F0 = 0, F1 = 1, Fn+2 = Fn+1 + Fn, n ≥0. The generating function P n≥0 Fnxn is x/(1 −x −x2). Therefore, the coeﬃcient FN = [xN] x 1−x−x2 is equal to [xN] x(1 + x −x2) 1 −3x2 + x4 = ( [x N 2 ] x 1−3x+x2 , if N is even, [x N−1 2 ] 1−x 1−3x+x2 , else. The computation of FN is reduced to that of a coeﬃcient of the form [xN] a + bx 1 −cx + x2 = [xN] (a + bx)(1 + cx + x2) 1 −(c2 −2)x2 + x4 which is equal to ( [x N 2 ] a+(bc+a)x 1−(c2−2)x+x2 , if N is even, [x N−1 2 ] (ac+b)+bx 1−(c2−2)x+x2 , else. This yields Algorithm 3 for the computation8 of FN. 8Notice that the same algorithm can be used to compute ef-ﬁciently the N-th Fibonacci polynomial, or the N-th Cheby-shev polynomial. Fibonacci polynomials in R[t] are deﬁned by Fn+2(t) = t · Fn+1(t) + Fn(t) with F0(t) = 1 and F1(t) = 1. It is suﬃcient to initialize c to t2+2 (instead of 3) and b to t when N is even (instead of 0). The complexity of this algorithm is O(M(N)) operations in R, which is quasi-optimal. Copyright c ⃝2021 by SIAM Unauthorized reproduction of this article is prohibited Algorithm 3 (NewFibo) Input: N Output: FN Assumptions: N ≥2 1: c ←3 2: if N is even then 3: [a, b] ←[0, 1] 4: else 5: [a, b] ←[1, −1] 6: N ←⌊N/2⌋ 7: while N > 1 do 8: if N is even then 9: b ←a + b · c 10: else 11: a ←b + a · c 12: c ←c2 −2 13: N ←⌊N/2⌋ 14: return b + a · c In the direct application of Algorithm 1 to the Fibonacci sequence, the condition N > 1 in line 7 should be N ≥1 and b + a · c in line 14 should be a. But, in that case, there is a redundant product c2 in line 12 when N = 1. This modiﬁcation saves one multiplication and one subtraction. This saving is crucial over the integers, when we consider bit complexity, since c grows exponentially in the iterations, and the bit complexity of the redundant multiplication c2 occupies a constant factor of the whole bit complexity. A close inspection reveals that this algorithm com-putes FN by a recursive use of the formula FN = L2⌊log N⌋· FN−2⌊log N⌋+ (−1)N · F21+⌊log N⌋−N, which is a particular instance of the classical formula Fn+m = LmFn + (−1)nFm−n relating the Fibonacci numbers and the Lucas numbers Ln = Fn+1 + Fn−1. When N is a power of 2, then Algorithm 3 degener-ates into Algorithm 4. This is equivalent to Algorithm fib(n) in [18, Fig. 6]9. It uses 2 log(N)−3 products (of which log(N)−2 are squarings) and log(N)−2 subtrac-tions. When N is arbitrary, Algorithm 3 has essentially the same cost: it uses at most 2 log(N) −1 products (of which at most log(N)−1 are squarings) and 2⌊log(N)⌋− 1 additions/subtractions. In contrast, [18, Fig. 6] uses a more complex algorithm, with higher cost. A nice feature of Algorithm 3 is not only that it is simple and natural, but also that its arithmetic and bit 9This algorithm had also appeared before, in Knuth’s book [45, p. 552, second solution]. Algorithm 4 Input: N Output: FN Assumptions: N ≥2 and N is a power of 2 1: [b, c] ←[1, 3] 2: N ←⌊N/2⌋ 3: while N > 2 do 4: b ←b · c 5: c ←c2 −2 6: N ←⌊N/2⌋ 7: return b · c complexity match the complexities of the state-of-art algorithms for computing Fibonacci numbers . 3 The MSB-ﬁrst algorithm and applications We present in §3.1 a “most signiﬁcant bit” (MSB) variant (Algoritm 5) of Algoritm 1. Then we discuss various applications of Algorithms 1 and 5. In §3.2 we design a faster algorithm for polynomial modular exponentiation, that we use in §3.3 to design a faster Fiduccia-like algorithm for computing a slice of d terms of indices N −d+1, . . . , N in ∼2 M(d) log N operations. 3.1 The MSB-ﬁrst algorithm In Fiduccia’s algo-rithm (§1.3), the N-th coeﬃcient uN in the power se-ries expansion P i≥0 uixi of P/Q is given by the inner product ⟨xN mod Γ(x), v0⟩, where Γ is the reversal poly-nomial of Q and v0 is the vector of initial coeﬃcients u0 · · · ud−1 . Here, xN mod Γ(x) depends only on the linear recurrence equation (1.1), and is independent of the initial terms v0. Hence, if we want to compute the N-th terms of k diﬀerent linearly recurrent sequences that share the same linear recurrence equation (1.1), we can ﬁrst determine ρ(x) := xN mod Γ(x), and then ⟨ρ, v(i) 0 ⟩for i = 1, . . . , k, where v(i) 0 denotes the vector of d initial terms of the i-th sequence. The total arithmetic complexity of this algorithm is O(M(d) log N +kd); this is faster than Fiduccia’s algorithm repeated indepen-dently k times, with cost O(k M(d) log N). On the other hand, in Algorithm 1 we iteratively update both the denominator and numerator, and each new numerator depends on the original numerator P(x) which encodes the initial d terms of the sequence. Hence, it is not a priori clear how to obtain with Algorithm 1 the good feature of Fiduccia’s algorithm. In this section, we present an algorithm that com-putes uN with complexity equal to that of Algorithm 1 and which, in addition, achieves the cost O(M(d) log N+ kd) for the above problem with k sequences. While Algorithm 1 looks at N from the least sig-niﬁcant bit (LSB), the main algorithm presented in this section (Algoritm 5) looks at N from the most signif-Copyright c ⃝2021 by SIAM Unauthorized reproduction of this article is prohibited Algorithm 5 (SliceCoeﬀ) Input: Q(x), N Output: FN,d(1/Q(x)) Assumptions: Q(0) invertible and deg(Q) =: d 1: function SliceCoeﬀ(N, Q(x)) 2: if N = 0 then 3: return xd−1/Q(0) 4: A(x) ←Q(x)Q(−x) ▷A = P2d i=0 Aixi 5: V (x) ←Pd i=0 A2ixi 6: W(x) ←SliceCoeﬀ(⌊N/2⌋, V (x)) 7: if N is even then 8: S(x) ←xW(x2) 9: else 10: S(x) ←W(x2) 11: B(x) ←Q(−x)S(x) ▷B = P3d−1 i=0 Bixi 12: return Pd−1 i=0 Bd+ixi icant bit (MSB). In fact, Algoritm 5 is in essentially equivalent to “the transposition” of Algorithm 1 in the sense of . This is the reason why this MSB-ﬁrst al-gorithm has the same cost as Algorithm 1. However, in order to keep the presentation self-contained, we are not going to appeal here to the general machinery of algorithmic transposition tools, but rather derive the transposed algorithm “by hand”, by a direct reasoning. In order to compute [xN] P(x)/Q(x), it is suﬃcient to compute the (N −d + 1)-th term to the N-th term of 1/Q(x) since the degree of P(x) is at most d −1. Let FN,d(P i≥0 aixi) := Pd−1 i=0 aN−d+1+ixi. Our goal is to compute FN,d(1/Q(x)). We have the equalities FN,d 1 Q(x) = FN,d Q(−x) Q(x)Q(−x) = FN,d Q(−x)xN−2d+1FN,2d 1 Q(x)Q(−x) = F2d−1,d Q(−x)FN,2d 1 V (x2) , where V (x2) := Q(x)Q(−x). In the second equality, we ignore the terms of 1/V (x2) except for the (N −2d + 1)-th term to the N-th term. In the third equality, we use the fact that FN,d(xA(x)) = FN−1,d(A(x)). Let now W(x) := F⌊N/2⌋,d(1/V (x)). Then, it is easy to see that FN,d 1 Q(x) = F2d−1,d (Q(−x)S(x)) , where S(x) := ( xW(x2), if N is even W(x2), else. Algorithm 6 (OneCoeﬀT) Input: P(x), Q(x), N Output: [xN] P (x) Q(x) Assumptions: Q(0) invertible, deg(P) < deg(Q) =: d 1: U ←FN,d(1/Q(x)) using Algorithm 5 ▷ U = uN−d+1 + · · · + uNxd−1 2: return p0uN + · · · + pd−1uN−d+1 ▷P = Pd−1 i=0 pixi Algorithm 7 Input: P1, . . . , Pk, Q, N Output: [xN] Pj Q , j = 1, . . . , k Assumptions: Q(0) invertible, deg(P) < deg(Q) =: d 1: U ←FN,d(1/Q(x)) using Algorithm 5 ▷ U = uN−d+1 + · · · + uNxd−1 2: return p(j) 0 uN + · · · + p(j) d−1uN−d+1, j = 1, . . . , k ▷ Pj = Pd−1 i=0 p(j) i xi The resulting method for computing FN,d(1/Q(x)) is summarized in Algorithm 5, and its applications to the computation of [xN]P/Q, and to [xN]P (i)/Q for several i = 1, . . . , k, are displayed in Algorithms 6 and 7. Let us analyze the complexity of Algorithms 5 and 6 more carefully. At each step, Algorithm 5 com-putes Q(x)Q(−x) and Q(−x)S(x), where the degrees of Q(x) and S(x) are d and at most 2d −1, respectively. Hence a direct analysis concludes that its complexity is 3 M(d) log N operations in R. However, an improve-ment comes from the remark that not all coeﬃcients of Q(−x)S(x) are needed: it is suﬃcient to compute the d-th coeﬃcient to the (2d−1)-th coeﬃcient of Q(−x)S(x). This operation is known as “the middle product”, and can be performed with the same arithmetic complexity as the standard product of two polynomials of degree d, plus a few d additional operations [35, 8]. Therefore, if steps 11 and 12 of Algorithm 5 are performed “at once” using a middle product, then the arithmetic com-plexity drops to (2 M(d) + d) log N. This complexity is also inherited by Algorithm 6, which uses at most 2d additional operations in the last step. It should be obvious at this point that the slight variant Algorithm 7 of Algorithm 6 achieves arithmetic complexity O(M(d) log N + kd) for the aforementioned problem with k sequences, and more precisely its cost is of at most (2 M(d) + d) log N + 2kd operations in R. In conclusion, Algorithm 5 achieves the same arith-metic complexity as Algorithm 1 and it extends to Al-gorithms 6 and 7. All algorithmic techniques speciﬁc to the FFT setting, that we will describe in Section 4, can also be applied to Algorithms 5, 6 and 7, yielding the same complexity gains. Copyright c ⃝2021 by SIAM Unauthorized reproduction of this article is prohibited Algorithm 8 (NewModExp) Input: Γ(x), N Output: xN mod Γ(x) Assumptions: lcoeﬀ(Γ) invertible, Γ(0) ̸= 0 and deg(Γ) =: d 1: Q(x) ←xdΓ(1/x) 2: u(x) ←FN,d(1/Q(x)) ▷using Algorithm 5 3: v(x) ←u(x)Q(x) mod xd 4: return v(1/x)xd−1 3.2 Faster modular exponentiation The algo-rithms of §3.1 are not only well-suited to compute the N-th terms of several sequences satisfying the same re-currence. In this section, we show that they also per-mit a surprising application to the computation of poly-nomial modular exponentiations. This fact has many consequences, since modular exponentiation is a central algorithmic task in algebraic computations. In §3.3, we will discuss a ﬁrst application in relation with the main topic of our article. Namely, we will design a new Fiduccia-style algorithm for the computation of the N-th term, and actually of a whole slice of k ≥d terms, in (2 M(d) + d) log N + O((k + d)M(d)/d) ring operations. More consequences will be separately discussed in §3.4. Assume we are given a polynomial Γ(x) ∈R[x] of degree d, an integer N, and that we want to compute ρ(x) := xN mod Γ(x). Without loss of generality, we may assume Γ(0) ̸= 0. Let Q(x) ∈R[x] be the reversal of Γ(x), that is Q(x) := xdΓ(1/x). Let us denote the power series expansion of 1/Q by P i≥0 aixi. Then, equation (1.3) implies that (3.5) aN · · · aN+d−1 = r × H, where r = r0 · · · rd−1 with ρ = Pd−1 i=0 rixi and H is the Hankel matrix H :=      a0 · · · ad−1 a1 · · · ad . . . ad−1 · · · a2d−2     . Note that the matrix H is invertible, as its determinant is equal (up to a sign) to ([xd]Q)d−1 = Γ(0)d−1. Therefore, r (and thus ρ) can be found by (1) computing aN · · · aN+d−1 using Algorithm 5; (2) solving the Hankel linear system (3.5). Step (1) has arithmetic cost (2 M(d) + d) log(N + d), while step (2) has negligible cost O(M(d) log d) us-ing , see also [6, Ch. 2, §5]. It is actually possible to improve a bit more on this algorithm, by using the next lemma. Lemma 2. Let N ∈N and let Γ(x) ∈R[x] be of degree d with Γ(0) ̸= 0. Let Q(x) ∈R[x] be its reversal, Q(x) := xdΓ(1/x). Denote its reciprocal 1/Q by P i≥0 aixi, and let u(x) be FN,d(1/Q(x)) = aN−d+1+ · · ·+aNxd−1. Deﬁne v(x) to be u(x)Q(x) mod xd. Then xN mod Γ(x) = v(1/x)xd−1. Proof. Write the Euclidean division xN = L(x) · Γ(x) + ρ(x), where deg(L) = N −d and ρ(x) = r0 + · · · + rd−1xd−1. Replacing x by 1/x on both sides, and then multiplying by xN yields 1 = Lrev(x) · Q(x) + xN−d+1 · ˜ ρ(x), where Lrev(x) = xN−d · L(1/x) and ˜ ρ(x) = xd−1 · ρ(1/x). In other words 1 Q(x) = Lrev(x) + xN−d+1 · ˜ ρ(x) Q(x). Since Lrev(x) has degree at most N −d, it follows that u(x) = ˜ ρ(x) Q(x) mod xd. Therefore, ˜ ρ(x) is equal to v(x), and the conclusion follows. The merit of Lemma 2 is that it shows that computing xN mod Γ(x) can be reduced to comput-ing FN,d(1/Q(x)), plus a few additional operations with negligible cost M(d). The resulting method is presented in Algorithm 8, whose complexity is (2 M(d) + d) log N + M(d). This proves Theorem 4. Note that Algorithm 8 is simpler, and faster by a factor of 1.5, than the classical algorithm based on bi-nary powering in the quotient ring R[x]/(Γ(x)). Algo-rithm 5, and hence also Algorithm 8, admits a special-ization into the FFT setting (Algorithm 13 in §4) with complexity ∼ 2 3 M(d) log N. Similarly to the case of Algorithm 11 in §4, this FFT variant of Algorithm 8 is faster by a factor of 2.5 than Shoup’s (compara-tively simple) algorithm [70, §7.3], and by a factor of 1.625 than the (much more complex) algorithm of Mihăilescu . This speed-up might be beneﬁcial for instance in applications to polynomial factoring in Fp[x], where one time-consuming step to factor f ∈Fp[x] is the computation of xp mod f, see [27, Algorithms 14.3, 14.8, 14.13, 14.15, 14.31, 14.33 and 14.36], and also [70, 48]. It is also so in point-counting methods such as Schoof’s algorithm and the Schoof-Elkies-Atkin (SEA) algorithm [7, Ch. VII], the second one being the best known method for counting the number of points of el-liptic curves deﬁned over ﬁnite ﬁelds of large character-istic. Indeed, the bulks of these algorithms are computa-tions of xq modulo the “division polynomial” fℓ(x) and of xq modulo the “modular polynomial” Φℓ(x), where ℓ= O(log(q)) and deg(fℓ) = O(ℓ2), deg(Φℓ) = O(ℓ). As a ﬁnal remark, note that while Fiduccia’s al-gorithm shows that computing the terms of indices Copyright c ⃝2021 by SIAM Unauthorized reproduction of this article is prohibited Algorithm 9 Input: rec. (1.1), u0, . . . , ud−1, N Output: uN, . . . , uN+d−1 Assumptions: Γ(x) = xd −Pd−1 i=0 cixi with c0 ̸= 0 1: ρ(x) ←xN mod Γ(x) ▷using Algorithm 8 2: U(x) ←u0 + · · · + u2d−2x2d−2 ▷using Algorithm in [69, p. 18] 3: V (x) ←U(x) · (xd · ρ(1/x)) ▷V = Pd−1 i=0 vixi 4: return [vd, . . . , v2d−1] N, . . . , N + d −1 of a linearly recurrent sequence of order d can be reduced to polynomial modular expo-nentiation (xN mod Γ(x)), Algorithm 8 shows that the converse is also true: polynomial modular exponentia-tion can be reduced to computing the terms of indices N, . . . , N + d −1 of a linearly recurrent sequence of or-der d. Therefore, these two problems are computation-ally equivalent. To our knowledge, this important fact seems not to have been noticed before. 3.3 A new Fiduccia-style algorithm We conclude this section by discussing a straightforward application of Algorithm 8. This is based on the next equality, generalizing (3.5) to any k ≥1: (3.6) uN · · · uN+k−1 = r × Hk, where as before r = r0 · · · rd−1 is the coeﬃcients vector of ρ = Pd−1 i=0 rixi, with ρ = xN mod Γ(x), and Hk is the Hankel matrix Hk :=      u0 · · · ud−1 · · · · · · uk−1 u1 · · · ud · · · · · · uk . . . . . . . . . . . . ud−1 · · · u2d−2 · · · · · · · · · uk+d−2     . The matrix Hk is built upon the ﬁrst terms of the sequence (un)n≥0 satisfying recurrence (1.1) with char-acteristic polynomial Γ = xd −Pd−1 i=0 cixi, or equiva-lently, from the power series expansion of the rational function P/Q with Q(x) = xd Γ(1/x). Note that the entries of Hk can be computed either from P and Q, or from the recurrence (1.1) together with the initial terms u0, . . . , ud−1, using O((k + d) M(d)/d) arithmetic operations, by the algo-rithm in [69, Thm. 3.1], see also [8, §5]. To compute uN, . . . , uN+k−1 it thus only remains to perform the vector-matrix product (3.6). When k = 1, the product r×H1 costs 2d operations and it yields the term uN. When k ≥d, the product r × Hk can be reduced to the polynomial multiplication of rd−1 + · · · + r0xd−1 by Pk+d−2 i=0 uixi, and this can be performed using ⌈k+d d ⌉M(d) arithmetic operations. As a consequence, the whole slice of coeﬃcients uN+i = [xN+i]P/Q for i = 0, . . . , k −1, can be computed using Algorithm 8 and eq. (3.6) for a total cost of arithmetic operations of (2 M(d) + d) log N + O k + d d M(d) . When k = d, this proves Theorem 3. The corresponding algorithm is given as Algorithm 9. We emphasize that this variant of Fiduccia’s algo-rithm is diﬀerent from Algorithm 1. It is actually a bit slower than Algorithm 1 when k = 1. However, when k > 1 terms are to be computed, it should be preferred to repeating k times Algorithm 1. It also compares fa-vorably with Fiduccia’s original algorithm, whose adap-tion to k terms has arithmetic complexity 3 M(d) log N + O d log(N) + k + d d M(d) . 3.4 Applications In this section, we discuss more applications of the MSB-ﬁrst algorithms (Algorithm 5 and 8) presented in §3.1 and §3.2. We deal with the case of multiplicities (§3.4.1), and explain a new way to speed up computations in that case. Then, we address another application, to faster powering of matrices (§3.4.2). To simplify matters, we assume in this section that R = K is a ﬁeld. 3.4.1 The case with multiplicities Hyun and his co-authors [42, 41] addressed the following question: is it possible to compute faster the N-th term of a linearly recurrent sequence when the characteristic polynomial of the recurrence has multiple roots? By the Chinese Remainder Theorem, it is suﬃcient to focus on the case where the characteristic polynomial is a pure power of a squarefree polynomial. In other words, the main step of [42, Algorithm 1] is to compute xN mod Q, where Q = (Q⋆)m and Q⋆is the squarefree part of Q. Under suitable invertibility conditions, the problem is solved in [42, 41] in O(M(d⋆) log N + M(d) log d) operations in K, where d⋆= deg(Q⋆) and d = deg(Q) = m · d⋆. This cost is obtained using an algorithm based on bi-variate computations, using the isomorphisms between K[x]/(Q) and K[y, x]/(Q⋆(y), (x −y)m) made eﬀective by the so-called tangling / untangling operations. We now propose an alternatively fast, but simpler, algo-rithm with the same cost. Let us explain this on an example, for “multiple-Fibonacci numbers”, that is when Q has the form (Q⋆)m, with Q⋆= 1 −x −x2 and d⋆= 2, d = 2m. Assume we want to compute the N-th coeﬃcient uN in the power series expansion of (x/(1 −x −x2))m. Copyright c ⃝2021 by SIAM Unauthorized reproduction of this article is prohibited The cost of Fiduccia’s algorithm, and also of our new algorithms, is O(M(m) · log N). Let us explain how we can lower this to O(log N + M(m) log m). The starting point is the observation that, by the structure theorem of linearly recurrent sequences [15, §A.(I)] (see also [62, §2]), uN is of the form wm(N)φN +vm(N)ψN, where φ and ψ are the two roots of 1+x = x2 and wm, vm are polynomials in K[x] of degree less than m. By an easy liner algebra argument, uN is thus equal to Wm(N)FN + Vm(N)FN+1, where Wm(x) and Vm(x) are polynomials in K[x] of degree less than m. These polynomials can be computed by (structured) linear algebra from the ﬁrst 2m values of the sequence (un), in complexity O(M(m) log m). For instance, when d = 2, we have U2(x) = −(x + 1)/5 and V2(x) = 2x/5. Once Um and Vm are determined, it remains to compute FN and FN+1 using Algorithm 9 in O(log N) operations in K, then to return the value Wm(N) · FN + Vm(N) · FN+1. The arguments extend to the general case and yields an algorithm of arithmetic complexity (2 M(d⋆) + d⋆) log N + O(M(d) log d). 3.4.2 Faster powering of matrices Assume we are given a matrix M ∈Md(K), an integer N, and that we want to compute the N-th power M N of M. The arithmetic complexity of binary powering in Md(K) is O(dθ log N) operations in K, where as before θ ∈[2, 3] is any feasible exponent for matrix multiplica-tion in Md(K). A better algorithm consists in ﬁrst computing the characteristic polynomial Γ(x) of the matrix M, then the remainder ρ(x) := xN mod Γ(x), and ﬁnally eval-uating the polynomial ρ(x) at M. By the Cayley-Hamilton theorem, ρ(M) = M N. The most costly step is the computation of ρ, which can be done as ex-plained in §3.2 using ∼2 M(d) log(N) operations in K. The cost of the other two steps is independent of N, and it is respectively O(dθ log d) and O(dθ+ 1 2 ), this last cost being achieved using the Paterson-Stockmeyer baby-step/giant-step algorithm . The total cost of this algorithm is (2 M(d) + d) log(N) + O(dθ+ 1 2 ). Note that a faster variant (w.r.t. d), of cost (2 M(d) + d) log(N) + O(dθ log d), can be obtained us-ing [28, Corollary 7.4]. The corresponding algorithm is based on the computation of the Frobenius (block-companion) form of the matrix M, followed by the pow-ering of companion matrices, which again reduces to modular exponentiation. 4 Analysis under the FFT multiplication model In this section, we specialize, optimize and analyze the generic Algorithm 1 to the FFT setting, in which polynomial products are assumed to be performed using the discrete Fourier transform (DFT), and its inverse. To do this, we will assume that the base ring R possesses roots of unity of suﬃciently high order. To simplify the exposition, R will be supposed to be a ﬁeld, but the arguments also apply without this assumption, modulo some technical complications, see [27, §8.2]. 4.1 Discrete Fourier Transform for polynomial products Let K be a ﬁeld with a primitive n-th root ωn of unity. Let A ∈K[x] be a polynomial of degree at most d ≤n −1. The DFT b A of A is deﬁned by b Ay := A(ω−y n ) = n−1 X i=0 Aiω−yi n for y = 0, 1, . . . , n −1. Here, Ay = 0 for y > d. It is classical that the DFT map is an invertible K-linear transform from Kn to itself, and that the polynomial A can be retrieved from its DFT b A using the formulas Ai = 1 n n−1 X y=0 b Ayωyi n for i = 0, 1, . . . , n −1. For computing the polynomial multiplication C(x) = A(x)B(x) for given A(x), B(x) ∈K[x] of degree at most d, it is suﬃcient to compute the DFT of C(x) for n ≥2d+1. Since b Cy = C(ω−y n ) = A(ω−y n )B(ω−y n ) = b Ay b By, the polynomial C(x) can be computed using two DFTs and one inverse DFT. Let E(n) be an arithmetic complexity for computing a DFT of length n. Then the cost of polynomial multiplication in K[x] is governed by M(d) = 3 E(2d) + O(d). 4.2 Fast Fourier Transform In this subsection, we brieﬂy recall the Fast Fourier Transform (FFT), which gives the quasi-linear estimate E(n) = O(n log n). Assume n is even. Then, for y = 0, 1, . . . , n/2 −1 we have b Ay = n/2−1 X i=0 A2iω−y(2i) n + n/2−1 X i=0 A2i+1ω−y(2i+1) n = n/2−1 X i=0 A2iω−yi n/2 + ω−y n · n/2−1 X i=0 A2i+1ω−yi n/2 = c Aey + ω−y n c Aoy for Ae(x) := Pn/2 i=0 A2ixi and Ao(x) := Pn/2 i=0 A2i+1xi. Similarly, we have b An/2+y = b Ae y −ω−y n b Ao y. We Copyright c ⃝2021 by SIAM Unauthorized reproduction of this article is prohibited therefore obtain the following matrix equation " b Ay b An/2+y # = 1 1 1 −1 1 0 0 ω−y n " b Ae y b Ao y # (4.7) for y = 0, 1, . . . , n/2 −1. Thus, computing a DFT in size n reduces to two DFTs in size n/2. More precisely, E(n) ≤2 E(n/2)+(3/2)n. If n is a power of two, n = 2k, and if the ﬁeld K contains a primitive 2k-th root of unity (as is the case for instance when K = C, K = Fp for a prime number p satisfying 2k \| p −1), this reduction can be repeated k = log n times, and it yields the estimate E(n) = 3 2n log n. The corresponding algorithm is called the decimation-in-time Cooley–Tukey fast Fourier transform , see also [5, §2]. By the arguments of §4.1, we conclude that polyno-mial multiplication in K[x] can be performed in arith-metic complexity M(d) = 9 d log d + O(d). 4.3 Eﬃciently doubling the length of a DFT In the FFT setting, it is useful for many applications to compute eﬃciently a DFT of length 2n starting from DFT of length n. Assume n ≥d + 1 and we have at our disposal the DFT b A of A, of length n. Assume that we want to compute the DFT b A(2n) of length 2n. The simplest algorithm is to apply the inverse DFT of length n to obtain A, and then to apply the DFT of length 2n to A. This costs E(n) + E(2n) arithmetic operations, that is 9 2n log n + 3n operations in K. This algorithm can be improved using the following formulas b A(2n) 2y = 2n−1 X i=0 Aiω−2yi 2n = n−1 X i=0 Aiω−yi n = b Ay, b A(2n) 2y+1 = 2n−1 X i=0 Aiω−(2y+1)i 2n = n−1 X i=0 ω−i 2nAiω−yi n = b By, where Bi := ω−i 2nAi for i = 0, 1, . . . , n −1. We obtain Algorithm 10 with arithmetic complexity 2 E(n)+n, i.e. 3n log n+n, [5, §12.8], see also [4, 52]10. Compared with the direct algorithm, the gain is roughly a factor of 3/2. 4.4 Algorithm 1 in the FFT setting Recall that our main objective is, given P, Q in K[x] with d = 10This “FFT doubling” trick is sometimes attributed to R. Kramer (2004), but we were not able to locate Kramer’s paper. Algorithm 10 Input: DFTn(A) Output: DFT2n(A) 1: function DoubleDFT( b A) 2: A ←IDFTn( b A) 3: Bi ←ω−i 2nAi for i = 0, 1, . . . , n −1 4: b B ←DFTn(B) 5: b A(2n) 2y ←b Ay for y = 0, 1, . . . , n −1 6: b A(2n) 2y+1 ←b By for y = 0, 1, . . . , n −1 7: return b A(2n) deg(Q) > deg(P), to compute the N-th coeﬃcient uN in the series expansion of P/Q. Let k be the minimum integer satisfying 2k ≥2d+1. Assume that there exists a primitive 2k-th root of unity in K. In this case, we can employ an FFT-based polynomial multiplication in K[x]. In each iteration of Algorithm 1, it is suﬃcient to compute P(x)Q(−x) and Q(x)Q(−x). Here, only two FFTs and two inverse FFT of length 2k are needed since b Q− y = b Q¯ y for Q−(x) := Q(−x) where ¯ y := y + 2k−1 if y < 2k−1 and ¯ y := y −2k−1 if y ≥2k−1. Hence, the arithmetic complexity S(d) for a single step in Algorithm 1 satisﬁes S(d) ≤4 E(2k) + O(2k). In the following we will show the improved estimate S(d) ≤4 E(2k−1) + O(2k). Before entering the while loop in Algorithm 1, the DFTs b P and b Q of P(x) and Q(x) of length 2k are computed, respectively. Inside the while loop, b P and b Q are updated. The recursive formula (??) for the decimation-in-time Cooley–Tukey FFT is equivalent to " b Ae y b Ao y # = 1 2 1 0 0 ωy n 1 1 1 −1 " b Ay b A2k−1+y # for y = 0, 1, . . . , 2k−1. By using this formula, b Ae (or b Ao) can be computed with O(2k) operations from b A. By using Algorithm 10, we obtain the updated b P from b Ae or b Ao. The algorithm is summarized in Algorithm 11. In each step, DoubleDFT is called twice. Hence, the total arithmetic complexity of Algorithm 11 is (4 E(2k−1) + O(2k)) · log N. When d is of the form 2ℓ−1, then11 one can take k = ℓ+ 1 and the cost simpliﬁes to T(N, d) = 4 E(d) log N + O(d log N), 11In the general case, it might be useful to use the Truncated Fourier Transform (TFT), which smoothes the “jumps” in com-plexity exhibited by FFT algorithms [38, 37, 3]. Copyright c ⃝2021 by SIAM Unauthorized reproduction of this article is prohibited Algorithm 11 (OneCoeﬀ-FFT) Input: P(x), Q(x), N Output: [xN] P (x) Q(x) 1: b P ←DFT2k(P) 2: b Q ←DFT2k(Q) 3: while N ≥1 do 4: b Uy ←b Py b Q¯ y for y = 0, 1, . . . , 2k −1 5: if N is even then 6: b U e y ← (b Uy + b Uy+2k−1)/2 for y = 0, 1, . . . , 2k−1 −1 7: b P ←DoubleDFT(b U e y) 8: else 9: b U o y ← ωy 2k(b Uy −b Uy+2k−1)/2 for y = 0, 1, . . . , 2k−1 −1 10: b P ←DoubleDFT(b U o y ) 11: b Ay ←b Qy b Q¯ y for y = 0, 1, . . . , 2k−1 −1 12: b Q ←DoubleDFT( b A) 13: N ←⌊N/2⌋ 14: P(0) ←P2k−1 y=0 b Py 15: Q(0) ←P2k−1 y=0 b Qy 16: return P(0)/Q(0) or, equivalently T(N, d) = 6 d log d log N + O(d log N). The (striking) conclusion of this analysis is that, in the FFT setting, our (variant of the) algorithm for comput-ing the N-th term of P/Q uses much less operations than in the general case, namely (4.8) T(N, d) ∼2 3 M(d) log N, while for a generic multiplication algorithm the cost is ∼2 M(d) log N. This proves Theorem 2. Note that the complexity bound (4.8) compares favorably with Fidducia’s algorithm combined with the best algorithms for modular squaring. For instance, Shoup’s algorithm [70, §7.3] computes one modular squaring in the FFT setting using ∼5 3 M(d) arithmetic operations, while Mihăilescu’s algorithm [53, Table 1] (based on Montgomery’s algorithm ) uses roughly ∼ 13 12 M(d) arithmetic operations. Our bound (4.8) is better by a factor of 2.5 than Shoup’s (comparatively simple) algorithm, and by a factor of 1.625 than the (much more complex) algorithm by Mihăilescu. Let us point out that all the other algorithms admit similarly fast versions in the FFT setting. We will however not give them in full detail here, mainly for space reasons. 4.5 Algorithm 5 in the FFT setting In this sec-tion, we present Algorithm 5 in the FFT setting with Algorithm 12 Input: DFT2k(A) Output: DFT2k−1(Asec) 1: function SecondHalfDFT( b A) 2: b By ←b A2y+1 for y = 0, 1, . . . , 2k−1 −1 3: B ←IDFT2k−1( b B) 4: Ci ←ωi 2kBi for i = 0, 1, . . . , 2k−1 −1 5: b C ←DFT2k−1(C) 6: b Dy ←( b A2y −b Cy)/2 for y = 0, 1, . . . , 2k−1 −1 7: return b D arithmetic complexity ((2/3)M(d) + O(d)) log N. Algo-rithm 5 includes one standard product Q(x)Q(−x) and one middle-product Q(−x)S(x). The standard product can be performed by two FFTs of length 2k−1 as in Al-gorithm 11. For the middle-product, we ﬁrst compute the cyclic product. i.e., the product in K[x]/(x2k −1), between Q(−x) and S(x), and then extract the second half of the cyclic product. For computing the DFT of the second half Asec(x) := P2k−1−1 i=0 Ai+2k−1xi, we use the formula for the decimation-in-frequency FFT. b A2y = 2k−1 X i=0 ω−2iy 2k Ai = 2k−1−1 X i=0 ω−iy 2k−1(Ai + Ai+2k−1) b A2y+1 = 2k−1 X i=0 ω−i(2y+1) 2k Ai = 2k−1−1 X i=0 ω−iy 2k−1ω−i 2k (Ai −Ai+2k−1). This formula gives Algorithm 12 which transforms b A to d Asec. The arithmetic complexity of Algorithm 12 is 2E(2k−1) + O(2k). The whole algorithm is shown in Algorithm 13. The input of the algorithm is N and b Q that is the DFT of Q of length 2k. The output of the algorithm is the DFT of FN,2k−1(1/Q(x)) of length 2k−1 rather than length 2k since the DFT of W(x2) of length 2k can be obtained eﬃciently from the DFT of W(x) of length 2k−1. Algorithm 13 calls one DoubleDFT and one SecondHalfDFT. Hence, similarly to Algorithm 11, the arithmetic complexity of Algorithm 13 is (4E(2k−1) + O(2k)) log N. 5 Conclusion We have proposed several algorithmic contributions to the classical ﬁeld of linearly recurrent sequences. Firstly, we have designed a simple and fast algo-rithm for computing the N-th term of a linearly recur-rent sequence of order d, using ∼2 M(d) log N arith-metic operations, which is faster by a factor of 1.5 than the state-of-the-art 1985 algorithm due to Fiduccia . When combined with FFT techniques, the algorithm has even better arithmetic complexity ∼2 3 M(d) log N Copyright c ⃝2021 by SIAM Unauthorized reproduction of this article is prohibited Algorithm 13 (SliceCoeﬀ-FFT) Input: DFT2k(Q), N Output: DFT2k−1(FN,2k−1(1/Q(x))) 1: function SliceCoeﬀ-FFT(N, b Q) 2: if N = 0 then 3: Q(0) ←P2k−1 y=0 b Qy 4: return [1 ω2k ω2 2k · · · ω2k−1 2k ]/Q(0) 5: b Ay ←b Qy b Q¯ y for y = 0, 1, . . . , 2k−1 −1 6: b V ←DoubleDFT( b A) 7: c W ←SliceCoeﬀ-FFT(⌊N/2⌋, b V ) 8: if N is even then 9: b Sy ←ω−y 2k c Wy mod 2k−1 for y = 0, 1, . . . , 2k−1 10: else 11: b Sy ←c Wy mod 2k−1 for y = 0, 1, . . . , 2k −1 12: b By ←b Sy b Q¯ y for y = 0, 1, . . . , 2k −1 13: return SecondHalfDFT( b B) which is faster than the fastest variant of Fiduccia’s al-gorithm in the FFT setting by a factor of 1.625. The new algorithms are based on a new method (Algo-rithm 1) for computing the N-th coeﬃcient of a rational power series. Secondly, using algorithmic transposition tech-niques, we have derived from Algorithm 1 a new method (Algorithm 5) for computing simultaneously the coeﬃ-cients of indices N −d + 1, . . . , N in the power series expansion of the reciprocal of a degree-d polynomial, us-ing again ∼2 M(d) log N arithmetic operations. Using Algorithm 5, we have designed a new algorithm for com-puting the remainder of xN modulo a given polynomial of degree d, using ∼2 M(d) log N arithmetic operations as well. This is better by a factor of 1.5 than the pre-vious best algorithm for modular exponentiation, with an even better speed-up in the FFT setting, as for Al-gorithm 1. Combined with the basic idea of Fiduccia’s algorithm, our new algorithm for modular exponentia-tion yields a faster Fiduccia-like algorithm (by the afore-mentioned constant factors) that computes a slice of d consecutive terms (of indices N −d + 1, . . . , N) of a lin-early recurrent sequence of order d using ∼2 M(d) log N arithmetic operations. Thirdly, we have discussed applications of the new algorithms to a few other algorithmic problems, includ-ing powering of matrices and the computation of terms of linearly recurrent sequences when the recurrence has roots with (high) multiplicities. As future work, we plan to investigate further the full power of our technique. To which extent can it be generalized to larger classes of power series? For instance, although it perfectly works for bivariate ra-tional power series U(x, y), the corresponding method does not directly provide a O(log N)-algorithm for com-puting the (N, N)-th coeﬃcient uN,N, the reason being that the log N new bivariate recurrences produced by the Graeﬀe process do not have constant orders, as in the univariate case. This is disappointing, but after all not surprising, because the generating function of the sequence (un,n)n is known to be algebraic, but not ra-tional anymore . As of today, no algorithm is known for computing the N-th coeﬃcient of an algebraic power series faster than in the P-recursive case (P), namely in a number of ring operations almost linear in √ N. Acknowledgements. We are grateful to the three reviewers for their kind and constructive remarks. 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15719	https://www.effortlessmath.com/wp-content/uploads/2020/04/Simplifying-Ration.pdf?srsltid=AfmBOopQEs2AFhCTi8mOq-I3YKz6-11XYC1k4f3iDP-WzBnT04qJhCLn	Math Worksheets Name: ____ Date: _____ … So Much More Online! Please visit: www.EffortlessMath.com Simplifying Ratios  Reduce each ratio. 1) 12 : 8 = _: _ 2) 2: 20 = _: _ 3) 3: 36 = _: _ 4) 8: 16 = _: _ 5) 6: 100 = _: _ 6) 10 : 60 = _: _ 7) 21 : 49 = _: _ 8) 20 : 40 = _: _ 9) 10 : 50 = _: _ 10) 14 : 18 = _: _ 11) 45 : 27 = _: _ 12) 49 : 21 = _: _ 13) 100 : 10 = _: _ 14) 35 : 45 = _: _ 15) 8: 20 = _: _ 16) 25 : 35 = _: _ 17) 21 : 27 = _: _ 18) 52 : 82 = _: _ 19) 12 : 36 = _: _ 20) 24 : 3 = _: _ 21) 15 : 30 = _: _ 22) 14 : 63 = _: _ 23) 68 : 80 = _: _ 24) 8: 80 = _: _  Write each ratio as a fraction in simplest form. 25) 2: 4 = 26) 6: 20 = 27) 5: 35 = 28) 10 : 55 = 29) 8: 24 = 30) 9: 42 = 31) 12 : 48 = 32) 6: 40 = 33) 15 : 36 = 34) 18 : 82 = 35) 22 : 26 = 36) 8: 36 = 37) 16 : 128 = 38) 14 : 77 = 39) 12 : 180 = 40) 36 : 108 = 41) 24 : 42 = 42) 18 : 120 = 43) 44 : 82 = 44) 60 : 240 = 45) 36 : 180 =Math Worksheets Name: ____ Date: _____ … So Much More Online! Please visit: www.EffortlessMath.com Answers 1) 3 ∶ 2 2) 1 ∶ 10 3) 1 ∶ 12 4) 1 ∶ 2 5) 3 ∶ 50 6) 1 ∶ 6 7) 3 ∶ 7 8) 1 ∶ 2 9) 1 ∶ 5 10) 7 ∶ 9 11) 5 ∶ 3 12) 7 ∶ 3 13) 10 ∶ 1 14) 7 ∶ 9 15) 2 ∶ 5 16) 5 ∶ 7 17) 7 ∶ 9 18) 26 ∶ 41 19) 1 ∶ 3 20) 8 ∶ 1 21) 1 ∶ 2 22) 2 ∶ 9 23) 17 ∶ 20 24) 1 ∶ 10 25) 1 2 26) 3 10 27) 1 7 28) 2 11 29) 1 3 30) 3 14 31) 1 4 32) 3 20 33) 5 12 34) 9 41 35) 11 13 36) 2 9 37) 1 8 38) 2 11 39) 1 15 40) 1 3 41) 4 7 42) 3 20 43) 22 41 44) 1 4 45) 1
15720	https://www.lehman.edu/faculty/dgaranin/Mathematical_Physics/Mathematical_physics-14-Eigenvalue%20problems.pdf	Eigenvalue problems Main idea and formulation in the linear algebra The word "eigenvalue" stems from the German word "Eigenwert" that can be translated into English as "Its own value" or "Inherent value". This is a value of a parameter in the equation or system of equations for which this equation has a nontriv-ial (nonzero) solution. Mathematically, the simplest formulation of the eigenvalue problem is in the linear algebra. For a given square matrix A one has to find such values of l, for which the equation (actually the system of linear equations) (1) A.X λX has a nontrivial solution for a vector (column) X. Moving the right part to the left, one obtains the equation HA −λIL.X 0, where I is the identity matrix having all diagonal elements one and nondiagonal elements zero. This matrix equation has nontrivial solutions only if its determinant is zero, Det@A −λID 0. This is equivalent to a Nth order algebraic equation for l, where N is the rank of the mathrix A. Thus there are N different eigenvalues ln (that can be complex), for which one can find the corresponding eigenvectors Xn. Eigenvectors are defined up to an arbitrary numerical factor, so that usually they are normalized by requiring Xn T∗.Xn 1, where X T is the row transposed and complex conjugate to the column X. It can be proven that eigenvectors that belong to different eigenvalues are orthogonal, so that, more generally than above, one has Xm T∗.Xn δmn. Here dmn is the Kronecker symbol, δmn = µ 1, m n 0, m ≠n. An important class of square matrices are Hermitean matrices that satisfy AT∗ A. Eigenvalues of Hermitean matrices are real. A real Hermitean matrix is just a symmetric matrix, AT ã A. For such matrices eigenvectors can be chosen real. ü Matrix eigenvalue problem in Mathematica Mathematica offers a solver for the matrix eivenvalue problem. If one is interested in eigenvalues only, one can use the command Eigenvalues[...]. Eigenvectors are computed by Eigenvectors[...], while both eigenvalues and eigenvectors are computed by the command Eigensystem[...]. Let us illustrate how it works for a real symmetric matrix A = K a b b −a O; Its eigenvalues are given by Eigenvalues@AD :− a2 + b2 , a2 + b2 > Its eigenvectors are given by Eigenvectors@AD ::− −a + a2 + b2 b , 1>, :− −a − a2 + b2 b , 1>> that is, Xi_ := Eigenvectors@AD@@iDD These two eigenvectors are orthogonal to each other X1.X2 êê Simplify 0 However, they are not normalized X1.X1 êê Expand êê Factor − 2 −a2 −b2 + a a2 + b2 b2 To see which eigenvector corresponds to each eigenvalue, one has to use the command Eigensystem ESys = Eigensystem@AD ::− a2 + b2 , a2 + b2 >, ::− −a + a2 + b2 b , 1>, :− −a − a2 + b2 b , 1>>> The first part of this List are eigenvalues and the second part are eigenvectors. One can better see the correspondence in the form TableForm@Transpose@ESysDD − a2 + b2 − −a+ a2+b2 b 1 a2 + b2 − −a− a2+b2 b 1 Mathematica also solves matrix eigenvalue problems numerically, that is the only way to go for big matrices. For instance, ESys = Eigensystem@A ê. 8a →1., b →2.<D 88−2.23607, 2.23607<, 880.525731, −0.850651<, 8−0.850651, −0.525731<<< The numerical eigenvectors Xi_ := ESys@@2DD@@iDD are orthonormal 2 Mathematical_physics-14-Eigenvalue problems.nb X1.X1 X1.X2 1. 0. For a complex Hermitean matrix eigenvalues are indeed real, although eigenvectors are complex TableFormBTransposeBEigensystemBK 2 −2 OFFF − 3 J−2 + 3 N 1 3 − J2 + 3 N 1 Eigenvalue problem for systems of linear ODEs on time The importance of the eigenvalue problem in physics (as well as in engineering and other areas) is that it arises on the way of solution of systems of linear ordinary differential equations with constant coefficients. We have already obtained the solution for the harmonic oscillator on this way in the chapter on differential equations. Every linear ODE or a system of ODEs can be represented in the basic matrix form with a constant matrix A X'@tD + A.X@tD 0, X being a vector. (We drop the inhomogeneous term.) Searching for the solution in the form X@tD = X0 −λt, one arrives at the eigenvalue problem Eq. (1) with X ﬂX0. After finding eigenvalues ln and normalized eivenvectors X0 n by linear algebra, one can write down the general solution of the equation as a linear superposition of all these solutions, X@tD = ‚ n=1 N Cn X0 n −λn t, where Cn are integration constants that can be found from the initial conditions. Eigenvalue problems for PDE In physical problems described by partial differential equations, eigenvalue problems usually arise due to boundary condi-tions. ü Standing waves in a pipe Consider, as an example, the wave equation for the pressure change (see Waves) in the 1d region 0 § x § L, ∂t 2δP −c2 ∂x 2δP. If we consider a pipe with both ends open to the atmosphere, the boundary conditions are dP@0, tD ã 0 , dP@L, tD ã 0 because the pressure at the open end (practically) merges with the constant atmospheric pressure. Searching for dP in the form δP@x, tD = ψ@xD Cos@ωt + φ0D Mathematical_physics-14-Eigenvalue problems.nb 3 one obtains the stationary wave equation ∂x 2ψ + k2 ψ 0, k = ω ê c, k being the wave vector. This is an eigenvalue problem because this equation has nontrivial solutions that satisfy the bound-ary conditions only for some values of k and thus of w. The general solution of the ODE above is ψ@xD = C1 Sin@kxD + C2 Cos@kxD. Since Cos@kxD does not satisfy the BC at x = 0, the solution simplifies to ψ@xD = C Sin@kxD that describes a standing wave. Next, the BC at x = L requires Sin@kLD = 0 from which one obtains the eigenvalues of the wave vector Sin@kLD = 0 k = kn = πn L , n = 1, 2, 3, — In terms of the wave length of the standing wave l = 2 pêk one has λn = 2 L n , n = 1, 2, 3, — Standing wave with n = 1 is called fundamental wave, whereas those with n = 2, 3,— are called overtones or harmonics. For the frequencies of these waves f = wêH2 pL = cêl one has fn = cn 2 L , n = 1, 2, 3, — The general solution of the wave equation for a pipe with both ends open is a linear superposition of all these solutions, (2) δP@x, tD = ‚ n=1 ∞ Cn Sin@kn xD Cos@ωn t + φnD, kn = ωn c = πn L , n = 1, 2, 3, — The coefficients Cn and phases fn in this solution are arbitrary. Similar results can be obtained for a pipe with both ends closed (such as flute). If one end is closed and one is open (clarinet), the solution is somewhat different and only odd overtones exist. (Excersize). While the phases fn are irrelevant for our ears (Ohm's phychoacoustical law), the amplitudes Cn define the quality of the sound via the relative weight of the overtones. This depends on how the music instrument is con-structed and how it is played. Eigenfunctions corresponding to different eigenvalues are orthogonal, ‡ 0 L Sin@km xD Sin@kn xD x = L 2 δmn. One can define the normalized eigenfunctions ψn@xD = 2 L Sin@kn xD that satisfy ‡ 0 L ψm@xD ψn@xD x = δmn. Several lowest eigenfunctions are plotted below. 4 Mathematical_physics-14-Eigenvalue problems.nb L = 1; ψnx@n_, x_D = 2 L SinB π n L xF; Plot@Table@ψnx@n, xD, 8n, 1, 4<D, 8x, 0, L<D 0.2 0.4 0.6 0.8 1.0 -1.0 -0.5 0.5 1.0 ü General formulation of the eigenvalue problem for PDE In general, the eigenvalue problem for PDE can be formulated in the form L ˆ ψ@rD λψ@rD, where L is a differential operator. The best example is the stationary Schrödinger equation for a quantum particle (3) H y ã Ey, H = -Ñ2 D 2 m + U@rD, where H is the Hamilton operator. For a particle bound to a potential well, this equation has solutions satisfying boundary conditions only for select discrete values of the energy, E = En, n = 0, 1, 2, — These energy eigenvalues are labeled by the index n in the increasing order, and n = 0 corresponds to the minimal energy, the so-called ground-state energy. When a quantum particle undergoes a transition from the energy level (energy eigenstate) m to a lower energy level n, it emits a quantum of electromagnetic radiation (a photon), to conserve the total energy. The energy of the photon ¶ is related to its frequency w as ¶ = Ñw (Max Planck), and from the energy conservation follows Ñw = Em - En. Spectroscopic investiga-tions of the light emitted (and absorbed) by simple quantum systems such as hydrogen atoms have shown that only particular discrete frequencies obeying certain simple laws are present in the spectrum. This was at a contradiction with classical mechanics and electrodynamics that predicted a continuous energy spectrum. Physicists started to look for a mathematical tool that describes discreteness and made use of the eigenvalue problem. Werner Heisenberg proposed his matrix quantum mechanics operating with eigenvalues of matrices describing quantum systems. Erwin Schrödinger obtained his famous Schrödinger equation in which discrete energy levels of a quantum system arise as eigenvalues of a differential operator. The approaches of Heisenberg and Schrödinger are equivalent. Solving the eigenvalue problem for a general differential operator, especially in 2d and 3d, is not an easy task, and there is no Mathematica solver for this problem at the moment. If the differential operator and boundary conditions possess special properties such as symmetry (e.g., spherical or cylindrical symmetry), the solution can be searched for in the form of a product of functions of different variables, so that the solution simplifies. This case is called "separation of variables". Similarly to eigenvectors of matrices, eigenfunctions of differential operators can be complex. It can be shown that eigenfunc-tions corresponding to different eigenvalues are orthogonal. Also eigenfunctions corresponding to discrete eigenvalues can be normalized. Allowing for complex eigenfunctions, the orthonormality conditions takes the form (4) ‡ ψm ∗@rD ψn@rD V = δmn, Mathematical_physics-14-Eigenvalue problems.nb 5 where integration is carried out over the volume. An important mathematical theorem states that the whole set of eigenfunc-tions yn@rD of a differential operator satisfying boundary conditions forms a complete basis, so that any function f @rD satisfying these BC can be expanded over the eigenfunctions as (5) f @rD = ‚ n=1 ¶ cn yn@rD. The expansion coefficients can be found using the orthonormality condition that yields (6) ‡ ym @rD f @rD „V = ‡ ‚ n=1 ¶ cn ym @rD yn@rD „V = ‚ n=1 ¶ cn dmn = cm. ü Quantum particle in a one-dimensional potential box The stationary Schrödinger equation (6) for a quantum particle in a one-dimensional potential box 0 § x § L has the form ∂x 2ψ + k2 ψ 0, k = 2 mE ë —. Everywhere in the box U = 0 while outside the box U = ¶, so that the wave function outside the box is zero and by continu-ity it has to be zero at the box walls, y@0, tD ã 0 , y@L, tD ã 0. Mathematically this eigenvalue problem is equivalent to the problem of standing waves in a pipe considered above, except for another definition of the wave vector k. It has the same solution for the normalized eigenfunctions ψn@xD = 2 L Sin@kn xD, kn = πn L , n = 1, 2, 3, — The corresponding energy eigenvalues are given by En = —2 kn 2 2 m = π2 —2 n2 2 mL2 . These energy levels are discrete, and a transition m z n should be accompanied by emission of a photon of frequency ωmn = Em −En — = π2 — 2 mL2 Im2 −n2M. Measuring all these frequencies, one can figure out the energy levels. ü Quantum particle in a three-dimensional potential box and density of states ü Energy levels Let us consider now a quantum particle in 3d potential box 0 § x § Lx, 0 § y § Ly, 0 § z § Lz. One can easily check that the solution with separated variables ψ@x, y, zD ∝Sin@kx xD SinAky yE Sin@kz zD satisfies the stationary Schrödinger equation ∆ψ + k2 ψ 0 since kx 2 + ky 2 + kz 2 = k2. Also this solution satisfies the BC at x, y, z = 0, where y vanishes. Taking into account three BC at the other sides of the box finally yields 6 Mathematical_physics-14-Eigenvalue problems.nb ψnx ny nz@x, y, zD = 23ê2 Lx Ly Lz Sin@kxnx xD SinAkyny yE Sin@kznz zD, where the quantized wave vectors are given by kαnα = π Lα nα, nα = 1, 2, 3, —, α = x, y, z. The discrete energy eigenvalues are parametrized by three integers, (7) Enx ny nz = π2 —2 2 m nx 2 Lx 2 + ny 2 Ly 2 + nz 2 Lz 2 . Another important case of separation of variables is the hydrogen atom which possesses a spherical symmetry. Rewriting Schrödinger equation in spherical coordinates allows to separate variables r, q, and f and find analytical solutions for the energy levels and corresponding wave functions. ü Density of states For potential boxes of sufficiently large sizes the quantum energy levels are very close to each other, so that one can consider the density of states r@ED as the number of energy levels „ NE in the energy interval „ E, NE = ρ@ED E. For the 3 d box the density of states is given by ρ@ED = V H2 πL2 2 m —2 3ê2 E , V = Lx Ly Lz. It is instructive to plot a hystogram of the discrete energy levels given by Eq. (7) and to compare it with the formula above. Mathematical_physics-14-Eigenvalue problems.nb 7 Lx = 91; Ly = 100; Lz = 110; V = Lx Ly Lz; nxMax = Lx; nyMax = Ly; nzMax = Lz; nLevels = nxMax nyMax nzMax — = 1; M = 1.; ρ@EE_D = V H2 πL2 2 M —2 3ê2 EE ; Enxnynz@nx_, ny_, nz_D = π2 —2 2 M nx2 Lx2 + ny2 Ly2 + nz2 Lz2 ; EMin = 0; EMax = 1 3 Enxnynz@nxMax, nyMax, nzMaxD; NE = 100; H∗Number of energy bins E ∗L dE = HEMax −EMinL ê NE; H∗Width of energy bin E ∗L EList = Flatten@Table@Enxnynz@nx, ny, nzD, 8nx, 0, nxMax<, 8ny, 0, nyMax<, 8nz, 0, nzMax<DD; H∗List of energy eigenvalues ∗L EBinCounts = 1 dE BinCounts@EList, 8EMin, EMax, dE<D; H∗Number of eigenvalues in energy bins E divided by E ∗L DOSList = Table@8nE dE, EBinCounts@@nEDD<, 8nE, 0, NE<D; H∗Same with the energy values for the energy bins ∗L Show@ ListPlot@DOSList, PlotStyle →8Orange, PointSize@0.02D<, AxesLabel →8"E", "ρ@ED"<D, Plot@ρ@EED, 8EE, EMin, EMax<, PlotStyle →8Black, Thick<, AxesLabel →8"E", "ρ@ED"<D D 1 001 000 1 2 3 4 5 E 50000 100000 150000 r@ED ü Quantum harmonic oscillator Consider a one-dimensional quantum harmonic oscillator that is described by the Hamilton function (8) H = p2 2 m + U@xD, U@xD = mω0 2 x2 2 where w0 is the oscillator's frequency. The stationary Schrödinger equation Ey ã H y for the oscillator reads —2 2 m 2 x2 +E −U@xD ψ 0. 8 Mathematical_physics-14-Eigenvalue problems.nb This is a second-order linear ODE with a variable coefficient. Boundary conditions require that y decreases to zero at x =±¶. Solution of such differential equations is a more complicated task than for those with constant coefficients, and the analytical solution is usually expressed via special functions, if they are known. In this case an involved analytical solution yields the energy levels En = —ω0 Hn + 1 ê 2L and normalized eigenfunctions (9) ψn@xD = 1 2n n! mω0 π— 1ê4 ExpB− mω0 x2 2 — F Hn mω0 — x , where Hn@xD are Hermite polynomials defined by Hn@xD = H−1Ln x2 n xn −x2. In Mathematica one can calculate Hermite polynomials straightforwardly and fast enough as f@x_D = −x2; MyHermiteH@n_, x_D := H−1Ln x2 Derivative@nD@fD@xD êê Simplify MyHermiteH@0, xD MyHermiteH@1, xD MyHermiteH@2, xD MyHermiteH@3, xD 1 2 x −2 + 4 x2 4 x I−3 + 2 x2M etc., although there is a built-in function HermiteH[n, x] HermiteH@0, xD HermiteH@1, xD HermiteH@2, xD HermiteH@3, xD 1 2 x −2 + 4 x2 −12 x + 8 x3 One can see that eigenfunctions of the quantum harmonic oscillator are, in fact, elementary functions. Still, Hermite and many other polynomials are put into the class of special functions by tradition. Let us plot the energy eigenvalues and eighefunctions together with the potential. Mathematical_physics-14-Eigenvalue problems.nb 9 Parameters = 8m →1, — →1, ω0 →1<; U@x_D = m ω0 2 x2 2 ê. Parameters En@n_D = — ω0 Hn + 1 ê 2L ê. Parameters ψnx@n_, x_D = 1 2n n! m ω0 π — 1ê4 ExpB− m ω0 x2 2 — F HermiteHBn, m ω0 — xF ê. Parameters x2 2 1 2 + n − x2 2 HermiteH@n, xD π1ê4 2n n! xMax = 6; nMax = 16; Show@ Plot@U@xD, 8x, −xMax, xMax<, PlotStyle →8Black, Thick<D, H∗Potential energy ∗L Plot@Table@En@nD, 8n, 0, nMax<D, 8x, −xMax, xMax<, PlotStyle →DashedD, H∗Energy eigenvalues ∗L Plot@Evaluate@Table@0.7 ψnx@n, xD + En@nD, 8n, 0, nMax<DD, 8x, −xMax, xMax<, PlotStyle →BlackD H∗Eigenfunctions at the height of their eigenvalues ∗L D -6 -4 -2 2 4 6 5 10 15 It is convenient to put eigenfunctions at the height of their eigenvalues for better assosiation and separation. The numeric factor at eigenfunctions has been introduced to avoid overlap. Command Evaluate makes plotting faster by first evaluating eigenfunctions and then plotting them. Otherwise they are evaluated a every plot point. One can see that even and odd eigenfunctions alternate. Similarly to plane quantum waves discussed in the chapter on Schrödinger equation, wave functions oscillate in the classically allowed region E - U@xD > 0 and exponentially decrease in the classically prohibited region E - U@xD 0. As we know, a general solution of a second-order differential equation is a superposition of two independent functions. As in the case of a plane wave, in the classically prohibited region one of these functions is exponentially increasing while the other is exponentially decreasing. In this and similar problems, eigenfunctions decrease on both sides of the potential well. Other functions increase in the prohibited region and have to be rejected. It turns out that the solutions of the stationary Schrödiger equation that decrease on both sides exist only if E = En. For other energy values, both solutions of the DE increase at least on one side of the well and thus violate the boundary conditions. 10 Mathematical_physics-14-Eigenvalue problems.nb This fact can be used to numerically find eigenvalues and eigenfunctions by the shooting method. One can start at some xleft left from the well setting any boundary conditions, say y@xleftD = 1 and y'@xleftD = 0 and solve the DE for any E to the right until some xright. If yAxrightE is exponentially small, E is one of the energy eigenvalues, and if yAxrightE is exponentially large, E is not an eigenvalue. Parameters = 8m →1, — →1, ω0 →1<; U@x_D = m ω0 2 x2 2 ê. Parameters ; En@n_D = — ω0 Hn + 1 ê 2L ê. Parameters xRight = 10; xLeft = −xRight; IConds = 8ψ@xLeftD 1, ψ'@xLeftD 0<; Sol@EE_D := NDSolveB JoinB: —2 2 m ψ''@xD + HEE −U@xDL ψ@xD 0 ê. Parameters >, ICondsF, ψ, 8x, xLeft, xRight<F; ψx@EE_ ?NumericQ, x_D := ψ@xD ê. Sol@EED@@1DD H∗Without ?NumericQ FindMinimum below crashes ∗L 1 2 + n Plotting shows that for our parameter choice the energy eigenvalues indeed are 0.5, 1.5, 2.5, etc., that is, En = n + 1ê2. Plot@Log@Abs@ψx@EE, xRightDDD, 8EE, 0, 4<D 1 2 3 4 80 85 90 95 100 Now one can plot eigenfunctions for the found eigenvalues Mathematical_physics-14-Eigenvalue problems.nb 11 xMax = 5; xMin = −xMax; Plot@Evaluate@ψx@0.5, xDD, 8x, xMin, xMax<, PlotRange →AllD Plot@Evaluate@ψx@1.5, xDD, 8x, xMin, xMax<, PlotRange →AllD Plot@Evaluate@ψx@2.5, xDD, 8x, xMin, xMax<, PlotRange →AllD -4 -2 2 4 5.0µ1020 1.0µ1021 1.5µ1021 2.0µ1021 2.5µ1021 -4 -2 2 4 -1.5µ1020 -1.0µ1020 -5.0µ1019 5.0µ1019 1.0µ1020 1.5µ1020 -4 -2 2 4 -1.0µ1019 -5.0µ1018 5.0µ1018 1.0µ1019 1.5µ1019 These are essentially the same eigenfunctions as the analytical result considered above, only they are not normalized. One could find the exact eigenvalues by minimizing Log[Abs[yx[EE, xRight]]] starting from different values of EE. FindMinimum@Log@Abs@ψx@EE, xRightDDD, 8EE, 0<D FindMinimum::sdprec: Line search unable to find a sufficient decrease in the function value with MachinePrecision digit precision. à 876.1993, 8EE →0.5<< 12 Mathematical_physics-14-Eigenvalue problems.nb Finding exact eigenvalues by numerical minimization is important because this shooting method can be applied to potentials that do not allow for an analytical solution. For instance, if one replaces x2 ﬂx4 in U@xD, one obtains, for the same parame-ters, Plot@Log@Abs@ψx@EE, xRightDDD, 8EE, 0, 5<D 1 2 3 4 5 660 665 These nonequidistant energy eigenvalues are not described by any analytical formula, so that finding them numerically is important. Note that power potentials U@xD ∂J x x0 N n approach the potential box -x0 x x0 in the limit n Ø ¶, where the analytical results have been found above. ü Tunneling of a quantum particle Let us now consider the potential U@xD = -ax2 + bx4 with a, b > 0 that has two symmetric minima at x = ≤x0 = ≤ aêH2 bL . It is convenient to represent this potential in the form (10) U@xD = m ω0 2 8 Ix2 −x0 2M 2 x0 2 . Its second derivative at the minima is given by U@x_D = m ω0 2 8 Ix2 −x02M 2 x02 ; ∂x,x U@xD ê. x →x0 m ω0 2 that is the same as for a harmonic oscillator with frequency w0. A classical particle of mass m would perform harmonic oscillations with this frequency near the minima of this potential. Now the solution Mathematical_physics-14-Eigenvalue problems.nb 13 Parameters = 8m →1, — →1, ω0 →1, x0 →3.5<; U@x_D = m ω0 2 8 Ix2 −x0 2M 2 x0 2 ê. Parameters ; xRight = 7; xLeft = −xRight; IConds = 8ψ@xLeftD 1, ψ'@xLeftD 0<; Sol@EE_D := NDSolveB JoinB: —2 2 m ψ''@xD + HEE −U@xDL ψ@xD 0 ê. Parameters >, ICondsF, ψ, 8x, xLeft, xRight<F; ψx@EE_ ?NumericQ, x_D := ψ@xD ê. Sol@EED@@1DD; H∗Without ?NumericQ FindMinimum below crashes ∗L xMax = 6; xMin = −xMax; EMax = 2; Plot@U@xD, 8x, xMin, xMax<D; Plot@Log@Abs@ψx@EE, xRightDDD, 8EE, 0, EMax<, PlotRange →AllD 0.5 1.0 1.5 2.0 10 15 20 25 reveals that low-lying energy eigenvalues form doublets. To see the lowest doublet better one can zoom in: Plot@Log@Abs@ψx@EE, xRightDDD, 8EE, 0.475, 0.479<, PlotRange →AllD 0.476 0.477 0.478 0.479 2 4 6 8 10 One can find both eigenvalues 14 Mathematical_physics-14-Eigenvalue problems.nb E0 = EE ê. FindMinimum@Log@Abs@ψx@EE, xRightDDD, 8EE, 0<D@@2DD E1 = EE ê. FindMinimum@Log@Abs@ψx@EE, xRightDDD, 8EE, 0.5<D@@2DD FindMinimum::sdprec: Line search unable to find a sufficient decrease in the function value with MachinePrecision digit precision. à 0.476188 FindMinimum::sdprec: Line search unable to find a sufficient decrease in the function value with MachinePrecision digit precision. à 0.478131 And their spitting ∆= E1 −E0 0.00194363 To normalize the eigenfunctions, one at first has to calculate the norms of both eigenfunctions N0 = NB‡ xLeft xRight ψx@E0, xD2 xF; N1 = NB‡ xLeft xRight ψx@E1, xD2 xF; ψx0@x_D = 1 N0 ψx@E0, xD; ψx1@x_D = 1 N1 ψx@E1, xD; Below the two normalized eigenfunctions are plotted together with the scaled potential energy and a energy levels. xMax = 7; xMin = −xMax; Show@ Plot@Evaluate@8ψx0@xD, ψx1@xD<D, 8x, xMin, xMax<, PlotRange →All, PlotStyle →8Blue, Red<D, Plot@0.1 U@xD, 8x, −xMax, xMax<, PlotStyle →8Black, Thick<, PlotRange →80, 1<D, H∗Potential energy ∗L Plot@0.1 E0, 8x, −xMax, xMax<, PlotStyle →8Black, Dashed<D, Plot@0.1 E1, 8x, −xMax, xMax<, PlotStyle →8Black, Dashed<D D -6 -4 -2 2 4 6 -0.5 0.5 1.0 Mathematical_physics-14-Eigenvalue problems.nb 15 One can see that the eigenfunctions of the ground-state doublet are even and odd. According to a quantum-mechanical theorem, the eigenfunction of the ground state y0@xD does not change its sign on x. They are localized in the wells with little probalility to be under the barrier, x~0. These two eigenfunctions can be interpreted as even and odd linear combinations of the states in the left and right wells that are hybridized via tunneling, an essentially quantum phenomenon. With increasing the width or/and the height of the barrier, the separation between energy eigenvalues in tunneling doublets (tunneling splitting) decreases very fast, and it becomes difficult to calculate the splitting numerically. Fortunately, in this case tunnel-ing splittting can be calculated analytically with the help of the quasiclassical approximation. ü Matrix quantum mechanics Eigenfunctions cn@rD of a Hamiltonian satisfying boundary conditions form a complete basis that can be used to expand the solution of Schrödinger equation. The Hamiltonian to which the eigenfunctions belong can be the Hamiltonian of the problem or any other Hamiltonian. Let us call it H 0. The eigenvalue expansion has the form (11) Ψ@r, tD = ‚ n=0 ∞ an@tD χn@rD, similarly to the pressure expansion for the standing waves in a pipe, Eq. (2). This expansion transforms the time-dependent Schrödinger equation to a system of ODEs for the coefficients an@tD. Substituting Eq. (11) into the Schrödinger equation, one obtains (12) ‚ n=0 ∞ — a n@tD χn@rD ‚ n=0 ∞ an@tD H ˆ χn@rD, Multiplicating this equation χm ∗@rDand integrating it over the space taking into account orthonormality of the eigenfunctions, one obtains (13) — a m@tD ‚ n=0 ∞ Hmn an@tD, Hmn ≡‡ χm ∗@rD H ˆ χn@rD V. This is a system of a (generally) infinite number of linear homogeneous ODEs that are coupled to each other via the matrix elements Hmn. If H = H 0, one has Hmn = em dmn, where em are eigenvalues corresponding to the eigenfunctions cm. In this case equations decouple, ÂÑ a ° m @tD ã em am@tD, and the solution is am@tD =am@0D ExpB- Âem Ñ tF. However, this is a trivial case, and usually the Hamiltonian of the system H is more complicated than H 0, so that the matrix Hmn is non-diagonal. Eq. (13) can be written in the matrix form (14) — A @tD H.A@tD , where A@tD is the vector composed of am@tD. If matrix H is time-independent, its solution is the matrix exponential A@tD = ExpB− H — tF.A@0D describing time evolution of the initial state A@0D. This formula can be used to obtain the solution with the help of Mathemat-ica. It should be noted that Heisenberg proposed his matrix quantum mechanics independently of Schrödinger equation. Stationary states are described by the time dependence A@tD = ExpB− E — tF C. Substituting this into Eq. (14), one arrives at the matrix eigenvalue problem (15) EC H.C 16 Mathematical_physics-14-Eigenvalue problems.nb or, explicitly, (16) Ecm ‚ n=0 ∞ Hmn cn for the energy eigenvalues E that is equivalent to Eq. (1). If matrix H is infinite, there is an infinite number of eigenvalues. Practically, the functional basis used to build matrices is cut, so that only N lowest basis eigenfunctions of H 0 are used and the matrices are of rank N + 1. Some quantum-mechanical problems such that spin problems are described by finite matrices from the very beginning. An advantage of the matrix formalism is that it allows to obtain all eigenvalues and eigenfunctions as a result of a single computation. ü Particle in a double - well potential Let us apply the matrix formalism to the problem of the quartic double-well potential, Eq. (10), and choose the basic formed by the eigenstates of the Hamiltonian H 0 of the harmonic oscillator, Eq. (8). The Hamiltonian of the problem can be written as H ˆ = H ˆ 0 + V ˆ, V ˆ = m ω0 2 2 Ix2 −x0 2M 2 4 x0 2 −x2 . Such a separation of the Hamiltonian into two parts is typical, and in the case V is small, there are perturbative analytical methods based on expansion in powers of V . However, in our case V is not small and the problem will be solved numeri-cally. With the help of H 0 mn = em dmn Eq. (16) becomes (17) HE −εmL cm ‚ n=0 ∞ Vmn cn, where em = Ñw0Hm + 1ê2L are the energy eigenvalues of the harmonic oscillator and Vmn ≡‡ −∞ ∞ χm@xD V ˆ χn@xD x, where cn@xD are given by Eq. (9). There is an analytical expression for these matrix elements for general m and n, however, Mathematica 7 cannot find it. Still, Mathematica can analytically calculate Vmn for any particular values of m and n that we need to build the Hamiltonian matrix. We will be labeling the eigenvalues of the problem with the index n, so that En are energy eigenvalues, cnn are the basis expansion coefficients, and ψν@xD ≡‚ n=0 ∞ cνn χn@xD are eigenfunctions of the problem. Below is Mathematica code that solves the matrix eigenvalue problem for a quartic potential using the harmonic oscillator basis. Mathematical_physics-14-Eigenvalue problems.nb 17 H∗Definitions ∗L NBasis = 20; H∗Number of the lowest oscillator eigenfunctions in the basis ∗L M = 1; H∗Mass of the particle ∗L — = 1; H∗Planck's constant ∗L ω0 = 1; H∗Frequency of oscillation near the minima of U@xD ∗L x0 = 3.5; H∗Positions of the minima of U@xD ∗L U@x_D = M ω0 2 8 Ix2 −x0 2M 2 x0 2 ; H∗Potential energy of the particle ∗L U0@x_D = M ω0 2 x2 2 ; H∗Potential energy of the harmonic oscillator ∗L εn@n_D = — ω0 Hn + 1 ê 2L; H∗Energy levels of the harmonic oscillator ∗L χnx@n_, x_D = 1 2n n! M ω0 π — 1ê4 ExpB− M ω0 x2 2 — F HermiteHBn, M ω0 — xF; H∗Eigenfunctions of the harmonic oscilator ∗L V@xD = U@xD −U0@xD; H∗Perturbation Hamiltonian ∗L Vmn@m_, n_D := IfBAbs@m −nD 0 »» Abs@m −nD 2 »» Abs@m −nD 4, ‡ −∞ ∞ χnx@m, xD V@xD χnx@n, xD x, 0F; H∗Most of matrix elements of V@xD are zero, so do not calculate them ∗L Hmn@m_, n_D := εn@nD KroneckerDelta@m, nD + Vmn@m, nD; H∗Elements of Hamiltonian matrix ∗L H∗Creating Hamiltonian matrix and solving matrix eigenvalue problem ∗L Timing@ HMatrix = Table@Hmn@m, nD, 8m, 0, NBasis<, 8n, 0, NBasis<D; D TableForm@Take@HMatrix, 9, 9DD H∗Show part of the Hamiltonian matrix ∗L Timing@ ES = Eigensystem@HMatrixD; D ES = Transpose@Sort@Transpose@ESDDD; H∗Eigensystem sorted in accending order ∗L EVal = First@ESD H∗Eigenvalues ∗L EVecT = Last@ESD; H∗This is a matrix of eigenvectors lying horizontally ∗L EVec = Transpose@EVecTD; H∗This is a matrix of eigenvectors standing vertically ∗L TableForm@Chop@Take@EVecT.EVec, 9, 9DDD H∗Check that eigenvectors are orthonormal ∗L H∗Finalizing the solution ∗L Eν@ν_D := EVal@@ν + 1DD; H∗Energy eigenvalues of the problem ∗L ∆= Eν@1D −Eν@0D; H∗Ground−state splitting ∗L c@ν_, n_D := EVecT@@ν + 1, n + 1DD H∗Expansion coefficients of the eigenfunctions of the problem over the harmonic oscillator basis ∗L ψνx@ν_, x_D := ‚ n=0 NBasis c@ν, nD χnx@n, xD H∗Eigenfunctions of the problem ∗L 823.922, Null< 1.6639 0 −0.508684 0 0.0124974 0 0 0 0 0 1.94452 0 −0.856072 0 0.027945 0 0 0 −0.508684 0 2.25574 0 −1.17532 0 0.0484022 0 0 0 −0.856072 0 2.59758 0 −1.4717 0 0.0739356 0 0.0124974 0 −1.17532 0 2.97003 0 −1.74656 0 0.104561 0 0.027945 0 −1.4717 0 3.37309 0 −2.00043 0 0 0 0.0484022 0 −1.74656 0 3.80676 0 −2.23354 0 0 0 0.0739356 0 −2.00043 0 4.27105 0 0 0 0 0 0.104561 0 −2.23354 0 4.76594 80., Null< 18 Mathematical_physics-14-Eigenvalue problems.nb 80.476188, 0.478132, 1.2695, 1.35141, 1.8394, 2.21869, 2.71446, 3.255, 3.84311, 4.47189, 5.13829, 5.84678, 6.57393, 7.45094, 8.21213, 9.51985, 10.3086, 12.3669, 13.2037, 16.5672, 17.4787< 1. 0 0 0 0 0 0 0 0 0 1. 0 0 0 0 0 0 0 0 0 1. 0 0 0 0 0 0 0 0 0 1. 0 0 0 0 0 0 0 0 0 1. 0 0 0 0 0 0 0 0 0 1. 0 0 0 0 0 0 0 0 0 1. 0 0 0 0 0 0 0 0 0 1. 0 0 0 0 0 0 0 0 0 1. The splitting of the ground-state doublet ∆= Eν@1D −Eν@0D 0.00194448 is in an excellent accordance with the value obtained above by the shooting method. The eigenfunctions of the ground-state doublet, plotted together with the scaled potential energy and a group of low-lying energy levels xMax = 7; xMin = −xMax; Show@ Plot@Evaluate@8ψνx@0, xD, ψνx@1, xD<D, 8x, xMin, xMax<, PlotRange →All, PlotStyle →8Blue, Red<D, Plot@0.1 U@xD, 8x, −xMax, xMax<, PlotStyle →8Black, Thick<, PlotRange →80, 1<D, H∗Potential energy ∗L Plot@Table@0.1 Eν@νD, 8ν, 0, 12<D, 8x, −xMax, xMax<, PlotStyle →8Black, Dashed<D D -6 -4 -2 2 4 6 -0.5 0.5 1.0 are the same as above, except for the irrelevant signs. The splitting of the ground-state doublet is not seen in this plot, whereas the spliting of the second doublet is seen. The states above the barrier are not grouped into doublets, and their separation increases with energy. Mathematical_physics-14-Eigenvalue problems.nb 19 ü Sparce matrices We have seen above that the main time of the calculation goes to fill the Hamiltonian matrix with elements, whereas finding its eigensystem is very fast. Filling a matrix of rank N requires N2 operations, even if many its elements are zero. In our Hamiltonian matrix, only the elements with m - n = 0,2,4 are nonzero. Such matrices in which most of the elements are zero are called sparce matrices. In Mathematica a special fast method of filling sparce matrices is employed. First, a matrix with all elements equal to zero is created by one step. Then nonzero elements are inserted at their places. Actually, it is only nonzero elements and the matrix dimensions that are stored in the memory. Thus, both memory usage and speed are dramati-cally improved. In the case of nearly-diagonal matrices, there are ~N nonzero elements, so that the time to fill such a sparce matrix if proportional to N rather than to N2. Mathematica command to create a sparce matrix object is SparceArray. Let us create a 5×5 matrix with diagonal elements equal to 1 and subdiagonal elements equal to 2 . MySparceMatrix = SparseArrayB:8i_, j_< ê; Abs@i −jD 0 →1, 8i_, j_< ê; Abs@i −jD 1 → 2 >, 85, 5<F SparseArray@<13>, 85, 5<D Here /; means "under the condition". The output object is named but its content is hidden because it is not a standard list. It can be visualized with the command MatrixForm: MatrixForm@MySparceMatrixD 1 2 0 0 0 2 1 2 0 0 0 2 1 2 0 0 0 2 1 2 0 0 0 2 1 One can apply all standard commands such as Eigensystem to spare matrix objects. Let us now compare the time needed to create the Hamiltonian matrix above as a standard (dense) matrix and sparse matrix. (The preceding code must be run before so that Mathematica knows definitions.) For NBasis = 100 one obtains NBasis = 40; Timing@ HMatrix = Table@Hmn@m, nD, 8m, 0, NBasis<, 8n, 0, NBasis<D; D MatrixForm@Take@HMatrix, 5, 5DD Timing@ HMatrix = SparseArray@ 88m_, n_< ê; Abs@m −nD 0 »» Abs@m −nD 2 »» Abs@m −nD 4 →Hmn@m −1, n −1D<, 8NBasis + 1, NBasis + 1<D; D MatrixForm@Take@HMatrix, 5, 5DD 897.156, Null< 1.6639 0 −0.508684 0 0.0124974 0 1.94452 0 −0.856072 0 −0.508684 0 2.25574 0 −1.17532 0 −0.856072 0 2.59758 0 0.0124974 0 −1.17532 0 2.97003 8118.86, Null< 1.6639 0 −0.508684 0 0.0124974 0 1.94452 0 −0.856072 0 −0.508684 0 2.25574 0 −1.17532 0 −0.856072 0 2.59758 0 0.0124974 0 −1.17532 0 2.97003 20 Mathematical_physics-14-Eigenvalue problems.nb For higher NBasis the gain in using sparce matrices will theoretically increase. This example is not the best, however, because the difference in speed between the two methods is marginal. The reason is that the standard method also uses the optimized definition of the matrix elements, in which their lengthy calculation via the integral is done only in the case they are nonzero, otherwise zeros are being put into the Hamiltonian matrix. A real gain in using sparce matrix objects can be achieved only for really big matrices in which calculation of nonzero elements is fast. ü Quantum rotator and pendulum Let us apply matrix quantum mechanics to the problem of a quantum pendulum that is described by the Hamiltonian H ˆ = —2 l ˆ2 2 I + U@φD, l ˆ ≡− ∂ ∂φ , U@φD = Mga H1 −Cos@φDL, where Ñ l is the quantum-mechanical angular momentum operator for a rotation around a fixed axis by the angle f, I is the moment of inertia, M is the mass, and a is the distance between the center of mass and the pivot point. Classical pendulum performs oscillations for E 2 Mga and rotations in positive or negative direction for E > 2 Mga. The frequency of small oscillations near the minimum of the potential energy is w0 = Mga I . In the case U@fD = 0 pendulum simplifies to rotator. Since adding 2 pn to the angle does not change the physical position of the pendulum, the wave function satisfies the periodic-ity condition Y@f, tD = Y@f + 2 pn, tD, n = ≤1, ≤2, — Thus we consider the wave functions in the interval -p § f § f and require Y@-p, tD = Y@p, tD. A suitable complete set of basis functions satisfying this boundary conditions consists of the eigenfunctions of the angular momentum l cm = mcm that are given by cm = 1 2 p ‰Âmf, m = 0, ≤1, ≤2, — One can check that these functions are orthonormal. They are eigenfunctions of the stationary Schrödinger equation Eψ H ˆ ψ for the rotator with the rotational energies Em = Ñ2 m2 2 I . For the pendulum, these functions are not eigenfunctions but can be used as a basis to formulate the matrix Schrödinger equation (16). The matrix elements of the Hamiltonian are given by Hmn = ‡ -p p cm @fD H ` cn@fD „f = ‡ -p p ‰-imf -Ñ2 2 I ∑2 ∑f2 +MgaH1 - Cos@fDL ‰inf „f = Ñ2 m2 2 I + Mga dmn -Mga 2 p ‡ -p p ‰-imf Cos@fD ‰inf „f. Using Cos@fD = I‰Âf + ‰-ÂfMë2, for the last term one obtains MgaIdm,n+1 + dm,n-1Më2, so that the matrix element becomes Mathematical_physics-14-Eigenvalue problems.nb 21 Hmn = Ñ2 m2 2 I + Mga dmn -Mga 2 Idm,n+1 + dm,n-1M. Below is a Mathematica code for solving the eigenvalue problem for the quantum pendulum. Among the parameters, at least one number has to be real (such as M=1.), so that Mathematica does not try to find eigenvalues analytically. H∗Definitions ∗L NBasis = 300; H∗Number of the lowest rotator eigenfunctions in the basis ∗L M = 1.; H∗Mass of the pendulum ∗L II = 300; H∗Moment of inertia of the pendulum ∗L — = 1; H∗Planck's constant ∗L g = 1; H∗Gravity acceleration ∗L a = 1; H∗Lever arm of the gravity force ∗L ω0 = M g a II ; H∗Frequency of small oscillations ∗L χmφ@m_, φ_D = 1 2 π m φ; H∗Eigenfunctions of the rotator ∗L εn@m_D = —2 m2 2 II + M g a ; Vmn@m_, n_D = − M g a 2 HKroneckerDelta@m, n + 1D + KroneckerDelta@m, n −1DL; Hmn@m_, n_D := εn@mD KroneckerDelta@m, nD + Vmn@m, nD; H∗Elements of Hamiltonian matrix ∗L H∗Creating Hamiltonian matrix and solving matrix eigenvalue problem ∗L Timing@ HMatrix = Table@Hmn@m, nD, 8m, −NBasis ê 2, NBasis ê 2<, 8n, −NBasis ê 2, NBasis ê 2<D; D Print@"Part of the Hamiltonian matrix:"D TableForm@Take@HMatrix, 9, 9DD H∗Show part of the Hamiltonian matrix ∗L Timing@ ES = Eigensystem@HMatrixD; D ES = Transpose@Sort@Transpose@ESDDD; H∗Eigensystem sorted in accending order ∗L EVal = First@ESD ; H∗Eigenvalues ∗L EVecT = Last@ESD; H∗This is a matrix of eigenvectors lying horizontally ∗L EVec = Transpose@EVecTD; H∗This is a matrix of eigenvectors standing vertically ∗L TableForm@Chop@Take@EVecT.EVec, 9, 9DDD ; H∗Check that eigenvectors are orthonormal ∗L H∗Finalizing the solution ∗L Eµ@µ_D := EVal@@µ + 1DD; H∗Energy eigenvalues of the problem ∗L c@µ_, m_D := EVecT@@µ + 1, m + 1 + NBasis ê 2DD H∗Expansion coefficients of the eigenfunctions of the problem over the rotator basis ∗L ψµφ@µ_, φ_D := ‚ m=−NBasisê2 NBasisê2 c@µ, mD χmφ@m, φD H∗Eigenfunctions of the problem ∗L Print@"Energy levels:"D LevelsToShow = 70; LevelsToShow = Min@ LevelsToShow, NBasisD; Take@EVal, LevelsToShowD Show@ ListPlot@Take@EVal, LevelsToShowD, AxesLabel →8"µ", "Eµ"<D, Plot@2 M g a, 8x, 0, LevelsToShow<D D 80.765, Null< 22 Mathematical_physics-14-Eigenvalue problems.nb Part of the Hamiltonian matrix: 38.5 −0.5 0 0 0 0 0 0 0 −0.5 38.0017 −0.5 0 0 0 0 0 0 0 −0.5 37.5067 −0.5 0 0 0 0 0 0 0 −0.5 37.015 −0.5 0 0 0 0 0 0 0 −0.5 36.5267 −0.5 0 0 0 0 0 0 0 −0.5 36.0417 −0.5 0 0 0 0 0 0 0 −0.5 35.56 −0.5 0 0 0 0 0 0 0 −0.5 35.0817 −0.5 0 0 0 0 0 0 0 −0.5 34.6067 80.063, Null< Energy levels: 80.028763, 0.0860783, 0.14297, 0.199433, 0.255463, 0.311053, 0.3662, 0.420896, 0.475137, 0.528915, 0.582225, 0.635059, 0.68741, 0.739272, 0.790635, 0.841492, 0.891834, 0.941652, 0.990935, 1.03967, 1.08786, 1.13547, 1.1825, 1.22894, 1.27477, 1.31998, 1.36454, 1.40844, 1.45166, 1.49417, 1.53595, 1.57696, 1.61719, 1.65658, 1.6951, 1.7327, 1.76931, 1.80486, 1.83926, 1.8724, 1.90401, 1.93429, 1.96081, 1.99032, 2.0026, 2.04529, 2.04672, 2.10579, 2.10586, 2.17306, 2.17306, 2.24618, 2.24618, 2.32446, 2.32446, 2.40747, 2.40747, 2.49492, 2.49492, 2.5866, 2.5866, 2.68235, 2.68235, 2.78207, 2.78207, 2.88565, 2.88565, 2.99302, 2.99302, 3.10412< 10 20 30 40 50 60 70 m 0.5 1.0 1.5 2.0 2.5 3.0 Em One can see that the low-lying energy levels are equidistant, as the levels of the harmonic oscillator. Levels with Em 2 Mga are oscillatory. Levels above this energy correspond to rotations over the barrier. These levels form doublets because of the two posisble directions of the rotation. However, these doublets are split because of the essentially quantum-mechanical effect, the overbarrier reflection. For instance, Eµ@50D −Eµ@49D Eµ@52D −Eµ@51D Eµ@54D −Eµ@53D 2.0047 × 10−6 4.45956 × 10−8 7.72255 × 10−10 Splitting of level doublets, as the overbarrier reflection, decreases very fast with the energy. At high energies one approxi-mately has Em = Ñ2Hmê2L2 2 I , Mathematical_physics-14-Eigenvalue problems.nb 23 the energy levels of the rotator. The eigenfunctions of the problem are similar to those of the harmonic oscillator at low energies. However, at high energies the probability density °ym• 2 is oscillating because eigenfunctions describe superposi-tions of rotations in different directions. PlotAEvaluateA9Abs@ψµφ@0, π xDD2, Abs@ψµφ@20, π xDD2, Abs@ψµφ@60, π xDD2=E, 8x, −1, 1<, PlotRange →All, PlotPoints →30, PlotStyle →88Black, Thick<, 8Black, Thin<, 8Red, Thin<<, AxesLabel →9"φêπ", "†ψµ@φD§2"=E -1.0 -0.5 0.5 1.0 fêp 0.5 1.0 1.5 2.0 °ym@fD•2 To better see mixing of rotations in different directions, one can plot the coefficients cmm for one the doublets 24 Mathematical_physics-14-Eigenvalue problems.nb Eµ@54D Eµ@53D Eµ@54D −Eµ@53D Plot@8c@53, Round@mDD, c@54, Round@mDD<, 8m, −LevelsToShow, LevelsToShow<, PlotRange →All, PlotStyle →88Blue, Dashed<, 8Black<<, AxesLabel →8"m", "cµm"<D 2.32446 2.32446 7.72255 × 10−10 -60 -40 -20 20 40 60 m -0.2 -0.1 0.1 0.2 cmm One can see that for one of the eigenfunctions the coefficients cmm are even in m (solid line) while for the other they are odd in m (dashed line), that is, both of them are mixtures of rotations in different directions. Mathematical_physics-14-Eigenvalue problems.nb 25
15721	https://math.stackexchange.com/questions/372454/using-the-constant-function-theorem-to-prove-the-increasing-function-theorem	calculus - Using the Constant Function Theorem to prove the Increasing Function Theorem - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Using the Constant Function Theorem to prove the Increasing Function Theorem Ask Question Asked 12 years, 5 months ago Modified9 years, 8 months ago Viewed 2k times This question shows research effort; it is useful and clear 5 Save this question. Show activity on this post. I quote Thomas W.Tucker: ... By the way, I view the Constant Function Theorem as even more basic than the IFT. It would be nice to use it as our theoretical cornerstone, but I know of no way to use it to get the IFT. ... from Rethinking Rigor in Calculus: The Role of the Mean Value Theorem - The American Mathematical Monthly, Vol. 104, No. 3 (Mar., 1997), page 233 where IFT means Increasing Function Theorem. For convenience: Increasing function f f means that if c<d c<d, then f(c)≤f(d)f(c)≤f(d); IFT: if f′(x)≥0 f′(x)≥0 on [a,b][a,b], then f f is increasing on [a,b][a,b]; CFT: if f′(x)=0 f′(x)=0 on [a,b][a,b], then f f is constant on [a,b][a,b]. Is it impossible to get the IFT from the CFT ? calculus real-analysis derivatives education Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Jan 24, 2016 at 20:59 user147263 asked Apr 25, 2013 at 13:22 Tony PiccoloTony Piccolo 4,496 2 2 gold badges 15 15 silver badges 26 26 bronze badges 3 3 In the paper referred to, the author in the discussion refers to what he calls Theorem 1c as SIFT, strictly increasing function theorem. Then later he (I think mistakenly) refers again to Theorem 1c as the constant function theorem (CFT). However I think it more likely that CFT refers to his Theorem 1b, which says if f′f′ is 0 on [a,b][a,b] then f f is constant on [a,b][a,b]. It would indeed be tricky to derive IFT from this.coffeemath –coffeemath 2013-04-25 14:15:54 +00:00 Commented Apr 25, 2013 at 14:15 1 @Tony Piccolo : I'd think it good if you were to include exact quotes of IFT and CFT in your posted question. I had to look them up in Tucker's paper, and even there had a doubt (see my previous comment). In my opinion the posed question would be better if it were self-contained.coffeemath –coffeemath 2013-04-25 14:38:34 +00:00 Commented Apr 25, 2013 at 14:38 @coffeemath: you are right, there is a misprint; CFT refers to Theorem 1b.Tony Piccolo –Tony Piccolo 2013-04-25 15:06:09 +00:00 Commented Apr 25, 2013 at 15:06 Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 3 Save this answer. Show activity on this post. It would be nice to have a clear statement of what it means to "get from CFT to IFT". In the most standard sense of logical implication, we have two true statements A A and B B, so the statement "A⟹B A⟹B" is certainly true. But surely something other than this trivial observation is meant. One nice way of construing statements like this is developed in Jim Propp's article Real Analysis in Reverse. Namely, both assertions are meaningful when applied to an arbitrary ordered field, so one can explore the class of ordered fields in which they hold. It turns out that many, but not all, of the interesting theorems in calculus imply the Dedekind completeness of the ordered field, so only hold in the real numbers. That is the case here, as is treated in Propp's article: a gap in an ordered field leads to a locally constant function which is not constant, so CFT implies Dedekind completeness (hence so does IFT). Does the above argument -- i.e., showing that CFT implies Dedekind completeness and then giving, say, the usual proof of the Mean Value Theorem and then deducing IFT -- count as "getting from CFT to IFT"? I would guess not. But then, as I pointed out above, it's less than clear what this "getting from" business really means. Note that exactly this issue comes up in §5.1§5.1 of this note of mine on real induction, in which I suggest a rather inadequate meaning for equivalence of theorems like this by saying that each one "immediately implies" the other...whatever that means! Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited May 4, 2013 at 16:38 answered May 4, 2013 at 6:54 Pete L. ClarkPete L. Clark 101k 11 11 gold badges 238 238 silver badges 383 383 bronze badges 4 I actually find the acceptance of this answer somewhat disappointing. I suspect there's more going on in the question than my answer addresses.Pete L. Clark –Pete L. Clark 2013-05-04 17:03:14 +00:00 Commented May 4, 2013 at 17:03 you say that the question is not well posed ... and I agree.Tony Piccolo –Tony Piccolo 2013-05-04 20:59:25 +00:00 Commented May 4, 2013 at 20:59 @Tony: okay, but...just because something is not well-posed doesn't mean that it couldn't be made so. I guess I feel like you may be giving up too easily.Pete L. Clark –Pete L. Clark 2013-05-05 02:14:18 +00:00 Commented May 5, 2013 at 2:14 I promise I'll think how to modify the question.Tony Piccolo –Tony Piccolo 2013-05-05 05:31:43 +00:00 Commented May 5, 2013 at 5:31 Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. IFT to CFT: consider f(x)f(x) and g(x):=−f(x)g(x):=−f(x) s.t. f(x)f(x) is differentiable... (the standard set-up). Now, f′(x)≥0 f′(x)≥0, g′(x)≤0 g′(x)≤0, so f f is increasing, and g g is decreasing. (All that jazz^ to show that we get decreasing functions if g′(x)≥0 g′(x)≥0) Now, if h(x)h(x) also differentiable and so on, h′(x)=0 h′(x)=0, we have h′(x)=0⟹h′(x)≥0⟹h is increasing h′(x)=0⟹h′(x)≥0⟹h is increasing, i.e, h(x)≥h(y)h(x)≥h(y) if x>y x>y h′(x)=0⟹h′(x)≤0⟹h is decreasing h′(x)=0⟹h′(x)≤0⟹h is decreasing, i.e., h(x)≤h(y)h(x)≤h(y) if x>y x>y so if x>y x>y, putting both inequalities together: h(x)=h(y)h(x)=h(y) (note this is for all x x and y y), so h h must be constant as it equals a constant value h(x)h(x) for all values of x x. I know this isn't the most rigorous answer, but you can fill in the details. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Oct 17, 2014 at 18:07 Cristhian Gz 2,597 1 1 gold badge 18 18 silver badges 24 24 bronze badges answered Oct 17, 2014 at 17:46 CharcCharc 1 Add a comment\| You must log in to answer this question. 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15722	https://artofproblemsolving.com/wiki/index.php/Trigonometric_identities?srsltid=AfmBOoqc9kSrNquu-qOPuIE34amp78YrSF9oRu7FYy667rnALyEDzWT6	Art of Problem Solving Trigonometric identities - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Trigonometric identities Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Trigonometric identities In trigonometry, trigonometric identities are equations involving trigonometric functions that are true for all input values. Trigonometric functions have an abundance of identities, of which only the most widely used are included in this article. Contents [hide] 1 Pythagorean identities 2 Angle addition identities 3 Double-angle identities 3.1 Cosine double-angle identity 4 Half-angle identities 5 Product-to-sum identities 6 Sum-to-product identities 7 Other identities 7.1 Triple-angle identities 7.2 Even-odd identities 7.3 Conversion identities 7.4 Euler's identity 7.5 Miscellaneous 8 Resources 9 See also Pythagorean identities The Pythagorean identities state that Using the unit circle definition of trigonometry, because the point is defined to be on the unit circle, it is a distance one away from the origin. Then by the distance formula, . To derive the other two Pythagorean identities, divide by either or and substitute the respective trigonometry in place of the ratios to obtain the desired result. Angle addition identities The trigonometric angle addition identities state the following identities: There are many proofs of these identities. For the sake of brevity, we list only one here. Euler's identity states that . We have that By looking at the real and imaginary parts, we derive the sine and cosine angle addition formulas. To derive the tangent addition formula, we reduce the problem to use sine and cosine, divide both numerator and denominator by , and simplify. as desired. Double-angle identities The trigonometric double-angle identities are easily derived from the angle addition formulas by just letting . Doing so yields: Cosine double-angle identity Here are two equally useful forms of the cosine double-angle identity. Both are derived via the Pythagorean identity on the cosine double-angle identity given above. In addition, the following identities are useful in integration and in deriving the half-angle identities. They are a simple rearrangement of the two above. Half-angle identities The trigonometric half-angle identities state the following equalities: The plus or minus does not mean that there are two answers, but that the sign of the expression depends on the quadrant in which the angle resides. Consider the two expressions listed in the cosine double-angle section for and , and substitute instead of . Taking the square root then yields the desired half-angle identities for sine and cosine. As for the tangent identity, divide the sine and cosine half-angle identities. Product-to-sum identities The product-to-sum identities are as follows: They can be derived by expanding out and or and , then combining them to isolate each term. Sum-to-product identities Substituting and into the product-to-sum identities yields the sum-to-product identities. Other identities Here are some identities that are less significant than those above, but still useful. Triple-angle identities All of these expansions can be proved using trick and perform the angle addition identities. Same for and for . Even-odd identities The functions , , , and are odd, while and are even. In other words, the six trigonometric functions satisfy the following equalities: These are derived by the unit circle definitions of trigonometry. Making any angle negative is the same as reflecting it across the x-axis. This keeps its x-coordinate the same, but makes its y-coordinate negative. Thus, and . Conversion identities The following identities are useful when converting trigonometric functions. All of these can be proven via the angle addition identities. Euler's identity Euler's identity is a formula in complex analysis that connects complex exponentiation with trigonometry. It states that for any real number , where is Euler's constant and is the imaginary unit. Euler's identity is fundamental to the study of complex numbers and is widely considered among the most beautiful formulas in math. Similar to the derivation of the product-to-sum identities, we can isolate sine and cosine by comparing and , which yields the following identities: They can also be derived by computing and . These expressions are occasionally used to define the trigonometric functions. Miscellaneous These are the identities that are not substantial enough to warrant a section of their own. Resources Table of trigonometric identities List of Trigonometric Identities See also Trigonometry Trigonometric substitution Proofs of trig identities Retrieved from " Category: Trigonometry Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
15723	https://www.britannica.com/science/topology/Homeomorphism	Homeomorphism Basic concepts of general topology Our editors will review what you’ve submitted and determine whether to revise the article. An intrinsic definition of topological equivalence (independent of any larger ambient space) involves a special type of function known as a homeomorphism. A function h is a homeomorphism, and objects X and Y are said to be homeomorphic, if and only if the function satisfies the following conditions. The notion of two objects being homeomorphic provides the definition of intrinsic topological equivalence and is the generally accepted meaning of topological equivalence. Two objects that are isotopic in some ambient space must also be homeomorphic. Thus, extrinsic topological equivalence implies intrinsic topological equivalence. Topological structure In its most general setting, topology involves objects that are abstract sets of elements. To discuss properties such as continuity of functions between such abstract sets, some additional structure must be imposed on them. Topological space One of the most basic structural concepts in topology is to turn a set X into a topological space by specifying a collection of subsets T of X. Such a collection must satisfy three axioms: (1) the set X itself and the empty set are members of T, (2) the intersection of any finite number of sets in T is in T, and (3) the union of any collection of sets in T is in T. The sets in T are called open sets and T is called a topology on X. For example, the real number line becomes a topological space when its topology is specified as the collection of all possible unions of open intervals—such as (−5, 2), (1/2, π), (0, Square root of√2), …. (An analogous process produces a topology on a metric space.) Other examples of topologies on sets occur purely in terms of set theory. For example, the collection of all subsets of a set X is called the discrete topology on X, and the collection consisting only of the empty set and X itself forms the indiscrete, or trivial, topology on X. A given topological space gives rise to other related topological spaces. For example, a subset A of a topological space X inherits a topology, called the relative topology, from X when the open sets of A are taken to be the intersections of A with open sets of X. The tremendous variety of topological spaces provides a rich source of examples to motivate general theorems, as well as counterexamples to demonstrate false conjectures. Moreover, the generality of the axioms for a topological space permit mathematicians to view many sorts of mathematical structures, such as collections of functions in analysis, as topological spaces and thereby explain associated phenomena in new ways. A topological space may also be defined by an alternative set of axioms involving closed sets, which are complements of open sets. In early consideration of topological ideas, especially for objects in n-dimensional Euclidean space, closed sets had arisen naturally in the investigation of convergence of infinite sequences (see infinite series). It is often convenient or useful to assume extra axioms for a topology in order to establish results that hold for a significant class of topological spaces but not for all topological spaces. One such axiom requires that two distinct points should belong to disjoint open sets. A topological space satisfying this axiom has come to be called a Hausdorff space. Continuity An important attribute of general topological spaces is the ease of defining continuity of functions. A function f mapping a topological space X into a topological space Y is defined to be continuous if, for each open set V of Y, the subset of X consisting of all points p for which f(p) belongs to V is an open set of X. Another version of this definition is easier to visualize, as shown in the figure. A function f from a topological space X to a topological space Y is continuous at p ∊ X if, for any neighbourhood V of f(p), there exists a neighbourhood U of p such that f(U) ⊆ V. These definitions provide important generalizations of the usual notion of continuity studied in analysis and also allow for a straightforward generalization of the notion of homeomorphism to the case of general topological spaces. Thus, for general topological spaces, invariant properties are those preserved by homeomorphisms.
15724	https://math.stackexchange.com/questions/4754436/number-of-colourings-of-the-necklace	combinatorics - Number of colourings of the necklace. - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR == Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… 1. 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You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Number of colourings of the necklace. Ask Question Asked 2 years ago Modified2 years ago Viewed 351 times This question shows research effort; it is useful and clear 3 Save this question. Show activity on this post. I want to count the number of ways to color beads of a necklace green and red, such that two adjacent beads cannot both be red. The necklace cannot be turned or reflected, the beads are labelled. Initially, I tried for a fixed number of red beads. Let there be n n beads in total, r r of which are red, so n−r n−r are green. Every bead can have a red bead to its right, so the places for red beads are n−r n−r. Hence, the number of possibilities is (n−r r)(n−r r) Summing up for all the numbers of red beads, we get ∑r(n−r r)=F n+1∑r(n−r r)=F n+1 However, this is not correct, mainly because for n=5 n=5, and r=1 r=1, the result is 4 4 instead of 5 5. Writing (n−r+1 r)(n−r+1 r) doesn't fix the situation. How can these colorings be correctly counted? combinatorics discrete-mathematics combinations necklace-and-bracelets Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Aug 18, 2023 at 0:47 RobPratt 51.6k 4 4 gold badges 32 32 silver badges 69 69 bronze badges asked Aug 17, 2023 at 15:44 Aleksander WojszAleksander Wojsz 675 3 3 silver badges 13 13 bronze badges 2 If r=1 r=1 then there's only one possible necklace, right? –Karl Commented Aug 17, 2023 at 16:15 Necklaces cannot be turned over (if they can, then they are called _bracelets_) but they can be rotated –Henry Commented Aug 17, 2023 at 16:21 Add a comment\| 3 Answers 3 Sorted by: Reset to default This answer is useful 6 Save this answer. Show activity on this post. Your argument counts the solutions for beads in a straight line so we can solve this problem in 2 parts by first looking at beads in a straight line and then worrying about beads in a loop. It is easy to see (using your logic or recursion) that if there are n beads that are in a straight line and do not loop around at the end the number of combinations is a Fibonacci number (F n+1 F n+1). Your argument works fine but a simpler proof is what follows: For n=1 there are 2 ways, for n=2 there are 3 ways, and in general the first bead is either red followed by a green or green so we sum the previous 2 cases. For example, for n = 3 we can either start with green which gives us 3 situations for the remaining 2 or start with red which forces a green after, leaving us with 2 possibilites for the remaining bead. So the total is 5 possibilites or 3+2. This is the Fibonacci sequence of course, 2, 3, 5, 8, 13... If we use F 0=1 F 0=1 and F 1=1 F 1=1 then the n beads case has F n+1 F n+1 solutions. Now for a necklace, notice that, any individual bead can either be green which reduces to n-1 beads with no loop (F n F n solutions) or be red and a green on either side of it which reduces to n-3 with no loop (F n−2 F n−2 solutions). So in general, the number of solutions when n>3 n>3 is F n+F n−2 F n+F n−2. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Aug 17, 2023 at 16:42 answered Aug 17, 2023 at 16:05 TheJackTheJack 492 2 2 silver badges 13 13 bronze badges 10 Thanks, I understand your answer. Although your solution gives 7 7 for n=5 n=5. When one counts by hand all the cases: GGGGG, RGGGG and four similar ones, RGRGG, RGGRG, GRGRG, GRGGR, GGRGR, it comes out 11 11, which suggests F n+2−F n−2 F n+2−F n−2 –Aleksander Wojsz Commented Aug 17, 2023 at 16:42 1 @Michał I am using F 0=1,F 1=1 F 0=1,F 1=1 so it is 8+3=11. Sorry if I confused you (some people start from 1; there are several notations) –TheJack Commented Aug 17, 2023 at 16:45 The 7 you're getting is 5+2 which is when n=4 (GGGG RGGG GRGG GGRG GGGR RGRG GRGR RGGR) –TheJack Commented Aug 17, 2023 at 17:00 @TheJack I think it would be better to write the recurrence relation in the form of F?=F?+F?..F?=F?+F?.. instead of just showing F n+F n−2 F n+F n−2. I think the last part is not clear, is your answer F n=F n+F n−2 F n=F n+F n−2 for n>3 n>3 –Not a Salmon Fish Commented Aug 17, 2023 at 18:50 1 @Michał Oh, sorry, I thought you meant the looped case. Yes you can get it that way, this is equivalent to what I said in my answer except that the shift is 2 instead of 1 since you are using F 0=0 F 0=0 and F 1=1 F 1=1 instead of F 0=1 F 0=1 and F 1=1 F 1=1 –TheJack Commented Aug 18, 2023 at 18:02 \|Show 5 more comments This answer is useful 3 Save this answer. Show activity on this post. I want to answer this question using a special form of generating functions called Smirnov words. Look at Analytic combinatorics example I I I.24 I I I.24. Lets find the number of arrangementss in a line such that no two consecutive red beads allowed, but no restriction over green beads. In our case, the generating function for the number of Smirnov words over our alphabet (i.e red and green strings) is equal to (1−x 1+x−x 1−x 1+x 1−x)−1=1+2 x+3 x 2+5 x 3+8 x 4+13 x 5+21 x 6+34 x 7+..+144 x 10+..(1−x 1+x−x 1−x 1+x 1−x)−1=1+2 x+3 x 2+5 x 3+8 x 4+13 x 5+21 x 6+34 x 7+..+144 x 10+.. As you realize it match with Fibonacci numbers. However, this generating functions counts the cases in a straight line ,and so it includes some cases that breaks the rule when this line become a circle. The rule breakers are the strings which starts and ends with red beads. So, if we discard these rule breakers, we can reach the correct numbers. The rule breakers strings are those which starts with rg and ends with gr in a desired Smirnov words, because when these Smirnov words are made a circle,the reds will merge and break the rule. When i come to the reason why we also put "g" after and before our red beads is that if we just calculate the number of Smirnov words of length n−2 n−2, then it would also count those which starts and ends with red beads. So, it would cause r,r,...,r,r r,r,...,r,r, but we do not want it. We just want those which obeys the rule but starts and ends with red beads. . Now ,lets assume that we want to calculate the number of rule breaker strings of size 7 7. Then, we should find the number of Smirnov words of length 3 3 calculated above. After that , subtract those rule breakers from the total ,i.e 34 34. Example: Lets calculate the number of necklaces of size 5 5: Step 1−)1−) Find the coefficient of x 5 x 5 in the generating function of desired Smirnov words, which is 13 13 Step 2−)2−) Find the coefficient of x 5−4 x 5−4 in the generating function of desired Smirnov words, which is 2 2 Step 3−)3−) Subtract those rule breakers from the total, so 13−2=11 13−2=11. We found that there are 11 11 necklace that do not contain two consecutive reds g,g,g,g,g r,g,g,g,g g,r,g,g,g g,g,r,g,g g,g,g,r,g g,g,g,g,r r,g,r,g,g r,g,g,r,g g,r,g,r,g g,r,g,g,r g,g,r,g,r Example 2: Lets calculate the number of necklaces of size 6 6: Step 1−)1−) Find the coefficient of x 5 x 5 in the generating function of desired Smirnov words, which is 21 21 Step 2−)2−) Find the coefficient of x 5−4 x 5−4 in the generating function of desired Smirnov words, which is 3 3 Step 3−)3−) Subtract those rule breakers from the total, so 21−3=18 21−3=18. We found that there are 18 18 necklace that do not contain two consecutive reds gggggg rggggg grgggg ggrggg gggrgg ggggrg gggggr rgrggg rggrgg rgggrg grgggr grggrg grgrgg ggrggr ggrgrg gggrgr grgrgr rgrgrg NOTE: Our solution works because of the beads are labelled. If they are not, then we needs Burnsides' lemma to get rid of overcounting. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Aug 18, 2023 at 13:28 answered Aug 17, 2023 at 19:41 Not a Salmon FishNot a Salmon Fish 5,920 2 2 gold badges 11 11 silver badges 31 31 bronze badges Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. For 5 5 beads of which 1 1 is red there is only 1 1 way to color the necklace when considering each 'bead slot' of the necklace as indistinguishable. We can think of it this way: for every red bead, there is exactly one adjacent clump of green beads to the left (or right, just pick one WLOG). I am choosing to think of it this way because when you get to the 'last' red bead, a group to the right of it will actually be part of the first clump of green beads since the necklace loops around. Proceeding with this line of thinking, we have r r clumps of green beads, x 1,x 2,...,x r x 1,x 2,...,x r where each clump must contain 1 1 or more green beads, (otherwise at least one pair of red beads will be adjacent) and ∑r i=1 x i=n−r∑i=1 r x i=n−r. The number of ways this can be done is just Theorem 1 1 of Stars and Bars: (n−r−1 r−1)(n−r−1 r−1) For n=5,r=1 n=5,r=1, this gives us (3 0)(3 0) or 1 1. If you are considering each slot of the necklace as unique, this becomes a different problem and requires a different approach, I think TheJack's answer addresses this case. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Aug 17, 2023 at 16:05 answered Aug 17, 2023 at 15:53 AlborzAlborz 1,273 4 4 silver badges 15 15 bronze badges 1 Sorry, that's the identity part of the question involving the Burnside lemma. So no rotations there. I edited my post. –Aleksander Wojsz Commented Aug 17, 2023 at 16:14 Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions combinatorics discrete-mathematics combinations necklace-and-bracelets See similar questions with these tags. 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15725	https://www.quora.com/What-is-the-difference-between-arc-length-and-chord-length	Something went wrong. Wait a moment and try again. Largest Chord Arc Length PLANE GEOMETRY Chord Construction Chord Length Arc (geometry) Concept of Geometry 5 What is the difference between arc length and chord length? · Arc length and chord length are two different measurements related to a circle or a circular segment. Here’s a breakdown of the differences: Arc Length Definition: Arc length is the distance along the curved line of a segment of a circle. Calculation: The arc length L can be calculated using the formula: L=rθ where r is the radius of the circle and θ is the angle in radians that the arc subtends at the center of the circle. Example: If you have a circle with a radius of 5 units and an angle of 60 degrees (which is π3 radians), the arc length would be: L=5×\fr Arc length and chord length are two different measurements related to a circle or a circular segment. Here’s a breakdown of the differences: Arc Length Definition: Arc length is the distance along the curved line of a segment of a circle. Calculation: The arc length L can be calculated using the formula: L=rθ where r is the radius of the circle and θ is the angle in radians that the arc subtends at the center of the circle. Example: If you have a circle with a radius of 5 units and an angle of 60 degrees (which is π3 radians), the arc length would be: L=5×π3≈5.24 units Chord Length Definition: Chord length is the straight-line distance between the two endpoints of the arc. Calculation: The length of a chord C can be calculated using the formula: C=2rsin(θ2) where r is the radius and θ is the angle in radians. Example: Using the same circle with a radius of 5 units and an angle of 60 degrees (or π3 radians), the chord length would be: C=2×5×sin(π6)=2×5×12=5 units Arc Length: Measures the distance along the curve. Chord Length: Measures the straight-line distance between the endpoints of the arc. In general, the arc length is always greater than or equal to the chord length unless the angle is zero, in which case both lengths are zero. Shardul Vinay Khanang AISSCEcertified with94.2% in PCM-EG & English (language), DAV Public School, HUDCO, Bhilai (Graduated 2021) · Author has 315 answers and 329K answer views · 4y Originally Answered: What are the differences between a radius and a chord? · Radius is a distance of any point on the circle from its center. Whereas, Chord is a distance between any two points on the circle. Refer to the following diagram. O is the center. A, B and C are points on the circle. So, OA representing red line is Radius and BC representing blue line is Chord. Hope this will help you. Please upvote my answer if you like it and be my follower if possible. Radius is a distance of any point on the circle from its center. Whereas, Chord is a distance between any two points on the circle. Refer to the following diagram. O is the center. A, B and C are points on the circle. So, OA representing red line is Radius and BC representing blue line is Chord. Hope this will help you. Please upvote my answer if you like it and be my follower if possible. Promoted by Grammarly Grammarly Great Writing, Simplified · Aug 18 Which are the best AI tools for students? There are a lot of AI tools out there right now—so how do you know which ones are actually worth your time? Which tools are built for students and school—not just for clicks or content generation? And more importantly, which ones help you sharpen what you already know instead of just doing the work for you? That’s where Grammarly comes in. It’s an all-in-one writing surface designed specifically for students, with tools that help you brainstorm, write, revise, and grow your skills—without cutting corners. Here are five AI tools inside Grammarly’s document editor that are worth checking out: Do There are a lot of AI tools out there right now—so how do you know which ones are actually worth your time? Which tools are built for students and school—not just for clicks or content generation? And more importantly, which ones help you sharpen what you already know instead of just doing the work for you? That’s where Grammarly comes in. It’s an all-in-one writing surface designed specifically for students, with tools that help you brainstorm, write, revise, and grow your skills—without cutting corners. Here are five AI tools inside Grammarly’s document editor that are worth checking out: Docs – Your all-in-one writing surface Think of docs as your smart notebook meets your favorite editor. It’s a writing surface where you can brainstorm, draft, organize your thoughts, and edit—all in one place. 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Try these features and more for free at Grammarly.com and get started today! Jayanta Mukherjee B Tech IEE in Instrumentation Engineering, Jadavpur University (Graduated 1990) · Author has 42.4K answers and 11M answer views · 4y Originally Answered: What are the differences between a radius and a chord? · The radius of a circle is any line segment connecting the centre of the circle to any point of the circle. The chord of a circle is a line segment joining any two points on the circle. The differences are: I) A particular radius intersects the circle at exactly one point. While, a chord intersects the circle at exactly two points. II) All radii of a given circle pass through the centre of the circle. While, all chords of the given circle does not necessarily pass through the centre of the circle. III) All radii of a given circle are of equal length. Whereas, all chords of a given circle are not of The radius of a circle is any line segment connecting the centre of the circle to any point of the circle. The chord of a circle is a line segment joining any two points on the circle. The differences are: I) A particular radius intersects the circle at exactly one point. While, a chord intersects the circle at exactly two points. II) All radii of a given circle pass through the centre of the circle. While, all chords of the given circle does not necessarily pass through the centre of the circle. III) All radii of a given circle are of equal length. Whereas, all chords of a given circle are not of equal length. If r be the length of radius of a given circle; and, c is expressed as the length of its chord, then 2r ≥ c > 0. Related questions What is the relationship between diameter, arc length, and chord length in circles? How is arc length related to chord length in circles, ellipses and parabolas? How can we draw an arc of specific length? What is the relationship between arc and chord? What is the relation between the arc length and the power factor in electric arc furnaces? Vishal Gupta Bookings Clerk (2019–present) · 2y So, the difference between chord length and arc length is that chord length gives us the length between two points and an arc length gives us the total portion covered between two points. Complete step-by-step answer: Both chord length and arc length are terms used for circles. Srinivasan conversion of radius into diameter and diameter into radius · Author has 4K answers and 4.8M answer views · 4y Originally Answered: What are the differences between a radius and a chord? · The line segment joining the center and any point on the circumference of a circle is known as radius. The length of a radius is same in a circle. All circles are drawn only for the length of their radius. Diameter of a circle is equals to twice the radius The line segment joining any two points on the circumference of a circle is known as a chord. The length of the chords may be different. Diameter is the biggest chord in a circle Sponsored by OrderlyMeds Is Your GLP-1 Personalized? Find GLP-1 plans tailored to your unique body needs. Subhasish Debroy Former SDE at Bharat Sanchar Nigam Limited (BSNL) · Author has 6.6K answers and 5.8M answer views · 4y Originally Answered: What are the differences between a radius and a chord? · A straight line connected between any two points on circumference of a circle is a chord of the circle and from this we can conclude that the diameter of a circle is also a chord as well as biggest chord of a circle. Any straight line connected between the centre of a circle and any point on it's circumference is termed as radius of the circle. Related questions What is the normal arc length in welding? On a 12 kilometer arc length between two points on a theoretically perfectly spherical earth, what is the difference between the chord length and the arc length? What do you mean by development length? Is a string an ideal tool to measure the length of a circle's arc, if one can prove the arc is of a certain length? What is the chord length of an airfoil? Bob Collier Former EE Designed Specialized Computers for 33 Years. · Author has 3.1K answers and 1.6M answer views · 2y Originally Answered: What is the difference between an arc length and a chord length? How do you find their values for different curves? · Arcs curve. Chords don’t. Read your text book or Google. Don’t be surprised if there is closed formula only for circles - it happens. For others, you may get to integrate along the ‘curve’ for a to b. That happens, too. Sponsored by Mutual of Omaha Comparing Medicare plans this fall? Answer a few questions to find a Medicare Plan that meets your needs. Mohammad Shakil Siddiqui Works at Indian Railways · Author has 878 answers and 276.4K answer views · 4y Originally Answered: What are the differences between a radius and a chord? · Chord is line between two points on a circumference of a circle. Radius is a line between a point on a circumference and center point of a circle. Shreyas Shenoy (श्रेयस शनोय) Potterhead, House Stark · Updated 4y Related What is the ratio of the chord length to the arc length? Isn't the derivation remarkably simple? I just hope I didn't make any silly mistakes due to my over-confidence. P.S. Not a fan of the codes on Quora for mathematical formulae! :-D Do check out the Sinc function - Wikipedia. Isn't the derivation remarkably simple? I just hope I didn't make any silly mistakes due to my over-confidence. P.S. Not a fan of the codes on Quora for mathematical formulae! :-D Do check out the Sinc function - Wikipedia. Sponsored by Project-Management Quickly find the Project Management software you need. Enhance team collaboration and streamline your projects with powerful Project Management tools. Adrian Giles Studied Physics & Mathematics (Graduated 1985) · Author has 3.5K answers and 784.2K answer views · Aug 1 Arc length is a curved length usually of part of a circle. In that case it is the distance between two points along the circle. A chord is a straight line between two points on a circle. Chuck Biehl Mathematics Education Specialist (2015–present) · Author has 239 answers and 84.9K answer views · 2y Arc length is a piece of the circle itself, chord length is the straight segment across the circle. Orlando Del Carmen Data Privacy Advocate, C\|EH, Infosec Personnel · Author has 1K answers and 2.1M answer views · 5y Related What is the height of a chord h inside an R75 circle, when the chord length and the chord arc length is equal to a 10% difference? To get the needed values, the following code will be of help. CODE: import mathradius = 75print("ANGLE",'\t',"ARC",'\t',"CHORD")print("----------------------")for angle in range(0,180,5): if(angle>0): degrees = (angle/180) math.pi arc = 75 degrees chord = 2 (math.sin(degrees)radius) ratio = round(arc/chord,2) if(round(ratio,1)==1.1): print(angle,'\t',int(arc),'\t',int(chord)) print("----------------------") Arc Measure Formula \| How to Find Angle Measure of an Arc OUTPUT: From the above code, we were able to generate the arc angle, arc length, and the chord length. To get the needed values, the following code will be of help. CODE: import mathradius = 75print("ANGLE",'\t',"ARC",'\t',"CHORD")print("----------------------")for angle in range(0,180,5): if(angle>0): degrees = (angle/180) math.pi arc = 75 degrees chord = 2 (math.sin(degrees)radius) ratio = round(arc/chord,2) if(round(ratio,1)==1.1): print(angle,'\t',int(arc),'\t',int(chord)) print("----------------------") Arc Measure Formula \| How to Find Angle Measure of an Arc OUTPUT: From the above code, we were able to generate the arc angle, arc length, and the chord length. It shows that the difference between the arc length and the chord length is 15. I just assumed though that that 10% difference is based from the arc length: So, 15 of 150 = 10% Given the values, we can now solve the height of the chord. Let’s denote it as H. H2=Radius2−(C/2)2 H=√752−(1352)2 H=√5625−4556.25=√1068.75 Height of the chord=32.69 Devanshu Gupta Btech in Electronics and Communication Engineering, Birla Institute of Technology, Mesra (Expected 2026) · 4y Related What are the differences between a chord and a diameter? Chord is a line passing through any two points in a curve Diameter is also a chord but the only difference is…the chord which passes through the center of the circle is called diameter Source: Chord Walls Chord is a line passing through any two points in a curve Diameter is also a chord but the only difference is…the chord which passes through the center of the circle is called diameter Source: Chord Walls Related questions What is the relationship between diameter, arc length, and chord length in circles? How is arc length related to chord length in circles, ellipses and parabolas? How can we draw an arc of specific length? What is the relationship between arc and chord? What is the relation between the arc length and the power factor in electric arc furnaces? What is the normal arc length in welding? On a 12 kilometer arc length between two points on a theoretically perfectly spherical earth, what is the difference between the chord length and the arc length? What do you mean by development length? Is a string an ideal tool to measure the length of a circle's arc, if one can prove the arc is of a certain length? What is the chord length of an airfoil? What is the formula for the length of an arc? How can we calculate the height between the arc length and chord length? We don't know the diameter. What is an arc length? What is the difference between 'diameter & length'? What is the formula for the chord length and the arc length? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
15726	https://zerodebikes.com/pages/unsprung-mass	Unsprung Mass – Zerode Bikes Skip to content Close menu Bikes Katipo 29er Taniwha Mullet G3 Downhill Sizing Shifting Options How To Get One Demo Shop Bikes Spare Parts Upgrades Apparel Contact Contact Us Bike Enquiry Workshop Support Simply Better About Behind The Bikes History Zerodes In Action Ambassadors Learn Why A Gearbox? Tech details Pinion Gearbox What About Drag? Manuals and Guides Unsprung Mass Our Warranty 9 Speed vs 12 Speed Instagram Facebook YouTube Cart Close cart Order note Discounts Subtotal $0.00 Shipping, taxes, and discount codes calculated at checkout. Check out One or more of the items in your cart is a recurring or deferred purchase. By continuing, I agree to the cancellation policy and authorize you to charge my payment method at the prices, frequency and dates listed on this page until my order is fulfilled or I cancel, if permitted. Your cart is currently empty. 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15727	https://math.libretexts.org/Bookshelves/Arithmetic_and_Basic_Math/Basic_Math_(Grade_6)/07%3A_Rational_Numbers/36%3A_Negative_Numbers_and_Absolute_Value/36.03%3A_Comparing_Positive_and_Negative_Numbers	Skip to main content 36.3: Comparing Positive and Negative Numbers Last updated : Mar 27, 2022 Save as PDF 36.2: Points on the Number Line 36.4: Ordering Rational Numbers Page ID : 40820 ( \newcommand{\kernel}{\mathrm{null}\,}) Lesson Let's compare numbers on the number line. Exercise 36.3.1: Which One Doesn't Belong: Inequalities Which inequality doesn't belong? 54<2 8.5>0.95 8.5<7 10.00<100 Exercise 36.3.2: Comparing Temperatures Here are the low temperatures, in degrees Celsius, for a week in Anchorage, Alaska. \| day \| Mon \| Tues \| Weds \| Thurs \| Fri \| Sat \| Sun \| --- --- --- --- \| \| temperature \| 5 \| −1 \| −5.5 \| −2 \| 3 \| 4 \| 0 \| Table 36.3.1 Plot the temperatures on a number line. Which day of the week had the lowest low temperature? The lowest temperature ever recorded in the United States was -62 degrees Celsius, in Prospect Creek Camp, Alaska. The average temperature on Mars is about -55 degrees Celsius. Which is warmer, the coldest temperature recorded in the USA, or the average temperature on Mars? Explain how you know. Write an inequality to show your answer. On a winter day the low temperature in Anchorage, Alaska, was -21 degrees Celsius and the low temperature in Minneapolis, Minnesota, was -14 degrees Celsius. Jada said, “I know that 14 is less than 21, so -14 is also less than -21. This means that it was colder in Minneapolis than in Anchorage.” Do you agree? Explain your reasoning. Are you ready for more? Another temperature scale frequently used in science is the Kelvin scale. In this scale, 0 is the lowest possible temperature of anything in the universe, and it is -273.15 degrees in the Celsius scale. Each 1K is the same as 1∘C, so 10K is the same as −263.15∘C. Water boils at 100∘C. What is this temperature in K? Ammonia boils at −35.5∘C. What is the boiling point of ammonia in K? Explain why only positive numbers (and 0) are needed to record temperature in K. Exercise 36.3.3: Rational Numbers on a Number Line Plot the numbers -2, 4, -7, and 10 on the number line. Label each point with its numeric value. Decide whether each inequality statement is true or false. Be prepared to explain your reasoning. −2<4 −2<−7 4>−7 −7>10 Drag each point to its proper place on the number line. Use your observations to help answer the questions that follow. Andre says that 14 is less than −34 because, of the two numbers, 14 is closer to 0. Do you agree? Explain your reasoning. Answer each question. Be prepared to explain how you know. Which number is greater: 14 or 54? Which number is farther from 0: 14 or 54? Which number is greater: −34 or 58? Which number is farther from 0: −34 or 58? Is the number that is farther from 0 always the greater number? Explain your reasoning. Summary We use the words greater than and less than to compare numbers on the number line. For example, the numbers -2.7, 0.8, and -1.3, are shown on the number line. Because -2.7 is to the left of -1.3, we say that -2.7 is less than -1.3. We write: −2.7<−1.3 In general, any number that is to the left of a number n is less than n. We can see that -1.3 is greater than -2.7 because -1.3 is to the right of -2.7. We write: −1.3>−2.7 In general, any number that is to the right of a number is greater than We can also see that 0.8>−1.3 and 0.8>−2.7. In general, any positive number is greater than any negative number. Glossary Entries Definition: Negative Number A negative number is a number that is less than zero. On a horizontal number line, negative numbers are usually shown to the left of 0. Definition: Opposite Two numbers are opposites if they are the same distance from 0 and on different sides of the number line. For example, 4 is the opposite of -4, and -4 is the opposite of 4. They are both the same distance from 0. One is negative, and the other is positive. Definition: Positive Number A positive number is a number that is greater than zero. On a horizontal number line, positive numbers are usually shown to the right of 0. Definition: Rational Number A rational number is a fraction or the opposite of a fraction. For example, 8 and -8 are rational numbers because they can be written as 81 and −81. Also, 0.75 and -0.75 are rational numbers because they can be written as 75100 and −75100. Definition: Sign The sign of any number other than 0 is either positive or negative. For example, the sign of 6 is positive. The sign of -6 is negative. Zero does not have a sign, because it is not positive or negative. Practice Exercise 36.3.4 Decide whether each inequality statement is true or false. Explain your reasoning. −5>2 3>−8 −12>−15 −12.5>−12 Exercise 36.3.5 Here is a true statement: −8.7<−8.4. Select all of the statements that are equivalent to −8.7<−8.4. -8.7 is further to the right on the number line than -8.4. -8.7 is further to the left on the number line than -8.4. -8.7 is less than -8.4. -8.7 is greater than -8.4. -8.4 is less than -8.7. -8.4 is greater than -8.7. Exercise 36.3.6 Plot each of the following numbers on the number line. Label each point with its numeric value. 0.4,−1.5,−1710,−1110 (From Unit 7.1.2) Exercise 36.3.7 The table shows five states and the lowest point in each state. \| state \| lowest elevation (feet) \| --- \| \| California \| −282 \| \| Colorado \| 3350 \| \| Louisiana \| −8 \| \| New Mexico \| 2842 \| \| Wyoming \| 3099 \| Table 36.3.2 Put the states in order by their lowest elevation, from least to greatest. (From Unit 7.1.4) Exercise 36.3.8 Each lap around the track is 400 meters. How many meters does someone run if they run: 2 laps? 5 laps? x laps? If Noah ran 14 laps, how many meters did he run? If Noah ran 7,600 meters, how many laps did he run? (From Unit 6.1.6) Exercise 36.3.9 A stadium can seat 16,000 people at full capacity. If there are 13,920 people in the stadium, what percentage of the capacity is filled? Explain or show your reasoning. What percentage of the capacity is not filled? (From Unit 3.4.7) 36.2: Points on the Number Line 36.4: Ordering Rational Numbers
15728	https://s4be.cochrane.org/blog/2017/08/29/regression/	Linear regression: a practical introduction - Students 4 Best Evidence En español – ExME Em português – EME A network for students interested in evidence-based health care Menu Home About What is Evidence-Based Medicine? What are the key steps in EBM? Who are S4BE? What’s the aim of S4BE? Website accessibility Topics Bias Critical thinking Health in the media Introductions to Evidence-Based Practice Key Concepts – Assessing treatment claims Paramedic Physiotherapy Screening Searching for evidence Shared decision making Statistics Study design and Research methods Systematic reviews Resources Library Cochrane: participate, connect and learn Blogging Tips Suggested blog topics Register Contact Log In Linear regression: a practical introduction Posted on 29th August 2017 by Saul Crandon Tutorials and Fundamentals The following blog article has been written to provide an overview of linear regression. It is suitable for those with little to no experience of this type of analysis. This is not a guide on how to conduct your own analysis, but instead will serve as a taster to some of the key terms and principles of regression. Further reading resources will be provided at the end for those who wish to further their knowledge. Correlation vs Regression Like regression, there are different types of correlation. The main two types of correlation are Pearson’s correlation and Spearman’s correlation. Pearson’s correlation focuses on describing a linear relationship between two variables, whereas Spearman’s correlation is more concerned with the rank-order of the points, regardless of where exactly they lie. Furthermore, it is important to distinguish correlation from regression: Correlation is a statistical technique that describes the strength of a relationship between two variables, given as a correlation coefficient, r. It doesn’t matter which variable is x and y, the correlation coefficient is always the same. The value of r has a possible range from -1 to +1 (1). The following plots illustrate the appearance of different correlation coefficients. Regression also describes relationships between variables. However, in regression, a change in one variable is associated with changes in another variable. The result of linear regression is described using R 2. Regression analysis involves creating a line of best fit. This is described mathematically as y = a + bx. The value of ‘a’ is the y intercept (this is the point at which the line would intersect the y axis), and ‘b’ is the gradient (or steepness) of the line. This is useful as once this line is created, you can predict values of y (or the dependent variable) based on values of x (or the independent variable). This line of best fit is used to summarise the relationship between the variables as shown below. A few similarities between correlation and regression should also be noted. Most importantly, neither can affirm causality(2). This is because although a change in x might result in a change in y, it is possible that both variables are related to a third confounding variable not described in the analysis. Secondly, the square of Pearson’s correlation coefficient (r) is the same value as the R 2 in simple linear regression. Simple vs Multiple Linear Regression Simple Linear Regression One variable is dependent and the other variable is independent. Regression results are given as R 2 and a p-value. R 2 represents the strength of association, whereas the p-value is testing the null hypothesis, that there is no association between the variables. Therefore, if the resulting p-value is low (usually defined as < 0.05), we can assume that the relationship between the variables is unlikely the result of chance and so, we reject the null hypothesis. Multiple Linear Regression One dependent variable and two or more independent variables. Multiple linear regression follows the same concept as simple linear regression. However, the difference is that it investigates how a dependent variable changes based on alterations in a combination of multiple independent variables. This reflects a more ‘real-life’ scenario as variables are usually influenced by a number of factors. The general process of multiple linear regression is as follows (a backwards elimination method is used in this example): An outcome (or dependent) variable is defined Two or more independent variables are defined All variables enter the model and the analysis is run Non-significant variables (P > 0.05) are removed and the analysis is re-run To validate the final model, residuals are calculated and plotted to assess linearity of the data If the assumption of linearity is met then the model is complete The model may be adjusted to optimise its suitability The removal of non-significant variables is important and can be done most commonly with either forwards or backwards elimination methods (2). Forwards elimination means you remove these variables before the analysis is run whereas backwards elimination does this after the results are obtained. The non-significant variables can be removed if their absence from the model would not significantly decrease the effectiveness of the model. Removing these improves the models ‘goodness-of-fit’. Variables with low p-values (P < 0.05) remain because this implies they are meaningful additions to the model and that changes in them are associated with changes in the dependent variable. A model is finalised when no more variables can be removed without a statistically significant loss in fit. Residuals are the difference between the observed and predicted value of the dependent variable (y). Residuals are crucial, as this allows the model to be validated. Residuals must be calculated for each data point, plotted and then inspected to verify the suitability of the model. The points on a residual plot should have an even distribution around the horizontal axis of zero. If so, the assumption of linearity of the data is true and it is suitable for its intended purpose. A bad residual plot is when there is clear pattern to the data. In other words, the residual data points are skewed. The following graphs show a good and bad example of residual plots. If the residual plot looks less than ideal, data transformations can be conducted. The methods of doing so are beyond the scope of the article, but in short, it allows the non-linear data to be used more effectively with the linear regression models. It is important to always state if any data transformations were performed. Once the model is complete, it should be tested on further independent data sets to check its suitability (i.e. data that was not used to construct the model itself). A developed model can be shown to be robust if it is still effective with independent data. Finally, it is always important to state clearly if there was any missing data. Complex methods do exist to handle missing data sets in linear regression, but these will not be discussed here. An example of a multiple linear regression model is shown below in the form of both a table and the resultant equation. Firstly, the table shows all the necessary components of the model. The R 2 value is given here; however it is good practice to use the adjusted R 2 value, as it accounts for the sample size used. The equation shows the line of best fit. It expands upon the y = a + bx formula to account for the multiple independent variables involved. A few assumptions needed for Linear Regression The number of observations should be greater than the number of x (independent) variables. There should be little to no multicollinearity (this will be discussed in the next section). The residual plots are evenly distributed around the horizontal axis of zero, i.e. the mean of the residuals is close to zero. This means the data is linear. A few problems with Regression Poor fit (over-fitting, under-fitting etc.) of a model means it does not serve it intended purpose effectively. There are ways that the model can be adjusted to optimise its usefulness. Multicollinearity exists when two or more of the independent variables in the model are highly correlated with one another. This poses a problem as these related variables offer much the same information to the model. To deal with multicollinearity, one of the variables must be removed (usually the least significant one/highest p-value). The Variance Inflation Factor (VIF) is calculated for each independent variable and is used to determine whether multicollinearity is present. Basically, a high VIF means the variable is explained by other independent variables, whereas a low VIF means it is not. A low VIF is good and indicates no significant multicollinearity. It’s important to clarify that no universal cut-off point exists and so it requires a subjective, but educated decision from the researchers conducting the analysis to determine if the value of VIF is acceptable or not. Regression is an artful science and so, requires informed judgement and experience to optimise a model for its intended purpose. Summary Linear regression is a useful method to predict changes in a dependent variable based on alterations in independent variables. It is hoped that this blog can act as a gentle introduction to this type of analysis. For more information about linear regression, the following web resources are recommended: Student 4 Best Evidence (S4BE) website– This site has a number of relevant student-written articles. YouTube – There is a vast library of useful videos on YouTube from basic introductions, to detailed step-by-step guides to conducting your own regression across a range of programs. BMJ website– The BMJ is a reliable source of high quality articles regarding statistics. Penn State Eberly College of Science: STAT101 course– This website provides detailed guides on all types of statistics. The following links provide thorough tutorials on simple linear regression: and multiple linear regression: References BMJ.com. (2017).11. Correlation and regression \| The BMJ. [online] Available at: [Accessed 22 Aug. 2017]. Schneider A, Hommel G, Blettner M. Linear Regression Analysis. Deutsches Artzeblatt International. 2010 Nov; 107(44): 776–782. Tags: regression Saul Crandon Saul is an Internal Medicine Trainee at NHS Greater Glasgow & Clyde. He also holds Honorary Clinical Lecturer status at the University of Glasgow as part of his role within the Glasgow Academic Training Environment (GATE). He has previously worked as a doctor in both the UK and Australia. He graduated from the University of Liverpool in 2018 after being awarded an intercalated Masters of Research degree (Distinction) from the University of Leeds in 2017. Saul is currently serving his fourth year on the committee for the Cochrane UK & Ireland Trainees Advisory Group. Before joining CUKI-TAG, he was a Students4BestEvidence Pioneer, producing a catalogue of articles and tutorials on research methodology. He has a strong interest in the education and promotion of evidence-based medicine. View more posts from Saul Leave a Reply Cancel reply Your email address will not be published.Required fields are marked Comment Name Email [x] Save my name, email, and website in this browser for the next time I comment. Δ No Comments on Linear regression: a practical introduction Taneisha Gilles This is very informative information help me to understand more about the topic, thanks for sharing. 20th February 2024 at 3:18 pm Reply to Taneisha Robin Rai what a wonderful guide on linear regression, very helpful and informative blog. Thank you for sharing 17th July 2021 at 7:31 am Reply to Robin Kamrunnaher Very informative 13th April 2019 at 5:55 pm Reply to Kamrunnaher Subscribe to our newsletter --------------------------- You will receive our monthly newsletter and free access to Trip Premium. Popular Articles Learning Resources A 20 Minute Introduction to Cluster Randomised Trials Another 20 minute tutorial from Tim. Tutorials and Fundamentals A beginner’s guide to interpreting odds ratios, confidence intervals and p-values The nuts and bolts 20 minute tutorial from Tim. Learning Resources Free information and resources for pupils and students interested in evidence-based health care This new webpage from Cochrane UK is aimed at students of all ages. What is evidence-based practice? What is ‘best available research evidence’? Which resources will help you understand evidence and evidence-based practice, and search for evidence? 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15729	https://www.youtube.com/watch?v=_Gnke2x3vT8	How To Solve Simple Harmonic Motion Problems In Physics The Organic Chemistry Tutor 9790000 subscribers 13544 likes Description 1069688 views Posted: 17 Nov 2017 This physics video tutorial provides a basic introduction into how to solve simple harmonic motion problems in physics. It explains how to calculate the frequency, period, spring constant and the angular frequency of a mass-spring system. It also explains how to find the amplitude and frequency from a displacement cosine equation. Physics Video Lessons: Final Exam and Test Prep Videos: Intro to Pressure: Fluid Pressure Review: Simple Harmonic Motion: Energy In a Simple Harmonic Oscillator: Simple Harmonic Motion Review: The Simple Pendulum: The Physical Pendulum: Mechanical Waves Problems: Wave Speed on a String: Standing Waves on a String: Speed of Sound In Matter: Sound Intensity Physics Problems: Sound Intensity Level: Beat Frequency: Standing Waves In Organ Pipes: Doppler Effect Physics Problems: Sound Waves Review: Physics PDF Worksheets: 334 comments Transcript: Horizontal Spring in this video we're going to talk about how to solve some basic simple harmonic motion problems in physics so let's start with number one a horizontal spring with a mass of 0.75 kilograms attached to it is undergoing simple harmonic motion calculate the period frequency and angular frequency of this oscillator so if you want to try this problem feel free to pause the video and work on it the formula that we need to calculate the period is this equation it's equal to 2 pi times the square root of the mass divided by the spring constant which we have all of that in this problem so the mass is 0.75 kilograms and the spring constant is 300 newtons per meter so let's go ahead and type that into the calculator so the period is .314 two seconds so that's how you could find the period of a simple harmonic oscillator all you need is the mass and the spring constant now part b what is the frequency frequency is 1 divided by the period so it's 1 divided by 0.3142 seconds so the frequency in this problem is 3.183 hertz now let's move on to part c let's calculate the angular frequency the angular frequency represented by the symbol omega it's 2 pi times the frequency so that's going to be 2 pi times 3.183 hertz and so that's going to be about 20 radians per second and that's it for this problem number two a force of 500 newtons is used to stretch a spring with a 0.5 kilogram mass attached to it by 0.35 meters what is the value of the spring constant and calculate the frequency of the oscillator so let's say this is a wall and we have a spring attached to it and there's a mass now we're going to stretch the spring using a force and so this is a 0.5 kilogram mass and at this point we're applying a force of 500 newtons to stretch it by point 35 meters so how can we calculate the spring constant well we know that the force is equal to kx based on hooke's law so the spring constant k is the ratio between the applied force and the amount that the length of this spring changes so it's going to be 500 newtons divided by 35 meters and so that's going to be 14 28.6 newtons per meter so that's how you can calculate the spring constant of a spring it's simply the force divided by the change in life now let's calculate the frequency of the oscillator the frequency of this spring is going to be one over two pi times the square root of k over m so it's 1 over 2 pi times the square root of the spring constant which is 1428.6 divided by the mass of 0.5 so the frequency is 8.51 hertz and so that's the answer to part b of that problem Spring Constant number three a spring with a constant of 100 newtons per meter vibrates at 25 hertz what is the frequency of vibration of a spring with a constant of 400 newtons per meter so what happens to the frequency if we increase the spring constant as the spring constant increases the frequency increases and we know that the frequency of a spring is one over two pi times the square root of k over m so notice that the frequency is proportional to the square root of k and in this example the spring constant increases from 100 to 400 so it increases by factor four and the square root of four is two so therefore the frequency should increase by a factor of two so the answer is 50 hertz now if you want a formula for this type of problem here's how you could derive it since we're dealing with two frequencies let's write a ratio between the two frequencies f2 and f1 so f2 is going to be 1 over 2 pi times the square root of k2 divided by m f one is going to be one over two pi times the square root of k one over m now i didn't write a subscript for m because the mass doesn't change in this problem so we could cancel it and at the same time we could cancel one over two pi so thus we have this expression f2 divided by f1 is equal to the square root of k2 divided by the square root of k1 or we could just simply write it like this f2 over f1 is simply the square root of k2 over k1 all within the single fraction so now let's calculate f2 so f1 in this problem that's 25 hertz and that corresponds to a a spring constant of 100 so k1 is 100 in this problem we're trying to find a new frequency at this new k value so k2 is going to be 400 so 400 divided by 100 is 4 and the square root of 4 is 2. so f2 over 25 is equal to 2. now let's cross multiply so f2 times 1 is simply f2 and then we have 25 times 2 which is 50. and so that's the new frequency so if you have a problem that relates frequency to the spring constant you could use that formula Example now let's work on this problem a 0.75 kilogram mass vibrates according to the equation x is equal to 0.65 cosine 7.35 t determine the amplitude frequency period and the spring constant so x represents the position of the oscillator so let's say the oscillator is at equilibrium it's right here so right now the position is x equals zero it can oscillate this way and that way so here x could be one and here x could be negative one it could be two it could be three it can vary the amplitude is the maximum displacement for the spring so let's say if the most that the spring will stretch to is up to this line and let's say that's 1.5 and so the most that you can stretch to in the other direction will be here so in this case the amplitude the maximum x value is 1.5 now x could be anywhere between negative 1.5 and positive 1.5 so x represents the current displacement a the amplitude is the maximum displacement so make sure you understand the difference between the two now you need to know the generic form of that formula the current displacement is equal to the maximum displacement times cosine omega t now sometimes you may have a phase angle but we don't have it for this problem so all we have is this formula so the amplitude is whatever number you see in front of cosine so the maximum displacement is 0.65 which means that x can be anywhere between negative 0.65 and positive 0.65 so if you graph this cosine equation the amplitude is 0.65 and negative 0.65 on the graph and cosine starts at the top so the cosine wave is going to look something like this and it's going to keep oscillating between these two points now what does this number represent 7.35 notice that the variable in front of t is omega so the 7.35 represents the angular frequency of this oscillator so it's 7.35 radians per second and if we know the angular frequency we can now find the frequency the angular frequency is equal to 2 pi f and so 7.35 radians per second is equal to 2 pi times the frequency so the regular frequency is just 7.35 divided by 2 pi and so you should get 1.17 hertz now how can we calculate the period if you know the frequency then you know the period the period is simply 1 over the frequency so in this example it's 1.17 hertz it's 1 over 1.17 hertz so 1 over 1.17 that's 0.855 seconds so we have the amplitude we have the frequency and we have the period so the last thing we need to calculate is the spring constant so what formula can help us to do that well we know that frequency is 1 over 2 pi times the square root of k over m so the first thing i'm going to do is multiply both sides by 2 pi so i can get rid of this so on the left i have 2 pi f and that's equal to the square root of k over m now i'm going to take the square of both sides to get rid of the radical on the right so i have 2 pi f squared and when you combine a square and a square root those two will cancel so on the right all i have is k over m so i'm going to multiply both sides by m so let's get rid of that so the spring constant is equal to the mass times 2 pi times the frequency squared so if you have the mass and the frequency you could use that equation to calculate the spring constant now keep in mind that omega is 2 pi f so we can replace 2 pi f with omega so the spring constant is the mass times the square of the angular frequency which we know it's 7.35 so we have a mass of 0.75 kilograms multiplied by 7.35 radians per second squared and so the spring constant is 40.5 newtons per perimeter and so that's the final answer for this problem so hopefully this video gave you a decent understanding of all the formulas that you need for solving basic simple harmonic motion problems thanks for watching you
15730	https://www.aft.com/support/product-tips/what-does-head-hgl-mean-for-submerged-pumps-and-exit-pressures	What Does “Head (HGL)” Mean for Submerged Pumps and Exit Pressures? - AFT Blog This website stores cookies on your computer. These cookies are used to collect information about how you interact with our website and allow us to remember you. We use this information in order to improve and customize your browsing experience and for analytics and metrics about our visitors both on this website and other media. To find out more about the cookies we use, see our Privacy Policy. Accept Contact Us Help Site Products Software Overview Fathom Arrow Impulse xStream CHEMCAD Chempak Database Educational Use How to Order Consulting Nuclear Validation (NVV) Support Support Request Support Installation Support Upgrade & Maintenance Services Problems We Solve Consulting Nuclear Validation (NVV) Downloads Current Versions Fathom Arrow Impulse xStream Chempak Previous Versions Free Demos Learning Center Training Seminar Case Studies Technical Papers Tips and Tricks Application Topics Video Tutorials Webinars Channel Partners About Contact Us Mission and History Industries We Serve Events Calendar News Blog Customer Testimonials Careers Directions and Lodging AFT Privacy Policy Fathom Student Latest Release Home Support Product Tips Ben Keiser What Does “Head (HGL)” Mean for Submerged Pumps and Exit Pressures? AFT Blog Welcome to the Applied Flow Technology Blog where you will find the latest news and training on how to use AFT Fathom, AFT Arrow, AFT Impulse, AFT xStream and other AFT software products. Home Categories Tags Authors Categories: All CategoriesSearch... Suggested keywords Search x Search Font size:+– Print 3 minutes reading time(608 words) What Does “Head (HGL)” Mean for Submerged Pumps and Exit Pressures? Tips and Tricks Ben Keiser Wednesday, 28 December 2016 12361 Hits 0 Comments In AFT Fathom and AFT Impulse, it is possible to model a submerged pump where a short and possibly frictionless suction pipe for the pump’s inlet does not need to be modeled. When modeling a submerged pump, there are two options available for specifying the system inlet boundary condition at the pump suction. As shown in Figure 1 below, the Submerged Pump’s Suction Pressure can either be specified as “Head (HGL)” or “Pressure”. Figure 1 – Pump Property window with Submerged Pump option selected. No upstream pipe required. Suction Pressure must be entered as boundary condition in terms of pressure or Head (HGL). Modeling a submerged pump is not the only time where the “Head (HGL)” or “Pressure” choices will arise. If an Exit Valve (i.e., a valve that discharges into ambient conditions with no downstream pipe), Exit Orifice, Spray Discharge Nozzle, Exit or Inline Exit Relief Valve is modeled, then an “Exit Pressure” value like that shown in Figure 2 will need to be specified with the same type of “Head (HGL)” or “Pressure” choices. This blog will clarify how to specify either option appropriately. Consider the system in Figure 3 where a pump is submerged in a large tank. The pump is at an elevation of 75 meters (246 ft), liquid surface elevation of the tank is 100 meters (328 ft) and is open to the atmosphere, therefore, the liquid column above the pump suction is 25 meters (82 ft). Modeling this system in AFT Fathom is very simple. Figure 4 contains three identical systems on the same AFT Fathom Workspace. The top system includes a reservoir junction that represents the tank from Figure 3 as well as a short, frictionless connector pipe. The top system serves as a reference to help determine the inlet suction pressure for the middle system and allows for a consistency comparison amongst the three systems. The middle system models a single pump junction with no suction pipe and represents the submerged pump in Figure 3. The middle system uses the “Pressure” option for the Submerged Pump Suction Pressure option while the bottom system uses the “Head (HGL)” option. After running the model, the pump inlet pressure in the top system is determined to be 3.458 bar (stagnation). This value can then be entered into the middle system when the “Submerged Pump” feature is chosen with the “Pressure” option selected for the Suction Pressure. The suction pressure can also easily be calculated from the relationship shown below the image in Figure 5. Figure 5 – Suction Pressure calculation details to use for Submerged Pump’s “Pressure” option for Suction Pressure, based upon 25 m (82 ft) liquid column above pump suction. style= The bottom system in Figure 4 uses the “Head (HGL)” option for the Suction Pressure specification on the submerged pump. One may be tempted to enter the value of the liquid column height above the pump suction. However, this is incorrect. When specifying the Suction Pressure with the “Head (HGL)” option, the liquid surface elevation of the reservoir or tank in which the pump is submerged is what needs to be specified. When the liquid surface elevation of the reservoir in which the pump is submerged in is specified, then AFT Fathom and AFT Impulse will be able to determine the amount of liquid head above the pump suction based upon the pump’s inlet elevation and the “Head (HGL)” value. This will provide the correct boundary condition at the inlet of the pump, as shown in Figure 6. After running the model for the three systems in Figure 4, you can see from the results shown in Figure 7 that the systems are identical. This should help clarify that the “Head (HGL)” value that needs to be specified is the liquid surface elevation of the reservoir or tank that the pump is submerged within, NOT the height of the liquid column above the pump suction. If the pump is submerged into a pit which is below zero elevation, the same concept applies, just be sure to specify the elevations correctly with the appropriate negative values. The pump suction is 170 m (558 ft) below sea level, therefore the “Inlet Elevation” for the pump will be -170 meters. However, the liquid surface elevation of the pit in this case is 150 meters (492 ft) below sea level, therefore, the “Head (HGL)” that is specified will be -150 meters. This establishes the 20 meters (66 ft) of liquid column above the pump suction. As mentioned previously in this blog, the “Head (HGL)” and “Pressure” options can exist for other junction types such as exit valves, spray discharge nozzles, etc. This is the “Exit Pressure” that must be defined for the case of an Exit Valve that was shown in Figure 2. The “Exit Pressure” is simply the ambient conditions in which the fluid is discharging into through the valve, spray discharge nozzle, etc. If the fluid is discharging directly into the atmosphere, then it is easiest to choose the “Pressure” option for the “Exit Pressure” and then simply enter the atmospheric pressure value. However, if the valve or spray discharge nozzle is submerged into a tank or reservoir and that is the medium in which the fluid is discharged into, then it may be easier to use the “Head (HGL)” option. Again, specify the elevation of the component in the junction property window with reference to the zero point of the model. Then for the “Head (HGL)” option, specify the liquid surface elevation of the tank or reservoir in which the component is submerged. In conclusion, this discussion should now have clarified that the value that needs to be entered for the “Head (HGL)” option for the Suction Pressure on a submerged pump or Exit Pressure for an Exit valve is the liquid surface elevation of the tank/reservoir in which the component is submerged. NOT the liquid column of fluid above the component. Based upon the junction’s inlet elevation and the “Head (HGL)” that is entered, AFT Fathom and AFT Impulse will properly take into account the liquid column of fluid above that component. Share Tweet Tags: AFT FathomAFT ImpulseSubmerged PumpHead (HGL)PumpEfficient Modeling Features America's Wacky New President...and Technology Seventy-Five Years After Pearl Harbor, Which Was 3... About the author Ben Keiser Ben Keiser Ben is AFT's Technical Sales Manager. He can be found teaching many of AFT's technical seminars, stopping over for lunch and learns with our customers and managing the AFT booth at trade shows. Prior to joining AFT, Ben worked as a contract engineer for Eaton Corporation and WellbornYX Corporation. He holds a Bachelor's of Science in Chemical Engineering (2009) from Colorado School of Mines. Author's recent posts More posts from author Wednesday, 24 July 2024 How Pipe Flow Analysis Enhances Chemical Processing Monday, 12 July 2021 What Is Transient Analysis? Sunday, 17 January 2021 Know Your Pump & System Curves - Part 3 Related Posts The Opposite of Good Vibrations - Evaluating FIV, AIV, FIP Guidelines from AFT Fathom and AFT Arrow Tips and Tricks Don't throw a fit over your Fitting K Factors - Part 1: Fittings & Losses in AFT Fathom Tips and Tricks AFT Impulse: A Quarter Century as a World Leader President's Perspective Know Your Pump & System Curves - Part 3 Tips and Tricks Know Your Pump & System Curves – Part 2B Tips and Tricks Comments No comments made yet. Be the first to submit a comment Monday, 29 September 2025 Submit Your Comment Log in © 1996 - 2025 Applied Flow Technology Mobile Nav Products AFT Software Overview AFT Fathom AFT Arrow AFT Impulse AFT xStream Chempak Database AFT Flow Expert Package Educational Use Pricing & How to Order Support Services Problems We Solve Consulting Software Development Nuclear Validation (NVV) Support Installation Instructions License Information USB Key Support Support Upgrade & Maintenance (SUM) Downloads Current Versions AFT Fathom AFT Arrow AFT Impulse AFT xStream Chempak Previous Versions Free Demos Free Viewers Learning Center Training Seminar Case Studies Technical Papers White Papers Tips and Tricks Application Topics Video Tutorials Webinars Upcoming Webinars Webinar Library Channel Partners Channel Partner Locations About Contact Us Vision & Mission Events Calendar Platinum Pipe Award News Blog Customer Testimonials Careers History Directions and Lodging AFT Privacy Policy Login to your account Don't have an account yet? 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15731	https://math.stackexchange.com/questions/1130023/what-can-the-floor-of-a-square-root-be-rewritten-as	What can the floor of a square root be rewritten as? - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more What can the floor of a square root be rewritten as? Ask Question Asked 10 years, 8 months ago Modified10 years, 8 months ago Viewed 6k times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. For example: Let n n be natural (positive) number. ⌊n+1−−−−−√⌋⌊n+1⌋ Floor: round down to nearest integer. What can I rewrite this as? Can I break it down or apart? ceiling-and-floor-functions Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Feb 2, 2015 at 8:07 Henrik supports the community 7,077 13 13 gold badges 28 28 silver badges 34 34 bronze badges asked Feb 2, 2015 at 7:55 UserUser 405 1 1 gold badge 8 8 silver badges 19 19 bronze badges 4 I'm not sure that I understand what you have in mind (the expression looks already rather simplified to my eyes).Fabian –Fabian 2015-02-02 08:23:32 +00:00 Commented Feb 2, 2015 at 8:23 Can I expand it, like how floor(x+1) = floor(x) + floor(1) ?User –User 2015-02-02 08:28:14 +00:00 Commented Feb 2, 2015 at 8:28 Why do you ask? Do you want a fast way to compute it?Lehs –Lehs 2015-02-02 10:13:40 +00:00 Commented Feb 2, 2015 at 10:13 I was thinking something with log2User –User 2015-02-02 15:37:32 +00:00 Commented Feb 2, 2015 at 15:37 Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 2 Save this answer. Show activity on this post. The expression ⌊n−−√⌋⌊n⌋ represents the largest integer m m such that m 2≤n m 2≤n. There is no commonly known, specific symbol or notation to describe such a number, other than ⌊n−−√⌋⌊n⌋. That said, we could define such a number in advance; e.g., "For a positive integer n n, let f(n)f(n) be the largest integer such that f(n)2≤n f(n)2≤n." Then n=f(n)2+r n=f(n)2+r, for some r∈{0,1,2,…,max(0,n−1)}r∈{0,1,2,…,max(0,n−1)}. This specifies the decomposition of n n into the sum of its largest integer square and a "remainder" term r r. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Feb 2, 2015 at 9:27 answered Feb 2, 2015 at 8:31 heropupheropup 145k 15 15 gold badges 114 114 silver badges 201 201 bronze badges 6 or "let f(n)f(n) be given by (f(n))2≤n<(f(n)+1)2(f(n))2≤n<(f(n)+1)2." user57159 –user57159 2015-02-02 08:47:16 +00:00 Commented Feb 2, 2015 at 8:47 Also, that decomposition doesn't work for n=0 n=0. user57159 –user57159 2015-02-02 08:47:46 +00:00 Commented Feb 2, 2015 at 8:47 @RickyDemer f(0)=0 f(0)=0, with 0=0 2+0 0=0 2+0, and ⌊0–√⌋=0⌊0⌋=0. I don't see what fails in the decomposition, and even if it did fail, I did quite clearly specify "For a positive integer n n,..."heropup –heropup 2015-02-02 08:53:42 +00:00 Commented Feb 2, 2015 at 8:53 What would fail in the decomposition is that 0,1,2,…,n−1 0,1,2,…,n−1 would be the empty range for n=0 n=0. However, your positivity specification does handle that. user57159 –user57159 2015-02-02 09:05:23 +00:00 Commented Feb 2, 2015 at 9:05 So are you suggesting that the remainder of 0 0 when divided by some positive integer n n is undefined? I find your objection pedantic. I will change the post accordingly.heropup –heropup 2015-02-02 09:26:29 +00:00 Commented Feb 2, 2015 at 9:26 \|Show 1 more comment You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions ceiling-and-floor-functions See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Related 37How do the floor and ceiling functions work on negative numbers? 2Prove property of floor function (one with square roots) 3Square root of floor of square root of sum ⌊\|x+y\|−−−−−−√⌋−−−−−−−−−√⌊\|x+y\|⌋ 0Question about floor function ⌊n x⌋⌊n x⌋ and what it equals 0If you take any natural number greater than three, take the square root and round it off ... 1How to Explicitly Round to a Given Number of Significant Digits 7What is the integral of the floor function and any power of the floor function? 3Is the floor function defined for complex values? 1Floor and square root function equation : ⌊x−−√⌋=⌊x⌋−−−√⌊x⌋=⌊x⌋ Hot Network Questions Repetition is the mother of learning Vampires defend Earth from Aliens Lingering odor presumably from bad chicken Making sense of perturbation theory in many-body physics Bypassing C64's PETSCII to screen code mapping Weird utility function Is direct sum of finite spectra cancellative? 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15732	https://resources.pcb.cadence.com/blog/2020-ac-peak-voltage-vs-peak-to-peak-voltage-vs-rms-voltage	Skip to main content AC Peak Voltage vs. Peak-to-Peak Voltage vs. RMS Voltage Published Date Author Cadence PCB Solutions Updated for 2022. Key Takeaways Learn how to calculate AC peak voltage. Gain a greater understanding of the importance of AC peak voltage to overall circuit design. Learn how to differentiate between AC peak voltage, peak-to-peak voltage, and RMS voltage. Checking the AC peak voltage at a distribution substation. The credit for discovering the electrical charge goes to the Greeks, and this discovery dates back to 2600 years ago. An electrical charge is also called static electricity or inert electricity. For several millennia, humans have possessed an unquenchable fascination with lightning and electricity. From Benjamin Franklin's kite experiment in 1752 to Volta's invention of the battery in 1800 to the electric light bulb invention in 1879 by Thomas Edison, this fascination is undeniable. Fast-forward to present day and there is still a desire to understand, harness, and efficiently use electricity. This is understandable since nearly every device we utilize relies on some form of power or electrical charge. However, not all sources of electrical energy are compatible with every device design. As our knowledge and understanding of electricity has increased, so have our power requirements. With this in mind, our designs rely on our ability to assess both electrical limits and power requirements accurately. This includes being able to calculate parameters like AC peak voltage. RMS voltage and RMS value or voltage value are representations of average power, average value, and peak value for AC voltage or AC waveform power spectrums. Additional voltage elements like sinusoidal voltage, calculating for instantaneous values or effective value toward the RMS measurement, and maximum value can all help in understanding true RMS for your next AC circuit or AC power supply. What Is Voltage? Voltage is the electric potential within a circuit that provides the potential for current to flow. But, the mere presence of voltage within a circuit does not indicate that there is current present in the circuit. For current to flow within a circuit, the circuit must be complete (closed path). Therefore, voltage provides the potential for current to be present in a circuit, but current only flows if there is a completed or closed path. This is because voltage provides the force which pushes or moves electrons within the circuit when the path is completed. For example, an electrical outlet does not have current flow if there is no device plugged in. However, the electric potential or voltage is still present. Once we plug in a device and complete the circuit by turning it on, the voltage is active and there is current flow in the device. AC Peak Voltage Every electronic device requires a power source that is fully compatible with its design. Some devices utilize DC, whereas other devices use AC. There are also devices, such as personal computers, that utilize DC converted from AC electrical outlets. In every case, there are parameters to which this electrical power source must adhere to for the device to function. The need to differentiate between maximum or peak voltage and RMS (root-mean-squared) or average voltage is paramount to both design and functionality. One such parameter is the AC peak voltage. As you might imagine, utilizing a voltage that exceeds a device's design will undoubtedly lead to catastrophic failure and possibly injury or death. So, what is AC peak voltage? As the name implies, AC peak voltage is the maximum or peak voltage the source can or will achieve. Peak voltage, which we designate as VP, is measured from the horizontal axis (at 0 reference height) to the top of the waveform or crest. AC Peak Voltage vs. AC Peak-to-Peak Voltage Keep in mind that AC stands for alternating current, and this also means that the voltage alternates (changes polarity) a set number of times in a given period. Take, for example, a 60Hz, 120 volt AC signal: The 60Hz designation indicates that the signal will alternate from negative (peak) to positive (peak) voltage 60 times within one second. This particular voltage parameter is called peak-to-peak or VPP, and it is not interchangeable with peak voltage. Understandably, these parameters affect the application that a particular voltage is compatible with and, thus, affect overall functionality. Calculating the AC peak voltage, peak-to-peak voltage, and RMS voltage is critical. Calculating AC Peak Voltage We can calculate peak voltage (VP) by using the peak-to-peak voltage (VPP), RMS voltage, or the average voltage. The formulas for calculating the VP for sinusoidal AC signals are as follows: If you acquire the peak-to-peak voltage (VPP) value, you can calculate the peak voltage (VP) by using the following formula: VP = VPP x 0.5 If you acquire the RMS voltage value, you can calculate the peak voltage (VP) by using the following formula: If you acquire the average voltage value, you can calculate the peak voltage by using the following formula: VP = average voltage x (π ÷ 2) or VP = average voltage x 1.57 AC peak voltage, like a myriad of other parameters we find in the field of electronics, is beneficial to overall design and functionality. The understanding of an electrical source's maximum, average, and minimum potential greatly enhances your design accuracy, functionality, and device performance. An AC peak voltage that exceeds the power strips capacity will lead to catastrophic failure. As the image above illustrates, when designing electronic circuits, it is important to be able to accurately determine the parameters of your AC signals to ensure adequate protection is included. This allows you to prevent contingencies that may damage equipment or threaten users. While this article pertains mostly to AC current, don't neglect DC circuit, DC signal, or DC power designs. You'll still need to understand power output to ensure security, whether that means looking into RMS AC, peak voltage value, or any additional alternating waveform elements. Cadence’s PCB Design and Analysis software leads the industry by providing comprehensive and integrated schematic capture, board layout and simulation and analysis capabilities. For simulation, which includes the ability to view various representations of AC waveforms, PSpice is the standard. If you’re looking to learn more about how Cadence has the solution for you, talk to us and our team of experts.
15733	https://uk.ixl.com/maths/year-6/percent-of-a-number	IXL - Percent of a number (Year 6 maths practice) IXL uses cookies to ensure that you get the best experience on our website. See our privacy policy to learn more. [x] - [x] IXL Learning Sign in- [x] Remember Sign in now Join now IXL Learning Learning Maths Skills Games English Skills Games Science Recommendations Recommendations wall Skill plans National curriculum Additional curricula Test prep Awards Student awards Diagnostic Analytics Learning All Learning Maths English Science Recommendations Skill plans Learning Skill plans National curriculum Additional curricula Test prep Awards Diagnostic Analytics Membership Sign in Maths Maths English English Science Science Recommendations Recommendations Skill plans Skill plans National curriculum Additional curricula Test prep Awards Awards Year 6 T.8 Percent of a number F2P Share skill Copy the link to this skill share to facebook share to twitter Time to get in the zone! Your teacher would like you to focus on skills in . Let's pick a skill from these categories. Let's go! Incomplete answer You did not finish the question. Do you want to go back to the question? Go back Submit Learn with an example What is 80% of 55? Submit reference_doc_toggle_button.back_to_practice ref_doc_title. reference_doc_toggle_button.back_to_practice Learn with an example Learn with an example question What is 50% of 58? key idea A percent is 1 part out of 100. 1% is equal to 1 100 or 0.01. Proportions can be used to solve percent problems. part whole=percent 100 solution The whole is 58. The percent is 50%. Let n represent the part. Write a proportion to solve for n. n 58=50 100 100 n=50 · 58 100 n=2,900 100 n ÷ 100=2,900 ÷ 100 n=29 50% of 58 is 29. You could also find the answer by writing and solving an equation: part=percentage of whole n=50% · 58 n=0.5 · 58 n=29 50% of 58 is 29. Learn with an example Excellent! You got that right! Continue Learn with an example Jumping to level 1 of 1 Excellent! Now entering the Challenge Zone—are you ready? Questions answered Questions 0 Time elapsed Time 00 00 02 hr min sec SmartScore out of 100 IXL's SmartScore is a dynamic measure of progress towards mastery, rather than a percentage grade. It tracks your skill level as you tackle progressively more difficult questions. Consistently answer questions correctly to reach excellence (90), or conquer the Challenge Zone to achieve mastery (100)! Learn more 0 You met your goal! Need a break? Work it out Not feeling ready yet? These can help:T.4 Convert between percents, fractions and decimalsT.4 Convert between percents, fractions and decimals - Year 6 V7PJ.8 Multiply a decimal by a one-digit whole numberJ.8 Multiply a decimal by a one-digit whole number - Year 6 9HVN.8 Multiply fractions by whole numbersN.8 Multiply fractions by whole numbers - Year 6 5HTX.12 Solve one-step equations with whole numbersX.12 Solve one-step equations with whole numbers - Year 6 H79 Company \| Membership \| Blog \| Help centre \| User guides \| Tell us what you think \| Testimonials \| Careers \| Contact us \| Terms of service \| Privacy policy © 2025 IXL Learning. All rights reserved. Follow us First time here? 17 million children and teens use IXL for academic help and enrichment. Reception to Year 13 Sign up nowKeep exploring
15734	https://www.quora.com/Number-Theory-Let-gcd-a-b-d-Then-a-dx-b-dy-Prove-gcd-x+y-xy-1	Number Theory: Let gcd (a,b) =d. Then a=dx, b=dy Prove gcd (x+y, xy) =1? - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Mathematics Greatest Common Factor ( ... Logical Proofs GCD Calculate Divisors (math) Greatest Common Divisor Number Theory Mathematical Proof Mathematical Sciences 5 Number Theory: Let gcd (a,b) =d. Then a=dx, b=dy Prove gcd (x+y, xy) =1? All related (34) Sort Recommended Vardhan Thigle Author has 751 answers and 1M answer views ·Updated 11y Originally Answered: Let gcd (a,b) =d. Then a=dx, b=dy Prove gcd (x+y, xy) =1? · EDIT:: I had forgotten about Euclidean algorithm which some other answers on Number Theory reminded me of. Adding another answer based on that Answer 1 From the question, we can directly infer. Since x x and y y are the products of factors left uncommon between a and b. g c d(x,y)=1 g c d(x,y)=1 g c d(x+y,x)=g c d(x,((x+y)m o d x))=g c d(x,(y m o d x))=g c d(x,y)=1 g c d(x+y,x)=g c d(x,((x+y)m o d x))=g c d(x,(y m o d x))=g c d(x,y)=1 g c d(x+y,y)=g c d(y,((x+y)m o d y)=g c d(y,(y m o d x))=g c d(y,x)=1 g c d(x+y,y)=g c d(y,((x+y)m o d y)=g c d(y,(y m o d x))=g c d(y,x)=1 g c d(x+y,x y)=1 g c d(x+y,x y)=1 since g c d(x+y,x)=1 g c d(x+y,x)=1 and g c d(x+y,y)=1 g c d(x+y,y)=1 Answer 2 The below answer's scope is restricted to integers. From the question, we can directly infer. Since x x and y y Continue Reading EDIT:: I had forgotten about Euclidean algorithm which some other answers on Number Theory reminded me of. Adding another answer based on that Answer 1 From the question, we can directly infer. Since x x and y y are the products of factors left uncommon between a and b. g c d(x,y)=1 g c d(x,y)=1 g c d(x+y,x)=g c d(x,((x+y)m o d x))=g c d(x,(y m o d x))=g c d(x,y)=1 g c d(x+y,x)=g c d(x,((x+y)m o d x))=g c d(x,(y m o d x))=g c d(x,y)=1 g c d(x+y,y)=g c d(y,((x+y)m o d y)=g c d(y,(y m o d x))=g c d(y,x)=1 g c d(x+y,y)=g c d(y,((x+y)m o d y)=g c d(y,(y m o d x))=g c d(y,x)=1 g c d(x+y,x y)=1 g c d(x+y,x y)=1 since g c d(x+y,x)=1 g c d(x+y,x)=1 and g c d(x+y,y)=1 g c d(x+y,y)=1 Answer 2 The below answer's scope is restricted to integers. From the question, we can directly infer. Since x x and y y are the products of factors left uncommon between a and b. g c d(x,y)=1 g c d(x,y)=1 Let x=Π P e i i x=Π P i e i Let y=Π P e k k y=Π P k e k, where P i P i and P k P k are the prime factors of x x and y y respectively Since g c d(x,y)=1 g c d(x,y)=1, P i≠P k P i≠P k for any i,k i,k. So x y=(Π P e i i)(Π P e k k)x y=(Π P i e i)(Π P k e k) Since P i≠P k P i≠P k , For any i i, x+y≡y(m o d P i)x+y≡y(m o d P i), y\nequiv 0(m o d P i)y\nequiv 0(m o d P i) For any k k, x+y≡x(m o d P k)x+y≡x(m o d P k), x\nequiv 0(m o d P k)x\nequiv 0(m o d P k) So no prime factor of x y x y is a factor of x+y x+y. Hence g c d(x+y,x y)=1 g c d(x+y,x y)=1 Upvote · 9 3 Promoted by The Hartford The Hartford We help protect over 1 million small businesses ·Updated Sep 19 What is small business insurance? Small business insurance is a comprehensive type of coverage designed to help protect small businesses from various risks and liabilities. It encompasses a range of policies based on the different aspects of a business’s operations, allowing owners to focus on growth and success. The primary purpose of small business insurance is to help safeguard a business’s financial health. It acts as a safety net, helping to mitigate financial losses that could arise from the unexpected, such as property damage, lawsuits, or employee injuries. For small business owners, it’s important for recovering quickl Continue Reading Small business insurance is a comprehensive type of coverage designed to help protect small businesses from various risks and liabilities. It encompasses a range of policies based on the different aspects of a business’s operations, allowing owners to focus on growth and success. The primary purpose of small business insurance is to help safeguard a business’s financial health. It acts as a safety net, helping to mitigate financial losses that could arise from the unexpected, such as property damage, lawsuits, or employee injuries. For small business owners, it’s important for recovering quickly and maintaining operations. Choosing the right insurance for your small business involves assessing your unique needs and consulting with an advisor to pick from comprehensive policy options. With over 200 years of experience and more than 1 million small business owners served, The Hartford is dedicated to providing personalized solutions that help you focus on growth and success. Learn about our coverage options! Upvote · 999 557 9 1 9 3 Related questions More answers below If GCD (a,b) = xa+yb, then how can I find x and y? How do I finish this proof in proving that gcd(a,b)=1,gcd(a,c)=1⟹gcd(a,b c)=1 gcd(a,b)=1,gcd(a,c)=1⟹gcd(a,b c)=1? How do you prove rove$\gcd (a_1,\ldots, a_m) \gcd (b_1,\ldots, b_n) =\gcd (\text {all products $a_ib_j$}) $ (elementary number theory, gcd and LCM, math)? How can I prove that : if gcd(a b) \|gcd(b c) and c\|a then gcd(a b) = gcd(b c)? Thanks How do I prove that gcd (a, b) = gcd (ab, LCM (a, b))? Amitabha Tripathi five decades of high school Algebra · Author has 4.7K answers and 13.9M answer views ·8y Any prime divisor p p of both x x and y y would mean that d p d p divides both a a and b b. This would contradict gcd(a,b)=d gcd(a,b)=d. Hence no prime divides both x x and y y, so that gcd(x,y)=1 gcd(x,y)=1. Suppose p p is a prime divisor of x y x y. Then p p divides either x x or y y, but not both. Hence p p does not divide x+y x+y. So no prime divisor of x y x y can divide x+y x+y. Therefore gcd(x+y,x y)=1 gcd(x+y,x y)=1. Upvote · 9 2 Drake Way Mathematics Hobbyist · Author has 4.2K answers and 3.3M answer views ·11mo There exists sdx+tdy = d by Bezout’s thereom. Therefore sx+ty = 1, therefore gcd(x,y) = 1. Because x and y are coprime, no divisor of x can divide x+y, nor can one of y. Because x and y share no divisors, all the divisors of xy cannot divide x+y. Therefore, gcd(x+y,xy)=1. Upvote · Promoted by Coverage.com Johnny M Master's Degree from Harvard University (Graduated 2011) ·Updated Sep 9 Does switching car insurance really save you money, or is that just marketing hype? This is one of those things that I didn’t expect to be worthwhile, but it was. You actually can save a solid chunk of money—if you use the right tool like this one. I ended up saving over $1,500/year, but I also insure four cars. I tested several comparison tools and while some of them ended up spamming me with junk, there were a couple like Coverage.com and these alternatives that I now recommend to my friend. Most insurance companies quietly raise your rate year after year. Nothing major, just enough that you don’t notice. They’re banking on you not shopping around—and to be honest, I didn’t. Continue Reading This is one of those things that I didn’t expect to be worthwhile, but it was. You actually can save a solid chunk of money—if you use the right tool like this one. I ended up saving over $1,500/year, but I also insure four cars. I tested several comparison tools and while some of them ended up spamming me with junk, there were a couple like Coverage.com and these alternatives that I now recommend to my friend. Most insurance companies quietly raise your rate year after year. Nothing major, just enough that you don’t notice. They’re banking on you not shopping around—and to be honest, I didn’t. It always sounded like a hassle. Dozens of tabs, endless forms, phone calls I didn’t want to take. But recently I decided to check so I used this quote tool, which compares everything in one place. It took maybe 2 minutes, tops. I just answered a few questions and it pulled up offers from multiple big-name providers, side by side. Prices, coverage details, even customer reviews—all laid out in a way that made the choice pretty obvious. They claimed I could save over $1,000 per year. I ended up exceeding that number and I cut my monthly premium by over $100. That’s over $1200 a year. For the exact same coverage. No phone tag. No junk emails. Just a better deal in less time than it takes to make coffee. Here’s the link to two comparison sites - the one I used and an alternative that I also tested. If it’s been a while since you’ve checked your rate, do it. You might be surprised at how much you’re overpaying. Upvote · 999 485 999 103 99 17 Ram Kushwah Up and coming Most viewed writer · Author has 6.8K answers and 15.7M answer views ·5y Related Let a,b a,b be positive integers such that a 2+b 2 1+a b a 2+b 2 1+a b is an integer. Let d=g c d(a,b)d=g c d(a,b). How do you prove that a 2+b 2 1+a b=d 2 a 2+b 2 1+a b=d 2? The Solution is at... Continue Reading The Solution is at... Upvote · 9 5 Assistant Bot · 1y To prove that gcd(x+y,x y)=1 gcd(x+y,x y)=1 given gcd(a,b)=d gcd(a,b)=d and a=d x a=d x, b=d y b=d y, we can proceed as follows: Step 1: Understand the given conditions Given: gcd(a,b)=d gcd(a,b)=d a=d x a=d x b=d y b=d y From the definition of the greatest common divisor, we know that x x and y y must be coprime (i.e., gcd(x,y)=1 gcd(x,y)=1). This is because d d is the greatest common divisor of a a and b b, and any common factor of x x and y y would also be a common factor of a a and b b. Step 2: Prove gcd(x+y,x y)=1 gcd(x+y,x y)=1 To show that gcd(x+y,x y)=1 gcd(x+y,x y)=1, we need to prove that x+y x+y and x y x y share no common factors other than 1. Assume there exists a common divisor Continue Reading To prove that gcd(x+y,x y)=1 gcd(x+y,x y)=1 given gcd(a,b)=d gcd(a,b)=d and a=d x a=d x, b=d y b=d y, we can proceed as follows: Step 1: Understand the given conditions Given: gcd(a,b)=d gcd(a,b)=d a=d x a=d x b=d y b=d y From the definition of the greatest common divisor, we know that x x and y y must be coprime (i.e., gcd(x,y)=1 gcd(x,y)=1). This is because d d is the greatest common divisor of a a and b b, and any common factor of x x and y y would also be a common factor of a a and b b. Step 2: Prove gcd(x+y,x y)=1 gcd(x+y,x y)=1 To show that gcd(x+y,x y)=1 gcd(x+y,x y)=1, we need to prove that x+y x+y and x y x y share no common factors other than 1. Assume there exists a common divisor k k Let k=gcd(x+y,x y)k=gcd(x+y,x y). Thus, k k divides both x+y x+y and x y x y. Analyze k k dividing x y x y Since k k divides x y x y, we can express x y x y as: x y=k⋅m for some integer m.x y=k⋅m for some integer m. Analyze k k dividing x+y x+y Since k k divides x+y x+y, we can express x+y x+y as: x+y=k⋅n for some integer n.x+y=k⋅n for some integer n. Consider x x and y y modulo k k From the equation x+y≡0 mod k x+y≡0 mod k, we have: y≡−x mod k.y≡−x mod k. Substituting y y in x y x y: x y=x(−x)=−x 2≡0 mod k.x y=x(−x)=−x 2≡0 mod k. This implies that k k divides x 2 x 2. Since k k divides x y x y and y≡−x mod k y≡−x mod k, we can substitute y y back into the equation: k∣x(−x)⇒k∣−x 2.k∣x(−x)⇒k∣−x 2. Conclude with coprimality Now, since k k divides both x+y x+y and x y x y, we also have: k k divides x+y x+y k k divides x 2 x 2 If k k divides x x, then it must also divide y y since y≡−x mod k y≡−x mod k. This contradicts our assumption that gcd(x,y)=1 gcd(x,y)=1, which implies that the only common divisor can be 1. Thus, the only possibility left is that k=1 k=1, indicating that: gcd(x+y,x y)=1.gcd(x+y,x y)=1. Conclusion Therefore, we conclude that: gcd(x+y,x y)=1.gcd(x+y,x y)=1. Upvote · Related questions More answers below How do I prove if gcd (a, b) = 1 then gcd (a+b, b) = 1? If gcd (a, b)=1 and gcd (c, d)=1 and also a/b=c/d, then is it true that a=c and b=d? How do I prove if gcd (a, b) =gcd (a, c) = d, then gcd(c,b) =1? How do you prove if gcd(a,b) =1, then gCD (a^n, b^n) =1? How do you prove that if gcd(a,b)=1 gcd(a,b)=1 then gcd(a c,b)=gcd(c,b)gcd(a c,b)=gcd(c,b) (elementary number theory, gcd and lcm, math)? Brad Ballinger Math teacher · Author has 344 answers and 350.2K answer views ·3y Related How do I prove GCD(a,b) = 1, ab\|c iff a\|c and b\|c? First, I recommend taking the time to write out the claim properly. The way you've got it written now might be taken to mean that GCD(a,b) = 1 and ab\|c iff a\|c and b\|c. That's incorrect, as shown by the example a=b=2, c=4. How would I write it? Like this: Suppose GCD(a,b) = 1. How do I prove ab\|c iff a\|c and b\|c? The point here is to clearly get the GCD assumption out of the iff. Now we continue. It's a little difficult to say how you can prove it without knowing what facts you have at your disposal, but one useful fact here is that ab=GCD(a,b)LCM(a,b). Since we are given that GCD(a,b)=1, this im Continue Reading First, I recommend taking the time to write out the claim properly. The way you've got it written now might be taken to mean that GCD(a,b) = 1 and ab\|c iff a\|c and b\|c. That's incorrect, as shown by the example a=b=2, c=4. How would I write it? Like this: Suppose GCD(a,b) = 1. How do I prove ab\|c iff a\|c and b\|c? The point here is to clearly get the GCD assumption out of the iff. Now we continue. It's a little difficult to say how you can prove it without knowing what facts you have at your disposal, but one useful fact here is that ab=GCD(a,b)LCM(a,b). Since we are given that GCD(a,b)=1, this implies ab=LCM(a,b). This can help with both directions of your iff, particularly if you know that every common multiple of a and b is also a multiple of their LCM. Upvote · 9 2 Sponsored by Amazon Business Solutions and supplies to support learning. Save on essentials and reinvest in students and staff. Sign Up 99 85 Manan Shah AIME Qualifier, Math Kangaroo National Champion ·1y Related If a and b are relatively prime, then gcd (a, a+b) = 1. Why? Proof by contradiction: Let’s suppose that gcd (a, a+b) > 1. Necessarily, there exists an integer n such that n > 1, a is divisible by n, and (a+b) is divisible by n. Therefore a = (nx) for some natural number x, and (a+b) = (ny) for some natural number y. Therefore, (a+b)-a = (ny)-(nx). This simplifies to b = n(y-x). Since we know that y and x are each natural numbers, it follows that (y-x) is an integer. Therefore, b is divisible by n. We have now shown that a and b are each divisible by n, so a and b are not relatively prime. This contradicts our initial assumption. Therefore, gcd (a, a+b Continue Reading Proof by contradiction: Let’s suppose that gcd (a, a+b) > 1. Necessarily, there exists an integer n such that n > 1, a is divisible by n, and (a+b) is divisible by n. Therefore a = (nx) for some natural number x, and (a+b) = (ny) for some natural number y. Therefore, (a+b)-a = (ny)-(nx). This simplifies to b = n(y-x). Since we know that y and x are each natural numbers, it follows that (y-x) is an integer. Therefore, b is divisible by n. We have now shown that a and b are each divisible by n, so a and b are not relatively prime. This contradicts our initial assumption. Therefore, gcd (a, a+b) = 1. Upvote · 9 2 Alex Eustis Ph.D. in Mathematics, University of California, San Diego (Graduated 2013) · Author has 4.6K answers and 23.8M answer views ·1y Related If gcd (d, n) = 1, how do you prove ad ≡ bd mod n if and only if a ≡ b mod n? If gcd(d,n)=1 gcd(d,n)=1, then d d has a multiplicative inverse in the ring of integers mod n. So, given a d=b d a d=b d in Z/n Z Z/n Z, just multiply both sides by d−1 d−1 and you’re done. How do we know this multiplicative inverse exists? Well, this follows from the Euclidean Algorithm which allows you to write gcd(a,b)gcd(a,b) in the form a x+b y a x+b y for some integers x,y x,y. In particular, if gcd(d,n)=1 gcd(d,n)=1 then we have d x+n y=1 d x+n y=1 which means that d x≡1 mod n d x≡1 mod n, i.e. x x is the multiplicative inverse of d d mod n n. Upvote · 9 6 Enrico Gregorio Associate professor in Algebra · Author has 18.4K answers and 16M answer views ·4y Related What is the gcd of (x+y, 3xy) if the gcd of (x,y) = 1? Suppose a prime p p divides x+y x+y and 3 x y 3 x y. There are three possibilities: p=3 p=3 p∣x p∣x p∣y p∣y By symmetry, we can look just one of the last two options. In case p∣x p∣x, then p p does not divide y y, so it does not divide x+y x+y: contradiction. So the only possibility is that p=3 p=3. Now, suppose 9 9 divides x+y x+y and 3 x y 3 x y: then 3 3 has to divide x x or y y, but then we get a contradiction like in the other case. Hence gcd(x+y,3 x y)gcd(x+y,3 x y) can only be 1 1 or 3 3. The latter happens when x+y x+y is divisible by 3 3, for instance when x=1,y=2 x=1,y=2. Otherwise the gcd is 1 1. Upvote · 9 3 Angelos Tsirimokos M.A. in Mathematics, Harvard University (Graduated 1976) · Author has 1.9K answers and 2.7M answer views ·Updated 4y Related Let d = gcd (a, b) = (a, b) and suppose that x and y are integers such that xa + yb = m. (a) If m = 1, show that d = 1. (b) Suppose that xa + yb = 15. What are the possible values of (a, b)? Since d divides a and b, it will also divide xa and yb, hence also xa+yb=m. If m=1, then d divides 1; what else could it be but 1? If m=15, then d divides 15. So it can only be 1, 3, 5 or 15. And all those possibilities can actually occur: (7, 8)=1 and 17+18=15 (6, 9)=3 and 16+19=15 (5, 10)=5, and 15+110=15 (30, 45)=15 and 230–145=15. And for any pair of integers (a, b) whose gcd is 1, 3, 5 or 15 integers x and y (not necessarily positive) can be found such that xa+yb = 15. On the other hand, if the gcd is not 1, 3, 5 or 15, then no such integers can be found, since the gcd would have to div Continue Reading Since d divides a and b, it will also divide xa and yb, hence also xa+yb=m. If m=1, then d divides 1; what else could it be but 1? If m=15, then d divides 15. So it can only be 1, 3, 5 or 15. And all those possibilities can actually occur: (7, 8)=1 and 17+18=15 (6, 9)=3 and 16+19=15 (5, 10)=5, and 15+110=15 (30, 45)=15 and 230–145=15. And for any pair of integers (a, b) whose gcd is 1, 3, 5 or 15 integers x and y (not necessarily positive) can be found such that xa+yb = 15. On the other hand, if the gcd is not 1, 3, 5 or 15, then no such integers can be found, since the gcd would have to divide 15. Upvote · 9 2 Daniel Claydon Learning mathematics · Upvoted by Nathan Hannon , Ph. D. Mathematics, University of California, Davis (2021) · Author has 779 answers and 4.3M answer views ·6y Related How does one prove that if a x+b y=gcd(a,b)a x+b y=gcd(a,b) then gcd(x,y)=1 gcd(x,y)=1? Write a=d A a=d A and b=d B b=d B, where d=gcd(a,b)d=gcd(a,b). Then your equation simplifies to A x+B y=1 A x+B y=1. If some number h>1 h>1 divides x x and y y then it must also divide the left side, so it divides the right side too, which is 1 1. That's a contradiction, so x x and y y have no non-unit common divisor, which is to say gcd(x,y)=1 gcd(x,y)=1. Upvote · 99 20 9 1 Alon Amit PhD in Mathematics; Mathcircler. · Upvoted by Jeremy Collins , M.A. Mathematics, Trinity College, Cambridge and Aditya Garg , M.Sc. Mathematics, Indian Institute of Technology, Delhi (2013) · Author has 8.8K answers and 173.8M answer views ·5y Related Let a,b a,b be positive integers such that a 2+b 2 1+a b a 2+b 2 1+a b is an integer. Let d=g c d(a,b)d=g c d(a,b). How do you prove that a 2+b 2 1+a b=d 2 a 2+b 2 1+a b=d 2? This is a slight rephrasing of a famous IMO problem. See Alon Amit's answer to Letaandbbe positive integers such thatab+1dividesa^2+b^2. Why must\frac{a^2+b^2}{ab+1}be a perfect square? Your response is private Was this worth your time? This helps us sort answers on the page. Absolutely not Definitely yes Upvote · 99 27 9 4 Gabriele Scopel Bachelors Degree in Mathematics, Tor Vergata University of Rome (Graduated 2025) · Author has 628 answers and 185.6K answer views ·2y Related Let d = gcd (a, b) = (a, b) and suppose that x and y are integers such that xa + yb = m. (a) If m = 1, show that d = 1. (b) Suppose that xa + yb = 15. What are the possible values of (a, b)? Suppose a=k⋅d a=k⋅d and b=h⋅d b=h⋅d, where k k and h h are integers. We can always write a a and b b this way since both of them are multiples of d d. Thus m=x a+y b=x(k d)+y(h d)=d(x k+y h)m=x a+y b=x(k d)+y(h d)=d(x k+y h). This means that m m is a multiple of d d, because x k+y h x k+y h is an integer. Thus, if m=1 m=1, then d=1 d=1 is the only possible diviWith sor (also −1−1 is, but the gcd gcd is defined to be positive) With the same reasoning, d d must be a positive divisor of 15 15. This means that d d can be 1,3,5,15 1,3,5,15 Upvote · 9 1 Related questions If GCD (a,b) = xa+yb, then how can I find x and y? How do I finish this proof in proving that gcd(a,b)=1,gcd(a,c)=1⟹gcd(a,b c)=1 gcd(a,b)=1,gcd(a,c)=1⟹gcd(a,b c)=1? How do you prove rove$\gcd (a_1,\ldots, a_m) \gcd (b_1,\ldots, b_n) =\gcd (\text {all products $a_ib_j$}) $ (elementary number theory, gcd and LCM, math)? How can I prove that : if gcd(a b) \|gcd(b c) and c\|a then gcd(a b) = gcd(b c)? Thanks How do I prove that gcd (a, b) = gcd (ab, LCM (a, b))? How do I prove if gcd (a, b) = 1 then gcd (a+b, b) = 1? If gcd (a, b)=1 and gcd (c, d)=1 and also a/b=c/d, then is it true that a=c and b=d? How do I prove if gcd (a, b) =gcd (a, c) = d, then gcd(c,b) =1? How do you prove if gcd(a,b) =1, then gCD (a^n, b^n) =1? How do you prove that if gcd(a,b)=1 gcd(a,b)=1 then gcd(a c,b)=gcd(c,b)gcd(a c,b)=gcd(c,b) (elementary number theory, gcd and lcm, math)? How can I prove that if a=b then gcd(a,b) =LCM(a,b)? How do we prove that gcd (a, a+1) = 1? And find integers x and y such that nx+(n+1) y = 1? Let a a and b b be positive integers such that a b+1 a b+1 divides a 2+b 2 a 2+b 2. Why must a 2+b 2 a b+1 a 2+b 2 a b+1 be a perfect square? If gcd(a,m) =gcd(b,m)=1, then how do you prove that GCD (ab, m) = 1? How do I prove that gcd (a, b, c) = gcd (a, gcd (b, c))? Note that if you want to use any facts about gcd (a, b, c) beyond the definition, you will need to prove them. Related questions If GCD (a,b) = xa+yb, then how can I find x and y? How do I finish this proof in proving that gcd(a,b)=1,gcd(a,c)=1⟹gcd(a,b c)=1 gcd(a,b)=1,gcd(a,c)=1⟹gcd(a,b c)=1? How do you prove rove$\gcd (a_1,\ldots, a_m) \gcd (b_1,\ldots, b_n) =\gcd (\text {all products $a_ib_j$}) $ (elementary number theory, gcd and LCM, math)? How can I prove that : if gcd(a b) \|gcd(b c) and c\|a then gcd(a b) = gcd(b c)? Thanks How do I prove that gcd (a, b) = gcd (ab, LCM (a, b))? How do I prove if gcd (a, b) = 1 then gcd (a+b, b) = 1? Advertisement About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
15735	https://en.wikipedia.org/wiki/Barber%27s_pole	Jump to content Barber's pole العربية 閩南語 / Bn-lm-gí Català Deutsch Español Français 한국어 Italiano עברית Nederlands 日本語 Türkçe Tiếng Việt 中文 Edit links From Wikipedia, the free encyclopedia Type of sign A barber's pole is a type of sign used by barbers to signify the place or shop where they perform their craft. The trade sign is, by a tradition dating back to the Middle Ages, a staff or pole with a helix of colored stripes (often red and white in many countries, but usually red, white and blue in Canada, Japan, the Philippines, South Korea, Vietnam, Hungary, and the United States). The pole may be stationary or may rotate, often with the aid of an electric motor. The consistent use of this advertising symbol can be seen as analogous to an apothecary's show globe, a tobacconist's cigar store Indian and a pawn broker's three gold balls. A "barber's pole" with a helical stripe is a familiar sight, and is used as a secondary metaphor to describe objects in many other contexts. For example, if the shaft or tower of a lighthouse has been painted with a helical stripe as a daymark, the lighthouse could be described as having been painted in "barber's pole" colors. Origin in barbering and surgery [edit] During the Middle Ages, barbers performed surgery on customers, as well as tooth extractions. The original pole had a brassen wash basin at the top (representing the vessel in which leeches were kept) and bottom (representing the basin that received the blood). The pole itself represents the staff that the patient gripped during the procedure to encourage blood flow, and the twined pole motif is likely related to the Caduceus, the staff of the Greek god of speed and commerce Hermes, evidenced for example by early physician van Helmont's description of himself as "Francis Mercurius Van Helmont, A Philosopher by that one in whom are all things, A Wandering Hermite." Under Pope Innocent II edicts were given forth (Council of Clermont 1130, of Rheims 1131, Second Council of the Lateran 1139) against medical practice by ecclesiastics. At the Council of Tours in 1163, the Roman Catholic clergy was banned from the practice of surgery. From then, physicians were clearly separated from the surgeons and barbers. Later, the role of the barbers was defined by the College de Saint-Côme et Saint-Damien, established by Jean Pitard in Paris circa 1210, as academic surgeons of the long robe and barber surgeons of the short robe. In Renaissance-era Amsterdam, the surgeons used the colored stripes to indicate that they were prepared to bleed their patients (red), set bones or pull teeth (white), or give a shave if nothing more urgent was needed (blue). After the formation of the United Barber Surgeon's Company in England, a statute required the barber to use a red and white pole and the surgeon to use a red pole. In the Kingdom of France, surgeons used a red pole with a basin attached to identify their offices. Blue often appears on poles in the United States, possibly as a homage to its national colors. Another, more fanciful interpretation of these barber pole colors is that red represents arterial blood, blue is symbolic of venous blood, and white depicts the bandage. In any event, the barber pole became emblematic of the profession. Manufacture of barber poles [edit] Prior to 1950, there were four manufacturers of barber poles in the United States. In 1950, William Marvy of St. Paul, Minnesota, started manufacturing barber poles. Marvy made his 50,000th barber pole in 1967, and, by 2010, over 82,000 had been produced. The William Marvy Company is now the sole manufacturer of barber poles in North America, and sells only 500 per year (compared to 5,100 in the 1960s). In recent years, the sale of spinning barber poles has dropped considerably, since few barber shops are opening, and many jurisdictions prohibit moving signs. Koken of St. Louis, Missouri, manufactured barber equipment such as chairs and assorted poles in the 19th century. Use in barbering [edit] The barber pole used to indicate the practice of a "barber-surgeon" who also performed tooth extraction, cupping, leeching, bloodletting, enemas, amputations, etc. Today's barber poles represent only barbershops where hair is cut and beards are shaved. As early as 1905, use of the poles was reported to be "diminishing" in the United States. Barber poles have become a topic of controversy in the hairstyling business. In some American states, such as Michigan in March 2012, legislation has emerged proposing that barber poles should only be permitted outside barbershops, but not traditional beauty salons. Barbers and cosmetologists have engaged in several legal battles claiming the right to use the barber pole symbol to indicate to potential customers that the business offers haircutting services. Barbers claim that they are entitled to exclusive rights to use the barber pole because of the tradition tied to the craft, whereas cosmetologists think that they are equally capable of cutting men's hair (though many cosmetologists are not permitted to use razors, depending on their state's laws). Use in prostitution [edit] In South Korea, barber's poles are used both for actual barbershops and for brothels. Brothels disguised as barbershops, referred to as 이발소 (ibalso) or 미용실 (miyongsil), are more likely to use two poles next to each other, often spinning in opposite directions, though the use of a single pole for the same reason is also quite common. Actual barbershops, or 미용실 (miyongsil), are more likely to be hair salons; to avoid confusion, they will usually use a pole that shows a picture of a woman with flowing hair on it with the words hair salon written on the pole. Visual illusion [edit] See also: Barberpole illusion and Strange loop A spinning barber pole creates a visual illusion, in which the stripes appear to be traveling up or down the length of the pole, rather than around it. Other uses of the term [edit] Visual similarity [edit] The Swan portion of M17, the Omega Nebula in the Sagittarius nebulosity is said to resemble a barber's pole. Barber pole-like structures have been observed at the cellular level. The effects, origins and causes are controversial, and are subject to intense research. Matthew Walker's knot is a decorative knot said to vaguely resemble a section of a barber's pole.[A] Sinosauropteryx (meaning "Chinese reptilian wing", in Chinese 中华龙鸟: zhonghua longniao) is the first genus of non-avian dinosaur found with the fossilized impressions of feathers, as well as the first non-avian dinosaur where coloration has been determined. It lived in China during the early Cretaceous period and was a close relative of Compsognathus. It was the first non-avialan dinosaur genus discovered from the famous Jehol Biota of Liaoning Province. Zhang found "that the filaments running down its back and tail may have made the dinosaur look like an orange-and-white-striped barber pole. Such a vibrant pattern suggest that 'feathers first arose as agents for color display,' Benton says." Referential naming [edit] Animal husbandry [edit] Haemonchus contortus, or "barber's pole worm", is the parasitic nematode responsible for anemia, bottle jaw, and death of infected sheep and goats, mainly during summer months in warm, humid climates. Humans may become infected by the worms. Crustacea [edit] Stenopus hispidus is a shrimp-like popcorn kernel decapod crustacean sometimes called the "barber pole shrimp". See also Stenopodidea. Entomology [edit] In the insect world, there is the barber pole grasshopper, Dactylotum bicolor. It is also known as the "painted grasshopper" and is said to be the "most beautiful" grasshopper. Ichthyology [edit] Because of its bright bands and colors, the redbanded rockfish Sebastes babcocki is referred to as "barber pole". Other pseudonyms include bandit, convict, canary, Hollywood, and Spanish flag. Candy [edit] The old-fashioned American stick candy is sometimes also referred to as "barber pole candy" due to its colorful, swirled appearance. (See also candy cane.) "Candy stripe" is a generic description of the candy cane color scheme. Among many other names, the candy has been called Polkagris. Computer science [edit] In user interface design, a barber pole-like pattern is used in progress bars when the wait time is indefinite. It is intended to be used like a throbber to tell the user that processing is continuing, although it is not known when the processing will complete.[citation needed] Barber pole is also sometimes used to describe a text pattern where a line of text is rolled left or right one character on the line below. The CHARGEN service generates a form of this pattern. It is used to test RAM, hard disks and printers. A similar pattern is also used in secure erasure of media.((cn)) Electronics [edit] The strength and direction of magnetic fields and electric currents can be measured using a "magnetoresistive barber-pole sensor" (also called a "hermetic proximity sensor"), and its performance can be depicted using a mathematical formula. Such a sensor interleaves a series of permanent magnet strips with a series of magnetoresistive strips. The "conductive barberpole strips are canted across the sensor and connect one magnetoresistive strip, over a permanent magnet strip, to another magnetoresistive strip." This is said to provide a "uniform magnetic field throughout the sensor" thereby enhancing its resistance to external magnetic fields. The technology is used in wireless sensor networks which "have gathered a lot of attention as an important research domain" and were "deployed in many applications, e.g., navigation, military, ambient intelligence, medical, and industrial tasks. Context-based processing and services, in particular location-context, are of key interest ..." (See Music (acoustic illusion), infra.) Aviation and space flight [edit] The term on the barber pole or keep it on the barber pole is pilot jargon that refers to flying an aircraft at the maximum safe velocity. The airspeed Indicator on aircraft capable of flying at altitude features a red/white striped needle resembling a barber pole. This needle displays the VMO (Maximum Operating Velocity) or—at altitude—the MMO (Mach Limit Maximum Operating Speed) of the aircraft. This needle also indicates the maximum operating Mach number above the VMO/MMO changeover level. As the aircraft increases in altitude and the air decreases in density and temperature, the speed of sound also decreases. Close to the speed of sound, an aircraft becomes susceptible to buffeting caused by shock waves produced by flying at transonic speeds. Thus, as the speed of sound decreases, so the maximum safe operating speed of the aircraft is reduced. The "barber pole" needle moves to indicate this speed. Flying "on the barber pole" therefore means to be flying the aircraft as fast as is safe to do so in the current conditions. Barberpole is a phrase used to describe the striped output of indicators used during the Apollo and Shuttle programs. Typically an indicator was positioned below a switch. When the switch was activated and the activation indeed performed, the resulted activation was talked back via a separated electrical line to the barberpole indicator to show a grey and white striped pattern, thus verifying the action to the astronaut. Such switches with barberpole indicators were called talkback switches. Various indicators in the Apollo Command Modules indicated barberpole when the corresponding system was inactive. Astronaut Jim Lovell can also be found describing system indications as "barber poled" in the transcript of radio transmissions during the Apollo 13 accident. The phrase barberpole continues to be found in many subsystem descriptions in the Space Shuttle News Reference Manual, as well as the NASA/KSC Acronym List. During World War I and World War II, the pattern has also been used as an insignia for aircraft identification. Spad XIIIs of the 94th Aero Squadron USAS in early 1919 used variations on barber pole patterns, including: "Barber Pole" of Lieutenant Dudley "Red" Outcault; S.16546 "Flag Bus" of Captain Reed Chambers; and "Rising Sun" of Lieutenant John Jeffers. Flyfishing [edit] Used in flyfishing, Au Sable River guide Earl Madsen's "Madsen's Barber pole" is a traditional Michigan fly in the form of a "Stonefly" imitation "with grizzly hackle tip wings tied in a downwing fashion". Photo of Madsen's Barber Pole Fly, parachute form. Gambling [edit] The phrase barber pole is derisive jargon in craps, and refers to the commingling of "gaming cheques of different denominations". Wagers that combine different denominations are "supposed to be stacked with the highest denomination at the bottom". Parachuting [edit] The Screaming Eagles 101st Airborne Division (Air Assault) Command Parachute Demonstration Team, which operates out of Fort Campbell, Kentucky, executes a "barber pole maneuver" (also known as "the Baton Pass") during demonstrations.[unreliable source?] Two jumpers leave the aircraft and fly their bodies together to link while in free fall. "Once together they will then exchange a wooden baton ... [and] maneuver their bodies ... to create the illusion of a giant barber pole in the sky." Alternatively, a "Four Man Star" can "Hook Up" and then the formation rotates to the right, creating a "Barber Pole" effect with use of trailing smoke. Another parachuting use of the term describes having a mess of lines tangled "behind your head and you have to cut away your main chute and pull your reserve." Meteorology [edit] According to the National Oceanic and Atmospheric Administration, barber pole is a slang term used by weather and storm spotters to describe "a thunderstorm updraft with a visual appearance including cloud striations that are curved in a manner similar to the stripes of a barber pole. The structure typically is most pronounced on the leading edge of the updraft, while drier air from the rear flank downdraft often erodes the clouds on the trailing side of the updraft." See Supercell. Supercell/barber's pole photograph. A lynchpin of the NOAA National Hurricane Research Laboratory's hurricane-research fleet is the Lockheed WP-3D Orion (P-3). It has two barber-pole samplers (named for their red-and-white stripes) which protrude from the aircraft's front, a tail Doppler weather radar, and other unique-looking instruments hanging from the wing. Booksellers [edit] Red or rubric posts were sometimes used by booksellers in England prior to 1800. William Roberts reports in The Book Hunter in London that certain 18th-century bookshops in the Little Britain district of London sported such poles: A few years before Nichols published [in 1816] his Literary Anecdotes, two booksellers used to sport their rubric posts close to each other here in Little Britain, and these rubric posts were once as much the type of a bookseller's shop as the pole is of a barber's ... Sewell, Cornhill, and Kecket and De Hondt, Strand, were among the last to use these curious trade signs. Border and lane markers [edit] Among the Fortifications of the inner German border, 2622 barber pole-styled markers were placed about 500 feet (150 m) apart to demarcate the no-man's land between East Germany and West Germany.[citation needed] The 41 Combat Engineer Regiment, a part of the Canadian Military Engineers, produced and delivered over 16,000 distinctive barber pole lane markers during World War II.[citation needed] Canadian Naval group [edit] The famous Barber Pole Group was originally a group of 120 Flower-class corvettes built in Canada during World War II, and charged primarily with protecting freighter convoys. The original group was Escort Group C-3. This group of ships, with its red and white barber pole stripes painted on the funnel, is still represented in the current Royal Canadian Navy: all Atlantic fleet ships wear this insignia. HMCS Sackville is the last remaining Flower-class corvette. Daymarks as a navigational aid [edit] A barber pole motif has been used as a daymark and navigational aid for lighthouses. The White Shoal Light is the only "barber pole" lighthouse in the United States, and has been used in Michigan's "Save our Lights" license plate. However, black and white helical daymarks do appear on other lights, such as Cape Hatteras Light and St. Augustine Light. Barber pole channel markers are sometimes used as they are in the Tamaki River. Hockey [edit] In the 1896–97 season, the Ottawa Senators first adopted the "barber pole" design for their hockey jersey, with which the team became identified. The design featured strong horizontal stripes of red, black and white; white pants; and red, white and black striped stockings. This basic design would be used for the rest of the organization's existence, except for the 1909–1910 season. In that season, the stripes were vertical and Montreal fans nicknamed the team derisively as les suisses, a slang term for chipmunk. In the 1929–30 season, the club added the "O" logo to the chest of the jersey. The "barber-pole" uniform was later adopted by the Ottawa 67's junior hockey team. The National Hockey League's Montreal Canadiens had a barber pole or "barber shop" design jersey for the year 1912–1913.[unreliable source?] In the 1920s and 1930s, beginning in the 1927–28 season, the Senators, Boston Bruins, Montreal Maroons, Chicago Blackhawks, Detroit Cougars, and Toronto Maple Leafs had a barber's pole variation in their jerseys. Meanwhile, the New York Americans, wore "basically ... the United States flag as a jersey." The style endured, but in the 1938–39 season, the Blackhawks were the last to have a barber pole jersey in the traditional sense. The Hawks retired their barber pole at the end of the 1954–55 season. In junior ranks, the Chicoutimi Saguenéens and the Ottawa 67's use them in the Quebec Major Junior Hockey League (QMJHL) and Ontario Hockey League (OHL). During their existence in the Pacific Coast Hockey Association, the Seattle Metropolitans wore a red/white/green striped design: this has occasionally been brought back by the Seattle Thunderbirds of the Western Hockey League (WHL) to honor the history of hockey in the city. The style remained dormant until the National Hockey League's 75th anniversary, when Chicago wore replicas of their barber-pole sweaters as part of the league's celebrations. Since then, Montreal has also worn barber-pole replicas during their centenary season, and the design has become popular with amateur teams. See NHL uniform and Throwback uniform. Music (acoustic illusion) [edit] See also: Shepard tone, tritone paradox, and pitch circularity The Shepard tone has been described as "a sonic barber pole" and an auditory illusion.[unreliable source?] "Barberpole flanging", also known as "infinite flanging" sonic illusion, is similar to the Shepard tone effect. "Barberpole Flanger" is one such, open source VST audio plug-in, implementation (with four different algorithms). The sweep of the "flanged sound seems to move in only one direction ("up" or "down") infinitely, instead of sweeping back-and-forth." "Barberpole phaser". Roger Shepard's original work used a computer program written by Max Mathews. However, the same type of effect can be accomplished using an analog synthesizer controlled by a gadget which may be called a "Shepard Function Generator". Harald Bode (popularizer of the Moog vocoder) invented a rack-mounted device called a "barberpole phaser" which was marketed in the 1980s.[unreliable source?] See also Buchla 200 series Electric Music Box and Buchla 200e. Trademark [edit] Barbasol cans use a barber pole motif. The can's motif is a registered trademark of Barbasol.[B] See also [edit] Glossary of nautical terms (disambiguation) Signage Footnotes [edit] ^ Used to keep the end of a rope from fraying and said to resemble a barber's pole. Though highly decorative, and historically one of the most common knots, on a modern yacht it is almost unused and unknown. ^ "Barbasol Co. v. Jacobs. No. 8969" (full text). 7th Circuit Court of Appeals, 160 F.2d 336. 1947. Retrieved 15 December 2010. As the court noted: 'Plaintiff's shaving cream product is identified by the word trade mark "Barbasol" and by the distinctive package design trade mark hereinafter referred to. Said product is displayed and offered for sale in two types of cartons. One of said cartons is of rectangular shape, the length of which is about 3½ times its width and the depth is a little less than the width. In addition to the word "Barbasol" being provided thereon, the carton has a striped border of blue, white and red diagonal stripes surrounding a rectangular panel or field in blue color. The shaving cream disposed in this type of carton is packed in an elongated soft metal tube, which is received and housed within said carton. The other type of carton is approximately square and it also has the word trade mark "Barbasol" provided thereon, and the entire carton is provided with diagonal colored stripes of red, white and blue, said stripes forming a border for a blue field, on which appears the word "Barbasol" and other printed matter. The shaving cream is disposed in a glass jar of octangular cross section and upon which the word "Barbasol" appears. The jar is white and is disposed within the square carton.' Citations [edit] ^ "Barber Pole". Webster's New World College Dictionary. Cleveland: Wiley Publishing. 2010. Retrieved 14 November 2010. ^ Smith, Kate. "Why Barber Poles are Red and White". Sensational Color. Archived from the original on 29 November 2010. Retrieved 14 November 2010. ^ The Freelance Star (Sep 11, 1982) Cigar Store Indian was Early Advertising Sign Retrieved 2010 May 3 ^ Andrews, William (1904). At the Sign of the Barber's Pole: Studies in Hirsute History. Cottingham, Yorkshire: J. R. Tutin. p. 1. ^ van Helmont, Jean Baptiste (1644). Van Helmont's Workes, containing his ... Philosophy, Physick, Chirurgery, Anatomy. pp. 504–516. ^ Van Helmont's works: containing his most excellent philosophy, anatomy ^ AMUNDSEN, DARREL W. (1978). "Medieval Canon Law on Medical and Surgical Practice by the Clergy". Bulletin of the History of Medicine. 52 (1). The Johns Hopkins University Press: 22–44. JSTOR 44450442. PMID 352450. Retrieved 20 December 2024. ^ MacNalty, Sir Arthur Salusbury (1 December 1945). "The Renaissance and its Influence on English Medicine, Surgery and Public Health". British Medical Journal. 2 (4430). London: British Medical Association: 755–759. doi:10.1136/bmj.2.4430.755. JSTOR 20364730. PMC 2060364. PMID 20786422. ^ Quesnay, François; Bellial des Vertus, François (1749). Histoire de l'origine et des progrès de la chirurgie en France [History of the Origin and Progress of Surgery in France] (in French). Paris: Ganeau. p. 41. ^ Wallace, Robert (1968). The World of Rembrandt: 1606–1669. New York: Time-Life Books. p. 62. ^ Nix, Elizabeth (25 June 2014). "Why are barber poles red, white and blue?". History.com. Retrieved 15 January 2017. ^ Journeyman Barber (9 September 2021) . Prize Contest No 4, Prize Contest No. 5 (Paperback or Softback). Vol. 1. Journeyman Barber Association, Legare Street Press. p. 119. ISBN 9781014587930. {{cite book}}: \|work= ignored (help) ^ "About Us". William Marvy Company. 2010. Archived from the original on 14 November 2011. Retrieved 14 November 2010. ^ Censky, Annalyn (2007). "Last of Their Kind: From Barber Poles to Limburger Cheese, These 5 Companies are the Last Left in America Making Iconic Products now in their Twilight". CNNMoney.com. Archived from the original on 11 October 2010. Retrieved 14 November 2010. ^ "History of the Barber - Barber Schools". barber-schools.org. 8 March 2012. ^ Tunis, Edwin. (1905). Colonial Craftsmen and the Beginnings of American Industry. Cleveland and New York: World Publishing. p. 42. ISBN 978-0-8018-6228-1. Retrieved 14 November 2010. {{cite book}}: ISBN / Date incompatibility (help) ^ "Banning Barber Poles at Michigan Salons? - Barber Schools". barber-schools.org. 26 March 2012. ^ Moon, Katharine Hyung-Sun (1997). Sex Among Allies: Military Prostitution in U.S.–Korea Relations. New York: Columbia University Press. p. 45. ISBN 978-0-231-10642-9. Retrieved 14 November 2010. ^ Trecker, Jamie (2007). Love and Blood: At the World Cup with the Footballers, Fans, and Freaks. Orlando: Houghton Mifflin Harcourt. p. 13. ISBN 978-0-15-603098-4. Retrieved 14 November 2010. prostitution barber. ^ "Barber Pole Illusion". sandlotscience.com. Archived from the original on 28 November 2010. Retrieved 14 November 2010. ^ Massaro, Dominic W., ed. (Spring 2007). "Book Reviews: What Are Musical Paradox and Illusion?" (PDF). American Journal of Psychology. 120 (1). University of California, Santa Cruz: 123–170, 124, 132. doi:10.2307/20445384. JSTOR 20445384. ^ Coe, Steven R. (2007). Nebulae and how to observe them. Phoenix, Arizona: Springer Science Media. p. 116. ISBN 978-1-84628-482-3. Retrieved 1 December 2010. ^ Powell, Kendall (15 July 2009). "Cell biology: Ahead of the curve – Cellular life is all slopes, arcs and circles – but there is much debate about how these curves are built" (PDF). Nature. 460 (7253): 460, 318–320. doi:10.1038/460318a. PMID 19606122. S2CID 205047888. Retrieved 15 December 2010. ^ "Double Matthew Walker Knot" (animation). animated knots by Grog. Retrieved 24 March 2013. ^ Ji, Q; Ji, S (1996). "On discovery of the earliest bird fossil in China and the origin of birds". Chinese Geology. 10 (233): 30–33. ^ Stone, Richard (December 2010). "Dinosaurs' Living Descendants". Smithsonian. Smithsonian Institution: 60. Archived from the original on 1 December 2010. Retrieved 14 December 2010. ^ "Barber's Pole Worm (Haemonchus contortus)". Sydney: Australian Wool Innovation. 2010. Archived from the original on 30 April 2012. Retrieved 14 November 2010. ^ Burke, Joan (February 2005). "Management of Barber Pole Worm in Sheep and Goats in the Southern U.S" (PDF). Small Farms Research Update. Booneville, Arkansas: U.S. Department of Agriculture Dale Bumpers Small Farms Research Center. Archived from the original (PDF) on 5 March 2009. Retrieved 4 March 2010. ^ Schoenian, Susan (15 February 2005). "Diseases: Worms". Sheep 101. ^ "Barberpole worms in Humans". Daily Puppy.com. 23 November 2010. Archived from the original on 8 July 2011. Retrieved 23 November 2010. ^ Marlos, Daniel (16 September 2010). "Barber Pole Grasshopper". What's That Bug?. Retrieved 14 November 2010. ^ Thiret, Beth (9 September 2010). "Ugh: What to do about Grasshoppers". The Recorder. Berthoud, Colorado. Archived from the original on 28 December 2010. Retrieved 14 November 2010.. ^ "Redbanded rockfish". Alaska Fisheries Science Center. National Marine Fisheries Service National Oceanic and Atmospheric Administration. Retrieved 14 December 2010. ^ "Candy Cane". The Titi Tudorancea Bulletin, English Edition. 7 October 2010. Retrieved 1 December 2010. ^ Tuman'skia, S.; Stabrowskia, M. (August 1985). "The Optimization and Design of Magnetoresistive Barber-Pole Sensors". Sensors and Actuators. 7 (4). Lausanne, Switzerland: NATO Advanced Study Institute on Chemically Sensitive Electronic Devices: 285–295. doi:10.1016/0250-6874(85)80008-1. ISSN 0250-6874. ^ US patent 5737156, Bonyhard, Peter I., "Barberpole MR sensor having interleaved permanent magnet and magnetoresistive segments", published 7 April 1998, assigned to Seagate Technology, Inc. ^ Carrella, Stefano; Iswandy, Kuncup; Lutz, Kai; King, Andreas (18–19 May 2010). "3D-Localization of Low-Power Wireless Sensor Nodes Based on AMR-Sensors in Industrial and AmI Applications" (CD-ROM). Sensoren und Messsysteme 2010. Vorträge der 15. ITG/GMA-Fachtagung. Nuremberg: Berlin Offenbach. ISBN 978-3-8007-3260-9. Retrieved 14 November 2010. ^ "McGraw Hill Dictionary of Aviation". Answers.com. McGraw Hill. Retrieved 28 December 2010. ^ Avery, Rob. "The Conventional Airspeed Indicator". Archived from the original on 4 May 2010. Retrieved 14 November 2010. ^ "Detailed Chronology of Events Surrounding the Apollo 13 Accident". 17 June 2011. Retrieved 20 March 2016. ^ NSTS 1988 News Reference Manual. Kennedy Space Center: National Aeronautic and Space Administration. 31 August 2000 [Original print date September 1988]. ^ Grinter, Kay; Rybe, Jeanne (20 February 2009). "NASA/KSC Acronym List". Kennedy Space Center: National Aeronautic and Space Administration. Archived from the original on 6 November 2010. Retrieved 14 November 2010. ^ "North American P-51 Mustang/F-51 Cavalier—USA". Wing's Palette. Russia. 12 August 2010. Archived from the original on 19 October 2010. Retrieved 14 November 2010. ^ Pearson, Bob (September 2000). "The "Showbirds" of Spad XIIIs of the 94th Aero Squadron USAS". Internet Modeler. ^ "A 'Michigan Original,' Tier Jerry Regan brings AuSable Lore to Livonia" (PDF). The Evening Hatch. Michigan Fly Fishing Club: 1–2. May 2005. Archived from the original (PDF) on 19 July 2011. Retrieved 21 November 2010. ^ Cameron, Scott (2010). "Craps Lingo". CasinoDealers. Archived from the original on 10 July 2011. Retrieved 14 November 2010. ^ "Crap Dictionary". World Casino Directory. 2010. Retrieved 14 November 2010. ^ Jump up to: a b "101st Airborne Division (Air Assault) Command Parachute Demonstration Team "Screaming Eagles"". Globalsecurity.org. Retrieved 14 November 2010. ^ "British Virgin Islands Airshow, 2009. Retrieved 29 November 2010". Archived from the original on 8 July 2011. Retrieved 23 November 2010. ^ Scott, Royce E. "Bo" (July–August 1988). "Jump School at Fort Benning" (PDF). Screaming Eagle Magazine. Archived from the original (PDF) on 30 November 2010. Retrieved 14 November 2010. ^ "A Comprehensive Glossary of Weather Terms for Storm Spotters". NOAA Technical Memorandum NWS SR-145. Norman, Oklahoma: National Oceanic and Atmospheric Administration, National Weather Service Weather Forecast Office. Retrieved 14 November 2010. ^ "Glossary". Tornado Chaser.net. Archived from the original on 30 November 2010. Retrieved 15 December 2010. ^ "The Majestic Supercell – Barber pole photograph". Verden, Oklahoma: Roger's Sky Pix. 3 April 2003. Retrieved 1 December 2010. ^ Krohn, Dennis (March 2008). "USGS Extreme Storm Team Receives Christmas Week Tour of NOAA Aircraft Facility". Archived from the original on 28 August 2017. Retrieved 14 December 2010. ^ Roberts, William (1895). The Book-Hunter in London: Historical and Other Studies of Collectors and Collecting. London: Elliot Stock. p. 176. Retrieved 14 November 2010. ^ "HMCS Sackville History". Halifax: Canadian Navy Memorial Trust. Archived from the original on 19 May 2011. Retrieved 14 November 2010. ^ Pearson, Bob; Banyai-Riepl, Chris (2002). "HMCS Sackville: The Last Flower: 1941–2000". History in Illustration. ^ "A Hundred Years of Naval Service Captured in Ice". Ottawa Start. 17 February 2010. Archived from the original on 29 November 2010. Retrieved 14 November 2010. ^ Wobser, David. "White Shoal Light". Lighthouses of the Great Lakes. Great Lakes & Seaway Shipping. Archived from the original on 11 July 2011. Retrieved 14 November 2010. ^ "White Shoal Lighthouse, Mackinaw City, Michigan". Michigan Lighthouse Conservancy. 23 August 2003. Archived from the original on 25 November 2010. Retrieved 14 November 2010. ^ "Save our Lights" license plate ^ Kirby, Doug; Smith, Ken; Wilkins, Mike (2010). "Buxton, North Carolina—America's Tallest Lighthouse—Climb It". Novato, California: Roadside America. Retrieved 14 November 2010. ^ Kitchen (2008), p. 167 ^ Kitchen (2008), p. 303 ^ Jump up to: a b Wazz, Scotty (6 November 2009). "Defending the Barber-Pole". The Strangest One of All. Internet Radio. ^ "Jerseys and Logos: From 1909 to 1946". Our History. Montreal Canadiens. 2008. Retrieved 14 November 2010.[permanent dead link] ^ "1912–1913 Season". Our History. Montreal Canadiens. 2008. Retrieved 14 November 2010.[permanent dead link] ^ Cycleback, David Rudd. "The Sonic Barber Pole: Shepard's Scale". cycleback.com. Retrieved 14 November 2010. ^ Budde, Christian. "Barberpole Flanger". KVR Audio Plugin Resources. Retrieved 14 November 2010. ^ Simonton, John (February 1983). "Shepard Function (Barberpole) CV Generator". Polyphony. Retrieved 14 November 2010. ^ Irwin, Michael. "Frequency shifters add an exotic dimension to the world of modular synthesis signal processing". Frequency Shifter electronic design. Archived from the original on 14 July 2011. Retrieved 20 November 2010. ^ "Barberpole phaser". Ampage Tube Amps /Music Electronics Forum. Archived from the original on 11 August 2011. Retrieved 14 November 2010. ^ "Barbasol Taps Gary Hall Jr. as First-Ever "Real Man"; Campaign Designed to ." Business Wire. Dublin, Ohio. 26 April 2005. Retrieved 24 November 2010. ^ "Barbasol-American-Original" (PDF). www.perio-inc.com. Archived from the original (PDF) on 15 July 2011. Retrieved 11 January 2022. ^ "Barbasol Shave Cream: An American Original" (PDF). Archived from the original (PDF) on 15 July 2011. Retrieved 1 December 2010. ^ "The Barbasol Company vs. Jacobs". 35 T.M. Rep. 135. 7th CCA. 28 February 1947. Retrieved 15 December 2010. Further reading [edit] Andrews, William (1904). At the Sign of the Barber's Pole: Studies in Hirsute History. Cottingham, Yorkshire: J.R. Tutin. pp. 1–8. Barber pole birmingham. John H. Lienhard (2001). "The Sign of the Barber Pole". The Engines of Our Ingenuity. Episode 1635. University of Houston School of Engineering. NPR. KUHF-FM. The Sign of the Barber Pole. and audio. "Blood, Bandages and Barber Poles". The Guide to Life, The Universe and Everything. BBC. 29 November 2002. Archived from the original on 8 January 2006. Retrieved 27 December 2005. Roberts, William (1895). The Book Hunter in London. London: Eliot Stock. p. 176. ISBN 978-1-112-48298-4. Roberts, William (1895). The Book Hunter in London. {{cite book}}: ISBN / Date incompatibility (help) at Google books and at Project Gutenberg. External links [edit] An Animated Ad Pillar In Red And Blue: The Barber's Pole (21 March 2008) at Pingmag. Archived 8 April 2010 at the Wayback Machine Barber Poles at National Library Board, Singapore. Sullivan, Pat. Adventures of Felix the Cat – Felix Pinches the Pole video newsreel film – Felix the Cat steals a barber's pole. British Pathé. Issue Date: 31 July 1924 Canister: EP 058 Film ID: 854.04 Sort number: EP 058 Tape: PM0854 Retrieved 21 November 2010. Barber-Schools.org – Barber Industry, Careers, Education & Licensing Information See Pseudo-barberpole and Barberpole patterns at ConwayLife.com \| Human hair \| \| --- \| \| Classification \| \| \| \| --- \| \| by type \| Lanugo Body Terminal Vellus \| \| by location \| Body Ear Nose Eyebrow + unibrow Eyelash Underarm Chest Abdominal Pubic Leg \| \| \| Head hairstyles (list of hairstyles) \| Afro Afro puffs Asymmetric cut Bald Bangs Beehive Big hair Blowout Bob cut Bouffant Bowl cut Braid Broccoli haircut Brush, butch, burr cut Bun (odango) Bunches Businessman cut Butterfly haircut Buzz cut Caesar cut Chignon Chonmage Comb over Conk Cornrows Crew cut Crochet braids Croydon facelift Curly hair Curtained hair Czupryna Devilock Dido flip Digital perm Dreadlocks Ducktail Edgar cut Eton crop Extensions Fauxhawk Feathered hair Finger wave Flattop Fontange French braid French twist Fringe Frosted tips Hair crimping Hair twists High and tight Hime cut Historical Christian hairstyles Hi-top fade Induction cut Ivy League, Harvard, Princeton cut Japanese women Jewfro Jesus look Jheri curl Kinky hair Kiss curl Laid edges Layered hair Liberty spikes Long hair Lob cut Lovelock Marcelling Mod cut Mohawk Mullet 1950s 1980s Pageboy Part Payot Pigtail Pixie cut Pompadour Ponytail Punch perm Professional cut Queue Quiff Rachel Rattail Razor cut Regular haircut Ringlets Shag Shape-up Shikha Shimada Short back and sides Short brush cut Short hair Spiky hair Straight hair Standard haircut Step cut Surfer hair Taper cut Temple fade Titus cut Tonsure Updo Undercut Victory rolls Waves Widow's peak Wings \| \| Facial hair (list) \| Beard + Chinstrap + Goatee + Ned Kelly + Shenandoah + Soul patch + Van Dyke Moustache + Fu Manchu + handlebar + horseshoe + pencil + toothbrush + walrus Designer stubble Sideburns \| \| Hair subtraction \| \| \| \| --- \| \| cosmetic \| Removal + waxing + threading + plucking + chemical + electric + laser + IPL Shaving + head + leg + cream + brush + soap Razor + electric + safety + straight \| \| disorders \| Alopecia + areata + totalis + universalis + Frictional alopecia Pattern hair loss Hypertrichosis Management Trichophilia Trichotillomania Pogonophobia \| \| \| Haircare products \| Brush Clay Clipper Comb Conditioner Dryer Gel Hairstyling products Hot comb Iron Mousse Pomade Relaxer Rollers Shampoo Spray Volumizer Wax Hair texture powder \| \| Haircare techniques \| Backcombing Hair coloring Crimping Curly Girl Method Hair cutting Perm Shampoo and set Straightening \| \| Health and medical \| Greying of hair Hair follicle Hair growth Trichology \| \| Related \| Andre Walker Hair Typing System Beard and haircut laws by country Bearded lady Barber (pole) Eponymous hairstyle Frizz Good hair Hairdresser + list Hair fetishism Wig \| Retrieved from " Categories: English inventions Hairdressing Professional symbols Signage Hidden categories: CS1 French-language sources (fr) CS1 errors: periodical ignored CS1 errors: ISBN date CS1: unfit URL All articles with dead external links Articles with dead external links from October 2016 Articles with permanently dead external links Articles with short description Short description is different from Wikidata Use dmy dates from July 2015 All articles with unsourced statements Articles with unsourced statements from December 2024 All articles lacking reliable references Articles lacking reliable references from November 2010 Commons category link is on Wikidata Webarchive template wayback links Articles containing video clips
15736	https://en.wikipedia.org/wiki/Primary_transcript	Jump to content Primary transcript العربية Bosanski Čeština 日本語 Slovenčina Српски / srpski Türkçe Edit links From Wikipedia, the free encyclopedia RNA produced by transcription A primary transcript is the single-stranded ribonucleic acid (RNA) product synthesized by transcription of DNA, and processed to yield various mature RNA products such as mRNAs, tRNAs, and rRNAs. The primary transcripts designated to be mRNAs are modified in preparation for translation. For example, a precursor mRNA (pre-mRNA) is a type of primary transcript that becomes a messenger RNA (mRNA) after processing. Pre-mRNA is synthesized from a DNA template in the cell nucleus by transcription. Pre-mRNA comprises the bulk of heterogeneous nuclear RNA (hnRNA). Once pre-mRNA has been completely processed, it is termed "mature messenger RNA", or simply "messenger RNA". The term hnRNA is often used as a synonym for pre-mRNA, although, in the strict sense, hnRNA may include nuclear RNA transcripts that do not end up as cytoplasmic mRNA. There are several steps contributing to the production of primary transcripts. All these steps involve a series of interactions to initiate and complete the transcription of DNA in the nucleus of eukaryotes. Certain factors play key roles in the activation and inhibition of transcription, where they regulate primary transcript production. Transcription produces primary transcripts that are further modified by several processes. These processes include the 5' cap, 3'-polyadenylation, and alternative splicing. In particular, alternative splicing directly contributes to the diversity of mRNA found in cells. The modifications of primary transcripts have been further studied in research seeking greater knowledge of the role and significance of these transcripts. Experimental studies based on molecular changes to primary transcripts and the processes before and after transcription have led to greater understanding of diseases involving primary transcripts. Production [edit] Main article: Transcription (genetics) The steps contributing to the production of primary transcripts involve a series of molecular interactions that initiate transcription of DNA within a cell's nucleus. Based on the needs of a given cell, certain DNA sequences are transcribed to produce a variety of RNA products to be translated into functional proteins for cellular use. To initiate the transcription process in a cell's nucleus, DNA double helices are unwound and hydrogen bonds connecting compatible nucleic acids of DNA are broken to produce two unconnected single DNA strands. One strand of the DNA template is used for transcription of the single-stranded primary transcript mRNA. This DNA strand is bound by an RNA polymerase at the promoter region of the DNA. In eukaryotes, three kinds of RNA—rRNA, tRNA, and mRNA—are produced based on the activity of three distinct RNA polymerases, whereas, in prokaryotes, only one RNA polymerase exists to create all kinds of RNA molecules. RNA polymerase II of eukaryotes transcribes the primary transcript, a transcript destined to be processed into mRNA, from the antisense DNA template in the 5' to 3' direction, and this newly synthesized primary transcript is complementary to the antisense strand of DNA. RNA polymerase II constructs the primary transcript using a set of four specific ribonucleoside monophosphate residues (adenosine monophosphate (AMP), cytidine monophosphate (CMP), guanosine monophosphate (GMP), and uridine monophosphate (UMP)) that are added continuously to the 3' hydroxyl group on the 3' end of the growing mRNA. Studies of primary transcripts produced by RNA polymerase II reveal that an average primary transcript is 7,000 nucleotides in length, with some growing as long as 20,000 nucleotides in length. The inclusion of both exon and intron sequences within primary transcripts explains the size difference between larger primary transcripts and smaller, mature mRNA ready for translation into protein.[citation needed] Regulation [edit] A number of factors contribute to the activation and inhibition of transcription and therefore regulate the production of primary transcripts from a given DNA template.[citation needed] Activation of RNA polymerase activity to produce primary transcripts is often controlled by sequences of DNA called enhancers. Transcription factors, proteins that bind to DNA elements to either activate or repress transcription, bind to enhancers and recruit enzymes that alter nucleosome components, causing DNA to be either more or less accessible to RNA polymerase. The unique combinations of either activating or inhibiting transcription factors that bind to enhancer DNA regions determine whether or not the gene that enhancer interacts with is activated for transcription or not. Activation of transcription depends on whether or not the transcription elongation complex, itself consisting of a variety of transcription factors, can induce RNA polymerase to dissociate from the Mediator complex that connects an enhancer region to the promoter. Inhibition of RNA polymerase activity can also be regulated by DNA sequences called silencers. Like enhancers, silencers may be located at locations farther up or downstream from the genes they regulate. These DNA sequences bind to factors that contribute to the destabilization of the initiation complex required to activate RNA polymerase, and therefore inhibit transcription. Histone modification by transcription factors is another key regulatory factor for transcription by RNA polymerase. In general, factors that lead to histone acetylation activate transcription while factors that lead to histone deacetylation inhibit transcription. Acetylation of histones induces repulsion between negative components within nucleosomes, allowing for RNA polymerase access. Deacetylation of histones stabilizes tightly coiled nucleosomes, inhibiting RNA polymerase access. In addition to acetylation patterns of histones, methylation patterns at promoter regions of DNA can regulate RNA polymerase access to a given template. RNA polymerase is often incapable of synthesizing a primary transcript if the targeted gene's promoter region contains specific methylated cytosines— residues that hinder binding of transcription-activating factors and recruit other enzymes to stabilize a tightly bound nucleosome structure, excluding access to RNA polymerase and preventing the production of primary transcripts. R-loops [edit] R-loops are formed during transcription. An R-loop is a three-stranded nucleic acid structure containing a DNA-RNA hybrid region and an associated non-template single-stranded DNA. Actively transcribed regions of DNA often form R-loops that are vulnerable to DNA damage. Introns reduce R-loop formation and DNA damage in highly expressed yeast genes. Transcription stress [edit] DNA damages arise in each cell, every day, with the number of damages in each cell reaching tens to hundreds of thousands, and such DNA damages can impede primary transcription. The process of gene expression itself is a source of endogenous DNA damages resulting from the susceptibility of single-stranded DNA to damage. Other sources of DNA damage are conflicts of the primary transcription machinery with the DNA replication machinery, and the activity of certain enzymes such as topoisomerases and base excision repair enzymes. Even though these processes are tightly regulated and are usually accurate, occasionally they can make mistakes and leave behind DNA breaks that drive chromosomal rearrangements or cell death. RNA processing [edit] Main article: Post-transcriptional modification Transcription, a highly regulated phase in gene expression, produces primary transcripts. However, transcription is only the first step which should be followed by many modifications that yield functional forms of RNAs. Otherwise stated, the newly synthesized primary transcripts are modified in several ways to be converted to their mature, functional forms to produce different proteins and RNAs such as mRNA, tRNA, and rRNA.[citation needed] Processing [edit] The basic primary transcript modification process is similar for tRNA and rRNA in both eukaryotic and prokaryotic cells. On the other hand, primary transcript processing varies in mRNAs of prokaryotic and eukaryotic cells. For example, some prokaryotic bacterial mRNAs serve as templates for synthesis of proteins at the same time they are being produced via transcription. Alternatively, pre-mRNA of eukaryotic cells undergo a wide range of modifications prior to their transport from the nucleus to cytoplasm where their mature forms are translated. These modifications are responsible for the different types of encoded messages that lead to translation of various types of products. Furthermore, primary transcript processing provides a control for gene expression as well as a regulatory mechanism for the degradation rates of mRNAs. The processing of pre-mRNA in eukaryotic cells includes 5' capping, 3' polyadenylation, and alternative splicing.[citation needed] 5' capping [edit] Main article: Five-prime cap Shortly after transcription is initiated in eukaryotes, a pre-mRNA's 5' end is modified by the addition of a 7-methylguanosine cap, also known as a 5' cap. The 5' capping modification is initiated by the addition of a GTP to the 5' terminal nucleotide of the pre-mRNA in reverse orientation followed by the addition of methyl groups to the G residue. 5' capping is essential for the production of functional mRNAs since the 5' cap is responsible for aligning the mRNA with the ribosome during translation. Polyadenylation [edit] Main article: Polyadenylation In eukaryotes, polyadenylation further modifies pre-mRNAs during which a structure called the poly-A tail is added. Signals for polyadenylation, which include several RNA sequence elements, are detected by a group of proteins which signal the addition of the poly-A tail (approximately 200 nucleotides in length). The polyadenylation reaction provides a signal for the end of transcription and this reaction ends approximately a few hundred nucleotides downstream from the poly-A tail location. Alternative splicing [edit] Main article: Alternative splicing Eukaryotic pre-mRNAs have their introns spliced out by spliceosomes made up of small nuclear ribonucleoproteins. In complex eukaryotic cells, one primary transcript is able to prepare large amounts of mature mRNAs due to alternative splicing. Alternative splicing is regulated so that each mature mRNA may encode a multiplicity of proteins. The effect of alternative splicing in gene expression can be seen in complex eukaryotes which have a fixed number of genes in their genome yet produce much larger numbers of different gene products. Most eukaryotic pre-mRNA transcripts contain multiple introns and exons. The various possible combinations of 5' and 3' splice sites in a pre-mRNA can lead to different excision and combination of exons while the introns are eliminated from the mature mRNA. Thus, various kinds of mature mRNAs are generated. Alternative splicing takes place in a large protein complex called the spliceosome. Alternative splicing is crucial for tissue-specific and developmental regulation in gene expression. Alternative splicing can be affected by various factors, including mutations such as chromosomal translocation. In prokaryotes, splicing is done by autocatalytic cleavage or by endolytic cleavage. Autocatalytic cleavages, in which no proteins are involved, are usually reserved for sections that code for rRNA, whereas endolytic cleavage corresponds to tRNA precursors. Research [edit] 5-Fluorouracil (FUra) exposure in methotrexate-resistant KB cells led to a two-fold reduction in total dihydrofolate reductase (DHFR) mRNA levels, while the level of DHFR pre-mRNA with certain introns remained unaffected. The half-life of DHFR mRNA or pre-mRNA did not change significantly, but the transition rate of DHFR RNA from the nucleus to the cytoplasm decreased, suggesting that FUra may influence mRNA processing and/or nuclear DHFR mRNA stability. In Drosophila and Aedes, hnRNA (pre-mRNA) size was larger in Aedes due to its larger genome, despite both species producing mature mRNA of similar size and sequence complexity. This indicates that hnRNA size increases with genome size. In HeLa cells, spliceosome groups on pre-mRNA were found to form within nuclear speckles, with this formation being temperature-dependent and influenced by specific RNA sequences. Pre-mRNA targeting and splicing factor loading in speckles were critical for spliceosome group formation, resulting in a speckled pattern. Recruiting pre-mRNA to nuclear speckles significantly increased splicing efficiency and protein levels, indicating that proximity to speckles enhances splicing efficiency. Related diseases [edit] Research has also led to greater knowledge about certain diseases related to changes within primary transcripts. One study involved estrogen receptors and differential splicing. The article entitled, "Alternative splicing of the human estrogen receptor alpha primary transcript: mechanisms of exon skipping" by Paola Ferro, Alessandra Forlani, Marco Muselli and Ulrich Pfeffer from the laboratory of Molecular Oncology at National Cancer Research Institute in Genoa, Italy, explains that 1785 nucleotides of the region in the DNA that codes for the estrogen receptor alpha (ER-alpha) are spread over a region that holds more than 300,000 nucleotides in the primary transcript. Splicing of this pre-mRNA frequently leads to variants or different kinds of the mRNA lacking one or more exons or regions necessary for coding proteins. These variants have been associated with breast cancer progression. In the life cycle of retroviruses, proviral DNA is incorporated in transcription of the DNA of the cell being infected. Since retroviruses need to change their pre-mRNA into DNA so that this DNA can be integrated within the DNA of the host it is affecting, the formation of that DNA template is a vital step for retrovirus replication. Cell type, the differentiation or changed state of the cell, and the physiological state of the cell, result in a significant change in the availability and activity of certain factors necessary for transcription. These variables create a wide range of viral gene expression. For example, tissue culture cells actively producing infectious virions of avian or murine leukemia viruses (ASLV or MLV) contain such high levels of viral RNA that 5–10% of the mRNA in a cell can be of viral origin. This shows that the primary transcripts produced by these retroviruses do not always follow the normal path to protein production and convert back into DNA in order to multiply and expand. See also [edit] cis-splicing Outron trans-splicing Transcription (biology) Transcriptome References [edit] ^ Jump up to: a b c Strachan T, Read AP (January 2004). Human Molecular Genetics 3. Garland Science. pp. 16–17. ISBN 978-0-8153-4184-0. ^ Jump up to: a b Alberts B (1994). "RNA Synthesis and RNA Processing". Molecular Biology of the Cell (3rd ed.). New York: Garland Science – via NCBI.{{cite book}}: CS1 maint: publisher location (link) ^ Griffiths AJ (2000). "An Introduction to Genetic Analysis". NCBI. New York: W.H. Freeman. Archived from the original on January 8, 2020. ^ Jump up to: a b c Gilbert SF (15 July 2013). Developmental Biology. Sinauer Associates, Incorporated. pp. 38–39, 50. ISBN 978-1-60535-173-5. ^ Brown TA (2002). "Assembly of the Transcription Initiation Complex". Genomes (2nd ed.). Oxford: Wiley-Liss. ^ Lodish H (2008). Molecular Cell Biology. W. H. Freeman. pp. 303–306. ISBN 978-0-7167-7601-7. ^ Bonnet A, Grosso AR, Elkaoutari A, Coleno E, Presle A, Sridhara SC, et al. (August 2017). "Introns Protect Eukaryotic Genomes from Transcription-Associated Genetic Instability". Molecular Cell. 67 (4): 608–621.e6. doi:10.1016/j.molcel.2017.07.002. PMID 28757210. ^ Jump up to: a b c Milano L, Gautam A, Caldecott KW (January 2024). "DNA damage and transcription stress". Mol Cell. 84 (1): 70–79. doi:10.1016/j.molcel.2023.11.014. PMID 38103560. This article incorporates text available under the CC BY 4.0 license. ^ Jump up to: a b c d e f g h i j k Cooper GM (2000). "RNA Processing and Turnover". The Cell: A Molecular Approach (2nd ed.). Sunderland (MA): Sinauer Associates; 2000. ^ Weaver RF (2005). Molecular Biology. New York, NY: McGraw-Hill. pp. 432–448. ISBN 0-07-284611-9. ^ Wahl MC, Will CL, Lührmann R (February 2009). "The spliceosome: design principles of a dynamic RNP machine". Cell. 136 (4): 701–18. doi:10.1016/j.cell.2009.02.009. hdl:11858/00-001M-0000-000F-9EAB-8. PMID 19239890. S2CID 21330280. ^ Will CL, Dolnick BJ (December 1989). "5-Fluorouracil inhibits dihydrofolate reductase precursor mRNA processing and/or nuclear mRNA stability in methotrexate-resistant KB cells". The Journal of Biological Chemistry. 264 (35): 21413–21. doi:10.1016/S0021-9258(19)30096-1. PMID 2592384. ^ Lengyel J, Penman S (July 1975). "hnRNA size and processing as related to different DNA content in two dipterans: Drosophila and Aedes". Cell. 5 (3): 281–90. doi:10.1016/0092-8674(75)90103-8. PMID 807333. S2CID 39038640. ^ Melčák I, Melčáková Š, Kopsky V, Večeřová J, Raška I (February 2001). "Prespliceosomal assembly on microinjected precursor mRNA takes place in nuclear speckles". Molecular Biology of the Cell. 12 (2): 393–406. CiteSeerX 10.1.1.324.8865. doi:10.1091/mbc.12.2.393. PMC 30951. PMID 11179423. ^ Bhat P, Chow A, Emert B, Ettlin O, Quinodoz SA, Strehle M, et al. 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External links [edit] Scienceden.com RNA Article \| v t e Post-transcriptional modification \| \| --- \| \| Nuclear \| \| \| \| --- \| \| Precursor mRNA 5′ cap formation Polyadenylation + CPSF + CstF + PAP + PAB2 + CFI + CFII Poly(A)-binding protein RNA editing Polyuridylation \| \| \| RNA splicing \| \| \| \| --- \| \| Intron / Exon snRNP Spliceosome + minor + U1 Alternative splicing \| \| \| pre-mRNA factors \| PLRG1 + PRPF3 + PRPF4 + PRPF4B + PRPF6 + PRPF8 + PRPF18 + PRPF19 + PRPF31 + PRPF38A + PRPF38B + PRPF39 + PRPF40A + PRPF40B \| \| \| \| Cytosolic \| 5′ cap methylation mRNA decapping + DCP1A + DCP1B + DCP2 + DCPS + EDC3 + EDC4 \| \| v t e Types of nucleic acids \| \| --- \| \| Constituents \| Nucleobases Nucleosides Nucleotides Deoxynucleotides \| \| Ribonucleic acids (coding, non-coding) \| \| \| \| --- \| \| Translational \| Messenger + precursor, heterogenous nuclear modified Messenger Transfer Ribosomal Transfer-messenger \| \| Regulatory \| Interferential + Micro + Small interfering + Piwi-interacting Antisense Processual + Small nuclear + Small nucleolar + Small Cajal Body RNAs + Y RNA Enhancer RNAs \| \| Others \| Guide Ribozyme Small hairpin Small temporal Trans-acting small interfering Subgenomic messenger \| \| \| Deoxyribonucleic acids \| Organellar + Chloroplast + Mitochondrial Complementary Deoxyribozyme Genomic Hachimoji Multicopy single-stranded \| \| Analogues \| Xeno + Glycol + Threose + Hexose + Locked + Peptide + Morpholino Phosphorothioate \| \| Cloning vectors \| Phagemid Plasmid Lambda phage Cosmid Fosmid Artificial chromosomes + P1-derived + Bacterial + Yeast + Human \| \| \| \| Retrieved from " Category: RNA Hidden categories: CS1 maint: publisher location Articles with imported Creative Commons Attribution 4.0 text Articles with short description Short description matches Wikidata All articles with unsourced statements Articles with unsourced statements from July 2024
15737	https://study.com/skill/learn/finding-domain-and-range-of-tangent-inverse-functions-explanation.html	Finding Domain and Range of Tangent Inverse Functions \| Trigonometry \| Study.com Log In Sign Up Menu Plans Courses By Subject College Courses High School Courses Middle School Courses Elementary School Courses By Subject Arts Business Computer Science Education & Teaching English (ELA) Foreign Language Health & Medicine History Humanities Math Psychology Science Social Science Subjects Art Business Computer Science Education & Teaching English Health & Medicine History Humanities Math Psychology Science Social Science Art Architecture Art History Design Performing Arts Visual Arts Business Accounting Business Administration Business Communication Business Ethics Business Intelligence Business Law Economics Finance Healthcare Administration Human Resources Information Technology International Business Operations Management Real Estate Sales & Marketing Computer Science Computer Engineering Computer Programming Cybersecurity Data Science Software Education & Teaching Education Law & Policy Pedagogy & Teaching Strategies Special & Specialized Education Student Support in Education Teaching English Language Learners English Grammar Literature Public Speaking Reading Vocabulary Writing & Composition Health & Medicine Counseling & Therapy Health Medicine Nursing Nutrition History US History World History Humanities Communication Ethics Foreign Languages Philosophy Religious Studies Math Algebra Basic Math Calculus Geometry Statistics Trigonometry Psychology Clinical & Abnormal Psychology Cognitive Science Developmental Psychology Educational Psychology Organizational Psychology Social Psychology Science Anatomy & Physiology Astronomy Biology Chemistry Earth Science Engineering Environmental Science Physics Scientific Research Social Science Anthropology Criminal Justice Geography Law Linguistics Political Science Sociology Teachers Teacher Certification Teaching Resources and Curriculum Skills Practice Lesson Plans Teacher Professional Development For schools & districts Certifications Teacher Certification Exams Nursing Exams Real Estate Exams Military Exams Finance Exams Human Resources Exams Counseling & Social Work Exams Allied Health & Medicine Exams All Test Prep Teacher Certification Exams Praxis Test Prep FTCE Test Prep TExES Test Prep CSET & CBEST Test Prep All Teacher Certification Test Prep Nursing Exams NCLEX Test Prep TEAS Test Prep HESI Test Prep All Nursing Test Prep Real Estate Exams Real Estate Sales Real Estate Brokers Real Estate Appraisals All Real Estate Test Prep Military Exams ASVAB Test Prep AFOQT Test Prep All Military Test Prep Finance Exams SIE Test Prep Series 6 Test Prep Series 65 Test Prep Series 66 Test Prep Series 7 Test Prep CPP Test Prep CMA Test Prep All Finance Test Prep Human Resources Exams SHRM Test Prep PHR Test Prep aPHR Test Prep PHRi Test Prep SPHR Test Prep All HR Test Prep Counseling & Social Work Exams NCE Test Prep NCMHCE Test Prep CPCE Test Prep ASWB Test Prep CRC Test Prep All Counseling & Social Work Test Prep Allied Health & Medicine Exams ASCP Test Prep CNA Test Prep CNS Test Prep All Medical Test Prep College Degrees College Credit Courses Partner Schools Success Stories Earn credit Sign Up Copyright Finding Domain and Range of Tangent Inverse Functions Trigonometry Skills Practice Click for sound 3:53 You must c C reate an account to continue watching Register to access this and thousands of other videos Are you a student or a teacher? I am a student I am a teacher Try Study.com, risk-free As a member, you'll also get unlimited access to over 88,000 lessons in math, English, science, history, and more. Plus, get practice tests, quizzes, and personalized coaching to help you succeed. Get unlimited access to over 88,000 lessons. Try it risk-free It only takes a few minutes to setup and you can cancel any time. It only takes a few minutes. Cancel any time. Already registered? Log in here for access Back What teachers are saying about Study.com Try it risk-free for 30 days Already registered? Log in here for access 0:09 Example 1 2:30 Example 2 (pause here) Jump to a specific example Speed Normal 0.5x Normal 1.25x 1.5x 1.75x 2x Speed Andrew Noble Andrew Noble Andrew has taught early algebra through advanced calculus to students for over 10 years. He has a bachelor's and master's degree in electrical engineering from Colorado State University. He also has 6 years of experience as a software developer. View bio Example SolutionsPractice Questions Steps for Finding Domain and Range of Tangent Inverse Functions Step 1: We begin with a graph of y=tan⁡(x). From the graph, we can see the domain of y=tan⁡(x) is (−π 2,π 2) and the range is (−∞,∞). Then, we consider the graph of the inverse, y=arctan⁡(x). The domain of y=arctan⁡(x) is (−∞,∞) and the range is (−π 2,π 2). The domain and range of an inverse function are swapped from the domain and range of the original. Now that we understand the domain and range of y=arctan⁡(x), we explore the standard form of an inverse tangent function. An inverse tangent function in standard form is given by y=a arctan⁡(b x+c)+d. Given an equation in this form, we must re-arrange it such that it is of the form y−d a=arctan⁡(b x+c). Step 2: To find the domain of the tangent inverse function, substitute the term in parentheses, b x+c, into the inequality −∞≤x≤∞ and simplify. The values of b and c may impact the domain of the inverse tangent function, but will not impact the range. Step 3: To find the range of the tangent inverse function, substitute y−d a into the inequality −π 2<y<π 2 and simplify. The values of a and d will impact the range of the inverse tangent function, but will not impact the domain. Definitions and Equations for Finding Domain and Range of Tangent Inverse Functions Domain: The domain of a function is the set of all x-values (or inputs) in which the function is defined. Range: The range of a function is the set of all y-values (or outputs) in which a function is defined. Tangent Inverse Function: The tangent inverse function, y=arctan⁡(x), is, as the name implies, the inverse of a tangent function y=tan⁡(x). Since the functions are inverses tan⁡(arctan⁡(x))=arctan⁡(tan⁡(x))=x. Standard Form of Tangent Inverse Function: The standard form of the tangent inverse function is y=a arctan⁡(b x+c)+d. Let's work through 2 examples to find the domain and range of tangent inverse functions. Example Problem 1 - Finding Domain and Range of Tangent Inverse Functions What are the domain and range of the function y=2 arctan⁡(x+1)+5? Step 1: First, we rearrange the given function so that it is in the form y−d a=arctan⁡(b x+c). y=2 arctan⁡(x+1)+5 y−5=2 arctan⁡(x+1)y−5 2=arctan⁡(x+1) Step 2: Now that we have rearranged the equation, we can substitute the values that impact the domain into the domain inequality for y=arctan⁡(x) and simplify. −∞<b x+c<∞−∞<x+1<∞−∞−1<x<∞−1−∞<x<∞ Since subtracting 1 from ±∞ still gives ±∞, the domain is unchanged by the vertical shift and remains (−∞,∞). Step 3: We substitute the values that impact the range of the function into the inequality for the range of y=arctan⁡(x). −π 2<y+d a<π 2−π 2<y−5 2<π 2−π<y−5<π 5−π<y<5+π Therefore, the range of the given tangent inverse function is (5−π,5+π). Example Problem 2 - Finding Domain and Range of Tangent Inverse Functions What are the domain and range of the function y=1 6 arctan⁡(2 x−5)+4? Step 1: We rearrange the equation such that the values that impact the range are on the left and the values that impact the domain are on the right. y=1 6 arctan⁡(2 x−5)+4 y−4=1 6 arctan⁡(2 x−5)6(y−4)=arctan⁡(2 x−5)6 y−24=arctan⁡(2 x−5) Step 2: We substitute the term (b x+c) into the domain inequality for y=arctan⁡(x) and simplify. −∞<b x+c<∞−∞<2 x−5<∞−∞+5 2<x<∞+5 2−∞<x<∞ Since we are dealing with ∞ in the domain, we once again arrive at a domain of (−∞,∞). Step 3: We substitute the y−d a portion of our equation into the range inequality for arctan⁡(x) and simplify. −π 2<y−d a<π 2−π 2<6 y−24<π 2−π 2+24<6 y<π 2+24−π 12+4<y<π 12+4 The range of the tangent inverse function is (−π 2+4,π 12+4). Get access to thousands of practice questions and explanations! 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15738	https://mathcs.holycross.edu/~dbd/old_courses/analysis_seminar/assignments/discussion8/node2.html	The Probability of Return to the Origin Next:About this document ...Up:No TitlePrevious:Introduction The Probability of Return to the Origin We had asked for the probability with which a random walk in the line will pass through a given position. If this position is the origin, we are asking for the probability of returning to the origin. Let us make an attempt at answering this question. Suppose x is a point in the integer lattice on the real line (fancy language for x is an integer) and suppose n is the number of steps in a random walk. We begin by considering n-step random walks. Compute the number N n,x of n-step random walks that are at location x after n steps. That is, calculate the number n-step random walks with . (It may help to write n = p + q where p is the number of +1's appearing in the walk and q is the number of -1's appearing in the walk up to this point.) Find the probability p n,x for an infinite random walk to satisfy . The case where x = 0 is of particular use in this argument, so let us give it a separate label, u n. Notice, n must be even in this case. Why? As in the gambler's ruin problem, it will be useful to keep track of the first return to the origin. Let f 2 k denote the probablity of the first return occurring at step 2 k, that is f 2 k is the probability of . Show that the probability that a random walk returns to the origin is Show that the _u_'s and the _f_'s are related by This will not be immediately useful to us, but we will be able to use it further on in the discussion of random walks. In order to proceed with the calculation of the probability of return to the origin, we must express the f's in a simple fashion. This will be accomplished in several steps. First, it will be useful to visualize a random walk as a piecewise-linear graph of a function in the plane. The graph is obtained by connecting the points . Show that for , the number of random walks from (x,y) to (x',y'), x' >x, that touch or cross the x-axis is the same as the number of random walks from (x,-y) to (x',y'). (We are concerned only with the number segments of random walks of length x' - x steps here.) Hint: Think geometrically. Draw a generic picture for this situation as a piecewise-linear plot in the plane and show that each random walk of the first type can be altered to produce a random walk of the second type. Show that the number of n-step random walks with and for i> 1 is the same as the number of n-step random walks with , , and for i> 1. Use this observation and the previous geometric fact to compute the number of n-step random walks from the origin to (n,x) with . Show that the probability that , is twice the probability that , . Show that the probability of satisfying is the same as the sum over all r>0 of the probabilities that Show that the previous sum of probablities is equal to u 2 k. What does this say in terms of probabilities of random walks. Use this to express f 2 k in terms of the u's. What is the probability that an infinite step random walk will return to the origin? Next:About this document ...Up:No TitlePrevious:Introduction 2000-03-07
15739	https://fastspring.com/blog/how-to-format-30-currencies-from-countries-all-over-the-world/	Skip to content Product Global Online Payments Easily transact across borders Fraud Prevention Intelligent alerts, fewer chargebacks Subscription Management + Billing Flexible recurring billing tools Affiliate Marketing Find engaged buyers for your product Checkout Localized, Embedded, and Branded Digital Invoicing Pain-free B2B digital sales Tax Compliance We handle all taxes including VAT Interactive Quotes Streamline custom SaaS deals Reporting and Analytics Understand Everything about Subscriptions and Revenue Merchant of Record We handle the complexities of global selling for you Developer Tools Review the most recent resources for building your store experience Explore all features Use Cases SaaS Accelerate your SaaS revenue Video Games Sell games or in-game purchases Digital Products Coverage from checkout to fulfillment Downloadable Software Save time while increasing conversions Mobile Apps Subscription or one-time payments for mobile apps eLearning and Courses Easily sell courses on your own site B2B Close PLG and sales-assisted prospects faster Customer Stories How FastSpring sellers succeed Developers Documentation Learn how to use FastSpring Store Builder Library JavaScript tools for commerce API Reference Endpoints, methods, and examples Code Samples Examples of common patterns Webhooks Handle important events Changelog See what's new per release Integrations Connect your stack to FastSpring Status Check in on the platform Resources Customer Stories They succeeded. We helped! Blog SaaS trends and best practices Documentation Learn how to use FastSpring Events Hang out with the FastSpring team [#### [Customer Story] Why TestDome Considers FastSpring a Real Partner TestDome has loved FastSpring's product and support for a long time. FIND OUT WHY →]( Support (opens in new tab) Sell SaaS, software, and digital goods Sign In B2B sales quotes to close deals fast Sign In Question a Charge Try Now Schedule Demo Loading search... Show results from: fastspring.com Docs Results from fastspring.com Get more than a payment gateway, right out of the gate. Go farther faster with FastSpring. Schedule Demo or Try Now November 19th, 2019 How to Format 30+ Currencies from Countries All Over the World Caitie Gonzalez Estimated read time: 8 minutes, 42 seconds It has never been easier to sell products to customers all over the world. And if you sell only digital products in your online store, there are even fewer limitations. But selling to international customers involves a lot more than converting dollar amounts and slapping a different currency symbol in front of it. There are four main elements you need to be aware of when formatting prices on your website. And yes, you need to alter all four elements to correctly localize your pricing. If you localize halfway, customers will wonder what else you’re doing only halfway. So, if you’re going to display different currencies based on your customer’s geographic location, do it right. Note, the formatting may vary depending on the language and region settings of international shoppers’ browsers. Here is the list of four formatting elements to consider when creating localized prices and a comprehensive list of countries and their currency formats. 1. Symbol Placement In the US, we’re used to the dollar symbol being to the left of the dollar amount. But currency symbol placement changes based on geographic location. For example, countries in the European Union place the euro symbol to the right of the dollar amount. But to make things interesting, Canada does it both ways. In English Canada, the dollar symbol goes to the left of the amount, while in French Canada, the dollar sign goes to the right. As a US business, you may be tempted to keep all currency symbols to the left of the dollar amount to keep things consistent. But don’t do it. While it may make sense for you, changing the placement of the currency symbol will be confusing for your foreign customers. 2. Spacing When you look at price tags, you probably don’t think about spacing very much. Do you even know if the US adds a space between the dollar symbol and the amount? I don’t think I ever really paid attention. The answer is, no, there is no space between the dollar symbol and the amount for US currency. On the other hand, there is a non-breaking space between the dollar amount and the euro sign for EU countries. But don’t assume spacing is determined by the placement of the currency symbol. There are no hard-and-fast rules when it comes to spacing; you just have to check the pricing format for individual countries and locations. 3. Commas vs. Decimal Points In the US, dollars and cents are separated by a decimal point and thousands are separated by a comma. But in many countries the opposite is true—they use commas for fractional separators and decimals for thousand separators. To determine how countries use commas and decimals in their pricing format, you have to check on each location. But again, Canada likes to mix it up. They us a decimal as the fractional separator in the English-speaking parts of the country, and a comma as the fractional separator in the French-speaking parts of the country. 4. Abbreviations vs. Symbols Each country has a three-letter code for their currency. These codes are usually meant for the international banking business and data interchange between computers. Which means most people aren’t familiar with the three-letter codes. However, you can use these banking codes as a last resort if your website’s font can’t handle specific currency symbols. If you need to decide between the currency’s abbreviation/code or symbol, choose only one. While using both the abbreviation and symbol seems to make sense, it looks weird. Once you decide between the symbol or code, be consistent across all currencies. Now that you know what to look for in currency formats, let’s look at how some of the most popular countries format their pricing. Popular Country Currencies Argentina Currency: Argentine Peso Abbreviation/Code: ARS Symbol: $ Format: Symbol to the left of the amount, non-breaking space, comma is fractional separator, decimal is thousand separator Example: $ 1.234,56 Australia Currency: Australian Dollar Abbreviation/Code: AUD Symbol: $ Format: Symbol to the left of the amount, non-breaking space, decimal is fractional separator, comma is thousand separator Example: $ 1,234.56 Brazil Currency: Brazilian Real Abbreviation/Code: BRL Symbol: R$ Format: Symbol to the left of the amount, non-breaking space, comma is fractional separator, decimal is thousand separator Example: R$ 1.234,56 Canada Currency: Canadian Dollar Abbreviation/Code: CAD Symbol: $ Format: Symbol to the left of the amount, non-breaking space, decimal is fractional separator, comma is thousand separator Example: $ 1,234.56 Canada (French) Currency: Canadian Dollar Abbreviation/Code: CAD Symbol: $ Format: Symbol to the right of amount, non-breaking space, comma is fractional separator, decimal is thousand separator Example: 1.234,56 $ Chile Currency: Chilean Peso Abbreviation/Code: CLP Symbol: $ Format: Symbol to the left of the amount, non-breaking space, decimal is fractional separator, comma is thousand separator Example: $ 1,234.56 China Currency: Yuan Renminbi Abbreviation/Code: CNY Symbol: ¥ Format: Symbol to the left of the amount, non-breaking space, decimal is fractional separator, comma is thousand separator Example: ¥ 1,234.56 Columbia Currency: Colombian Peso Abbreviation/Code: COP Symbol: $ Format: Symbol to the left of amount, non-breaking space, decimal is fractional separator, comma is thousand separator Example: $ 1,234.56 Czech Republic Currency: Czech Koruna Abbreviation/Code: CZK Symbol: Kč Format: Symbol to the right of the amount, non-breaking space, comma is fractional separator, decimal is thousand separator Example: 1.234,56 Kč Denmark Currency: Danish Krone Abbreviation/Code: DKK Symbol: kr. Format: Symbol to the left of the amount, non-breaking space, comma is fractional separator, decimal is thousand separator Example: kr. 1.234,56 European Union (Austria, Belgium, Bulgaria, Croatia, Republic of Cyprus, Czech Republic, Denmark, Estonia, Finland, France, Germany, Greece, Hungary, Ireland, Italy, Latvia, Lithuania, Luxembourg, Malta, Netherlands, Poland, Portugal, Romania, Slovakia, Slovenia, Spain, and Sweden.) Currency: Euro Abbreviation/Code: EUR Symbol: € Format: Symbol to the right of the amount, non-breaking space, comma is fractional separator, decimal is thousand separator Example: €1.234,56 Hong Kong Currency: Hong Kong Dollar Abbreviation/Code: HKD Symbol: HK$ Format: Symbol to the left of the amount, non-breaking space, decimal is fractional separator, comma is thousand separator Example: HK$ 1,234.56 Hungary Currency: Hungarian Forint Abbreviation: HUF Symbol: Ft Format: Symbol to the right of the amount, non-breaking space, comma is fractional separator, decimal is thousand separator Example: 1.234,56 Ft India Currency: Indian Rupee Abbreviation/Code: INR Symbol: ₹ Format: Symbol to the left of the amount, non-breaking space, decimal is fractional separator, comma is thousand separator Example: ₹ 1,234.56 Israel Currency: New Israeli Shekel Abbreviation/Code: ILS Symbol: ₪ Format: Symbol to the left of the amount, non-breaking space, comma is fractional separator, decimal is thousand separator Example: ₪ 1.234,56 Japan Currency: Yen Abbreviation/Code: JPY Symbol: ¥ Format: Symbol to the left of the amount, non-breaking space, decimal is fractional separator, comma is thousand separator Example: ¥ 1,234.56 Korea, Republic of Currency: Won Abbreviation/Code: KRW Symbol: ₩ Format: Symbol to the left of the amount, non-breaking space, decimal is fractional separator, comma is thousand separator Example: ₩ 1,234.56 Malaysia Currency: Malaysian Ringgit Abbreviation/Code: MYR Symbol: RM Format: Symbol to the left of the amount, non-breaking space, decimal is fractional separator, comma is thousand separator Example: RM 1,234.56 Mexico Currency: Mexican Peso Abbreviation/Code: MXN Symbol: $ Format: Symbol to the left of the amount, non-breaking space, decimal is fractional separator, comma is thousand separator Example: $ 1,234.56 Morocco Currency: Moroccan Dirham Abbreviation/Code: MAD Symbol: .د.م. Format: Symbol to the right of the amount, non-breaking space, decimal is fractional separator, comma is thousand separator Example: 1,234.56 .د.م. New Zealand Currency: New Zealand Dollar Abbreviation/Code: NZD Symbol: $ Format: Symbol to the left of the amount, non-breaking space, decimal is fractional separator, comma is thousand separator Example: $ 1,234.56 Norway Currency: Norwegian Krone Abbreviation/Code: NOK Symbol: kr Format: Symbol to the left of the amount, non-breaking space, decimal is fractional separator, comma is thousand separator Example: kr 1,234.56 Philippines Currency: Philippine Peso Abbreviation/Code: PHP Symbol: ₱ Format: Symbol to the left of the amount, non-breaking space, decimal is fractional separator, comma is thousand separator Example: ₱ 1,234.56 Poland Currency: Zloty Abbreviation/Code: PLN Symbol: zł Format: Symbol to the right of the amount, non-breaking space, comma is fractional separator, decimal is thousand separator Example: 1.234,56 zł Russian Federation Currency: Russian Ruble Abbreviation/Code: RUB Symbol: p. Format: Symbol to the right of the amount, non-breaking space, comma is fractional separator, decimal is thousand separator Example: 1.234,56 p. Saudi Arabia Currency: Saudi Riyal Abbreviation/Code: SAR Symbol: ﷼ Format: Symbol to the right of the amount, non-breaking space, decimal is fractional separator, comma is thousand separator Example: 1,234.56 ﷼ Singapore Currency: Singapore Dollar Abbreviation/Code: SGD Symbol: $ Format: Symbol to the left of the amount, no space, decimal is fractional separator, comma is thousand separator Example: $1,234.56 South Africa Currency: Rand Abbreviation/Code: ZAR Symbol: R Format: Symbol to the left of the amount, non-breaking space, decimal is fractional separator, comma is thousand separator Example: R 1,234.56 Sweden Currency: Swedish Krona Abbreviation/Code: SEK Symbol: kr Format: Symbol to the right of the amount, non-breaking space, comma is fractional separator, decimal is thousand separator Example: 1.234,56 kr Switzerland Currency: Swiss Franc Abbreviation/Code: CHF Symbol: fr. Format: Symbol to the left of the amount, non-breaking space, comma is fractional separator, decimal is thousand separator Example: fr. 1.234,56 Taiwan Currency: New Taiwan Dollar Abbreviation/Code: TWD Symbol: 元 Format: Symbol to the left of the amount, non-breaking space, decimal is fractional separator, comma is thousand separator Example: 元 1,234.56 Thailand Currency: Baht Abbreviation/Code: THB Symbol: ฿ Format: Symbol to the right of the amount, non-breaking space, decimal is fractional separator, comma is thousand separator Example: 1,234.56 ฿ Turkey Currency: Turkish Lira Abbreviation/Code: TRY Symbol: ₺ Format: Symbol to the right of the amount, non-breaking space, decimal is fractional separator, comma is thousand separator Example: 1,234.56 ₺ UK/Great Britain Currency: Pound Sterling Abbreviation/Code: GBP Symbol: ‎£ Format: Symbol to the left of the amount, no space, decimal is fractional separator, comma is thousand separator Example: £1,234.56 United States Currency: US Dollar Abbreviation/Code: USD Symbol: $ Format: Symbol to the left of the amount, no space, decimal is fractional separator, comma is thousand separator Example: $1,234.56 Vietnam Currency: Dong Abbreviation/Code: VND Symbol: ₫ Format: Symbol to the right of the amount, non-breaking space, comma is fractional separator, decimal is thousand separator Example: 1.234,56 ₫ We’ve gone through just about every country and their currency, but now I have a secret to tell you. There’s an easier way to keep your currencies consistent with all geo-locations. All you have to do is get a full-service ecommerce partner, like FastSpring, to do it for you. There’s no need to manually change currency formats to keep international customers happy when FastSpring can handle all localization for you. 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15740	https://courses.lumenlearning.com/suny-microbiology/chapter/agglutination-assays/	Agglutination Assays \| Microbiology Skip to main content Microbiology Laboratory Analysis of the Immune Response Search for: Agglutination Assays Learning Objectives Compare direct and indirect agglutination Identify various uses of hemagglutination in the diagnosis of disease Explain how blood types are determined Explain the steps used to cross-match blood to be used in a transfusion In addition to causing precipitation of soluble molecules and flocculation of molecules in suspension, antibodies can also clump together cells or particles (e.g., antigen-coated latex beads) in a process called agglutination (Figure 7 in Overview of Specific Adaptive Immunity). Agglutination can be used as an indicator of the presence of antibodies against bacteria or red blood cells. Agglutination assays are usually quick and easy to perform on a glass slide or microtiter plate (Figure 1). Microtiter plates have an array of wells to hold small volumes of reagents and to observe reactions (e.g., agglutination) either visually or using a specially designed spectrophotometer. The wells come in many different sizes for assays involving different volumes of reagents. Figure 1. Microtiter plates are used for conducting numerous reactions simultaneously in an array of wells. (credit: modification of work by “Microrao”/Wikimedia) Agglutination of Bacteria and Viruses The use of agglutination tests to identify streptococcal bacteria was developed in the 1920s by Rebecca Lancefield working with her colleagues A.R. Dochez and Oswald Avery. She used antibodies to identify M protein, a virulence factor on streptococci that is necessary for the bacteria’s ability to cause strep throat. Production of antibodies against M protein is crucial in mounting a protective response against the bacteria. Lancefield used antisera to show that different strains of the same species of streptococci express different versions of M protein, which explains why children can come down with strep throat repeatedly. Lancefield classified beta-hemolytic streptococci into many groups based on antigenic differences in group-specific polysaccharides located in the bacterial cell wall. The strains are called serovars because they are differentiated using antisera. Identifying the serovars present in a disease outbreak is important because some serovars may cause more severe disease than others. Figure 2. Antibodies against six different serovars of Group A strep were attached to latex beads. Each of the six antibody preparations was mixed with bacteria isolated from a patient. The tiny clumps seen in well 4 are indicative of agglutination, which is absent from all other wells. This indicates that the serovar associated with well 4 is present in the patient sample. (credit: modification of work by American Society for Microbiology) The method developed by Lancefield is a direct agglutination assay, since the bacterial cells themselves agglutinate. A similar strategy is more commonly used today when identifying serovars of bacteria and viruses; however, to improve visualization of the agglutination, the antibodies may be attached to inert latex beads. This technique is called an indirect agglutination assay (or latex fixation assay), because the agglutination of the beads is a marker for antibody binding to some other antigen (Figure 2). Indirect assays can be used to detect the presence of either antibodies or specific antigens. To identify antibodies in a patient’s serum, the antigen of interest is attached to latex beads. When mixed with patient serum, the antibodies will bind the antigen, cross-linking the latex beads and causing the beads to agglutinate indirectly; this indicates the presence of the antibody (Figure 3). This technique is most often used when looking for IgM antibodies, because their structure provides maximum cross-linking. One widely used example of this assay is a test for rheumatoid factor (RF) to confirm a diagnosis of rheumatoid arthritis. RF is, in fact, the presence of IgM antibodies that bind to the patient’s own IgG. RF will agglutinate IgG-coated latex beads. In the reverse test, soluble antigens can be detected in a patient’s serum by attaching specific antibodies (commonly mAbs) to the latex beads and mixing this complex with the serum (Figure 3). Figure 3. (a) Latex beads coated with an antigen will agglutinate when mixed with patient serum if the serum contains IgM antibodies against the antigen. (b) Latex beads coated with antibodies will agglutinate when mixed with patient serum if the serum contains antigens specific to the antibodies. Agglutination tests are widely used in underdeveloped countries that may lack appropriate facilities for culturing bacteria. For example, the Widal test, used for the diagnosis of typhoid fever, looks for agglutination of Salmonella enterica subspecies typhi in patient sera. The Widal test is rapid, inexpensive, and useful for monitoring the extent of an outbreak; however, it is not as accurate as tests that involve culturing of the bacteria. The Widal test frequently produces false positives in patients with previous infections with other subspecies of Salmonella, as well as false negatives in patients with hyperproteinemia or immune deficiencies. In addition, agglutination tests are limited by the fact that patients generally do not produce detectable levels of antibody during the first week (or longer) of an infection. A patient is said to have undergone seroconversion when antibody levels reach the threshold for detection. Typically, seroconversion coincides with the onset of signs and symptoms of disease. However, in an HIV infection, for example, it generally takes 3 weeks for seroconversion to take place, and in some instances, it may take much longer. Similar to techniques for the precipitin ring test and plaque assays, it is routine to prepare serial two-fold dilutions of the patient’s serum and determine the titer of agglutinating antibody present. Since antibody levels change over time in both primary and secondary immune responses, by checking samples over time, changes in antibody titer can be detected. For example, a comparison of the titer during the acute phase of an infection versus the titer from the convalescent phase will distinguish whether an infection is current or has occurred in the past. It is also possible to monitor how well the patient’s immune system is responding to the pathogen. Watch this video that demonstrates agglutination reactions with latex beads. Think about It How is agglutination used to distinguish serovars from each other? In a latex bead assay to test for antibodies in a patient’s serum, with what are the beads coated? What has happened when a patient has undergone seroconversion? Hemagglutination Agglutination of red blood cells is called hemagglutination. One common assay that uses hemagglutination is the direct Coombs’ test, also called the direct antihuman globulin test (DAT), which generally looks for nonagglutinating antibodies. The test can also detect complement attached to red blood cells. The Coombs’ test is often employed when a newborn has jaundice, yellowing of the skin caused by high blood concentrations of bilirubin, a product of the breakdown of hemoglobin in the blood. The Coombs’ test is used to determine whether the child’s red blood cells have been bound by the mother’s antibodies. These antibodies would activate complement, leading to red blood cell lysis and the subsequent jaundice. Other conditions that can cause positive direct Coombs’ tests include hemolytic transfusion reactions, autoimmune hemolytic anemia, infectious mononucleosis (caused by Epstein-Barr virus), syphilis, and Mycoplasma pneumonia. A positive direct Coombs’ test may also be seen in some cancers and as an allergic reaction to some drugs (e.g., penicillin). The antibodies bound to red blood cells in these conditions are most often IgG, and because of the orientation of the antigen-binding sites on IgG and the comparatively large size of a red blood cell, it is unlikely that any visible agglutination will occur. However, the presence of IgG bound to red blood cells can be detected by adding Coombs’ reagent, an antiserum containing antihuman IgG antibodies (that may be combined with anti-complement) (Figure 4). The Coombs’ reagent links the IgG attached to neighboring red blood cells and thus promotes agglutination. There is also an indirect Coombs’ test known as the indirect antiglobulin test (IAT). This screens an individual for antibodies against red blood cell antigens (other than the A and B antigens) that are unbound in a patient’s serum (Figure 4). IAT can be used to screen pregnant women for antibodies that may cause hemolytic disease of the newborn. It can also be used prior to giving blood transfusions. More detail on how the IAT is performed is discussed below. Figure 4. Click for a larger image. The steps in direct and indirect Coombs’ tests are shown in the illustration. Antibodies that bind to red blood cells are not the only cause of hemagglutination. Some viruses also bind to red blood cells, and this binding can cause agglutination when the viruses cross-link the red blood cells. For example, influenza viruses have two different types of viral spikes called neuraminidase (N) and hemagglutinin (H), the latter named for its ability to agglutinate red blood cells (see Viruses). Thus, we can use red blood cells to detect the presence of influenza virus by direct hemagglutination assays (HA), in which the virus causes visible agglutination of red blood cells. The mumps and rubella viruses can also be detected using HA. Most frequently, a serial dilution viral agglutination assay is used to measure the titer or estimate the amount of virus produced in cell culture or for vaccine production. A viral titer can be determined using a direct HA by making a serial dilution of the sample containing the virus, starting with a high concentration of sample that is then diluted in a series of wells. The highest dilution producing visible agglutination is the titer. The assay is carried out in a microtiter plate with V- or round-bottomed wells. In the presence of agglutinating viruses, the red blood cells and virus clump together and produce a diffuse mat over the bottom of the well. In the absence of virus, the red blood cells roll or sediment to the bottom of the well and form a dense pellet, which is why flat-bottomed wells cannot be used (Figure 5). Figure 5. A viral suspension is mixed with a standardized amount of red blood cells. No agglutination of red blood cells is visible when the virus is absent, and the cells form a compact pellet at the bottom of the well. In the presence of virus, a diffuse pink precipitate forms in the well. (credit bottom: modification of work by American Society for Microbiology) A modification of the HA assay can be used to determine the titer of antiviral antibodies. The presence of these antibodies in a patient’s serum or in a lab-produced antiserum will neutralize the virus and block it from agglutinating the red cells, making this a viral hemagglutination inhibition assay (HIA). In this assay, patient serum is mixed with a standardized amount of virus. After a short incubation, a standardized amount of red blood cells is added and hemagglutination is observed. The titer of the patient’s serum is the highest dilution that blocks agglutination (Figure 6). Figure 6. In this HIA, serum containing antibodies to influenzavirus underwent serial two-fold dilutions in a microtiter plate. Red blood cells were then added to the wells. Agglutination only occurred in those wells where the antibodies were too dilute to neutralize the virus. The highest concentration at which agglutination occurs is the titer of the antibodies in the patient’s serum. In the case of this test, Sample A shows a titer of 128, and Sample C shows a titer of 64. (credit: modification of work by Evan Burkala) Think about It What is the mechanism by which viruses are detected in a hemagglutination assay? Which hemagglutination result tells us the titer of virus in a sample? Animals in the Laboratory Much of what we know today about the human immune system has been learned through research conducted using animals—primarily, mammals—as models. Besides research, mammals are also used for the production of most of the antibodies and other immune system components needed for immunodiagnostics. Vaccines, diagnostics, therapies, and translational medicine in general have all been developed through research with animal models. Consider some of the common uses of laboratory animals for producing immune system components. Guinea pigs are used as a source of complement, and mice are the primary source of cells for making mAbs. These mAbs can be used in research and for therapeutic purposes. Antisera are raised in a variety of species, including horses, sheep, goats, and rabbits. When producing an antiserum, the animal will usually be injected at least twice, and adjuvants may be used to boost the antibody response. The larger animals used for making antisera will have blood harvested repeatedly over long periods of time, with little harm to the animals, but that is not usually the case for rabbits. Although we can obtain a few milliliters of blood from the ear veins of rabbits, we usually need larger volumes, which results in the deaths of the animals. We also use animals for the study of disease. The only way to grow Treponema pallidum for the study of syphilis is in living animals. Many viruses can be grown in cell culture, but growth in cell culture tells us very little about how the immune system will respond to the virus. When working on a newly discovered disease, we still employ Koch’s postulates, which require causing disease in lab animals using pathogens from pure culture as a crucial step in proving that a particular microorganism is the cause of a disease. Studying the proliferation of bacteria and viruses in animal hosts, and how the host immune system responds, has been central to microbiological research for well over 100 years. While the practice of using laboratory animals is essential to scientific research and medical diagnostics, many people strongly object to the exploitation of animals for human benefit. This ethical argument is not a new one—indeed, one of Charles Darwin’s daughters was an active antivivisectionist (vivisection is the practice of cutting or dissecting a live animal to study it). Most scientists acknowledge that there should be limits on the extent to which animals can be exploited for research purposes. Ethical considerations have led the National Institutes of Health (NIH) to develop strict regulations on the types of research that may be performed. These regulations also include guidelines for the humane treatment of lab animals, setting standards for their housing, care, and euthanization. The NIH document “Guide for the Care and Use of Laboratory Animals” makes it clear that the use of animals in research is a privilege granted by society to researchers. The NIH guidelines are based on the principle of the three R’s: replace, refine, and reduce. Researchers should strive to replace animal models with nonliving models, replace vertebrates with invertebrates whenever possible, or use computer-models when applicable. They should refine husbandry and experimental procedures to reduce pain and suffering, and use experimental designs and procedures that reduce the number of animals needed to obtain the desired information. To obtain funding, researchers must satisfy NIH reviewers that the research justifies the use of animals and that their use is in accordance with the guidelines. At the local level, any facility that uses animals and receives federal funding must have an Institutional Animal Care and Use Committee (IACUC) that ensures that the NIH guidelines are being followed. The IACUC must include researchers, administrators, a veterinarian, and at least one person with no ties to the institution, that is, a concerned citizen. This committee also performs inspections of laboratories and protocols. For research involving human subjects, an Institutional Review Board (IRB) ensures that proper guidelines are followed. Visit this site to view the NIH Guide for the Care and Use of Laboratory Animals. Blood Typing and Cross-Matching In addition to antibodies against bacteria and viruses to which they have previously been exposed, most individuals also carry antibodies against blood types other than their own. There are presently 33 immunologically important blood-type systems, many of which are restricted within various ethnic groups or rarely result in the production of antibodies. The most important and perhaps best known are the ABO and Rh blood groups (see Figure 3 in Hypersensitivities). When units of blood are being considered for transfusion, pretransfusion blood testing must be performed. For the blood unit, commercially prepared antibodies against the A, B, and Rh antigens are mixed with red blood cells from the units to initially confirm that the blood type on the unit is accurate. Once a unit of blood has been requested for transfusion, it is vitally important to make sure the donor (unit of blood) and recipient (patient) are compatible for these crucial antigens. In addition to confirming the blood type of the unit, the patient’s blood type is also confirmed using the same commercially prepared antibodies to A, B, and Rh. For example, as shown in Figure 7, if the donor blood is A-positive, it will agglutinate with the anti-A antiserum and with the anti-Rh antiserum. If no agglutination is observed with any of the sera, then the blood type would be O-negative. Following determination of the blood type, immediately prior to releasing the blood for transfusion, a cross-match is performed in which a small aliquot of the donor red blood cells are mixed with serum from the patient awaiting transfusion. If the patient does have antibodies against the donor red blood cells, hemagglutination will occur. To confirm any negative test results and check for sensitized red blood cells, Coombs’ reagent may be added to the mix to facilitate visualization of the antibody-red blood cell interaction. Under some circumstances, a minor cross-match may be performed as well. In this assay, a small aliquot of donor serum is mixed with patient red blood cells. This allows the detection of agglutinizing antibodies in the donor serum. This test is rarely necessary because transfusions generally use packed red blood cells with most of the plasma removed by centrifugation. Red blood cells have many other antigens in addition to ABO and Rh. While most people are unlikely to have antibodies against these antigens, women who have had multiple pregnancies or patients who have had multiple transfusions may have them because of repeated exposure. For this reason, an antibody screen test is used to determine if such antibodies are present. Patient serum is checked against commercially prepared, pooled, type O red blood cells that express these antigens. If agglutination occurs, the antigen to which the patient is responding must be identified and determined not to be present in the donor unit. Figure 7. This sample of a commercially produced “bedside” card enables quick typing of both a recipient’s and donor’s blood before transfusion. The card contains three reaction sites or wells. One is coated with an anti-A antibody, one with an anti-B antibody, and one with an anti-Rh antibody. Agglutination of red blood cells in a given site indicates a positive identification of the blood antigens: in this case, A and Rh antigens for blood type A-positive. Think about It If a patient’s blood agglutinates with anti-B serum, what is the patient’s blood type? What is a cross-match assay, and why is it performed? Table 1 summarizes the various kinds of agglutination assays discussed in this section. \| Table 1. Mechanisms of Select Antibody-Antigen Assays \| \| Type of Assay \| Mechanism \| Example \| \| Agglutination \| Direct: Antibody is used to clump bacterial cells or other large structures \| Serotyping bacteria \| \| Indirect: Latex beads are coupled with antigen or antibody to look for antibody or antigen, respectively, in patient serum \| Confirming the presence of rheumatoid factor (IgM-binding Ig) in patient serum \| \| Hemagglutination \| Direct: Some bacteria and viruses cross-link red blood cells and clump them together \| Diagnosing influenza, mumps, and measles \| \| Direct Coombs’ test (DAT): Detects nonagglutinating antibodies or complement proteins on red blood cells in vivo \| Checking for maternal antibodies binding to neonatal red blood cells \| \| Indirect Coombs’ test (IAT): Screens an individual for antibodies against red blood cell antigens (other than the A and B antigens) that are unbound in a patient’s serum in vitro \| Performing pretransfusion blood testing \| \| Viral hemagglutination inhibition: Uses antibodies from a patient to inhibit viral agglutination \| Diagnosing various viral diseases by the presence of patient antibodies against the virus \| \| Blood typing and cross-matching: Detects ABO, Rh, and minor antigens in the blood \| Matches donor blood to recipient immune requirements \| Key Concepts and Summary Antibodies can agglutinate cells or large particles into a visible matrix. Agglutination tests are often done on cards or in microtiter plates that allow multiple reactions to take place side by side using small volumes of reagents. Using antisera against certain proteins allows identification of serovars within species of bacteria. Detecting antibodies against a pathogen can be a powerful tool for diagnosing disease, but there is a period of time before patients go through seroconversion and the level of antibodies becomes detectable. Agglutination of latex beads in indirect agglutination assays can be used to detect the presence of specific antigens or specific antibodies in patient serum. The presence of some antibacterial and antiviral antibodies can be confirmed by the use of the direct Coombs’ test, which uses Coombs’ reagent to cross-link antibodies bound to red blood cells and facilitate hemagglutination. Some viruses and bacteria will bind and agglutinate red blood cells; this interaction is the basis of the direct hemagglutination assay, most often used to determine the titer of virus in solution. Neutralization assays quantify the level of virus-specific antibody by measuring the decrease in hemagglutination observed after mixing patient serum with a standardized amount of virus. Hemagglutination assays are also used to screen and cross-match donor and recipient blood to ensure that the transfusion recipient does not have antibodies to antigens in the donated blood. Multiple Choice We use antisera to distinguish between various __ within a species of bacteria. isotypes serovars subspecies lines Show Answer Answer b.We use antisera to distinguish between various serovars within a species of bacteria. When using antisera to characterize bacteria, we will often link the antibodies to __ to better visualize the agglutination. latex beads red blood cells other bacteria white blood cells Show Answer Answer a.When using antisera to characterize bacteria, we will often link the antibodies to latex beads to better visualize the agglutination. The antibody screening test that is done along with pretransfusion blood typing is used to ensure that the recipient does not have a previously undetected bacterial or viral infection. is not immunocompromised. actually does have the blood type stated in the online chart. is not making antibodies against antigens outside the ABO or Rh systems. Show Answer Answer d. This test is done to ensure that the recipient is not making antibodies against antigens outside the ABO or Rh systems. The direct Coombs’ test is designed to detect when people have a disease that causes them to have an excessively high fever. quit making antibodies. make too many red blood cells. produce antibodies that bind to their own red blood cells. Show Answer Answer d. The direct Coombs’ test is designed to detect when people have a disease that causes them to produce antibodies that bind to their own red blood cells. Viral hemagglutination assays only work with certain types of viruses because the virus must be able to cross-link red blood cells directly. the virus must be able to lyse red blood cells. the virus must not be able to lyse red blood cells. other viruses are too dangerous to work with in a clinical lab setting. Show Answer Answer a.Viral hemagglutination assays only work with certain types of viruses because the virus must be able to cross-link red blood cells directly. Fill in the Blank In the major cross-match, we mix __ with the donor red blood cells and look for agglutination. Show Answer In the major cross-match, we mix patient serum with the donor red blood cells and look for agglutination. Coombs’ reagent is an antiserum with antibodies that bind to human __. Show Answer Coombs’ reagent is an antiserum with antibodies that bind to human immunoglobulins, antibodies, and/or complement. Think about It Explain why the titer of a direct hemagglutination assay is the highest dilution that still causes hemagglutination, whereas in the viral hemagglutination inhibition assay, the titer is the highest dilution at which hemagglutination is not observed. Why would a doctor order a direct Coombs’ test when a baby is born with jaundice? Critical Thinking When shortages of donated blood occur, O-negative blood may be given to patients, even if they have a different blood type. Why is this the case? If O-negative blood supplies were depleted, what would be the next-best choice for a patient with a different blood type in critical need of a transfusion? Explain your answers. Candela Citations CC licensed content, Shared previously OpenStax Microbiology. Provided by: OpenStax CNX. Located at: License: CC BY: Attribution. License Terms: Download for free at Lancefield, Rebecca C., "The Antigenic Complex of Streptococcus haemoliticus. I. Demonstration of a Type-Specific Substance in Extracts of Streptococcus haemolyticus," The Journal of Experimental Medicine 47, no. 1 (1928): 91-103. ↵ Licenses and Attributions CC licensed content, Shared previously OpenStax Microbiology. Provided by: OpenStax CNX. Located at: License: CC BY: Attribution. License Terms: Download for free at PreviousNext Privacy Policy
15741	https://www.hmhco.com/blog/teaching-ratios-and-unit-rates-in-math?srsltid=AfmBOorO-l-3EiUOzf1TiA6Sxp0kFgUfvl0PLUHuO70Sv6EFS9EZn3OV	Teaching Ratios & Unit Rates in Math \| HMH Curriculum Literacy Core Curriculum Into Literature, 6-12 Into Reading, K-6 See all Literacy Performance Suite Intervention English 3D, K-12 Read 180, 3-12 See all Reading Intervention Assessment Supplemental A Chance in the World SEL, 8-12 Amira Learning, PreK–8 Classcraft, K-8 JillE Literacy, K-3 Waggle, K-8 Writable, 3-12 See all Reading Supplemental Personalized Path Math Core Curriculum Arriba las Matematicas, K-8 Go Math!, K-6 Into Algebra 1, Geometry, Algebra 2, 8-12 Into Math, K-8 Math in Focus, K-8 See all Math Performance Suite Supplemental Classcraft, K-8 Waggle, K-8 See all Supplemental Math Personalized Path Intervention Math 180, 3-12 See all Math Intervention See all Assessment Science Core Curriculum Into Science, K-5 Into Science, 6-8 Science Dimensions, K-12 See all Science Readers ScienceSaurus, K-8 Social Studies Core Curriculum HMH Social Studies, 6-12 See all Social Studies Supplemental Writable Professional Development For Teachers Coachly Teacher's Corner Live Online Courses Program-Aligned Courses See all Professional Development For Leaders The Center for Model Schools More AI Tools on HMH Ed Assessment Early Learning English Language Development Homeschool Intervention Literacy Mathematics Professional Development Science School Improvement Social Studies Special Education Summer School See all Solutions Resources Browse Resources AI Tools Classroom Activities Customer Resource Hub Customer Success Stories Digital Samples Events Grants & Funding Family & Caregiver Resources International Product Updates Research Library Shaped, HMH Blog Webinars Customer Support Contact Sales Customer Service & Technical Support Portal Platform Login Company Learn about us About Corporate Responsibility Leadership News Announcements Our Culture Our Legacy Join us Careers Educator Input Panel Suppliers Divisions Center for Model Schools Heinemann NWEA Shop Support Platform Login From setup to support, HMH will ensure your back-to-school season runs smoothly. 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The numbers or measurements being compared are sometimes called the terms of the ratio. For example, if a store sells 6 red shirts and 8 green shirts, the ratio of red to green shirts is 6 to 8. You can write this ratio as 6 red/8 green, 6 red:8 green—or when writing fast or trying to make a point—simply 6/8 or 6:8. Both expressions mean that there are 6 red shirts “for every” 8 green shirts. Notice how you can rewrite 6/8 as 3/4, no different from any other time a math concept can appear as a fraction. A rate is a special ratio in which the two terms are in different units. For example, if a 12-ounce can of corn costs 69¢, the rate is 69¢ for 12 ounces. This is not a ratio of two like units, such as shirts. This is a ratio of two unlike units: cents and ounces. The first term of the ratio (69¢) is measured in cents, and the second term (12) in ounces. You can write this rate as 69¢/12 ounces or 69¢:12 ounces. Both expressions mean that you pay 69¢ “for every” 12 ounces of corn, and similar to the shirt ratio, can enter calculations as the fraction 69/12. But notice that this time, a new unit is created: cents per ounce. Rates are used by people every day, such as when they work 40 hours per week or earn interest every year at a bank. When rates are expressed as a quantity of 1, such as 2 feet per second (that is, per 1 second) or 5 miles per hour (that is, per 1 hour), they can be defined as unit rates. You can write any rate as a unit rate by reducing the fraction so it has a 1 as the denominator or second term. As a unit rate example, you can show that the unit rate of 120 students for every 3 buses is 40 students per bus. 120/3 = 40/1 You could also find the unit rate by dividing the first term of the ratio by the second term. 120 ÷ 3 = 40 When a price is expressed as a quantity of 1, such as $25 per ticket or $0.89 per can, it is called a unit price. If you have a non-unit price, such as $5.50 for 5 pounds of potatoes, and want to find the unit price, divide the terms of the ratio. $5.50 ÷ 5 pounds = $1.10 per pound The unit price of potatoes that cost $5.50 for 5 pounds is $1.10 per pound. Rates in the Real World Rate and unit rate are used to solve many real-world problems. Look at the following problem. “Tonya works 60 hours every 3 weeks. At that rate, how many hours will she work in 12 weeks?” The problem tells you that Tonya works at the rate of 60 hours every 3 weeks. To find the number of hours she will work in 12 weeks, write a ratio equal to 60/3 that has a second term of 12. 60/3 = ?/1260/3 = 240/12 Removing the units makes the calculation easier to see. However, it is important to remember the units when interpreting the new ratio. Tonya will work 240 hours in 12 weeks. You could also solve this problem by first finding the unit rate and multiplying it by 12. 60/3 = 20/1 20 × 12 = 240 When you find equal ratios, it is important to remember that if you multiply or divide one term of a ratio by a number, then you need to multiply or divide the other term by that same number. Let's take a look at a problem that involves unit price. “A sign in a store says 3 Pens for $2.70. How much would 10 pens cost?” To solve the problem, find the unit price of the pens, then multiply by 10. $2.70 ÷ 3 pens = $0.90 per pen $0.90×10 pens = $9.00 Finding the cost of one unit enables you to find the cost of any number of units. What Is a Unit Rate in Math? Your students have no doubt encountered rates and ratios before (have they seen a speed limit sign?), but it may help them to review these concepts before solving problems that use them. Standard:Understand the concept of a unit rate a/b associated with a ratio a:b with b ≠ 0. (6.RP.A.2) Prerequisite Skills and Concepts: Students should have a basic understanding of ratios, how to write them, and an ability to simplify a ratio. Students should also have an ability to work with fractions and find equivalent fractions. Say:Today we are going to look at a special type of ratio called a rate. Does anyone know what I mean by a rate? Students may say that a rate is a ratio in which the quantities being compared use different units such as dollars and ounces or miles and hours. Students may use common English synonyms for rate such as speed. If so, point out that speed means calculating how fast something is going by comparing distance to time, such as miles to hours. If necessary, explain what a rate is. Say:Rates are commonly found in everyday life. The prices in grocery stores and department stores are often rates. Rates are also used in pricing gasoline or tickets, measuring speed, or paying hourly wages and monthly fees. Have students think of other examples of rates. In addition to common real-world examples, encourage silly or unusual rates such as hip-hop artists per zip code or diamond collars per chihuahua. Say:Two important ideas are unit rates and unit prices. What is the difference between rate and unit rate? Or price and unit price? Does anybody have any ideas? Students will probably not know what a unit rate is, so provide them with the following explanation to explain rate vs. unit rate. Say:A unit means one of something. A unit rate means a rate for one of something. We write this as a ratio with a denominator of one. For example, if you ran 70 yards in 10 seconds, you ran on average 7 yards in 1 second. Both of the ratios, 70 yards in 10 seconds and 7 yards in 1 second, are rates, but the 7 yards in 1 second is a unit rate. Ask:Now that you know what a unit rate is, what do you think a unit price is? Students will say that it is the price of one item. If they don't, tell them what it is. Ask:What is the unit price of 10 pounds of potatoes that costs $2.80? Help students calculate that the unit price is $0.28 per pound by dividing the price by 10. Share the following problem: “One store has carrots on sale for $1.14 for 3 pounds, while another store has carrots on sale for $0.78 for two pounds. Which store has the better deal?” Ask:What are we trying to find in this problem? Students should say that we are trying to find out which is the better deal for carrots when thinking about the cost of each carrot. Ask:What would help us find the better deal? Students should say that if we find the unit price for the carrots at each store, we would know which was the better deal. Say:Find the unit prices for the carrots at both stores and then we will discuss what you did. Have students independently calculate the unit price and answer which store has the better deal. Compare how different students did the calculations, and have students discuss similarities and differences among the models they used and solutions they found. Allow for varying answers such as “the second store had the better deal for me because I would only want two carrots anyway.” If time permits, have students solve the following problem as well. “One animal can run 60 feet in 4 seconds, while another animal can run 100 feet in 8 seconds. Which animal runs faster?” (The first animal runs faster at 15 feet per second.) Developing the Concept: Rates Now that students know how to find a unit rate, they will learn how to find an equivalent ratio using unit rates. Finding equivalent ratios uses the same thought process as finding equivalent fractions. Standard:Use ratio and rate reasoning to find equivalent ratios and solve real-world problems (6.RP.A.3) Say:Before we learned how to find a unit rate. Now we are going to learn how to use that unit rate to solve problems. Look at this problem. Share the following problem: “Yesterday Ebony ran 18 laps around the track in 12 minutes. If she runs at that rate for 30 laps, how long will it take her?” (Tip: You can substitute the context with anything that would interest your students.) Ask:What are we trying to find in this problem? We are trying to find out how long it takes Ebony to run 30 laps. Ask:What information do we know that will help us solve this problem? We know that Ebony can run 18 laps in 12 minutes. We also know she is going to run at that same rate for 30 laps. Ask:How far does Ebony run in one minute? Have students try to figure it out individually. Compare student solutions, and discuss why Ebony runs 1.5 laps in one minute. Say:Let's make a table to list the information we know. Make the following table but leave "Minutes" blank. Fill it in by soliciting class input. LapsMinutes 1.5 1 3 2 6 4 12 8 18 12 24 16 30 20 Ask:Does anyone see another way we could have found the time of 20 minutes without writing down the whole table? Compare the different ideas students offer. Some students might recognize that if they divided 30 by 1.5, they would get 20 minutes. Discuss this strategy with the class. Write the following problem on the board: “Maria wants to buy a pencil for everyone in her class. It costs $0.78 for 3 pencils. How much would Maria have to spend if she bought a pencil for each of her 24 classmates?” Say:I'd like you to solve this problem on your own, and then we'll discuss what you did. Have students individually share their solutions. Some students may have solved it by finding the unit price or completing a table. Others may have solved it by noticing that 3 pencils cost $0.78, and 24 = 3 × 8. So they multiplied $0.78 by 8 to get $6.24. Looking for more free math lessons and activities for elementary school students? 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15742	https://gdaymath.com/lessons/explodingdots/10-3-unusual-mathematics-unusual-numbers/	Published Time: Tue, 26 Aug 2025 12:12:16 GMT 10.3 Some Unusual Mathematics for Unusual Numbers \| G'Day Math Courses Resources Essays About Contact Exploding Dots 10.3 Some Unusual Mathematics for Unusual Numbers Lesson materials located below the video overview. Previous Next Course Home It is possible to develop an arithmetic system of numbers for which a quantity like …9999…9999 actually is meaningful (and by what we proved in the previous section has value −1−1). Here’s one such system, one that requires we change our sense of distance between numbers on the number line. It is a system that will allow us to say, for instance, that the numbers 9 9, 99 99, 999 999, … are indeed marching closer and closer to −1−1 on the number line despite what our geometric training says! Warping Normal Distance on the Number Line We usually say the number 5 5, for instance, is a distance five from 0 0 on the number line because 5 5 is five unit lengths away from the zero. (We usually use absolute value notation for this distance: \|5\|=5\|5\|=5.) And 3.7 3.7 is a distance 3.7 3.7 from 0 0, \|3.7\|=3.7\|3.7\|=3.7, because three-and-seven-tenths of a unit fit between 0 0 and 3.7 3.7 on the number line. And so on. This is a very additive way of thinking about distance: adding five 1 1 s gets you from 0 0 to 5 5; adding 3.7 3.7 1 1 s get you from 0 0 to 3.7 3.7; and so on. We can say that the distance of a point a a on the number line, in this thinking, is the number of 1 1 s that go additively into a a. But much of mathematics is not only concerned with the additive properties of numbers, but also the multiplicative properties of numbers. For example, many people are interested in the prime factorizations of numbers (for example, 1000=2 3 5 3 1000=2 3 5 3 and 105=3⋅5⋅7 105=3⋅5⋅7). There are so many unanswered questions about the prime numbers and prime factorizations still in mathematics today. These questions are, in general, very hard! Is there are way to bring the geometry of the number line into play to possibly help with multiplicative questions? Is there a way to think about the number line itself as perhaps structured multiplicatively rather than additively? To think about this, rather than focus on all possible factor of numbers, let’s focus on one possible factors of numbers. And to keep matters relevant to our base-ten arithmetic thinking, let’s focus on the number 10 10. In our additive thinking for distance on the number line we use the unit of 1 1 and ask how many ones (additively) go into each number for its distance from 0 0. We now want to use the unit of 10 10 and ask how many times ten goes multiplicatively into each number. What could that mean? In the world of integers the number 0 0 is the most divisible number of all: it can be divided by any integer any number of times and still give an integer result (namely 0 0) each and every time. Focusing on our chosen factor of ten, we can divide 0 0 by ten once, or twice, or thirty-seven times, and still have an integer. The number 40 40 is a little bit “zero-like” in this sense in that we can divide it by ten once and still have an integer. The number 1700 1700 is more zero-like as it can be divided by ten twice and still give an integer result. A googol is very much more zero-like: it can be divided by ten one hundred times and still stay an integer. The integer 5 5 is not very zero-like at all: one can’t divide it by ten even once and stay an integer. In this setting the more times ten “goes into” into a number multiplicatively, the more zero-like it is. So in this sense, a googol is much closer to zero than 5 5 is. So let’s develop a distance formula that regards numbers with large powers of ten as factors as closer to zero than numbers with less counts of powers of ten as factors. There are a number of ways one might think to do this, but let’s try to mimic the additive properties of the number line we are familiar with. Normally we would say that 850 850 is further from zero than 85 85 is, and, in fact, we might even say 850 850 is ten times further from zero as 85 85 is. In our multiplicative thinking,850 850 is now closer to zero than 85 85 is and it would be natural to have it as ten times closer. The following formula seems a natural way to have this happen. If a a can be divided by ten a maximum of k k times and remain an integer, then set \|a\|t e n=1 10 k\|a\|t e n=1 10 k. For example, then, \|850\|t e n=1 10 1=0.1\|850\|t e n=1 10 1=0.1 and \|8500\|t e n=1 10 2=0.01\|8500\|t e n=1 10 2=0.01 and \|8500000\|t e n=1 10 5=0.00001\|8500000\|t e n=1 10 5=0.00001. Also, since \|85\|t e n=1 10 0=1\|85\|t e n=1 10 0=1. we see, indeed, that 850 850 is ten times closer to zero than 85 85 is. We can also measure the distance between any two numbers in this multiplicative way. For example, the distance between 3 3 and 33 33 is \|33−3\|t e n=\|30\|t e n=1 10 1=0.1\|33−3\|t e n=\|30\|t e n=1 10 1=0.1. With this new way to measure distance, we see that 1 1, 10 10, 100 100, 1000 1000, …… is a sequence of numbers getting closer and closer to zero. We have \|1\|t e n=1\|1\|t e n=1 and \|10\|t e n=0.1\|10\|t e n=0.1 and \|100\|t e n=0.01\|100\|t e n=0.01 and \|1000\|t e n=0.001\|1000\|t e n=0.001, and so on, indeed approaching a distance of zero from 0 0. In terms of values in a 1←10 1←10 machine, we see that boxes far to the left in the machine, representing high powers of ten, are representing values very close to zero. (Before, in our additive thinking, boxes to the far right for decimals represented values very closer and closer to zero.) Mathematicians call this way of viewing distances between the non-negative integers ten-adic arithmetic. (The suffix adic means “a counting of operations” and here we are counting factors of ten.) It is fun to think how to extend this notion of distance to fractions too, and then to all real numbers. The number …9999…9999 Let’s look now at the sequence of numbers 9 9 and 99 99 and 999 999 and so on marching off to the right on the number line. Could they possibly be marching closer and closer to the value −1−1? Yes, if by “closer” we mean this new multiplicative way to think of distance. We have The numbers 9 9, 99 99, 999 999, 9999 9999, … are indeed approaching the value 0−1=−1 0−1=−1. Comment:We can now justify the (very) long addition computation given below. We first compute 9+1=10 9+1=10, and then we add 90 90 to this to obtain 100 100, and then we add 900 900 to obtain 1000 1000, and so on. The further along we go with the computation the closer our results are to the number zero. You can intuitively see this in the 1←10 1←10 machine: when you add one more dot to this loaded machine and perform the explosions, one clears away dots, pushing what remains further and further to the left where boxes have less and less significant value. Computation of this next (very) long multiplication is justified in a similar way. We first compute 3×7=21 3×7=21, and then add to this 3×60 3×60 to get 201 201, and then add to this 3×600 3×600 to get 2001 2001, and so on. Now the numbers 20 20, 200 200, 2000 2000, … are getting closer and closer to zero, so the numbers 21=20+1 21=20+1, 201=200+1 201=200+1, 2001=2000+1 2001=2000+1, … are getting closer and closer to 1 1. The further along we go with this computation, the closer our answers are to the number 1 1. Again, we can intuitively see this reasoning at play by tripling all the values in this loaded 1←10 1←10 machine. And did you discover that ⋯\|3\|3\|3\|3\|4⋯\|3\|3\|3\|3\|4 behaves like the fraction 2 3 2 3? Question:Doubling⋯\|3\|3\|3\|3\|4⋯\|3\|3\|3\|3\|4 gives ⋯\|6\|6\|6\|6\|8⋯\|6\|6\|6\|6\|8 which is one more than ⋯\|6\|6\|6\|6\|7⋯\|6\|6\|6\|6\|7, which is 1 3 1 3. Is this consistent? Challenge:Show that in a 2←3 2←3 machine that ⋯\|1\|1\|1\|1\|2⋯\|1\|1\|1\|1\|2 is negative one! Show that ⋯\|0\|1\|0\|1\|0\|1\|0\|2⋯\|0\|1\|0\|1\|0\|1\|0\|2 when multiplied by 5 5 gives 1 1, and so represents 1 5 1 5. (What measure of distance might we be using on the number line this time for these “numbers” to make sense?) Constructing Negative Integers In our base-ten thinking with our multiplicative notion of distance on the number line, we set \|a\|t e n=1 10 k\|a\|t e n=1 10 k where k k is the largest count of times a a can be divided by ten and remain an integer. And we have made sense of ⋯9999⋯9999 as a meaningful number with value −1−1. So what’s −2−2 in this unusual system of arithmetic? Let’s think in terms of a 1←10 1←10 machine. Since −1=⋯9\|9\|9\|9−1=⋯9\|9\|9\|9 , and −2−2 is double −1−1, we should have −2=⋯18\|18\|18\|18−2=⋯18\|18\|18\|18. With explosions we get −2=⋯18\|18\|18\|18−2=⋯18\|18\|18\|18 =⋯18\|18\|19\|8=⋯18\|18\|19\|8 =⋯18\|19\|9\|8=⋯18\|19\|9\|8 =⋯19\|9\|9\|8=⋯19\|9\|9\|8 =⋯9\|9\|9\|8=⋯9\|9\|9\|8. And one can check that this long addition does give zero. We can see now how to readily construct any negative integer. For example, we can see that adding 47 47 to ⋯999953⋯999953 will give zero and so this latter quantity must be −47−47, and that adding 3000 3000 to ⋯9997000⋯9997000 gives zero and so this quantity must be −3000−3000. Challenge:What is −2−2 in a 2←3 2←3 machine? What is −5−5? Constructing Fractions We saw that ⋯66667⋯66667 is the fraction 1 3 1 3: multiply this quantity by three and you get 1 1. The 1←10 1←10 machine provides a natural way to compute such fractions. For example, let’s find the ten-adic representation of 4 7 4 7 . That is, let’s find a number x x such that 7×x=4 7×x=4. Start by writing x=⋯h\|g\|f\|e\|d\|c\|b\|a x=⋯h\|g\|f\|e\|d\|c\|b\|a as for a 1←10 1←10 machine. Then 7 x=⋯7 h\|7 g\|7 f\|7 e\|7 d\|7 c\|7 b\|7 a 7 x=⋯7 h\|7 g\|7 f\|7 e\|7 d\|7 c\|7 b\|7 a. We want 7 a 7 a ,after explosions, to leave a 4 4. So we need a multiple of 7 7 four greater than a multiple of 10 10. We see that 7 a=14 7 a=14 is good. So let’s set a=2 a=2. x=⋯h\|g\|f\|e\|d\|c\|b\|2 x=⋯h\|g\|f\|e\|d\|c\|b\|2 7 x=⋯7 h\|7 g\|7 f\|7 e\|7 d\|7 c\|7 b\|14 7 x=⋯7 h\|7 g\|7 f\|7 e\|7 d\|7 c\|7 b\|14 =⋯7 h\|7 g\|7 f\|7 e\|7 d\|7 c\|7 b+1\|4=⋯7 h\|7 g\|7 f\|7 e\|7 d\|7 c\|7 b+1\|4 Now we want 7 b+1 7 b+1 to be a multiple of 10 10 so that all dots in that box explode to leave zero behind. This suggests b=7 b=7. x=⋯h\|g\|f\|e\|d\|c\|7\|2 x=⋯h\|g\|f\|e\|d\|c\|7\|2 7 x=⋯7 h\|7 g\|7 f\|7 e\|7 d\|7 c\|50\|4 7 x=⋯7 h\|7 g\|7 f\|7 e\|7 d\|7 c\|50\|4 =⋯7 h\|7 g\|7 f\|7 e\|7 d\|7 c+5\|0\|4=⋯7 h\|7 g\|7 f\|7 e\|7 d\|7 c+5\|0\|4 Now we need 7 c+5 7 c+5 a multiple of 10 10. Choose c=5 c=5. x=⋯h\|g\|f\|e\|d\|5\|7\|2 x=⋯h\|g\|f\|e\|d\|5\|7\|2 7 x=⋯7 h\|7 g\|7 f\|7 e\|7 d\|40\|0\|4 7 x=⋯7 h\|7 g\|7 f\|7 e\|7 d\|40\|0\|4 =⋯7 h\|7 g\|7 f\|7 e\|7 d+4\|0\|0\|4=⋯7 h\|7 g\|7 f\|7 e\|7 d+4\|0\|0\|4 Now choose d=8 d=8. x=⋯h\|g\|f\|e\|8\|5\|7\|2 x=⋯h\|g\|f\|e\|8\|5\|7\|2 7 x=⋯7 h\|7 g\|7 f\|7 e\|60\|0\|0\|4 7 x=⋯7 h\|7 g\|7 f\|7 e\|60\|0\|0\|4 =⋯7 h\|7 g\|7 f\|7 e+6\|0\|0\|0\|4=⋯7 h\|7 g\|7 f\|7 e+6\|0\|0\|0\|4 And then e=2 e=2. x=⋯h\|g\|f\|2\|8\|5\|7\|2 x=⋯h\|g\|f\|2\|8\|5\|7\|2 7 x=⋯7 h\|7 g\|7 f\|20\|0\|0\|0\|4 7 x=⋯7 h\|7 g\|7 f\|20\|0\|0\|0\|4 =⋯7 h\|7 g\|7 f+2\|0\|0\|0\|0\|4=⋯7 h\|7 g\|7 f+2\|0\|0\|0\|0\|4 And f=4 f=4. x=⋯h\|g\|4\|2\|8\|5\|7\|2 x=⋯h\|g\|4\|2\|8\|5\|7\|2 7 x=⋯7 h\|7 g\|30\|0\|0\|0\|0\|4 7 x=⋯7 h\|7 g\|30\|0\|0\|0\|0\|4 =⋯7 h\|7 g+3\|0\|0\|0\|0\|0\|4=⋯7 h\|7 g+3\|0\|0\|0\|0\|0\|4 And g=1 g=1. x=⋯h\|1\|4\|2\|8\|5\|7\|2 x=⋯h\|1\|4\|2\|8\|5\|7\|2 7 x=⋯7 h\|10\|0\|0\|0\|0\|0\|4 7 x=⋯7 h\|10\|0\|0\|0\|0\|0\|4 =⋯7 h+1\|0\|0\|0\|0\|0\|0\|4=⋯7 h+1\|0\|0\|0\|0\|0\|0\|4 And now I am doing the same work as I did for a value b b, making 7 b+1 7 b+1 a multiple of 10 10. We are in a cycle and so x=4 7 x=4 7 is represented as ⋯142857 142857 142857 2=142857¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯2⋯142857 142857 142857 2=142857¯2. Challenge:This process felt reminiscent of the task of writing 4 7 4 7 as a decimal in ordinary arithmetic using a 1←10 1←10 machine with decimals. We argued there too that the decimal represent had to fall into a cycle. Can you argue that the fraction 2 13 2 13 will also have a repeating ten-adic expansion? Challenge:What is the ten-adic expansion of−4 7−4 7 ? One approach: Write −4 7−4 7 as −1\|−4\|−2\|−8\|−5\|−7¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯\|−2−1\|−4\|−2\|−8\|−5\|−7¯\|−2 and add some dots and antidot pairs to make all the terms positive. −1\|−4\|−2\|−8\|−5\|−7¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯\|−2−1\|−4\|−2\|−8\|−5\|−7¯\|−2 =−1\|−4\|−2\|−8\|−5\|−7¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯\|−2+(−8+8)=−1\|−4\|−2\|−8\|−5\|−7¯\|−2+(−8+8) =−1\|−4\|−2\|−8\|−5\|−7−1¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯\|8=−1\|−4\|−2\|−8\|−5\|−7−1¯\|8 =⋯=⋯ A Glitch Let’s try to compute the ten-adic representation of the fraction 1 2 1 2. Here we seek a number x=⋯h\|g\|f\|e\|d\|c\|b\|a x=⋯h\|g\|f\|e\|d\|c\|b\|a so that 2 x=⋯2 h\|2 g\|2 f\|2 e\|2 d\|2 c\|2 b\|2 a 2 x=⋯2 h\|2 g\|2 f\|2 e\|2 d\|2 c\|2 b\|2 a equals 1 1. This means we a number a a so that, after explosions, 2 a 2 a leaves a single dot. That is, we need 2 a 2 a to be one more than a multiple of ten. This is not possible! Challenge:Contemplate the ten-adic expansions for 1 5 1 5 and 3 10 3 10 and 2 35 2 35. In general, which fractions p q p q seem to be problematic? Challenge:Develop a general theory that if p q p q is a reduced fraction with q q sharing no factor in common with ten (other than 1 1), then it is for certain possible to express p q p q as a ten-adic number ⋯h g f e d c b a⋯h g f e d c b a. Show further that its expression is sure to fall into a repeating cycle. Broadening our Definition a Tad It seems we have defined a ten-adic value to be an expression of the form ⋯e d c b a⋯e d c b a with each digit one of the standard digits 0 0 through 9 9, allowing for non-zero digits to appear infinitely far to the left. In this system we have the ordinary positive integers, eg 5 5 is ⋯00005⋯00005, the negative numbers eg −5−5 is ⋯99995⋯99995 and some fractions eg 1 3 1 3 is ⋯66667⋯66667. But not all fractions. It turns out that the troublesome fractions are the ones p q p q which, when written in reduced form, have a denominator a multiple of 2 2 or 5 5 or both. We can obviate this problem if we allow a ten-adic number to extend finitely far into the decimal places on the right. That is, set a ten-adic expression to be one of the form ⋯e d c b a.x y…z⋯e d c b a.x y…z with each digit one of the standard digits 0 0 through 9 9, allowing for non-zero digits to appear infinitely far to the left of the decimal point, and only finitely far to its right. (After all, we do the analogous thing in ordinary arithmetic by writing 33.3333⋯33.3333⋯, for example, for thirty-three and a third.) Now we have 1 2=0.5 1 2=0.5 is ⋯0000.5⋯0000.5 and 23 100=0.23 23 100=0.23 is ⋯0000.23⋯0000.23. We can also handle 2 35 2 35 by thinking of this as 2 7×5=2×2 7×10=4/7 10 2 7×5=2×2 7×10=4/7 10. Since 4 7 4 7 is 142857¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯2 142857¯2 we must have 2 35=142857¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯.2 2 35=142857¯.2. Challenge:Show that 1 6=⋯3333.5 1 6=⋯3333.5 and hence find the ten-adic expression for 5 12=1 6+1 4 5 12=1 6+1 4. What is the ten-adic expression for 1 12 1 12? Challenge:Explain why every fraction is now sure to have a ten-adic representation. Challenge:Show that 53¯¯¯¯¯23=⋯53535323 53¯23=⋯53535323 is the number −1000 33−1000 33 in ten-adic arithmetic. (Hint: Multiply the quantity by 100 100 and subtract.) One can use the technique of this question to show that every ten-adic number that eventually falls into a cycle going leftwards is a rational number. Challenge:In ordinary arithmetic, the quantity 0.a b c a b c a b c⋯=0.a b c¯¯¯¯¯¯¯0.a b c a b c a b c⋯=0.a b c¯ is the fraction a b c 999 a b c 999. We see this by setting x=0.a b c a b c a b c⋯x=0.a b c a b c a b c⋯ and noticing that 1000 x=a b c.a b c a b c a b c⋯1000 x=a b c.a b c a b c a b c⋯. Subtracting then yields 999 x=a b c 999 x=a b c. Show that the same algebra applied to the ten-adic number ⋯a b c a b c a b c=a b c¯¯¯¯¯¯¯⋯a b c a b c a b c=a b c¯ shows that it this number has value −a b c 999−a b c 999. In fact, prove the following general result. Suppose b 1 b 1, b 2 b 2, …, b k b k are single digits. If 0.b 1 b 2…b k¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯0.b 1 b 2…b k¯ is the fraction p q p q in ordinary arithmetic, then b 1 b 2…b k¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯b 1 b 2…b k¯ is the fraction −p q−p q in ten-adic arithmetic, and vice versa. Challenge:Explore a theory of “3/2-adic” representations of fractions using a 2←3 2←3 machine. Please join the conversation on Facebook and Twitter and kindly share this page using the buttons below. Facebook Twitter Previous Next Course Home Resources Books ----- Take your understanding to the next level with easy to understand books by James Tanton. BROWSE BOOKS Guides & Solutions ------------------ Dive deeper into key topics through detailed, easy to follow guides and solution sets. 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15743	https://people.bath.ac.uk/gt223/MA30237/lnotes.pdf	Lecture Notes in Group Theory Gunnar Traustason (Autumn 2016) 0 0 Introduction. Groups and symmetry Group Theory can be viewed as the mathematical theory that deals with symmetry, where symmetry has a very general meaning. To illustrate this we will look at two very diﬀerent kinds of symmetries. In both case we have ‘transformations’ that help us to capture the type of symmetry we are interested in. We then have the ‘objects’ that we are analysing and to each object we will associate a ‘symmetry group’ that captures the symmetric properties of the object in precise mathematical terms. I. Isometric symmetry in R2 Transformations: Isometries. An isometry on the plane is a bijection f : R2 →R2 that preserves distances. Objects: Figures in the place (that is subsets of the plane). The symmetry group of a ﬁgure A: For any ﬁgure(subset) A of the plane, we let GA be the set of all isometries that preserve the ﬁgure (as a set). This is a group with composition as the group multiplication. We call it the symmetry group of A. Example → ↑ 3·············1 ··········· 2 · · · · · · · · · · · A For the equilateral triangle A, GA consists of three rotations r, r2 and r3 = e = id, with r being a counterclockwise rotation of 120 degrees around the center of A, and three reﬂections s1, s2 and s3 with respect to the three symmetry axes of A, through the points 1, 2 and 3 respectively. We can now write a multiplication table for GA: 1 e r r2 s1 s2 s3 e e r r2 s1 s2 s3 r r r2 e s3 s1 s2 r2 r2 e r s2 s3 s1 s1 s1 s2 s3 e r r2 s2 s2 s3 s1 r2 e r s3 s3 s1 s2 r r2 e Every equilateral triangle in the plane has a group G of isometries that contains three rotations and three reﬂections as above. It depends on the triangle what exactly these rotations and reﬂections are but the algebraic structure is always going to be as in the multiplication table above. So the symmetry is captured in the algebraic structure of G. In fact the group above is ismorphic to S3, the group of all permutations of 1, 2, 3. This is because the 6 elements in GA permute the corner points of the triangle and all the 6 = 3! permutations of S3 occur: r and r2 correspond to (1 2 3) and (1 3 2) and the three reﬂections s1, s2 and s3 correspond to the (2 3), (1 3) and (1 2). The following questions now arise naturally: (Q1) What symmetries are out there? (Q2) What are their properties? Or, translating these into formal mathematics questions: (q1) What groups are there? (Classiﬁcation) (q2) What is their structure like? (Structure theory) The symmetry we have just looked at is of geometric nature and groups and geome-try have some strong links. For example, one can think of Euclidean geometry in the plane as the theory that studies properties that are invariant under isometries (i.e. angle, length, area, triangle, ...). During the 19th century there was a development of a number of diﬀerent geometries (i.e. aﬃne geometry, projective geometry, hyperbolic geometry, ....) and Felix Klein (1872) made the general observation that, like Euclidean geometry can be characterised by the group of isometries, each geometry can be characterised by some group of transformations. The origin of abstract group theory goes however further back to Galois (1811-1832) and the problem of solving polynomial equations by algebraic methods. This we turn to next. II. Arithmetic symmetry in C. The origin of group theory. Transformations: Automorphisms. A automorphism on C is a bijective function f : C →C that preserves the addition and the multiplication: f(a + b) = f(a) + f(b) f(ab) = f(a)f(b). 2 Claim. Any automorphism f ﬁxes all the elements in Q. Proof. Firstly f(0) = 0 and f(1) = 1 as f(0) + 0 = f(0) = f(0 + 0) = f(0) + f(0) f(1) · 1 = f(1) = f(1 · 1) = f(1) · f(1). and cancellation gives what we want. Notice that we can cancel by f(1) as it can’t be 0 (f is bijective and 0 is already taken as a value). Next suppose that n ≥1 is an integer. Then f(n) = f(1 + 1 + · · · + 1 \| {z } n ) = f(1) + f(1) + · · · + f(1) \| {z } n = 1 + 1 + · · · + 1 \| {z } n = n and f(n) = n for all positve integers n. Before going further we observe that f has the property that f(−a) = −f(a) and also that f(1/a) = 1/f(a) whenever a ̸= 0. The reason for this is the following f(a) + f(−a) = f(a + (−a)) = f(0) = 0 f(a) · f(1/a) = f(a · 1/a) = f(1) = 1. Using this we can now ﬁnish the proof of the claim. Firstly for n > 0 we have f(−n) = −f(n) = −n which shows that f ﬁxes any integer. Finally if q = a/b for some integers a, b, where b ̸= 0, then f(q) = f(a · 1/b) = f(a) · f(1/b) = f(a) · 1/f(b) = a/b = q and we have proved the claim. 2 Objects: Polynomials in Q[x]. Let P = anxn + an−1xn−1 + · · · + a0 be a polynomial over Q with distinct roots x1, . . . , xn. Claim. Any automorphism f permutes the complex roots of P. Proof. We need to show that if t is a root then f(t) is also a root. But this follows from 0 = f(0) = f(antn + an−1tn−1 + · · · + a0) = f(antn) + f(an−1tn−1) + · · · + f(a0) = f(an)f(t)n + f(an−1)f(t)n−1 + · · · + f(a0) = anf(t)n + an−1f(t)n−1 + · · · + a0 = P(f(t)) where the 2nd last equality follows from the fact that the coeﬃcients are rational num-bers. 2 3 We have seen that any isomorphism f must permute the roots x1, . . . , xn of P. Hence f induces a permutation in Sn (if we identify 1, 2, . . . , n with x1, . . . , xn). The symmetry group of the polynomial P. (Also called the Galois group of P): We let GP = {σ ∈Sn : σ is induced by an isomorphism }. GP is then the symmetry group of P. (By saying that σ ∈Sn is induced by the automorphism f : C →C means that σ(i) = j if and only if f(xi) = xj). Example 1. Determine GP where P = x2 −3x + 2. Solution. P = x2 −3x + 2 = (x −1)(x −2) has only rational roots so every iso-morphism must ﬁx these and thus induce the trivial permutation on the roots. Thus GP = {id}. Example 2. Determine GP where P = x4 −1. Solution. The polynomial P = x4 −1 has the roots x1 = 1, x2 = −1, x3 = i and x4 = −i. Here all isomorphisms must ﬁx 1 and −1. This leaves the possibility of swap-ping i and −i, and the isomorphism f on C that maps z to ¯ z does that (recall that a + b = a + b and ab = a · b which implies that f is a isomorphism). Thus GP = {α, id} where α = x1 x2 x3 x4 x1 x2 x4 x3 or, under the identiﬁcation of 1, 2, 3, 4 with x1, x2, x3, x4, α = 1 2 3 4 1 2 4 3 i.e. α swaps x3 and x4 (or 3 and 4). Remark. In general GP is a subgroup of Sn and thus thas at most n! elements (in fact \|GP\| divides \|Sn\| = n! by Lagrange’s Theorem). We say that a polynomial P is solvable by radicals if its roots can be expressed using only the coeﬃcients, the arithmetic operations and extracting roots. That any quadratic ax2 + bx + c is solvable by radicals is for example a consequence of the formula: x = −b ± √ b2 −4ac 2a . Such formulas for solving the cubics and the quartics were discovered during the 16th cen-tury but despite much eﬀort the quintic continued to remain a challenge. The question was not settled until 1824 when the Norwegian mathematican Niels Henrik Abel demon-strated that the quintic is not in general solvable by radicals. The French mathematician 4 ´ Evariste Galois (1811-1832) proved this independently and went further by ﬁnding a suf-ﬁcient and necessary condition under which a given polynomial is solvable by radicals. In doing so he developed a new mathematical theory of symmetry, namely group theory. His famous theorem is the following: Theorem (Galois). A polynomial P is solvable by radicals iﬀGP is solvable. For a group to be solvable means having a structure of a special kind. You will see the precise deﬁnition later in the course. Fact. For each positive integer n there exists a polynomial Pn of degree n such that GPn = Sn (all the permutations of the n roots). Theorem. Sn is solvable iﬀn ≤4. (We will prove this later in the course). Corollary. For any n ≥5 there exists a polynomial of degree n (namely Pn) that is not solvable by radicals. 5 1 Deﬁnitions and basic properties I. The group axioms and some examples of groups. We start by recalling the deﬁnition of a group. Deﬁnition. A group is a pair (G, ∗), where G is a set, ∗is a binary operation and the following axioms hold: (a) (The associative law) (a ∗b) ∗c = a ∗(b ∗c) for all a, b, c ∈G. (b) (Existence of an identity) There exists an element e ∈G with the property that e ∗a = a and a ∗e = a for all a ∈G. (c) (The existence of an inverse) For each a ∈G there exists an element b ∈G such that a ∗b = b ∗a = e. Remark. Notice that ∗: G × G →G is a binary operation and thus the ‘closure axiom’: a, b ∈G ⇒a ∗b ∈G is implicit in the deﬁnition. Deﬁnition. We say that a group (G, ∗) is abelian or commutative if a ∗b = b ∗a for all a, b ∈G. Remarks.(1) Recall that the identity e is the unique element in G with the property given in (b). To see this suppose we have another identity f. Using the fact that both of these are identities we see that f = f ∗e = e. we will usually denote this element by 1 (or by 0 if the group operation is commutative). (2) the element b ∈G as in (c) is unique. To see this suppose that c is another in-verse to a. Then c = c ∗e = c ∗(a ∗b) = (c ∗a) ∗b = e ∗b = b. We call this unique element b, the inverse of a. It is often denoted a−1 (or −a when the group operation is commutative). (3) If it is clear from the context what the group operation ∗is, one often simply refers to the group G rather then the pair (G, ∗). 6 Some examples of groups. (1) Let X be a set and let Sym (X) be the set of all bijective maps from X to itself. Then Sym (X) is a group with respect to composition, ◦, of maps. This group is called the symmetric group on X and we often refer to the elements of Sym (X) as permutations of X. When X = {1, 2, · · · , n} the group is often denoted Sn and called the symmetric group on n letters. (2) Let (R, +, ·) be any ring. Then (R, +) is an abelian group. This includes for ex-ample the group of integers (Z, +) and the ﬁelds Q, R, C with repect to addition. It also includes, for any positive integer n, the group of integers modulo n (Zn, +). (3) Let again (R, +, ·) be any ring with unity 1. Then the set of all invertible elements (the units), R∗, is a group with respect to the ring multiplication ·. This group is referred to as the group of units of R. This includes Q∗, R∗, C∗and Z∗ n for any positive integer. (4) Let V be a ﬁnite dimensional vector space over a ﬁeld K. Consider the ring End (V ) of all linear operators α : V →V . Here the group of units is denoted GL(V ) and called the general linear group on V . (5) Let K be a ﬁeld and let Mn(K) be the ring of all n×n matrices over K. The group of units here is denoted GLn(K) and called the general linear group of n×n matrices over K. Remarks. (1) We will see later that any group G can be viewed as a subgroup of some group of permutations Sym (X). (2) One can see that any group G can be viewed as a subgroup of the group of units of some ring R. We will see this later at least in the case when G is ﬁnite. II. Subgroups and Lagrange’s Theorem. Deﬁnition. Let G be a group with a subset H. We say that H is a subgroup of G if the following two conditions hold. (a) 1 ∈H, (b) If a, b ∈H then ab, a−1 ∈H. Recall. One can replace (a) and (b) with the more economical: (a)’ H ̸= ∅, (b)’ If a, b ∈H then ab−1 ∈H. Remark. It is not diﬃcult to see that one could equivalently say that H is a sub-group of G if H is closed under the group multiplication ∗and that H with the induced multiplication of ∗on H is a group in its own right. So subgroups are groups contained within G that inherit the multiplication from G. Notation. We write H ≤G or G ≥H for ‘H is a subgroup of G’. 7 Cosets as equivalence classes. Suppose G is a group with a subgroup H. We de-ﬁne a relation ≃on G as follows: x ≃y iﬀx−1y ∈H. This relation is an equivalence relation. To see this we need to see that it is reﬂexive, symmetric and transitive. Firstly it is reﬂexive as x−1x = 1 ∈H implies that x ≃x. To see that it is symmetric suppose x ≃y. Then x−1y ∈H and as H is a subgroup it follows that y−1x = (x−1y)−1 ∈H and thus y ≃x. Finally to see that the relation is transitive notice that if x ≃y and y ≃z then x−1y, y−1z ∈H. Being a subgroup, H is closed under the group multiplication and thus x−1z = (x−1y)·(y−1z) ∈H. Thus x ≃z. Notice that x ≃y if and only if x−1y ∈H if and only if y ∈xH. Hence the equiva-lence class of x is [x] = xH, the left coset of H in G. Theorem 1.1 (Lagrange) Let G be a ﬁnite group with a subgroup H. Then \|H\| divides \|G\|. Proof Using the equivalence relation above, G gets partitioned into pairwise disjoint equivalence classes, say G = a1H ∪a2H ∪· · · ∪arH and adding up we get \|G\| = \|a1H\| + \|a2H\| + · · · + \|arH\| = r · \|H\|. Notice that the map from G to itself that takes g to aig is a bijection (the inverse is the map g 7→a−1 i g) and thus \|aiH\| = \|H\|. 2 Remark. If we had used instead the relation x ≃y iﬀxy−1 ∈H, we would have had [x] = Hx. Hence G also partions into a pairwise disjoint union of right cosets. (Re-call that in general the partions into right cosets and into left cosets are diﬀerent). Examples. (1) The subsets {1} and G are always subgroups of G. (2) The subset Cn = {a ∈C : an = 1} is a subgroup of (C, ·). In fact 1n = 1 and if a, b ∈Cn then (ab)n = anbn = 1 and (a−1)n = (an)−1 = 1. Thus both the subgroup criteria (a) and (b) hold. (3) H = {id, (1, 2)} is a subgroup of S3. Clearly (a) holds as id ∈H and direct in-spection shows that (b) holds as well. Deﬁnition. Let G be a group and a ∈G. The cyclic subgroup generated by a is ⟨a⟩= {an : n ∈Z}. Remark. We have that 1 = a0 ∈⟨a⟩. We also have that ⟨a⟩is closed under the group multiplication and taking inverses since an · am = an+m and (an)−1 = a−n. Hence ⟨a⟩is a subgroup of G. It is clearly the smallest subgroup of G that contains a. Deﬁnition. We say that a group G is cyclic if there exists an element a ∈G where G = ⟨a⟩. 8 Deﬁnition. Let G be a group and a ∈G. The order of a, denoted o(a), is deﬁned as follows. If there is a positive integer m such that am = 1 then o(a) is the smallest such integer. If there is on the other hand no such positive integer we say that a is of inﬁnite order and write o(a) = ∞. Remarks.(1) If o(a) = n < ∞, then ⟨a⟩= {1 = a0, a1, . . . , an−1} where the elments 1, a, a2, . . . , an−1 are distinct. To see why the elements are diﬀerent suppose for a contraction that ar = as for some 0 ≤r < s ≤n −1. But then as−r = 1 where 0 < s −r ≤n −1 < n. This however contradicts the fact that n = o(a) is the smallest positive integer where an = 1. (2)Thus o(a) = n = \|⟨a⟩\|. Note also that am = 1 iﬀn\|m. It follows that ar = as if and only if n\|(r −s). (The structure of the group is just like that of Zn). (3) Let G be a ﬁnite group and a ∈G. As o(a) = \|⟨a⟩\| that divides \|G\| by Lagrange, we have from Remark (2) that a\|G\| = 1. Let G = ⟨a⟩be a ﬁnite cyclic group. By Lagrange any subgroup has a order d that is a divisor of n. For cyclic groups there is conversely exactly one subgroup of order d for each divisor d. Proposition 1.2 Let G = ⟨a⟩be a ﬁnite cyclic group of order n and let d be a divisor of n. The subgroup ⟨an/d⟩is the unique subgroup of order d. Proof. Let H be a subgroup of order d. As ⟨an/d⟩has also d elements it suﬃces to show that H ⊆⟨an/d⟩. Let am ∈H. By Remark (3) above we have 1 = am\|H\| = amd and, by Remark (2), it follows that n = o(a) divides md. Hence n/d divides m, say m = r · (n/d), and am = (an/d)r ∈⟨an/d⟩. 2 Proposition 1.3 Let p be a prime number and G be a group such that \|G\| = p. The group G is cyclic. Proof As p ≥2 there has to be some element a ̸= 1 in G. Then \|⟨a⟩\| ≥2 and (by Lagrange’s Theorem) \|⟨a⟩\| divides \|G\| = p. As p is a prime we must have \|⟨a⟩\| = p and thus ⟨a⟩= G. 2. III. Congruences and quotient groups. Deﬁnition. Let G be a group. A congruence on G is an equivalence relation ≃on G that satisﬁes: a1 ≃a2, b1 ≃b2 ⇒a1b1 ≃a2b2. Remark. This extra condition is needed to introduce a well deﬁned multiplication on the equivalence classes [a] · [b] = [ab]. 9 Lemma 1.4 Let G be a group with congruence ≃. Then N = is a subgroup of G that satisﬁes: g−1Ng ⊆N for all g ∈G. Furthermore a ≃b if and only if a−1b ∈N. Proof. To see that N is a subgroup, we go through the subgroup criteria. As ≃is reﬂexive we have 1 ≃1 and thus 1 ∈N = . It remains to see that N is closed under group multiplication and taking inverses. For the ﬁrst of these, notice that of a, b ∈N then a, b ≃1 and the congruence property gives us that ab ≃1 · 1 = 1. Thus ab ∈N. To see that N is closed under taking inverses, suppose that a ∈N then a ≃1 and the congruence property gives us that 1 = a−1a ≃a−1 · 1 = a−1. This shows that a−1 ∈N. It remains to see that N has the requested extra property. So suppose a ∈N. Then a ≃1 and the congruence property implies that g−1ag ≃g−1 ·1·g = 1. Hence g−1ag ∈N. Finally we have a ≃b iﬀ1 = a−1a ≃a−1b iﬀa−1b ∈ = N. 2 Deﬁnition. A subgroup H of G is said to be a normal subgroup if g−1Hg ⊆H ∀g ∈G. Notation. We write H G or G H for ‘H is a normal subgroup of G’ Lemma 1.5 Let G be a group with a normal subgroup N and deﬁne a relation ≃on G by x ≃y if and only if x−1y ∈N. Then ≃is a congruence on G and [a] = aN. In particular = N. Proof We have seen in the proof of Lagrange’s Theorem that ≃is an equivalence relation and that [a] = aN. It remains to see that the congruence property holds. So suppose that a1 ≃a2 and b1 ≃b2. This means that a−1 1 a2, b−1 1 b2 ∈N. We want to show that a1b1 ≃a2b2. But this follows from (a1b1)−1(a2b2) = b−1 1 (a−1 1 a2)b2 = (b−1 1 b2) · b−1 2 (a−1 1 a2)b2. As N is normal we have that b−1 2 (a−1 1 a2)b2 ∈N and thus the equation above shows that (a1b1)−1a2b2 is a product of two elements from N. As N is a subgroup of G, this product is in N. Hence a1b1 ≃a2b2. 2 Remark. It follows from Lemmas 1.4 and 1.5 that there is a 1-1 correspondence be-tween congruences on G and normal subgroups of G. Remarks. (1) We write often more shortly Ha instead of a−1Ha and call it a conju-gate of H by a. Similarly if x ∈G then xa = a−1xa is a conjugate of x by a. (2) Let x, a, b ∈G and 1 be the identity element in G. Then xab = (ab)−1xab = b−1(a−1xa)b = (xa)b x1 = 1−1 · x · 1 = x. It follows then that if H ≤G we also have Hab = (Ha)b and H1 = H. 10 (3) Notice that Ha is a subgroup of G: ﬁrstly 1 = a−1 · 1 · a = 1a ∈Ha and then xaya = a−1xaa−1ya = a−1(xy)a = (xy)a and (xa)−1 = (a−1xa)−1 = a−1x−1a = (x−1)a. In fact the group Ha has the same structure as H. (The conjugation by a is a bit like a renaming or an ornament). Lemma 1.6 The following are equivalent: (a) H G, (b) Ha = H for all a ∈G, (c) Ha = aH for all a ∈G. Proof (b)⇒(a) is obvious. To prove (a)⇒(b), notice that (a) implies in particular that for any a ∈G we have Ha−1 ⊆H and therefore H = He = Ha−1a = (Ha−1)a ⊆Ha. This gives Ha = H. It now only remains to show that (b)⇔(c). But this is easy a−1Ha = H ⇔a · a−1Ha = aH ⇔Ha = aH. This ﬁnishes the proof. 2 Deﬁnition. Let G be a group with a subgroup H. The number of left cosets of H in G is called the index of H in G and is denoted [G : H]. Remark. Suppose that G is ﬁnite. Recall from the proof of Lagrange’s Theorem that we get a partition of G into a union of pairwise disjoint union of left cosets G = a1H ∪a2H ∪· · · ∪anH. As each of the cosets have order \|H\|, it follows that \|G\| = r · \|H\|. Hence [G : H] = r = \|G\|/\|H\|. (Likewise we have that G can be written as a pairwise disjoint union of right cosets and the same reasoning shows that their number is also \|G\|/\|H\|. Examples. (1) Every subgroup N of an abelian group G is normal (since then obvi-ously aN = Na for all a ∈G). (2) The trivial subgroup {1} and G itself are always normal subgroups of G. (3) If H is a subgroup of G such that [G : H] = 2 then H G (since the left cosets are H, G \ H which are also the right cosets. Hence the right cosets are the same as the left cosets). The quotient group G/N. Let G be a group with a congruence ≃and a corresponding normal subgroup N. Let G/N = { [a] = aN : a ∈G} with a binary operation [a] · [b] = [ab] (that is aN · bN = abN). Notice that this is well deﬁned as ≃is a congruence. To see that G/N is a group with respect to this binary operation we check that the three group axioms hold. Firstly there is an identity element, namely = N as · [a] = [1 · a] = [a] and 11 [a] · = [a · 1] = [a]. Secondly every element [a] ∈G/N has an inverse, namely [a−1] since [a]·[a−1] = [a·a−1] = and [a−1] · [a] = [a−1 · a] = . Finally associativity in G/N follows from associativity in G: [a] · ([b] · [c]) = [a] · [bc] = [a(bc)] = [(ab)c] = [ab] · [c] = ([a] · [b]) · [c]. Remark. That the binary operation on G/N is well deﬁned followed from the fact that ≃is a congruence. There is another way of seeing this using the fact that N is normal in G. First we introduce set products in the natural way. So if X, Y ⊆G then we let X · Y = {xy : x ∈X, y ∈Y }. Then, using this set product as the action on G/N, we get [a] · [b] = aN · bN = abNN = abN = [ab]. Hence the binary operation (being the same as the set multiplication) is well deﬁned. Notice that we used the fact that N is normal when applying Nb = bN. Also N · N ⊆N as N is a subgroup and N = N · {1} ⊂N · N as 1 ∈N. Thus N · N = N. Remark. Notice that the size of the group G/N is [G : N] and when G is ﬁnite this is the same as \|G\|/\|N\|. Examples. (1) We always have G G. The congruence with respect to the normal subgroup G is x ≃y ⇔x−1y ∈G. As the latter holds for any x, y ∈G we are identifying all the elements. Hence G/G = {} = {G} is the trivial group with only one element. (2) The trivial subgroup N = {1} is always normal in G. The congruence in this case is given by x ≃y ⇔x−1y ∈N ⇔x−1y = 1 ⇔y = x. Thus G/N = {{a} : a ∈G}. The structure is just like the structure of G: {a} · {b} = {ab}. (The curly bracket is there just as a decoration). (3) Let G = S3 = {id, (1 2), (1 3), (2 3), (1 2 3), (1 3 2)} and N = A3 = {id, (1 2 3), (1 3 2)}. Here [G : N] = 2 and thus G/N has two elements. Notice that these are N = {id, (1 2 3), (1 3 2)} = [id] = [(1 2 3)] = [(1 3 2)] and (1 2)N = {(1 2), (2 3), (1 4)} = [(1 2)] = [(1 3)] = [(2 3)]. (So here we have identiﬁed all the even permutations and likewise all the odd permuta-tions). G/N = {1 = [id], a = [(1 2)]}. This is the unique group structure with 2 elements: 1 · a = a · 1 = a, 1 · 1 = 1 and a · a = 1. 12 IV. Homomorphisms and isomorphisms Deﬁnition. Let G, H be groups. A map φ : G →H is a homomorphism if φ(ab) = φ(a)φ(b) for all a, b ∈G. Furthermore φ is an isomorphism if it is bijective. A group G is said to be isomor-phic to H if there is an isomorphism φ : G →H. We then write G ∼ = H. Remarks. (1) If φ : G →H and ψ : H →K are homomorphisms then their com-position ψ ◦φ : G →K is also a homomorphism. This is simply because ψ(φ(ab)) = ψ(φ(a) · φ(b)) = ψ(φ(a)) · ψ(φ(b)). In particular if G ∼ = H and H ∼ = K then G ∼ = K. (2) If φ : G →H is an isomorphism then φ−1 : H →G is also an isomorphism. To see this let a = φ(x), b = φ(y) ∈H. Then φ−1(a · b) = φ−1(φ(x) · φ(y)) = φ−1(φ(xy)) = xy = φ−1(a) · φ−1(b). In particular if G ∼ = H then also H ∼ = G. (3) If G ∼ = H then there is no structural diﬀerence between G and H. You can think of the isomorphism φ : G →H as a renaming function. If ab = c then φ(a), φ(b), φ(c) are the new a, b, c. We want the new c to be the product of the new a and b. This means φ(ab) = φ(a)φ(b). Lemma 1.7 Let φ : G →H be a homomorphism, then (a) φ(1G) = 1H, (b) φ(a−1) = φ(a)−1. Proof (a) We have 1H · φ(1G) = φ(1G) = φ(1G · 1G) = φ(1G) · φ(1G) and cancellation gives 1H = φ(1G). (b) Using (a) we have φ(a−1)φ(a) = φ(a−1a) = φ(1G) = 1H and φ(a)φ(a−1) = φ(aa−1) = φ(1G) = 1H. Hence φ(a−1) is the inverse of φ(a). 2 Examples (1) Let N be a normal subgroup of G. The map φ : G →G/N, a 7→[a] = aN is a homomorphism as φ(ab) = [ab] = [a] · [b] = φ(a) · φ(b). (2) Let R+ be the set of all the postive real numbers. There is a (well known) iso-morphism φ : (R, +) →R+, ·) given by φ(x) = ex. (As ex+y = exey. This is a bijective 13 homomorphism). Lemma 1.8 Let φ : G →H be a homomorphism, then (a) A ≤G ⇒φ(A) ≤H. (b) B ≤H ⇒φ−1(B) ≤G. (c) B H ⇒φ−1(B) G. Proof To prove (a) and (b) we apply the usual three subgroup criteria, i.e. the subset in question needs to contain the identity and be closed under multiplication and taking inverses. For (a) this follows from 1H = φ(1G), φ(x)φ(y) = φ(xy) and φ(x)−1 = φ(x−1). Notice that, as A ≤G, we have 1G ∈A and xy, x−1 ∈A whenever x, y ∈A. Sim-ilarly for proving (b), it is ﬁrst clear that 1G ∈φ−1(B) as φ(1G) = 1H ∈B (since B ≤H). Furthermore, if x, y ∈φ−1(B), then φ(x), φ(y) ∈B. As B ≤H, it follows that φ(xy) = φ(x)φ(y) ∈B and φ(x−1) = φ(x)−1 ∈B. This shows that xy, x−1 ∈φ−1(B). For the proof of part (c) suppose furthermore that the subgroup B of H is normal. Let x ∈φ−1(B) and g ∈G. Then φ(g−1xg) = φ(g)−1φ(x)φ(g) ∈φ(g)−1Bφ(g) ⊆B. Hence g−1xg ∈φ−1(B). This shows that φ−1(B) is normal in G. 2. V. The Isomorphism Theorems Let N G and consider the homomorphism φ : G →G/N, a 7→[a] = aN. Let SN(G) = {H : N ≤H ≤G} and S(G/N) = {R : R ≤G/N}. Consider the map Ψ : SN(G) →S(G/N), Ψ(H) = φ(H) = H/N. Remark. Thus Ψ(H) is the set φ(H) = {φ(a) : a ∈H}. Theorem 1.9 (Correspondence Theorem). Ψ is a bijection and furthermore H G iﬀ Ψ(H) G/N. Proof. (Ψ is injective). Let N ≤H, K ≤G and suppose that Ψ(H) = H/N is equal to Ψ(K) = K/N. Then H = [ xN∈H/N xN = [ xN∈K/N xN = K. (Ψ is surjective). Let R be a subgroup of G/N. Then, by Lemma 1.8, φ−1(R) is a subgroup of G (that clearly contains N as all the elements in N map to the identity element of G/N that is in R) and as φ is surjective, we have Ψ(φ−1(R)) = φ(φ−1(R)) = R. (Notice that N ⊆H implies that N = g−1Ng ⊆g−1Hg and thus the subgroup g−1Hg is also in SN(G)). This shows that Ψ is a bijection. Finally we are going to use Ψ(g−1Hg) = φ(g−1Hg) = φ(g)−1φ(H)φ(g) = φ(g)−1Ψ(H)φ(g). 14 We have that H G iﬀg−1Hg = H for all g ∈G. As Ψ is a bijection this holds iﬀΨ(g−1Hg) = Ψ(H) for all g ∈G. In view of the identity above this holds iﬀ φ(g)−1Ψ(H)φ(g) = Ψ(H) for all g ∈G. But as φ is surjective this is true iﬀr−1Ψ(H)r = Ψ(H) for all r ∈G/N that is iﬀΨ(H) G/N. 2 The picture that is good to keep in mind is the following. . . . H aN bN eN = N Ψ(H) = H/N          bN . . . aN eN          Ψ(H) is the collection of all the cosets of N in H and H is the pairwise disjoint union of these cosets. Thus if we know H we get Ψ(H) as the cosets of N in H and if we know Ψ(H) we get H as the union of the cosets in Ψ(H). Deﬁnition. Let φ : G →H be a group homomorphism. The image of φ is im φ = {φ(g) : g ∈G} and the kernel of φ is ker φ = {g ∈G : φ(g) = 1}. Notice that as G ≤G, it follows from Lemma 1.8 that im φ = φ(G) is a subgroup of H. Also, as {1} H it follows from Lemma 1.8 that ker φ = φ−1({1}) is a normal subgroup of G. Theorem 1.10 (1st Isomorphism Theorem). Let φ : G →H be a homomorphism. Then Im φ ≤H, Ker φ G and G/Ker φ ∼ = Im φ. Proof As we have noted previously, it follows from Lemma 1.8 that Im φ ≤H and Ker φ G. Deﬁne a map Φ : G/Ker φ →Im φ by setting Φ([a]) = φ(a). This map is clearly surjective. We next show that it is well deﬁned and injective. This follows from Φ([a]) = Φ([b]) ⇔ φ(a) = φ(b) ⇔ φ(a−1b) = φ(a)−1φ(b) = 1 ⇔ a−1b ∈Ker φ ⇔ [a] = [b] To show that Φ is an isomorphism, it remains to show that Φ is a homomorphism. This follows from Φ([a] · [b]) = Φ([ab]) = φ(ab) = φ(a)φ(b) = Φ([a]) · Φ([b]). 15 This ﬁnishes the proof. 2 Theorem 1.11 (2nd Isomorphism Theorem). Let H ≤G and N G. Then HN ≤G, H ∩N H and H/(H ∩N) ∼ = HN/N. Proof We apply the 1st Isomorphism Theorem. Consider the homomorphism φ : G →G/N, a 7→aN. Let ψ be the restriction of φ on H. This gives us a homomorphism ψ : H →G/N. By the 1st Isomorphism Theorem we have that Im ψ = {hN : h ∈H} is a subgroup of G/N. By the correspondence theorem we have that this subgroup is of the form U/N, where U is a subgroup of G that is given by U = [ h∈H hN = HN. Thus Im ψ = HN/N. It remains to identify the kernel. The identity of G/N is the coset eN = N. Then for h ∈H, we have ψ(h) = N ⇔ hN = N ⇔ h ∈N. As h ∈H this shows that the kernel of ψ is H∩N. Thus by the 1st Isomorphism Theorem, H ∩N H and H/H ∩N = H/Ker ψ ≃Im ψ = HN/N This ﬁnishes the proof. 2 Theorem 1.12 (3rd Isomorphism Theorem). Suppose that H, N G and N ≤H. Then H/N G/N and (G/N)/(H/N) ∼ = G/H. Proof Again we apply the 1st Isomorphism Theorem. This time on the map φ : G/N →G/H aN 7→aH. Let us ﬁrst see that this is well deﬁned. If aN = bN then a−1b ∈N ⊆H and thus aH = bH. It is also a homomorphism as φ(aN · bN) = φ(abN) = abH = aH · bH = φ(aN) · φ(bN). We clearly have that Im φ = G/H and it remains to identify the kernel. The identity in G/H is the coset eH = H and then φ(aN) = H ⇔ aH = H ⇔ a ∈H. The kernel thus consists of the cosets aN of G/N where a ∈H. That is the kernel is H/N. The 1st Isomorphism Theorem now gives us that H/N G/N (that we had proved already in the proof of the correspondence theorem anyway) and that (G/N)/(H/N) = (G/N)/Ker φ ∼ = Im φ = G/H. This ﬁnishes the proof. 2 16 2 Direct products and abelian groups I. Direct products. Closure properties for the set of normal subgroups of G. (1) If H, K G then H ∩K G. To see that this is a subgroup notice that 1 ∈H and 1 ∈K as both are subgroups and hence 1 ∈H ∩K. Now let a, b ∈H ∩K. As H ≤G and a, b ∈H we know that ab, a−1 ∈H. Similarly as K is a subgroup, containing a, b, we have ab, a−1 ∈K. Thus ab, a−1 ∈H ∩K. To see H ∩K is normal notice that for g ∈G, we have (H∩K)g ⊆Hg = H and (H∩K)g ⊆Kg = K and thus (H∩K)g ⊆H∩K. (2) We also have that if H, KG then HKG: It follows from the 2nd Isomorphism Theo-rem that HK ≤G. To see that HK is normal notice that we have (HK)g = HgKg = HK for g ∈G. Normal products. We have seen that if H, K G then HK G. Inductively it follows that if H1, . . . , Hn G, then H1 · · · Hn G. Since HiHj = HjHi for 1 ≤i < j ≤n, we have Hσ(1) · · · Hσ(n) = H1 · · · Hn for all σ ∈Sn. Lemma 2.1 Let H and K be ﬁnite subgroups of G where K is normal. Then \|HK\| = \|H\| · \|K\| \|H ∩K\| . Proof By the 2nd Isomorphism Theorem, we have HK/K ∼ = H/H ∩K. Taking the orders on both sides gives. \|HK\|/\|K\| = \|H\|/\|H ∩K\|. The result follows immediately from this. 2. Remark. In particular it follows that \|HK\| = \|H\| · \|K\| if and only if H ∩K = {1}. Deﬁnition Let H1, . . . , Hn G. The product H1 · · · Hn is said to be an (internal) direct product of H1, . . . , Hn if Hi ∩ Y j̸=i Hj = {1} for i = 1, . . . , n. Remark. Suppose 1 ≤i < j ≤n. As Hj ≤Q k̸=i Hk, we know in particular that 17 Hi ∩Hj = {1} it follows from Exercise 1 on sheet 2 that all the elements in Hi commute with all the elements in Hj. So if xi ∈Hi then xσ(1) · · · xσ(n) = x1 · · · xn for all σ ∈Sn. Proposition 2.2 Let H1, . . . , Hn G and suppose that H1H2 · · · Hn is an internal direct product. (a) Every element a ∈H1 · · · Hn is of the form a = x1x2 · · · xn for unique xi ∈Hi, i = 1, . . . , n. (b) If xi, yi ∈Hi for i = 1, . . . , n then x1 · · · xn · y1 · · · yn = (x1y1) · · · (xnyn). Proof (a) If x1 · · · xn = y1 · · · yn for some xi, yi ∈Hi, then for each 1 ≤i ≤n xi Y j̸=i xj = yi Y j̸=i yj and thus y−1 i xi = ( Y j̸=i yj) · ( Y j̸=i xj)−1 and thus y−1 i xi is in Hi ∩Q j̸=i Hj = {1} and xi = yi. (b) Using the fact that yi commutes with xj when j > i we have x1x2 · · · xny1y2 · · · yn = x1y1x2 · · · xny2 · · · yn . . . = (x1y1)(x2y2) · · · (xnyn). This ﬁnishes the proof. 2 Remarks. (1) The last Proposition shows that the structure of the internal direct prod-uct H1H2 · · · Hn only depends on the structure of H1, . . . , Hn. Each element is like an n-tuple (x1, . . . , xn) and we multiply two such componentwise. Later we will formalise this when we introduce the external direct product. (2) Notice that it follows from part (a) of last proposition that for an internal direct product H1H2 · · · Hn, we have \|H1 · · · Hn\| = \|H1\| · · · \|Hn\|. The internal direct products are useful for helping us sorting out the structure of a given group. Next we discuss external direct products that are useful for constructing new 18 groups from old groups. Deﬁnition. Let H1, . . . , Hn be groups. The (external) direct product of H1, . . . , Hn is the cartesian set product H1 × · · · × Hn with multiplication (a1, . . . , an) · (b1, . . . , bn) = (a1b1, . . . , anbn). Remark. Since each Hi is a group it is immediate that the direct product is also a group with identity (1H1, . . . , 1Hn). The inverse of (a1, a2, . . . , an) is (a−1 1 , a−1 2 , . . . , a−1 n ). The associatative law follows from the fact that it holds in each component. Next result tells us that the internal direct product is the same as the external direct product. Lemma 2.3 Suppose G is the internal direct product of H1, . . . , Hn. Then G ∼ = H1 × · · · × Hn. Proof (See sheet 4) II. Abelian groups. In this section, we will use additive notation. Thus we use + for the group operation, −a for the inverse of a and 0 for the group identity. We also talk about direct sums rather than direct products. Notice that every subgroup of an abelian group G is normal. Thus for subgroups H1, H2, . . . , Hn of G we have that H1 + · · · + Hn is an internal direct sum of H1, . . . , Hn if Hi ∩ X j̸=i Hj = {0} for i = 1, . . . , n. The external direct sum of H1, . . . , Hn is also denoted H1 ⊕H2 ⊕· · · ⊕Hn instead of H1 × H2 × · · · × Hn. The cyclic group generated by a, ⟨a⟩= {na : n ∈Z}, will often be denoted Za. Deﬁnition. Let G be any abelian group and let p be a prime. The subset Gp = {x ∈G : o(x) is a power of p} is called the p-primary subgroup of G. Lemma 2.4 Gp is a subgroup of G. 19 Proof As the order of 0 is 1 = p0, it is clear that 0 ∈Gp. Now let x, y ∈Gp with orders pn, pm. Then pmax{n,m}(x + y) = pmax{n,m}x + pmax{n,m}y = 0 + 0 = 0 and thus o(x + y) divides pmax{n,m} and is thus also a power of p. Hence x + y ∈Gp and as o(−x) = o(x) = pn we also have that −x ∈Gp. Hence Gp ≤G. 2 Remark. If G is ﬁnite then \|Gp\| must be a power of p. This follows from Exercise 4(a) on sheet 3. If there was another prime q ̸= p that divided \|Gp\| then by this exercise we would have an element in Gp of order q but this contradicts the deﬁnition of Gp. Deﬁnition. An abelian group is said to be a p-group if G = Gp. Next lemma reduces the study of ﬁnite abelian groups to the study of ﬁnite abelian groups of prime power order. Lemma 2.5 Let G be a ﬁnite abelian group where \|G\| = pr1 1 · · · prn n for some positive integers r1, . . . , rn. Then G is the internal direct sum of Gp1, Gp2, . . . , Gpn. Furthermore \|Gpi\| = pri i . Proof Let x ∈G. Then by Lagrange’s Theorem o(x) divides \|G\|, say o(x) = ps1 1 · · · psn n . The numbers q1 = o(x) ps1 1 , . . . , qn = o(x) psn n are then coprime and we can ﬁnd integers a1, . . . , an such that a1q1 +· · ·+anqn = 1. Thus x = (a1q1 + · · · + anqn)x = a1q1x + · · · + anqnx and as psi i (aiqix) = aio(x)x = 0 we have that aiqix ∈Gpi. Thus G = Gp1 + · · · + Gpn. To see that the sum is direct let x ∈Gpi ∩P j̸=i Gpj, say x = xi = X j̸=i xj where the order of xk is pek. Then pei i x = 0 and also (Q j̸=i p ej j )x = 0 and the order of x divides two coprime numbers. Hence o(x) = 1 and thus x = 0. This shows that the intersection is trivial and hence we have a direct sum. By the remark made before the Lemma, we know that \|Gpi\| = psi i for some integer si. Since G is the direct sum of Gp1, . . . , Gpn, we have pr1 1 · · · prn n = \|G\| = n Y i=1 \|Gpi\| = ps1 1 · · · psn n . Comparison of the two sides gives si = ri, i = 1, . . . , n. 2 Remark. Thus G ∼ = Gp1 ⊕· · · ⊕Gpn. And the study of ﬁnite abelian groups reduces to understanding the ﬁnite abelian p-groups. Deﬁnition. Let G be a ﬁnite group. The exponent of G is the smallest positive in-teger n such that xn = 1 for all x ∈G. (Or with additive notation nx = 0 for all x ∈G). 20 Abelian groups of exponent p as vector spaces. Let G be a ﬁnite abelian group of exponent p. Then px = 0 for all x ∈G and the group addition induces a scalar multipli-cation from the ﬁeld Zp as follows. For [m] = m + Zp we let [m]x = mx = x + · · · + x \| {z } m . This is well deﬁned and turns G into a vector space over Zp. One also has that a subset H of G is a subgroup of the group G if and only if H is a subspace of the vector space G. (See Sheet 5, exercise 1 for the details). Lemma 2.6 Let G be a ﬁnite abelian group of exponent p. Then G can be written as an internal direct sum of cyclic groups of order p. Proof Viewing G as a vector space over Zp we know that it has a basis x1, . . . , xn as all these elements are non-trivial and as the exponent of G is p, they must all be of order p. To say that these elements form a basis for the vector space G is the same as saying that we have a direct sum of one dimensional subspaces G = Zpx1 + · · · + Zpxn. This happens if and only if Zpxj ∩ X k̸=j Zpxk = {0} for j = 1, . . . , n. But as Zpxk = Zxk, this is the same as saying that Zxj ∩ X k̸=j Zxk = {0} for j = 1, . . . , n which is the same as saying that G = Zx1 + · · · + Zxr is an internal direct sum of cyclic subgroup of order p. 2. Remark. If we have the direct sum G = Zx1 + · · · + Zxn then \|G\| = pn. The num-ber of direct summands is thus unique and is logp(\|G\|). Lemma 2.7 We have that sum H1 +· · ·+Hn is direct if and only if for any xi ∈Hi, i = 1, . . . , n we have x1 + · · · + xn = 0 ⇒x1 = . . . = xn = 0. Proof To prove this, notice ﬁrst that a direct sum would have this property by Proposition 2.2. Conversely, suppose that this property holds and take some xi = P j̸=i(−xj) in Hi ∩P j̸=i Hj. Then x1 + · · · + xn = 0 and thus x = xi = 0 by the property. So the intersection is trivial and the sum is direct. 2. Proposition 2.8 Let G be a ﬁnite abelian p-group. G can be written as an internal direct sum of non-trivial cyclic groups. Furthermore the number of cyclic summands of any given order is unique for G. Proof (See later). From Lemma 2.5 and Proposition 2.8 we can derive the main result of this chapter. 21 Theorem 2.9 (The Fundamental Theorem for ﬁnite abelian groups). Let G be a ﬁnite abelian group. G can be written as an internal direct sum of non-trival cyclic groups of prime power order. Furthermore the number of cyclic summands for any given order is unique for G. Remark. Suppose that G = Zx1 + Zx2 + · · · + Zxn is a direct sum of cyclic group of prime power order. Notice that G = Zxσ(1) + Zσ(2) + · · · + Zσ(n) for all σ ∈Sn. Convention. We order the cyclic summands as follows. First we order them with respect to the primes involved in ascending order. Then for each prime we order the summands in ascending order. Example. If G is ﬁnite abelian group written as an internal direct sum G = Zx1 + Zx2 + Zx3 + Zx4 + Zx5 of cyclic groups of orders 9, 2, 4, 3, 4, then we order the summands so that they come instead in orders 2, 4, 4, 3, 9. Notice then that G is isomorphic to Z2 ⊕Z4 ⊕Z4 ⊕Z3 ⊕Z9. Remarks. (1) This discussion shows that any ﬁnite abelian group is isomorphic to a unique external direct sum Zpe1 1 ⊕· · · ⊕Zper r where p1 ≤p2 ≤· · · ≤pr and if pi = pi+1 then ei ≤ei+1. (2) Finding all abelian groups of a given order n = pm1 1 · · · pmr r , where p1 < p2 < · · · < pr are primes, reduces then to the problem of ﬁnding, for i = 1, . . . , r, all possible partitions (pe1 i , . . . , pel i ) of the number pmi i . This means that 1 ≤e1 ≤e2 ≤. . . ≤el and e1 + · · · + el = mi. Example. Find (up to isomorphism) all abelian groups of order 72. Solution. We have 72 = 23 · 32. The possible partitions of 23 are (8), (2, 4), (2, 2, 2) whereas the possible partions for 32 are (32), (3, 3). We then have that the abelian groups of order 72 are Z8 ⊕Z9, Z2 ⊕Z4 ⊕Z9, Z2 ⊕Z2 ⊕Z2 ⊕Z9, Z8 ⊕Z3 ⊕Z3, Z2 ⊕Z4 ⊕Z3 ⊕Z3, Z2 ⊕Z2 ⊕Z2 ⊕Z3 ⊕Z3. We now turn to the proof of Proposition 2.8. First as a preparation here are two sub-groups that will play an important part in the proof. Some useful subgroups. Let G be a ﬁnite abelian group. The following subgroups are going to play an important role in the proof of our next main result. That these are subgroups is shown on exercise sheet 3 (using multiplicative notation). 22 PG = {px : x ∈G}, G[p] = {x ∈G : px = 0}. As G[p] is of exponent p it can be viewed as a vector space over Zp. Proof of Proposition 2.8 First we deal with the existence of such a decomposition into a direct sum. Let the exponent of G be pn. We prove the proposition by induction on n. If n = 1 then the result holds by Lemma 2.6. Now suppose that n ≥2 and that the result holds for smaller values of n. The exponent of pG is pn−1 and by the induction hypothesis we have that pG is a direct sum of non-trivial cyclic groups, say pG = Zpx1 + · · · + Zpxr. (1) Suppose the order of xi is pmi (notice that mi ≥2 as pxi ̸= 0). Then pm1−1x1, . . . , pmr−1xr are in G[p]. As G[p] is of exponent p, it can be viewed as a vector space over Zp and we can then extend to a basis (pm1−1x1, . . . , pmr−1xr, xr+1, . . . , xs) for G[p]. It follows that we have a direct sum G[p] = Zpmi−1x1 + · · · + Zpmr−1xr + Zxr+1 + · · · + Zxs. (2) We now want to show that G = Zx1 + · · · + Zxs is a direct sum. First we show that x1, . . . , xs generate G. Let x ∈G. Then by (1) px = a1px1 + · · · + arpxr for some integers a1, . . . , ar. Thus x −(a1x1 + · · · + arxr) is in G[p] and thus by (2) in Zx1 + · · · + Zxs. Hence x is also in Zx1 + · · · + Zxs. It remains to see that the sum G = Zx1 + · · · + Zxs is direct. Suppose that a1x1 + · · · + asxs = 0. We want to show that a1x1 = . . . = asxs = 0. Now multiplying by p we get a1px1 + · · · + arpxr = 0 and since the Zpx1 + · · · + Zpxr is direct, it follows that pa1x1 = . . . = parxr = 0. Thus pmj−1 divides aj for j = 1, . . . , r, say aj = bjpmj−1. So we have b1pm1−1x1 + · · · + brpmr−1xr + ar+1xr+1 + . . . + asxs = 0. As G[p] = Zpm1−1x1 + · · · + Zpmr−1xr + Zxr+1 + · · · + Zxs is direct we must have b1pm1−1x1 = . . . = brpmr−1xr = ar+1xr+1 = . . . = asxr = 0. That is a1x1 = . . . = asxs = 0. This ﬁnishes the inductive proof. To deal with uniqueness part, write G as a direct sum of cyclic groups of p-power or-der G = Za1 + · · · + Zar + Zb1 + · · · + Zbs 23 where a1, . . . , ar have order at most pm−1 whereas b1, . . . , bs have order at least pm (notice that as G is a p-group the orders of all these elements are powers of p). Then \|pm−1G pmG \| = \|Zpm−1b1\| · · · \|Zpm−1bs\| \|Zpmb1\| · · · \|Zpmbs\| = o(pm−1b1) o(pmb1) · · · o(pm−1bs) o(pmbs) . Notice that, in a ﬁnite abelian p-group, we have that if a ̸= 0 then o(pa) = 1 po(a) (If pl is the order of a then pl−1 is the order of pa). The formula above thus implies that \|pm−1G pmG \| = ps and thus the number of summands of order at least pm is logp \|pm−1G pmG \|. Similarly the number of summands of order at least pm+1 is logp \| pmG pm+1G\|. The number of summands of order exactly pm is thus the diﬀerence logp \|pm−1G pmG \| −logp \| pmG pm+1G\|. This shows that the number of summands of order exactly pm is an invariant that does not depend on what the decomposition is. 2 24 3 Composition series and solvable groups I. Simple groups. The primes of group theory. We now introduce an important notion, namely that of a simple group. These can be thought of as the atoms or the primes of group theory. Deﬁnition. A group G is simple if G ̸= {1} and the only normal subgroups of G are {1} and G. Example. The abelian simple groups are the cyclic groups of prime order. See exer-cise sheet 6. Remark. Look at Exercise 5 on sheet 4. According to this exercise we have that if G is a direct product of non-abelian simple groups, then the simple factors are unique up to order (and not only up to isomorphism!). Thus we had here something analogous to a unique prime factorisation of a number. When we also allow for abelian simple factors the result would be similar and we get that the factors are unique (this time up to iso-morphism). The problem is that not all ﬁnite groups can be written as direct products of simple groups. Example is S3 and Z4. It turns out that any ﬁnite group can still in a diﬀerent sense been built out of simple groups. To describe what this means we need to talk ﬁrst about composition series. Deﬁnition. Let G be a group. (1) A subnormal series of G is a series {1} = H0 ≤H1 ≤· · · ≤Hn = G where Hi−1 Hi, i = 1, . . . , n. The quotient groups H1/H0, H2/H1, . . . , Hn/Hn−1 are called the factors of the series. (2) A subnormal series is called a composition series if H1/H0, H2/H1, . . . , Hn/Hn−1 are simple groups, called the composition factors. Example. Let G = Za be a cyclic group of order 6. Then the subgroup 3G is of order 2 and index 3 and we get a subnormal series {0} ≤3G ≤G 25 with factors 3G/{0} ∼ = Z2 and G/3G ∼ = Z3. Similarly the subgroup 2G is a subgroup of order 3 and index 2 that gives us another subnormal series {0} ≤2G ≤G with factors 2G/{0} ∼ = Z3 and G/2G ∼ = Z2. In fact these are both composition series as the factors are simple. Notice that the composition factors turn out to be the same (up to order). In fact this is always true. The Jordan-H¨ older Theorem. Suppose that a group G has composition series {1} = H0 < H1 < . . . < Hn = G and {1} = K0 < K1 < . . . < Km = G. Then n = m and the composition factors H1/H0, . . . , Hn/Hn−1 are the same (up to order) as K1/K0, . . . , Kn/Kn−1. Remarks. (1) Let G be a group with a normal subgroup N. It follows from the corre-spondence theorem that G/N is simple iﬀG ̸= N and there is no normal subgroup M in G such that N < M < G. (2) Suppose that for some group G we have a subnormal series {1} = H0 < H1 < . . . < Hn = G that is not a composition series. Then some quotient Hm/Hm−1 is not simple and by remark (1) there exists some subgroup K of G such that Hm−1 < K < Hm where K is normal in Hm. Notice also that (as Hm−1 is normal in Hm) Hm−1 is normal in K. By adding K, we thus get a subnormal series that is longer. (3) Let G be a ﬁnite group. It has a subnormal series (for example {1} < G). Ap-plying remark (2) we can continue adding terms while the series is not a composition series. Each time we get a longer series and as G is ﬁnite, this procedure must terminate in a composition series for G. Hence every ﬁnite group has a composition series. Examples (1) G be an internal direct product of S1, . . . , Sn where Si is simple. The map φ : S1 · · · Sn →Sn, a1a1 · · · an 7→an is a group homomorphism with kernel S1 · · · Sn−1. By the ﬁrst Isomorphism Theorem we have S1 · · · Sn S1 · · · Sn−1 ∼ = Sn we thus get a compostion series {0} < S1 < S1S2 < . . . < S1 · · · Sn = G with composition factors S1···Si S1···Si−1 ∼ = Si. This shows that there exists at least one group with S1, . . . , Sn as composition factors. (We can take S1 × S2 × · · · × Sn). 26 (2) Let n be a positive intger. All ﬁnite abelian groups of order n have the same com-position factors (Sheet 6). So normally there are a number of diﬀerent groups that have some given composition factors S1, . . . , Sn. The Jordan H¨ older theorem suggests the following possible strategy for ﬁnding all ﬁ-nite groups. (a) Find all the simple groups. (b) For any given choice S1, . . . , Sr of simple groups ﬁnd all the possible groups G whose composition factors are S1, . . . , Sr. Remarks. (1) Classifying all ﬁnite groups is generally concidered too hard. These are too rich and for a given choice of simple groups S1, . . . , Sn there is a great variety of ways of obtaining a group G with these as composition factors. As the number, n, of simple factors increases this becomes more and more complicated. (2) On the other hand (a) is done! This is one of the real triumphs of 20th century mathematics. The classiﬁcation result was announced in 1981. The proof is a collection of a number of journal articles by many diﬀerent mathematicians and runs over 10000 journal pages! According to the classiﬁcation of ﬁnite simple groups, these are (1) The cyclic groups of prime order, Zp, (2) The alternating groups, An, n ≥5, (3) The simple groups of Lie type (a number of inﬁnite families that crop up in a geomet-rical context) (4) Twenty six exceptional groups that do not belong to any of the inﬁnite families above. The groups in (1) are dealt with on sheet 6. In the next chapter we deal with (2). II. Solvable groups Deﬁnition. We say that a group is solvable if it has a subnormal series with abelian factors. Examples (1) Every abelian group is solvable. (2) We have that S3 has a composition series {1} < A3 < S3 with factors A3/{1} ∼ = Z3 and S3/A3 ∼ = Z2. As the factors are abelian S3 is solvable. Remark. We will see on sheet 6 that S4 is solvable. In next chapter we will how-ever see that Sn is not solvable for n ≥5. This is the underlying reason for the fact that we can’t solve the quintic by radicals. Proposition 3.1 A ﬁnite group G is solvable if and only its composition factors are cyclic of prime order. 27 Proof (⇐). A composition series with abelian factors is a subnormal series with abelian factors. (⇒). Suppose G is ﬁnite solvable group with subnormal series {1} = H0 < H1 < . . . < Hn = G where the factors are abelian. If this series is not a composition series, then some factor Hi/Hi−1 is not simple and we can insert some K, such that Hi−1 < K < Hi, to get a longer series. Notice that K/Hi−1 ≤Hi/Hi−1 and thus abelian. Also we have by the 3rd Isomorphism Theorem that Hi/K ∼ = Hi/Hi−1 K/Hi−1 that is a quotient of the abelian group Hi/Hi−1 and thus abelian. Thus the new longer series also has abelian factors. Continuing adding terms until we get a composition series, gives us then a composition series with abelian factors and thus factors that are cyclic of prime order. 2 How common are ﬁnite solvable groups? In fact surprisingly common. We mention two famous results. Theorem A (Burnside’s (p,q)-Theorem, 1904) Let p, q be prime numbers. Any group of order pnqm is solvable. Theorem B. (The odd order Theorem, Feit-Thompson, 1963). Any group of odd or-der is solvable. (This is really a magniﬁcent result. The proof is almost 300 pages and takes up a whole issue of a mathematics journal. Thompson received the Field’s medal for his contribu-tion). 28 4 Permutation groups and group actions I. Permutation groups and the simplicity of An, n ≥5 Convention. We will work with permutations from right to left. So if α, β ∈Sn then for αβ, we apply β ﬁrst and then α. Lemma 4.1 Let α ∈Sn. Then α(i1 i2 . . . im)α−1 = (α(i1) α(i2) . . . α(im)). Proof First suppose that k = α(j) is not in {α(i1), α(i2), · · · , α(im)}. Then j is not in {i1, i2, · · · , im} and α(i1 i2 . . . im)α−1(α(j)) = α(i1 i2 . . . im)(j) = α(j). This shows that α(i1 i2 . . . im)α−1 ﬁxes the elements outside {α(i1), α(i2), · · · , α(im)}. It remains to show that this map cyclically permutes α(i1), α(i2), · · · , α(im). But α(i1 i2 · · · im)α−1(α(ir)) = α(i1 i2 · · · im)(ir) = α(ir+1) where im+1 is interpreted as i1. This ﬁnishes the proof. 2 Orbits. Let i ∈{1, . . . , n}. Recall that the α-orbit containing i is the subset {αr(i) : r ∈ Z} and that {1, . . . , n} partitions into a pairwise disjoint union of α-orbits. Cycle structure. Suppose that the orbits of α ∈Sn are O1, O2, . . . , Or of sizes l1 ≥ l2 ≥· · · ≥lr. We then say that α has a cycle structure of type (l1, . . . , lr). Example. Let α = 1 2 3 4 5 6 7 8 3 5 4 2 1 7 6 8 = (1 3 4 2 5)(6 7)(8). Then α is of type (5, 2, 1). Deﬁnition. Let G be a group and x ∈G. The conjugacy class of G containing x is xG = {xg : g ∈G}. On sheet 6, we see that G is a pairwise disjoint union of its conjugacy classes. By Lemma 4.1, we have that if α is a permutation of some type (l1, . . . , lr), then the conjugacy class αSn consists of all permutations of that type. It follows also that if a normal subgroup N contains a permutation of type (l1, l2, . . . , lr) then it contains all per-mutations of that type. 29 Example. [(1 2)(3 4)]S4 = {(1 2)(3 4), (1 3)(2 4), (1 4)(2 3)}. Remarks. We have the following formula (check it) (i1 i2 · · · im) = (i1 im)(i1 im−1) · · · (i1 i2). (3) Remark As every permutation in Sn can be written as a product of disjoint cycles, this formula implies that every permutation in Sn can be written as a product of 2-cycles. Recall. A permutation α ∈Sn is said to be even/odd if it can be written as a product of even/odd number of 2-cycles. We also know that no permutation is both even and odd and thus Sn gets partitioned into even and odd elements. We denote by An the collection of all even elements. This is a subgroup that contains half the elements of Sn and for any odd element a in Sn, we have Sn = An ∪aAn. In particular An is of index 2 in Sn and is thus normal. Remark. By (3) we have that (i1 · · · im) is a even/odd permutation if and only if m is odd/even. Remark Any even permutation in An can be written as a product of even number of 2-cycles. So every permutation in An is a product of elements of one the following forms (for i, j, r and s distinct) (i j)(i r) = (i r j) (i j)(r s) = (i j)(i r)(r i)(r s) = (i r j)(r s i). It follows that any permutation in An can be written as a product of 3-cycles. Lemma 4.2 (a) If N Sn contains a 2-cycle then N = Sn. (b) If N An contains a 3-cycle then N = An. Proof (a) Let (i1 i2) be a 2-cycle of N. Let (j1 j2) be any other 2-cycle of Sn. Let α be a permutation that maps ik to jk. By Lemma 4.1 we have that (j1 j2) = α(i1 i2)α−1 which being a conjugate of (i1 i2) is also in N. So every 2-cycle is in N and as Sn is generated by 2-cycles it follows that N = Sn. (b) The proof is similar. Let (i1 i2 i3) be a 3-cycle of N and let (j1 j2 j3) be any other 3-cycle of An. Let α ∈Sn be a permutation that maps ik to jk. If α ∈An then (j1 j2 j3) = α(i1 i2 i3)α−1 is in N as before. If α on the other hand is odd then consider ﬁrst instead β = (j1 j2)α ∈An. The element (j2 j1 j3) = β(i1 i2 i3)β−1 is then in N and then also (j1 j2 j3) = (j2 j1 j3)−1. So all the 3-cycles are contained in N and as An is generated by the 3-cycles, it follows that N = An. 2 30 Lemma 4.3 Suppose n ≥5 and that {id} ̸= N An. Then \|N\| > n. Proof As N ̸= {id}, we have some id ̸= x ∈N. It suﬃces to show then xAn has at least n elements since then N would contain these elements plus the identity and thus more than n elements. Write x as a product of disjoint cycles and suppose that the longest cycle in the product has length m. There are two possibilities. Case 1. m ≥3. Here x is of the form x = (i j k · · ·)y where (i j k · · ·) is one of the cycles of longest length and y is the product of the remaining cycles. Now take any distinct r, s, t, u, v ∈{1, 2, . . . , n}. Let α ∈Sn such that α(i) = r, α(j) = s and α(k) = t. Notice that by Lemma 4.1, we have xα−1 = (r s t ...)yα−1. The same is true if α is replaced by (u v)α (notice that we are using n ≥5 here), so we can assume that α is even. It follows that we can choose r, s, t to be any elements in {1, 2, . . . , n} that we like. We can now easily ﬁnd at least n elments in xAn. For example we can take the elements (1 2 3 · · ·)y1, (1 2 4 · · ·)y2, (1 3 2 · · ·)y3, (1 4 2 · · ·)y4, · · · , (1 n 2 · · ·)yn Case 2. m = 2. As x is even we have to have at least two 2-cycles in the product. It follows that x = (i j)(k l)y where (i j), (k l) are two of the 2-cycles and y is the product of the remaining cycles. Now take any distinct r, s, t, u ∈{1, 2, . . . , n}. Let α ∈Sn such that α(i) = r, α(j) = s, α(k) = t and α(l) = u. Notice that xα−1 = (r s)(t u)yα−1 and the same holds when α is replaced by (r s)α (as (s r) = (r s)). We can therefore again suppose that α is even. As r, s, t, u can be chosen arbitrarily we can now again easily ﬁnd at least n elments in xAn. For example we can take these to be (1 2)(3 4)y1, (1 2)(3 5)y2, (1 3)(2 4)y3, (1 4)(2 3)y4, · · · , (1 n)(2 3)yn. So in both cases we have at least n elments in xAn and as N also contains the identity element, it follows that N has at least n + 1 elements. 2 Theorem 4.4 The group An is simple for n ≥5. Proof We prove this by induction on n ≥5. The induction basis, n = 5, is dealt with on Sheet 7. Now for the induction step, suppose n ≥6 and that we know that An−1 is simple. Let G(n) = {α ∈An : α(n) = n}. Notice that G(n) ∼ = An−1 and thus simple by 31 induction hypothesis. Now let {id} ̸= N An, we want to show that N = An. Step 1. N ∩G(n) ̸= {id}. We argue by contradiction and suppose that N ∩G(n) = {id}. This means that the only element in N that ﬁxes n is id. Now take α, β ∈N and suppose that α(n) = β(n). Then α−1β(n) = α−1(α(n)) = n and by what we have just said it follows that α−1β = id or α = β. Hence, a permutation α in N is determined by α(n) and since there are at most n values, we have that \|N\| ≤n. But his contradicts Lemma 4.3. Step 2. N = An. Now {id} ̸= N ∩G(n) G(n) (by the 2nd Isomorphism Theorem) and since G(n) is simple by induction hypothesis, it follows that N ∩G(n) = G(n). In particular, N con-tains a 3-cycle and thus N = An by Lemma 4.2. 2 II. Group actions Theorem 4.5 (Cayley). Any group G is isomorphic to a subgroup of Sym (G). Proof For a ∈G consider the map La : G →G, x 7→ax. Notice that La is bijective with inverse La−1 and thus La ∈Sym (G). Now consider the map φ : G →Sym (G), a 7→La. Notice that (La ◦Lb)(x) = abx = Lab(x) and thus φ(ab) = Lab = La ◦Lb = φ(a) ◦φ(b). Thus φ is a homomorphism. This homomorphism is injective since if φ(a) = φ(b) then a = a · 1 = La(1) = Lb(1) = b · 1 = b. Thus G is ismorphic to im φ where the latter is a subgroup of Sym (G). 2 Deﬁnition. Let X be a set and G a group. We say that X is a G-set if we have a right multiplication from G, i.e. a map φ : X × G →X, (x, g) 7→x · g satisfying (a) x · 1 = x ∀x ∈X (b) (x · a) · b = x · (ab) ∀a, b ∈G and x ∈X. Remark. One also says that G acts on X. Notice that x · g is just a notation for φ(x, g). Notice also that for every a ∈G we have that the map X →X : x 7→x · a is a permutation with inverse X →X : x 7→x · a−1. Examples. (1) Let X = G be a group. We can consider this as a G-set with re-spect to the natural right group multiplication x ∗g = xg. Clearly x ∗1 = x1 = x and (x ∗a) ∗b = (xa)b = x(ab) = x ∗(ab) by the associativity in G. 32 (2) Let H ≤G and let X be the collection of all the right cosets of H in G. We can again consider X as a G-set with respect to the natural right group multiplications Hg∗a = Hga again it is easy to see that Hg∗1 = Hg and (Hg∗a)∗b = Hg∗(ab) = Hgab. (3) Let G be a group and X = G. We deﬁne a group action by G on X by letting x ∗a = a−1xa = xa. Then X becomes a G-set as x1 = x and (xa)b = xab. (4) Let X be the collection of all the subgroups of G. We can consider X as a G-set with respect to the conjugation action. That is the right multiplication is given by H ∗g = g−1Hg = Hg. Again X is a G-set. Deﬁnition. Let X be a G-set. The stabilizer of x ∈X is Gx = {g ∈G : x · g = x} and the G-orbit of x ∈X is x · G = {x · g : g ∈G}. Lemma 4.6 Gx ≤G Proof Firstly by condition (a) we have 1 ∈Gx. Now suppose that a, b ∈Gx. Using condition (b) we then have x · (ab) = (x · a) · b = x · b = x and ab ∈Gx. It remains to show that Gx is closed under taking inverses. But this follows from x = x · 1 = x · (aa−1) = (x · a) · a−1 = x · a−1. This ﬁnishes the proof. 2 Theorem 4.7 (The Orbit Stabilizer Theorem). Let X be a G-set and x ∈X. Let H be the collection of all the right cosets of Gx in G. The map Ψ : H →x · G, Gxa 7→x · a is a bijection. In particular \|x · G\| = \|H\| = [G : Gx]. (In other words the cardinality of the G-orbit generated by x is the same as the cardinality of the collection of the right cosets of Gx in G). Proof Ψ is well deﬁned and injective. We have x · a = x · b ⇔x · ab−1 = x ⇔ab−1 ∈Gx ⇔Gxb = Gxa. As Ψ is clearly surjective, this ﬁnishes the proof. 2 Proposition 4.8 Let X be a G-set. The relation x ∼y if y ∈x · G is an equivalence relation on X and the equivalence classes are the G-orbits. 33 Proof As x = x · 1 it is clear that x ∼x and we have that ∼is reﬂexive. Now suppose that y = x·a. Then x = y ·a−1. This shows that ∼is symmetric. It now remains to show that ∼is transitive. But if y = x · a and z = y · b then x · (ab) = (x · a) · b = y · b = z. Hence we get x ∼z from x ∼y and y ∼z and this shows that ∼is transitive and thus an equivalence relation. Finally x ∼y iﬀy ∈x·G. Hence the equivalence class containing x is the G-orbit x·G. 2 Corollary 4.9 Suppose that the G-orbits of X are are xi · G, i ∈I. Then \|X\| = X i∈I [G : Gxi]. Proof We have that X = ∪i∈IxiG where the union in pairwise disjoint. Thus \|X\| = X i∈I \|xi · G\| = X i∈I [G : Gxi]. Where the ﬁnal equality follows from the Orbit Stabilizer Theorem. 34 5 Finite groups and Sylow Theory Deﬁnition. Let G be a group and x ∈G. The centralizer of x in G is CG(x) = {g ∈G : gx = xg}. Remark. We are going to see shortly that CG(x) is a stabilizer of x with respect to a certain action. Hence it will follow that CG(x) is a subgroup of G. This we can also see more directly. Conjugacy action and the class equation. Let G be a ﬁnite group. We can then think of G as a G-set where the right multiplication is deﬁned by x ∗g = xg = g−1xg. The G-orbit x ∗G is then {x ∗g = xg : g ∈G} = xG, the conjugacy class of x, and the stabilizer of x is Gx = {g ∈G : x = x ∗g = g−1xg} = {g ∈G : xg = gx} = CG(x). The orbit- stabiliser theorem thus tells us that \|xG\| = [G : CG(x)] We next write G as a disjoint union of G-orbits, that is conjugacy classes: G = aG 1 ∪aG 2 ∪· · · ∪aG r \| {z } each of size ≥2 ∪bG 1 ∪bG 2 ∪· · · bG s \| {z } each of size 1 Recall that Z(G) is the set of all those elements that commute with every element of G and that this is a normal subgroup of G. Now x ∈Z(G) if and only if x = g−1xg = xg for all g ∈G. It follows that x ∈Z(G) if and only if it’s conjucacy class {xg : g ∈G} consists only of one element x. Therefore Z(G) = {b1, . . . , bs} and G = Z(G) ∪aG 1 ∪aG 2 ∪· · · ∪aG r . and \|G\| = \|Z(G)\| + Pr i=1 \|aG i \|. Using the Orbit-Stabilizer Theorem we can deduce from this the class equation \|G\| = \|Z(G)\| + r X i=1 [G : CG(ai)] where the sum is taken over the r conjugacy classes with more than one element (so each [G : CG(ai)] > 1). Deﬁnition. Let p be a prime. A ﬁnite group G is said to be a p-group if \|G\| = pm for some m ≥0. Remark. The trivial group G = {1} is a p-group for any prime p. 35 Theorem 5.1 If G is a non-trivial ﬁnite p-group, then Z(G) is non-trivial. Proof We use the class equation \|G\| = \|Z(G)\| + r X i=1 [G : CG(ai)] \| {z } each ≥2 . Since 1 ̸= \|G\| is of p-power order it follows that \|G\| and each index [G : CG(ai)] are divisible by p. From the class equation it then follows that \|Z(G)\| is divisible by p. In particular it has at least two elements. 2 Example. The result above does not hold for ﬁnite groups in general. For example Z(S3) = {1} . Theorem 5.2 (Cauchy). Let G be a ﬁnite group with order that is divisible by a prime p. Then G contains an element of order p. Remark. From exercise 4 on sheet 3, we know that this is true when G is abelian. Proof We prove this by induction on \|G\|. If \|G\| = 1 then the result is trivial (\|G\| is then not divisible by any prime p so the statement will not get contradicted). Now suppose that \|G\| ≥2 and that the result holds for all groups of smaller order. Consider the class equation \|G\| = \|Z(G)\| + r X i=1 [G : CG(ai)] \| {z } each ≥2 . If any of the \|CG(ai)\| is divisible by p, then, as \|CG(ai)\| < \|G\|, we can use the induc-tion hypothesis to conclude that CG(ai) contains an element of order p (and thus G as well). Thus we can assume that none of \|CG(ai)\| are divisible by p. But then, as \|G\| = [G : CG(ai)] · \|CG(ai)\|, all the indices [G : CG(ai)] are divisible by p and the class equation implies that \|Z(G)\| is divisible by p. But Z(G) is abelian so it follows from the remark that it then contains an element of order p. 2 Theorem 5.3 Let G be a ﬁnite p-group and suppose that \|G\| = pn. There exist a chain of normal subgroups of G {1} = H0 < H1 < . . . < Hn = G where \|Hi\| = pi for i = 0, 1, . . . , n. Proof. We use induction on \|G\| = pn. If n = 0 then {1} = H0 = G is the chain we want. Now suppose that n ≥1 and that the result holds for all p-groups of smaller order. By Theorem 5.1, we have that Z(G) is non-trivial and by Cauchy’s Thoerem (the abelian version suﬃces) we know that there is a subgroup H1 of Z(G) such that \|H1\| = p. Notice that H1 G (as all the elements of H1 commute with all the elements of G and thus 36 gH1 = H1g for all g ∈G). Now \|G/H1\| = pn−1 and by induction hypothesis, there is a normal chain of subgroups {1} = K0 < K1 < · · · < Kn−1 = G/H1. By the Correspondence Theorem this chain corresponds to a normal chain of intermediate subgroups between H1 and G H1 < H2 < · · · < Hn = G where Ki−1 = Hi/H1. Then \|Hi\| = \|Ki−1\| · \|H1\| = pi−1 · p = pi and the chain {1} = H0 < H1 < · · · < Hn = G is the chain we want. 2. Remark. In particular this last result tells us that the converse to Lagrange’s Theo-rem holds when G is a p-group. To see the converse of Lagrange’s Theorem doesn’t hold in general consider the group A5. This is a simple group with 60 elements that has no subgroup with 30 elements. This is because a subgroup with 30 elements would have index 2 and thus be normal contradicting the simplicity of A5. Deﬁnition. Let G be a ﬁnite group of order pn · m where p does not divide m. A subgroup of order pn is called a Sylow p-subgroup of G. Remark. A more elegant way of saying that H is a Sylow p-subgroup of G is to say that H is a p-group such that [G : H] is not divisible by p. We are now going to prove a number of very nice and useful results about these. In particular we will see that these subgroups always exist and are (for a given group G) all isomorphic. We will also get some information about the number of the Sylow p-subgroups. These results, known collectively as the Sylow theorems, are going to be an important tool to understand the structure of the larger group G. Theorem 5.4 (1st Sylow Theorem) Let G be a ﬁnite group and p a prime number. There exists a Sylow p-subgroup of G. Proof We prove this by induction on \|G\|. If \|G\| = 1 then {1} is the Sylow p-subgroup for any prime p and thus the Sylow p-subgroups exist trivially in this case. Suppose now that \|G\| ≥2, and that the result holds for groups of smaller order. Let p be any prime and suppose that \|G\| = pnm where p ̸ \|m. If n = 0 then the trivial subgroup {1} would be a Sylow p-subgroup. We can thus assume that n ≥1. We use the class equation \|G\| = \|Z(G)\| + r X i=1 [G : CG(ai)] \| {z } each ≥2 . Suppose ﬁrst that some of [G : CG(ai)] is not divisible by p. Notice that \|G\| = [G : CG(ai)] · \|CG(ai)\| and as p does not divide [G : CG(ai)], whereas pn divides \|G\|, it follows that pn divides \|CG(ai)\|. But \|CG(ai)\| < \|G\| and thus by induction hypothesis CG(ai) 37 contains a Sylow p-subgroup that is of order pn and thus a Sylow p-subgroup of G as well. We are then left with the case when all of the indices [G : CG(ai)] are divisible by p. Then \|G\| is divisible by p and from the class equation it then follows that p divides the order of \|Z(G)\|. By Cauchy’s Theorem (we only need the abelian version) we know that Z(G) has a subgroup N of order p which has to be normal in G, since Ng = gN for all g ∈G. By induction hypothesis, G/N contains a Sylow p-subgroup that is a subgroup of order pn−1. By the Correspondence Theorem this subgroup is of the form P/N for some N ≤P ≤G. Notice that \|P\| = \|N\| · \|P/N\| = p · pn−1 = pn and thus P is a Sylow p-subgroup of G. 2 Corollary 5.5 Let G be a group of ﬁnite order and let pr be any power of a prime that divides the order of G. Then there exists a subgroup of order pr. Proof Suppose that \|G\| = pnm where p ̸ \|m. By the ﬁrst Sylow theorem there is a subgroup P of order pn and by Theorem 5.3 we know that P has a subgroup of order pr. 2 For the proof of the 1st Sylow Theorems we used arguments that involved counting the ele-ments of G. For our proofs of the other Sylow theorems we will be counting cosets instead. Counting cosets If H, K ≤G and let X be the set of all right cosets of H in G. Then K acts naturally on X through right multiplication: Ha ∗x = Hax. This turns X into a K-set. The next Lemma gives us a useful formula of counting the number of cosets that belongs to any given K-orbit. Lemma 5.6 The number of cosets in the K-orbit containing Ha are \|Ha ∗K\| = [K : K ∩Ha]. Proof We apply the Orbit-Stablizer Theorem. We need to determine the stablizer of the coset Ha in K. Now Hak = Ha ⇔Haka−1 = H ⇔aka−1 ∈H ⇔k ∈a−1Ha = Ha As k was in K to start with, this shows that the stablizer of Ha is K ∩Ha and the Orbit-Stabilizer Theorem tells us that \|Ha ∗K\| = [K : K ∩Ha]. 2 A formula for counting cosets. As before we let X be the set of all right H-cosets that we consider as a K-set. Suppose that X = Ha1 ∗K ∪Ha2 ∗K ∪· · · ∪Ham ∗K is the partition of X into disjoint K-orbits. Using Lemma 5.6 this implies that [G : H] = \|X\| = \|Ha1 ∗K\| + \|Ha2 ∗K\| + · · · + \|Ham ∗K\| = [K : K ∩Ha1] + [K : K ∩Ha2] + · · · + [K : K ∩Ham]. 38 Remark. We know from Theorem 5.3 that any Sylow p-subgroup contains a subgroup of an order that is an arbitrary p-power divisor of \|G\|. Now we show that the converse is true. Every subgroup of p-power order is contained in some Sylow p-subgroup. In fact we prove something much stronger. Theorem 5.7 Let H ≤G where H is a subgroup of p-power order. Let P be any Sylow p-subgroup of G. Then H ≤P a for some a ∈G. Proof Suppose that \|G\| = pnm where p ̸ \|m. Let X be the collection of all the right P cosets that we consider as a H-set. By the formula for counting cosets, we have m = [G : P] = [H : H ∩P a1] + [H : H ∩P a2] + . . . + [H : H ∩P am] (4) for some a1, . . . , am ∈G. We claim that H ∩P ai = H for some i = 1, . . . , m. Otherwise all the indices on the RHS of (4) would be divisible by p and we would get the contradiction that m is divisible by p. Hence H ∩P ai = H for some i ∈{1, . . . , m} or equivalently H ⊆P ai. 2 The 2nd Sylow theorem is a direct consequence of this. Theorem 5.8 (2nd Sylow Theorem). Any two Sylow p-subgroup are conjugate. (So they form a single conjugacy class). Proof Let P and Q be Sylow subgroups of G. By last theorem we know that Q ⊆P a for some a ∈G. But these two groups have the same order. Hence we have Q = P a. 2 Remark. The map φ : P →P a, x 7→xa is an isomorphism and thus P and P a are isomorphic. So all the Sylow p-subgroups are isomorphic and up to isomorphism we can talk about the Sylow p-subgroup. We now move on to the third and the last of the Sylow theorems. This is going to give us some information on the number of Sylow p-subgroups that is immensely useful as we will see. Theorem 5.9 (3rd Sylow Theorem). Let G be a ﬁnite group and p a prime. The number n(p) of Sylow p-subgroups of G satisﬁes: (i) n(p) = 1 + pr, for some non-negative integer r. (ii) n(p) divides \|G\|. 39 Proof (See at the end of this chapter). Remarks. (1) Suppose that \|G\| = pnm whre p does not divide m. Let P be a Sy-low p-subgroup of G. As \|G\| = pnm = \|P\| · [G : P] and as n(p) = 1 + pr divides \|G\| while being coprime to p, we must have that n(p) divides m = [G : P]. (2) Let P be a Sylow p-subgroup of G. The Sylow p-subgroups form a single conjugacy class { a−1Pa : a ∈G} The number n(p) of these is one iﬀall of them are equal to P, i.e. iﬀP G. Example 1. Let G be a group of order 2 · pr where p is an odd prime and r ≥1. By the Sylow theorems there exist a subgroup of order pr that is then of index 2 and therefore normal. Hence G can’t be simple if it is of order 2pr. Example 2. Let G be a group of order pq where p and q are primes and p > q. Now the number n(p) of Sylow p-subgroups, satisﬁes n(p) = 1 + pr and n(p) divides \|G\|/p = q The only possible n(p) satisfying these criteria is n(p) = 1. It follows that there is only one subgroup of order p and this must then be normal in G. We have thus shown that there are no simple groups of order pq. Example 3. To demonstrate the usefulness of the Sylow theorems, let us see how we can use them to see that there is no simple group of order 12 = 3 · 22. Firstly we have by the 1st Sylow theorem (or Cauchy’s thm) that there is a subgroup of order 3 and the number n(3) of these satisﬁes n(3) = 1 + 3r and n(3) divides \|G\|/3 = 4. There are only two possibilities, n(3) = 1 or n(3) = 4. In the ﬁrst case there is a normal subgroup of order 3. Let us look at the latter case. We have 4 groups of order 3 and therefore 4 · 2 = 8 elements of order 3 (in each of the Sylow 3-subgroups there are two elements of order 3 and as the intersection of any two of these is {1} we get exactly 4 · 2 = 8 elements of order 3). There remain 4 elements that must form a unique Sylow 2-subgroup Q (which has order 4). Notice that none of the elements of order 3 can be in Q as 3 does not divide 4. As n(2) = 1 we now have that Q G. Example 4. Let p, q be distinct primes. We will see that there is no simple group of order p2q. We consider two cases. If p > q then n(p) = 1 + pr should divide q and as p > q this can only happen if n(p) = 1. But in this case we have a normal Sylow p-subgroup. We can thus assume that p < q. Now n(q) = 1 + qr divides \|G\|/q = p2. If n(q) = 1 we have a normal Sylow q-subgroup, so we can suppose that n(q) > 1. As q > p the only possibility is that n(q) = p2. We then have 1 + qr = p2 ⇔qr = p2 −1 = (p −1)(p + 1). 40 As the prime q is greater than p, it follows that q divides p + 1 and again as q > p, we must have q = p + 1. The only two primes that are one apart are 2 and 3. Thus p = 2 and q = 3 and \|G\| = p2 · q = 12. But by Example 3, there is no simple group of order 12 and we are done. Remark. We mentioned before a famous result of Burnside, the Burnside’s (p, q)- Theo-rem. This said that any group G of order pnqm is solvable. This means that there are not composition factors that are non-abelian. In particular G can’t be non-abelian simple. Later in the notes and on the exercise sheets we will apply the Sylow theorems to ﬁnd all groups of order up to and including 15. We will also see that there is no non-abelian simple group of order less than 60 (\|A5\| = 60). Before leaving this section we add another weapon to our list. This is Poincar´ e’s Lemma that is often of great help. Deﬁnition. Suppose H ≤G. The subgroup HG = \ g∈G Hg is called the core of H in G. Remarks. (1) As H = He is one of the conjugates of H it is clear that HG ≤H and we will see later that HG G as a part of Poincar´ e’s Lemma. This we can also see directly. Let a ∈G then Ha G = \ g∈G Hga = \ b∈G Hb = HG, where the last identity holds from the fact that Ga = G. (2) If N ≤H and N G then for all g ∈G we have N = N g ≤Hg. It follows that N ≤ \ g∈G Hg = HG. This shows that HG is the largest normal subgroup of G that is contained in H. Theorem 5.10 Suppose G is a group (possibly inﬁnite) and let H ≤G such that [G : H] = n < ∞. Then G/HG ∼ = K for some K ≤Sn. Proof Let X = {gH : g ∈G}. For each a ∈G, we get a map La : X →X, gH 7→agH. Notice that La is bijective with inverse La−1. Also notice that La ◦Lb(gH) = La(bgH) = abgH = Lab(gH). Now consider the map φ : G →Sym (X), a 7→La. We have just seen that Lab = La ◦Lb and this implies that φ(ab) = φ(a) ◦φ(b). Thus φ is a homomorphism. We next identify 41 the kernel. We have φ(a) = La = id ⇔ agH = gH for all g ∈G ⇔ g−1agH = H for all g ∈G ⇔ g−1ag ∈H for all g ∈G ⇔ a ∈gHg−1 for all g ∈G. Therefore the kernel is T g∈G Hg−1 = T a∈G Ha = HG. By the 1st Isomorphism Theorem we have that HG G and G/HG = G/ker φ ∼ = im φ where im φ ≤Sym (X). As \|X\| = n we have that Sym (X) ∼ = Sn and thus G/HG isomorphic to a subgroup of Sn. 2. Corollary 5.11 (Poincar´ e’s Lemma). Let G be a ﬁnite simple group with a subgroup H such that [G : H] = n > 1. Then G ∼ = K for some K ≤Sn. In particular \|G\| divides \|Sn\| = n!. Proof HG is a normal subgroup of G and as HG is contained in H we can’t have HG = G. Now G is simple and we conclude that HG = {1}. The result now follows from Theorem 5.11 as G/{1} ∼ = G. 2 Example 5. Let us give another proof of the fact that there is no simple group of order 12. We argue by contradiction and suppose that G is a simple groups with 12 elements. By the Sylow theorems we have a subgroup of order 4 and thus of index 3. By Corollary 5.12 if follows that 12 = \|G\| divides the 3! = 6. This is absurd. We end this section by proving the 3rd Sylow Theorem. We need ﬁrst some prelimi-nary work. Deﬁnition Let H ≤G. The normalizer of H in G is NG(H) = {g ∈G : Hg = H}. One can easily check that this is a subgroup of G (in fact it follows also from next remark as NG(H) turns out to be a stabiliser with respect to a certain G-action) and clearly H NG(H). Remarks. (1) Let X be the set of all subgroups of G. As we have seen before G acts naturally on X by conjugation and so we can think of X as a G-set with respect to this action. The stabilizer of the subgroup H is then NG(H) and the Orbit-Stabilizer theorem tells us that the number of conjugates of H, that is the size of the G orbit {Hg : g ∈G}, is [G : NG(H)]. (2) Let P be a Sylow p-subgroup of G. By the 2nd Sylow Theorem, we know that the Sylow p-subgroups form a single conjugacy class {P g : g ∈G}. By Remark (1) the total number of all Sylow p-subgroups is then n(p) = [G : NG(P)]. 42 Lemma 5.12 Let P be a Sylow p-subgroup of G. Then P is the unique Sylow p-subgroup of NG(P). Proof Let Q be any Sylow p-subgroup of NG(P). By the second Sylow Theorem we have Q = P a for some a ∈NG(P). But then Q = P a = P since a normalizes P. 2. Proof of the 3rd Sylow Theorem. Let P be a Sylow p-subgroup of G. Since the Sylow p-subgroups form a single conjugacy class {P a : a ∈G}, we know from the remark above that their number is n(p) = [G : NG(P)]. In particular n(p) divides \|G\|. This proves (ii). To prove (i) we need more work. Let N = NG(P) and let X be the collection of all the right N cosets of G that we consider as a P-set. Write X as a disjoint union of P-orbits, say X = Na1 ∗P ∪Na2 ∗P ∪· · · ∪Nam ∗P where we assume that the ﬁrst orbit Na1 ∗P is the one containing the coset N · 1 = N and we can also then assume that a1 = 1 From this we get that n(p) = \|Na1 ∗P\| + \|Na2 ∗P\| + · · · + \|Nam ∗P\| = [P : P ∩N a1] + [P : P ∩N a2] + · · · + [P : P ∩N am]. Now notice that P ∩N ai = P iﬀP ≤N ai iﬀP a−1 i ≤N. However, by Lemma 5.9, this happens iﬀP a−1 i = P that happens iﬀai ∈NG(P). But then Na = Nai ∈Nai ∗P and as the only orbit containing N is Na1 ∗P, it follows that i = 1. (Notice also that a1 ∈NGP and thus [P : P ∩N a1] = 1). We conclude from this that [P : P ∩N ai] is divisble by p for i = 2, . . . , m and that [P : P ∩N a1] = 1. Hence n(p) = 1 + pr for some non-negative integer r. 2 43 6 Semidirect products and groups of order ≤15 I. Semidirect products. We will now introduce a generalization of direct products that is very useful for de-scribing and constructing groups. As with direct products these come in two disguises internal and external semidirect products. Notation. Suppose that N is a group and φ : N →N is an automorphism. In this section we will use bφ for the value of b under φ (instead of φ(b)). This will actually make things look clearer. We will also operate a composition of two automorphisms from left to right. Thus bψ◦φ = (bψ)φ. Deﬁnition. Let G be a group and N G, H ≤G. We say that G is the internal semidirect product of N by H if G = HN and H ∩N = {1}. Remark. The deﬁnition is thus very similar to the deﬁnition of an internal direct prod-uct. The only diﬀerenct is that one of the groups H does not have to be normal in general. When H is normal as well then we get a direct product. Lemma 6.1 Let G be an internal semidirect product of N by H, then the following hold. (1) Every element g ∈G can be written uniquely as g = ab with a ∈H and b ∈N. (2) Let a1, a2 ∈H and b1, b2 ∈N. Then (a1b1) · (a2b2) = (a1a2) · (ba2 1 b2) Proof (1) If a1b1 = a2b2 then a−1 2 a1 = b2b−1 1 is in H ∩N and thus trivial. So a1 = a2 and b1 = b2). (2) We have (a1b1) · (a2b2) = (a1a2) · (a−1 2 b1a2b2) = (a1a2) · (ba2 1 b2). This ﬁnishes the proof. 2 Remark. Thus, like for internal direct products, we can treat elements like pairs ab where a is the H component and b is the N component. Furthermore multiplying two such elements a1b1 and a2b2 gives us a new element whose H compenent is a1a2 and whose 44 N component is ba2 1 b2. It follows that if we know the structure of H and N and if we know how H acts on N by conjugation, then we know the structure of the semidirect product G. If you for example had a multiplication table for H and N and you knew how H acts on N by conjugation then you could write down a multiplication table for G. Remark. For a ∈H, let φa : N →N be the conjugation by a. We have seen pre-viously that this map is an automorphism. Now xφab = xab = (xa)b = (xφa)φb = xφa◦φb. Consider the map Ψ : H →Aut (N), a 7→φa. As Ψ(ab) = φab = φa ◦φb = Ψ(a) ◦Ψ(b), this map is a homomorphism. Notice also a1b1 · a2b2 = (a1a2) · (ba2 1 b2) = (a1a2) · (b φa2 1 b2) = (a1a2) · (bΨ(a2) 1 b2) This motivates the following structure. Deﬁnition Let N, H be groups and let Ψ : H →Aut (N) be a homomorphism. The external semidirect product H ⋉Ψ N, of N by H with respect to Ψ, is the cartesian set product of H and N with the binary operation (a1, b1) · (a2, b2) = (a1a2, bΨ(a2) 1 b2) H ⋉Ψ N is a group. First let us check the associativity. Firstly [(a1, b1) · (a2, b2)] · (a3, b3) = (a1a2, bΨ(a2) 1 b2) · (a3, b3) = (a1a2a3, (bΨ(a2) 1 b2)Ψ(a3)b3) Since Ψ(a3) ∈Aut (N) and since Ψ is a homorphism, we get (bΨ(a2) 1 b2)Ψ(a3)b3 = bΨ(a2)Ψ(a3) 1 bΨ(a3) 2 b3 = bΨ(a2a3) 1 bΨ(a3) 2 b3. Then secondly (a1, b1) · [(a2, b2) · (a3, b3)] = (a1, b1) · (a2a3, bΨ(a3) 2 b3) = (a1a2a3, bΨ(a2a3) 1 bΨ(a3) 2 b3). This shows that the associative law holds. To see that (1, 1) is the identity. Notice that any automorphism maps 1 to itself and that Ψ(1) = id. Thus (1, 1) · (a, b) = (1 · a, 1Ψ(a)b) = (a, b) and (a, b) · (1, 1) = (a · 1, bΨ(1) · 1) = (a, bid · 1) = (a, b). Finally, the inverse of (a, b) is (a−1, (bΨ(a−1))−1) since (a, b) · (a−1, (bΨ(a−1))−1) = (a · a−1, bΨ(a−1)(bΨ(a−1))−1) = (1, 1) 45 and (a−1, (bΨ(a−1))−1) · (a, b) = (aa−1, ((bΨ(a−1))−1)Ψ(a)b) = (1, (bΨ(a−1)Ψ(a))−1b) = (1, (bΨ(1))−1b) = (1, (bid)−1b) = (1, 1). Remark. Consider an internal semidirect product of N by H and let Ψ : H →Aut (N) be the homomorphism that maps a to φa where the latter is the automorphism that takes b to ba. Using the data N, H and Ψ, we can also construct the external semidirect product H ⋉Ψ N. Not surprisingly, the two are isomorphic (see Exercise 1 on Sheet 10). II. Groups of order less than 16. In this section (and on the exercise sheets) we play with our new tools and ﬁnd all groups of order up to and including 15. We have already shown previously (Exercise 2 on sheet 9) that the only group of order 15 is Z15 and we have no diﬃculty with groups of order 1. When p is a prime, there is exactly one group of order p, the cyclic group Zp of order p. On exercise sheet 8 we also show that there are only two groups of order p2, namely Zp2 and Zp ⊕Zp. Semidirect products of cyclic groups. Suppose that G = HN is an internal semidi-rect product of a cyclic group N = ⟨a⟩by another cyclic group ⟨b⟩. We have that ab = ar (5) for some r ∈Z. Inductively it follows that abn = arn and then that (am)bn = (abn)m = amrn . Thus the structure of G is determined by (5) and the orders of a and b. We will now introduce an inﬁnite family of groups, many of which will crop up in the list of groups of orders 1 to 15. Example.(D2n, the dihedral group of order 2n). Consider a regular n-gon in the com-plex plane with corners 1, u, u2, . . . , un−1 where u = e2πi/n (draw a ﬁgure). The symmetry group of this regular n-gon is generated by a counter clockwise rotation a of 2π/n around the origin and the reﬂection b in the real axis. This can be described explicitly as follows: a(z) = e2πi/n · z b(z) = ¯ z. Let us calculate b−1ab(z) = bab(z) = ba(¯ z) = b(e2πi/n · ¯ z) = e−2πi/nz = a−1(z). 46 This means that the symmetry group is a group of order 2n that is a semidirect product of ⟨a⟩, a cyclic group of order n, and ⟨b⟩, a cyclic group of order 2. Furthermore the action of ⟨b⟩on ⟨a⟩is determined by ab = a−1. The unique group of order 2n with a normal cyclic subgroup ⟨a⟩of order n, and a cyclic subgroup ⟨b⟩where ab = a−1 is called the dihedral group of order 2n and is denoted D2n. Theorem 6.2 Let p be an odd prime. There are (up to isomorphism) exactly two groups of order 2p these are Z2p and D2p. Proof By the Sylow theorems (or Cauchy’s thm) there is a subgroup N = ⟨a⟩of order p. Since N is of index 2 it is normal. There is also a group H = ⟨b⟩of order 2. Clearly H ∩N = {1}, since it is a subgroup of both H and N and thus its order divides both 2 and p. So we have that G is a semidirect product of N by H. To determine the group structure it remains to see how H can act on N. Now b−1ab = ar for some 0 ≤r ≤p −1. Using the fact that b is of order 2, we see that a = b−1(b−1ab)b = b−1arb = (b−1ab)r = ar2. This implies that ar2−1 = 1 and thus p must divide r2 −1 = (r −1)(r + 1). The only possibilities for this to happen is when r = 1 or r = p −1. In the ﬁrst case the group is abelian and G = ⟨ba⟩is a cyclic group of order 2p. (Notice that (ba)2 = a2 ̸= 1 and (ba)p = b ̸= 1 so the order of ba is 2p by Lagrange’s theorem). In the latter case we have the relations ap = 1, b2 = 1, bab−1 = ap−1 = a−1 which gives us D2p as we have seen. 2 Remark. The only orders up to 15 that are not covered by 1, 15, p, p2 and 2p are 8 and 12. These are dealt with on the excercise sheets 9 and 10. We end by constructing a certain group of order 12, using the external semidirect product. Example. Let N = ⟨a⟩be a cyclic group of order 3 and H = ⟨b⟩be a cyclic group of order 4. The map φ : N →N, x 7→x−1 is in Aut (N). The map Ψ : H →Aut (N), br 7→φr is a homomorphism. It is well deﬁned as br = bs ⇒bs−r = 1 ⇒4\|(r −s) ⇒φs−r = id ⇒ φr = φs. Consider the external semidirect product T = H ⋉Ψ N. It is a group of order 12 with a cyclic normal Sylow 3-subgroup of order 3 and a cyclic Sylow 2-subgroup of order 4. From our study in this chapter and the exercise sheets we can conclude that the groups of order ≤15 are (up to isomorphism) 47 order groups 1 {1} 2 Z2 3 Z3 4 Z4, Z2 ⊕Z2 5 Z5 6 Z6, D6 7 Z7 8 Z8, Z4 ⊕Z2, Z2 ⊕Z2 ⊕Z2, D8, Q 9 Z9, Z3 ⊕Z3 10 Z10, D10 11 Z11 12 Z4 ⊕Z3, Z2 ⊕Z2 ⊕Z3, A4, D12, T 13 Z13 14 Z14, D14 15 Z15 48
15744	https://www.studocu.com/in/document/mahatma-gandhi-university/bsc-mathematics/arithmetic-progression-wikipedia/42803893	Arithmetic progression - Wikipedia - Sear Arithmetic progression An arithmetic progression or - Studocu Skip to document Teachers University High School Discovery Sign in Welcome to Studocu Sign in to access study resources Sign in Register Guest user Add your university or school 0 followers 0 Uploads 0 upvotes New Home My Library AI Notes Ask AI AI Quiz Chats Recent You don't have any recent items yet. My Library Courses You don't have any courses yet. Add Courses Books You don't have any books yet. Studylists You don't have any Studylists yet. Create a Studylist Home My Library Discovery Discovery Universities High Schools Teaching resources Lesson plan generator Test generator Live quiz generator Ask AI Arithmetic progression - Wikipedia About Arithmetic progression Course BSC Mathematics (MAM036H) 624 documents University Mahatma Gandhi University Academic year:2021/2022 Uploaded by: Anonymous Student Mahatma Gandhi University Recommended for you 33 MATH Practice QUIZ AND Answers BSC Mathematics Practice materials 100% (3) 31 A quick review of vector algebra from vector calculus BSC Mathematics Practice materials 100% (1) 6 Flywheel- moment of inertia BSC Mathematics Practice materials 100% (1) 7 Torsion pendulum - It's practical notes BSC Mathematics Practice materials 100% (1) 23 Mathongo.com- Ncert-Solutions-Class-12-Maths-Chapter-2-Inverse-Trigonometric-Functions BSC Mathematics Practice materials 100% (1) Comments Please sign in or register to post comments. Report Document Students also viewed Mathongo - Algebra (969) maths gyaan sutra jee main Maths combined question and answers Basic introduction to real Analysis previous year question papers previous year question paper 2021 Related documents Techniques of Counting Matrices assistant 1 Partial Differential Equations Linear algebra Previous year question 3 Mathongo - Differentiability Preview text Sear Arithmetic progression An arithmetic progression or arithmetic sequence (AP) is a sequence of numbers such that the difference between the consecutive terms is constant. For instance, the sequence 5, 7, 9, 11, 13, 15,... is an arithmetic progression with a common difference of 2. If the initial term of an arithmetic progression is and the common difference of successive members is , then the -th term of the sequence ( ) is given by: , If there are m terms in the AP, then represents the last term which is given by: . A finite portion of an arithmetic progression is called a finite arithmetic progression and sometimes just called an arithmetic progression. The sum of a finite arithmetic progression is called an arithmetic series. Sum The sum of the members of a finite arithmetic progression is called an arithmetic series. For example, consider the sum: This sum can be found quickly by taking the number n of terms being added (here 5), multiplying by the sum of the first and last number in the progression (here 2 + 14 = 16), and dividing by 2: In the case above, this gives the equation: This formula works for any real numbers and. For example: this Derivation 2 + 5 + 8 + 11 + 14 = 40 14 + 11 + 8 + 5 + 2 = 40 16 + 16 + 16 + 16 + 16 = 80 Computation of the sum 2 + 5 + 8 + 11 + 14. When the sequence is reversed and added to itself term by term, the resulting sequence has a single repeated value in it, equal to the sum of the first and last numbers (2 + 14 = 16). Thus 16 × 5 = 80 is twice the sum. The product of the members of a finite arithmetic progression with an initial element a 1 , common differences d, and n elements in total is determined in a closed expression where denotes the Gamma function. The formula is not valid when is negative or zero. This is a generalization from the fact that the product of the progression is given by the factorial and that the product for positive integers and is given by Derivation where denotes the rising factorial. By the recurrence formula , valid for a complex number , , , Product ... so that for a positive integer and a positive complex number. Thus, if , , and, finally, Examples Example 1 Taking the example , the product of the terms of the arithmetic progression given by up to the 50th term is Example 2 The product of the first 10 odd numbers is given by = 654,729, The standard deviation of any arithmetic progression can be calculated as ... Standard deviation Arithmetico-geometric sequence Inequality of arithmetic and geometric means Primes in arithmetic progression Linear difference equation Generalized arithmetic progression, a set of integers constructed as an arithmetic progression is, but allowing several possible differences Heronian triangles with sides in arithmetic progression Problems involving arithmetic progressions Utonality Polynomials calculating sums of powers of arithmetic progressions # 1. Duchet, Pierre (1995), "Hypergraphs", in Graham, R. L.; Grötschel, M.; Lovász, L. (eds.), Handbook of # combinatorics, Vol. 1, 2, Amsterdam: Elsevier, pp. 381–432, MR 1373663 (mathscinet.ams/m # athscinet-getitem?mr=1373663). See in particular Section 2, "Helly Property", pp. 393–394 ( # oks.google/books?id=5Y9NCwlx63IC&pg=PA393). # 2. Hayes, Brian (2006). "Gauss's Day of Reckoning" (americanscientist/article/gausss-day- # of-reckoning). American Scientist. 94 (3): 200. doi:10.1511/2006.59 (doi/10.1511%2F # 006.59). Archived (web.archive/web/20120112140951/americanscientist # rg/issues/id,y,no.,content,page,css.print/issue) from the original on 12 January # 2012. Retrieved 16 October 2020. # 3. Høyrup, J. The “Unknown Heritage”: trace of a forgotten locus of mathematical sophistication. Arch. Hist. # Exact Sci. 62, 613–654 (2008). doi/10.1007/s00407-008-0025-y # 4. Tropfke, Johannes (1924). Analysis, analytische Geometrie (books.google/books?id=9dJ_F4l # CXTQC). Walter de Gruyter. pp. 3–15. ISBN 978-3-11-108062-8. # 5. Tropfke, Johannes (1979). Arithmetik und Algebra (books.google/books?id=7UW0DwAAQ # BAJ). Walter de Gruyter. pp. 344–354. ISBN 978-3-11-004893-3. # 6. Problems to Sharpen the Young (jstor/stable/3620384) , John Hadley and David # Singmaster, The Mathematical Gazette, 76 , #475 (March 1992), pp. 102–126. References # 7. Ross, H. & Knott,B (2019) Dicuil (9th century) on triangular and square numbers, British Journal for the # History of Mathematics, 34:2, 79-94, doi/10.1080/26375451.2019. # 8. Sigler, Laurence E. (trans.) (2002). Fibonacci's Liber Abaci (archive/details/fibonaccislibera # 0sigl). Springer-Verlag. pp. 259 (archive/details/fibonaccislibera00sigl/page/n260) –260. # ISBN 0-387-95419-8. # 9. Katz, Victor J. (edit.) (2016). Sourcebook in the Mathematics of Medieval Europe and North Africa (http # s:books.google/books?id=39waDQAAQBAJ). Princeton University Press. pp. 91, 257. # ISBN 9780691156859. # 10. Stern, M. (1990). 74 A Mediaeval Derivation of the Sum of an Arithmetic Progression. The # Mathematical Gazette, 74(468), 157-159. doi:10/ "Arithmetic series" (encyclopediaofmath/index.php?title=Arithmetic_series) , Encyclopedia of Mathematics, EMS Press, 2001 Weisstein, Eric W. "Arithmetic progression" (mathworld.wolfram/ArithmeticProgre ssion). MathWorld. Weisstein, Eric W. "Arithmetic series" (mathworld.wolfram/ArithmeticSeries.htm l). MathWorld. Retrieved from "en.wikipedia/w/index.php? title=Arithmetic_progression&oldid=1121229340" External links Arithmetic progression - Wikipedia Download Download AI Tools Ask AI Multiple Choice Flashcards Quiz Video Audio Lesson 0 0 Save Arithmetic progression - Wikipedia Course: BSC Mathematics (MAM036H) 624 documents University: Mahatma Gandhi University Info More info Download Download AI Tools Ask AI Multiple Choice Flashcards Quiz Video Audio Lesson 0 0 Save Sear Arithmetic progression An arithmetic progr ession or arithmetic sequence (AP) is a sequence of numbers such that the differ ence between the consecutive terms is constant. For instance, the sequence 5, 7, 9, 11, 13, 15, . . . is an arithmetic progression with a common diff erence of 2. If the initial term of an arithmetic progression is and the common difference of successiv e members is , then the -th term of the sequence () is given by: , If there are m terms in the AP, then represents the last term which is giv en by: . A ﬁnite por tion of an arithmetic progression is called a ﬁnite arithmetic pr ogression and sometimes just called an arithmetic progression. The sum of a ﬁnite arithmetic pr ogression is called an arithmetic series. Sum The sum of the members of a ﬁnite arithmetic progression is called an arithmetic series. F or example, consider the sum: This sum can be found quickly by taking the number n of terms being added (here 5), multiplying by the sum of the ﬁrst and last number in the progr ession (here 2 + 14 = 16), and dividing by 2: In the case above, this gives the equation: This formula works for any real numbers and . For example: this Derivation 2+5+8+11+14=40 14+11+8+5+2=40 16+16+16+16+16=80 Computation of the sum 2 + 5 + 8 + 11 + 14. When the sequence is rev ersed and added to itself term by term, the resulting sequence has a single repeated value in it, equal t o the sum of the ﬁrst and last numbers (2 + 14 = 16). Thus 16 × 5 = 80 is twice the sum. T o derive the above formula, begin b y expressing the arithmetic series in two differ ent ways: Rewriting the terms in reverse or der: Adding the corresponding terms of both sides of the two equations and halving both sides: This formula can be simpliﬁed as: Furthermore, the mean value of the series can be calculated via: : The formula is very similar to the mean of a discrete uniform distribution. … Animated proof for the formula giving the sum of the ﬁrst integers 1+2+...+n. Too long to read on your phone? Save to read later on your computer Save to a Studylist The product of the members of a ﬁnite arithmetic progr ession with an initial element a 1, common differences d, and n elements in t otal is determined in a closed expression where denotes the Gamma function. The formula is not valid when is negative or zer o. This is a generalization fr om the fact that the product of the progression is given by the factorial and that the product for positive integers and is given by Derivation where denotes the rising factorial. By the recurrence formula , valid for a complex number , , , Product … so that for a positive integer and a positive complex number. Thus, if , , and, ﬁnally, Examples Example 1 T aking the example , the product of the terms of the arithmetic progression giv en by up to the 50th term is Example 2 The product of the ﬁrst 10 odd numbers is given by = 654,729,075 The standard deviation of any arithmetic pr ogression can be calculated as … Standard deviation where is the number of terms in the pr ogression and is the common difference between terms. The formula is very similar to the standard deviation of a discr ete uniform distribution. The intersection of any two doubly inﬁnite arithmetic progressions is either empty or another arithmetic progression, which can be found using the Chinese r emainder theorem. If each pair of progressions in a family of doubly inﬁnite arithmetic pr ogressions hav e a non-empty intersection, then there exists a number common to all of them; that is, inﬁnite arithmetic progressions form a Helly family. Howe ver, the intersection of inﬁnitely many inﬁnite arithmetic progressions might be a single number r ather than itself being an inﬁnite progression. According to an anecdote of uncertain reliability, young Carl F riedrich Gauss in primary school reinvented this method t o compute the sum of the integers from 1 through 100, by multiplying n 2 pairs of numbers in the sum by the values of each pair n + 1. Howev er, regar dless of the truth of this story, Gauss was not the ﬁrst to discov er this formula, and some ﬁnd it likely that its origin goes back to the Pythagoreans in the 5th century BC. Similar rules were known in antiquity to Archimedes, Hypsicles and Diophantus; in China to Zhang Qiujian; in India t o Aryabhata, Brahmagupta and Bhaskara II; and in medie val Europe t o Alcuin,Dicuil,Fibonacci, Sacrobosco and to anonymous commentat ors of T almud known as T osaﬁsts. Geometric progression Harmonic progression T riangular number Intersections History See also Document continues below Discover more from: BSC MathematicsMAM036HMahatma Gandhi University 624 documents Go to course 2 Model Exam: Foundation of Mathematics MM1CRT01 First Semester 2022 BSC Mathematics Other 100% (6) 1 1st Sem MM1CRT01 Foundation of Mathematics - Internal Assessment Notes BSC Mathematics Other 100% (4) 3 First Semester MM1CRT01 Foundation of Mathematics Question Paper - July 2022 BSC Mathematics Other 100% (3) 3 ST3CMT03 Probability Distributions Final Exam January 2023 BSC Mathematics Other 100% (3) 18 Phy- Motion Under Gravity: Historical Perspectives and Principles BSC Mathematics Lecture notes 100% (3) 2 First Order Differential Equations Flowchart Notes MATH 2420 BSC Mathematics Lecture notes 100% (3) Discover more from: BSC MathematicsMAM036HMahatma Gandhi University624 documents Go to course 2 Model Exam: Foundation of Mathematics MM1CRT01 First Semester 2022 BSC Mathematics 100% (6) 1 1st Sem MM1CRT01 Foundation of Mathematics - Internal Assessment Notes BSC Mathematics 100% (4) 3 First Semester MM1CRT01 Foundation of Mathematics Question Paper - July 2022 BSC Mathematics 100% (3) 3 ST3CMT03 Probability Distributions Final Exam January 2023 BSC Mathematics 100% (3) 18 Phy- Motion Under Gravity: Historical Perspectives and Principles BSC Mathematics 100% (3) 2 First Order Differential Equations Flowchart Notes MATH 2420 BSC Mathematics 100% (3) Arithmetico-geometric sequence Inequality of arithmetic and geometric means Primes in arithmetic progression Linear differ ence equation Generaliz ed arithmetic progression, a set of integers constructed as an arithmetic pr ogression is, but allowing sever al possible differences Heronian triangles with sides in arithmetic progr ession Problems involving arithmetic pr ogressions Utonality Polynomials calculating sums of powers of arithmetic progr essions Duchet, Pierre (1995), "Hypergr aphs", in Gr aham, R. L.; Grötschel, M.; Lo vász, L. (eds.), Handbook of combinatorics, V ol. 1, 2, Amsterdam: Elsevier, pp.381–432, MR1373663 ( athscinet-getitem?mr=1373663). See in particular Section 2.5, "Helly Property", pp.393–394 ( oks.google.com/books?id=5Y9NCwlx63IC&pg=P A393). Hayes, Brian (2006). "Gauss's Day of Reckoning" ( ticle/gausss-day- of-reckoning). American Scientist. 94 (3): 200. doi:10.1511/2006.59.200 ( g/10.1511%2F2 006.59.200). Archiv ed ( g/web/20120112140951/ rg/issues/id.3483,y.0,no.,content.true,page.1,css.print/issue.aspx) from the original on 12 January Retrieved 16 Oct ober 2020. Høyrup, J. The “Unknown Heritage”: trace of a for gotten locus of mathematical sophistication. Arch. Hist. Exact Sci. 62, 613–654 (2008). 007/s00407-008-0025-y T ropfke, Johannes (1924). Analysis, analytisc he Geometrie ( CXTQC). Walter de Gruyter. pp.3–15. ISBN978-3-11-108062-8. T ropfke, Johannes (1979). Arithmetik und Algebra ( BAJ). Walter de Gruy ter. pp.344–354. ISBN978-3-11-004893-3. Problems t o Sharpen the Y oung ( John Hadley and Davi d Singmaster, The Mathematical Gazette, 76, #475 (March 1992), pp. 102–126. References 1 out of 9 Share Download Download More from:BSC Mathematics(MAM036H) More from: BSC MathematicsMAM036HMahatma Gandhi University 624 documents Go to course 2 Model Exam: Foundation of Mathematics MM1CRT01 First Semester 2022 BSC Mathematics Other 100% (6) 1 1st Sem MM1CRT01 Foundation of Mathematics - Internal Assessment Notes BSC Mathematics Other 100% (4) 3 First Semester MM1CRT01 Foundation of Mathematics Question Paper - July 2022 BSC Mathematics Other 100% (3) 3 ST3CMT03 Probability Distributions Final Exam January 2023 BSC Mathematics Other 100% (3) More from: BSC MathematicsMAM036HMahatma Gandhi University624 documents Go to course 2 Model Exam: Foundation of Mathematics MM1CRT01 First Semester 2022 BSC Mathematics 100% (6) 1 1st Sem MM1CRT01 Foundation of Mathematics - Internal Assessment Notes BSC Mathematics 100% (4) 3 First Semester MM1CRT01 Foundation of Mathematics Question Paper - July 2022 BSC Mathematics 100% (3) 3 ST3CMT03 Probability Distributions Final Exam January 2023 BSC Mathematics 100% (3) 18 Phy- Motion Under Gravity: Historical Perspectives and Principles BSC Mathematics 100% (3) 2 First Order Differential Equations Flowchart Notes MATH 2420 BSC Mathematics 100% (3) Recommended for you 33 MATH Practice QUIZ AND Answers BSC Mathematics Practice materials 100% (3) 31 A quick review of vector algebra from vector calculus BSC Mathematics Practice materials 100% (1) 6 Flywheel- moment of inertia BSC Mathematics Practice materials 100% (1) 7 Torsion pendulum - It's practical notes BSC Mathematics Practice materials 100% (1) 33 MATH Practice QUIZ AND Answers BSC Mathematics 100% (3) 31 A quick review of vector algebra from vector calculus BSC Mathematics 100% (1) 6 Flywheel- moment of inertia BSC Mathematics 100% (1) 7 Torsion pendulum - It's practical notes BSC Mathematics 100% (1) 23 Mathongo.com- Ncert-Solutions-Class-12-Maths-Chapter-2-Inverse-Trigonometric-Functions BSC Mathematics 100% (1) 2 Speed distance and time in maths BSC Mathematics 100% (1) Students also viewed Mathongo - Algebra (969) maths gyaan sutra jee main Maths combined question and answers Basic introduction to real Analysis previous year question papers previous year question paper 2021 Related documents Techniques of Counting Matrices assistant 1 Partial Differential Equations Linear algebra Previous year question 3 Mathongo - Differentiability Get homework AI help with the Studocu App Open the App English India Company About us Studocu Premium Academic Integrity Jobs Blog Dutch Website Study Tools All Tools Ask AI AI Notes AI Quiz Generator Notes to Quiz Videos Notes to Audio Infographic Generator Contact & Help F.A.Q. 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15745	https://artofproblemsolving.com/wiki/index.php/Telescoping_series?srsltid=AfmBOoqEUky15psIteF5m5nYUTSC7yHG5nQNOYCb0VRcBoQ2r7ZltZQx	Art of Problem Solving Telescoping series - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Telescoping series Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Telescoping series In mathematics, a telescoping series is a series whose partial sums eventually only have a finite number of terms after cancellation. This is often done by using a form of for some expression . Contents 1 Example 1 2 Solution 1 3 Example 2 4 Solution 2 5 Problems 5.1 Introductory 5.2 Intermediate 5.3 Olympiad 6 See Also Example 1 Derive the formula for the sum of the first counting numbers. Solution 1 We wish to write for some expression . This expression is as . We then telescope the expression: . (Notice how the sum telescopes— contains a positive and a negative of every value of from to , so those terms cancel. We are then left with , the only terms which did not cancel.) Example 2 Find a general formula for , where . Solution 2 We wish to write for some expression . This can be easily achieved with as by simple computation. We then telescope the expression: . Problems Introductory When simplified the product becomes: (Source) The sum can be expressed as , where and are positive integers. What is ? (Source) Which of the following is equivalent to (Hint: difference of squares!) (Source) Intermediate Let denote the value of the sum can be expressed as , where and are positive integers and is not divisible by the square of any prime. Determine . (Source) Olympiad Find the value of , where is the Riemann zeta function See Also Algebra Summation Retrieved from " Category: Algebra Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
15746	https://www.sparknotes.com/math/geometry3/geometricproofs/problems_3/	My PLUS My PLUS Dashboard Please wait while we process your payment Reset Password Your password reset email should arrive shortly. If you don't see it, please check your spam folder. Sometimes it can end up there. Something went wrong If you don't see it, please check your spam folder. Sometimes it can end up there. Please wait while we process your payment Log in or Create account to start your free trial of SparkNotes Plus. By signing up you agree to our terms and privacy policy. Don’t have an account? Subscribe now Create Your Account Sign up for your FREE 7-day trial Ad-free experience Note-taking Flashcards & Quizzes AP® English Test Prep Plus much more By signing up you agree to our terms and privacy policy. Already have an account? Log in Your Email Choose Your Plan Individual Group Discount BEST VALUE Save over 50% with a SparkNotes PLUS Annual Plan! 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Ad-Free experience Easy-to-access study notes Flashcards & Quizzes AP® English test prep Plus much more Already have an account? Log in Topics Geometric Proofs Problems 3 Problems 3 Topics Geometric Proofs Problems 3 Prove the following statements indirectly. Problem : Given: Triangle ABC is an isosceles triangle with base AC.Segment BE is a median. Prove: Segment BE is an angle bisector. Problem : Given: Circle F is inscribed in triangle ABC. Prove: Angle DCF is congruent to angle ECF. Problem : Use indirect reasoning to explain why a triangle cannot have more than one obtuse angle. First, assume that a triangle does have more than one obtuse angle. The measure of an obtuse angle is greater than 90 degrees. Hence, the sum of the measures of two obtuse angles is greater than 180 degrees, and the sum of the measures of three obtuse angles is greater that 270 degrees. The sum of the angles of a triangle, however, equals 180 degrees. Therefore, only one angle in a triangle can be obtuse. 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15747	https://math.stackexchange.com/questions/3106595/important-olympiad-inequalities	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Important Olympiad-inequalities [duplicate] Ask Question Asked Modified 6 years, 7 months ago Viewed 6k times 17 $\begingroup$ As an olympiad-participant, I've had to solve numerous inequalities; some easy ones and some very difficult ones. Inequalities might appear in every Olympiad discipline (Number theory, Algebra, Geometry and Combinatorics) and usually require previous manipulations, which makes them even harder to solve... Some time ago, someone told me that Solving inequalities is kind of applying the same hundred tricks again and again And in fact, knowledge and experience play a fundamental role when it comes to proving/solving inequalities, rather than instinct. This is the reason why I wanted to gather the most important Olympiad-inequalities such as AM-GM (and the weighted one) Cauchy-Schwarz Jensen ... Could you suggest some more? This question was inspired by the fantastic contributions of @Michael Rozenberg on inequalities. inequality contest-math big-list Share edited Feb 10, 2019 at 2:39 Alexander Gruber♦ 28.2k3232 gold badges131131 silver badges220220 bronze badges asked Feb 9, 2019 at 19:24 Dr. MathvaDr. Mathva 9,03322 gold badges1818 silver badges4848 bronze badges $\endgroup$ 4 $\begingroup$ @Michael Rozenberg, I know that the question you've linked might look similar to mine; however, I wanted to emphasize the fact that I'm looking for $\mathbf{olympiad}$ inequalities, which has nothing to do with the inequalities you might require for the maths-degree for instance... $\endgroup$ Dr. Mathva – Dr. Mathva 2019-02-10 11:39:41 +00:00 Commented Feb 10, 2019 at 11:39 $\begingroup$ All these they are Olimpiad inequalities. I think these themes they are same. Remember, there is also IMC. $\endgroup$ Michael Rozenberg – Michael Rozenberg 2019-02-10 12:42:35 +00:00 Commented Feb 10, 2019 at 12:42 $\begingroup$ @Michael Rozenberg What do you mean by IMC? $\endgroup$ Dr. Mathva – Dr. Mathva 2019-02-10 14:22:02 +00:00 Commented Feb 10, 2019 at 14:22 $\begingroup$ See here: imc-math.org.uk $\endgroup$ Michael Rozenberg – Michael Rozenberg 2019-02-10 14:25:54 +00:00 Commented Feb 10, 2019 at 14:25 Add a comment \| 2 Answers 2 Reset to default 17 $\begingroup$ Essential reading: Olympiad Inequalities, Thomas J. Mildorf All useful inequalities are clearly listed and explaind on the first few pages. Mildorf calls them "The Standard Dozen": EDIT: If you look for a good book, here is my favorite one: The book covers in extensive detail the following topics: Also a fine reading: A Brief Introduction to Olympiad Inequalities, Evan Chen Share edited Feb 10, 2019 at 19:31 answered Feb 9, 2019 at 19:48 OldboyOldboy 17.3k11 gold badge2525 silver badges6565 bronze badges $\endgroup$ 1 5 $\begingroup$ While the reference you provide might be useful, displaying text-as-images makes your answer inaccessible to screenreaders, and hampers the searchability of your answer. It would be preferable if you could summarize (in text) the major important results. $\endgroup$ Xander Henderson – Xander Henderson ♦ 2019-02-11 18:00:32 +00:00 Commented Feb 11, 2019 at 18:00 Add a comment \| 9 $\begingroup$ I did not find a link, but I wrote about this theme already. I'll write something again. There are many methods: Cauchy-Schwarz (C-S) AM-GM Holder Jensen Minkowski Maclaurin Rearrangement Chebyshov Muirhead Karamata Lagrange multipliers Buffalo Way (BW) Contradiction Tangent Line method Schur Sum Of Squares (SOS) Schur-SOS method (S-S) Bernoulli Bacteria RCF, LCF, HCF (with half convex, half concave functions) by V.Cirtoaje E-V Method by V.Cirtoaje uvw Inequalities like Schur pRr method for the geometric inequalities and more. In my opinion, the best book it's the inequalities forum in the AoPS: Just read it! Also, there is the last book by Vasile Cirtoaje (2018) and his papers. An example for using pRr. Let $a$, $b$ and $c$ be sides-lengths of a triangle. Prove that: $$a^3+b^3+c^3-a^2b-a^2c-b^2a-b^2c-c^2a-c^2b+3abc\geq0.$$ Proof: It's $$R\geq2r,$$ which is obvious. Actually, the inequality $$\sum_{cyc}(a^3-a^2b-a^2c+abc)\geq0$$ is true for all non-negatives $a$, $b$ and $c$ and named as the Schur's inequality. Share edited Feb 10, 2019 at 12:51 answered Feb 9, 2019 at 20:50 Michael RozenbergMichael Rozenberg 208k3131 gold badges171171 silver badges294294 bronze badges $\endgroup$ 6 $\begingroup$ Similar to this, which was also written by OP $\endgroup$ user574848 – user574848 2019-02-10 01:54:55 +00:00 Commented Feb 10, 2019 at 1:54 $\begingroup$ This is the link. Thank you! $\endgroup$ Michael Rozenberg – Michael Rozenberg 2019-02-10 03:17:33 +00:00 Commented Feb 10, 2019 at 3:17 $\begingroup$ Not that it's very important, but you're missing a dot after 16 which has kind of given a bad look to the list. :P Also, what is the pRr method? I googled it but ended up with results in biology which I don't think are very relevant. It's hard to find relevant results about some of the acronyms you used on Google. Don't even get me started on "Bacteria". :P $\endgroup$ stressed out – stressed out 2019-02-10 12:16:08 +00:00 Commented Feb 10, 2019 at 12:16 $\begingroup$ @stressed out I added something. See now. About Bacteria see here: math.stackexchange.com/questions/2903914 $\endgroup$ Michael Rozenberg – Michael Rozenberg 2019-02-10 12:52:17 +00:00 Commented Feb 10, 2019 at 12:52 1 $\begingroup$ @stressed out Yes, of course! But it's a semi-perimeter. Sometimes it's very useful. $\endgroup$ Michael Rozenberg – Michael Rozenberg 2019-02-10 12:58:40 +00:00 Commented Feb 10, 2019 at 12:58 \| Show 1 more comment Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions inequality contest-math big-list See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Linked What are the most popular techniques of proving inequalities? Which things can be used directly in Math Olympiads without needing to be proved? 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15748	https://www.wyzant.com/resources/answers/745649/given-tan-x-4-3-and-sin-x-is-positive-find-the-sin-2x	Given tan(x)=4/3 and sin(x) is positive, find the sin(2x) \| Wyzant Ask An Expert Log inSign up Find A Tutor Search For Tutors Request A Tutor Online Tutoring How It Works For Students FAQ What Customers Say Resources Ask An Expert Search Questions Ask a Question Wyzant Blog Start Tutoring Apply Now About Tutors Jobs Find Tutoring Jobs How It Works For Tutors FAQ About Us About Us Careers Contact Us All Questions Search for a Question Find an Online Tutor Now Ask a Question for Free Login WYZANT TUTORING Log in Sign up Find A Tutor Search For Tutors Request A Tutor Online Tutoring How It Works For Students FAQ What Customers Say Resources Ask An Expert Search Questions Ask a Question Wyzant Blog Start Tutoring Apply Now About Tutors Jobs Find Tutoring Jobs How It Works For Tutors FAQ About Us About Us Careers Contact Us Subject ZIP Search SearchFind an Online Tutor NowAsk Ask a Question For Free Login TrigonometrySum Identities Kimmale A. asked • 03/13/20 Given tan(x)=4/3 and sin(x) is positive, find the sin(2x) Follow •1 Add comment More Report 1 Expert Answer Best Newest Oldest By: JACQUES D.answered • 03/14/20 Tutor 4.9(257) Physics, Chemistry, Math - No problem! About this tutor› About this tutor› Make a triangle with an opposite side of 4 and adjacent side of 3 (hypotenuse is 5). You can now work out all trig functions for that angle. including sin(x) Use the double angle formula to find sin(2x) Upvote • 0Downvote Add comment More Report Still looking for help? Get the right answer, fast. Ask a question for free Get a free answer to a quick problem. Most questions answered within 4 hours. OR Find an Online Tutor Now Choose an expert and meet online. 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15749	https://www.academia.edu/50914657/Organic_Chemistry_Morrison_and_Boyd	Academia.edu no longer supports Internet Explorer. To browse Academia.edu and the wider internet faster and more securely, please take a few seconds to upgrade your browser. Log In Sign Up About Press Papers Terms Privacy Copyright We're Hiring! Help Center Outline Title Organic Chemistry- Morrison & Boyd Rafaela Pere Sign up for access to the world's latest research checkGet notified about relevant papers checkSave papers to use in your research checkJoin the discussion with peers checkTrack your impact Loading Preview Sorry, preview is currently unavailable. You can download the paper by clicking the button above. Related papers LECTURE Irene Olala downloadDownload free PDFView PDFchevron_right Organic chemistry Stevens Morlu The success achieved by this book's forerunners, Basic Principles of Organic Chemistry and Modern Organic Chemistry, was to a considerable extent due to the rigor with which the subject of organic chemistry was presented. In the present work we have tried to paint an interesting, relevant, and up-to-date picture of organic chemistry while retaining the rigorous approach of the earlier books. Organic chemistry sometimes appears to be enormously complex to the beginning student, particularly if he must immediately grapple with the subjects of structural isomerism and nomenclature. We have attempted to avoid this difficulty in the following way. Chapter 1 briefly relates carbon to its neighbors in the Periodic Table and reviews some fundamental concepts. Chapter 2 deals with the four C, and C , hydrocarbons-methane, ethane, ethene, and ethyne-and discusses their conformational and configurational properties and some of their chemical reactions. The reader thus makes an acquaintance with the properties of someimportant organiccompoundsbefore dealing in an open-ended way with families of compounds-alkanes, alcohols, etc. A heavy emphasis on spectroscopy is retained but the subject is introduced somewhat later than in the earlier books. Important additions are chapters dealing with enzymic processes and metabolism and with cyclization reactions. Many of the exercises of the earlier books have been retained and have been supplemented with drill-type problems. It seems a shame to burden the mind of the beginning student with trivial names, some of them quite illogical, and throughout we have stressed IUPAC nomenclature, which is both logical and easy to learn. The instructor, who may well carry lightly the excess baggage of redundant names, may occasionally find this irritating but we ask him to consider the larger good. As a further aid to the student, each chapter concludes with a summary of important points. The simple introduction to the subject and the emphasis on relevance, particularly to living systems, should make the book appealing to the general student. At the same time we hope that the up-to-date and more advanced topics that are included-the effect of orbital symmetry on cyclization reactions, for example-will also appeal to the chemistry specialist. We should like to acknowledge the help of many persons who read all or parts of the manuscript and offered sound advice. Professor George E. Hall read the manuscript at several stages of revision and we are particularly downloadDownload free PDFView PDFchevron_right ADVICES FOR STUDYING ORGANIC CHEMISTRY Sabrina Islam downloadDownload free PDFView PDFchevron_right Principles of organic chemistry Chris Gonzales downloadDownload free PDFView PDFchevron_right Explore Papers Topics Features Mentions Analytics PDF Packages Advanced Search Search Alerts Journals Academia.edu Journals My submissions Reviewer Hub Why publish with us Testimonials Company About Careers Press Help Center Terms Privacy Copyright Content Policy 580 California St., Suite 400 San Francisco, CA, 94104 © 2025 Academia. All rights reserved
15750	https://www.mytutor.co.uk/answers/55621/A-Level/Maths/Integrate-4-x-2/	Integrate 4/x^2 \| MyTutor Find a tutorHow it worksPricesResourcesFor schoolsBecome a tutor +44 (0) 203 773 6024 Log inSign up Answers>Maths>A Level>Article Integrate 4/x^2 First of all, I would ask the tutee if they were familiar with the law of integration. i.e. that you increase the power of the variable (x) by 1, then divide by the new power, and if not then I would clarify this with them.Then, we should manipulate the fraction into a format that we know how to integrate more easily.We know that we can rewrite 1/x^2 as x^-2Hence we are simply multiplying this by 4 i.e. we can see this as 4 1/x^2Finally, we apply the law of integration to 4x^-2 by adding 1 to the power, then dividing by the new powerSo the answer would be -4x^-1 AE Answered by Ahmed E. •Maths tutor 4533 Views See similar Maths A Level tutors Need help with Maths? One-to-one online tuition can be a great way to brush up on your Maths knowledge. Have a Free Meeting with one of our hand picked tutors from the UK's top universities Find a tutor Download MyTutor's free revision handbook? This handbook will help you plan your study time, beat procrastination, memorise the info and get your notes in order. 8 study hacks, 3 revision templates, 6 revision techniques, 10 exam and self-care tips. Download handbook Free weekly group tutorials Join MyTutor Squads for free (and fun) help with Maths, Coding & Study Skills. Sign up Related Maths A Level answers All answers ▸ #### The parametric equations of a curve are: x = cos2θ y = sinθcosθ. Find the cartesian form of the equation. Answered by Amelia N. #### FP2 (old specification) - How do you find the derivative of arsinhx? Answered by Lois M. #### Show that: [sin(2a)] / [1+cos(2a)] = tan(a) Answered by George H. #### Evaluate the indefinite integral: ∫ (e^x)sin(x) dx Answered by ShenZhen N. We're here to help Contact us+44 (0) 203 773 6020 Company Information CareersBlogSubject answersBecome a tutorSchoolsSafeguarding policyFAQsUsing the Online Lesson SpaceTestimonials & pressSitemapTerms & ConditionsPrivacy Policy Popular Requests Maths tutorChemistry tutorPhysics tutorBiology tutorEnglish tutorGCSE tutorsA level tutorsIB tutorsPhysics & Maths tutorsChemistry & Maths tutorsGCSE Maths tutors ##### CLICK CEOP Internet Safety Payment Security Cyber Essentials MyTutor is part of the IXL family of brands: ##### IXL Comprehensive K-12 personalized learning ##### Rosetta Stone Immersive learning for 25 languages ##### Wyzant Trusted tutors for 300 subjects ##### Education.com 35,000 worksheets, games, and lesson plans ##### Vocabulary.com Adaptive learning for English vocabulary ##### Emmersion Fast and accurate language certification ##### Thesaurus.com Essential reference for synonyms and antonyms ##### Dictionary.com Comprehensive resource for word definitions and usage ##### SpanishDictionary.com Spanish-English dictionary, translator, and learning resources ##### FrenchDictionary.com French-English dictionary, translator, and learning ##### Ingles.com Diccionario ingles-espanol, traductor y sitio de apremdizaje ##### ABCya Fun educational games for kids © 2025 by IXL Learning
15751	https://www.unsiap.or.jp/sites/default/files/pdf/e-learning_el_material_5_agri_rap2ag_02_04_l2.1_samplingdesigns.pdf	Basic Concepts of Sampling- Brief Review: Sampling Designs Dr. A.C. Kulshreshtha U.N. Statistical Institute for Asia and the Pacific (SIAP) Second RAP Regional Workshop on Building Training Resources for Improving Agricultural & Rural Statistics Sampling Methods for Agricultural Statistics-Review of Current Practices SCI, Tehran, Islamic Republic of Iran 10-17 September 2013 1. Determining survey objectives and data requirements 2. The population of interest or the target population 3. Reference period; Geographic and demographic boundaries 4. Sampling frame and sampling unit 5. Sample design 6. Selection of the sample (at different stages) 7. Survey management and field procedures 8. Data collection 9. Summary and analysis of the data 10. Dissemination 2 Survey Design – Issues involved Sept.2013 Sample is a subset of the population on which observations are taken for obtaining information about the population. Since studying a sample we wish to draw valid conclusions about the population, sample should desirably be ‘representative’ of the target population. Sampling design specifies how to select the part of the population to be surveyed 3 Introduction to Survey Design Sept.2013 The most common sampling techniques used for official surveys are : • Simple Random Sampling • Systematic Sampling • Stratified Sampling • Probability Proportional to Size (PPS) sampling • Cluster Sampling • Multi-Stage Sampling All are examples of probability sampling 4 Probability Sampling Designs Sept.2013 Basic Sampling Schemes Simple random sampling (SRS): is a probability selection scheme where each unit in the population is given an equal probability of selection Systematic sampling: A method in which the sample is obtained by selecting every kth element of the population, where k is an integer > 1. Often the units are ordered with respect to that auxiliary data Stratified sampling: Uses auxiliary information (stratification variables) to divide the sampling units of the population into groups called ‘strata’ and increase the efficiency of a sample design 5 Probability Sampling Designs (Contd.) Sept.2013 Basic Sampling Schemes (Contd.) Probability Proportional to Size (PPS): The procedure of sampling in which the units are selected with probability proportional to a given measure of size The size measure is the value of an auxiliary variable X related to the characteristic Y under study 6 Probability Sampling Designs (Contd.) Sept.2013 • SRS is simplest method of probability sampling • SRS is special type of equal probability selection method (epsem) • Rarely used in practice for large scale surveys but is theoretical basis for other sample designs • SRS selection can be made – With Replacement (SRSWR) or – Without Replacement (SRSWOR) • Selection probability: the probability that a population unit is selected at any given draw is the same, namely for both SRSWR and SRSWOR N: number of units in the population (Population size) 7 Simple Random Sampling (SRS) Sept.2013 N 1 What is SRS ? Selection Procedure- Steps involved:  Get a list (sampling frame) which uniquely identifies each unit in the population  Allocate a serial number to each unit of the frame  Generate random numbers [in the range of 1 to N] using Random Number Table/ Random Number Generator on computer:  For SRSWR: select the units with the serial numbers same as the first n random numbers generated, even if there be repetitions  For SRSWOR: select the units with the serial numbers same as the first n distinct random numbers generated 8 SRS (Contd.) Sept.2013 9 Estimators of Mean, Variance, Standard Error, CV in SRS Estimator of variance of sample mean Where, sample mean , and estimated sample variance is Estimated standard error of will be Estimated CV of sample mean Estimator of variance of proportion p: ∑ = = n i i n y y 1 1 ) ( 1 2 2 − − = ∑ = n y y s n i i y n s s y y 2 2 = 1 ) 1 ( ) 1 ( ) ( − − − = n p p f p v 100 ) 1 ( ) ( ×         − = ∧ y s n f y CV In SRSWR and in SRSWOR ( ) n s f y v 2 1 ) ( − = y ) (y v SRS (Contd.) • Systematic Sampling (SYS), like SRS, involves selecting n sample units from a population of N units • Instead of randomly choosing the n units in the sample, a skip pattern is run through a list (frame) of the N units to select the sample • The skip or sampling interval, k = N/n 10 Systematic Sampling Sept.2013 Linear Systematic Sampling k ..... N r 1 1 2 3 n-2 n-1 k Random start Sampling interval, k 11 Systematic Sampling (Contd.) Sept.2013 Steps involved: • Form a sequential list of population units • Decide on a sample size n and compute the skip (sampling interval), k = N/n • Choose a random number, r (random start) between 1 and k (inclusive) • Add “k” to selected random number to select the second unit and continue to add “k” repeatedly to previously selected unit number to select the remainder of the sample 12 Selection Procedure - Linear Systematic Sampling Systematic Sampling (Contd.) Sept.2013 If N is a multiple of n, then the number of units in each of the k possible systematic samples is n In this case systematic sampling amounts to grouping the N units into k samples of exactly n units each in a systematic manner and selecting one of them with probability 1/k In this case, the sampling scheme is epsem But, if N/n is not an integer, then the number of units selected systematically with the sampling interval k [ = nearest integer to N/n] – no longer epsem This problem may be overcome by adopting a device, known as circular systematic sampling 13 Problem - Linear Systematic Sampling Systematic Sampling (Contd.) Sept.2013 Circular Systematic Sampling K=5/2=2.5 a) If k=2 possible samples are: ac; bd; ce; da and eb b) If k=3 possible samples are: ad; be; ca; db and ec 14 Systematic Sampling (Contd.) Sept.2013 Circular Systematic selection • Useful when N/n is not integer • Determine the interval k – rounding down to the integer nearest to N/n [If N = 15 and n = 4, then k is taken as 3 and not 4] • Take a random start between 1 and N • Skip through the circle by k units each time to select the next unit until n units are selected • Thus there could be N possible distinct samples instead of k • This method is termed Circular Systematic Sampling (CSS) 15 Systematic Sampling (Contd.) Sept.2013 Systematic Sampling – Important Features • Often used as an alternative to SRS • Requires ordering of the population units – Ordering enables SYS sample to be more representative – Ordering done by geographical location (say of dwellings) ensures fair spread of sample – Ordering done by industry type ensures fair representation of industries • Ensures each population unit equal chance of being selected into sample 16 Systematic Sampling (Contd.) Sept.2013 Advantages and Disadvantages Advantages: • Operationally convenient - easier to draw a sample. • SYS distributes the sample more evenly over the population – thus likely to be more efficient than SRSWOR, particularly when the ordering of the units in the list is related to characteristics of the variable of interest Disadvantages : • Requires complete list of the population • A bad arrangement of the units may produce a very inefficient sample • Variance estimates cannot be obtained from a single systematic sample 17 Systematic Sampling (Contd.) Sept.2013 18 Row Number Random start (cluster number) 1 2 …. r …. k 1 1 2 …. r …. k 2 1+k 2+k …. r+k …. 2k . . . …. . …. . . . . …. . …. . . . . …. . …. . j 1+(j-I)k 2+(j-I)k …. r+(j-I)k …. jk . . . …. . …. . . . . …. . …. . n 1+(n-I)k 2+(n-1)k …. r+(n-I)k …. nk The above table shows serial numbering of the nk population units. There are k possible samples (clusters). We select one cluster by a random start between 1 and k. Conceptual Framework (General) Systematic Sampling (Contd.) Sept.2013 19 Estimation of Variance of Systematic Sample • SYS is a random sample of one cluster only. • No estimate of variance can be formed from the sample • Approximately treating the SYS as a random sample of n units, estimate of variance would be • where is the mean square within the selected cluster (systematic sample) among units • This is, however, not an unbiased estimator swc 2 swc sy nk n y v 2 1 1 ) (       − = Systematic Sampling (Contd.) Sept.2013 • Design Variance of Sample Mean in Systematic Sampling is where denotes the mean square between the clusters • Variance of systematic sample mean is the between sample (cluster) means variance • Correlation coefficient between pairs of sampling units in a ‘cluster’ of study variable Y is, • Sampling variance of the systematic sample mean can also be expressed in terms of S c k r r b sys k k Y y k y V 2 2 1 2 1 ) ( 1 ) ( − = − = = ∑ = σ S c 2 σ σ σ 2 2 2 w b − = 2 1 1 1 2 1 1 2 2 ) ( 1 ) ( 1 ) ( 1 r k r n j rj k r r k r n j rj y y nk Y y k Y y nk − + − = − = ∑∑ ∑ ∑∑ = = = = = σ 2 1 1 ) 1 ( ) )( ( σ ρ − − − = ∑∑ = = ′ ≠ ′ n kn Y y Y y k r n j j j r rj c [ ] ρ σ σ c b n n ) 1 ( 1 2 2 − + = 2 b σ ρ c Sept.2013 20 Systematic Sampling (Contd.) Design Variance of Systematic Sample 21 • Systematic sampling compared to simple random sampling with replacement is – More efficient, if – Equally efficient, if ρ = 0 – Less efficient, if 0 < ρ < 1 ( ) ( ) 1 ( 1) ( ) = = + − ρ sys sys srswr V y DEFF y n V y 1 0 1 − < ρ < − n Sept.2013 Systematic Sampling (Contd.) Design Efficiency (DEFF) • Divides the population into a number of distinct groups (strata) based on auxiliary information - referred to as stratification variables - relating to study variable(s) • Each stratum is composed of units that satisfy the condition set by the values of the stratifying variable • Main objectives: – Improve the sample estimations, i.e. to reduce the standard error of the estimates or higher efficiency for given per unit of cost – Provide separate estimates required for each sub-division of the population – “domain” estimates – Using different sampling procedures for different sub-population, to (i) increase efficiency of the estimates (ii) organize the field work 22 Stratified Sampling- Stratification Sept.2013 Stratification – Mutually Exclusive subsets X X X X X X X 1 2 h L NH Nh N2 N1 Stratum no. Stratum size 23 Stratified Sampling (Contd.) Sept.2013 Stratified sampling involves: • division or stratification of the population into homogeneous (similar) groups called strata • selecting the sample using a selection procedure – like SRS or systematic sampling or PPS within each stratum – independent of the other strata • Sampling in each stratum is carried out independently – Sampling fractions may differ – Selection procedures may also be different • The total sample size is distributed over all the strata – allocation • At the end of the survey, the stratum results are combined to provide an estimate for entire population 24 Stratified Sampling (Contd.) Sept.2013 Stratified Sampling – in practice • In most surveys - household or establishment surveys - stratification is used • Stratification can be used with any type of sampling design • Stratified sampling can be used in – Single - stage designs – Multi - stage designs 25 Stratified Sampling (Contd.) Sept.2013 Defining Strata 1. Choice of stratification variables (location, output, etc): – Homogeneous within strata; Heterogeneous across strata – Highly correlated with study variables (output with profit or number of employees, etc.) 2. Number of strata – Depends on availability of stratifying information in sampling frame: less information, fewer strata – At least two sampling units per stratum to be able to compute sampling error 26 Clustering and Stratification Sept.2013 X X X X X X X 1 2 h H NH Nh N2 N1 n n1 n2 nH nh • Given a total sample size, “n”, how should this be allocated among the strata? Maximize precision for fixed cost OR Minimize cost for required precision Allocation Sample over Strata Stratification - Allocation Sept.2013 27 Sample Allocation to Strata Alternatives Methods: – Proportionate allocation • Uniform or equal allocation – Disproportionate allocation • Optimum allocation (minimum variance), fixed sample size • Cost optimum allocation Stratification - Allocation Sept.2013 28 Sample Allocation to Strata • In proportionate stratification, an uniform sampling fraction is applied to each strata; that is, the sample size selected from each stratum is made proportionate to the population size of the stratum • In disproportionate stratification, different sampling rates are used deliberately in different strata Stratification - Allocation Sept.2013 29 Proportionate Allocation • In proportionate stratification, Stratification - Allocation is specified to be the same for each stratum N n N n N n h h = h h N n This implies that the overall sampling fraction is The number of elements taken from the hth stratum is ( ) N n N n h h = Sept.2013 30 Proportionate Allocation Thus, for proportionate stratified deff < 1 For a given total variability in the population, the gain is greater if: the strata mean are more heterogeneous (more unequal strata mean) OR the element values within the strata are more homogeneous Stratification - Allocation prop SRS V V ≥ Sept.2013 31 Optimum Allocation • Uses widely different sampling rates for the various strata • Objective: to achieve the least variance for the overall mean for the given sample size (Neyman’s allocation); as well as given per unit of cost in different strata • Without cost consideration, the allocation is • This gives better efficiency as compared to proportionate allocation: ∑ = h h h h h N N n n σ σ Stratification - Allocation opt prop SRS V V V ≥ ≥ Sept.2013 32 Implicit Stratification • This refers to a systematic sampling with the units arranged in a certain order • Prior to sample selection, all the units are sorted with respect to one or more variables that are deemed to have a high correlation with the variable of interest • Implicit stratification guarantees that the sample of units will be spread across the categories of the stratification variables 33 Stratified Sampling (Contd.) Sept.2013 34 Variance of Stratified Sample Mean Stratified sample mean, Variance of the estimate is, If sampling fractions fh are negligible, h h h h st n S W f y V 2 2 ) 1 ( ) ( ∑ − = h H h h st y W y ∑ = = 1 st y h h h st n S W y V 2 2 ) ( ∑ = Stratified Sampling (Contd.) Sept.2013 35 Estimated Variance of Stratified Sample Mean If a SRS is taken in each stratum, an unbiased estimate of is , Estimated variance of the estimate is, If sampling fractions fh are negligible, h h h h st n s W f y v 2 2 ) 1 ( ) ( ∑ − = st y 2 h S 2 h s 2 1 2 ) ( ) 1 ( 1 h n i hi h h y y n s h − − = ∑ = h h h H h st n s W y v 2 2 1 ) ( ∑ = = Stratified Sampling (Contd.) Sept.2013 36 Precision of proportionate stratified sample over SRS For a given total variability in the population, the gain is greater if the strata mean are more heterogeneous (more unequal strata mean) or equivalently the more homogeneous are the element values within the strata. The variance of a stratified sample is given by The variance for stratified sample under optimum allocation h h h n h h H h h h h h st n S W f n S f W y V h 2 2 1 1 2 2 ) 1 ( ) 1 ( ) ( ∑ ∑ = = − = − = 2 2 2 2 2 2 1 ) ( 1 1 ) ( 1 h h h h h h h h opt S N N S N nN S W N S W n V ∑ ∑ ∑ − Σ = − = opt prop ran V V V ≥ ≥ Stratified Sampling (Contd.) Sept.2013 37 • Probability of selection is related to an auxiliary variable, Z, that is a measure of “size” • Example Information on Number of households, Area of farms • “Larger” units are given higher chance of selection than “smaller” units • Selection probability of ith unit is , i = 1,2, … , N Selection Procedures: • Cumulative total method: with replacement • Cumulative total method: without replacement • PPS systematic sampling • Lahiri’s method Probability Proportional to Size (PPS) Sampling ∑ = = N i i i i Z Z p 1 Sept.2013 38 Cumulative Total Method • Sampling unit: village • Measure of size: number of households in village • Selection probability: Select a sample of 5 villages using varying probability WR sampling, the size being the number of households Solution PPS Sampling (Contd.)……PPS Sample Selection pi Sept.2013 • Write down cumulative total for the sizes Zi, i=1,2..N • Choose a random number r such that 1 ≤ r ≤ Z • Select ith population unit if • Ti-1 ≤ r ≤ Ti where Ti-1 = Z1 + Z2 + .. + Zi-1 and Ti = Z1 + Z2 + .. + Zi Cumulative Total Method (Contd.) PPS Sampling (Contd.)……PPS Sample Selection Sept.2013 39 40 • To select a village, a random number r, 1 ≤ r ≤ 700, is selected • Suppose r = 259, Since 231 ≤ 259 ≤ 288, the 7th village is therefore selected. The next 4 random numbers to be considered are 548, 170, 231, 505 Hence the required sample selected using PPS with replacement are 16th, 5th, 7th, 15th Note: The 7th village is selected twice Cumulative Total Method (Contd.) PPS Sampling (Contd.)……PPS Sample Selection Sept.2013 41 • For a PPSWR selection therefore the sample would be: 16th, 5th, 7th, 15th , with 7th village repeated. • For a PPSWOR selection, we have to continue further to get 5 distinct units in the sample. • Suppose the next random selected is r = 375, The required PPSWOR sample would be 16th, 5th, 7th, 15th & 11th . Cumulative Total Method (Contd.) PPS Sampling (Contd.)……PPS Sample Selection Sept.2013 42 • Derive cumulative totals for the sizes Zi, i=1,2..N, and allot random numbers to different units. • Calculate interval k = ZN /n (in this case 700/5 = 140) • Select a random number r (say 101) from 1 to k; and obtain r+k, r+2k, r+3k, …, r+(n-1)k • In this case, the selected cumulative sizes are 101, 241, 382, 523 & 664. PPP Systematic PPS Sampling (Contd.)……PPS Sample Selection Sept.2013 43 • Thus the selected units are: 3rd (for 101), 7th (for 241), 11th (for 382), 15th (for 523) & 20th (for 664) • Note: If any unit has size greater than k, it may be selected more than once. PPS Systematic (Contd.) PPS Sampling (Contd.)……PPS Sample Selection Sept.2013 44 Lahiri’s Method • A procedure which avoids the need of calculating cumulative totals for each unit has been given by Lahiri (1951) Steps involved: 1. Select a random number i from 1 to N 2. Select another random number j, such that 1 ≤ j ≤ M, where M is either equal to the maximum of sizes Zi, i =1,2,.. N, or is more than the maximum size in the population. 3. If j ≤ Zi , the ith unit is selected, otherwise, the pair (i, j) of random numbers is rejected and another pair is chosen by repeating the steps (1) and (2) PPS Sampling (Contd.)……PPS Sample Selection Sept.2013 45 Select a sample of 2 villages using varying probability WR sampling, the size being the number of households Solution • N =20, n=2 , M =58 • Select a random number i, 1 ≤ i ≤ 14, • Then a second random number j, 1 ≤ j ≤ 58, • Suppose the 1st pair of random number is (2, 30). Since 30 ≤ 45 thus 2nd village is selected . PPS Sampling (Contd.)……PPS Sample Selection Lahiri’s Method Sept.2013 46 Solution (continued) • Similarly we find the next pair of random number (12, 47) since 47 >30, the 12th village is not selected The 3rd pair ot random numbers (7, 40) results in the selection of 7th village since 40 ≤ 58 • Hence, the selected sample are 2nd and 7th villages PPS Sampling (Contd.)……PPS Sample Selection Lahiri’s Method Sept.2013 47 N i X X p N i i i i , , 1 , 1  = = ∑ = i.e. Sampling is done with a varying probability In epsem the probability of ith unit being in the sample is:            = =      − − = M m p N n p N p i i n i cluster, Simple SRSWOR, 1 1 1 SRSWR, fixed probabilities But in PPS, pi depends on Xi , size of the unit which is variable The larger the Xi , the more probable ith unit to be chosen PPS Sampling (Contd.)……Probability of unit in the sample Sept.2013 48 • If values were known before sampling and sampling is carried out with probability proportional to , • If instead of drawing a unit with probability proportional to its actual value, we draw with probability proportional to an auxiliary variable whose size ( ) is related to by relation, where k is positive constant, the probability of selection remains the same, and would give the same results as pps of y ∑ = = N i i i i y y p 1 i y i x i y i y i i ky x = i N i i N i i i N i i i i ppy y y y k ky x x ppx i = = = = ∑ ∑ ∑ = = = 1 1 1 PPS Sampling (Contd.)……Probability of selection unit in sample Sept.2013 49 In SRS from a population of size N , population total Y • Probability of selection of a unit at any draw is 1/N • Unbiased estimator of Y from ith unit is N. yi Or, • Similarly, the unbiased estimator of the population total Y in varying probability sampling from the ith draw is , where pi (which varies from unit to unit in the universe) is probability of selection of yi N y y N i i Y 1 . = = ∧ i i pps p y Y = ∧ PPS Sampling (Contd.)……Estimator of Y from ith unit Sept.2013 50 • If we draw a sample of n units with replacement out of N units, with the initial probability of selection of the ith unit as pi ,the combined unbiased estimator of Y is • The sampling variance of this estimator is • Estimated variance of is v( ) = = pps Y ∧ pps Y ∧ 2 1 ) ( ) 1 ( 1 pps n i i i Y p y n n ∧ = − −∑ 2 1 2 ) 1 ( 1 PPS n i i i Y n p y n n       −         − ∧ = ∑         − =         − = ∑ ∑ = = N i i i i N i i i pps y Y p Y n p Y p Y n 1 2 2 2 1 2 1 1 σ ∑ = = n i i i pps p y n Y 1 1 ˆ PPS Sampling (Contd.)……Estimated variance of estimator of Y Sept.2013 51 estimator Unbiased ; 1 ˆ ) ( ˆ 1 ∑ = = = n i i i PPS PPS p y n Y Y T [ ] lue unknown va an ) ( 1 ) ( ˆ var 2 1 =       − = ∑ = j N j j j PPS p Y T p Y n Y T ( ): ) ( of Estimator Unbiased PPS Y T Var [ ] ( ) ( ) ( ) ( ) PPS n i PPS i i n i PPS i i Y T Y T n p y n n Y T p y n n ˆ var ˆ ) 1 ( 1 ˆ ) 1 ( 1 (Y) T ˆ var 1 2 2 2 1 pps ∧ = = =         −         − =         − − = ∑ ∑ Sept.2013 PPS Sampling (Contd.)……Estimation in PPSWR Sampling 52 Therefore, unbiased estimator of population mean is Y ( )PPS PPS Y T N y ˆ 1 = with variance ( ) [ ] ( ) 2 1 2 2 1 ˆ var 1 ) ( ∑ =         − = = N j j i PPS PPS Y T p Y nN Y T N y Var whose unbiased estimator is ( ) ( ) ( ) ( ) ( ) 2 1 2 2 ˆ 1 1 var 1 var ∑ =         − − = = n i PPS i i PPS PPS Y T p y N n n Y T N y Sept.2013 PPS Sampling (Contd.)…Estimated variance of estimator of Y (Contd.) 53 Strata and Clusters • Both stratification and clustering involve subdividing the population into mutually exclusive groups • Sub-divisions of the population are called ‘clusters’ or ‘strata’ depending upon the sampling procedure adopted • The term ‘cluster’ is used in the context of cluster sampling and multi-stage (cluster) sampling • To understand the application of these in different situations, let us take a simple example Clustering and Stratification Sept.2013 54 Naturally occurring clusters Clusters are usually defined as groups of units that are found naturally ‘clustered’ together - by location or socially defined entities like households or by institutions like schools and enterprises Cluster Population Unit Census Enumeration Area Dwelling household Person Day Hour School Student Employer Employee Clustering and Stratification Sept.2013 55 • Typically, sample surveys by NSOs involve subdividing population into strata and clusters • Technique of stratifying clusters and then further stratifying the units within clusters are applied to obtain the final sample • Sampler’s objective is to get right combination of stratification and clustering to get required estimates at desired level of accuracy with given resources • Reliability or precision of estimates depend on degree to which sample is clustered • Generally, clustering increases sampling variance considerably • Usually, stratification is applied to decrease the sampling variance, but its effect is often not significant • Effects of clustering and stratification is measured by the design effect, or deff • Primarily, deff indicates, how much clustering there is in the survey sample Clustering and Stratification in Sample Design Sept.2013 56 • Cluster sampling - selection of a sample of clusters and survey all the units of each selected clusters • This is also called ‘Single-stage cluster sampling’ • ‘Multi-stage cluster sampling’ or simply ‘multi-stage sampling’: Instead of doing survey of all the units of selected clusters, only a sample of units are taken from each selected clusters Cluster Sampling Sept.2013 57 Selecting a (single-stage) cluster sample • Require sampling frame: list of all the clusters • From the list, a sample of clusters is selected - using a selection scheme (e.g., SRS, Systematic) • All population units within the selected clusters are listed • The information is then collected from all the units of the selected clusters Cluster Sampling (Contd.) Sept.2013 58 Main advantage • Exact knowledge of the size of the sub-divisions (clusters) not required, unlike that for stratified sampling • Often a complete list of clusters - defined by location or as social entities or by institutions – is available, but frame of population units is not available or is costly to obtain • Reduced cost if personal interviews, particularly when the survey cost increases with the distance separating the sampled units Main disadvantage • Increased sampling error due to a less representative sample • in practice, units are homogeneous within normally defined clusters • composition of clusters can not be altered, as they are pre-defined Cluster Sampling- Advantages/ disadvantage Sept.2013 Clusters of equal size- Estimates • Assume a SRS of n clusters from a population of N clusters ∑ = = n 1 i i y n N Y ˆ N = # clusters in pop n = # clusters in sample M i y = # units in cluster = ith cluster total Total ∑ = = n 1 i i y nM 1 Y ˆ ∑ = = n 1 i i y n 1 i y = ith cluster mean Mean Variance of estimators 2 b N S ) n 1 ( n 1 ) Y ˆ V( − = ∑ = − − = N 1 i i 2 1 ) y y ( N S 2 b ( ) ( ) ρ 1 1 2 − + M M S = Where is intra class correlation and is population mean square 2 S ρ Sept.2013 59 Clusters of equal size- Estimates (contd.) Estimating the variance 2 b 2 s N M N ) n 1 ( n ) Y ˆ ( V ˆ 2 − = ∑ = − − = n 1 i i 2 1 ) y ˆ y ( n 2 b s where 2 b s N) n 1 ( n 1 ) Y ˆ ( V ˆ − = 2 2 2 ) 1 ( ) 1 ( ) 1 ( b w S N M S M N S NM − + − ≡ − 2 2 2 w b S S S + ≅ 2 2 2 2 ) 1 ( ) 1 ( ) 1 ( b w w b S N M S M N NS S N M − + − − − = ρ 2 2 2 2 ) 1 ( b w w b MS S M S MS + − − ≅ Sept.2013 60 Clusters of unequal size- Estimating Assume that a SRS of n clusters from a population of N clusters is selected ∑ = = n 1 i i y n N Y ˆ N = # clusters in pop n = # clusters in sample i y = # units in ith cluster = it cluster total Estimating Total i M Sept.2013 61 Estimating the variance Total ) N n 1 ( n N ) Y ˆ ( V ˆ 2 2 b s′ − = 2 1 1 1 1 1 ∑ ∑ = =       − − = ′ n i n i i i y n y n 2 b s Where, Clusters of unequal size (contd.) Estimating Mean Total = ∑ ∑ = = = N i i i N i i y M y 1 1 Mean = ∑ ∑ = = = N i i N i i i M y M Y 1 1 ; ∑ = = n i i y n Y 1 1 1 ˆ ∑ ∑ = = = = n i i i n i i i y u n y M M n Y 1 1 2 1 1 ˆ ∑ ∑ = = = n i i n i i i M y M Y 1 1 3 ˆ M M u M N M i i N i i = = ∑ = , 1 1 ; Sept.2013 62 Clusters of unequal size (contd.) ( ) 2 1 1 Y ˆ 1 -n 1 ; ) N n 1 ( n 1 ) Y ˆ ( V ˆ ∑ − = − = i y 2 b 2 b s s ( ) 2 2 2 Y ˆ 1 -n 1 ; ) N n 1 ( n 1 ) Y ˆ ( V ˆ ∑ − = ′ ′ − = i i y u 2 b 2 b s s ( ) 2 3 2 2 3 Y ˆ 1 -n 1 ; ) N n 1 ( n 1 ) Y ˆ ( V ˆ ∑ − = ′ ′ ′ ′ − = i n i y M M 2 b 2 b s s Estimating variances Where ∑ = i n M n M 1 Sept.2013 63 64  In a single-stage cluster sampling, a sample of cluster is selected and all the population units of each selected cluster are surveyed  When clusters are to large to cover all their population units in the survey, a sample of population units from each selected cluster is surveyed  Such a design is ‘Multi-stage cluster’ / ‘Multi-stage’ sampling  Multi-stage sampling involves multiple stages of sampling  The number of stages can be numerous, although it is rare to have more than 3 stages  We will concentrate only on two-stage sampling  Process of selecting a sample of population units from selected clusters is known as Sub-sampling Multi-stage (Cluster) Sampling Sept.2013 65 Stage-wise Selection  Stage sampling is an extension of cluster sampling. For a two-stage sampling  we select the clusters at the first stage – selected clusters are called first stage units (FSUs) or primary stage units (PSUs)  then select a sample of units from within each selected cluster – selected units are called second stage units (SSUs) Multi-stage (Cluster) Sampling (Contd.) Sept.2013 66 Sampling Units at different Stages Examples of two-stage sampling Stage 1 Stage 2 villages households Dwellings People Hospitals Patients Businesses Employees Coconut trees Coconuts Multi-stage (Cluster) Sampling (Contd.) Sept.2013 67 Advantages of multistage sampling • Sampling frames normally available at higher stages, may be prepared at lower stages • Cost considerations • Flexibility in choice of sampling units and methods of selection at different stages • Contributions of different stages towards sampling variance may be estimated separately Multi-stage (Cluster) Sampling (Contd.) Sept.2013 68 Sampling at two stages • In practice, many multi-stage designs involve complex sub-sampling of and within PSUs • The selection at the two stages are done independently and may employ different sampling schemes like: – SRSWOR – Systematic – Probability Proportional to Size (PPS) Multi-stage (Cluster) Sampling (Contd.) Sept.2013 Different methods of selection Method 1 Stage Selection I II N th i i M srs srs i m n Sept.2013 69 Multi-stage (Cluster) Sampling (Contd.) ∑ ∑ = = = = n 1 i i n 1 i i M n N Y ˆ n N ˆ i y Y Estimating Total ( ) 2 1 2 2 2 1 1 1 1 ˆ i i i N i i b S M m M n N S N n N Y V         − + ′       − = ∑ = 2 1 2 2 1 2 1 1 1 1 1 1 ∑ ∑ ∑ ∑ = =         − − =       − − = ′ i M j ij i ij i i N i i i b Y M Y M S and Y N Y N S Method 1 (contd.) Sept.2013 70 Multi-stage (Cluster) Sampling (Contd.) Method 1 (contd.) Estimating Variance ( ) 2 1 2 2 2 1 1 1 1 ˆ ˆ i i i n i i b s M m M n N s N n N Y V         − + ′       − = ∑ = 2 1 2 2 1 2 1 1 1 ˆ 1 ˆ 1 1 ∑ ∑ ∑ ∑ = =         − − =       − − = ′ i m j ij i ij i i n i i i b y M y m s and Y n Y n s Sept.2013 71 Multi-stage (Cluster) Sampling (Contd.) Method 1 (contd.) Estimating Mean (equal PSU’s) ∑ = = n i i y n Y 1 1 ˆ ( ) 2 2 1 1 1 1 1 ˆ w b S M m n S N n Y V       − +       − = Where, ( ) ∑ ∑ ∑ ∑ = = = − − = =       − − = M j i ij i N i i w N i i i b Y y M S S N S and Y N Y N S 1 2 2 1 2 2 2 1 2 1 1 ; 1 1 1 1 ; Sept.2013 72 Multi-stage (Cluster) Sampling (Contd.) Estimating variance (equal PSU’s) Method 1 (contd.) ( ) 2 2 1 1 1 1 1 ˆ ˆ w b s M m N s N n Y V       − +       − = ( ) ( ) ∑ ∑ ∑ = = = − − = = − − = m j i ij i n i i w n i i b y y m s s n s and Y y n s 1 2 2 1 2 2 2 1 2 1 1 ; 1 ˆ 1 1 Sept.2013 73 Multi-stage (Cluster) Sampling (Contd.) Estimation of population mean Unequal PSU’s ∑ = = n i i i y M nM N Y 1 0 1 ˆ 2 1 2 2 2 1 1 ) 1 1 ( 1 ) 1 1 ( ) ˆ ( i i N i i b S Mi m M M nN S N n Y V − + − = ∑ = 2 1 2 2 2 1 1 ) 1 1 ( 1 ) 1 1 ( ) ˆ ( ˆ i i i n i i b s M m M M nN s N n Y V − + − = ∑ = Sept.2013 74 Multi-stage (Cluster) Sampling (Contd.) Estimation of population mean Unequal PSU’s (Contd.) ∑ = = n i i y n Y 1 2 1 ˆ i N i i Y M M M N Y B ) ( 1 ) ˆ ( 1 2 ∑ = − − = 2 1 2 2 2 ) 1 1 ( 1 ) 1 1 ( ) ( i i N i i b S M m nN S N n Y V − + − = ∑ = 2 1 2 2 2 ) 1 1 ( 1 ) 1 1 ( ) ( ˆ i i N i i b s M m nN s N n Y V − + − = ∑ = Sept.2013 75 Multi-stage (Cluster) Sampling (Contd.) Estimating Population mean unequal clusters (Contd.) ∑ ∑ = = = n i i n i i i M y M Y 1 1 3 ˆ 2 1 2 2 2 3 3 ) 1 1 ( 1 ) 1 1 ( ) ˆ ( i i i N i i b S M m M nN M S N n Y V − + − = ∑ = 2 1 2 2 2 3 3 ) 1 1 ( 1 ) 1 1 ( ) ˆ ( ˆ i i i n i i b s M m M nN M s N n Y V − + − = ∑ = Sept.2013 76 Multi-stage (Cluster) Sampling (Contd.) Different methods of selection Method 2 Stage Selection I II N th i M srs systematic m n Sept.2013 77 Multi-stage (Cluster) Sampling (Contd.) Method 2 (cont) Estimating Mean ∑ = = n i i y n Y 1 1 ˆ ( ) ( ) ( ) ∑ = − +      − +       − = N i i i b m S N M nm s N n Y V 1 2 2 1 1 1 1 1 1 1 1 ˆ ρ ( ) n s Y V b 2 ˆ ˆ ≅ Sept.2013 78 Multi-stage (Cluster) Sampling (Contd.) Different methods of selection Method 3 Stage Selection I II N th i i M ppswr srs i m n Sept.2013 79 Multi-stage (Cluster) Sampling (Contd.) Method 3 (Contd) Estimating Total i i i n i i i y M Y p Y n Y = = ∑ = ˆ ; ˆ 1 ˆ 1 ( ) ( )∑ =         − − = n i i i Y p Y n n Y V 1 2 ˆ ˆ 1 1 ˆ ˆ Sept.2013 80 Multi-stage (Cluster) Sampling (Contd.) Optimum sample sizes • Equal PSU’s • SRSWOR at both the stages • Variance function • Cost function nm S n S V w b 2 2 + ≅ 2 1 0 nmc nc C C + + = Sept.2013 81 Multi-stage (Cluster) Sampling (Contd.) Optimum sample sizes (Contd.) • Minimize V subject to given cost C • Optimum m and n are obtained as 2 1 2 2 2 1 ) / ( b w S c S c m = ) ( ) ( 2 2 2 1 1 2 0 w b b S c S c c S C C n + − = Sept.2013 82 Multi-stage (Cluster) Sampling (Contd.) Two stage: Optimum Sample Sizes (1) • Goal: get the most information (and hence, more statistically efficient) for the least cost • Illustrative example: PSUs with equal sizes, SRSWOR at both stages Sept.2013 83 Multi-stage (Cluster) Sampling (Contd.) Two stage: Optimum Sample Sizes (2) • Variance function 2 2 b w S S V a ab ≅ + 0 1 2 C C ac abc = + + • Cost function • Problem: Minimize V subject to given cost C Sept.2013 84 Multi-stage (Cluster) Sampling (Contd.) Two stage: Optimum Sample Sizes (3) • Minimize V subject to given cost C • Optimum a=a and b=b 2 2 1 2 / w b b c S c S = 2 0 1 2 2 1 2 ( ) b b w S C C c a c S c S − = + Sept.2013 85 Multi-stage (Cluster) Sampling (Contd.) Two stage: Optimum Sample Sizes (4) • Optimum b=b 2 1 1 2 2 2 1 w b S c c roh b c S c roh − = = Sept.2013 86 Multi-stage (Cluster) Sampling (Contd.) Thanks 87
15752	https://courses.lumenlearning.com/cuny-hunter-collegealgebra/chapter/rational-equations/	8.4 – Rational Equations Learning Objectives (8.4.1) – Solving rational equations by clearing denominators Identify extraneous solutions in a rational equation (8.4.1) – Solving rational equations by clearing denominators Equations that contain rational expressions are called rational equations. For example, 2x+14=7x is a rational equation. Rational equations can be useful for representing real-life situations and for finding answers to real problems. In particular, they are quite good for describing a variety of proportional relationships. One of the most straightforward ways to solve a rational equation is to eliminate denominators with the common denominator, then use properties of equality to isolate the variable. This method is often used to solve linear equations that involve fractions as in the following example: Solve 23x−56=34 by clearing the fractions in the equation first. 23x−56=34 Multiply each term by LCD,122(12)3x−5(12)6=3(12)4 Reduce fractions2(4)x−5(2)=3(3) Multiply8x−10=9 Solve+10+10–––––––––– Add10 to both sides8x=19 Divide both sides by8¯¯¯8¯¯¯8x=198 Our Solution We could have found a common denominator and worked with fractions, but that often leads to more mistakes. We can apply the same idea to solving rational equations. The difference between a linear equation and a rational equation is that rational equations can have polynomials in the numerator and denominator of the fractions. This means that clearing the denominator may sometimes mean multiplying the whole rational equation by a polynomial. In the next example, we will clear the denominators of a rational equation with a terms that has a polynomial in the numerator. Example Solve the equation: 3x−12=1x Show Solution 3x−12=1x Multiply each term by LCD,(2x)3x(2x)−(2x)2=(2x)x Reduce fractions6x2−x=2 Distribute−2−2–––––––– Subtract2 from both sides6x2−x−2=0 Factor(3x−2)(2x+1)=0 Set each factor equal to zero3x−2=0 or2x+1=0 Solve each equation+2+2––––––––––−1−1––––––––3x=22x=−1x=23 orx=−12 Check solutions, LCD can′t be zero2(32)=2(−12)=−1 Neither make LCD zero, both are solutionsx=23 orx=−12 Our Solution Answer x=23 orx=−12 In the next two examples, we show how to solve a rational equation with a binomial in the denominator of one term. We will use the common denominator to eliminate the denominators from both fractions. Note that the LCD is the product of both denominators because they don’t share any common factors. Example Solve the equation 8x+1=43. Show Solution Clear the denominators by multiplying each side by the common denominator. The common denominator is 3(x+1) since 3 and x+1 don’t have any common factors. 3(x+1)(8x+1)=3(x+1)(43) Simplify common factors. 3(x+1)(8x+1)=3(x+1)(43)24=4(x+1)24=4x+4 Now this looks like a linear equation, and we can use the addition and multiplication properties of equality to solve it. 24=4x+4−4–––−4–––20=4xx=5 Check the solution in the original equation. 8(x+1)=438(5+1)=4386=43 Reduce the fraction 86 by simplifying the common factor of 2: 2⋅42⋅3=43 Answer x=5 Example Solve the equation: 5x+5x+2+3x=x2x+2 Show Answer 5x+5x+2+3x=x2x+2 Multiply each term by LCD,(x+2)(5x+5)(x+2)x+2+3x(x+2)=x2(x+2)x+2 Reduce fractions5x+5+3x(x+2)=x2 Distribute5x+5+3x2+6x=x2 Combine like terms3x2+11x+5=x2 Make equation equal zero−x2−x2–––––––––– Subtractx2 from both sides2x2+11x+5=0 Factor(2x+1)(x+5)=0 Set each factor equal to zero2x+1=0 orx+5=0 Solve each equation2x=−1 orx=−5x=−12 or−5 Check solutions, LCD can′t be zero−12+2=32−5+2=−3 Neither make LCD zero, both are solutionsx=−12 or−5 Our Solution Answer x=−12 orx=−5 In the video that follows we present two ways to solve rational equations with both integer and variable denominators. Identify Excluded Values and Extraneous Solutions Some rational expressions have a variable in the denominator. When this is the case, there is an extra step in solving them. Since division by 0 is undefined, you must exclude values of the variable that would result in a denominator of 0. These values are called excluded values. Let’s look at an example. Example Solve the equation 2x−5x−5=15x−5. Show Solution Determine any values for xthat would make the denominator 0. 2x−5x−5=15x−5 5 is an excluded value because it makes the denominator x−5 equal to 0. Since the denominator of each expression in the equation is the same, the numerators must be equal. Set the numerators equal to one another and solve for x. 2x−5=152x=20x=10 Check the solution in the original equation. 2x−5x−5=15x−52(10)−510−5=1510−520−510−5=1510−5155=155 Answer x=10 In the following video we present an example of solving a rational equation with variables in the denominator. You’ve seen that there is more than one way to solve rational equations. Because both of these techniques manipulate and rewrite terms, sometimes they can produce solutions that don’t work in the original form of the equation. These types of answers are called extraneous solutions. That’s why it is always important to check all solutions in the original equations—you may find that they yield untrue statements or produce undefined expressions. Example Solve the equation 16m+4=m2m+4. Show Solution Determine any values for mthat would make the denominator 0. −4 is an excluded value because it makes m+4 equal to 0. Since the denominator of each expression in the equation is the same, the numerators must be equal. Set the numerators equal to one another and solve for m. 16=m20=m2−160=(m+4)(m−4) 0=m+4or0=m−4m=−4orm=4m=4,−4 Check the solutions in the original equation. Since m=−4 leads to division by 0, it is an extraneous solution. 16m+4=m2m+416−4+4=(−4)2−4+4160=160 −4 is excluded because it leads to division by 0. 164+4=(4)24+4168=168 Answer m=4 Sometimes, solving a rational equation results in a quadratic. When this happens, we continue the solution by simplifying the quadratic equation by one of the methods we have seen. It may turn out that there is no solution. Example: Solving a Rational Equation Leading to a Quadratic Solve the following rational equation: −4xx−1+4x+1=−8x2−1. Solution We want all denominators in factored form to find the LCD. Two of the denominators cannot be factored further. However, x2−1=(x+1)(x−1). Then, the LCD is (x+1)(x−1). Next, we multiply the whole equation by the LCD. (x+1)(x−1)[−4xx−1+4x+1]=−8(x+1)(x−1)(x−1)−4x(x+1)+4(x−1)=−8−4x2−4x+4x−4=−8−4x2+4=0−4(x2−1)=0−4(x+1)(x−1)=0x=−1x=1 In this case, either solution produces a zero in the denominator in the original equation. Thus, there is no solution. Example Solve 3x+2x−2+1x=−2x2−2x. Answer x=−1, (x=0 is not a solution). Try It Try It Candela Citations CC licensed content, Original Revision and Adaptation. Provided by: Lumen Learning. License: CC BY: Attribution CC licensed content, Shared previously Unit 15: Rational Expressions, from Developmental Math: An Open Program. Provided by: Monterey Institute of Technology and Education. License: CC BY: Attribution Solve Basic Rational Equations. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. Located at: License: CC BY: Attribution Solve Rational Equations with Like Denominators. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. Located at: License: CC BY: Attribution College Algebra. Authored by: Abramson, Jay et al.. Located at: License: Public Domain: No Known Copyright. License Terms: Download for free at : Question ID#44861. Authored by: Kulinsky,Carla. License: CC BY: Attribution Question ID 3496. Authored by: Shawn Triplett. License: CC BY: Attribution Rational Expressions - Equations examples. Authored by: Tyler Wallace. Located at: Project: Beginning and Intermediate Algebra. License: CC BY: Attribution Licenses and Attributions CC licensed content, Original Revision and Adaptation. Provided by: Lumen Learning. License: CC BY: Attribution CC licensed content, Shared previously Unit 15: Rational Expressions, from Developmental Math: An Open Program. Provided by: Monterey Institute of Technology and Education. License: CC BY: Attribution Solve Basic Rational Equations. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. Located at: License: CC BY: Attribution Solve Rational Equations with Like Denominators. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. Located at: License: CC BY: Attribution College Algebra. Authored by: Abramson, Jay et al.. Located at: License: Public Domain: No Known Copyright. License Terms: Download for free at : Question ID#44861. Authored by: Kulinsky,Carla. License: CC BY: Attribution Question ID 3496. Authored by: Shawn Triplett. License: CC BY: Attribution Rational Expressions - Equations examples. Authored by: Tyler Wallace. Located at: Project: Beginning and Intermediate Algebra. License: CC BY: Attribution
15753	https://andrewzigerelli.com/2022/11/15/cycle-detection-using-floyds-tortoise-and-hare/	Cycle Detection using Floyd's Tortoise and Hare =================== == Andrew's Site == Start. Posts. Work. Documentation. Cycle Detection using Floyd's Tortoise and Hare 15 Nov 2022algo Finding cycles in graphs is an important problem in computer science. I’m ripping straight from wikipedia, but you can use it to check for infinite loops (e.g. function calls accidentally loop, which is a cycle in the call graph) or checking cryptographic hash functions (since they should be close to random functions, we can calculate statistics of random functions, and if the hash function we are testing is too far off, we have problems)1. One algorithm to find cycles is Floyd’s tortoise and hare algorithm. It is attributed to Floyd, even though this may be folklore. One way to state the problem is “functionally”, which is done in the wikipedia article. Then, you can prove something using iterations of the function f. In my opinion, this proof doesn’t “translate” well to finding cycles in linked lists. I think you can do better with some simple modular arithmetic. To be precise, for linked lists, the output for the cycle finding algorithm is NULL if there is no cycle, or the address of the first node of the cycle. The idea of the tortoise and hare algorithm is to have a slow pointer (the tortoise) and fast pointer (the hare), and iterate through a linked list. On each iteration, the tortoise advances one node, and the hare advances two nodes. If there is not a cycle, the fast pointer will reach NULL first. If there is a cycle, eventually fast and slow will be at the same node. Then, with a little more thought, we can calculate the first node in the cycle. Obviously, since the fast pointer moves faster, it will reach NULL first if there is no cycle. So, we only have to consider the case there is a cycle. Let’s label nodes starting from 1. Let node k be the node where the cycle begins, and assume the cycle length is j. Thus, the full set of nodes is {1,2,…,k,k+1,…,k+j-1}. Each node i’s next pointer points to node i+1, with the exception that node k+j-1’s next pointer points to node k. Thus, consider a infinite walk starting at node 1. The node visit sequence looks like: (1,2,3,…,k,k+1,k+2,…,k+j-1,k,k+1,k+2,….,k+j-1,k,k+1,k+2,…) After the first iteration, the slow pointer points to node 1, and the fast points at node 2. After the second iteration, slow points to node 2, and fast node 4 (or the equivalent one in the cycle). Continuing like this, consider the state after k iterations. The slow pointer is at node k. The fast pointer is at the 2k position in the sequence, which is inside the cycle, specifically the k mod j node inside the cycle. In other words, we can relabel nodes k,k+1,….k+j-1 as 0,1,…j. Then the slow pointer is at 0, the fast is at k mod j. Now, if we continue iterating, when will they meet? After i iterations, the slow pointer is at i mod j, and the fast pointer is at k+2i mod j. Solving for i, they will meet after -k mod j iterations. So we have a proof they meet, but how do we get the first node of the cycle? Well, if we iterate k more iterations after they meet, we will be at node -k + k = 0 mod j, which, before relabeling, is at node k, which is where the cycle begins! But in a linked list, we don’t actually know what k is (or else why would we even walk the linked list to find the cycle). However, we can actually count k more iterations if we think carefully. Remember, the linked list starts at node 1, and the cycle starts at node k by assumption. So, if we save the head pointer (node 1), and then walk (k-1) iterations, the head pointer will point to node k, which is the start of the cycle. And, from above, if we walk k iterations from the slow node, after the slow node and fast node meet, we will be at the start of the cycle, node k. Thus, we iterate once for the slow node, and then iterate both the slow and head nodes until they are the same, we will meet at the start of the cycle. So the basic algorithm is clear. Iterate the slow and fast pointers (by 2 and 1, respectively) until they meet. Once they meet, iterate the slow pointer once, and then iterate both the slow and head pointer until they meet, which we proved is the start of the cycle. Here’s some code. struct ListNode { int val; ListNode next; ListNode(int x) : val(x), next(NULL) {} }; ListNode detectCycle(ListNode head) { if (head == nullptr) { return nullptr; } ListNode slow = head; ListNode fast = slow->next; while(fast != nullptr) { // safe since slow lags behind fast slow = slow->next; fast = fast->next; if(fast) fast = fast->next; if (fast == slow) { //there is a cycle //slow node needs one more iteration slow = slow->next; //head and slow will meet at cycle start while(head != slow) { slow = slow->next; head = head->next; } return slow; } } return fast; } LATEST POSTS Cycle Detection using Floyd's Tortoise and Hare Can't Stop Them All Don't forget the null bytes Play with Cpp features Read generatingfunctionology by Herbert Wilf © 2023 Andrew's Site. Tags.
15754	https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_12?srsltid=AfmBOoqNsbcWR7lmLNHF3g3H_W_7GlKT8-TqNNwTntHAbzITg4aYBxeY	Art of Problem Solving 2018 AIME I Problems/Problem 12 - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. 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If is chosen at random among all subsets of , the probability that is divisible by is , where and are relatively prime positive integers. Find . Solution 1 The question asks us for the probability that a randomly chosen subset of the set of the first 18 positive integers has the property that the sum of its elements is divisible by 3. Note that the total number of subsets is because each element can either be in or not in the subset. To find the probability, we will find the total numbers of ways the problem can occur and divide by . To simplify the problem, let’s convert the set to mod 3: Note that there are six elements congruent to 0 mod 3, 6 congruent to 1 mod 3, and 6 congruent to 2 mod 3. After conversion to mod three, the problem is the same but we’re dealing with much simpler numbers. Let’s apply casework on this converted set based on , the sum of the elements of a subset of . In this case, we can restrict the subsets to subsets that only contain 0. There are six 0’s and each one can be in or out of the subset, for a total of subsets. In fact, for every case we will have, we can always add a sequence of 0’s and the total sum will not change, so we will have to multiply by 64 for each case. Therefore, let’s just calculate the total number of ways we can have each case and remember to multiply it in after summing up the cases. This is equivalent to finding the number of ways you can choose such subsets without including the 0's. Therefore this case gives us way. In this case and each of the subsequent cases, we denote the number of 1’s in the case and the number of two’s in the case as respectively. Then in this case we have two subcases; This case has ways. This case has ways. In total, this case has ways. In this case, there are 4 subcases; This case has way. This case has ways. This case has ways. This case has ways. In total, this case has ways. Note that for each case, the # of 1’s goes down by 2 and the # of 2’s goes up by 1. This is because the sum is fixed, so when we change one it must be balanced out by the other. This is similar to the Diophantine equation and the total number of solutions will be . From here we continue our casework, and our subcases will be coming out quickly as we have realized this relation. In this case we have 3 subcases: This case has ways. This case has ways. This case has ways. In total, this case has ways. In this case we have 4 subcases: This case has ways. This case has ways. This case has ways. This case has way. Therefore the total number of ways in this case is . Here we notice something interesting. Not only is the answer the same as Case 3, the subcases deliver the exact same answers, just in reverse order. Why is that? We discover the pattern that the values of of each subcase of Case 5 can be obtained by subtracting the corresponding values of of each subcase in Case 3 from 6 ( For example, the subcase 0, 6 in Case 5 corresponds to the subcase 6, 0 in Case 3). Then by the combinatorial identity, , which is why each subcase evaluates to the same number. But can we extend this reasoning to all subcases to save time? Let’s write this proof formally. Let be the number of subsets of the set (where each 1 and 2 is distinguishable) such that the sum of the elements of the subset is and is divisible by 3. We define the empty set as having a sum of 0. We claim that . if and only if there exists that satisfy , , , and is the set of the pairs . This is because for each pair , there are six elements of the same residue mod(3) to choose and numbers from, and their value sum must be . Let all , satisfy and . We can rewrite the equation Therefore, since and and , . As shown above, and since and 18 are divisible by 3, 18 - is divisible by 3. Therefore, by the fact that , we have that; , proving our claim. We have found our shortcut, so instead of bashing out the remaining cases, we can use this symmetry. The total number of ways over all the cases is . The final answer is There are no more 2’s left to factor out of the numerator, so we are done and the answer is . ~KingRavi Solution 2 Consider the elements of modulo Ignore the 's because we're gonna multiply at the end. Let be the and be the . The key here is that so the difference between the number of and is a multiple of . Counted twice because and can be switched: Counted once: ... By Vandermonde's Identity on the second case, this is Solution 3 (Triple Vandermonde's) Elements can either be included or excluded, for a total of . Then, like the previous solution, let be the number of elements and be the number of elements . Then, . Since , there are 3 cases: Then, we have, Using the substitution on every second binomial coefficient, it is clear that the bottom numbers sum to , and the ones above sum to . Apply Vandermonde's, we obtain For , using the same substitution and applying Vandermonde's, we get: Then, for , it is completely symmetric with . We get We have We get . ~CrazyVideoGamez Solution 4 (Symmetry) Note that in general, the answer will be around of . We will use this to our advantage. Partition into disjoint subsets . Within each of these subsets, there are 3 possible remainders depending on what elements we choose to include into : (Using set for demonstration purposes) Remainder of zero: Include Remainder of one: Include Remainder of two: Include Suppose all subsets are of the form . Then, since both choices are , we can choose either one for all six subsets, giving us combinations. On the other hand, suppose there exists a subset that isn't of the form . Then, for , it is equally likely that a remainder of zero, one, or two is chosen, since each remainder has ways to be achieved, i.e. there is a chance for each. Consider the sum of the other subsets . The sum is either . Whatever that remainder might be, we can always choose as the remainder for our set , giving us a total of . The probability we choose remainder is as previously mentioned. So, we get total combinations. Therefore, we get ~CrazyVideoGamez Solution 5 Rewrite the set after mod 3 as above 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 All 0s can be omitted Case 1 No 1 No 2 Case 2 Case 3 Case 4 Case 5 Case 6 Case 7 Case 8 Case 9 Case 10 Case 11 Case 12 Case 13 Case 14 Case 15 Case 16 Case 17 Total ANS= By SpecialBeing2017 Solution 6 Consider the numbers . Each of those are congruent to . There is way to choose zero numbers ways to choose and so on. There ends up being possible subsets congruent to . There are possible subsets of these numbers. By symmetry there are subsets each for and . We get the same numbers for the subsets of . For , all subsets are . So the probability is: Solution 7 Notice that six numbers are , six are , and six are . Having numbers will not change the remainder when is divided by , so we can choose any number of these in our subset. We ignore these for now. The number of numbers that are , minus the number of numbers that are , must be a multiple of , possibly zero or negative. We can now split into cases based on how many numbers that are are in the set. CASE 1- , , or integers: There can be , , or integers that are . We can choose these in CASE 2- or integers: There can be or integers that are . We can choose these in CASE 3- or integers: There can be or integers that are . We can choose these in Adding these up, we get that there are ways to choose the numbers such that their sum is a multiple of three. Putting back in the possibility that there can be multiples of in our set, we have that there are subsets with a sum that is a multiple of . Since there are total subsets, the probability is so the answer is . Solution 8 (Generating Functions) We use generating functions. Each element of has two choices that occur with equal probability--either it is in the set or out of the set. Therefore, given , the probability generating function is Therefore, in the generating function the coefficient of represents the probability of obtaining a sum of . We wish to find the sum of the coefficients of all terms of the form (where ). If is a cube root of unity, then it is well know that for a polynomial , will yield the sum of the coefficients of the terms of the form . (This is known as the third Roots of Unity Filter.) Then we find To evaluate the last two products, we utilized the facts that and . Therefore, the desired probability is Thus the answer is . Solution 9 Define a set that "works" to be a set for which the sum of the terms is mod . The given set mod is Let be the number of subsets of the first terms of this set that "work." Consider just the first terms: There are total subsets, and (the subsets are and ). Now consider the first terms: To find , we set aside the last terms as a "reserve" that we can pair with subsets of the first terms (which we looked at earlier). First, all subsets of the first terms can either be paired with a or a (or nothing) from the "reserve" terms so that they "work," creating subsets that "work" already. But for each of these, we have the option to add a from the reserve, so we now have subsets that "work." For each of the subsets of the first terms that "work", we can also add on a or a from the reserves, creating new subsets that "work." And that covers all of them. With all of this information, we can write as The same process can be used to calculate then etc. so we can generalize it to Thus, we calculate with this formula: Because there are total possible subsets, the desired probability is so the answer is Solution 10 Try smaller cases and find a pattern. Using similar casework as in Solution 1, we can easily find the desired probability if is of a small size. If , then out of subsets work, for a probability of . If , then out of subsets work, for a probability of . If , then out of subsets work, for a probability of . Let be the numerator of the desired probability if is of size . Then and . Noticing that the denominators are multiplied by each time, we can conclude that the pattern of the numerators is , so . Solution 11 (Quick guesswork for about 2 minutes remaining) We conjecture that the difference from the probability will be as small as possible from (The value approached as , where is the number of terms in the largest subset). We also see that there are subsets and know the denominator will be a power of (since the numerator is an integer). We essentially want to guess that the greatest integer satisfying can be plugged in to get the solution of round . We see that this occurs at , and get round round()= . Solution 12 (Very quick recursion) The total number of subsets is simply Now we need to find the number of subsets that have a sum divisible by Ignore the 6 numbers in the list that are divisible by 3. We look only at the number of subsets of then multiply by at the end. This is because adding a multiple of to the sum will not change whether it is divisible by or not and for each of the multiples, we have two options of whether to add it to the subset or not. Now, let be the number of subsets of that have a sum divisible by . There are numbers in the set. Let us look at the first numbers: all subsets of the first numbers will have a residue or mod If it is add both and to the subset. If it is add just to the set, and if it is add just We don't have to compute all three cases separately; since there is a 1 to 1 correspondence, we only need the sum of the three cases (which we know is ). Now, this counts all the subsets except for those that include neither nor However, this is just Thus, and the base case is We have, Multiplying this by we have, The cancels out with the denominator. However, the second term in the product is obviously odd and so does not cancel further with the denominator, which is just a power of Thus, we want to find which equals Note: While the explanation may seem a bit complicated, the concept behind it is pretty simple. And once you get the solution, computing the answer doesn't take much time. Solution 13 The total number of subsets is . If , the sum of the elements divides 3. We can rewrite the set as 6 0s, 6 1s, and 6 2s. We can ignore the zeros for now, since they won't influence the sum so we focus on each configuration of the 6 1s and 6 2s such that the sum is divisible by 3 and then multiply by at the end. We proceed with casework on the number of values that are equivalent to as follows: Case 1: There are zero elements: Then, there are only configurations of the values congruent to 1 mod 3. There are ways to assign values in the set that are congruent to 1 mod 3 such that the sum of those values divides 3. Therefore there are configurations for this case. Case 2: There is one element: There are ways to choose which element is included in the subset and ways to assign values to the numbers congruent to 1 mod 3 in the set. Therefore there are configurations for this case. Case 3: There are two elements: There are ways to choose the 2 elements in the set that are congruent to 2 mod 3 and possible ways to assign values to the numbers congruent to 1 mod 3. Therefore there are configurations for this case. Case 4: There are three elements: There are ways to choose three elements in the set that are congruent to 2 mod 3. There are ways to assign values to the numbers congruent to 1 mod 3. Therefore, there are configurations for this case. Case 5, Case 6, and Case 7 are symmetric to Case 3, Case 2, and Case 1 respectively so we can multiply by 2 for those cases. Putting all the cases together we obtain . Multiplying by gives . Since , . Therefore, ~Magnetoninja Video Solution (3b1b video on a very similar problem) See Also 2018 AIME I (Problems • Answer Key • Resources) Preceded by Problem 11Followed by Problem 13 1•2•3•4•5•6•7•8•9•10•11•12•13•14•15 All AIME Problems and Solutions These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 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15755	https://www.youtube.com/watch?v=KVSO8-IyPFo	Que es el circulo unitario Autodidacta 9380 subscribers 2927 likes Description 123521 views Posted: 8 Mar 2022 Mis estimados Autodidacta el día de hoy vamos aprender todo acerca del circulo unitario. Vamos a responder las siguientes preguntas: ¿Qué es el circulo unitario? ¿Para que sirve el circulo unitario? ¿Cómo hacer el circulo unitario? ¿Cómo recordar el circulo unitario? Esperemos que les guste! No olviden que pueden descargar el formulario: Formulario: 0:00 Introducción 0:20 ¿Qué es el circulo unitario? 2:08 Relación del circulo unitario y las funciones trigonometricas 5:22 Coordenadas en el circulo unitario 6:12 ¿Cómo construir el circulo unitario? 9:09 Triangulo especifico 30 - 60 - 90 10:10 Triangulo especifico 45 - 45 - 90 11:34 Manera sencilla de escribir el circulo unitario 13:54 Como construir los cuadrantes faltantes 20:24 Conclusion 21:34 Despedida Redes Sociales 🔴 Facebook: 🔴 Instagram: 🔴 Tiktok: Listas de Reproducción: Trigonometría: Ángulos: Grados Sexagesimales: Factorización: Torre de Pascal: Productos Notables: Introducción al Álgebra: Calculo Vectorial: 68 comments Transcript: Introducción mis autodidactas a continuación vamos a aprender uno de los temas más interesantes dentro de la trigonometría siendo el círculo unitario en donde vamos a aprender cómo escribirlo y de igual manera vamos a aprender cómo memorizarlo de una manera muy sencilla y muy práctica entonces a continuación podemos observar que vamos a tener un fondo blanco con el fin de que no se me pierdan y básicamente el círculo ¿Qué es el circulo unitario? unitario es un círculo con un radio que es igual a 1 el cual está dentro del plano coordinado con el centro en el origen es decir las coordenadas 0 0 como lo podemos observar en la siguiente imagen básicamente este círculo nos permite extender el dominio tanto del seno como del coseno a todos los números reales y de igual manera podemos determinar tanto el seno y el coseno para cualquier ángulo theta entonces básicamente podemos observar las primeras cuatro coordenadas las cuales de cajón nos tenemos que aprender ahora sabemos que estamos en un plano coordinado lo cual nos quiere decir que tenemos coordenadas tanto en x como en ya sabemos que acá nuestra línea va a ser igual allí y nuestra línea horizontal va a ser igual a equis ahora como tenemos el radio de la unidad sabemos que esta línea cuando toca por acá tenemos el lado 10 es decir avanzamos una unidad en x pero no avanzamos nada en y ahora lo mismo pasa con este otro puntito por acá no avanzamos nada en x es por eso que tenemos el cero y avanzamos una unidad de y es por eso que tenemos la coordenada 01 ahora lo mismo pasa con estas otras dos coordenadas sin embargo sabemos que esos cuadrantes son negativos es decir acá tenemos menos 10 avanzamos menos una unidad en x pero no avanzamos nada en y acá tenemos el 0 como menos 1 en donde no avanzamos nada en x pero disminuimos una unidad en 10 entonces estas son las cuatro primeras coordenadas que tenemos que aprendernos y a continuación van a observar la relación que van a tener sin embargo ustedes se han de preguntar qué tiene que ver un círculo unitario con las Relación del circulo unitario y las funciones trigonometricas funciones trigonométricas qué tiene que ver con un triángulo pues bueno de manera muy sencilla y cómo lo vamos a observar dentro de toda nuestra formación académica podemos observar que a partir de esta circunferencia podemos trazar una vertical en donde el radio toca la circunferencia es decir a partir de este punto podemos trazar una vertical y dentro de esta vertical podemos observar que vamos a formar un ángulo teta entonces podemos observar que la vertical al ángulo que acabamos de formar va a ser el cateto opuesto y la línea que vamos a tener por debajo va a ser el cateto adyacente entonces a continuación podemos observar que hemos formado un triángulo a partir del radio y sabemos que el radio de este círculo unitario va a ser igual a la unidad entonces a partir de eso es que formamos el círculo unitario entonces a continuación vamos a observar qué más podemos hacer con este círculo unitario básicamente sabemos que la hipotenusa siempre va a ser igual a la unidad debido a que si nos paramos por aquí o nos ponemos por acá o nos ponemos por acá el radio va a ser exactamente lo mismo lo único que va a estar cambiando va a ser ángulo que vamos a estar de notando y a partir de ese ángulo pues podemos trazar diferentes triángulos para obtener tanto el cateto puesto y el cateto adyacente entonces aquí podemos observar que tenemos un triángulo recta do formado a partir de la vertical que acabamos de trazar en donde nuevamente el lado que es opuesto al ángulo teta va a ser el cateto opuesto y tenemos que recordar claramente que función trigonométricas trabaja con el cateto opuesto básicamente la función que trabaja con el cuarto puesto es el seno de teta y si recordamos que la hipotenusa es igual a radio y ésta es igual a la unidad podemos observar que el seno de teta va a ser igual al khater opuesto sobre la unidad de tal manera que podemos observar que en jeff dado que nos estaríamos moviendo en 10 siendo el cateto opuesto pues las coordenadas en llevan a estar en función del seno de teta y el seno de teta bastad en función del cateto opuesto sin embargo esta parte del caso propuesto no es tan relevante porque vamos a estar trabajando con los triángulos especiales que previamente hemos aprendido entonces a partir de esta observación es importante recalcar que las coordenadas en el plano coordenada y van a depender del seno de teta por otro lado dentro de este mismo triángulo podemos observar que tenemos el cateto adyacente el cual claramente va a ser el lado adyacente al ángulo teta ahora si recordamos que la función trigonométricas que relaciona el cateto adyacente es el coseno y nuevamente recordamos que la hipotenusa es igual al radio y este es igual a la unidad podemos observar que el coche no detecta va a ser igual al cateto adyacente sobre la unidad y entonces esto nos dice que el movimiento en donde nos vamos a estar desplazando en x en el plano coordinado va a depender de la función del coseno dt está siendo igual al cateto adyacente pero nuevamente debido a que vamos a estar trabajando con triángulos especiales pues esta parte del cateto adyacente no nos importa lo importante es entender que dentro del plano coordinado lo que nos vamos a estar moviendo en jay y lo que nos vamos a estar moviendo en x va a depender de las funciones del seno y del co 0 entonces es lo que vamos a ver a continuación podemos observar que acá tenemos nuestro plano coordinado y las Coordenadas en el circulo unitario coordenadas en el círculo unitario siendo xy van a depender del coste no de teta y van a depender del seno en donde como podemos observar acá podemos observar y el co seno y el seno de teta se van a estar desplazando de menos uno a uno y de menos uno a uno esto por qué porque podemos observar en morado que acá tenemos la línea que nos representa el movimiento que va a tener el coseno dt está dentro del círculo unitario y podemos observar que claramente va de menos 1 a 1 lo mismo pasa con el seno de teta acá podemos observar el desplazamiento que vamos a tener en el plano coordinado y nuevamente va a ir de menos 1 a 1 siendo igual a menos 1 o igual a 1 sin embargo nunca mayor entonces a partir de eso es que vamos a estar formando nuestro círculo unitario entonces vamos a empezar a aprender ahora ya que hemos definido todo lo que ¿Cómo construir el circulo unitario? forma el círculo unitario y ya sabemos que acá tenemos las coordenadas x las cuales van a depender del coseno de teta y el seno de teta podemos empezar a darle forma en primer lugar vamos a partir de cero después tendríamos el grado de 90 los grados a 180 y 270 ahora estos son los primeros cuatro grados que tenemos de una manera muy exacta entonces básicamente acá podemos observar que tenemos los grados sexagesimal es acá vamos a tener los rangers es debido a que usualmente lo tenemos que aprender con los radiales y acá tenemos los valores de las coordenadas en la equis y ahora esto es muy importante y a continuación vamos a explicar por qué sabemos que tenemos las coordenadas x y las cuales 10 sin embargo esto a qué se refiere con el coseno y el seno pues es muy sencillo esto nos quiere decir que el coseno de 0 grados o el coseno de 0 radiales va a ser igual a la unidad y de igual manera esto es lo indica con el seno el seno de 0 grados o el seno de 0 radiales va a ser igual a 0 siendo esta relación lo primero que nos tenemos que aprender a partir de las relaciones trigonométricas ahora nuevamente que tenemos 90 grados el cual es igual a pi medios ya que sabemos la relación que tienen los radiales con los grados sexagesimal es básicamente esto nos indica que el cocino de 90 grados va a ser igual a 0 sin embargo el seno de 90 grados o de pi medios va a ser igual entonces como ustedes ya saben que es un círculo unitario y sabemos desplazarnos tanto en xy ya tenemos los primeros cuatro resultados a cero grados a 90 grados a 180 grados y a 270 grados es decir el coseno de 180 podemos observar que va a ser igual a menos 1 y el seno de 180 va a ser igual a 0 lo mismo pasa con 270 el coseno de 270 base de igual a 0 y el seno de 270 va a ser igual a menos 1 ahora que tendríamos el 2 pi ahora a continuación que tenemos que aprender pues bueno vamos a comenzar marcando los siguientes ángulos en primer lugar vamos a marcar con línea azul todos los múltiplos de pi sextos y sextos por dos por tres por cuatro por cinco por seis etcétera es decir tenemos 30 grados y son múltiplos de 30 grados por dos por tres por cuatro por cinco etcétera básicamente yo les recomendaría a aprenderse únicamente el primer cuadrante debido a que a partir del primer cuadrante podemos formar los otros tres básicamente como les he mencionado anteriormente en azul vamos a marcar los múltiplos y sextos o 30 grados y es importante hacer la observación que a partir de estos vamos a estar formando el ángulo de 90 182 100 70 sin embargo como estos previamente ya los hemos aprendido pues podemos omitir los y entonces podemos observar que si trazamos la vertical acá estaríamos formando únicamente nuestro triángulo Triangulo especifico 30 - 60 - 90 especial 30 60 90 en donde como previamente ya hemos visto en otro vídeo pues básicamente ya sabemos los valores de los catetos a partir de este triángulo siendo igual a 1 la raíz de 3 y pues la hipotenusa va a ser de 2 porque porque claramente este es un triángulo especial que previamente ya hemos aprendido entonces como previamente ya hemos aprendido ya sabemos los valores del coseno de 30 grados y el seno de 30 grados y sabemos que el coste no va a ser el desplazamiento de x y el seno va a ser el desplazamiento de y entonces básicamente esto nos indica las coordenadas del plano es decir x va a ser igual a la raíz de 3 sobre 2 y en que va a ser igual a un medio entonces básicamente a partir de esto podemos colocar las nuevas ordenadas siendo raíz de tres sobre dos un medio ahora vamos a trabajar con nuestro siguiente grado el cual va a ser 45 grados api cuartos entonces tenemos que recordar nuevamente que cuando tenemos este 45 grados estamos hablando nuevamente de un Triangulo especifico 45 - 45 - 90 triángulo especial 145 45 90 y como previamente ya lo hemos discutido ya sabemos los valores tanto del coseno de 45 grados y del seno de 45 grados entonces nuevamente este resultado siendo raíz de 2 sobre 2 y raíz de 2 sobre 2 tanto en x como el que nos indican el desplazamiento en el plano coordinado de x entonces nuevamente vamos a colocar estas coordenadas x siendo las mismas como podemos observar a continuación y a continuación vamos a trabajar con 60 grados siendo el múltiplo de 30 si lo multiplicamos por 2 ahora nuevamente podemos observar que estaríamos recordando nuestro triángulo especial 30 60 90 sin embargo esta vez vamos a estar trabajando con los 60 grados nuevamente sabemos ya los valores 960 y del seno de 60 el cual es un medio y la raíz de 3 sobre 2 entonces podemos observar que ya tenemos el desplazamiento en x y el desplazamiento en tiempo recuerden que básicamente lo que estamos haciendo es obtener precisamente la posición de estos puntos en el plano coordinado utilizando las coordenadas que tenemos tanto en x como jeff entonces podemos observar que las últimas coordenadas que vamos a tener va a ser un medio como la raíz de 3 sobre 2 entonces a continuación podemos observar Manera sencilla de escribir el circulo unitario que ya tenemos las coordenadas tanto en x como en yemen en donde acá tenemos 1,0 siendo esta primera coordenada después la raíz de tres sobre dos como un medio siendo la segunda coordenada y así nos seguimos sin embargo es evidente observar que tenemos una relación muy evidente podemos observar que el coseno de cero va a ser igual al seno de 90 grados entonces podemos observar que la relación se repite es decir mientras una crece otra decrece tenemos 11 raíz de 3 sobre 2 raíz de 2 sobre 2 y después tenemos un medio y después tenemos los ceros de tal manera que podemos observar que existe una relación entre estos es decir mientras uno crece la otra decrece entonces usualmente es común trabajar con el seno lo cual vamos a hacer a continuación una manera de escribir el cero va a ser la raíz de 0 sobre 2 debido a que vamos a aprender una forma en la cual podemos aprendernos el círculo unitario para escribirlo de una manera muy rápida y sencilla a continuación una manera de escribir en un medio va a ser la raíz de 1 sobre 2 de tal manera que esto claramente nos daría un medio ahora los siguientes 2 nos vamos a dejar constantes es decir la raíz de 2 sobre 2 lo escribimos igual y la raíz de 3 sobre 2 lo escribimos igual finalmente para obtener la unidad pues podemos observar que podemos escribir la raíz de 4 sobre 2 de tal manera que al aprendernos la siguiente relación podemos observar que podemos construir el círculo unitario de una forma muy sencilla sin embargo tenemos que aprender que estamos trabajando con el seno es decir tendríamos los valores en y siendo 0 un medio la raíz de 2 sobre 2 la raíz de 3 sobre 2 y la unidad de tal manera que lo único que tenemos que hacer es en primer lugar escribir los valores de iu y después escribir los valores de x es decir es escribir estos mismos valores sin embargo empezando por el último siendo 1 la raíz de 3 sobre 3 la raíz de 2 sobre 2 un medio y 0 es también que con eso ya tenemos la formación de todo el círculo unitario bueno únicamente de la primera parte a continuación vamos a aprender cómo escribir la segunda parte de este mismo Como construir los cuadrantes faltantes círculo unitario entonces es evidente observar que en esta ocasión tenemos todos los valores de x negativos es decir vamos a tener menos el coseno de teta sin embargo eso no nos va a impedir aprenderlo de una manera muy rápida y sencilla podemos observar que tenemos este mismo círculo unitario sin embargo tenemos nuestro primer valor el cual va a ser 150 grados y podemos observar que es el mismo resultado que vamos del grado de 30 grados o pi sextos y esto porque lo sería pues básicamente como les había mencionado antes podemos observar que este ángulo que vamos a formar básicamente es un ángulo a 30 grados sin embargo en otro cuadrante y lo único que va a cambiar pues va a ser que estamos posicionados en valores negativos de x es decir por eso tenemos menos la raíz de 3 sobre 2 entonces podemos observar que lo único que va a cambiar el segundo cuadrante es que vamos a tener los mismos resultados sin embargo los valores de x van a ser negativos entonces básicamente lo único que tenemos que hacer es hacer lo mismo que hemos hecho en este primer cuadrante en primer lugar vamos a trazar el ángulo a 30 grados y vamos a sacar su correspondiente en el segundo cuadrante siendo 150 entonces a continuación tendríamos que sacar la correspondiente a 45 grados opi cuartos es decir si multiplicamos 45 por 3 podemos obtener ese grado el cual sería 135 grados o 3 pi y nuevamente podemos observar que tendríamos menos la raíz de 2 sobre 2 y la raíz de 2 sobre 2 esto porque pues claramente podemos observar que nuevamente los valores de x van a ser negativos mientras que los valores de llevan a ser positivos y finalmente estaríamos trazando el grado correspondiente a 60 grados el cual claramente va a ser igual a 120 y nuevamente podemos observar es el mismo resultado que hemos obtenido en 60 grados sin embargo ahora podemos observar que tenemos únicamente el valor de x negativo y el valor de ye positivo entonces podemos observar que si trazamos una recta de aquí acá pues claramente estaremos teniendo el mismo valor del ángulo lo mismo pasaría con el ángulo de 135 y 45 y con el ángulo de 30 grados y 150 básicamente lo que tienen que aprender es que acá estaríamos teniendo un ángulo de 30 aquí estaríamos teniendo un ángulo de 45 y a partir de aquí estaríamos teniendo un ángulo de 60 grados entonces esta es una manera muy sencilla en la cual ustedes pueden aprenderse todos los valores del círculo unitario entonces a continuación vamos a comenzar a trabajar con la tercera región del círculo unitario en donde podemos observar que a partir de esta tercera región tenemos valores negativos tanto de x como de ye entonces para aprender esta tercera región existen muchos tipos y son los tipos que a continuación vamos a aprender entonces en primer lugar aquí ya tenemos trazados todos los ángulos entonces el primer tip que yo les daría si están aprendiendo a hacer este círculo unitario desde cero podemos observar que este primer ángulo siendo el de 30 grados que acabamos de trazar en la primera región lo podemos extrapolar a la tercera región de tal manera que lo extrapolamos hasta aquí y podemos observar que en este punto claramente vamos a tener los mismos valores del ángulo de 30 grados sin embargo acá tendríamos menos la raíz de tres y menos un medio porque bueno claramente mencionamos anteriormente que acá las coordenadas tanto de x como de i son negativas porque estamos en el tercer cuadrante ahora nuevamente esto lo podemos observar a partir del ángulo de 45 grados es decir podemos extrapolar este ángulo de la primera región a la tercera región que tenemos por acá con el fin de poder observar que tenemos el mismo resultado sin embargo tenemos los dos signos negativos y lo mismo pasa con el ángulo de 60 grados si lo extrapolamos nuevamente por acá podemos observar que vamos a formar el ángulo a 240 grados y vamos a tener los mismos valores que tenemos en el ángulo de 60 grados ahora esto mismo pasa a partir de la segunda región como yo hicimos anteriormente lo único que tenemos que hacer es trazar una recta en donde obtuvimos el valor a 150 hacia la tercera región y podemos observar que vamos a obtener el mismo valor sin embargo con el valor negativo en yen porque claramente estamos hablando de que en la tercera región los valores tanto de x como de ye son negativos lo mismo podemos hacer a partir de este ángulo lo extrapolamos hacia la tercera región y podemos observar que coincide nuevamente con este mismo ángulo y lo mismo con 120 grados sin embargo lo que también podemos hacer es que podemos trazar estos mismos ángulos utilizando un mismo compás básicamente lo que queremos observar es que vamos a obtener los mismos valores de la primera región sin embargo lo único que va a estar cambiando van a ser los signos ya que en la tercera región los dos signos son negativos finalmente lo último que tenemos que aprender es la cuarta región básicamente en la cuarta región volvemos a tener valores positivos de xy valores negativos de jeff y como pueden observar acá en nuevamente ya tenemos trazados todos los ángulos en donde podemos observar tanto los radiales como los grados sexagesimal es ahora nuevamente podemos trazar una vertical a partir de este ángulo y podemos observar que vamos a obtener los mismos valores del ángulo de 30 grados sin embargo como sabemos que estamos en la cuarta región los valores de llevan a ser negativos como lo podemos observar con este menos un medio ahora nuevamente podemos hacerlo eso con el ángulo de 45 grados o con el ángulo de 60 grados y como previamente lo hemos aprendido con la región 3 a partir de la región 2 podemos extrapolar este y podemos observar que vamos a tener los mismos valores sin embargo con diferentes signos de igual manera si extrapolamos este con este vamos a obtener este mismo ángulo sin embargo nuevamente con los signos diferentes y lo mismo pasa con el ángulo a 150 grados ahora nuevamente podemos aprenderlo con la tercera región si es que necesitan otra forma de aprenderlo y de igual manera si trazamos una horizontal a partir de este ángulo hacia la cuarta región podemos observar que nuevamente tenemos el mismo resultado sin embargo con un signo completamente diferente ahora lo mismo pasa con estos otros dos ángulos entonces finalmente a modo de Conclusion conclusión tenemos que recordar lo siguiente ya sabemos cómo formar el primer cuadrante y ya sabemos que vamos a tener los resultados de la raíz de 0 sobre 2 la raíz de 1 sobre 2 la raíz de 3 sobre 2 la raíz de 4 sobre 2 etcétera entonces a partir de esos resultados podemos formar el seno siendo la función que dependen y a partir de eso si lo ponemos de una manera inversa podemos tener los resultados del coseno en donde vamos a obtener los valores que vamos a en x entonces ya hemos aprendido a formar la primera región del círculo unitario y ya sabemos que las líneas azules siempre van a ser múltiplos de pi sextos o del ángulo de 30 grados en donde a partir de este ángulo podemos formar todo el círculo unitario en donde podemos observar que 30 es múltiplo de 30 siendo la multiplicación de 30 por 2 ahora sí tenemos 120 a que sería 30 por 4 150 sería 30 por 5 y así nos podemos ir con todos los valores que tenemos en azul ahora nuevamente el 45° básicamente a partir de este ángulo de 45 grados podemos formar un cuadrado dentro de este círculo unitario y bueno mis Despedida autodidactas eso sería todo por el vídeo del día de hoy espero que les haya gustado espero que les haya servido no olviden checar las listas de reproducción porque ahí hay temas que pueden ser de su interés no olviden suscribirse al canal síguenos en nuestras redes sociales y dejar su magnífico me gusta así que sin más nos vemos en un nuevo vídeo
15756	https://otexts.com/fpp2/stochastic-and-deterministic-trends.html	Forecasting: Principles and Practice (2nd ed) 9.4 Stochastic and deterministic trends There are two different ways of modelling a linear trend. A deterministic trend is obtained using the regression model yt=β0+β1t+ηt, where ηt is an ARMA process. A stochastic trend is obtained using the model yt=β0+β1t+ηt, where ηt is an ARIMA process with d=1. In the latter case, we can difference both sides so that y′t=β1+η′t, where η′t is an ARMA process. In other words, yt=yt−1+β1+η′t. This is similar to a random walk with drift (introduced in Section 8.1), but here the error term is an ARMA process rather than simply white noise. Although these models appear quite similar (they only differ in the number of differences that need to be applied to ηt), their forecasting characteristics are quite different. Example: International visitors to Australia autoplot(austa) + xlab("Year") + ylab("millions of people") + ggtitle("Total annual international visitors to Australia") Figure 9.9: Annual international visitors to Australia, 1980–2015. Figure 9.9 shows the total number of international visitors to Australia each year from 1980 to 2015. We will fit both a deterministic and a stochastic trend model to these data. The deterministic trend model is obtained as follows: ``` trend <- seq_along(austa) (fit1 <- auto.arima(austa, d=0, xreg=trend)) > Series: austa > Regression with ARIMA(2,0,0) errors > > Coefficients: > ar1 ar2 intercept xreg > 1.113 -0.380 0.416 0.171 > s.e. 0.160 0.158 0.190 0.009 > > sigma^2 = 0.0298: log likelihood = 13.6 > AIC=-17.2 AICc=-15.2 BIC=-9.28 ``` This model can be written as yt=0.416+0.171t+ηtηt=1.113ηt−1−0.380ηt−2+εtεt∼NID(0,0.030). The estimated growth in visitor numbers is 0.17 million people per year. Alternatively, the stochastic trend model can be estimated. ``` (fit2 <- auto.arima(austa, d=1)) > Series: austa > ARIMA(0,1,1) with drift > > Coefficients: > ma1 drift > 0.301 0.173 > s.e. 0.165 0.039 > > sigma^2 = 0.0338: log likelihood = 10.62 > AIC=-15.24 AICc=-14.46 BIC=-10.57 ``` This model can be written as yt−yt−1=0.173+η′t, or equivalently yt=y0+0.173t+ηtηt=ηt−1+0.301εt−1+εtεt∼NID(0,0.034). In this case, the estimated growth in visitor numbers is also 0.17 million people per year. Although the growth estimates are similar, the prediction intervals are not, as Figure 9.10 shows. In particular, stochastic trends have much wider prediction intervals because the errors are non-stationary. fc1 <- forecast(fit1, xreg = length(austa) + 1:10) fc2 <- forecast(fit2, h=10) autoplot(austa) + autolayer(fc2, series="Stochastic trend") + autolayer(fc1, series="Deterministic trend") + ggtitle("Forecasts from trend models") + xlab("Year") + ylab("Visitors to Australia (millions)") + guides(colour=guide_legend(title="Forecast")) Figure 9.10: Forecasts of annual international visitors to Australia using a deterministic trend model and a stochastic trend model. There is an implicit assumption with deterministic trends that the slope of the trend is not going to change over time. On the other hand, stochastic trends can change, and the estimated growth is only assumed to be the average growth over the historical period, not necessarily the rate of growth that will be observed into the future. Consequently, it is safer to forecast with stochastic trends, especially for longer forecast horizons, as the prediction intervals allow for greater uncertainty in future growth.
15757	https://artofproblemsolving.com/wiki/index.php/Binomial_Theorem?srsltid=AfmBOopPCpUWwjPsb1ESDzVXYTV6SUduKdcHtvpKzP4qBjJMt2wMRq8n	Art of Problem Solving Binomial Theorem - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Binomial Theorem Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Binomial Theorem The Binomial Theorem states that for real or complex, , and non-negativeinteger, where is a binomial coefficient. In other words, the coefficients when is expanded and like terms are collected are the same as the entries in the th row of Pascal's Triangle. For example, , with coefficients , , , etc. Contents [hide] 1 Proof 1.1 Proof via Induction 1.2 Proof using calculus 2 Generalizations 2.1 Proof 3 Usage 4 See also Proof There are a number of different ways to prove the Binomial Theorem, for example by a straightforward application of mathematical induction. The Binomial Theorem also has a nice combinatorial proof: We can write . Repeatedly using the distributive property, we see that for a term , we must choose of the terms to contribute an to the term, and then each of the other terms of the product must contribute a . Thus, the coefficient of is the number of ways to choose objects from a set of size , or . Extending this to all possible values of from to , we see that , as claimed. Similarly, the coefficients of will be the entries of the row of Pascal's Triangle. This is explained further in the Counting and Probability textbook [AoPS]. Proof via Induction Given the constants are all natural numbers, it's clear to see that . Assuming that , Therefore, if the theorem holds under , it must be valid. (Note that for ) Proof using calculus The Taylor series for is for all . Since , and power series for the same function are termwise equal, the series at is the convolution of the series at and . Examining the degree- term of each, which simplifies to for all natural numbers. Generalizations The Binomial Theorem was generalized by Isaac Newton, who used an infiniteseries to allow for complex exponents: For any real or complex, , and , . Proof Consider the function for constants . It is easy to see that . Then, we have . So, the Taylor series for centered at is Usage Many factorizations involve complicated polynomials with binomial coefficients. For example, if a contest problem involved the polynomial , one could factor it as such: . It is a good idea to be familiar with binomial expansions, including knowing the first few binomial coefficients. See also Combinatorics Multinomial Theorem Retrieved from " Categories: Theorems Combinatorics Algebra Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
15758	https://svs.gsfc.nasa.gov/5187/	NASA SVS \| Moon Phase and Libration, 2024 Scientific Visualization Studio Galleries Help Search 5191: Global Temperature Graph 1880-20245188: Moon Phase and Libration, 2024 South Up A newer version of this visualization is available. 5415: Moon Phase and Libration, 2025 Moon Phase and Libration, 2024 Released Thursday, November 16, 2023 ID: 5187 Visualizations by: Ernie Wright Scientific consulting by: Noah Petro Produced by: David Ladd View full credits Download ###### Images moon.6512.jpg (730x730) comp.6512.tif (5760x3240) Enter a time to see what the moon looked like (or will look like) at that time. Time (UTC) [x] North Up \| Time (UTC) \| 9/29/2025, 7:00:00 AM \| \| Obscuration \| 00.0% \| \| Phase \| 43.31% (7d 11h 5m) \| \| Diameter \| 1788.3 arcseconds \| \| Distance \| 400780 km (31.41 Earth diameters) \| \| J2000 Right Ascension, Declination \| 17h 51m 41s, -28° 35' 33" \| \| Sub-Solar Longitude, Latitude \| 93.032°, -0.481° \| \| Sub-Earth Longitude, Latitude \| -4.652°, 6.715° \| \| Position Angle \| 1.122° \| The data in the table for all of 2024 can be downloaded as a JSON file or as a text file. Click on the image to download a high-resolution version with feature labels and additional graphics. Hover over the image to reveal the animation frame number, which can be used to locate and download the corresponding frame from any of the animations on this page, including unlabeled high-resolution Moon images. The animation archived on this page shows the geocentric phase, libration, position angle of the axis, and apparent diameter of the Moon throughout the year 2024, at hourly intervals. Until the end of 2024, the initial Dial-A-Moon image will be the frame from this animation for the current hour. More in this series: Moon Phase and Libration Gallery Lunar Reconnaissance Orbiter (LRO) has been in orbit around the Moon since the summer of 2009. Its laser altimeter (LOLA) and camera (LROC) are recording the rugged, airless lunar terrain in exceptional detail, making it possible to visualize the Moon with unprecedented fidelity. This is especially evident in the long shadows cast near the terminator, or day-night line. The pummeled, craggy landscape thrown into high relief at the terminator would be impossible to recreate in the computer without global terrain maps like those from LRO. The Moon always keeps the same face to us, but not exactly the same face. Because of the tilt and shape of its orbit, we see the Moon from slightly different angles over the course of a month. When a month is compressed into 24 seconds, as it is in this animation, our changing view of the Moon makes it look like it's wobbling. This wobble is called libration. The word comes from the Latin for "balance scale" (as does the name of the zodiac constellation Libra) and refers to the way such a scale tips up and down on alternating sides. The sub-Earth point gives the amount of libration in longitude and latitude. The sub-Earth point is also the apparent center of the Moon's disk and the location on the Moon where the Earth is directly overhead. The Moon is subject to other motions as well. It appears to roll back and forth around the sub-Earth point. The roll angle is given by the position angle of the axis, which is the angle of the Moon's north pole relative to celestial north. The Moon also approaches and recedes from us, appearing to grow and shrink. The two extremes, called perigee (near) and apogee (far), differ by as much as 14%. The most noticed monthly variation in the Moon's appearance is the cycle of phases, caused by the changing angle of the Sun as the Moon orbits the Earth. The cycle begins with the waxing (growing) crescent Moon visible in the west just after sunset. By first quarter, the Moon is high in the sky at sunset and sets around midnight. The full Moon rises at sunset and is high in the sky at midnight. The third quarter Moon is often surprisingly conspicuous in the daylit western sky long after sunrise. Celestial north is up in these images, corresponding to the view from the northern hemisphere. The descriptions of the print resolution stills also assume a northern hemisphere orientation. (There is also a south-up version of this page.) The phase and libration of the Moon for 2024, at hourly intervals. Includes supplemental graphics that display the Moon's orbit, subsolar and sub-Earth points, and the Moon's distance from Earth at true scale. Craters near the terminator are labeled, as are Apollo landing sites, maria and other albedo features in sunlight. Download ###### Movies phases_2024_fancy_720p30.mp4 (1280x720) [72.0 MB] phases_2024_fancy_1080p30.mp4 (1920x1080) [146.5 MB] phases_2024_fancy_360p30.mp4 (640x360) [24.2 MB] phases_2024_fancy_2160p30.mp4 (3840x2160) [509.2 MB] phases_2024_fancy_2160p30.mov (3840x2160) [15.1 GB] Frame sets fancy (1920x1080) [8785 Item(s)] fancy (5760x3240) [8785 Item(s)] fancy (3840x2160) [8785 Item(s)] Images comp.0001_print.jpg (1024x576) [109.8 KB] comp.0001_searchweb.png (320x180) [60.1 KB] comp.0001_thm.png (80x40) [6.0 KB] Hyperwall shows phases_2024_fancy_1080p30.mp4.hwshow [191 bytes] The phase and libration of the Moon for 2024, at hourly intervals. Includes supplemental graphics that display the Moon's orbit, subsolar and sub-Earth points, and the Moon's distance from Earth at true scale. Craters near the terminator are labeled, as are Apollo landing sites, maria and other albedo features in sunlight. The phase and libration of the Moon for 2024, at hourly intervals. Includes music, supplemental graphics that display the Moon's orbit, subsolar and sub-Earth points, and the Moon's distance from Earth at true scale. Craters near the terminator are labeled, as are Apollo landing sites, maria and other albedo features in sunlight.Music provided by Universal Production Music: “Go Win It” – Alexander HitchensThis video can also be viewed on the NASA Goddard YouTube channel. Download ###### Movies 2024_MoonPhaseNorthernHemisphere_YouTubeHD.mp4 (1920x1080) [568.3 MB] 2024_MoonPhaseNorthernHemisphere_YouTube4K.mp4 (3840x2160) [1.4 GB] 2024_MoonPhaseNorthernHemisphere_MASTER.mov (3840x2160) [15.2 GB] Captions 2024_MoonPhaseNorthernHemisphere_CAPTIONS.en_US.srt [46 bytes] 2024_MoonPhaseNorthernHemisphere_CAPTIONS.en_US.vtt [56 bytes] Images music.0001_print.jpg (1024x576) [121.9 KB] The phase and libration of the Moon for 2024, at hourly intervals. Includes music, supplemental graphics that display the Moon's orbit, subsolar and sub-Earth points, and the Moon's distance from Earth at true scale. Craters near the terminator are labeled, as are Apollo landing sites, maria and other albedo features in sunlight. Music provided by Universal Production Music: “Go Win It” – Alexander Hitchens This video can also be viewed on the NASA Goddard YouTube channel. The phase and libration of the Moon for 2024, at hourly intervals. The vertical (portrait) aspect ratio is targeted for viewing on mobile devices. Includes supplemental graphics that display the Moon's orbit, subsolar and sub-Earth points, and the Moon's distance from Earth at true scale. Craters near the terminator are labeled, as are Apollo landing sites, maria and other albedo features in sunlight. Download ###### Movies phases_2024_fancy_1920p30.mp4 (1080x1920) [140.9 MB] 2024_MoonPhaseNorthernHemisphere_Vertical.mp4 (1080x1920) [564.6 MB] Frame sets 1080x1920_9x16_30p (1080x1920) [8785 Item(s)] Images vertical.0001_print.jpg (576x1024) [121.9 KB] The phase and libration of the Moon for 2024, at hourly intervals. The vertical (portrait) aspect ratio is targeted for viewing on mobile devices. Includes supplemental graphics that display the Moon's orbit, subsolar and sub-Earth points, and the Moon's distance from Earth at true scale. Craters near the terminator are labeled, as are Apollo landing sites, maria and other albedo features in sunlight. The phase and libration of the Moon for 2024, at hourly intervals. The higher-resolution frames include an alpha channel. The .mov file is a ProRes 4444 movie with alpha. Download ###### Movies phases_2024_plain_1080p30.mp4 (1920x1080) [94.1 MB] phases_2024_plain_720p30.mp4 (1280x720) [41.4 MB] phases_2024_plain_360p30.mp4 (640x360) [11.2 MB] phases_2024_plain_2160p30.mp4 (3840x2160) [371.3 MB] phases_2024_plain_2160p30.mov (3840x2160) [31.1 GB] Frame sets plain (1920x1080) [8785 Item(s)] 730x730_1x1_30p (730x730) [8785 Item(s)] exr (5760x3240) [8785 Item(s)] plain (5760x3240) [8785 Item(s)] plain (3840x2160) [8785 Item(s)] 216x216_1x1_30p (216x216) [8785 Item(s)] Images moon.0001_print.jpg (1024x576) [46.8 KB] Hyperwall shows phases_2024_plain_1080p30.mp4.hwshow [191 bytes] The phase and libration of the Moon for 2024, at hourly intervals. The higher-resolution frames include an alpha channel. The .mov file is a ProRes 4444 movie with alpha. The Moon's Orbit The orbit of the Moon in 2024, viewed from the north pole of the ecliptic, with the vernal equinox to the right. The sizes of the Earth and Moon are exaggerated. Download ###### Movies orbit_2024_1080p30.mp4 (1080x1080) [36.1 MB] Frame sets 1080x1080_1x1_30p (1080x1080) [8785 Item(s)] 850x850_1x1_30p (850x850) [8785 Item(s)] 420x420_1x1_30p (420x420) [8785 Item(s)] Images orbit.0001_print.jpg (1024x1024) [108.9 KB] The orbit of the Moon in 2024, viewed from the north pole of the ecliptic, with the vernal equinox to the right. The sizes of the Earth and Moon are exaggerated. From this birdseye view, it's somewhat easier to see that the phases of the Moon are an effect of the changing angles of the Sun, Moon and Earth. The Moon is full when its orbit places it in the middle of the night side of the Earth. First and Third Quarter Moon occur when the Moon is along the day-night line on the Earth. The Sun's direction is indicated by the yellow arrow. The view here is perpendicular to the plane of the Earth's orbit around the Sun, called the ecliptic. The teal-colored ring is the plane of the Moon's orbit around the Earth, which is tilted about five degrees to the ecliptic. The thickness of the ring shows the range of the Moon's distance, and the darker half is the part below (south of) the ecliptic. The two points where the orbit crosses the ecliptic are the ascending and descending nodes. Also labeled are the perigee and apogee, the points along the orbit that are nearest to and farthest from Earth. The First Point of Aries is at the 3 o'clock position in the image. The Sun is in this direction at the March equinox. You can check this by freezing the animation at around the 1:03 mark, or by freezing the full animation with the time stamp near March 20. This direction serves as the zero point for both ecliptic longitude and right ascension. The north pole of the Earth is tilted 23.5 degrees toward the 12 o'clock position at the top of the image. The tilt of the Earth is important for understanding why the north pole of the Moon seems to swing back and forth. In the full animation, watch both the orbit and the "gyroscope" Moon in the lower left. The widest swings happen when the Moon is at the 3 o'clock and 9 o'clock positions. When the Moon is at the 3 o'clock position, the ground we're standing on is tilted to the left when we look at the Moon. At the 9 o'clock position, it's tilted to the right. The tilt itself doesn't change. We're just turned around, looking in the opposite direction. An animated diagram of the subsolar and sub-Earth points for 2024. The Moon's north pole, equator, and meridian are indicated. The frames include an alpha channel. Download ###### Movies subpnts_2024_960p30.mp4 (960x960) [37.5 MB] Frame sets 960x960_1x1_30p (960x960) [8785 Item(s)] 640x640_1x1_30p (640x640) [8785 Item(s)] 320x320_1x1_30p (320x320) [8785 Item(s)] Images globe.0001_print.jpg (1024x1024) [74.8 KB] An animated diagram of the subsolar and sub-Earth points for 2024. The Moon's north pole, equator, and meridian are indicated. The frames include an alpha channel. The subsolar and sub-Earth points are the locations on the Moon's surface where the Sun or the Earth are directly overhead, at the zenith. A line pointing straight up at one of these points will be pointing toward the Sun or the Earth. The sub-Earth point is also the apparent center of the Moon's disk as observed from the Earth. In the animation, the blue dot is the sub-Earth point, and the yellow cone is the subsolar point. The lunar latitude and longitude of the sub-Earth point is a measure of the Moon's libration. For example, when the blue dot moves to the left of the meridian (the line at 0 degrees longitude), an extra bit of the Moon's western limb is rotating into view, and when it moves above the equator, a bit of the far side beyond the north pole becomes visible. At any given time, half of the Moon is in sunlight, and the subsolar point is in the center of the lit half. Full Moon occurs when the subsolar point is near the center of the Moon's disk. When the subsolar point is somewhere on the far side of the Moon, observers on Earth see a crescent phase. An animated diagram of the Moon's distance from the Earth for 2024. The sizes and distances are true to scale, and the lighting and Earth tilt are correct. The frames include an alpha channel. Download ###### Movies distance_2024_720p30.mp4 (1280x720) [1.3 MB] distance_2024_1080p30.mp4 (1920x1080) [2.2 MB] distance_2024_360p30.mp4 (640x360) [653.0 KB] distance_2024_2160p30.mp4 (3840x2160) [6.3 MB] Frame sets distance (1920x1080) [8785 Item(s)] distance (5760x3240) [8785 Item(s)] distance (3840x2160) [8785 Item(s)] Images dist.0001_print.jpg (1024x576) [4.4 KB] Hyperwall shows distance_2024_1080p30.mp4.hwshow [187 bytes] An animated diagram of the Moon's distance from the Earth for 2024. The sizes and distances are true to scale, and the lighting and Earth tilt are correct. The frames include an alpha channel. The Moon's orbit around the Earth isn't a perfect circle. The orbit is slightly elliptical, and because of that, the Moon's distance from the Earth varies between 28 and 32 Earth diameters, or about 356,400 and 406,700 kilometers. In each orbit, the smallest distance is called perigee, from Greek words meaning "near earth," while the greatest distance is called apogee. The Moon looks largest at perigee because that's when it's closest to us. The animation follows the imaginary line connecting the Earth and the Moon as it sweeps around the Moon's orbit. From this vantage point, it's easy to see the variation in the Moon's distance. Both the distance and the sizes of the Earth and Moon are to scale in this view. In the HD-resolution frames, the Earth is 50 pixels wide, the Moon is 14 pixels wide, and the distance between them is about 1500 pixels, on average. Note too that the Earth appears to go through phases just like the Moon does. For someone standing on the surface of the Moon, the Sun and the stars rise and set, but the Earth doesn't move very much in the sky. It goes through a monthly sequence of phases as the Sun angle changes. The phases are the opposite of the Moon's. During New Moon here, the Earth is full as viewed from the Moon. Feature labels. Crater labels appear when the center of the crater is within 20 degrees of the terminator (the day-night line). They are on the western edge of the crater during waxing phases (before Full Moon) and to the east during waning phases. Mare, sinus, and lacus features are labeled when in sunlight. Apollo landing site labels are always visible. The frames include an alpha channel. Download ###### Movies labels_2024_1080p30.mp4 (1920x1080) [24.3 MB] labels_2024_720p30.mp4 (1280x720) [12.3 MB] labels_2024_360p30.mp4 (640x360) [3.8 MB] labels_2024_2160p30.mp4 (3840x2160) [91.3 MB] Frame sets labels (1920x1080) [8785 Item(s)] labels (5760x3240) [8785 Item(s)] labels (3840x2160) [8785 Item(s)] Images label.0001_print.jpg (1024x576) [9.4 KB] Hyperwall shows labels_2024_1080p30.mp4.hwshow [185 bytes] Feature labels. Crater labels appear when the center of the crater is within 20 degrees of the terminator (the day-night line). They are on the western edge of the crater during waxing phases (before Full Moon) and to the east during waning phases. Mare, sinus, and lacus features are labeled when in sunlight. Apollo landing site labels are always visible. The frames include an alpha channel. The Named Phases The following is a gallery containing examples of each of the Moon phases that have names in English. New, full, and quarter phases occur on specific days, while crescent and gibbous phases are the transitions between these points and span multiple days. The quarters are so named because they occur when the Moon is one fourth or three fourths of the way through its cycle of phases. Many people find this confusing, though, since visually they are half moons. It might be helpful to remember that the visible half of the Moon's disk is really only one quarter of its spherical surface. Download ###### Images phase_waxing_crescent.2028_print.jpg (1024x1024) [134.5 KB] phase_waxing_crescent.2028.tif (3240x3240) [10.7 MB] Waxing crescent. Visible toward the southwest in early evening. Download ###### Images phase_first_quarter.2091_print.jpg (1024x1024) [157.2 KB] phase_first_quarter.2091.tif (3240x3240) [10.3 MB] First quarter. Visible high in the southern sky in early evening. Download ###### Images phase_waxing_gibbous.2158_print.jpg (1024x1024) [163.6 KB] phase_waxing_gibbous.2158.tif (3240x3240) [9.3 MB] Waxing gibbous. Visible to the southeast in early evening, up for most of the night. Download ###### Images phase_full.1571_print.jpg (1024x1024) [184.7 KB] phase_full.1571.tif (3240x3240) [9.4 MB] Full Moon. Rises at sunset, high in the sky around midnight. Visible all night. Download ###### Images phase_waning_gibbous.2403_print.jpg (1024x1024) [192.9 KB] phase_waning_gibbous.2403.tif (3240x3240) [11.0 MB] Waning gibbous. Rises after sunset, high in the sky after midnight, visible to the southwest after sunrise. Download ###### Images phase_third_quarter.1755_print.jpg (1024x1024) [162.4 KB] phase_third_quarter.1755.tif (3240x3240) [11.2 MB] Third quarter. Rises around midnight, visible to the south after sunrise. Download ###### Images phase_waning_crescent.1810_print.jpg (1024x1024) [131.1 KB] phase_waning_crescent.1810.tif (3240x3240) [11.5 MB] Waning crescent. Low to the east before sunrise. Download ###### Images phase_new.1933_print.jpg (1024x1024) [91.8 KB] phase_new.1933.tif (3240x3240) [10.3 MB] New Moon. By the modern definition, New Moon occurs when the Moon and Sun are at the same geocentric ecliptic longitude. The part of the Moon facing us is completely in shadow then. Pictured here is the traditional New Moon, the earliest visible waxing crescent, which signals the start of a new month in many lunar and lunisolar calendars. Planets & Moons Albedo Elevation data HDTV Hyperwall Laser Altimeter LOLA LRO LROC Lunar Lunar Reconnaissance Orbiter Lunar Surface Lunar Topography Credits Please give credit for this item to: NASA's Scientific Visualization Studio Visualizer Ernie Wright (USRA) Scientist Noah Petro (NASA/GSFC) Producer David Ladd (Advocates in Manpower Management, Inc.) Missions This page is related to the following missions: LRO (Lunar Reconnaissance Orbiter) Series This page can be found in the following series: LRO - Animations The Moon Datasets used DEM (Digital Elevation Map) [LRO: LOLA] ID: 653 Sensor: LOLA See all pages that use this dataset DE421 (JPL DE421) ID: 752 Type: Ephemeris Collected by: NASA/JPL Planetary ephemerides This dataset can be found at: See all pages that use this dataset LROC WAC Color Mosaic (Natural Color Hapke Normalized WAC Mosaic) [Lunar Reconnaissance Orbiter: LRO Camera] ID: 1015 Type: Mosaic Sensor: LRO Camera Collected by: Arizona State University This natural-color global mosaic is based on the 'Hapke normalized' mosaic from LRO's wide-angle camera. The data has been gamma corrected, white balanced, and range adjusted to more closely match human vision. See all pages that use this dataset Note: While we identify the data sets used on this page, we do not store any further details, nor the data sets themselves on our site. Related ID: 5416 Visualization ### Moon Phase and Libration, 2025 South Up ID: 14551 Produced Video ### The Countdown Is On For The Historic Solar Eclipse On April 8th That Will Sweep Across the U.S. Are You Ready for It? ID: 14529 Produced Video ### First U.S. Commercial Provider Just Days From Landing NASA Science And Technology Instruments on the Moon ID: 5188 Visualization ### Moon Phase and Libration, 2024 South Up Newer Versions ID: 5415 Visualization ### Moon Phase and Libration, 2025 Older Versions ID: 5048 Visualization ### Moon Phase and Libration, 2023 Used as a Source In ID: 14886 Produced Video ### Artemis Science: Exploring the Moon’s South Pole Release date This page was originally published on Thursday, November 16, 2023. This page was last updated on Monday, June 23, 2025 at 12:16 AM EDT. 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15759	https://www.phyley.com/decompose-force-into-xy-components	Support Ukraine 🇺🇦 Help Ukrainian Army Humanitarian Assistance to Ukrainians Articles › Force and Motion › This article How to decompose a force into x and y components It is often useful to decompose a force into x and y components, i.e. find two forces such that one is in the x direction, the other is in the y direction, and the vector sum of the two forces is equal to the original force. Let's see how we can do this. Suppose we have a force F that makes an angle of 30° with the positive x axis, as shown below: And we want to decompose F into x and y components. The first thing we need to do is to represent the two components on the xy-plane. We do this by dropping two perpendiculars from the head of F: one to the x axis, the other to the y axis. Like this: And we join the origin of the xy-plane with the x-intercept to represent the x component of F: And again, we join the origin with the y-intercept to represent the y component of F: Fx and Fy are two vectors, i.e. they both have a magnitude and a direction. However, since Fx and Fy are in the directions of the x and y axes, they are commonly expressed by the magnitude alone, preceded by a positive or negative sign: positive when they point in the positive directions, and negative when they point in the negative directions of the x and y axes. In our example Fx and Fy are positive because both point in the positive directions of the x and y axes. The positive values of Fx and Fy can be found using trigonometry: Fx = F cos 30° Fy = F sin 30° To keep it simple, just remember that if a component is adjacent to the angle, then it is cos, otherwise it is sin. Often Fx will be the component adjacent to the angle, so it will be cos, and Fy will be sin. Let's now consider a force that has one of its components negative: In this case Fx is negative because it points in the negative direction of the x axis. Therefore: Fx = −F cos 15° Fy = F sin 15° Notice the minus sign before F cos 15° which we have added to make Fx negative. You have to be very careful if your angle is not between 0° and 90°, because the sin or(and) cos of that angle may be already negative, so the product is also negative and you don't need to add a minus sign. To be on the safe side, we recommend to always work with angles between 0° and 90°, so that the sin and cos are always positive, and therefore the product is also always positive. The bottom line We can summarize the process of decomposing a force F, as follows: Represent the x and y components of the force on the xy-plane by dropping perpendiculars from the head of the force to the x and y axes, and then joining the origin of the xy-plane with the two intercepts (the goal of graphically representing the components is to help you see which component is adjacent to the angle and what the signs of the two components are). Find the values of the x and y components: the component adjacent to the angle will be F cos θ and the other will be F sin θ. Components that point in the negative directions of the x and y axes are negative, therefore you will need to add a minus sign (given that you are working with θ between 0° and 90°, so that F cos θ and F sin θ are always positive). Forces with tail not in the origin Sometimes a force does not have the tale in the origin of the xy-plane. For example: In cases like this, we draw two straight-lines parallel to the x and y axis that pass through the tail of the force, and then we drop two perpendiculars from the head of the force to the straight-lines: Fx = F cos 30° Fy = F sin 30° Forces that are already in the x or y direction Often we deal with forces that are already in the x or y direction. In that case we can determine the x and y components in a simpler and more intuitive way, without using trigonometry. If, for example, we have a force F that is in the direction of the positive x axis: It is obvious that the y component of F is 0, and the x component is positive with magnitude equal to the magnitude of F: Fx = F Fy = 0 On the other hand, if we have a force F in the direction of the negative x axis: Then the y component is again 0, and the x component is negative (because it points in the negative direction of the x axis) and has the same magnitude as F: Fx = −F Fy = 0 The same can be shown for forces in the y direction: They will always have x component 0, and y component either positive or negative with magnitude equal to the magnitude of the force. To verify your understanding of the concept, do the exercises below. Exercises #1 A force of 19 N is in the direction of the negative x axis. Find the x and y components of the force. Solution Fx is negative, and has the same magnitude as the force (19 N). Fy is zero. Fx = −19 N Fy = 0 N #2 A force of 114 N makes an angle of 67° with the positive x axis. Decompose the force into x and y components. Solution Both the components are positive. Fx = F cos 67° = 44.5 N Fy = F sin 67° = 105 N #3 A force makes an angle of 221° with the positive x axis. Assuming the force has magnitude 3.1 × 103 N, find the x and y components. Solution Instead of dealing with the 221° angle, we want to deal with the 41° angle (221° − 180°) that the force makes with the negative x axis: As you can see, both Fx and Fy are negative: Fx = −F cos 41° = −2.3 × 103 N Fy = −F sin 41° = −2.0 × 103 N #4 A force of 4.5 × 105 N has the direction of the positive y axis. Determine its components. Solution This one is simple. Fx is zero. Fy is positive and has the same magnitude as the force. Fx = 0 N Fy = F = 4.5 × 105 N #5 A 90.0 N force makes an angle of 33° with the positive y direction. Calculate the x and y components of the force. Tip: Since the angle is +33°, it goes counterclockwise from the positive y axis. Solution In this case Fy is adjacent to the angle, therefore its magnitude is the force times the cos of the angle, while the magnitude of Fx is the force times the sin. Also notice that Fx is negative: Fx = −F sin 33° = −49.0 N Fy = F cos 33° = 75.5 N Alternatively you could have considered the 57° angle (90° − 33°) which F makes with Fx. That way Fx would have been adjacent to the angle. #6 A force that has magnitude 3.21 × 104 N makes an angle of −50° with the positive x axis. Determine the components. Tip: The angle is negative, meaning it goes clockwise from the positive x axis. Solution Fx = F cos 50° = 2.06 × 104 N Fy = −F sin 50° = −2.46 × 104 N Problems with solutions To further verify your understanding of force decomposition, see our force problems, which include problems where you need to decompose forces acting on objects that move horizontally, move up an incline, and hang from ropes. 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15760	https://www.cs.virginia.edu/~grimshaw/publications/Framework_Partitioning_Parallel_Computations_95.pdf	1 A Framework for Partitioning Parallel Computations in Heterogeneous Environments Jon B. Weissman Andrew S. Grimshaw Department of Computer Science University of Virginia Abstract In this paper we present a framework for partitioning data parallel computations across a heterogeneous meta-system at runtime. The framework is guided by program and resource information which are made available to the system. Three difficult problems are handled by the framework: processor selection, task placement, and heteroge-neous data domain decomposition. Solving each of these problems contributes to reduced elapsed time. In particular, processor selection determines the best grain size at which to run the computation, task placement reduces communi-cation cost, and data domain decomposition achieves processor load balance. We present results that indicate excel-lent performance is achievable using the framework. This paper extends our earlier work in partitioning data parallel computations across a single-level network of heterogeneous workstations1. 1.0 Introduction A great deal of recent interest has been sparked within academic, industry, and government circles in the emerging technology of metasystem-based high-performance computing. A metasystem is a shared ensemble of workstations, vector, and parallel machines connected by local- and wide-area networks, see Figure 1. The promise of on-line gigabit networks coupled with the tremendous computing power of the metasystem makes it very attractive for parallel computations. 1. This work has been partially funded by grants NSF ASC-9201822, JPL-959303, and NASA NGT-50970. M Shared Memory Workstations Vector Machine Mesh-based multicomputer Network backbone Figure 1. A typical metasystem Hypercube-based multicomputer 2 The potentially large array of heterogeneous resources in the metasystem offers an opportunity for delivering high performance on a range of parallel computations. Choosing the best set of available resources is a difficult problem and is the subject of this paper. Consider the set of machines in Table 1 and observe that they have different computation and communication capacities. Loosely-coupled parallel computations with infrequent communication would likely benefit by applying the fastest set of computa-tional resources (perhaps the DEC-Alpha cluster), and may benefit from distribution across many machines. On the other hand, more tightly-coupled parallel computations are best suited to machines that have a higher communication capacity (perhaps an Intel Paragon), but may also benefit from distribution across many machines if the computation granularity is sufficient. We address the latter problem in this paper. We present a framework that automates partitioning and placement of data parallel computations across metasystems such as in Figure 1. Partitioning is performed at runtime when the state of the metasys-tem resources are known. Three difficult problems are handled by the framework: processor selection, task placement, and heterogeneous data domain decomposition. Solving each of these problems contributes to reduced completion time. Processor selection chooses the best number and type of processors to apply to the computation. This is important because if too many processors are chosen the computation granularity will be too small, and if too few processors are selected the computation granularity will be too large. In either case, the elapsed time will be increased. Task placement is performed to limit communication costs, while data domain decomposition is done to achieve processor load balance. To solve each of these problems, we exploit information about the computation and communication structure of the data parallel program and the computation and communication capacities of the metasys-tem resources. The former is provided by a set of callback functions that are described. The partitioning framework is based on three runtime phases: processor availability, partitioning, and instantiation. Processor availability is a runtime monitoring process that determines the available set of processors based on the current usage of metasystem resources. Availability depends on the machine class. Machine Latency Bandwidth Processor speed iPSC/2 moderate moderate slow Sparc ethernet cluster high low moderate DEC-Alpha cluster on Gigaswitch high high very fast Paragon low very high fast Cray T3D low very high very fast Sequent Symmetry low low slow Table 1. A spectrum of heterogeneity 3 For example, workstation availability is based on system load, while multicomputer availability is based on partition or subcube availability. Partitioning determines the number and type of processors to use from the available set, the placement of tasks across processors, and a decomposition of the data domain. Instan-tiation is an application-dependent phase that starts the data parallel computation across processors selected by partitioning using the determined task placement. We begin by describing our metasystem model and the communication and processor resource infor-mation that supports the processor availability and partitioning phases. We discuss placement in the con-text of the metasystem communication model. Next we describe the data parallel computation model and the program information that is used to guide partitioning and placement. We then describe the partitioning method and an implementation to show how the resource and program information is used. Experimental and simulation results indicate that partitioning can be effectively automated for a large and useful class of data parallel computations. We also show that the partitioning can be done efficiently and excellent perfor-mance is achievable for these computations. 2.0 Framework 2.1 Metasystem Model The metasystem illustrated in Figure 2 contains heterogeneous machines organized in a hierarchy. The basis of this organization is the processor cluster denoted by the large circles. A processor cluster con-tains a homogeneous family of processors that may contain workstations, vector, or parallel machines. For example, processor clusters range from tightly-coupled multiprocessors such as a Sequent in which proces-sors communicate via shared-memory to distributed-memory multicomputers such as a Paragon or loosely-coupled workstations such as a Sun 4 cluster in which processors communicate via messages. Communication between processors in different processor clusters is accomplished by message-pass-ing. Taken as a whole, the metasystem is a multi-level distributed-memory MIMD machine. We will use the following notation throughout this section: Ni = the ith network cluster Ci = the ith processor cluster τ = application communication topology b = message size in bytes c1 .. c4 = communication cost coefficients f () = processor-dependent communication function r1, r2 = router cost constants e1 = coercion cost constant k = number of messages that cross through the router 4 A network cluster contains one or more processor clusters and is denoted by the boxes labelled N1, N2 and N3 in Figure 2. The key property of a network cluster is that it has private communication bandwidth with respect to other network clusters, and shared bandwidth with respect to the processor clusters it con-tains. For example, the total available bandwidth in the metasystem of Figure 2 is the sum of the bandwidth in N1, N2 and N3, but the available bandwidth in N1 is shared between the Sun 4 and SGI clusters. Network clusters are connected by one or more routers. For wider-area metasystems, we define network clusters hierarchically as shown in Figure 3. Here N4 is a network cluster that contains N1, N2 and N3. The metasystem organizations in Figure 2-3 are locality-based in which processors are grouped by physical connectivity. This has important implications for partitioning parallel computations in which min-imizing communication overhead is key. For example, communicating processors within a processor clus-ter pay no routing penalty, while processors in different processor clusters pay a routing penalty. If the processors are within the same network cluster than a single hop routing penalty is incurred. We will see later in this section that task placement exploits this locality information. In this paper, we focus on local-area metasystems such as in Figure 2 and restrict ourselves to multi-computers and workstations. Current network technology limits the benefit that tightly-coupled data paral-lel computations will receive by wider-area distribution due to high and often unpredictable communication costs. However, this is likely to change when the gigabit networks become more wide-spread. We also make the simplifying assumption of one processor cluster per network cluster. This assumption allows us to present a simpler communication model and partitioning method. We will refer only to processor clusters in the remainder of this paper. Figure 2. Cluster-based metasystem organization R SGI Paragon Sun4 Sequent N1 N2 N3 N4 N1 N2 N3 R N7 Figure 3. Wider-area metasystem organization R N5 N6 R 5 Processor clusters maintain important resource information that will be exploited by partitioning and placement (the is not implemented and the information in bold will be used in this paper): •Interconnection topology •Processors (total, avail) •Aggregate power (mflops, mips) •Communication functions •B/W (peak, effective, avail) •Latency (idle, effective) •Manager This information is stored by a processor we call the manager and is denoted by the shaded circle in Fig-ure 2. The topology refers to bus (ethernet), ring (FDDI), mesh (multicomputer), and hypercube (multi-computer) and may be different for different processor clusters. The peak bandwidth is the maximum communication bandwidth achievable for this cluster based on the processor type (assuming idle machines and network). Because this environment is dynamic and shared, both bandwidth and process-ing resources may be committed to other computations. The effective bandwidth may be less if proces-sors in the cluster are loaded or unavailable, and the available bandwidth is the amount available to us based on the current traffic profile. The relationship between these quantities is: peak < effective < avail-able. At present, we do not have a network monitor so information about the available bandwidth is not collected. The available processors in a cluster may be less than the total number due to sharing. For example, when a workstation gets highly loaded or multicomputer partition or subcube gets allocated, we mark these processors as unavailable. The managers monitor the system state to determine availability within a cluster. This is known as processor availability and is the first phase of partitioning. For workstations, a simple load threshold policy is used to determine availability. We treat all processors below this thresh-old as available and equal in terms of computation and communication capacity. A more general strategy that is being investigated is to allow all processors to be available, but with adjustments to both commu-nication capacity (i.e., effective bandwidth) and computation power (i.e., aggregate power) based on pro-cessor load. Operating system facilities are provided in Unix (uptime, kmem) and NX (pspart, cubeinfo) to determine processor availability. We do not discuss this process in any detail in this paper. The reader may assume that we can determine the available processor pool needed during partitioning. Aggregate power is the cumulative processing power based on the peak instruction rate for the processor type and the number of available processors. 6 2.2 Communication The communication functions determine the cost of communication within the cluster and are the cor-nerstone of partitioning and placement. Choosing the appropriate number and type of processors depends on the communication cost that results from this selection. For example, choosing too many processors results in a small computation granularity and increased elapsed time due to high communication costs. Partitioning uses a set of functions to estimate communication costs for candidate processor selections. In metasystems consisting of workstations and Intel multicomputers, communication is enabled by a reliable heterogeneous message-passing system (MMPS) . MMPS uses UDP datagrams for communi-cation among workstations and between processors in different clusters and NX for communication among processors in Intel multicomputer clusters. Partitioning requires that an accurate estimate of communica-tion costs in the metasystem be known. Consider the simple case where all communication occurs within a cluster Ci. The communication cost function for Ci depends on the application communication topology and the interconnection topology of Ci. The particular cost experienced by an application depends on two application-dependent parameters provided to this function: (1) the size of messages exchanged, and (2) the number of communicating pro-cessors. In the model we present in the next section, these parameters will be known at runtime. We are exploring three communication topologies often found in data parallel computations: 1-D, ring, and tree. The 1-D is common in scientific computing problems based on grids or matrices and is class of near-est neighbor topologies. In the 1-D topology processors simultaneously send to their north and south neighbors and then receive from their north and south neighbors. The ring topology is common to systolic algorithms and pipeline computations. In the ring topology communication is much more synchronous. A processor receives from its left neighbor and then sends to its right neighbor. The tree topology is used for global operations such as reductions. In the fan-in, fan-out tree topology communication occurs in two phases. In fan-in a parent processor receives from all of its children before sending to its parent, while chil-dren simultaneously send to their parent. Once the root receives from its children the process is repeated in reverse during fan-out. These communication topologies are synchronous in that all processors participate in the communica-tion collectively. The synchronous nature of the communication means that the communication cost expe-rienced by all processors is roughly the same and is determined by the processor experiencing the greatest cost. This observation has been verified by empirical data. A set of accurate communication cost functions can be constructed for each cluster by benchmarking a set of topology-specific communication programs. These cost functions determine the average communi-cation cost, measured as elapsed time, incurred by a processor during a single communication cycle. A 7 cycle corresponds to a single iteration of the computation. For example in a single cycle of a ring commu-nication, a processor receives one message from its left neighbor and sends one message to its right neigh-bor. For each cluster Ci and communication topology τ, we have a communication cost function of the form: Tcomm [Ci, τ] (b, p). The cost function is parameterized by p, the number of communicating processors within the cluster, and b, the number of bytes per message. For example suppose C1 refers to the SGI cluster in Figure 2. The cost function Tcomm [C1, 1-D] (b, p) refers to the average cost of sending and receiving a b byte message in a 1-D communication topology of p processors within the SGI cluster computed as elapsed time. The cost functions have a latency term that depends on p and a bandwidth term that depends on both p and b (c1 and c2 are latency constants and c3 and c4 are bandwidth constants): Tcomm [Ci, τ] (b, p) = c1+c2 f(p)+ b(c3+c4 f(p)) (EQ 1) The function f depends on the cluster interconnect and the communication topology. For example, on eth-ernet we often see f linear in p for all communication topologies due to contention for the single ethernet channel. On the other hand, richer communication topologies such as meshes and hypercubes have greater communication bandwidth that scales more easily with the number of processors. For example, we have observed that for tree communication on a mesh, f is log in p. Each communication cost function is bench-marked using different p and b values to derive the appropriate constants. A set of communicating tasks are mapped over the processors to perform the benchmarking. The placement of tasks depends on the commu-nication and interconnection topologies and is discussed later in this section. If we decide to use processors within a particular Ci only, then the cost function in (EQ 1) determines the communication cost. If processors in several clusters are used, then communication will cross cluster boundaries and two additional costs come into play: Trouter [Ci, Cj] (b) = r1+r2 b (EQ 2) Tcoerce [Ci, Cj] (b) = e1b The router cost includes a latency penalty r1 and a per byte penalty r2 that captures any delay or buffering required in routing a message from a processor in Ci to a processor in Cj. Since proces-sors in different clusters may support different data formats, coercion may be needed. Coercion is paid as a per byte processor cost e1 (e.g., endian conversion) by the sending or receiving proces-sor. These costs are determined by benchmarking. Suppose that processors in Ci and Cj are com-municating and no other clusters are used. The communication cost for processors in Ci becomes the sum of the previous cost equation in (EQ 1) plus several new terms (Cj may be written simi-larly): 8 Tcomm [Ci, τ] = Tcomm [Ci, τ] + k (Trouter [Ci, Cj] + Tcoerce [Ci, Cj]) where k is the number of messages that cross between Ci and Cj per cycle. The router also increases contention for communication bandwidth and this is discussed in . The total com-munication cost experienced by all processors, which we denote by Tcomm [τ], depends on the topology τ: Tcomm [1-D] = maxi {Tcomm [Ci, 1-D]} (EQ 3) Tcomm [ring] = sumi {Tcomm [Ci, ring]} Tcomm [tree] = Tcomm [Croot, tree] + maxi∈leaves {Tcomm [Ci, tree]} For the 1-D topology, all processors communicate simultaneously and the total communication cost is lim-ited by the slowest cluster (i.e., the cluster with the largest communication cost). For the synchronous ring topology, the communication cost is additive. The tree topology is more complicated. It has both concur-rent communication (e.g., the leaves communicate simultaneously), and synchronous communication (e.g., communication is ring-like along the critical path). The cost is defined recursively. Empirical evidence suggests that these cost functions are accurate. The benefit of this approach is that very accurate topology-specific communication costs can be estimated and these costs are key to making effective partitioning decisions. We are currently studying the impact of load on communication cost and how the communication cost functions may be adjusted at runtime to reflect reduced bandwidth. Once the cost functions are constructed they can be applied to other data parallel computations that contains these topologies. Placement The communication cost functions in (EQ 1-3) are benchmarked using a set of task placement strate-gies. Task placement assigns tasks to processors in a communication efficient manner. Reducing commu-nication costs is achieved by (1) maintaining communication locality (i.e., avoiding router crossings and potential coercion) and (2) effectively exploiting communication bandwidth within clusters. The former is achieved by inter-cluster placement and the objective is to minimize communication costs between clus-ters. The latter is achieved by intra-cluster placement and the objective is to minimize communication costs within clusters. Intra-cluster placement is also known as mapping or embedding and has been widely studied . Algorithms for both forms of placement are topology-specific. Inter-cluster placement depends on the communication topology. In Figure 4, we present several inter-cluster placement strategies for the 1-D, ring, and tree topologies across several clusters (the black boxes are routers). Notice that the number of router crossings or communication hops are minimized. 9 Intra-cluster placement is dependent on the communication topology (as is inter-cluster placement) and on the interconnection topology. Intra-cluster placement maps tasks to specific processors within a cluster. Two factors that contribute to intra-cluster communication costs are dilation (or hops) and conten-tion. Intra-cluster placement should keep the average dilation small and limit contention. For example, a grey-scale mapping of a 1-D topology onto a hypercube achieves minimal dilation and contention and has reduced communication overhead. On the other hand, a random placement suffices on a bus interconnect for any communication topology. High dilation and contention will tend to limit the exploitable communi-cation bandwidth. There is a rich literature on the mapping problem and many of the algorithms are well known. Traditionally these algorithms have been applied within a static compile-time scheduling frame-work. Our approach will apply these algorithms in a novel way: within in a runtime partitioning frame-work. The algorithms for inter- and intra-cluster placement are used to benchmark the communication functions and are canned for use at runtime during partitioning. 2.3 Data Parallel Computations A common approach for implementing data parallel computations in MIMD environments is SPMD . In the SPMD model, the computation is performed by a set of identical tasks, placed one per processor, each assigned a different portion of the data domain. We have adopted a dynamic SPMD model in which tasks are instantiated at runtime based on the processor selection. We assume that the task implementation has been provided either by the user or as a result of a compilation process. A task executable for each architecture type is assumed to exist. The data domain is decomposed into a number of primitive data units or PDUs, where the PDU is the smallest unit of data decomposition. The PDU is problem and application specific. For example, the PDU might be a row, column, or block of a matrix in a matrix-based problem, or a collection of particles in a particle simulation. The PDU is similar to the virtual processor and may arise from unstructured data domains. The partitioning method does not depend on the nature of the PDU but rather manipulates PDUs in the abstract. Two views of the data parallel computation are provided to the partitioning framework: task view and phase view. In the task view, the computation is represented as a set of communicating tasks. The task view provides important topology information that is needed by placement. In the phase view, the compu-Figure 4. Inter-cluster placement (a) 1-D and ring (b) tree C1 C2 C1 C2 C3 10 tation is represented as a sequence of alternating computation and communication phases . These phases are more tightly-coupled than the phases discussed in which require data redistribution. A communication phase contains a synchronous communication executed by all processors as discussed in Section 2.2. A computation phase contains only computation. Communication and computation phases may be overlapped. Most data parallel computations are iterative with the computation and communication phases repeating after some number of phases. This is known as a cycle. The phase view provides important information that is needed by partitioning. This information is provided by callbacks functions. The callbacks are a set of runtime functions that provide critical informa-tion about the communication and computation structure of the application that is used by the partitioning method. We discuss callback specification later in this section and present an implementation of callbacks in Section 2.5.1. Computation phase callbacks Each computation phase must have the following callbacks defined: • numPDUs • comp_complexity • arch_cost The number of PDUs manipulated during a computation phase, numPDUs, depends on problem parame-ters (e.g., problem size). The amount of computation performed on a PDU in a single cycle is known as the computation complexity, comp_complexity. It has two components: the number of instructions executed per PDU, and the number of instructions executed that are independent of the number of PDUs. The archi-tecture-specific execution costs associated with comp_complexity are captured by arch_cost, provided in units of usec/instruction. The arch_cost contains an entry for each processor type in the target metasystem. To obtain the arch_cost, the sequential code must be benchmarked on each processor type. Communication phase callbacks Each communication phase must have the following callbacks defined: • topology • comm_complexity • overlap The topology refers to the communication topologies discussed in Section 2.2. The amount of communica-tion between tasks is known as the communication complexity, comm_complexity. It is the number of bytes transmitted by a task in a single communication during a single cycle of the communication phase. Similar to comp_complexity, it has two components: the number of bytes transmitted per PDU and the number of bytes transmitted that are independent of the number of PDUs. It is used to determine the parameter b in the communication cost equations. If a communication phase is overlapped with a computa-11 tion phase, then the name of the computation phase is provided by the overlap callback. Among the com-putation and communication phases, two phases are distinguished. The dominant computation phase has the largest computation complexity, while the dominant communication phase has the largest communica-tion complexity2. A simple example that illustrates the callbacks for a regular NxN five-point stencil computation is given in Figure 5 (the arch_cost is omitted). These are functions that return the values indicated. For comp_complexity we show only the PDU-dependent cost and for comm_complexity we show the non-PDU-dependent message size. This computation has been implemented using a block-row decomposition of the grid. In this application, the PDU is a single row and the processors are arranged in a 1-D communi-cation topology. The stencil computation is iterative and consists of two dominant phases: a 1-D communi-cation to exchange north and south borders, and a simple computation phase that computes each grid point to be the average of its neighbors. Notice that the callback functions may depend on problem parameters (e.g., N) that are unknown until runtime. The callbacks associated with the dominant phases are used by the partitioning method. In partic-ular, the callbacks associated with the computation and communication complexity allow an estimate of the computation granularity to be computed at runtime. This estimate is used to determine the number of processors to use. The topology is used to select the appropriate communication function. The computation complexity is also used to determine a decomposition of the data domain, i.e., the number of PDUs to be assigned to each task. In a heterogeneous metasystem environment, processors may be assigned different numbers of PDUs for load balance. This information is contained in a structure known as the partition_map that is defined as follows: Ai = number of PDUs assigned to processor pi ∑Ai = numPDUs The partition_map has an entry for each processor and the association of its entries to processors is topol-ogy-dependent, see Figure 6. The topology-dependence reflects the data locality relationships in the prob-2. The dominant phases may be problem or runtime dependent, and this is discussed in Section 2.5.1. topology ⇒ 1-D comm_complexity ⇒ 4N (bytes) numPDUs ⇒ N comp_complexity ⇒ 5N (fp ops) Data domain (NxN) (a) Stencil computation (b) Callbacks for stencil Figure 5. Example: 1-D stencil computation processors 12 lem. This information is needed when the data domain is decomposed to processors. For example in the 1-D stencil problem of Figure 5, a 20x20 grid might be decomposed across four processors as shown Figure 6a (processor 1 gets the first 2 PDUs or rows, processor 2 gets the next 5 PDUs, and so on). The partition_map is a logical decomposition of the data domain and is computed at runtime by the partitioning method. The application is responsible for using it in a manner appropriate to the problem. In Section 2.5, we sketch an implementation in which the grid is physically decomposed using the partition_map and the pieces are passed to the appropriate tasks. Callback Specification The callback mechanism is very powerful and can be applied to data parallel computations less regu-lar than the five-point stencil. Since the callbacks may be arbitrary and complex functions, they can handle data-dependent computations by pre-processing the data domain. For example, the computation complex-ity for a sparse matrix problem depends on the non-zero structure of the matrix, and this is problem-depen-dent. But a simple callback can be written to capture this dependence. We are currently doing this with a finite-element code that we have written previously . For irregular or control-dependent data parallel computations, it is likely that off-line benchmarking of the sequential code is needed to determine average values for some callbacks. A simple example is Gaussian elimination in which the amount of computation and communication changes from cycle to cycle. For irregular, control- or data-dependent computations it is likely that the domain programmer will have to write the callback functions by-hand. For regular problems such as the stencil computation, we believe that current compiler technology may be able to generate the callbacks if sufficient language sup-port is provided. In the absence of compiler technology, we believe that domain programmers should be able to provide the callbacks. Another strategy is to provide libraries of callbacks for well-known computa-tional structures and this technique is discussed in Section 2.5. Current compiler technology also looks promising for generating the SPMD task implementation and automatically decomposing the data domain for regular problems . 2 5 5 8 2 5 5 8 2 5 5 8 (a) 1-D (b) 2-D (c) tree Figure 6. Topology-dependent partition_map (numPDUs = 20) 13 2.4 Partitioning Partitioning is the second stage in the three stage process introduced in Section 1.0. Once processor availability has determined the available set of processors (the first phase), the partitioning phase computes the best subset of processors to use and a distribution of the data domain across these processors. Choosing the subset of processors in known as processor selection and the objective is to determine the best grain size at which to run the computation. Dividing the data domain is known as data domain decomposition, and the objective is to achieve processor load balance. Data domain decomposition was introduced in Sec-tion 2.3. A load balanced decomposition and an appropriate computation granularity lead to reduced com-pletion time and this is the objective of partitioning. Partitioning is performed once at runtime. Dynamic repartitioning in the event of load imbalance could be accommodated with the framework, but it is outside the scope of this paper. We will use the following notation throughout this section: pi = a particular processor Ai = number of PDUs assigned to processor pi Vi = number of available processors within cluster Ci Pi = number of processors selected for Ci wi = relative processor weight for ith processor based on arch_cost m = number of clusters g() = the amount of computation as a function of Ai d(i) 1 = non-PDU cost constant for ith processor d(i) 2 = PDU cost constant for ith processor Tc = per cycle elapsed time Tstartup = startup overhead Tcomm = per cycle communication cost Tcomp = per cycle computation cost More formally, we define a processor configuration as a set of processors Pi (0≤Pi≤Vi, i=1 to m), where Vi is the number of processors available within Ci. Partitioning computes a processor configuration that yields an appropriate computation granularity and a decomposition of the data domain Ai associated with each pi (i.e., the partition_map) that load balances the processors. We discuss each of these problems in turn, beginning with data domain decomposition. 2.4.1 Data Domain Decomposition We compute a load balanced decomposition based on the dominant computation phase. The amount of time spent in a single cycle of the dominant computation phase, denoted by Tcomp, is defined as follows (shown for a processor pi): Tcomp [pi] = comp_complexity arch_cost g(Ai) The computation time depends on the problem and processor characteristics and on number of PDUs, Ai, given to each processor. In the general case, the dependence on Ai may be arbitrary function g of Ai. In 14 practice however, it is common that the dependence is linear for SPMD computations. Invoking callbacks for comp_complexity and arch_cost at runtime determines two cost constants that are problem-dependent: d1 is a non_PDU cost and d2 is a PDU cost. The form for Tcomp becomes (shown for processor i): Tcomp [pi] = d1 (i) + d2 (i)g(Ai) (EQ 4) Load balance requires that Tcomp be the same for all processors (P total processors): d1 (1) + d2 (1)g(A1) = d1 (2) + d2 (2)g(A2) = ... d1 (P) + d2 (P)g(AP) subject to ∑Ai = numPDUs If g is non-linear then this is a difficult system to solve. Fortunately, g is most often linear for SPMD com-putations in which the same computation is performed on each data element (i.e., PDU) independently. If g is linear, we can combine these equations easily (Pj comes in with the second constraint): (EQ 5) If we assume that the non_PDU cost is 0 (i.e., d1 (i)= 0), we get a simple equation for the partition_map: (EQ 6) This equation has the property that faster processors will receive a greater share of the data domain and processors in the same cluster will receive an equal share. Since Ai must be integral, the individual entries in the partition_map are rounded to the nearest integer. Computing the partition_map via (EQ 6) requires knowing the number of processors to be used (i.e., Pj). This is the subject of processor selection, discussed next. 2.4.2 Processor Selection The job of processor selection is to choose a subset of available processors that are best applied to the computation. Nearly all parallel computations reach a point of diminishing returns with respect to the num-ber of processors due to communication overhead. At this point, we have achieved the best computation grain for the problem and the elapsed time will be minimized. Because processor power and cluster com-munication capacities differ in the metasystem, locating this point is difficult. Our method is a heuristic that is guided by runtime cost estimation that uses information provided by the callbacks of Section 2.3. The elapsed time Telapsed, may be defined as follows: Ai wi wj Pj ⋅ ---------------j ∑         NumPDUs pj d1 j ( ) d1 i ( ) – d2 j ( ) -------------------------j i ≠ ∑ – wi d2 i ( ) max d2 j ( )       ---------------------------= , ⋅ = Ai wi wj Pj ⋅ --------------- NumPDUs ⋅ j ∑ = 15 or (EQ 7) , where (EQ 8) The startup overhead, Tstartup, may include initial data distribution costs or problem setup costs. The amount of time spent in the ith iteration or cycle is denoted by Tc[i] and the average over all Tc[i] is denoted by Tc. If Tstartup is small relative to the elapsed time, then minimizing Telapsed can be achieved by minimiz-ing Tc. Tc contains both computation and communication cost terms and is defined as follows: Tc = Tcomp + Tcomm or (EQ 9) Tc = max {Tcomp, Tcomm} if computation and communication are overlapped. Since Tc is assumed to be an average, Tcomp and Tcomm must be computed as averages in the event that they differ from cycle to cycle. An example of this is Gaussian elimination discussed in Section 3.1. Our heuris-tic computes Tcomp and Tcomm based on the dominant computation and communication phases3. The form for Tcomp was given in (EQ 4) and the form for Tcomm was given in (EQ 3). The appropriate Tcomm will be selected by the topology of the dominant communication phase. The function Tc is non-linear in the num-ber of processors and possibly non-convex (if max appears). The optimal solution therefore requires mini-mization of a non-linear, possibly non-convex objective function, subject to constraints, and heuristics must therefore be used. We describe two heuristics for processor selection, H1 and H2, that have yielded promising results. Both heuristics explore a series of processor configurations in an attempt to achieve a minimum Tc, hence minimized elapsed time. For each configuration explored, we compute Tc via (EQ 9). To do this we first compute the partition_map Ai via (EQ 6). Once Ai is determined, we can compute Tcomp and Tcomm easily by invoking the callbacks and selecting the appropriate communication function. All of these computations are simple and can be performed efficiently at runtime. For a given configuration, the placement heuristics are used to determine task placement and the expected communication costs that result using this place-ment are computed by Tcomm. It is not possible to explore all processor configurations since the space is exponential in both the number of processors and clusters. The first step of the processor selection heuristics are to order the clus-ters in order to reduce the search space. The idea is that the best clusters should be explored first. Heuristic H1 3. A more accurate and computationally expensive technique is to compute Tc for all phases. Telapsed Tstartup Tc i [ ] i 1 = cycles ∑ + = Telapsed Tstartup = cycles Tc ⋅ + Tc Tc i [ ] ∑ cycles -------------------= 16 Heuristic H1 has been designed for workstation network environments in which communication capacities are the same within the metasystem (e.g., ethernet-based clusters), and routing costs are high. The algorithm begins by ordering the clusters based on aggregate computation power. In this environment, the clusters with the highest aggregate computation power will often have the smallest communication costs. Using clusters with a higher aggregate computation power will lead to reduced completion time since Tcomp and Tcomm will both be smaller. The next stage of the algorithm explores the processor config-uration space in a greedy fashion. All processors of a cluster are selected before processors in the next clus-ter are considered thus avoiding router crossings if possible. This algorithm tries to maintain communication locality by avoiding the router penalty and potential coercion overhead. The algorithm ter-minates when adding processors in the current cluster causes Tc to increase, and is sketched in Figure 7. The algorithm computes two things in get_curr_config: the best processor configuration (and task placement) based on the previous configuration and the current cluster, and the partition_map. It has the property that once Pj is computed for cluster Cj, it is not modified in subsequent iterations. For each cluster it locates the best number of processors by a form of binary search, the details of which are provided in . The worst-case order of this algorithm is O(mlog2P) for m clusters and P total processors. The algo-rithm is scalable and the overhead in practice is small. In Section 3.1, we present some experimental results that indicate H1 is efficient and produces excellent results. A more general heuristic is needed when communication capacities differ in the metasystem or if router costs are not prohibitively high. The latter means that additional communication bandwidth may by effectively exploited across multiple clusters even with a router penalty. A greedy strategy will not work well in this case. Heuristic H2 Order clusters C1 .. Cm by aggregate computation power Initialize curr_config, min_cost For each cluster Ci { // Determine config that yields min Tc given previous Pj (j min_cost) break; else { curr_config = best_curr_config; min_cost = best_curr_config.cost; } } return best_curr_config; Figure 7. Heuristic H1 17 Heuristic H2 has been designed for general metasystems and relaxes the assumptions made in H1. Because communication capacities may be different, a simple cluster ordering strategy based solely on computation power will not always work well. For example, consider that a slow network of very fast machines such as a DEC-Alpha cluster might be chosen over a Paragon partition because the DEC-Alpha is faster than the i860. Clearly this may be a poor choice for some tightly-coupled parallel computations. Instead, we adopt a ordering heuristic based on Tc since Tc gives us a real measure of cost that includes both computation and communication. We apply H1 to each cluster in isolation to compute the minimum Tc. We then order the clusters by this value. A more complex two-phase strategy is adopted for exploring the processor configurations, see Figure 8. In phase 1, we add processors for the current cluster using the same algorithm as described in H1. It is guaranteed that adding processors will decrease the Tcomp component of Tc. The addition of processors will never decrease Tcomm, though it may remain unchanged. In phase 2, we try to reduce the Tcomm component of Tc. Recall from Section 2.2 that the total communication cost is a function of the communication cost contributed by each cluster based on intra-cluster placement, see (EQ 3). The cluster that contributes the maximum communication cost is targeted for reducing the overall communication cost. In phase 2, we add processors to the current cluster while removing processors from the cluster than contributed the largest communication cost. This is guaranteed to reduce Tcomm, but the impact on Tc is unpredictable since Tcomp may increase since we are trading potentially faster processors for slower ones. The cluster that contributes the largest cost may change during the course of phase 2 as processors are traded. The configuration that yields the minimum Tc after both phase 1 and phase 2 is stored. This is the starting configuration that used as the next cluster is considered, much like in H1. Unlike H1, H2 is conservative and does not terminate until all clusters are explored. The worst-case order of this algorithm is O(mP): O(mlog2P) for cluster ordering, O(mlog2P) for phase 1, and O(mP) for phase 2. In practice, the linear dependence on P is tolera-ble since the amount of computation performed for each configuration is very small. We are exploring heu-ristics to improve the time bound for phase 2. In Section 3.2, we present some simulation results that indicate that H2 is viable. 2.5 Implementation The final phase of the partitioning framework is instantiation. Instantiation initiates the data parallel computation across the processor configuration using the determined task placement. We present a partial implementation of the stencil computation implemented in Mentat to illustrate the instantiation process. Mentat is parallel object-oriented parallel processing system based on C++ . We also present a C++ callback interface that is currently being implemented. 18 2.5.1 Callback Specification The callbacks are encapsulated by a class called a domain, see Figure 9. The domain class can be tai-lored to the specific computation by means of derivation. For example, we have defined a stencil_domain derived from domain for stencil computations and we show the implementation of the comp_complexity callback for the single computation phase. This class is instantiated at runtime with a programmer-defined structure known as a parameter vector (PV) that contains any problem-dependent parameters that are used in the implementation of the callbacks (e.g., N for the stencil problem). Notice that the callback functions may depend on the number of processors, np, that are determined at runtime. Also observe that the domi-nant computation and communication phases are determined by callbacks since the dominant phases may depend on problem parameters. The domain instance is used by the partitioning framework at runtime. 2.5.2 Program Interface A Mentat code fragment for the stencil computation that uses the stencil_domain is given in Figure 10. The main program marshals the relevant program parameters into PV, instantiates the stencil_domain and calls the function partition that implements the partitioning heuristics of Section 2.4. This function Order clusters C1 .. Cm by Tc Initialize curr_config, min_cost For each cluster Ci { // Phase 1 -- Try to reduce Tcomp // Determine config that yields min Tc given previous Pj (j<i) best_curr_config = get_curr_config (curr_config, Ci) // min_cost is stored // Phase 2 -- Try to reduce Tcomm curr_config.Pi = 0 // Repeatedly trade processors in Ci with processors in Ck (k<i) // where Ck is the cluster with the largest communication cost // Ck may change during phase 2 -- if it is the current cluster, exit while ((curr_config.Pi <= Ni) && (k!=i)) { curr_config.Pi++; curr_config.Pk--; Tc = get_Tc (curr_config); if (Tc < min_cost) { best_curr_config = curr_config; min_cost = Tc; } } curr_config = best_curr_config; } return best_curr_config; Figure 8. Heuristic H2 19 returns a structure that contains the processor configuration and partition_map. In the Mentat implementa-tion, this structure is passed to a Mentat routine DP_create that instantiates a task (Mentat object) on each selected processor and passes the names of the neighboring Mentat objects to each Mentat object to enable communication. It is then up to application to start the data parallel computation and use the partition_map in an appropriate manner. In this code fragment, the grid is physically decomposed by the function 1D_carve, and the member function compute is called on each Mentat object. Each Mentat object is passed its portion of the problem via compute. The 1-D communication to exchange north and south borders occurs within the implementation of compute. 3.0 Results 3.1 Experimental We have obtained very promising results with H1 on several real parallel codes: the stencil computa-tion and Gaussian elimination with partial pivoting. The stencil code has been implemented on an hetero-class domain { char PV; public: virtual domain (char PV); virtual phase dominant_comp_phase (int np); virtual phase dominant_comm_phase (int np); virtual phase_rec num_phases (); virtual comp_rec comp_complexity (int np, phase comp_phase); virtual int numPDUs (phase comp_phase); virtual cost_rec arch_cost (proc_type proc, phase comp_phase); virtual comm_rec comm_complexity (int np, phase comm_phase); virtual phase overlap (phase comm_phase); virtual top topology (phase comm_phase); } class stencil_domain: domain { public: ... comp_rec comp_complexity (int np, phase comp_phase) { int N = atoi (PV); // extract problem size comp_rec CR = new comp_rec; CR->PDU_inst = 5N; // 5 fp operations per PDU in this problem CR->non_PDU_inst = 0; return CR; } ... } Figure 9. Domain class callback interface 20 geneous workstation network and Gaussian elimination on the Intel Gamma. In both environments, the communication capacities are the same (i.e., the workstation network is ethernet only and the Intel Gamma is a single machine with a hypercube interconnect), so H1 is appropriate. Stencil We have implemented two versions of the stencil code, STEN-1 and STEN-2 in which communica-tion is handled by the MMPS communication library discussed in Section 2.2. STEN-2 overlaps computa-tion and communication while STEN-1 does not. The callbacks for the stencil code were given in Figure 5b. The codes were run on two ethernet-connected clusters of Sun 4s joined by a router. Cluster C1 con-tains 6 Sparc2s and C2 contains 6 Sun 4 IPCs. The arch_cost for this problem was determined to be .3 usec and .6 usec for C1 and C2 respectively. Thus, the Sparc2s are about twice as fast as the Sun 4 IPCs and the partition_map will give each Sparc2 processor twice as many PDUs as each Sun 4 IPC via (EQ 6). The codes were implemented using a 1-D topology and the communication cost functions were derived by benchmarking. Heuristic H1 computes Tcomp and Tcomm by invoking callbacks with Tc computed to be: Tc [STEN-1] = Tcomp + Tcomm Tc [STEN-2] = max {Tcomp, Tcomm} We ran H1 off-line on a range of problem sizes for both STEN-1 and STEN-2: N=60, 300, 600, and 1200. When STEN-1 and STEN-2 were run on the network using the processor configuration and partition_map computed by H1, minimum elapsed times were obtained . The codes were run multiple times when the network and processors were lightly loaded and averages reported. We also timed H1 and found that the largest overhead was approximately 250 usec on a Sparc2, while the elapsed times for STEN-1 and STEN-main() { stencil_class workers, mo; stencil_domain dom; ... // Problem-specific code: (N and Grid are read from file) PV= itoa (N); // marshal PV for problem instance dom = new stencil_domain (PV); // instantiate domain PM = partition (dom); mclass workers = (mclass) DP_create (PM, mo); // Application-specific code 1D_grid = 1D_carve (Grid, PM.partition_map); for (int i=0; i 3.7% difference Figure 13. Sample simulator run 25 ment. Oregami differs from our framework in two significant ways: it is limited to homogeneous parallel machines and it is a compile-time approach. 5.0 Conclusion We described a framework for partitioning data parallel computations across heterogeneous metasys-tems at runtime. Three difficult problems, processor selection, task placement, and heterogeneous data domain decomposition, are handled automatically by the framework. The framework is applicable to a large class of data parallel computations and the early results indicate that excellent performance is achiev-able. Our approach exploits program and resource information in a novel way. A model of data parallel computations was presented in which program information is made available by a powerful callback mechanism. A C++ implementation of callbacks based on domain classes was given. A hierarchical meta-system organization was also presented in which resource information is collected and maintained by the system. A key part of this model is a set of topology-specific communication functions that are central to the processor selection process. We are currently implementing the framework in Legion , a heteroge-neous parallel processing system based on Mentat. 6.0 References A.L. Cheung, and A.P. Reeves, “High Performance Computing on a Cluster of Workstations,” Proceedings of the First Symposium on High-Performance Distributed Computing, Sept 1992. G. Fox et al, “Fortran D Language Specification,” TR90-141, Department of Computer Science, Rice Uni-versity, December 1990. R.F. Freund and H.J. Siegel, “Heterogeneous Processing,” IEEE Computer, June 1993. G.H. Golub and J.M. Ortega, Scientific Computing and Differential Equations, Academic Press, Inc., 1992. A.S. Grimshaw, D. Mack, and T. Strayer, “MMPS: Portable Message Passing Support for Parallel Comput-ing,” Proceedings of the Fifth Distributed Memory Computing Conference, April 1990. A.S. Grimshaw, J.B. Weissman, E.A. West, and E. Loyot, “Metasystems: An Approach Combining Parallel Processing And Heterogeneous Distributed Computing Systems,” Journal of Parallel and Distributed Com-puting, Vol. 21, No. 3, June 1994. A.S. Grimshaw, “Easy to Use Object-Oriented Parallel Programming with Mentat,” IEEE Computer, May 1993. A.S. Grimshaw et al, “Legion: The Next Logical Step Toward a Nationwide Virtual Computer,” Computer Science Technical Report, CS-94-21, University of Virginia, June, 1994. P.J. Hatcher, M.J. Quinn, and A.J. Lapadula, “Data-parallel Programming on MIMD Computers,” IEEE Transactions on Parallel and Distributed Systems, Vol 2, July 1991. F.T. Leighton, Introduction to Parallel Algorithms and Architectures: Arrays, Trees, Hypercubes, Morgan-Kaufmann Publishers, 1992. V.M. Lo et al, “OREGAMI: Tools for Mapping Parallel Computations to Parallel Architectures,” CIS-TR-89-18a, Department of Computer Science, University of Oregon, April 1992. N. Nedeljkovic and M.J. Quinn, “Data-Parallel Programming on a Network of Heterogeneous Worksta-tions,” Proceedings of the First Symposium on High-Performance Distributed Computing, Sept. 1992. D.M. Nicol and P.F. Reynolds, Jr., “Optimal Dynamic Remapping of Data Parallel Computations,” IEEE 26 Transactions on Computers, Vol. 39, No. 2, February 1990. V.S. Sunderam, “PVM: A framework for parallel distributed computing,” Concurrency: Practice and Experience, vol. 2(4), pp. 315-339, December 1990. J.B. Weissman, Andrew S. Grimshaw, and Robert R. Ferraro, “Parallel Object-Oriented Computa-tion Applied to a Finite Element Problem,” Journal of Scientific Programming, Vol. 2 No. 4, 1993. J.B. Weissman, “Multigranular Scheduling of Data Parallel Programs,” TR CS-93-38, Department of Computer Science, University of Virginia, July 1993. J.B. Weissman and A.S. Grimshaw, “Network Partitioning of Data Parallel Computations,” Pro-ceedings of the Third International Symposium on High-Performance Distributed Computing, August 1994.
15761	https://www.rcet.org.in/uploads/academics/rohini_92561697030.pdf	ROHINI COLLEGE OF ENGINEERING & TECHNOLOGY EC8651 TRANSMISSION LINES AND RF SYSTEMS 4.4 BESSEL’S DIFFERENTIAL EQUATION AND BESSEL FUNCTION & TM AND TE WAVES IN CIRCULAR WAVE GUIDES: A circular waveguide is a hollow metallic tube with circular cross section for propagating the electromagnetic waves by continuous reflections from the surfaces or walls of the guide. The circular waveguides are avoided because of the following reasons: a) The frequency difference between the lowest frequency on the dominant mode and the next mode is smaller than in a rectangular waveguide, with b/a= 0.5 b) The circular symmetry of the waveguide may reflect on the possibility of the wave not maintaining its polarization throughout the length of the guide. c) For the same operating frequency, circular waveguide is bigger in size than a rectangular waveguide. The possible TM modes in a circular waveguide are: TM01, TM02, TM11, TM12. The root values for the TM modes are: • (ha)01 = 2.405 for TM01 • (ha)02 = 5.53 for TM02 • (ha)11 = 3.85 for TM11 • (ha)12 = 7.02 for TM12 The dominant mode for a circular waveguide is defined as the lowest order mode having the lowest root value. The possible TE modes in a circular waveguide are: TE01, TE02, TE11, TE12. The root values for the TE modes are: • (ha)01 = 3.85 for TE01 • (ha)02 = 7.02 for TE02 • (ha)11 = 1.841 for TE11 ROHINI COLLEGE OF ENGINEERING & TECHNOLOGY EC8651 TRANSMISSION LINES AND RF SYSTEMS • (ha)12 = 5.53 for TE12 The dominant mode for TE waves in a circular waveguide is the TE11.v. Because it has the lowest root value of 1.841. Since the root value of TE11 is lower than TM01, TE11 is the dominant or the lowest order mode for a circular waveguide. RECTANGULAR AND CIRCULAR CAVITY RESONATORS: Resonator is a tuned circuit which resonates at a particular frequency at which the energy stored in the electric field is equal to the energy stored in the magnetic field. Resonant frequency of microwave resonator is the frequency at which the energy in the resonator attains maximum value. i.e., twice the electric energy or magnetic energy. At low frequencies upto VHF (300 MHz), the resonator is made up of the reactive elements or the lumped elements like the capacitance and the inductance. The inductance and the capacitance values are too small as the frequency is increased beyond the VHF range and hence difficult to realize. Transmission line resonator can be built using distributed elements like sections of coaxial lines. The coaxial lines are either opened or shunted at the end sections thus confining the electromagnetic energy within the section and acts as the resonant circuit having a natural resonant frequency. At very high frequencies transmission line resonator does not give very high quality factor Q due to skin effect and radiation loss. So, transmission line resonator is not used as microwave resonator. The performance parameters of microwave resonator are: (i) Resonant frequency ROHINI COLLEGE OF ENGINEERING & TECHNOLOGY EC8651 TRANSMISSION LINES AND RF SYSTEMS (ii)Quality factor (iii) Input impedance Quality Factor of a Resonator.: • The quality factor Q is a measure of frequency selectivity of the resonator. • It is defined as Q = 2 x Maximum energy stored / Energy dissipated per cycle = W / P Where, a. W is the maximum stored energy b. P is the average power loss The methods used for constructing a resonator: The resonators are built by, a)Using lumped elements like L and C b) Using distributed elements like sections of coaxial lines c) Using rectangular or circular waveguide There are two types of cavity resonators. a)Rectangular cavity resonator b)Circular cavity resonator Rectangular or circular cavities can be used as microwave resonators because they have natural resonant frequency and behave like a LCR circuit. Cavity resonator can be represented by a LCR circuit as: • The electromagnetic energy is stored in the entire volume of the cavity in the form of electric and magnetic fields. • The presence of electric field gives rise to a capacitance value and the presence of magnetic field gives rise to a inductance value and the finite ROHINI COLLEGE OF ENGINEERING & TECHNOLOGY EC8651 TRANSMISSION LINES AND RF SYSTEMS conductivity in the walls gives rise to loss along the walls giving rise to a resistance value. • Thus the cavity resonator can be represented by a equivalent LCR circuit and have a natural resonant frequency. Cavity resonators are formed by placing the perfectly conducting sheets on the rectangular or circular waveguide on the two end sections and hence all the sides are surrounded by the conducting walls thus forming a cavity. The electromagnetic energy is confined within this metallic enclosure and they acts as resonant circuits.
15762	https://math.stackexchange.com/questions/1858274/how-to-determine-if-two-lines-are-parallel-almost-parallel	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams How to determine if two lines are parallel/ almost parallel? Ask Question Asked Modified 6 years, 10 months ago Viewed 19k times $\begingroup$ I have $2$ lines in this form Line $1$: $(x_1,y_1)$ $(x_2,y_2)$ Line $2$: $(x_3,y_3)$ $(x_4,y_4)$ I want to detect if the two lines are parallel or almost parallel. My idea is to if the angle between the two lines is $\leq$ some threshold angle (like $10$ degrees or so), then they are almost parallel. But don't know how to compute angle between two lines. Please help me with the equations or if there is any idea to detect almost parallel lines. Thank you. edited Jul 13, 2016 at 15:46 snulty 4,48522 gold badges2828 silver badges5252 bronze badges asked Jul 13, 2016 at 15:05 helloansumanhelloansuman 13711 gold badge22 silver badges66 bronze badges $\endgroup$ 4 3 $\begingroup$ in the plane, two lines are parallel if they have the same slope (or put another way, if they make the same angle with one of the coordinate axes) $\endgroup$ costrom – costrom 2016-07-13 15:07:19 +00:00 Commented Jul 13, 2016 at 15:07 $\begingroup$ Thank you for your quick reply. Can you please bit more specific about the equations to be used? $\endgroup$ helloansuman – helloansuman 2016-07-13 15:11:47 +00:00 Commented Jul 13, 2016 at 15:11 2 $\begingroup$ Is there a specific application for checking if lines are almost parallel that you're interested in. Something you'd like to conclude from that? $\endgroup$ snulty – snulty 2016-07-13 15:50:16 +00:00 Commented Jul 13, 2016 at 15:50 $\begingroup$ I am applying this to know if two people are walking side-by-side to each other. I have their location. $\endgroup$ helloansuman – helloansuman 2016-07-14 06:53:01 +00:00 Commented Jul 14, 2016 at 6:53 Add a comment \| 12 Answers 12 Reset to default 13 $\begingroup$ A line of slope $m$ makes a signed angle with the $x$-axis equal to $\theta = \arctan(m)$ (expressed in radians), and this angle has value between $-\pi/2$ and $+\pi/2$. For two lines of slopes $m_1$, $m_2$, which make signed angles equal to $\theta_1 = \arctan(m_1)$ and $\theta_2 = \arctan(m_2)$, the "angular distance" between those two lines will be either $\|\theta_1 - \theta_2\|$ or $\pi - \|\theta_1 - \theta_2\|$, whichever is smaller. The alternative $\|\theta_1 - \theta_2\|$ is used when that number is $\le \pi/2$, and if that number is $> \pi/2$ then the other alternative $\pi - \|\theta_1 - \theta_2]$ will be $\le \pi/2$. Once you have computed the angular distance between the two lines, set it to be less than whatever threshold you desire. Share edited Nov 4, 2018 at 20:18 answered Jul 13, 2016 at 15:20 Lee MosherLee Mosher 137k77 gold badges105105 silver badges218218 bronze badges $\endgroup$ 2 $\begingroup$ Thank you for a more specific idea of angle comparison. I am feeling now the angular comparison will be better than the idea of subtraction of slopes. $\endgroup$ helloansuman – helloansuman 2016-07-13 15:30:50 +00:00 Commented Jul 13, 2016 at 15:30 $\begingroup$ I made a correction to my formula for the angular distance. $\endgroup$ Lee Mosher – Lee Mosher 2016-07-13 18:10:53 +00:00 Commented Jul 13, 2016 at 18:10 Add a comment \| 11 $\begingroup$ Suppose you have vectors $u = (u_1,u_2)$ and $v = (v_1,v_2),$ where $u_1$ is the x-component and $u_2$ is the y-component of $u.$ Recall that $\cos{\theta} = \frac{u \cdot v}{\|\|u\|\| \,\|\|v\|\|},$ where $u \cdot v = u_1v_1 + u_2v_2$ and $\|\|u\|\| = \sqrt{u \cdot u}$. Or, one could compute the slopes of each line, and say that if \|slope 1 - slope 2\| $< \epsilon$ threshold, then they are almost parallel. Share answered Jul 13, 2016 at 15:11 MerkhMerkh 3,7201717 silver badges3030 bronze badges $\endgroup$ 5 $\begingroup$ Thank you for a specific answer. I think \|m1-m2\|< threshold will work hopefully. $\endgroup$ helloansuman – helloansuman 2016-07-13 15:18:44 +00:00 Commented Jul 13, 2016 at 15:18 $\begingroup$ When the slopes are close the tangents of two angles are close but that doesn't tell you how close together the angles are. $\endgroup$ Ethan Bolker – Ethan Bolker 2016-07-13 15:19:13 +00:00 Commented Jul 13, 2016 at 15:19 14 $\begingroup$ Subtracting slopes is a terrible idea if the lines are nearly vertical. The difference of angles can be tiny while the difference of slopes is enormous. $\endgroup$ Lee Mosher – Lee Mosher 2016-07-13 15:21:22 +00:00 Commented Jul 13, 2016 at 15:21 $\begingroup$ Ethan @LeeMosher Thanks for identifying the problem in the solution. $\endgroup$ helloansuman – helloansuman 2016-07-13 15:26:26 +00:00 Commented Jul 13, 2016 at 15:26 $\begingroup$ It all depends on how one defines "almost parallel". From a visual perspective, using the angle between the lines would make most sense, as this is free from choosing a specific coordinate basis and computing the slopes relative to that basis, which may be a problem as Lee pointed out. However, there is no reason "almost parallel" can't be a difference in slopes, as this would give a measure of how much the two linear functions will vary later on down the line. So of course, it depends on what one is interested in. $\endgroup$ Merkh – Merkh 2016-07-13 16:28:51 +00:00 Commented Jul 13, 2016 at 16:28 Add a comment \| 6 $\begingroup$ The slope of the first line is $m_{1}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$ and the slope of the second is $m_{2}=\frac{y_{4}-y_{3}}{x_{4}-x_{3}}$. The lines are parallel if and only if $m_{1}=m_{2}$. Share answered Jul 13, 2016 at 15:11 aduhaduh 8,90433 gold badges2323 silver badges5757 bronze badges $\endgroup$ 3 $\begingroup$ Thanks for the reply. But I have to also detect two lines are almost parallel condition. $\endgroup$ helloansuman – helloansuman 2016-07-13 15:17:36 +00:00 Commented Jul 13, 2016 at 15:17 1 $\begingroup$ Say that the two lines are $\epsilon$-almost parallel if $\|m_{2} - m_{1}\| < \epsilon$. $\endgroup$ aduh – aduh 2016-07-13 15:20:19 +00:00 Commented Jul 13, 2016 at 15:20 3 $\begingroup$ According to @LeeMosher above, it is a terrible idea to subtract slopes. $\endgroup$ helloansuman – helloansuman 2016-07-13 15:27:48 +00:00 Commented Jul 13, 2016 at 15:27 Add a comment \| 6 $\begingroup$ Let $u = (a_1, b_1) = (x_2 - x_1, y_2 - y_1)$ and $v = (a_2, b_2) = (x_4 - x_3, y_4 - y_3)$ be the corresponding vectors. For detecting angles near 0, the best is to use the cross product: $$ \sin \theta = \frac{u \times v}{\|u\|\|v\|},$$ where $u \times v = \det(u, v) = a_1 b_2 - a_2 b_1$. So, for instance, if you want to test for $\|\theta\| < \theta_0$, you can just test for $$(u \times v)^2 < \varepsilon^2 \|u\|^2\|v\|^2,$$ with $\varepsilon = \sin\theta_0 \approx \theta_0$. Notice that this formula only involves addition/subtraction and multiplication. Share answered Jul 13, 2016 at 17:44 filiposfilipos 1,57277 silver badges88 bronze badges $\endgroup$ 2 $\begingroup$ Thank you for a really simple explanation. The final equation is so simple and computational efficient. $\endgroup$ helloansuman – helloansuman 2016-07-14 04:37:07 +00:00 Commented Jul 14, 2016 at 4:37 1 $\begingroup$ This also works in higher dimensions, if we use the wedge product: $$\lVert u\wedge v\rVert^2<\varepsilon^2\lVert u\rVert^2\lVert v\rVert^2$$ $$\sum_{1\leq i $\endgroup$ mr_e_man – mr_e_man 2020-01-08 04:11:00 +00:00 Commented Jan 8, 2020 at 4:11 Add a comment \| 2 $\begingroup$ Let $$\mathrm v_1 := \mathrm x_2 - \mathrm x_1 \qquad \qquad \qquad \mathrm v_2 := \mathrm x_4 - \mathrm x_3$$ be the direction vectors of each of the two lines. Compute the rank of the following matrix $$\begin{bmatrix} \| & \|\ \mathrm v_1 & \mathrm v_2\ \| & \|\end{bmatrix}$$ If the matrix has full column rank, then the two given lines are not parallel. If the matrix is rank-$1$, then the two given lines are parallel. Share answered Jul 13, 2016 at 15:15 Rodrigo de AzevedoRodrigo de Azevedo 23.4k77 gold badges4949 silver badges116116 bronze badges $\endgroup$ 3 $\begingroup$ Thanks for the reply. But I have to also detect two lines are almost parallel condition. $\endgroup$ helloansuman – helloansuman 2016-07-13 15:17:55 +00:00 Commented Jul 13, 2016 at 15:17 $\begingroup$ @helloansuman Then normalize each of the two columns of the matrix and then compute the angle formed by the two columns. $\endgroup$ Rodrigo de Azevedo – Rodrigo de Azevedo 2016-07-13 15:22:52 +00:00 Commented Jul 13, 2016 at 15:22 $\begingroup$ Thank you very much, sir. I will consider this solution too. As I am using MatLab to simulate a matrix representation will help. $\endgroup$ helloansuman – helloansuman 2016-07-13 15:33:07 +00:00 Commented Jul 13, 2016 at 15:33 Add a comment \| 2 $\begingroup$ Here's a method that may work for you. (The other answers so far will tell you whether the lines are parallel, but don't find the angle so don't give you a way to decide on "almost parallel".) Move both line segments to the origin. For the first one that would be $$ (x_2 - x_1, y_2 - y_1) . $$ Calculate the angle between the line segments Lots of websites tell you how. One is . Decide how small that angle should be to count as "nearly parallel". Share answered Jul 13, 2016 at 15:17 Ethan BolkerEthan Bolker 105k77 gold badges127127 silver badges223223 bronze badges $\endgroup$ 1 $\begingroup$ For actual numerics, I would not recommend the dot-product formula. See e.g. this. $\endgroup$ J. M. ain't a mathematician – J. M. ain't a mathematician 2016-07-13 16:55:02 +00:00 Commented Jul 13, 2016 at 16:55 Add a comment \| 2 $\begingroup$ I have a method I've used to determine "almost parallelism" that may work for you. It's something I've used computationally to determine parallelism within some very small tolerance. A consequence of Bezout's theorem is that two distinct lines will intersect at exactly 1 point. For parallel lines, they meet at infinity. "Almost parallel" lines will intersect at a point very far away. The question then becomes how to recognize a point very far away or at infinity. If you change the points into homogeneous coordinates (for this case specifically, by mapping a 2D point $(X,Y)$ as a 3D point $(X,Y,1)$ and converting a 3D point $(x,y,w)$ to a 2D point $(x/w,y/w)$), you can do some neat tricks. You can then also represent the lines as triples $(a,b,c)$ such that it satisfies the implicit line equation $aX+bY+c=0$. Using the dot product operator, you can express this equation as a dot product: $(a,b,c) \cdot (x,y,w) = 0$. Another cool trick is that you can similarly use the cross product operator to determine the implicit triple for the line that passes through the 2 points: $(x_1,y_1,w_1)\,\times\,(x_2,y_2,w_2) = (a,b,c)$. You can again use the cross product operator to determine the point at which 2 lines expressed as implicit triples intersect: $(a_1,b_1,c_1)\,\times\,(a_2,b_2,c_2)=(x,y,w)$. What's interesting about this is that it tells you where the point is, even at infinity. Because of the way we defined the mapping from 2D to 3D, if the $w$ component of the 3D point is small, the $x$ and $y$ components of the 2D point are large. If $w=0$, then the 2D point is at infinity. For my uses, I only test strictly for $w=0$ (the exactly parallel case), however you might be able to adapt this for your use case. Hope this helps! Share edited Jul 15, 2016 at 0:34 answered Jul 13, 2016 at 17:58 CADJunkieCADJunkie 21822 silver badges88 bronze badges $\endgroup$ 3 $\begingroup$ Got your trick. I have to just put a threshold on the final 'w' value. Thanks for contribution. I will have a look on that. $\endgroup$ helloansuman – helloansuman 2016-07-14 04:20:25 +00:00 Commented Jul 14, 2016 at 4:20 $\begingroup$ I think you'll also need some way to truly use w as a distance comparison. What you can do is a sort of normalization, where you force sqrt(x^2+y^2)=1 by dividing all components by sqrt(x^2+y^2). Otherwise, with a naive w test, (0.2,0.5,0.1) would seem "farther" than (2,5,1) when they really represent the same point. $\endgroup$ CADJunkie – CADJunkie 2016-07-15 00:31:59 +00:00 Commented Jul 15, 2016 at 0:31 $\begingroup$ Thank you all for helping me out with my problem. I think I just found 1000 ways of solving the same problem. But I am still confused which one to use. I am going to try every methods. $\endgroup$ helloansuman – helloansuman 2016-08-01 18:09:52 +00:00 Commented Aug 1, 2016 at 18:09 Add a comment \| 2 $\begingroup$ Quicker than finding the arctan of each slope, I'd use the slopes themselves; call them $a,b$. The tangent of the divergence angle is then $$\pm\frac{a-b}{1+ab}$$ This requires no transcendental operations, after you choose a threshold. EDIT: To lessen the possibility of overflow that J.M. points out, use instead $$\pm\frac{\Delta y_a \Delta x_b - \Delta x_a \Delta y_b}{\Delta x_a \Delta x_b + \Delta y_a \Delta y_b}$$ This expression is equivalent to the first, with numerator and denominator multiplied by $\Delta x_a \Delta x_b$. We may note in passing that it is the magnitude of the cross product divided by the dot product. LATER EDIT: Better yet, test whether $$ \|{\Delta y_a \Delta x_b - \Delta x_a \Delta y_b}\| < m \, \|{\Delta x_a \Delta x_b + \Delta y_a \Delta y_b}\| $$ where m = tan(maxangle); no divisions this way. Share edited Aug 4, 2016 at 8:52 answered Jul 14, 2016 at 8:35 Anton SherwoodAnton Sherwood 63833 silver badges1313 bronze badges $\endgroup$ 1 $\begingroup$ For nearly vertical lines, of course, multiplying the slopes can lead to overflow. $\endgroup$ J. M. ain't a mathematician – J. M. ain't a mathematician 2016-07-14 11:37:19 +00:00 Commented Jul 14, 2016 at 11:37 Add a comment \| 1 $\begingroup$ Two lines are parallel or not. I suppose that the notion of ''almost parallel'' refers, for you, to a situation in which the angle between the two lines is less than some fixed (small) value. If it is so, and if you want the angle between the two lines, note that the slope of the first line is $m_1=\frac{y_2-y_1}{x_2-x_1}$ and the slope of the second line is $m_2=\frac{y_4-y_3}{x_4-x_3}$, so you can find the angles between the lines and the $x$ axis as $\theta_1=\arctan m_1$ and $\theta_2=\arctan m_2$ and the angle between the lines is $\theta=\|\theta_1-\theta_2\|$. Share answered Jul 13, 2016 at 15:18 Emilio NovatiEmilio Novati 64.5k66 gold badges4949 silver badges128128 bronze badges $\endgroup$ 6 $\begingroup$ You got me right. That was my idea to calculate angle between them if it is zero then parallel or if angle is less than some threshold then almost parallel. $\endgroup$ helloansuman – helloansuman 2016-07-13 15:23:14 +00:00 Commented Jul 13, 2016 at 15:23 $\begingroup$ Yes. Note that you have some problem if $x_2=x_1$ or $x_4=x_3$, but you can treat this cases (lines parallel to the $y$ axis) separately. $\endgroup$ Emilio Novati – Emilio Novati 2016-07-13 15:34:38 +00:00 Commented Jul 13, 2016 at 15:34 $\begingroup$ That is not a problem as that case will not appear in my case all points are distinct. Thanks for pre-informing though. $\endgroup$ helloansuman – helloansuman 2016-07-13 15:43:48 +00:00 Commented Jul 13, 2016 at 15:43 $\begingroup$ In practice, one will of course use two-argument arctangent... $\endgroup$ J. M. ain't a mathematician – J. M. ain't a mathematician 2016-07-13 16:48:43 +00:00 Commented Jul 13, 2016 at 16:48 $\begingroup$ @J.M. Sorry can't got your point. Can you please elaborate? Thanks. $\endgroup$ helloansuman – helloansuman 2016-07-14 04:22:23 +00:00 Commented Jul 14, 2016 at 4:22 \| Show 1 more comment 1 $\begingroup$ Generally, the inner product or dot product of two vectors is used to find the angle between two lines and so whether they're parallel or almost parallel. Or you can skip to the bottom for just the formula. First find a corresponding vector for each line, which we'll write as the change in x and y coordinates respectively. For vector $\vec u$ representing the direction of line 1, $$\vec u = (u_x, u_y) = (x_2-x_1,y_2-y_1)$$ and for vector $\vec v$ representing the direction of line 2, $$\vec v = (v_x, v_y) = (x_4-x_3,y_4-y_3)$$ The inner product of two vectors, $\vec u \cdot \vec v$ (also written $\langle \vec u, \vec v \rangle$) can be calculated by summing the products of matching terms in the two vectors. Here, $$\vec u \cdot \vec v = u_x v_x + u_y v_y = (x_2-x_1)(x_4-x_3) + (y_2-y_1)(y_4-y_3)$$ To find the angle $\theta$ between $\vec u$ and $\vec v$ we use that $$\vec u \cdot \vec v = \lVert \vec u \rVert \lVert \vec v \rVert cos(\theta)$$ so $$ \theta = cos^{-1}(\frac{\vec u \cdot \vec v}{\lVert \vec u \rVert \lVert \vec v \rVert}) $$ where the norm or magnitude $\lVert \vec v \rVert$ of a vector $\vec v$ here is its length which we can find with the Pythagorean theorem. $$ \lVert \vec u \rVert = \sqrt{(u_x)^2 + (u_y)^2} = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$$ $$ \lVert \vec v \rVert = \sqrt{(v_x)^2 + (v_y)^2} = \sqrt{(x_4-x_3)^2 + (y_4-y_3)^2}$$ In the case of two lines, we can check if the angle is less than some value $\theta_{max}$ with $$ \lvert \vec u \cdot \vec v \rvert < \lVert \vec u \rVert \lVert \vec v \rVert cos(\theta_{max}) $$ where the absolute value accounts for how we could pick our vectors to go in either direction along each line. So for 10 degrees $$ \lvert \vec u \cdot \vec v \rvert < cos(10^\circ) \lVert \vec u \rVert \lVert \vec v \rVert$$ is approximately $$ \lvert \vec u \cdot \vec v \rvert < .985 \lVert \vec u \rVert \lVert \vec v \rVert $$ Or written out completely, lines given by $(x_1,y_1)$ $(x_2,y_2)$ and $(x_3,y_3)$ $(x_4,y_4)$ are almost parallel (here given as within about 10 degrees of each other) if and only if $$ \lvert (x_2 - x_1)(x_4 - x_3) + (y_2 - y_1)(y_4 - y_1) \rvert < .985 \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \sqrt{(x_4 - x_3)^2 + (y_4 - y_3)^2}$$ where .985 can be changed to the cosine of whichever angle you want to be the threshold. Share answered Jul 14, 2016 at 5:11 eblingebling 1111 bronze badge $\endgroup$ 1 $\begingroup$ The inequality must be in other direction, since if $\theta = 0$ then $\|u \cdot v\| = \lVert u \rVert \lVert v \rVert$. $\endgroup$ obareey – obareey 2023-03-20 17:26:36 +00:00 Commented Mar 20, 2023 at 17:26 Add a comment \| 1 $\begingroup$ The orthogonality of two vectors is detected by the scalar product, while the parallelity is detected by the cross product. Therefore put $${\bf u}:=(x_2-x_1,y_2-y_1),\quad {\bf v}:=(x_4-x_3,y_4-y_3)$$ and compute the (unsigned) angle $\phi$ between the two lines by means of $$\sin\phi={{\bf u}\wedge{\bf v}\over \|{\bf u}\|>\|{\bf v}\|}={\|u_1v_2-u_2v_1\|\over\sqrt{u_1^2+u_2^2}>\sqrt{v_1^2+v_2^2}}\ .$$ Share answered Aug 4, 2016 at 9:19 Christian BlatterChristian Blatter 233k1414 gold badges204204 silver badges490490 bronze badges $\endgroup$ Add a comment \| $\begingroup$ The others have responded on how to find angle between two lines. Difference between slopes or angles of inclination of two given lines to x-axis .. that is what you want to control: $$ \tan^{-1} \frac{y_2-y_1}{x_2-x_1}-\tan^{-1} \frac{y_4-y_3}{x_4-x_3} < \frac{10\, \pi}{180}$$ The smaller it is, the more parallel the lines are. If zero, they are completely parallel. Taking tan on both sides seems to bring it to an algebraic form, but I think it is really not better for numerical calculation. Share edited Jul 14, 2016 at 8:24 answered Jul 14, 2016 at 8:13 NarasimhamNarasimham 42.5k77 gold badges4646 silver badges112112 bronze badges $\endgroup$ 1 $\begingroup$ Thank you for your answer. So according to you what type of computation you are suggesting? $\endgroup$ helloansuman – helloansuman 2016-07-14 11:09:51 +00:00 Commented Jul 14, 2016 at 11:09 Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. 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15763	https://law.justia.com/codes/new-york/rpp/article-12-a/443/	New York Real Property Law § 443 (2024) - Disclosure Regarding Real Estate Agency Relationship; Form. :: 2024 New York Laws :: U.S. Codes and Statutes :: U.S. Law :: Justia Log InSign Up Find a Lawyer Ask a Lawyer Research the Law Law Schools Laws & Regs Newsletters Marketing Solutions Justia Connect Pro Membership Practice Membership Public Membership Justia Lawyer Directory Platinum Placements Gold Placements Justia Elevate SEO Websites Blogs Justia Amplify PPC Management Google Business Profile Social Media Justia Onward Blog Justia › U.S. Law › U.S. Codes and Statutes › New York Laws › 2024 New York Laws › RPP - Real Property › Article 12-A - Real Estate Brokers and Real Estate Salespersons › 443 - Disclosure Regarding Real Estate Agency Relationship; Form. Go to Previous Versions of the New York Laws 2024 N.Y. Laws (here) 2023 N.Y. Laws 2022 N.Y. Laws View All Versions 2024 New York Laws RPP - Real Property Article 12-A - Real Estate Brokers and Real Estate Salespersons 443 - Disclosure Regarding Real Estate Agency Relationship; Form. Universal Citation: NY Real Prop L § 443 (2024) Learn more This media-neutral citation is based on the American Association of Law Libraries Universal Citation Guide and is not necessarily the official citation. PreviousNext § 443. Disclosure regarding real estate agency relationship; form. 1. Definitions. As used in this section, the following terms shall have the following meanings: a. "Agent" means a person who is licensed as a real estate broker, associate real estate broker or real estate salesperson under section four hundred forty-a of this article and is acting in a fiduciary capacity. b. "Buyer" means a transferee in a residential real property transaction and includes a person who executes an offer to purchase residential real property from a seller through an agent, or who has engaged the services of an agent with the object of entering into a residential real property transaction as a transferee. c. "Buyer's agent" means an agent who contracts to locate residential real property for a buyer or who finds a buyer for a property and presents an offer to purchase to the seller or seller's agent and negotiates on behalf of the buyer. d. "Listing agent" means a person who has entered into a listing agreement to act as an agent of the seller or landlord for compensation. e. "Listing agreement" means a contract between an owner or owners of residential real property and an agent, by which the agent has been authorized to sell or lease the residential real property or to find or obtain a buyer or lessee therefor. f. "Residential real property" means real property used or occupied, or intended to be used or occupied, wholly or partly, as the home or residence of one or more persons improved by (i) a one-to-four family dwelling or (ii) condominium or cooperative apartments but shall not refer to unimproved real property upon which such dwellings are to be constructed. g. "Seller" means the transferor in a residential real property transaction, and includes an owner who lists residential real property for sale with an agent, whether or not a transfer results, or who receives an offer to purchase residential real property. h. "Seller's agent" means a listing agent who acts alone, or an agent who acts in cooperation with a listing agent, acts as a seller's subagent or acts as a broker's agent to find or obtain a buyer for residential real property. i. "Dual agent" means an agent who is acting as a buyer's agent and a seller's agent or a tenant's agent and a landlord's agent in the same transaction. j. "Designated sales agent" means a licensed real estate salesperson or associate broker, working under the supervision of a real estate broker, who has been assigned to represent a client when a different client is also represented by such real estate broker in the same transaction. k. "Broker's agent" means an agent that cooperates or is engaged by a listing agent, buyer's agent or tenant's agent (but does not work for the same firm as the listing agent, buyer's agent or tenant's agent) to assist the listing agent, buyer's agent or tenant's agent in locating a property to sell, buy or lease respectively, for the listing agent's seller or landlord, the buyer agent's buyer or the tenant's agent tenant. The broker's agent does not have a direct relationship with the seller, buyer, landlord or tenant and the seller, buyer, landlord or tenant can not provide instructions or direction directly to the broker's agent. Therefore, the seller, buyer, landlord or tenant do not have vicarious liability for the acts of the broker's agent. The listing agent, buyer's agent or tenant's agent do provide direction and instruction to the broker's agent and therefore the listing agent, buyer's agent or tenant's agent will have liability for the broker's agent. l. "Tenant" means a lessee in a residential real property transaction and includes a person who executes an offer to lease residential real property from a landlord through an agent, or who has engaged the services of an agent with the object of entering into a residential real property transaction as a lessee. m. "Landlord" means the lessor in a residential real property transaction, and includes an owner who lists residential real property for lease with an agent, whether or not a lease results, or who receives an offer to lease residential real property. n. "Tenant's agent" means an agent who contracts to locate residential real property for a tenant or who finds a tenant for a property and presents an offer to lease to the landlord or landlord's agent and negotiates on behalf of the tenant. o. "Landlord's agent" means a listing agent who acts alone, or an agent who acts in cooperation with a listing agent, acts as a landlord's subagent or acts as a broker's agent to find or obtain a tenant for residential real property. p. "Advance consent to dual agency" means written informed consent signed by the seller/landlord or buyer/tenant that the listing agent and/or buyer's agent may act as a dual agent for that seller/landlord and a buyer/tenant for residential real property which is the subject of a listing agreement. q. "Advance consent to dual agency with designated sales agents" means written informed consent signed by the seller/landlord or buyer/tenant that indicates the name of the agent appointed to represent the seller/landlord or buyer/tenant as a designated sales agent for residential real property which is the subject of a listing agreement. This section shall apply only to transactions involving residential real property. a. A listing agent shall provide the disclosure form set forth in subdivision four of this section to a seller or landlord prior to entering into a listing agreement with the seller or landlord and shall obtain a signed acknowledgment from the seller or landlord, except as provided in paragraph e of this subdivision. b. A seller's agent or landlord's agent shall provide the disclosure form set forth in subdivision four of this section to a buyer, buyer's agent, tenant or tenant's agent at the time of the first substantive contact with the buyer or tenant and shall obtain a signed acknowledgement from the buyer or tenant, except as provided in paragraph e of this subdivision. c. A buyer's agent or tenant's agent shall provide the disclosure form to the buyer or tenant prior to entering into an agreement to act as the buyer's agent or tenant's agent and shall obtain a signed acknowledgment from the buyer or tenant, except as provided in paragraph e of this subdivision. A buyer's agent or tenant's agent shall provide the form to the seller, seller's agent, landlord or landlord's agent at the time of the first substantive contact with the seller or landlord and shall obtain a signed acknowledgment from the seller, landlord or the listing agent, except as provided in paragraph e of this subdivision. d. The agent shall provide to the buyer, seller, tenant or landlord a copy of the signed acknowledgment and shall maintain a copy of the signed acknowledgment for not less than three years. e. If the seller, buyer, landlord or tenant refuses to sign an acknowledgment of receipt pursuant to this subdivision, the agent shall set forth under oath or affirmation a written declaration of the facts of the refusal and shall maintain a copy of the declaration for not less than three years. f. A seller/landlord or buyer/tenant may provide advance informed consent to dual agency and dual agency with designated sales agents by indicating the same on the form set forth in subdivision four of this section. a. For buyer-seller transactions, the following shall be the disclosure form: NEW YORK STATE DISCLOSURE FORM FOR BUYER AND SELLER THIS IS NOT A CONTRACT New York state law requires real estate licensees who are acting as agents of buyers or sellers of property to advise the potential buyers or sellers with whom they work of the nature of their agency relationship and the rights and obligations it creates. This disclosure will help you to make informed choices about your relationship with the real estate broker and its sales agents. Throughout the transaction you may receive more than one disclosure form. The law may require each agent assisting in the transaction to present you with this disclosure form. A real estate agent is a person qualified to advise about real estate. If you need legal, tax or other advice, consult with a professional in that field. DISCLOSURE REGARDING REAL ESTATE AGENCY RELATIONSHIPS SELLER'S AGENT A seller's agent is an agent who is engaged by a seller to represent the seller's interests. The seller's agent does this by securing a buyer for the seller's home at a price and on terms acceptable to the seller. A seller's agent has, without limitation, the following fiduciary duties to the seller: reasonable care, undivided loyalty, confidentiality, full disclosure, obedience and duty to account. A seller's agent does not represent the interests of the buyer. The obligations of a seller's agent are also subject to any specific provisions set forth in an agreement between the agent and the seller. In dealings with the buyer, a seller's agent should (a) exercise reasonable skill and care in performance of the agent's duties; (b) deal honestly, fairly and in good faith; and (c) disclose all facts known to the agent materially affecting the value or desirability of property, except as otherwise provided by law. BUYER'S AGENT A buyer's agent is an agent who is engaged by a buyer to represent the buyer's interests. The buyer's agent does this by negotiating the purchase of a home at a price and on terms acceptable to the buyer. A buyer's agent has, without limitation, the following fiduciary duties to the buyer: reasonable care, undivided loyalty, confidentiality, full disclosure, obedience and duty to account. A buyer's agent does not represent the interests of the seller. The obligations of a buyer's agent are also subject to any specific provisions set forth in an agreement between the agent and the buyer. In dealings with the seller, a buyer's agent should (a) exercise reasonable skill and care in performance of the agent's duties; (b) deal honestly, fairly and in good faith; and (c) disclose all facts known to the agent materially affecting the buyer's ability and/or willingness to perform a contract to acquire seller's property that are not inconsistent with the agent's fiduciary duties to the buyer. BROKER'S AGENTS A broker's agent is an agent that cooperates or is engaged by a listing agent or a buyer's agent (but does not work for the same firm as the listing agent or buyer's agent) to assist the listing agent or buyer's agent in locating a property to sell or buy, respectively, for the listing agent's seller or the buyer agent's buyer. The broker's agent does not have a direct relationship with the buyer or seller and the buyer or seller can not provide instructions or direction directly to the broker's agent. The buyer and the seller therefore do not have vicarious liability for the acts of the broker's agent. The listing agent or buyer's agent do provide direction and instruction to the broker's agent and therefore the listing agent or buyer's agent will have liability for the acts of the broker's agent. DUAL AGENT A real estate broker may represent both the buyer and the seller if both the buyer and seller give their informed consent in writing. In such a dual agency situation, the agent will not be able to provide the full range of fiduciary duties to the buyer and seller. The obligations of an agent are also subject to any specific provisions set forth in an agreement between the agent, and the buyer and seller. An agent acting as a dual agent must explain carefully to both the buyer and seller that the agent is acting for the other party as well. The agent should also explain the possible effects of dual representation, including that by consenting to the dual agency relationship the buyer and seller are giving up their right to undivided loyalty. A buyer or seller should carefully consider the possible consequences of a dual agency relationship before agreeing to such representation. A seller or buyer may provide advance informed consent to dual agency by indicating the same on this form. DUAL AGENT WITH DESIGNATED SALES AGENTS If the buyer and the seller provide their informed consent in writing, the principals and the real estate broker who represents both parties as a dual agent may designate a sales agent to represent the buyer and another sales agent to represent the seller to negotiate the purchase and sale of real estate. A sales agent works under the supervision of the real estate broker. With the informed consent of the buyer and the seller in writing, the designated sales agent for the buyer will function as the buyer's agent representing the interests of and advocating on behalf of the buyer and the designated sales agent for the seller will function as the seller's agent representing the interests of and advocating on behalf of the seller in the negotiations between the buyer and seller. A designated sales agent cannot provide the full range of fiduciary duties to the buyer or seller. The designated sales agent must explain that like the dual agent under whose supervision they function, they cannot provide undivided loyalty. A buyer or seller should carefully consider the possible consequences of a dual agency relationship with designated sales agents before agreeing to such representation. A seller or buyer may provide advance informed consent to dual agency with designated sales agents by indicating the same on this form. This form was provided to me by ____ (print name of licensee) of ________ (print name of company, firm or brokerage), a licensed real estate broker acting in the interest of the: ) Seller as a ) Buyer as a (check relationship below) (check relationship below) ) Seller's agent ) Buyer's agent ) Broker's agent ) Broker's agent ) Dual agent ) Dual agent with designated sales agents For advance informed consent to either dual agency or dual agency with designated sales agents complete section below: ) Advance informed consent dual agency. ) Advance informed consent to dual agency with designated sales agents. If dual agent with designated sales agents is indicated above: ________ is appointed to represent the buyer; and ________ is appointed to represent the seller in this transaction. (I)(We) acknowledge receipt of a copy of this disclosure form: Signature of { } Buyer(s) and/or { } Seller(s): Date:__ Date:__ b. For landlord-tenant transactions, the following shall be the disclosure form: NEW YORK STATE DISCLOSURE FORM FOR LANDLORD AND TENANT THIS IS NOT A CONTRACT New York state law requires real estate licensees who are acting as agents of landlords and tenants of real property to advise the potential landlords and tenants with whom they work of the nature of their agency relationship and the rights and obligations it creates. This disclosure will help you to make informed choices about your relationship with the real estate broker and its sales agents. Throughout the transaction you may receive more than one disclosure form. The law may require each agent assisting in the transaction to present you with this disclosure form. A real estate agent is a person qualified to advise about real estate. If you need legal, tax or other advice, consult with a professional in that field. DISCLOSURE REGARDING REAL ESTATE AGENCY RELATIONSHIPS LANDLORD'S AGENT A landlord's agent is an agent who is engaged by a landlord to represent the landlord's interest. The landlord's agent does this by securing a tenant for the landlord's apartment or house at a rent and on terms acceptable to the landlord. A landlord's agent has, without limitation, the following fiduciary duties to the landlord: reasonable care, undivided loyalty, confidentiality, full disclosure, obedience and duty to account. A landlord's agent does not represent the interests of the tenant. The obligations of a landlord's agent are also subject to any specific provisions set forth in an agreement between the agent and the landlord. In dealings with the tenant, a landlord's agent should (a) exercise reasonable skill and care in performance of the agent's duties; (b) deal honestly, fairly and in good faith; and (c) disclose all facts known to the agent materially affecting the value or desirability of property, except as otherwise provided by law. TENANT'S AGENT A tenant's agent is an agent who is engaged by a tenant to represent the tenant's interest. The tenant's agent does this by negotiating the rental or lease of an apartment or house at a rent and on terms acceptable to the tenant. A tenant's agent has, without limitation, the following fiduciary duties to the tenant: reasonable care, undivided loyalty, confidentiality, full disclosure, obedience and duty to account. A tenant's agent does not represent the interest of the landlord. The obligations of a tenant's agent are also subject to any specific provisions set forth in an agreement between the agent and the tenant. In dealings with the landlord, a tenant's agent should (a) exercise reasonable skill and care in performance of the agent's duties; (b) deal honestly, fairly and in good faith; and (c) disclose all facts known to the tenant's ability and/or willingness to perform a contract to rent or lease landlord's property that are not inconsistent with the agent's fiduciary duties to the buyer. BROKER'S AGENTS A broker's agent is an agent that cooperates or is engaged by a listing agent or a tenant's agent (but does not work for the same firm as the listing agent or tenant's agent) to assist the listing agent or tenant's agent in locating a property to rent or lease for the listing agent's landlord or the tenant agent's tenant. The broker's agent does not have a direct relationship with the tenant or landlord and the tenant or landlord can not provide instructions or direction directly to the broker's agent. The tenant and the landlord therefore do not have vicarious liability for the acts of the broker's agent. The listing agent or tenant's agent do provide direction and instruction to the broker's agent and therefore the listing agent or tenant's agent will have liability for the acts of the broker's agent. DUAL AGENT A real estate broker may represent both the tenant and the landlord if both the tenant and landlord give their informed consent in writing. In such a dual agency situation, the agent will not be able to provide the full range of fiduciary duties to the landlord and the tenant. The obligations of an agent are also subject to any specific provisions set forth in an agreement between the agent, and the tenant and landlord. An agent acting as a dual agent must explain carefully to both the landlord and tenant that the agent is acting for the other party as well. The agent should also explain the possible effects of dual representation, including that by consenting to the dual agency relationship the landlord and tenant are giving up their right to undivided loyalty. A landlord and tenant should carefully consider the possible consequences of a dual agency relationship before agreeing to such representation. A landlord or tenant may provide advance informed consent to dual agency by indicating the same on this form. DUAL AGENT WITH DESIGNATED SALES AGENTS If the tenant and the landlord provide their informed consent in writing, the principals and the real estate broker who represents both parties as a dual agent may designate a sales agent to represent the tenant and another sales agent to represent the landlord. A sales agent works under the supervision of the real estate broker. With the informed consent in writing of the tenant and the landlord, the designated sales agent for the tenant will function as the tenant's agent representing the interests of and advocating on behalf of the tenant and the designated sales agent for the landlord will function as the landlord's agent representing the interests of and advocating on behalf of the landlord in the negotiations between the tenant and the landlord. A designated sales agent cannot provide the full range of fiduciary duties to the landlord or tenant. The designated sales agent must explain that like the dual agent under whose supervision they function, they cannot provide undivided loyalty. A landlord or tenant should carefully consider the possible consequences of a dual agency relationship with designated sales agents before agreeing to such representation. A landlord or tenant may provide advance informed consent to dual agency with designated sales agents by indicating the same on this form. This form was provided to me by ___ (print name of licensee) of ______ (print name of company, firm or brokerage), a licensed real estate broker acting in the interest of the: ) Landlord as a ) Tenant as a (check relationship below) (check relationship below) ) Landlord's agent ( ) Tenant's agent ) Broker's agent) Broker's agent ) Dual agent ) Dual agent with designated sales agents For advance informed consent to either dual agency or dual agency with designated sales agents complete section below: ) Advance informed consent dual agency. ) Advance informed consent to dual agency with designated sales agents. If dual agent with designated sales agents is indicated above: _______ is appointed to represent the tenant; and _______ is appointed to represent the landlord in this transaction. (I) (We) _________ acknowledge receipt of a copy of this disclosure form: Signature of { } Landlord(s) and/or { } Tenant(s): ________ ________ Date: __ Date: __ This section shall not apply to a real estate licensee who works with a buyer, seller, tenant or landlord in accordance with terms agreed to by the licensee and buyer, seller, tenant or landlord and in a capacity other than as an agent, as such term is defined in paragraph a of subdivision one of this section. Nothing in this section shall be construed to limit or alter the application of the common law of agency with respect to residential real estate transactions. PreviousNext Disclaimer: These codes may not be the most recent version. New York may have more current or accurate information. We make no warranties or guarantees about the accuracy, completeness, or adequacy of the information contained on this site or the information linked to on the state site. Please check official sources. 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15764	https://www.quora.com/Can-the-sum-of-a-number-and-its-reciprocal-be-an-integer	Something went wrong. Wait a moment and try again. Rational Number Theory Number Theory 3 Can the sum of a number and its reciprocal be an integer? Reuven Harmelin Studied Mathematics at טכניון (Graduated 1978) · Author has 2.3K answers and 1.9M answer views · 1y You are asking about solutions of the equation where n is some integer. Since the function is an odd function, which is positive for every x>0, it is enough to consider the solutions over the positive half the real line. From the following algebraic inequality and the equality attained for, and only for x=1, so at least x=1 is an answer to your question Moreover, since it is enough to look for possible solutions x>1. Here we may use the intermediate property of continuous function, the fact that f(x) is continuous all over the positive (and the negative ) half of the real line, that this function is st You are asking about solutions of the equation where n is some integer. Since the function is an odd function, which is positive for every x>0, it is enough to consider the solutions over the positive half the real line. From the following algebraic inequality and the equality attained for, and only for x=1, so at least x=1 is an answer to your question Moreover, since it is enough to look for possible solutions x>1. Here we may use the intermediate property of continuous function, the fact that f(x) is continuous all over the positive (and the negative ) half of the real line, that this function is strictly increasing in the infinite interval (0,∞) and that the limit of f(x) at infinity is infinity, and deduce that for every natural number n>2 there must be exactly one x>1 such that If you want the explicit solution x>1 for any given natural number n>2, you can rewrite the above equation as follows and obtain from the formula of the roots of quadratic equations Notice that since for every natural number n>2 the discriminant could never be equal to a perfect square, its square root is always an irrational number, and therefore x=1 is the only positive integer for which the sum of it and its reciprocal could be an integer. Related questions What is the sum of a number and its reciprocal? Can a number be an integer without having an integer as its reciprocal? If the sum of a number and its reciprocal is 2, then what is the number? What is the sum of a positive number and its reciprocal? Can the reciprocal of an irrational number be an integer? Richard P MMath in Mathematics, Churchill College, Cambridge (Graduated 2011) · Author has 795 answers and 306.1K answer views · 1y Yes. [Solutions follow for x + 1/x = n, where n≥2 is an integer] An example entirely in integers is 1 + 1/1 = 2. For integer n>2, there are real solutions for x: x + 1/2 = n implies x^2 - nx + 1 = 0, with solutions x = (n/2) ± √(n^2/4 - 1) Examples: n=3, x = (3±√5)/2 n=4, x = 2±√3 n=5, x = (5±√21)/2 n=6, x = 3±2√2 … Dave Neary B.Sc. in Mathematics, National University of Ireland, Galway (Graduated 1996) · Author has 2.9K answers and 1.2M answer views · 1y Yes. Take any positive integer n, and solve for the equation: x+1x=n x2−nx+1=0 x=12(n±√n2−4) which has two real solutions for all positive integers n≥2. Jack Zinn MS in Statistics (academic discipline), Cornell University (Graduated 1990) · 1y Sure. The obvious answer is 1, which is its own reciprocal. 1 + (1/1) = 2. This is also true for -1, as -1 + (1/-1) = -2. There are no other values where z + (1/z) = N, where N is an integer, whether z is Real or Complex. Related questions What is the reciprocal of an integer? What is a number such that the sum of the number at its reciprocal is 8? Is the reciprocal of every number meaningful? If the sum of a number and its reciprocal is 14.Then, what is the value of sum of the cubes of the number and its reciprocal? What will be the sum reciprocal of 2 numbers if their sum is 36 and product is 248? Ted Hopp Math hobbyist and Ph.D. in CS · Upvoted by Michael Jørgensen , PhD in mathematics · Author has 2.4K answers and 3.5M answer views · Updated 5y Related What are all possible ways of writing 1/21 as the sum of the reciprocals of two positive integers? Let the two positive integers be x and y. We have 1x+1y=121 which can be rewritten as follows: x+yxy=121xy−21(x+y)=0xy−21(x+y)+212=212(x−21)(y−21)=212 Substitute u=x−21,v=y−21, we seek two integers greater than −21 such that uv=3272 where 3272 is the prime factorization of 212. Note that u and v are either both positive or both negative. Also, since addition is commutative, we can assume without loss of generality that \|u\|≤\|v\|. Then the possible positive Let the two positive integers be x and y. We have 1x+1y=121 which can be rewritten as follows: x+yxy=121xy−21(x+y)=0xy−21(x+y)+212=212(x−21)(y−21)=212 Substitute u=x−21,v=y−21, we seek two integers greater than −21 such that uv=3272 where 3272 is the prime factorization of 212. Note that u and v are either both positive or both negative. Also, since addition is commutative, we can assume without loss of generality that \|u\|≤\|v\|. Then the possible positive values for u and v are: u v1(=3070)441(=3272)3(=3170)147(=3172)7(=3071)63(=3271)9(=3270)49(=3072)21(=3171)21(=3171) There are no negative values for u or v for which both would be greater than −21, so this is a complete list. Adding 21 to each possible pair of values for u and v generates the following sums: 121=122+1462=124+1168=128+184=130+170=142+142 Note that if we relax the requirement that x and y be positive, we have several other solutions. We still require that u,v≠−21 (since neither x or y can be 0) but we can use the following factorizations: uv−1−441−3−147−7−63−9−49 These correspond to the following differences of reciprocals of two integers: 121=120−1420=118−1126=114−142=112−128 There are no other sums or differences of two integer reciprocals that equal 1/21. Amitabha Tripathi five decades of high school Algebra · Upvoted by Jeremy Collins , M.A. Mathematics, Trinity College, Cambridge · Author has 4.7K answers and 13.8M answer views · Updated 6y Related How can I determine all possible ways of writing 1/11 as the sum of the reciprocals of two positive integers? More generally, for prime p, 1p=1a+1b with . Note that . Thus , and so . Doctor Sachidanand Das PhD(Physics),PG (Maths),BSc(Gold Med),38yrs academic & govt. · Author has 11.3K answers and 15.9M answer views · 7y Related The sum of two integers is 10 and the sum of their reciprocal is 5/12. What is the largest of the integers? Answer: 6 Proof: Let the two integers be m and n. It is given that the sum of the two integers is 10. This gives the equation m+ n = 10………………….. ………………(1) It is also given that the sum of their reciprocal is 5/12. The reciprocal of m is 1/m and the reciprocal of n is 1/n. Therefore we get the second equation 1/m + 1/n = 5/12 Or, (m+n)/mn = 5/12 Substituting for m+n = 10 from (1), 10/mn = 5/12 Divide both sides by 5 to get 2/mn = 1/12 Taking reciprocals on both sides, mn/2 = 12 Cross-multiplying, mn = 24 This gives m = 24/n Substituting this value of m in (1), 24/n + n = 10 Taking LCM, (24 + n^2)/n = 10 Cross-mult Answer: 6 Proof: Let the two integers be m and n. It is given that the sum of the two integers is 10. This gives the equation m+ n = 10………………….. ………………(1) It is also given that the sum of their reciprocal is 5/12. The reciprocal of m is 1/m and the reciprocal of n is 1/n. Therefore we get the second equation 1/m + 1/n = 5/12 Or, (m+n)/mn = 5/12 Substituting for m+n = 10 from (1), 10/mn = 5/12 Divide both sides by 5 to get 2/mn = 1/12 Taking reciprocals on both sides, mn/2 = 12 Cross-multiplying, mn = 24 This gives m = 24/n Substituting this value of m in (1), 24/n + n = 10 Taking LCM, (24 + n^2)/n = 10 Cross-multiplying, 24 + n^2 = 10n Or, n^2 -10n + 24 = 0 Factorising, n^2 -6n -4n + 24 = 0 Or, n(n-6) - 4(n-6) =0 Or, (n-6) (n-4) = 0 This gives two solutions for n: 6 and 4 The corresponding solutions for m are obtained from (1) as 10-n and they are 4 and 6. Thus there are two pairs of solutions for (m,n), (4,6) and (6,4) It follows from above that The largest integer is 6. Ayush Anand B.Tech in Computer Science Engineering, Dr. B. R. Ambedkar National Institute of Technology, Jalandhar ( NITJ ) (Graduated 2023) · 4y Related How can you prove that the sum of a positive number and its reciprocal is never less than two? Well, there a rule or a property in algebra which says that the Arithmetic Mean (A.M) of a set of numbers is always greater than or equal to its Geometric Mean (G.M). More formally, AM >= GM Lets take the number to be x, so its reciprocal would be 1 / x. So now, so by AM >= GM we have the result, ( x + 1/x) / 2 >= 1 or, x + 1/x >= 2 Therefore, the sum of a number and its reciprocal is always greater than or equal to 2. Alternatively, if you know how to find maxima/minima of a function then you can assume your function f(x) = Well, there a rule or a property in algebra which says that the Arithmetic Mean (A.M) of a set of numbers is always greater than or equal to its Geometric Mean (G.M). More formally, AM >= GM Lets take the number to be x, so its reciprocal would be 1 / x. So now, so by AM >= GM we have the result, ( x + 1/x) / 2 >= 1 or, x + 1/x >= 2 Therefore, the sum of a number and its reciprocal is always greater than or equal to 2. Alternatively, if you know how to find maxima/minima of a function then you can assume your function f(x) = (x + 1/x). and then equate its derivative with zero, and also perform the single or double derivative test to know the critical points. With that you also get to know that if the number x is negative, then x + 1/x <= -2 So the final conclusion comes out to be x + 1/x >= 2, when x > 0 and x + 1/x <= -2 when x < 0 Hope I answered you!! David Harden Bachelor of Science in Mathematics, Massachusetts Institute of Technology (Graduated 2004) · Author has 1.3K answers and 428.3K answer views · 1y Related Is there a set of consecutive composite numbers whose reciprocals sum up to an integer? There isn’t such a set, and it’s not hard to see this using a little familiarity with the harmonic series. First of all, even before we remove primes, we must have a max/min ratio exceeding 2, since, for any positive integer , and therefore . Here, because 4 is the smallest composite number. Now, since you said these are consecutive composite numbers, we’re not allowed to skip any. Since the max/min ratio among them exceeds 2, there must be a power of 2 among them. Then let be There isn’t such a set, and it’s not hard to see this using a little familiarity with the harmonic series. First of all, even before we remove primes, we must have a max/min ratio exceeding 2, since, for any positive integer , and therefore . Here, because 4 is the smallest composite number. Now, since you said these are consecutive composite numbers, we’re not allowed to skip any. Since the max/min ratio among them exceeds 2, there must be a power of 2 among them. Then let be the largest power of 2 in the set; note that because is composite. Then, since is not in the set and these composite numbers are consecutive, is the only multiple of in the set. But then it’s easy to see that the sum of the reciprocals of these composite numbers must be expressible as for some nonnegative integers and . A number of this form can’t be an integer. Alberto Cid M.S.E. in Telecommunications Engineering & Data Transmission, Technical University of Madrid (Graduated 2008) · Author has 2K answers and 3.8M answer views · 1y Related Can the sum of a transcendental number and its reciprocal be algebraic? Let “t” be transcendental and “a” be algebraic. a = t + 1/t = (t²-1)/t The expression “a is algebraic” means there exists some polynomial P with integer coefficients such that: P(a) = 0 P(x) = c(n) x^n + c(n-1) x^(n-1) + … P(a) = c(n) a^n + c(n-1) a^(n-1) + … Let's see how any power of “a” is: a^n = (t + 1/t)^n = t^n + n•t^(n-2) + … + n•t^-(n-2) + t^-n Then x^n • a^n = Q(x) polynomial with integer coefficients Then x^n • P(x) = R(x) polynomial with integer coefficients Then in case “a” is algebraic then P(a) = 0 where coefficients of P(x) are integers, then there exists some R(x) with integer coefficien Let “t” be transcendental and “a” be algebraic. a = t + 1/t = (t²-1)/t The expression “a is algebraic” means there exists some polynomial P with integer coefficients such that: P(a) = 0 P(x) = c(n) x^n + c(n-1) x^(n-1) + … P(a) = c(n) a^n + c(n-1) a^(n-1) + … Let's see how any power of “a” is: a^n = (t + 1/t)^n = t^n + n•t^(n-2) + … + n•t^-(n-2) + t^-n Then x^n • a^n = Q(x) polynomial with integer coefficients Then x^n • P(x) = R(x) polynomial with integer coefficients Then in case “a” is algebraic then P(a) = 0 where coefficients of P(x) are integers, then there exists some R(x) with integer coefficients such that R(x) = 0… then x should not be transcendental. Then: NO, that's not possible. Furthermore: the sum of two powers with integer exponents (except exponent zero) of any transcendental number must be transcendental. Rik Bos Ph.D. Mathematics from Utrecht University (Graduated 1979) · Author has 1.3K answers and 1.3M answer views · 3y Related The sum of the reciprocals of three consecutive integers is 47/60. What is the sum of these three integers? Let’s generalize and drop the condition that the numbers are consecutive. Just suppose are positive integers satisfying . Clearly we can interchange , so let’s assume . Then . Also, if , then . Therefore or . In the first case, . Clearly and if , we have . Therefore . Now it’s easy to check that gives and gives . Moreo Let’s generalize and drop the condition that the numbers are consecutive. Just suppose are positive integers satisfying . Clearly we can interchange , so let’s assume . Then . Also, if , then . Therefore or . In the first case, . Clearly and if , we have . Therefore . Now it’s easy to check that gives and gives . Moreover, or don’t allow integer solutions for . So far we have found the solutions and . Now suppose . Then . If , then . Therefore or and it’s easy to check that gives and that doesn’t allow an integer solution for . So combining this with the case where we have three solutions, namely , and . It’s worth to mention here the idea of Egyptian fractions, which are sums of unit fractions (i.e. fractions of the form where is a positive integer). As illustrated by the analysis above, we can write as for each of the three solutions above. By the way, it is impossible to write as a sum of two unit fractions (by a similar analysis as above), but every positive rational can be written as an Egyptian fraction. In fact, there is an algorithm to find a particular Egyptian fraction for a given positive rational. To illustrate this algorithm, let’s look again at . If is the first summand and we want this term to be as large as possible, then is the smallest integer greater than . So and we now have to write as an Egyptian fraction. Repeating the previous step applied to this fraction we find as second term where is the smallest integer greater than . So . And finally we find . This algorithm can produce representations with many terms and there are many interesting questions about the minimum length of the representation. For instance, it is not known if the fraction can always be represented as the sum of three unit fractions, although it will be very hard to find counter examples; see [Erdős–Straus conjecture - Wikipedia On unit fractions adding to 4/n Unsolved problem in mathematics Does 4 n = 1 x + 1 y + 1 z {\displaystyle {\tfrac {4}{n}}={\tfrac {1}{x}}+{\tfrac {1}{y}}+{\tfrac {1}{z}}} have a positive integer solution for every integer n ≥ 2 {\displaystyle n\geq 2} ? The Erdős–Straus conjecture is an unproven statement in number theory . The conjecture is that, for every integer n {\displaystyle n} that is greater than or equal to 2, there exist positive integers x {\displaystyle x} , y {\displaystyle y} , and z {\displaystyle z} for which 4 n = 1 x + 1 y + 1 z . {\displaystyle {\frac {4}{n}}={\frac {1}{x}}+{\frac {1}{y}}+{\frac {1}{z}}.} In other words, the number 4 / n {\displaystyle 4/n} can be written as a sum of three positive unit fractions . The conjecture is named after Paul Erdős and Ernst G. Straus , who formulated it in 1948, but it is connected to much more ancient mathematics; sums of unit fractions, like the one in this problem, are known as Egyptian fractions , because of their use in ancient Egyptian mathematics . The Erdős–Straus conjecture is one of many conjectures by Erdős , and one of many unsolved problems in mathematics concerning Diophantine equations . Although a solution is not known for all values of n , infinitely many values in certain infinite arithmetic progressions have simple formulas for their solution, and skipping these known values can speed up searches for counterexamples . Additionally, these searches need only consider values of n {\displaystyle n} that are prime numbers , because any composite counterexample would have a smaller counterexample among its prime factors . Computer searches have verified the truth of the conjecture up to n ≤ 10 17 {\displaystyle n\leq 10^{17}} . If the conjecture is reframed to allow negative unit fractions, then it is known to be true. Generalizations of the conjecture to fractions with numerator 5 or larger have also been studied. Background and history [ edit ] When a rational number is expanded into a sum of unit fractions, the expansion is called an Egyptian fraction . This way of writing fractions dates to the mathematics of ancient Egypt , in which fractions were written this way instead of in the more modern vulgar fraction form a b {\displaystyle {\tfrac {a}{b}}} with a numerator a {\displaystyle a} and denominator b {\displaystyle b} . The Egyptians produced tables of Egyptian fractions for unit fractions multiplied by two, the numbers that in modern notation would be written 2 n {\displaystyle {\tfrac {2}{n}}} , such as the Rhind Mathematical Papyrus table ; in these tables, most of these expansions use either two or three terms. [ 1 ] These tables were needed, because the obvious expansion 2 n = 1 n + 1 n {\displaystyle {\tfrac {2}{n}}={\tfrac {1}{n}}+{\tfrac {1}{n}}} was not allowed: the Egyptians required all of the fractions in an Egyptian fraction to be different from each other. This same requirement, that all fractions be different, is sometimes imposed in the Erdős–Strau]( "en.wikipedia.org") Andy Albee BSSE in Mathematics & Physics, University of Idaho (Graduated 1992) · Author has 1K answers and 116K answer views · 1y Related Can an integer and its reciprocal add up to a perfect square if one of the numbers is negative? To add an integer and its reciprocal, you must have common denominators. Since every integer has a denominator of 1, they can be represented by p/1 and 1/p where p is an integer If you take p/1 and 1/p and add them together, you will get: taking the first ratio and multiplying it by results in a common denominator which can then be added since cannot be a perfect square and p is not a factor of then cannot be a perfect square. And your statement that “one of the numbers is negative” is not possible. If an integer is No. To add an integer and its reciprocal, you must have common denominators. Since every integer has a denominator of 1, they can be represented by p/1 and 1/p where p is an integer If you take p/1 and 1/p and add them together, you will get: taking the first ratio and multiplying it by results in a common denominator which can then be added since cannot be a perfect square and p is not a factor of then cannot be a perfect square. And your statement that “one of the numbers is negative” is not possible. If an integer is positive, its reciprocal will be positive. If an integer is negative, its reciprocal will be negative. Since a square cannot be negative, neither of the numbers being added can be negative. David Joyce Professor Emeritus of Mathematics at Clark Univerisity · Upvoted by Terry Moore , M.Sc. Mathematics, University of Southampton (1968) and Aditya Garg , M.Sc. Mathematics, Indian Institute of Technology, Delhi (2013) · Author has 9.9K answers and 68M answer views · 6y Related The sum of two integers is 10 and the sum of their reciprocal is 5/12. What is the largest of the integers? It’s a good idea to take advantage of symmetry when you see it. Since the sum of the two numbers is the larger one will be and the smaller where is some positive number. Their reciprocals sum to which should equal Clearing the denominators, we get the equation so Therefore, and so does Thus, the larger number is Related questions What is the sum of a number and its reciprocal? Can a number be an integer without having an integer as its reciprocal? If the sum of a number and its reciprocal is 2, then what is the number? What is the sum of a positive number and its reciprocal? Can the reciprocal of an irrational number be an integer? What is the reciprocal of an integer? What is a number such that the sum of the number at its reciprocal is 8? Is the reciprocal of every number meaningful? If the sum of a number and its reciprocal is 14.Then, what is the value of sum of the cubes of the number and its reciprocal? What will be the sum reciprocal of 2 numbers if their sum is 36 and product is 248? Which number does not have a reciprocal? The sum of a number and its reciprocal is 4. What is their difference? The sum of a number and 6 times its reciprocal is 7 . Find the number? Does dividing one by an integer reciprocal give an integer? What is the sum of two numbers and their reciprocal? Related questions What is the sum of a number and its reciprocal? Can a number be an integer without having an integer as its reciprocal? If the sum of a number and its reciprocal is 2, then what is the number? What is the sum of a positive number and its reciprocal? Can the reciprocal of an irrational number be an integer? What is the reciprocal of an integer? What is a number such that the sum of the number at its reciprocal is 8? Is the reciprocal of every number meaningful? If the sum of a number and its reciprocal is 14.Then, what is the value of sum of the cubes of the number and its reciprocal? 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15765	https://www.purplemath.com/modules/systlin1.htm	Published Time: Thu, 04 Jan 2024 19:57:46 GMT Systems of Linear Equations: Definitions \| Purplemath Skip to main content Home Lessons HW Guidelines Study Skills Quiz Find Local Tutors Demo MathHelp.com Join MathHelp.com Login Select a Course Below Standardized Test Prep ACCUPLACER Math ACT Math ALEKS Math ASVAB Math CBEST Math CLEP Math FTCE Math GED Math GMAT Math GRE Math HESI Math Math Placement Test NES Math PERT Math PRAXIS Math SAT Math TEAS Math TSI Math VPT Math + more tests K12 Math 5th Grade Math 6th Grade Math Pre-Algebra Algebra 1 Geometry Algebra 2 College Math College Pre-Algebra Introductory Algebra Intermediate Algebra College Algebra Homeschool Math Pre-Algebra Algebra 1 Geometry Algebra 2 Search Systems of Linear Equations: Definitions DefinitionsGraphingSpecial CasesSubstitutionElimination/AdditionGaussian EliminationMore Examples Purplemath A "system" of equations is a set or collection of equations that you deal with all together at once. Linear equations (ones that graph as straight lines) are simpler than non-linear equations, and the simplest linear system is one with two equations and two variables. Think back to when you were first learning about two-variable linear equations, often stated in the form "y = mx + b. For instance, consider the linear equation y = 3 x − 5. A "solution" to this equation was any (x, y)-point that "worked" in the equation. Content Continues Below MathHelp.com Solving Systems by Addition So, for instance, the point (2, 1) is a solution to the equation because plugging in 2 for x returns the correct value for y, as shown below: 3 x − 5 = 3(2) − 5 = 6 − 5 = 1 = y On the other hand, the point (1, 2) is not a solution to the equation, because plugging in 1 for x leads to a value which is not the given y-value, as we can see: 3 x − 5 = 3(1) − 5 = 3 − 5 = −2 Advertisement This value, −2, is not the value of y that we were given; namely, +2. So the point cannot be a solution to the equation. Affiliate Of course, in practical terms, you did not find solutions to an equation by picking random points, plugging them in, and checking to see if they "work" in the equation. Instead, you picked x-values and then calculated the corresponding y-values. And you used this same procedure to graph the equation. This points out an important fact: Every point on the graph was a solution to the equation, and any solution to the equation was a point on the graph. Now consider the following two-variable system of linear equations: y = 3 x − 2 y = −x − 6 Since the two equations above are in a system, we deal with them together at the same time. In particular, we can graph them together on the same axis system, like this: A solution for a single equation is any point that lies on the line for that equation. A solution for a system of equations is any point that lies on each line in the system. For example, the red point on the graph below is not a solution to the system, because it is not on either line: The blue point on the graph below is not a solution to the system, because it lies on only one of the lines, not on both of them: The purple point on the graph below is a solution to the system, because it does lie on both of the lines: Affiliate Affordable tutors for hire Find tutors In particular, this purple point marks the intersection of the two lines. Since this point is on both lines, it thus solves both equations, so it therefore solves the entire system of equations. This relationship is always true: For systems of equations, "solutions" of systems are "intersections" of lines. And you can confirm that the intersection point is the solution to the system by plugging that point back into each of the system's equations, and confirming that the point works in each equation. Content Continues Below Determine whether either of the points (−1, −5) and (0, −2) is a solution to the given system of equations. y = 3 x − 2 y = −x − 6 To check the given possible solutions, I just plug the x- and y-coordinates into the equations, and check to see if they work. checking (−1, −5): (−5) ?=? 3(−1) − 2 −5 ?=? −3 − 2 −5 = −5 (solution checks) (−5) ?=? −(−1) − 6 −5 ?=? 1 − 6 −5 = −5 (solution checks) Affiliate Since the given point works in each equation, it is a solution to the system. Now I'll check the other point (which we already know, from looking at the graph, is not a solution): checking (0, −2): (−2) ?=? 3(0) − 2 −2 ?=? 0 − 2 −2 = −2 (solution checks) So the solution does work in one of the equations. But to be a solution to the system of equations, it has to work in both of the equations in that system. Continuing the check: (−2) ?=? −(0) − 6 −2 ?=? 0 − 6 −2 ?=? −6 But −2 does not equal −6, so this "solution" does not check. Then the answer is: only the point (−1, −5) is a solution to the system URL: Page 1Page 2Page 3Page 4Page 5Page 6Page 7 You can use the Mathway widget below to practice checking if a given point is a solution to a given system. Try the entered exercise, or type in your own exercise. Then click the button to compare your answer to Mathway's. Please accept "preferences" cookies in order to enable this widget. (Click "Tap to view steps" to be taken directly to the Mathway site for a paid upgrade.) 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15766	https://artofproblemsolving.com/wiki/index.php/2015_USAMO_Problems/Problem_1?srsltid=AfmBOorLHLfGbYyeofCKe4xU5rCcpd_WuLLVx_WgS5GuwnzfqKPRatDh	Art of Problem Solving 2015 USAMO Problems/Problem 1 - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki 2015 USAMO Problems/Problem 1 Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search 2015 USAMO Problems/Problem 1 Problem Solve in integers the equation Solution We first notice that both sides must be integers, so must be an integer. We can therefore perform the substitution where is an integer. Then: is therefore the square of an odd integer and can be replaced with By substituting using we get: or Using substitution we get the solutions: Solution 2 Let . Thus, . We have Substituting for , we have Treating as a variable and as a constant, we have which turns into a quadratic equation. By the quadratic formula, which simplifies to Since we want and to be integers, we need to be a perfect square. We can factor the aforementioned equation to be for an integer . Since is always a perfect square, for to be a perfect square, has to be a perfect square as well. Since is odd, the square root of the aforementioned equation must be odd as well. Thus, we have for some odd . Thus, in which by difference of squares it is easy to see that all the possible values for are just , where is a positive integer. Thus, Thus, the general form for for a positive integer . (This is an integer since is an even perfect square (since is always even, as well as being always even) as established, and is always even as well. Thus, the whole numerator is even, which makes the quantity of that divided by always an integer.) Since , the general form for is just (This is an integer since is an even perfect square (since is always even, as well as being always even) as established, and is always even as well. Thus, the whole numerator is even, which makes the quantity of that divided by always an integer, which thus trivially makes an integer.) for a positive integer . Thus, our general in integers is -fidgetboss_4000 Retrieved from " Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
15767	https://www.youtube.com/watch?v=pv8URIRgCdo	Two Digit Subtraction with Regrouping - Common Core Super Schoolhouse 17200 subscribers 2974 likes Description 591229 views Posted: 17 Apr 2015 A simple and clear explanation, with drawings, of how to subtract two-digit numbers with regrouping. Transcript: Introduction hi everybody today we're going to talk about subtraction with two-digit numbers and regrouping so let's get started and look at a real subtraction problem let's Example say that we've got the number 42 minus 25 now before we get started let's make a drawing of the number 42 so that it's a little easier to see what we're doing so if we want to draw the number of 42 how many tens are there there are four tens so let's draw them one two three and four and how many ones have we got there are two ones so let's bring in one two ones so that's the number 42 drawn with the blue squares and we're ready to subtract or take away 25 now the first thing we want to do is subtract our ones but right away we're going to notice that we've got a problem we can't do 2 minus 5 we don't have enough ones at the top we can't take away away five ones from two ones so where are we going to get some more ones well the answer is we're going to go over to our tens we're going to take one of our tens right here and we're going to do something called regrouping we're going to turn this one group of tens into a group of 10 ones then we're going to bring those 10 ones over to our other ones this is still the number 42 but we've just regrouped so that we can do our subtra subaction now let's show our regrouping with our numbers we don't have four 10 anymore we've got 310 and we don't have two ones anymore we've got 12 ones now we have enough ones to do our subtraction what is 12 - 5 well let's take away five ones from our drawing 1 2 3 four five and we can see that we have seven left now let's move over to our tens and we take away two tens one two and we have just one 10 so the answer to our subtraction problem is 17 all right so let's see if we can do another problem like this without the drawing let's do for example 63 minus No Drawing 37 first thing we do is look at our ones can we do 3us 7 no we can't so we need to get more ones from our tens we're going to regroup one of our tens by taking away one of them so that we'll have just five left then we're moving that 10 over to our ones so that instead of three ones we have 13 1es now we can subtract our ones what is 13 minus 7 13 - 7 is 6 so we write that down in our ones column then we sub subract our 10 what is 5 - 3 easy 5 - 3 is 2 and now we know that our answer is 26 and that is subtraction with two-digit numbers and regrouping
15768	https://www.purplemath.com/modules/asymtote3.htm	Published Time: Thu, 13 Mar 2025 18:47:04 GMT How to find the slant (or oblique) asymptotes \| Purplemath Skip to main content Home Lessons HW Guidelines Study Skills Quiz Find Local Tutors Demo MathHelp.com Join MathHelp.com Login Select a Course Below Standardized Test Prep ACCUPLACER Math ACT Math ALEKS Math ASVAB Math CBEST Math CLEP Math FTCE Math GED Math GMAT Math GRE Math HESI Math Math Placement Test NES Math PERT Math PRAXIS Math SAT Math TEAS Math TSI Math VPT Math + more tests K12 Math 5th Grade Math 6th Grade Math Pre-Algebra Algebra 1 Geometry Algebra 2 College Math College Pre-Algebra Introductory Algebra Intermediate Algebra College Algebra Homeschool Math Pre-Algebra Algebra 1 Geometry Algebra 2 Search Slant (Oblique) Asymptotes VerticalHorizontalSlantExamples Purplemath In the previous section, covering horizontal asymptotes, we learned how to deal with rational functions where the degree of the numerator was equal to or less than that of the denominator. But what happens if the degree is greater in the numerator than in the denominator? Content Continues Below MathHelp.com Recall that, when the degree of the denominator was bigger than that of the numerator, we saw that the value in the denominator got so much bigger, so quickly, that it was so much stronger that it pulled the functional value to zero, giving us a horizontal asymptote of the x-axis. Reasonably, then, if the numerator has a power that is larger than that of the denominator, then the value of the numerator ought to be stronger, and ought to pull the graph away from the x-axis; that is, it ought to pull the graph away from the line y=0 and away from any other fixed y-value. To investigate this, let's look at the following function: y=−3 x 2+2 x−1 y = \dfrac{-3x^2 + 2}{x - 1}y=x−1−3 x 2+2 For reasons that will shortly become clear, I'm going to apply long polynomial division to this rational expression. My work looks like this: −3 x−3+2 x−1)−3 x 2+0 x 0 0 0+2‾−3 x 2+3 x+2‾−3 x+2−3 x+3‾−1\small{ \begin{array}{r} -3x\enspace -3\, \phantom{+2} \ x-1 \overline{\smash{\big)} \, -3x^2 + 0x\vphantom{0^{0^0}} + 2} \[0.5em] \underline{-3x^2 + 3x \phantom{+2}} \[0.5em] -3x+2 \[0.5em] \underline{-3x+3} \[0.5em] -1 \end{array} }−3 x−3+2 x−1)−3 x 2+0 x 0 0 0+2−3 x 2+3 x+2−3 x+2−3 x+3−1 Across the top is the quotient, being the linear polynomial expression −3 x−3. At the bottom is the remainder. This means that, via long division, I can convert the original rational function they gave me into something akin to what would be mixed-number format in numerical values: y=−3 x−3+−1 x−1 y = -3x - 3 + \dfrac{-1}{x - 1}y=−3 x−3+x−1−1 This is the exact same function. All I've done is rearrange it a bit. Why? You're about to see. First, take a look at the graph of the rational function they gave us: The parts in the middle, where the line goes pretty much straight up though the top and then comes pretty much straight up from the bottom, indicate where there's a vertical asymptote (at x=1). But what about the slanty parts, that seem to kind of line up with each other? Thinking back to the results of my long division, we know what the graph of y=−3 x−3 looks like; it's a decreasing straight line that crosses the y-axis at −3, having a slope of m=−3. Now take a look at this second graph of the same rational function, but with the line y=−3 x−3 superimposed on it: As you can see, apart from the middle of the plot near the origin (that is, apart from when the graph is close to the vertical asymptote), the graph hugs the line y=−3 x−3. Because of this "skinnying along the line" behavior of the graph, the line y=−3 x−3 is an asymptote. Clearly, though, it's not a horizontal asymptote. Instead, because this asymptote is slanted or, in fancy terminology, "oblique", this is called a slant (or oblique) asymptote. Affiliate Affordable tutors for hire Find tutors The graph above illustrates the fact that, if the degree of the numerator is exactly one more than the degree of the denominator (so that the polynomial fraction is an "improper" fraction), then the graph of the rational function will be, roughly, a slanty straight line with some fiddly bits in the middle. Why does it work like this? Because of the how the long division works out. The degree of the numerator is one more than the degree of the denominator, so the long division will give you a degree-1 (that is, a linear) polynomial part (in other words, something of the form m x+b) and a "proper" fractional part. We already know that, off to the sides, the fractional part will quickly go to the x-axis; that is, its value will quickly get, and stay, close to zero. So, off to the sides, the graph will pretty much boil down to (the linear polynomial part) plus (zero). In other words, most of the graph will be approximately equal to the linear polynomial part. Because the graph will be nearly equal to this slanted straight-line equivalent, the asymptote for this sort of rational function is called a slant (or "oblique") asymptote. The equation for the slant asymptote is the polynomial part of the rational that you get after doing the long division. (By the way, this relationship — between an improper rational function, its associated polynomial, and its graph — holds true regardless of the difference in the degrees of the numerator and denominator. However, in most textbooks, they only have you work with a degree-difference of one.) Content Continues Below To summarize: What is a slant (or oblique) asymptote? Suppose a rational function has a numerator whose degree is exactly 1 greater than the denominator's degree. The slant (or oblique) asymptote for that rational function is a straight (but not horizontal or vertical) line that shows where the graph goes, off to the sides. How do you find the slant (or oblique) asymptote? To find the slant asymptote, do the long division of the numerator by the denominator. The result will be a degree-2 polynomial part (across the top of the long division) and a proper fractional part (formed by dividing the remainder by the denominattor). The linear polynomial, when set equal to y, is the slant asymptote. What is the difference between slant, horizontal, and vertical asymptotes? Vertical asymptotes are caused by zeroes of the denominator; they indicate where the graph must never go, as this would cause division by zero. As a result, they can never be touched or crossed. Horizontal asymptotes are caused by the numerator having a degree that is smaller than, or equal to, the degree of the denominator; they indicate where the graph will be when it's off to the sides (away from vertical asymptotes, etc). Horizontal asymptotes can be touched and/or crossed. Slant asymptotes are caused by the numerator having a degree that is 1 greater than that of the denominator; they indicate where the graph will be when it's off to the sides. Slant asymptotes can be touched and/or crossed. Find the slant asymptote of the following function: y=x 2+3 x+2 x−2\small{ \boldsymbol{\color{green}{y = \dfrac{x^2 + 3x + 2}{x - 2} }}}y=x−2 x 2+3 x+2 To find the slant asymptote, I need to do the long division: x+5 x+2 x−2)x 2+3 x 0 0 0+0 2‾x 2−2 x+02‾5 x+0 2 5 x−10‾12\small{ \begin{array}{r} x\;+5\phantom{x} \phantom{+2} \ x-2 {\overline{\smash{\big)} x^2+3x\vphantom{0^{0^0}}+\phantom{0}2}} \[0.5em] \underline{x^2-2x\phantom{+02}} \[0.5em]5x+\phantom{0}2 \[0.5em] \underline{5x-10} \[0.5em] 12 \end{array} }x+5 x+2 x−2)x 2+3 x 0 0 0+0 2x 2−2 x+025 x+0 2 5 x−1012 I need to remember that the slant asymptote is the polynomial part of the answer (that is, the asymptote is the part across the top of the division, set equal to y), not the remainder (that is, not the last value at the bottom). Then my answer is: slant asymptote: y = x + 5 Affiliate Affordable tutors for hire Find tutors Find the slant asymptote of the following function: y=2 x 3+4 x 2−9 3−x 2\small{ \boldsymbol{\color{green}{y = \dfrac{2x^3 + 4x^2 - 9}{3 - x^2} }}}y=3−x 2 2 x 3+4 x 2−9 They've tried to trip me up here! They omitted a linear term in the polynomial on top, and they put the terms in the wrong order underneath. When I'm doing my long division, I'll need to be careful to include the missing linear term in the numerator, and I'll need to keep track of the signs when I reverse the terms in the denominator. −2 x 2−4 x 2+0 x−19−x 2+3)2 x 3+4 x 2+0 0 0 0 x+1 9‾2 x 3+4 x 2−6 x−19‾4 x 2+6 x−1 9 4 x 2+6 x−12‾6 x+1 3\small{ \begin{array}{r} -2x^2 - 4 \phantom{x^2+0x-19} \ -x^2+3 {\overline{\smash{\big)} \, 2x^3+4x^2+\vphantom{0^{0^0}}0x+\phantom{1}9}} \[0.5em] \underline{2x^3 \phantom{+4x^2} -6x \phantom{-19}} \[0.5em] 4x^2+6x-\phantom{1}9 \[0.5em] \underline{4x^2 \phantom{+6x} -12} \[0.5em] 6x+\phantom{1}3 \end{array} }−2 x 2−4 x 2+0 x−19−x 2+3)2 x 3+4 x 2+0 0 0 0 x+1 92 x 3+4 x 2−6 x−194 x 2+6 x−1 9 4 x 2+6 x−126 x+1 3 The slant asymptote is the polynomial part of the answer; I can ignore the remainder. So my answer is: slant asymptote: y = −2 x − 4 (If you're not comfortable with the long-division part of these exercises, then go back and reviewnow!) Anote for the curious regarding the horizontal and slant asymptote rules. Otherwise, continue on to the worked examples. URL: Page 1Page 2Page 3Page 4 Select a Course Below Standardized Test Prep ACCUPLACER Math ACT Math ALEKS Math ASVAB Math CBEST Math CLEP Math FTCE Math GED Math GMAT Math GRE Math HESI Math Math Placement Test NES Math PERT Math PRAXIS Math SAT Math TEAS Math TSI Math VPT Math + more tests K12 Math 5th Grade Math 6th Grade Math Pre-Algebra Algebra 1 Geometry Algebra 2 College Math College Pre-Algebra Introductory Algebra Intermediate Algebra College Algebra Homeschool Math Pre-Algebra Algebra 1 Geometry Algebra 2 Share This Page Terms of Use Privacy Contact About Purplemath About the Author Tutoring from PM Advertising Linking to PM Site licencing Visit Our Profiles © 2024 Purplemath, Inc.All right reserved.Web Design by
15769	https://www.youtube.com/watch?v=YJ4MfpyzSUc	Using Prime Factorization To Simplify Radicals mrmaisonet 68300 subscribers 68 likes Description 9803 views Posted: 15 Jul 2019 Watch and learn how to simplify radicals by breaking a number down into its prime factors. Follow Me At: 12 comments Transcript: hey this is mr. maisonette and in this tutorial what we're going to do is we're going to practice expressing the square root of a number in its simplest radical form and to accomplish this what we're going to do is we're going to do a prime factorization of that number so let's start by taking the number underneath our radical sign we want to express for 24 in simplest radical form so we go off to the side here and we write for 24 and we start by making a factor tree of that number and what we're going to do is we're going to take this number and we're gonna break it down into all prime numbers and remember a prime number is a number with exactly two factors which are only one and itself alright so let's start by multiplying two by two hundred twelve now as soon as you get a factor that is a prime factor or a prime number I like to go ahead and circle that number just to know that we're finished with that number we cannot break that down any further because it is a prime 212 we can produce by multiplying two times 106 and we go ahead and circle this two because it is prime we can break 106 down into two times 53 and two is prime and 53 is also prime we cannot break that down any further alright once you have taken your number and have broken it down into all prime elements what you should do is you go to the side and you make two radical symbols now the first radical symbol is going to be for all of your factors that make a perfect square and the second radical symbol is going to be for all of the factors that do not produce a perfect square now here's what I mean by this you go to all of your prime numbers that you have on the side here and you find pairs of prime numbers for example we have a two here in a two here so we would say this is a pair of primes and the reason that we pair them up is because if we multiply them together they would produce a perfect square for example two times two is four and we know that 4 is a perfect square because the square root of 4 is 2 so after you have identified a pair of numbers just multiply them together and write that result underneath the radical that is your perfect square radical now we have a 2 left over here and a 53 left over here each one of these numbers do not have a number to pair up with so we just write whatever we have remaining underneath our other radical symbol all right now what we do is we take what we have underneath this first radical and just write the answer we know the square root of 4 is 2 and then we take everything we have underneath this radical and just multiply them out two times 53 is 106 and we have to leave that trapped underneath our radical symbol so we would say that 2 times the square root of 106 is equal to the square root of 4 24 so the square root of 424 in simplest radical form is 2 times the square root of 106 all right let's go ahead and do another example all right with this example we have 936 so let's go ahead and break that down into all prime numbers so I'm gonna start by multiplying 9 times 104 to make 936 on right away I recognize that 9 can fit into itself once and 9 can fit into 36 exactly 4 times so that's why I knew that 9 times 104 produces 936 now 9 can be broken down into 3 times 3 and I have to circle each one of these threes because 3 is a prime number and now we can break down this 104 and 104 is even so I'm just gonna multiply 2 times 52 and I circle this 2 because its prime and we can break 52 down into 2 times 26 and 2 is a prime number so I go ahead and circle that and 26 can be broken down into the factors 2 and 13 and I circle each one of those factors because 2 is prime and thirteen is a prime number all right after you have broken your number down into all prime numbers you go off to the side here and you make two radical symbols one for the perfect squares and one for the factors that you have left over all right so now we have to identify pairs of numbers because remember when you multiply a number by itself it produces a perfect square so right here we have a pair of twos and we know that two times two is four so I'm gonna write four underneath my perfect square radical and we have a pair of threes and three times three is nine so we write a nine underneath that radical symbol and we have a two left over and a 13 left over so I like to take my leftovers and write them underneath the other radical symbol all right so we're gonna go ahead and multiply these perfect squares together 4 times 9 is 36 so we're gonna have to find the square root of 36 which is a perfect square and that would be 6 and we leave everything under here alone because they are not a perfect square and we multiply these two numbers together 2 times 13 is 26 but we have to leave that result trapped underneath the radical symbol because 26 does not have a factor that is a perfect square so we would say the square root of 936 in simplest radical form is 6 times the square root of 26 all right let's go ahead and do one more example okay so we have to take 1250 and break it down into all prime factors so to start off here because this number ends in a zero I know that ten can fit into 1250 equally so we're gonna do 10 times 125 and that we can break 10 down into 2 times 5 and 2 is prime so we Circle it as is 5 and 125 is composite so we can break that down further by multiplying 5 by 25 and 5 is prime so we Circle it and 25 we can break down into 5 times 5 and we circle both of those of course alright so we go off to the side here and we make our two radical symbols here and underneath the first one I like to write the perfect squares so we have to find pairs of prime numbers all right right here we have a pair of fives so we're gonna go ahead and take 5 times 5 which is 25 and write that underneath our first radical symbol and we have another pair of fives right here so 5 times 5 produces the perfect square of 25 so we write it underneath our first radical symbol and we have a 2 left over there's no other two to pair up with it so that's not going to produce any kind of perfect square so we write that underneath this radical symbol all right now notice underneath the first radical symbol we are multiplying the same number by itself which means that the square root of whatever that produces is going to be 25 so 25 times 25 is 625 but because we're multiplying the same number by itself we can just take that one number and just write it one time and then we just bring down the square root of 2 so the square root of 1250 in simplest radical form is 25 times the square root of 2 all right so that was just three quick examples showing how you can use a prime factorization to break a number down into simplest radical form please don't forget to hit the subscribe button and I hope to see you back in my channel soon [Music]
15770	https://pressbooks.lib.vt.edu/pulmonaryphysiology/chapter/lung-volumes-and-compliance/	Skip to content Want to create or adapt books like this? Learn more about how Pressbooks supports open publishing practices. 3 Lung Volumes and Compliance Learning objectives Describe the lung volumes that can be determined by spirometry. Describe the factors that determine lung compliance as the lung inflates from residual volume to total lung capacity. Describe the ramifications of dead space on the pattern of breathing during hyperpnea. Lung Volumes Introduction In this section we will look at some of the nomenclature for a variety of lung volumes and how these are clinically pertinent and can change in disease. We will also begin to look at the work of breathing and what factors affect how easy or hard the lung is to inflate, that is, lung compliance. We will then see how breathing pattern is generated to improve the efficiency of the lung and reduce the work of breathing. Lung Volumes First let us look at lung volumes. This trace from a spirometer (figure 3.1) shows the change in lung volume as a patient breathes normally and then performs some specific maneuvers. Let us work through the trace from left to right. The initial part of the trace shows resting or “tidal” breathing. The amount of volume inspired during each breath is referred to as tidal volume. Once a normal expiration is complete, however, the lung is far from empty, and when instructed, this patient (figure 3.1) breathes out as far as they can; this excess that comes out the lung is referred to as the expiratory reserve volume. Even at this point, however, some air remains in the lung, and this is referred to as residual volume. Even with maximal efforts, this volume cannot be exhaled, so at no point can the lung be fully emptied. This also means that residual volume can never be measured with a spirometer. Our patient (figure 3.1) returns to normal tidal breathing for two breaths before taking a full breath in, filling the lungs as much as they can. This extra volume into the lung after a normal tidal inspiration is referred to as inspiratory reserve volume. Related to this volume is the inspiratory capacity, which is the volume that can be taken into the lung after a normal expiration; inspiratory capacity is a useful clinical measurement that we will return to when we deal with some disease states. Another clinically valuable measurement is vital capacity, which is the volume of air that our patient can move out of the lung after a full inspiration, that is, the total lung capacity, minus the residual volume (remember: residual volume cannot be expelled). Forced vital capacity is a common measure taken in pulmonary function testing, and this is simply the volume that can be expelled from total lung capacity during a forceful expiration. The importance of this maneuver being forced will be dealt with when we look at airway compression (chapter 6). While the volumes we have just seen measured by spirometry in the pulmonary function lab provide valuable clinical information, we need to now look at some physiological variables that are also critical for our understanding of lung function and disease. Components of Tidal Breathing As you have seen, the volume of air inspired during a normal breath is tidal volume, and the size of this is dependent on body size, but in the example here is listed as 500 mL (a good approximation). Not all this 500 mL reaches the gas exchange surfaces in the respiratory zone, however, as some never gets further than the conducting zone (i.e., it stays in the anatomical dead space). From chapter 1 we know that this dead space has a volume of 150 mL, so the amount of air reaching the alveoli in the respiratory zone is our tidal volume (500 mL), minus the dead space volume, so alveolar volume is 350 mL. This brings us to an important point of clarification. Minute ventilation (denoted as Ve) is the volume of air exchanged in the lung within a minute. This is analogous to cardiac output, the volume of blood pumped by the heart in a minute. As such, minute ventilation is the average tidal volume (VT) multiplied by the number of breaths taken in a minute (RR). Equation 3.1 So if respiratory rate is 10 bpm and tidal volume is 500 mL, minute ventilation is 5,000 mL. Equation 3.2 Physiologically more important, however, is the alveolar minute ventilation (VA) that accounts for the “wasted” ventilation that never reached a gas exchange surface but remained in the anatomical dead space. So the calculation for VA is Equation 3.3 where VD is the anatomical dead space (approximately 150 mL). So for our previous example, alveolar minute ventilation is Equation 3.4 describing only the volume of air that reached the respiratory zone. So far the involvement of anatomical dead space might seem academic, as it remains constant. But let us consider a different breathing pattern (as often occurs in disease states). In our example above minute ventilation is 5,000 mL, but accounting for dead space we see that alveolar minute ventilation is 3,500 mL. Now let us consider another breathing pattern—one typical of a patient with restrictive lung disease where tidal volume is reduced and respiratory rate is increased. With a tidal volume of 250 mL and rate of 20, the minute ventilation remains the same, 5,000 mL. Equation 3.5 But calculating alveolar minute ventilation we see that a greater proportion of the reduced tidal volume is consumed by dead space. Equation 3.6 So despite maintaining the same minute ventilation, the second patient’s alveolar minute ventilation is reduced by 1,500 mL, which is significant given that this is the volume of air going to the gas exchange surfaces. This partially explains why increases in ventilation are initially achieved by increases in tidal volume; as shown in figure 3.2, as tidal volume increases during exercise intensity (represented by oxygen uptake) until it reaches a plateau. Only when this plateau is reached are further increases in minute ventilation achieved by increasing respiratory rate. So why not keep increasing tidal volume? At higher lung volumes the elastic limit of the lung is approached, and it takes more energy (muscular force) to expand, so it is more efficient and the work of breathing is less if the rate of breathing is increased to achieve higher levels of minute ventilation. This brings us to our next topic, lung compliance. Lung Compliance Introduction Lung compliance is a description of how easy the lung is to inflate, more specifically, how much volume will change for a given pressure differential. Figure 3.3 shows a typical and normal lung compliance curve. The lower line shows how volume changes as intrapleural pressure becomes more negative (as the chest wall and diaphragm expand the thorax). The upper curve is the compliance of the lung during expiration, and it is clearly different; this is an example of hysteresis, meaning that the relationship depends on direction, and we will see why this exists later. Lung Compliance During Inspiration You will notice at low lung volumes the slope of the compliance curve (figure 3.3) is shallower, meaning that it takes a relatively large pressure change to cause an increase in volume. This tells us at low lung volumes the lung is less distensible, or has low compliance. If we start to breathe at a higher lung volume, the slope of the curve is steeper, meaning that for a similar change in pressure there is a greater change in volume (i.e., the lung is more compliant). If we start breathing at a higher lung volume still, closer to total lung capacity, we see the slope of the compliance curve flatten out again, showing that at the lung volumes the compliance of the lung is low. As you might imagine, the normal range for breathing is in the middle range where the slope is steep and the lung compliant. This corresponds to an intrapleural pressure range of −5 to −10 cm H2O, which you should know is the normal range of intrapleural pressures during tidal breathing. This means we normally breathe at a lung volume at which the lung is most compliant and therefore takes less work to inflate. Too low a lung volume and compliance falls and work of breathing increases, likewise during breathing at high lung volumes, another contributing reason for why tidal volume plateaus during exercise. So now let us look at why compliance is low at high and low lung volumes, starting with the cause of low lung compliance at low volumes. Low compliance at low volumes—Surface tension: The reason why the lung takes more pressure to inflate at low volumes is surface tension. As mentioned in chapter 1 the alveoli have a thin layer of fluid lining their inner surface. As we saw in the pleural space, this causes surface tension. Unlike the surface tension in the pleural space, in the alveoli surface tension is a disadvantage. Surface tension is generated as water molecules cluster together to reduce their exposure to the gas in the alveolar space. As they gather together they drag the alveolar wall with them, producing a force that tends to pull the alveolar walls inward. The alveolar pressure opposes this force and should prevent the alveolus from collapsing (figure 3.4). The relationship between these two opposing forces is described by Laplace’s law that states the outward (alveolar) pressure needed to oppose the inwardly directed tension is proportionate to the tension (obviously), but also inversely related to the radius of the alveolus (i.e., the smaller the radius, the greater the inwardly acting force). This explains why compliance is low at low lung volumes. At low lung volumes the alveoli are smaller and thus have a smaller radius. Laplace’s law states that with a low radius the pressure needed to overcome the inward force will be greater, explaining why a larger alveolar (outward) pressure is needed to inflate the alveolus from a low starting volume. As lung volume increases, and thus alveolar radius increases, the pressure needed to overcome the inward acting force becomes less and the compliance of the lung increases. This explains why compliance is improved at the normal operating range of lung volumes. This also explains the hysteresis of the compliance curve. During expiration as alveoli are becoming progressively smaller, the inwardly acting force generated by surface tension becomes progressively greater. This phenomenon consequently assists expiration and contributes to expiration being a passive process. Low compliance at high lung volumes—Elastic limit: At high lung volumes the alveolar radius has increased further, suggesting that compliance should be further improved as the effect of surface tension will be much less. But surface tension is not the only factor involved, and the compliance curve flattens here, meaning a greater pressure is needed to achieve a volume change at high lung volumes. The low compliance at high lung volumes is caused by another phenomenon altogether. At high lung volumes expansion of the lung becomes limited by the elastic limit of the lung, a little like trying to further stretched an already stretch elastic band—it is harder to do. So with surface tension causing problems at low lung volumes and tissue elastic limit causing problems at high lung volumes, the compliance curve is steepest (i.e., most favorable) in the middle, as mentioned before, which is the operating volume of the lung. These principles are summarized in figure 3.5. Improving lung compliance with surfactant: So after that information on how surface tension is a problem for the lung, we now have to look at how it could be so much worse if the lung did not protect itself. Despite it having an effect, particularly at low lung volumes, the lung actually reduces the effect of alveolar surface tension by releasing “surfactant,” a molecule that disrupts surface tension. In brief, the surfactant molecule (dipalmitoyl phosphatidylcholine) has a similar structure to the phospholipids that make up cell membranes with a hydrophobic end and a hydrophilic end, allowing it to surround water and repel it at the same time, thus breaking up the interaction between water molecules. So as surfactant significantly reduces surface tension, it thereby increases lung compliance and the risk of alveolar collapse. It also helps keep the air space dry, as excessive surface tension tends to draw water into the space from the capillaries and interstitial spaces. Surfactant is released onto the alveolar inner surface by Type II alveolar cells (recall Type I cells are those making up the alveolar wall). Type II cells produce surfactant at a high rate and thus demand a constant and generous blood flow; therefore any condition that disrupts this blood supply will cause surfactant concentrations to decline and therefore put the alveolus at risk of collapse as surface tension is allowed to increase. A good illustration of the effect of surfactant is respiratory distress syndrome of the newborn. The underdeveloped lungs of infants born prematurely (at about twenty-eight weeks), cannot produce sufficient surfactant. Alveoli rapidly collapse (known as atelectasis), and pulmonary edema develops because of the excessive surface tension in the alveolar walls. References, Resources, and Further Reading Text Levitsky, Michael G. “Chapter 2: Mechanics of Breathing.” In Pulmonary Physiology, 9th ed. New York: McGraw Hill Education, 2018. West, John B. “Chapter 7: Mechanics of Breathing—How the Lung Is Supported and Moved.” In Respiratory Physiology: The Essentials, 9th ed. Philadelphia: Wolters Kluwer Health/Lippincott Williams and Wilkins, 2012. Widdicombe, John G., and Andrew S. Davis. “Chapter 2.” In Respiratory Physiology. Baltimore: University Park Press, 1983. Figures Figure 3.1: Lung volumes detected by spirometry. Grey, Kindred. 2022. CC BY 4.0. Figure 3.2: Changes in breathing tidal volume and respiratory rate with increasing levels of exercise. Grey, Kindred. 2022. CC BY 4.0. Figure 3.3: Lung compliance curve. Grey, Kindred. 2022. CC BY 4.0. Figure 3.4: Opposing forces of alveolar pressure and surface tension. Grey, Kindred. 2022. CC BY 4.0. Figure 3.5: Summary of lung volumes and compliance. Grey, Kindred. 2022. CC BY 4.0.
15771	https://www.testing.com/tests/immunoreactive-trypsinogen-irt/	Published Time: 2020-06-05T00:00:00+00:00 Immunoreactive Trypsinogen (IRT) - Testing.com Home Types of Tests Types of Tests At-Home Testing Allergy Testing Cancer Testing Cardiac and Cholesterol Testing Disease Detection and Monitoring Testing Drug Testing Genetic Testing Health and Wellness Testing Heavy Metals and Toxins Testing Hormone Testing Immunity Testing Infectious Disease Testing Inflammation and Autoimmune Testing Pregnancy and Fertility Testing STD Testing Thyroid Function Testing Vitamins and Nutritional Testing View All At-Home Tests At-Home Tests Best At-Home Sleep Apnea Tests Best At-Home Blood Tests Best At-Home Chlamydia Tests Best At-Home Drug Tests Best At-Home Flu Tests Best At-Home HIV Tests Best At-Home STD Tests Best At-Home UTI tests View All About About Medical Review Board Policies Editorial Policy Privacy Policy Terms of Use Accessibiliity Statement Contact Us FAQs Partners Resources Resources News Articles Glossary Conditions Find a Lab For Health Professionals Contact us Last modified on Jun 05, 2020 Immunoreactive Trypsinogen (IRT) Also Known As: IRT, Immunoreactive Trypsin, Serum Trypsinogen, Trypsin-like Immunoreactivity Board Approved At a Glance Why Get Tested? To screen for cystic fibrosis (CF), an inherited disease that affects mainly the lungs, pancreas, and sweat glands; sometimes to detect pancreatitis, which is inflammation of the pancreas When To Get Tested? As part of a newborn screening test; sometimes when you have symptoms of pancreatitis, such as severe abdominal pain that may be persistent or intermittent, nausea, vomiting, weakness, and jaundice Sample Required? A blood sample collected by pricking the heel of the infant and collecting a few drops of blood in a small tube or as a spot of blood on filter paper; a blood sample drawn from a vein Test Preparation Needed? None What is being tested? Trypsinogen is an inactive precursor produced by the pancreas that is converted to the enzyme trypsin. This test measures the amount of trypsinogen in the blood. Normally, trypsinogen is produced in the pancreas and transported to the small intestine. In the small intestine, it is activated and converted to trypsin. Trypsin is one of the enzymes responsible for breaking down the protein in food into smaller pieces called peptides. Without sufficient trypsinogen and trypsin, a person will not be able to properly digest and use proteins. Any condition that prevents trypsinogen from reaching the small intestine may cause an increase in trypsinogen in the blood. In people with cystic fibrosis (CF), mucus plugs can block the pancreatic ducts, preventing trypsinogen from reaching the small intestine, resulting in decreased breakdown of food proteins. Damage to the pancreas caused by other diseases, such as chronic pancreatitis and pancreatic cancer, may also cause blockages that prevent trypsinogen from reaching the small intestine. The cells that produce trypsinogen can also become damaged or be destroyed, decreasing the body’s supply. As part of a group of newborn screening tests, infants may be screened for CF using a test called immunoreactive trypsinogen (IRT). Newborns with CF may have elevated levels of IRT. Common Questions How is the test used? Immunoreactive trypsinogen (IRT) is used as part of some newborn screening programs to screen for cystic fibrosis (CF). It may be used in conjunction with a sweat chloride test and/or a cystic fibrosis gene mutation panel to help identify CF. IRT may also sometimes be used to help detect acute pancreatitis. When is it ordered? This test may be ordered soon after a baby is born as part of a newborn screen for cystic fibrosis. Cystic fibrosis screening is now a required part of newborn screening in all 50 states in the U.S. Immunoreactive trypsinogen (IRT) is one of the tests used by some states to screen for CF. An IRT test sometimes is ordered when a person has signs and symptoms of acute pancreatitis, such as: Severe pain in the upper abdomen that may radiate to the mid-back, which usually lasts at least several hours at a time Nausea, vomiting Weakness Yellowing of the skin and/or eyes (jaundice) Fever What does the test result mean? If an IRT level is elevated, a newborn may have cystic fibrosis (CF). If the IRT is elevated, a child or adult may have abnormal pancreatic enzyme production, pancreatitis, or pancreatic cancer. However, IRT testing is not diagnostic. There are a fair number of false positives and problems other than cystic fibrosis and pancreatitis that can cause an elevated IRT. Factors such as age at collection and race/ethnicity could affect IRT levels. An elevated level must be followed up with other testing. When diagnosing cystic fibrosis, this may include another IRT test in a month, CF gene mutation testing, and/or sweat chloride testing. If the IRT level is not elevated, then it is likely that the newborn does not have CF. However, if suspicion of CF is high and the infant has signs and symptoms consistent with CF, other testing for cystic fibrosis, such as sweat chloride and/or CF gene mutation testing, should be considered. What other tests might my healthcare practitioner do to check pancreas function? Your healthcare practitioner may order a stool test for fecal fat, fecal elastase, or chymotrypsinor a blood test for amylase or lipase to look at other aspects of pancreas and digestive function. Is there anything else I should know? In testing for cystic fibrosis (CF), the IRT test is only useful for screening. Additional testing is needed to establish a diagnosis. In those who do have CF, the degree of IRT elevation does not reflect the severity of the disease. Related Tests ChymotrypsinLearn More Sweat Chloride TestLearn More Cystic Fibrosis (CF) Gene Mutations TestingLearn More Amylase TestLearn More Lipase TestLearn More Cystic Fibrosis FoundationLearn More Cystic Fibrosis CanadaLearn More National Heart, Lung, and Blood Institute: Cystic FibrosisLearn More MedlinePlus: Pancreatic DiseasesLearn More MedlinePlus: PancreatitisLearn More NIDDK: PancreatitisLearn More National Pancreas Foundation: About Acute PancreatitisLearn More Sources Sources Used in Current Review 2020 review performed by Jayson V. Pagaduan, PhD, Clinical Chemist, Intermountain Healthcare. Therrell, B.L., Jr., et al. Immunoreactive Trypsinogen (IRT) as a Biomarker for Cystic Fibrosis: challenges in newborn dried blood spot screening. Mol Genet Metab, 2012. 106(1): p. 1-6. National Institute of Diabetes and Digestive and Kidney Diseases. Symptoms & Causes of Pancreatitis. Available online at Accessed on 4/20/2020. Weidler, S., et al. A product of immunoreactive trypsinogen and pancreatitis-associated protein as second-tier strategy in cystic fibrosis newborn screening. J Cyst Fibros, 2016. 15(6): p. 752-758. Castellani, C. and J. Massie.Emerging issues in cystic fibrosis newborn screening. C urr Opin Pulm Med, 2010. 16(6): p. 584-90. Sources Used in Previous Reviews Thomas, Clayton L., Editor (1997). Taber’s Cyclopedic Medical Dictionary. F.A. Davis Company, Philadelphia, PA [18th Edition]. Pagana, Kathleen D. & Pagana, Timothy J. (2001). Mosby’s Diagnostic and Laboratory Test Reference 5th Edition: Mosby, Inc., Saint Louis, MO. NIDDK (July 1997). Cystic Fibrosis Research Directions. The National Institute of Diabetes and Digestive and Kidney Diseases (NIDDK) [NIH Publication No. 97-4200]. Available online at CFF. Sweat Testing Procedure and Commonly Asked Questions. Cystic Fibrosis Foundation [On-line information]. Available online at Tait, J., et. al. (26 March 2001). Cystic Fibrosis. GENEReviews [On-line information]. Available online at Mayo Clinic (2001, February 09). What is Cystic Fibrosis? Mayo Clinic [On-line information]. Available online at MedlinePlus(2002, January 2, Updated). Cystic Fibrosis. MedlinePlus Health Information [On-line Information]. Available online at MedlinePlus (2002, January 2, Updated). Sweat Electrolytes. MedlinePlus Health Information [On-line Information]. Available online at MedlinePlus (2002, January 2, Updated). Trypsin and chymotrypsin in stool. MedlinePlus Health Information [On-line Information]. Available online at Mountain States Genetics (1999, September, Revised). Cystic Fibrosis (CF). Newborn Screening Practitioner’s Manual [On-line information]. Available online at National Institutes of Health (1995, November). Facts About Cystic Fibrosis. National Heart, Lung, and Blood Institute NIH Publication No. 95-3650 [On-line information]. Available online at Quest Diagnostics NEWS (2002, Winter). What is the Best Test to Screen for Cystic Fibrosis? [On-line serial]. PDF available for download at NIH (2002, January 2, Updated). Neonatal cystic fibrosis screening. MedlinePlus Health Information [On-line Information]. Available online at Drkoop (2001). Cystic fibrosis. Medical Encyclopedia [On-line information]. Available online at NIH (2002, January 2, Updated). Trypsinogen. MedlinePlus Health Information [On-line Information]. Available online at Biomedical Hypertext (1999 May 20, last update). Exocrine Secretions of the Pancreas. Colorado State University [On-line information]. Available online at UPCMD (1998 – 2002). Cystic Fibrosis. University Pathology Consortium, LLC [On-line information]. Available online at Sainato, D., (2002, March). Genetic Testing for CF Going Mainstream? Clinical Laboratory News [Journal], Vol 28. Peter Jacky, PhD, FACMG. Director of Cytogenetics and Molecular Genetics, Airport Way Regional Laboratory, Portland, OR. (Update October 15, 2007) Greco, F. MedlinePlus Medical Encyclopedia. Trypsinogen. Available online at Accessed January 2009. Tietz Textbook of Clinical Chemistry and Molecular Diagnostics. Burtis CA, Ashwood ER, Bruns DE, eds. St. Louis: Elsevier Saunders; 2006. P. 622. Cystic Fibrosis Foundation. Testing for Cystic Fibrosis. Available online at Accessed December 2008. Cleveland Clinic. Cystic Fibrosis. Available online at Accessed December 2008. State of Missouri Department of Health and Senior Services. Cystic Fibrosis. Available online at Accessed January 2009. Denise I. Quigley, PhD, FACMG. Co-Director Cytogenetics/Molecular Genetics, Airport Way Regional Laboratory, Portland, OR. Peter Jacky, PhD, FACMG. Director of Cytogenetics and Molecular Genetics, Airport Way Regional Laboratory, Portland, OR. Kaneshiro, N. (Updated 2012 May 16).Neonatal cystic fibrosis screening. MedlinePlus Medical Encyclopedia [On-line information]. Available online at Accessed September 2012. Grenache, D. et. al. (Updated 2012 April) Cystic Fibrosis – CF. ARUP Consult [On-line information]. Available online at Accessed September 2012. Dugdale, D. (Updated 2011 February 4). Trypsinogen test. MedlinePlus Medical Encyclopedia [On-line information]. Available online at Accessed September 2012. Bender, L. et. al. (2011 September 29). Kids in America Newborn Screening for Cystic Fibrosis. Medscape Today News from Lab Med v42 (10):595-601 [On-line information]. Available online at Accessed September 2012. Sharma, G. (2012 May 15). Cystic Fibrosis. Medscape Reference [On-line information]. Available online at Accessed August 2012. Kaneshiro, N. (2012 May 16). Cystic fibrosis. MedlinePlus Medical Encyclopedia [On-line information]. Available online at Accessed August 2012. Timothy S. Uphoff, Ph.D., D(ABMG), MLS(ASCP)CM. Section Head Molecular Pathology Laboratory, Marshfield Labs, Marshfield Clinic, Marshfield WI. Lal, S. (2015 February 4 Updated). Trypsinogen test. MedlinePlus Medical Encyclopedia. Available online at Accessed on 9/05/16. (© 2016). Diagnosing and Treating Cystic Fibrosis. American Lung Association. Available online at Accessed on 9/05/16. Sharma, G. (2016 June 08 Updated). Cystic Fibrosis. Medscape Drugs and Diseases. Available online at Accessed on 9/05/16. Newborn Screening for CF. Cystic Fibrosis Foundation. Available online at Accessed on 9/05/16. Earley, M. (2013). New Guidelines for Cystic Fibrosis Newborn Screening Laboratory Tests. CLSI Presentation. Available online at Accessed on 9/05/16. Rosenstein, B. (2016 January Revised). Cystic Fibrosis. Merck Manual for Professionals Available online at Accessed on 9/05/16. (©2016) Mayo Medical Laboratories. Trypsin-like Immunoreactivity. Available online at Accessed November 2016. See More See Less Table of Contents At a Glance What is being tested? 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15772	https://www.doubtnut.com/qna/1339800	Equation of the chord of the ellipse whose midpoint is (x1, y1) Download Allen App Courses IIT-JEE Class 11 Class 12 Class 12+ NEET Class 11 Class 12 Class 12+ UP Board Class 9 Class 10 Class 11 Class 12 Bihar Board Class 9 Class 10 Class 11 Class 12 CBSE Board Class 9 Class 10 Class 11 Class 12 Other Boards MP Board Class 9 Class 10 Class 11 Class 12 Jharkhand Board Class 9 Class 10 Class 11 Class 12 Haryana Board Class 9 Class 10 Class 11 Class 12 Himachal Board Class 9 Class 10 Class 11 Class 12 Chhattisgarh Board Class 9 Class 10 Class 11 Class 12 Uttarakhand Board Class 9 Class 10 Class 11 Class 12 Rajasthan Board Class 9 Class 10 Class 11 Class 12 Ncert Solutions Class 6 Maths Physics Chemistry Biology English Class 7 Maths Physics Chemistry Biology English Class 8 Maths Physics Chemistry Biology English Class 9 Maths Physics Chemistry Biology English Class 10 Maths Physics Chemistry Biology English Reasoning Class 11 Maths Physics Chemistry Biology English Class 12 Maths Physics Chemistry Biology English Exam IIT-JEE NEET CBSE UP Board Bihar Board Results Study with ALLEN JEE NEET Class 6-10 My Profile Logout Ncert Solutions English Medium Class 6 Maths Physics Chemistry Biology English Class 7 Maths Physics Chemistry Biology English Class 8 Maths Physics Chemistry Biology English Class 9 Maths Physics Chemistry Biology English Class 10 Maths Physics Chemistry Biology English Reasoning Class 11 Maths Physics Chemistry Biology English Class 12 Maths Physics Chemistry Biology English Courses IIT-JEE Class 11 Class 12 Class 12+ NEET Class 11 Class 12 Class 12+ UP Board Class 9 Class 10 Class 11 Class 12 Bihar Board Class 9 Class 10 Class 11 Class 12 CBSE Board Class 9 Class 10 Class 11 Class 12 Other Boards MP Board Class 9 Class 10 Class 11 Class 12 Jharkhand Board Class 9 Class 10 Class 11 Class 12 Haryana Board Class 9 Class 10 Class 11 Class 12 Himachal Board Class 9 Class 10 Class 11 Class 12 Chhattisgarh Board Class 9 Class 10 Class 11 Class 12 Uttarakhand Board Class 9 Class 10 Class 11 Class 12 Rajasthan Board Class 9 Class 10 Class 11 Class 12 Exam IIT-JEE NEET CBSE UP Board Bihar Board Study with ALLEN JEE NEET Class 6-10 Books Class 12 Class 11 Class 10 Class 9 Class 8 Class 7 Class 6 Online Class Blog Select Theme Class 11 MATHS Equation of the chord of the ellipse who... 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Updated on:21/07/2023 Class 11MATHSELLIPSE Similar Questions Equation of the chord of the ellipse whose midpoint is (x 1,y 1) View Solution The equation of the chord of the circle S=0 whose midpoint is (x 1,y 1) is View Solution Knowledge Check The equation of the chord joining two points (x 1,y 1) and (x 2,y 2) on the rectangular hyperbola x y=c 2 is A x x 1+x 2+y y 1+y 2=1 B x x 1−x 2+y y 1−y 2=1 C x y 1+y 2+y x 1+x 2=1 D x y 1−y 2+y x 1−x 2=1 Submit Chord of the ellipse x 2 25+y 2 16=1 whose middle point is (1 2,2 5) meets the minor axis at A and major axis at B, length of AB (in units) is : A√41 5 B 2√41 5 C 3√41 5 D 7√41 5 Submit Ellipse- Normal equation OF pair OF tangent \|\| Equation OF chord OF ellipse whose middle point is given View Solution The equation of the chord of the ellipse 2 x 2+3 y 2=6 having (1,−1) as its mid point is View Solution Find the equation of the chord of the circle x 2+y 2=9 whose middle point is (1,−2) View Solution The equation of the chord of x 2+y 2−4 x+6 y+3=0 whose mid point is (1,−2) is View Solution The equation of the chord of the ellipse 2 x 2+5 y 2=20 which is bisected at the point (2,1) is View Solution The equation of a circle is x 2+y?2=25 The equation of its chord whose middle point is (1,−2) is given by View Solution Recommended Questions Equation of the chord of the ellipse whose midpoint is (x1, y1) 05:22 Equation of chord whose midpoint is (x1,y1) 03:55 Equation of the chord of the ellipse whose midpoint is (x1, y1) 05:22 Equation of chord whose midpoint is (x1,y1) 04:19 Show that the equation of the chord of the parabola y^2 = 4ax through ... 05:48 दीर्घवृत्त x^2/a^2+y^2/b^2=1 के बिंदु (x1,y1) पर स्पर्श रेखा का समीक... 06:52 Position Of Point And Equation Of Chord Whose Midpoint Is Given 21:31 Equation Of Chord Whose Midpoint Is Given 27:41 Ellipse (x^(2))/25+(y^(2))/9=1 The equation of chord whose midpoint is... 02:15 Exams IIT JEE NEET UP Board Bihar Board CBSE Free Textbook Solutions KC Sinha Solutions for Maths Cengage Solutions for Maths DC Pandey Solutions for Physics HC Verma Solutions for Physics Sunil Batra Solutions for Physics Pradeep Solutions for Physics Errorless Solutions for Physics Narendra Awasthi Solutions for Chemistry MS Chouhan Solutions for Chemistry Errorless Solutions for Biology Free Ncert Solutions English Medium NCERT Solutions NCERT Solutions for Class 12 English Medium NCERT Solutions for Class 11 English Medium NCERT Solutions for Class 10 English Medium NCERT Solutions for Class 9 English Medium NCERT Solutions for Class 8 English Medium NCERT Solutions for Class 7 English Medium NCERT Solutions for Class 6 English Medium Free Ncert Solutions Hindi Medium NCERT Solutions NCERT Solutions for Class 12 Hindi Medium NCERT Solutions for Class 11 Hindi Medium NCERT Solutions for Class 10 Hindi Medium NCERT Solutions for Class 9 Hindi Medium NCERT Solutions for Class 8 Hindi Medium NCERT Solutions for Class 7 Hindi Medium NCERT Solutions for Class 6 Hindi Medium Boards CBSE UP Board Bihar Board UP Board Resources Unit Converters Calculators Books Store Seminar Results Blogs Class 12 All Books Class 11 All Books Class 10 All Books Class 9 All Books Class 8 All Books Class 7 All Books Class 6 All Books Doubtnut is No.1 Study App and Learning App with Instant Video Solutions for NCERT Class 6, Class 7, Class 8, Class 9, Class 10, Class 11 and Class 12, IIT JEE prep, NEET preparation and CBSE, UP Board, Bihar Board, Rajasthan Board, MP Board, Telangana Board etc NCERT solutions for CBSE and other state boards is a key requirement for students. 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15773	https://space.stackexchange.com/questions/37928/how-is-foot-pounds-of-energy-defined	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Space Exploration Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams How is foot-pounds of energy defined? Ask Question Asked Modified 5 years, 4 months ago Viewed 4k times 6 $\begingroup$ Tom Lodgson's classic textbook Orbital Mechanics: Theory and Applications frequently measures energy in foot-pounds, which I'd always thought to measure torque. The book doesn't define it. Is it the archaic foot pound-force [sic], or the gravitational potential energy of one pound hoisted one foot in a constant gravitational field, or something else? (Is this unit still in use in aerospace or in other fields?) orbital-mechanics history measurement unit-systems Share Improve this question edited May 22, 2020 at 21:35 DrSheldon 48.8k1313 gold badges171171 silver badges354354 bronze badges asked Aug 5, 2019 at 15:05 Camille GoudeseuneCamille Goudeseune 12.5k11 gold badge5656 silver badges9191 bronze badges $\endgroup$ 1 $\begingroup$ It's certainly true that the same set of units can describe multiple situations. Take a look at, e.g., some of the tables of constants in the NIST standard SP811 $\endgroup$ Carl Witthoft – Carl Witthoft 2019-08-06 13:27:55 +00:00 Commented Aug 6, 2019 at 13:27 Add a comment \| 3 Answers 3 Reset to default 9 $\begingroup$ Is it... the gravitational potential energy of one pound hoisted one foot in a constant gravitational field...? Yes indeed it is! To be energy, the pound has to be parallel to the foot. $$E = \int \mathbf{F} \cdot d \mathbf{s}$$ To be torque, the pound has to be perpendicular to the foot $$\tau = \mathbf{r} \times \mathbf{F}$$ 1 foot-pound (or pound foot) of energy is 9.81 (m/s^2) / 2.2 (kgf/lb) / 3.3 (ft/m) = 1.35 Joules. Share Improve this answer edited Aug 6, 2019 at 4:07 answered Aug 5, 2019 at 15:33 user12102user12102 $\endgroup$ 13 2 $\begingroup$ @RussellBorogove in physics they are, and have to be, the same thing. The order can not matter, so I'm forced to pound my foot and stick to my principles. $\endgroup$ user12102 – user12102 2019-08-05 15:38:57 +00:00 Commented Aug 5, 2019 at 15:38 1 $\begingroup$ The last line here emphasizes that my two guesses for the unit's meaning are equivalent. Bingo! $\endgroup$ Camille Goudeseune – Camille Goudeseune 2019-08-05 15:39:35 +00:00 Commented Aug 5, 2019 at 15:39 2 $\begingroup$ Delightful pun... and the unwashed rabble (of which I must thus be a member) also use lb-ft and ft-lb interchangeably. $\endgroup$ Camille Goudeseune – Camille Goudeseune 2019-08-05 16:17:14 +00:00 Commented Aug 5, 2019 at 16:17 1 $\begingroup$ Re, "parallel,"..."perpendicular",... That description somewhat masks the truth. To be energy, the object to which the pound of force is applied must move one foot in the direction of the force. To be torque, nothing needs to move, but the pound of force must be applied to a point that is one foot away from the axis of rotation, and the radius and the axis and the force all must be mutually perpendicular. $\endgroup$ Solomon Slow – Solomon Slow 2019-08-05 17:38:31 +00:00 Commented Aug 5, 2019 at 17:38 2 $\begingroup$ I guess that would be obvious to a physicist. Maybe not so obvious to somebody who is struggling to understand how torque and energy can appear to be specified with the same units (i.e., to the guy who asked the original question.) $\endgroup$ Solomon Slow – Solomon Slow 2019-08-06 15:09:26 +00:00 Commented Aug 6, 2019 at 15:09 \| Show 8 more comments 4 $\begingroup$ Foot-pound or pound-foot are synonymous, and represent the arithmetic product of pound (force) and foot (length). The pound (force) is the weight of one pound (mass) at the Earth's surface (somewhat imprecise because Earth's gravity field varies depending on your location, and the effective weight of an object will be influenced by the centrifugal force due to Earth's rotation, again dependent on location). As a unit of energy, it is the energy of applying a one pound force over a distance of one foot. It is equivalent to raising a one-pound mass one foot in height, well... because. As a unit of torque, it is the torque resulting from a one pound tangent force applied at a distance of one foot from the axis of rotation. The same could be said of the Newton-meter (or meter-Newton, but it's never expressed that way); as a unit of energy, it is a one Newton force applied over a distance of one meter; as a unit of torque, it is a one Newton tangent force applied at a distance of one meter from the axis of rotation, except that the Newton is specifically a unit of force with a precise definition where pound may be either force or mass and pound (force) lacks a precise definition. The measurement system which includes pounds and feet has a long history. When it developed, the variability of Earth's gravitational field and its impact on the weights and measures which depended on it wasn't understood, wasn't measurable, and/or wasn't significant for the engineering problems of the time. Culture, history, and familiarity keep these weights and measures in use despite the awkwardness and the advantages of metric. Share Improve this answer edited Aug 5, 2019 at 22:57 answered Aug 5, 2019 at 22:03 Anthony XAnthony X 18k11 gold badge6565 silver badges105105 bronze badges $\endgroup$ 0 Add a comment \| 3 $\begingroup$ According to Wikipedia foot-pound and foot-pound-force are synonymous: The foot pound-force (symbol: ftlbf or ftlb) is a unit of work or energy in the Engineering and Gravitational Systems in United States customary and imperial units of measure. It is the energy transferred upon applying a force of one pound-force (lbf) through a linear displacement of one foot. Its equal to 1.356 Joules. The torque unit is the pound-foot, not foot-pound. Share Improve this answer edited Aug 5, 2019 at 15:12 answered Aug 5, 2019 at 15:08 Russell BorogoveRussell Borogove 176k1515 gold badges622622 silver badges733733 bronze badges $\endgroup$ 3 $\begingroup$ But pound-foot is used for torques too. $\endgroup$ Uwe – Uwe 2019-08-05 15:10:31 +00:00 Commented Aug 5, 2019 at 15:10 $\begingroup$ Yes, I was mistaken about lb-ft vs ft-lb. I'll also admit that wikipedia's claim ft-lbf = ft-lb was true thirty years ago when Tom started lecturing. Let's see what others report... $\endgroup$ Camille Goudeseune – Camille Goudeseune 2019-08-05 15:28:24 +00:00 Commented Aug 5, 2019 at 15:28 5 $\begingroup$ Per @uhoh's answer, for both energy and torque, when you decompose the mathematical shorthand in the integral and the cross product, you wind up with the units being a force times a distance. Multiplication is commutative, so force times distance is equivalent to distance times force. Distinction between pound-foot and foot-pound isn't physics, it's semantics. $\endgroup$ Tom Spilker – Tom Spilker 2019-08-05 19:14:16 +00:00 Commented Aug 5, 2019 at 19:14 Add a comment \| Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions orbital-mechanics history measurement unit-systems See similar questions with these tags. The Overflow Blog The history and future of software development (part 1) Getting Backstage in front of a shifting dev experience Featured on Meta Spevacus has joined us as a Community Manager Introducing a new proactive anti-spam measure Related 6 How are base temperature units found on other planets? Will the foot print of Neil Armstrong erode? 5 How can I calculate the Characteristic Energy to Sun-Earth L1? Is specific orbital energy a constant? How does this affect the semi-major axis? 5 How much energy is needed to bring Phobos closer to Mars? 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15774	http://www.hsc.edu.kw/student/materials/Physics/website/hyperphysics%20modified/hbase/torq.html	Torque and Equilibrium Torque A torque is an influence which tends to change the rotational motion of an object. One way to quantify a torque is Torque = Force applied x lever arm The lever arm is defined as the perpendicular distance from the axis of rotation to the line of action of the force. Index Rotation concepts Torque concepts HyperPhysics Mechanics RotationR NaveGo Back Conditions for Equilibrium An object at equilibrium has no net influences to cause it to move, either in translation (linear motion) or rotation. The basic conditions for equilibrium are: Index HyperPhysics Mechanics RotationR NaveGo Back
15775	https://pmc.ncbi.nlm.nih.gov/articles/PMC12348179/	Investigation of the Adsorption and Reactions of Methyl Radicals on Transition Metal (M = Co, Ni, Pd, Pt) (111) Surfaces in Aqueous Suspensions - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Journal List User Guide View on publisher site Download PDF Add to Collections Cite Permalink PERMALINK Copy As a library, NLM provides access to scientific literature. 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Learn more: PMC Disclaimer \| PMC Copyright Notice Molecules . 2025 Jul 22;30(15):3065. doi: 10.3390/molecules30153065 Search in PMC Search in PubMed View in NLM Catalog Add to search Investigation of the Adsorption and Reactions of Methyl Radicals on Transition Metal (M = Co, Ni, Pd, Pt) (111) Surfaces in Aqueous Suspensions Pankaj Kumar Pankaj Kumar 1 Chemical Sciences Department, The Radical Reactions Research Center, Ariel University, Ariel 4070000, Israel; pankajkumar121091@gmail.com Find articles by Pankaj Kumar 1, Dan Meyerstein Dan Meyerstein 1 Chemical Sciences Department, The Radical Reactions Research Center, Ariel University, Ariel 4070000, Israel; pankajkumar121091@gmail.com 2 Chemistry Department, Ben-Gurion University, Beer-Sheva 8410501, Israel Find articles by Dan Meyerstein 1,2,, Amir Mizrahi Amir Mizrahi 3 Nuclear Research Centre Negev, Beer-Sheva 84190, Israel; amirmizrachi@gmail.com Find articles by Amir Mizrahi 3, Haya Kornweitz Haya Kornweitz 1 Chemical Sciences Department, The Radical Reactions Research Center, Ariel University, Ariel 4070000, Israel; pankajkumar121091@gmail.com Find articles by Haya Kornweitz 1, Editor: Takashiro Akitsu Author information Article notes Copyright and License information 1 Chemical Sciences Department, The Radical Reactions Research Center, Ariel University, Ariel 4070000, Israel; pankajkumar121091@gmail.com 2 Chemistry Department, Ben-Gurion University, Beer-Sheva 8410501, Israel 3 Nuclear Research Centre Negev, Beer-Sheva 84190, Israel; amirmizrachi@gmail.com Correspondence: danm@ariel.ac.il (D.M.); hayak@ariel.ac.il (H.K.) Roles Takashiro Akitsu: Academic Editor Received 2025 Jul 2; Revised 2025 Jul 17; Accepted 2025 Jul 17; Collection date 2025 Aug. © 2025 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license ( PMC Copyright notice PMCID: PMC12348179 PMID: 40807241 Abstract The DFT method was used to evaluate the adsorption of methyl radicals and the evolution of ethane on the M(111) (M = Co, Ni, Pd, Pt) surfaces, eight metal atoms, in aqueous medium. A maximum of five and four radicals can be adsorbed on Co(111) and Ni(111), respectively, and six on Pd(111) and Pt(111) (top site). The ethane evolution occurs via the Langmuir–Hinshelwood (LH) or Eley–Rideal (ER) mechanisms. The production of ethane through the interaction of two adsorbed radicals is thermodynamically feasible for high coverage ratios on the four surfaces; however, kinetically, it is feasible at room temperature only on Co(111) at a coverage of (5/5) and on Pd(111) at a coverage ratio of 4/6, 5/6, and 6/6. Ethane production occurs via the ER mechanism: a collision with solvated methyl radical produces either C 2 H 6 or C⁢H∗2+C⁢H 4⁢(a⁢q). On Pd(111) the product is only C 2 H 6, on Pt(111), both products (C 2 H 6 or C⁢H∗2) are plausible, and on Co(111) and Ni(111), only C⁢H∗2+C⁢H 4⁢(a⁢q) is produced. Further reactions of C⁢H∗2 with C⁢H∗2 or C⁢H∗3 to give C 2⁢H∗4 or C 2⁢H∗5 are thermodynamically plausible only on Pt(111); however, they are very slow due to high energy barriers, 1.48 and 1.36 eV, respectively. Keywords: DFT, CH 3 radical, ethane production, heterogeneous reaction, transition metals 1. Introduction In recent decades, radical chemistry has evolved into a significant and essential component of organic chemistry. Even though Gomberg identified the first instance of an organic radical in 1900, the progress in the field was quite gradual for the following decades, and radicals seldom found application in synthesis [2,3]. A common structural component found in many biological substances, medications, and materials is the methyl group (CH 3) [4,5,6,7]. A diverse range of well-known methylation agents has been created for the introduction of methyl radicals, comprising CH 3 OH, DMSO, CH 4, and HC(OCH 3)3, along with a variety of peroxides like t-BuOOH (TBHP), t-BuOOCOPh (TBPO), dicumyl peroxide (DCP), cumyl hydroperoxide (CRHP), and t-BuOOtBu (DTBP), among others . Radicals are often formed near surfaces, e.g., in electrochemistry (e.g., the Kolbe reaction [9,10,11], reduction of halo-organic compounds , etc.), photocatalytic process (e.g., on TiO 2 [13,14,15,16,17,18] and other semiconductors [19,20], where the added electron in the conduction band reduces a variety of substrates and the hole in the valence band oxidizes a variety of substrates), heterogeneous catalytic processes, e.g., catalytic hydrogenations and de-hydrogenations on Pt 0 , M 0-NPs catalysed de-halogenations by [22,23,24], heterogeneous catalysed Fenton-like processes [25,26], and photo-Fenton-like processes . The radicals thus formed often react with the surfaces, if not formed bound to the surface, as the reactions of radicals with metals and semiconductors are fast . Reactions of methyl radicals are of importance, as discussed above. Furthermore, their reactions are considered as models for the reactions of other alkyl radicals, though the bond strengths of the products formed often differ considerably. The easiest way to study the reactions of methyl radicals is to produce them radiolytically. Methyl radicals generated radiolytically in aqueous solutions effectively react with Cr 0, Mn 0, Fe 0, Ni 0, Cu 0, and Zn 0 powders submerged in the solution, thus reducing the concentration of radicals and diminishing the steady state concentrations of radicals in the system [29,30]. The final products of methyl radicals observed in these studies contain CH 4, C 2 H 4, C 2 H 6, C 3 H 6, and C 3 H 8 . The adsorption of CH, CH 2, CH 3, and CH 4 onto Ni(111) has been previously reported . The CH 2 and CH 3 radicals were identified using a threshold ionization method . The threefold hollow site is the best adsorption site for CH 3 on Ni(111) [33,34,35,36,37,38,39,40]. There are two widely used kinetic models for surface reactions: the Langmuir−Hinshelwood (LH) model, in which two adsorbed species react, and the Eley–Rideal (ER) mechanism, in which a gas-phase or solvated species collides and reacts with a surface-adsorbed species. The aforementioned studies have demonstrated that methyl radicals are adsorbed at the surface of a nanoparticle (NP) via a covalent bond [16,41] then, a catalytic dimerization occurs, according to the LH mechanism, resulting in the production of ethane [15,21,42,43]. This process is described in Equations (1)–(3): (1) (2) (3) Another mechanism for the production of ethane is via the ER mechanism, reaction 4: (4) The experimental outcomes supported reaction 3 but did not eliminate the possibility of reaction 4. However, a reaction analogous to reaction 4, though in homogeneous solutions, was reported . In this article, we investigate adsorption energy, charge transfer (CT), and activation energy barriers of methyl radicals on M(111) (M = Co, Ni, Pd, Pt) surfaces using density functional theory. We mainly focus on the production of ethane either by two adsorbed methyls (LH mechanism) or by one adsorbed methyl and one solvated methyl (ER mechanism), but other products are also considered. We previously investigated these processes on Cu(111), Ag(111), and Au(111) surfaces . In the earlier study, the development of ethane through the LH mechanism is thermodynamically feasible for all three surfaces (Cu(111), Ag(111), Au(111)), yet kinetically, at room temperature, ethane is produced solely on Au(111) and Ag(111) under full coverage. Ethane can also be produced on Au(111) and Ag(111) via the ER mechanism in a barrierless process. On Cu(111), the products of the ER mechanism are CH 4(aq) and an adsorbed CH 2, which reacts further with a non-adsorbed water molecule, producing adsorbed CH 3 OH . 2. Result Analysis 2.1. The Adsorption of Methyl Radicals on M(111) Surfaces The energetically best adsorption site for the methyl radical (CH 3·) and ethane (C 2 H 6) on M(111) (M = Co, Ni, Pd, Pt) surfaces in aqueous solution was determined. The metallic surfaces consisted of eight metal atoms per layer. In this section, the adsorption of one methyl radical is considered. Figure 1 illustrates the optimized ground state geometries of the methyl radical and ethane molecule adsorbed on M(111) surfaces. Table 1 presents the adsorption energies, CT, and binding distances. The best adsorption site for methyl radicals in the aqueous suspensions is the fcc site for Co(111) and Ni(111) surfaces, as was reported previously [33,34,35,36,37,38,39,40], while for Pd(111) and Pt(111) surfaces, it is the atop site. The best adsorption site for ethane is the hcp site for Ni(111), Pd(111), and Pt(111) surfaces, while for the Co(111) surface, it is the bridge site. Figure 1. Open in a new tab Optimized geometries, at the best adsorption site of adsorbed methyl radicals and ethane on different M(111) surfaces. Table 1. Values of adsorption energies (E ads), CT (e), and distance between M(111) and C atoms in an aqueous medium at room temperature. \| Metals \| Adsorbates \| Adsorption Sites \| Adsorption \| Charge Transfer a (e) \| M(111)-C Distance (Å) \| :---: :---: :---: \| \| Co \| \| fcc \| −3.34 \| 0.40 \| 2.03 \| \| \| bridge \| −0.55 \| 0.01 \| 2.91 \| \| Ni \| \| fcc \| −3.11 \| 0.37 \| 2.06 \| \| \| hcp \| −0.75 \| 0.03 \| 3.16 \| \| Pd \| \| top \| −2.73 \| 0.07 \| 2.03 \| \| \| hcp \| −0.68 \| −0.02 \| 2.89 \| \| Pt \| \| top \| −3.18 \| 0.04 \| 2.07 \| \| \| hcp \| −0.69 \| −0.04 \| 3.08 \| Open in a new tab a Positive value specifies CT from surface to the adsorbate, and vice versa. The determined values of the nearest binding distances, CT, and adsorption energies of CH 3 and C 2 H 6 adsorbed to M(111) surfaces in aqueous suspensions are presented in Table 1. The optimal adsorption of CH 3 shows a gradual decline in the following order: Co(111) > Pt(111) > Ni(111) > Pd(111), whereas for C 2 H 6, the decreasing trend is in the order of Pt(111) > Pd(111) > Ni(111) > Co(111). Tables S1 and S2 present the values for ZPE and adsorption energies of methyl radicals and ethane at optimal adsorption locations. The highest CT from surface to CH 3 is observed at the Co(111) surface (0.40 e), and then on Ni(111) (0.37 e), whereas the lowest CT to CH 3 occurs on Pt(111) (0.04 e), while on Pd, the value is similar (0.07 e). For ethane, the CT is very small, on Co and Ni the CT is from the surface to the molecule (0.01 and 0.03 e, respectively), while for Pd and Pt, surfaces act as an acceptor, exhibiting small negative values of CT (−0.02 and −0.04 e, respectively), meaning CT from molecule to surface. Tables S3 and S4 present the CT values for methyl radicals and ethane across all adsorption sites in the aqueous phase. The binding distances between M(111) and C atoms are similar for CH 3 across all metals, whereas for C 2 H 6, the closest interaction is observed on Pd(111). All the optimized structures of methyl radicals and ethane on M(111) surfaces at all adsorption sites are depicted in Table S5 for methyl radicals and in Table S6 for ethane. The nearest binding distances for various numbers of methyl radicals adsorbed on M(111) surfaces are depicted in Figure S1. 2.2. Adsorption of Methyl Radicals at Higher Coverage Figure 2 shows the total adsorption energies for n (where n = 1–6 for Co; n = 1–7 for Pd and Pt, and n = 1–5 for Ni) methyl radicals on the four M(111) surfaces. Table S7 displays all of the optimal structures for methyl radicals on the M(111) surfaces. As depicted in Figure 2, Ni and Pt surfaces show a monotonic decreasing trend Ni and Pt surfaces show a monotonic decreasing trend up to four (Ni) and six (Pt) methyls, and then, the value rises, suggesting that the adsorption of the fifth and seventh methyl radicals is unfavoured. For Co and Pd, there is a twist in the graph, indicating that the structures with six Co(111) and seven Pd(111) methyl radicals are distorted. The addition of another methyl radical onto a surface containing n − 1 (n = 1–7) pre-adsorbed methyl radicals result in n adsorbed radicals. The adsorption energy of the last methyl is given in Table S8. Reaction 5 was used to assess the thermodynamic limit of the coverage ratio. The free energy () for reaction 5 was determined for each surface: (5) Figure 2. Open in a new tab Adsorption energies () for n methyl radicals on the M(111) surfaces in an aqueous phase (using Equation (5)). The values of on the fcc hollow sites of Co(111) and Ni(111) surfaces, and at the top positions on the Pd(111) and Pt(111) surfaces, are gradually enlarged by sequentially adding one methyl radical at the closest fcc or top sites. The process of adding adsorbed methyl radicals persisted until the free energy of adsorption () for reaction 5 turned endergonic on Ni(111), Co(111), Pd(111), and Pt(111). No more than five radicals can be adsorbed on Co, because for six methyl radicals the structure is completely distorted as depicted in Table S7, for the fifth radical, the free energy of adsorption becomes almost zero (−0.06 eV). On Ni(111), only four CH 3 radicals are adsorbed spontaneously; the adsorption of the fifth methyl is endergonic. On Pd(111) and Pt(111), six methyl radicals can be adsorbed. For Pt, the adsorption of the seventh radical is endergonic; therefore, no more than six methyl radicals can be adsorbed spontaneously, while for Pd, the adsorption of the seventh radical distorts the structure, as depicted in Table S8; therefore, the adsorption of only six radicals is considered. The values are presented in Figure 3. The charge transfer for all the adsorbed methyl radicals is presented in Figures S2 and S3, respectively. Figure 3. Open in a new tab Gibbs free energies () for the adsorption of an additional methyl radical on the M(111) surfaces (using Equation (5)). 2.3. Projected Density of States The electronic properties of M(111) surfaces from the perspective of the electronic projected density of state are shown in Figure 4. Figure 4 displays the contribution of the projected density of states (PDOS) orbitals around the Fermi level of the M(111) surfaces. Most of the theoretical studies based on first principles concentrate on understanding how the adsorbate interacts with the d-electrons of the surfaces of transition metals [46,47,48,49,50,51]. The d-band center model created by Hammer and Nørskov over a decade ago is the most commonly used framework to analyze the function of d-electrons [52,53,54,55]. The d-band center () and p-band center () are determined through the equation provided below : (6) Figure 4. Open in a new tab The projected density of states for a pristine M(111) surface, M = Co, Ni, Pd, and Pt (a,b,e,f), and of the surfaces with an adsorbed CH 3 (c,d,g,h). The band center for each orbital (x = p, d) is denoted by , which corresponds to the p-band center () and d-band center (). Figure 4 {(a, c), (b, d), (e, g) and (f, h)} presents the projected density of states for pristine metallic surface and a surface with an adsorbed CH 3 of Co(111), Ni(111), Pd(111) and Pt(111), respectively. The purple vertical line represents the d-band center value in all the figures in the diagram of PDOS, and the green vertical line represents the pbc of the carbon of the adsorbed methyl. The values of the dbc and the pbc are given in Table 2. For all the metals, the dbc for the pristine metal and the dbc for the metal with the adsorbate are almost the same; only a small shift to a higher value is observed (0.02–0.04 eV). In principle, as the dbc is higher, the adsorption should be stronger, as fewer anti-bonding states are occupied; therefore, lower adsorption energies are expected on Co(111) and Ni(111) than on Pd(111) and Pt(111). Surprisingly, the adsorption energy on Pt(111) is very low (−3.18 eV). The reason is the minimal value (0.50 eV) of the dbc-pbc, as depicted in Table 2, indicating a very good overlap between the d-band of the metal and the p-band of the carbon (C) of the adsorbed CH 3. In addition, Co(111) has a lower dbc (−1.47 eV) than Ni(111) (−1.30 eV), but the adsorption energy is stronger (−3.34 eV) than on Ni(111) (−3.11 eV). The strong adsorption on Co(111) is attributed to the smaller valence electrons (9) than all the other metals (10); therefore, fewer electrons are available in the case of Co(111) to occupy the anti-bonding states. Table 2. The estimated values of d-band center (dbc) for pristine surfaces and adsorbed CH 3 (), p-band center (pbc), and difference of d- and p-band centers (dbc-pbc). \| Metal Surfaces \| \| pbc () (eV) \| dbc-pbc (eV) \| :---: :---: \| \| Pristine Surfaces \| \| \| Co(111) \| −1.47 \| −1.43 \| −5.31 \| 3.88 \| \| Ni(111) \| −1.30 \| −1.28 \| −4.33 \| 3.05 \| \| Pd(111) \| −1.87 \| −1.85 \| −3.06 \| 1.21 \| \| Pt(111) \| −2.62 \| −2.59 \| −3.09 \| 0.50 \| Open in a new tab 2.4. Production of Ethane via the LH Mechanism—A Reaction of Two out of n Adsorbed Radicals In this section, we examine the ethane evolution reaction at various coverage ratios up to 5/5 for Co(111), 6/6 for Pd(111) and Pt(111) surfaces, as well as 4/4 for the Ni(111) surface, according to the adsorption ratios observed in Section 2.2. The free energies of ethane evolution via the LH mechanism () are determined through reaction 7: (7) The standard reaction free energies for ethane formation (), based on reaction 7, are presented in Figure 5 for the aqueous suspensions. values and barrier heights for n = 6 (Pd, Pt), n = 5 (Co, Pd, Pt), and n = 4 (Co, Ni, Pd) are given in Table 3. Based on the findings in Table S9, ethane will not be formed on all M(111) surfaces with a coverage ratio of 2/n. Since the adsorption energy diminishes as the coverage ratio increases (Figure 3), the feasibility of ethane formation increases. Based on the findings shown in Figure 5, the highest exergonicity for this reaction is seen on Pd(111), followed by Co(111), while the values for Ni(111) and Pt(111) are considerably smaller. The exergonicity increases in magnitude (becomes more negative) with a rise in the coverage ratio. These values diminish from 1 to n (n = 5 for Co, 6 for Pd and Pt, and 4 for Ni), with Co(111) changing from 1.67 to −2.95 eV, Ni(111) from 1.20 to −0.17 eV, Pd(111) from 0.34 to −2.50 eV, and Pt(111) from 1.35 to −1.57 eV. All the reaction free energies ΔG 0(7) for the production of ethane on various M 0(111) surfaces at different surface coverage ratios are provided in Table S9. Figure 5. Open in a new tab Reaction-free energies () for the evolution of ethane on the M(111) surfaces (using Equation (7)). The dotted line indicates zero energy. Table 3. The evaluated reaction free energies (ΔG 0(7)), activation energy barrier (E a), C-C distance in the initial state (IS), and transition state (TS) on different M(111) surfaces. \| M(111) Surfaces \| No. of Methyl Radicals Adsorb \| \| \| C-C Bond (Å) in IS \| C-C Bond (Å) in TS \| :---: :---: :---: \| \| Co \| \| −0.39 \| 1.24 \| 3.46 \| 1.85 \| \| Ni \| \| −0.17 \| 1.75 \| 3.72 \| 1.93 \| \| Pd \| \| −0.49 \| 0.83 \| 3.21 \| 2.01 \| \| Co \| \| −2.95 \| 0.52 \| 3.21 \| 3.15 \| \| Pd \| \| −1.04 \| 0.65 \| 3.00 \| 2.04 \| \| Pt \| \| −0.18 \| 1.36 \| 3.05 \| 1.93 \| \| Pd \| \| −2.50 \| 0.41 \| 3.09 \| 2.48 \| \| Pt \| \| −1.57 \| 1.32 \| 3.09 \| 3.03 \| Open in a new tab According to the results presented in Table 3, the lowest barrier was found for Pd at high coverage (n = 6) (0.41 eV), then on Co at high coverage (n = 5) (0.52 eV). Moderate barriers were observed for Pd; n = 5 (0.65 eV) and n = 4 (0.83 eV). All other barriers are very high and exclude the formation of ethane via this mechanism. Table 3 also presents the carbon–carbon bond distances in both the initial and the transition states. Figure 6 illustrates the lowest barrier alongside the exergonic reaction, on M(111) surfaces, for n = 4, 5, 6. Additionally, the activation barrier for n = 3–6 CH 3 adsorbed on Pd(111) is presented in Figure S4. The activation energy barrier is lowered with an increase in the coverage ratio; for Co(111), it reduces from 1.24 to 0.52 eV (4–5 CH 3), for Ni(111) it is 1.75 eV (4 CH 3), for Pd(111) it decreases from 1.19 to 0.41 eV (3–6 CH 3), and for Pt(111) it is 1.36 to 1.32 eV (5–6 CH 3). These values suggest that the production of ethane via this mechanism is feasible exclusively on the Pd(111) and Co(111) surfaces. As the coverage ratio rises, the activation energy barrier values are shown in Table S10. The structures for the initial state (IS), TS, and final state (FS) are provided in Figure 7 and Figure 8 for 4–6 CH 3 adsorbed on M(111) surfaces, respectively. Figure 6. Open in a new tab The activation energy barrier (E a) for the evolution of ethane in the case of (a) 4CH 3 for Co, Ni, Pd; (b) 5CH 3 for Co, Pd, Pt; (c) 6CH 3 for Pd, Pt. The dotted line indicates zero energy. Figure 7. Open in a new tab The IS, TS, and FS for 4–5CH 3 are placed on the Co(111) and Ni(111) surfaces for evolution of ethane. Dark blue—Co, gray—Ni, brown—C, and pink—H. Figure 8. Open in a new tab The IS, TS, and FS of 5–6CH 3 for Pd(111) and Pt(111) surfaces for the evolution of ethane. blue—Pd, light blue—Pt, brown—C, and pink—H. 2.5. Ethane Production via the RE Mechanism—Reaction Between One Adsorbed Methyl Radical and Another One in Solution (8) (9) (10) The generation of ethane is examined via the ER mechanism—a methyl radical moves randomly within the aqueous solution, and while moving, it collides with an adsorbed methyl, and ethane is produced according to reaction 8 or methane according to reaction 9 followed by reaction 10. Reaction 8 and reaction 9 are competing reactions, constrained by the lifetime of the radical in solution. Both reactions may occur on Pt(111), Ni(111), and Co(111), while on Pd(111), only reaction 8 occurs, and is not formed according to reaction 9. Table 4 provides the reaction free energies and barriers for these reactions. The IS, TS, and FS figures are shown in Figure 9. For Pt(111), both reactions are barrierless; therefore, ethane and are both produced via the ER mechanism. On Ni(111) and Co(111), reaction 9 is barrierless, while there is a barrier for reaction 8 (0.16 eV and 0.56 eV, respectively); therefore, reaction 9 is predominant, and is produced. A small barrier is also observed for reaction 8 on Pd(111) (0.21 eV). The small barrier on Ni(111) and Pd(111) enables reaction 8 to occur if the steady state concentration of is small, e.g., in continuous photolysis or radiolysis. Formation of ethane via the RE mechanism on Co(111) is less likely. Table 4. The evaluated reaction free energies (ΔG 0) and activation energy barriers (E a) on M(111) surfaces for reactions 8 to 12 in aqueous suspensions. \| Reaction Numbers \| Reactions \| Co(111) (eV) \| Ni(111) (eV) \| Pd(111) (eV) \| Pt(111) (eV) \| :---: :---: :---: \| \| \| \| \| \| \| \| \| \| \| 8 \| \| −2.11 \| 0.56 \| −2.36 \| 0.16 \| −2.86 \| 0.21 \| −2.72 \| NB \| \| 9 \| \| −2.93 \| NB \| −2.89 \| NB - \| −2.69 \| NB \| \| 10 \| \| 0.82 \| NC \| 0.52 \| NC \| NC \| NC \| −0.03 \| 1.60 \| \| 11 \| \| 0.48 \| NC \| 0.13 \| NC \| NC \| NC \| −0.61 \| 1.48 \| \| 12 \| \| 0.67 \| NC \| 0.72 \| NC \| NC \| NC \| −0.13 \| 1.36 \| Open in a new tab NB: no barrier; NC: not calculated. Figure 9. Open in a new tab The IS, TS, and FS for the evolution of ethane according to reaction 8. dark blue—Co, grey—Ni, blue—Pd, light blue—Pt, brown—C, and pink—H. The formation of on Ni(111) was reported previously . On Co(111) and Ni(111), reaction 10 is endergonic (0.82 and 0.52 eV, respectively), preventing the production of ethane via this mechanism. On Pt(111), reaction 10 is exergonic (−0.03 eV), but is not formed via reaction 10 as the barrier for this reaction is very high, 1.60 eV. Reaction of and to form and on M(111) surfaces: (11) (12) Reactions 11 and 12 do not occur on Co(111) and Ni(111) surfaces as they are endergonic (for Co(111), ΔG 0 = 0.48 eV (reaction 11) and 0.67 eV (reaction 12); for Ni(111), the values are 0.13 and 0.72 eV, respectively); however, these reactions may occur on Pt(111) surfaces since they are exergonic reactions, −0.61 and −0.13 eV. These values and the values of the energy barriers are given in Table 4. These reactions are expected to be very slow due to their high barriers (1.48 eV, rate constant = 9.65 × 10−14 M−1 s−1 for reaction 11 and 1.36 eV, rate constant = 1.06 × 10−11 M−1 s−1 for reaction 12). We have assumed, based on Eyring theory , that the rate constant for a reaction between two radicals in aqueous solution is 10 12. and are not produced on Pd(111) as is not formed according to reaction 9 on this surface. The structures of IS, TS, and FS for these reactions are depicted in Tables S11–S14. The reaction of with according to the RE mechanism was studied as well. This reaction does not occur directly, is first adsorbed, and only then, it reacts with , this is an exergonic reaction only on Pt(111) (−3.31 eV). The high adsorption energy of on Pt(111) (−3.18 eV) is enough to overcome the high barrier of reaction 12 (1.36 eV), so C 2 H 5 is produced on Pt(111), as was determined experimentally . The formation of on Pt(111) can be followed by one of the following reactions: (13) (14) (15) According to reaction 13, splits into This is an exergonic reaction (ΔG 0 = −0.43 eV) with a barrier of 0.65 eV; was observed as a product of the reaction of with Pt 0 [29,31]. Reaction 14 is an isoenergetic reaction (ΔG 0 = 0.00 eV) with a high barrier of 1.81 eV; therefore, this reaction is not plausible on Pt(111) surfaces. The activation barrier of reaction 13 is depicted in Figure 10. Figure 10. Open in a new tab The IS, TS, and FS of splits into in solution for Pt(111) surface. light blue—Pt, brown—C, and pink—H. Surprisingly, recombination of an adsorbed methyl radical with an adsorbed ethyl radical, according to reaction 15, is not plausible, as this is an endergonic reaction (ΔG 0 = 0.51 eV). 2.6. Formation of Methanol (CH 3 OH) on M(111) Surfaces (16) The produced on the M(111) (M = Co, Ni, Pt) can interact with an adsorbed water molecule to yield methanol according to reaction 16. This reaction is not feasible on all the M(111) surfaces, as it is endergonic (the values are 1.24, 0.91, and 0.58 eV for Co, Ni, and Pt, respectively). Also, the reaction of with an adsorbed water molecule to form is not feasible on all the M(111) surfaces, as it is endergonic (ΔG 0 = 0.86 eV (Co) = 0.64 eV (Ni) = 0.32 eV (Pd) = 0.63 eV (Pt)). 2.7. The Diffusion of Adsorbed CH 2 and CH 3 on M(111) Surfaces Since it is anticipated that the CH 2 and CH 3 will be randomly adsorbed at their optimal adsorption locations on the M(111) surfaces, they are expected to move on the surface until they attain the required configuration for the reaction. Consequently, the energy barrier for the diffusion of an adsorbed CH 2 and CH 3 from one identical site to another was determined. The diffusion process of on M(111) surfaces occurs in two steps, and two barriers are involved. On Co(111), the first step refers to the shift from hcp to fcc, and the second refers to the shift from fcc to hcp. On Ni(111), the steps are fcc to hcp and then hcp to fcc. On Pt(111), the first step is the bridge to fcc, and then fcc to bridge. On Pd(111), only one barrier is observed, for the first shift from bridge to fcc, the second shift from fcc to bridge is barrierless. The barriers are depicted in Table 5. The values are on Co of 0.48 and 0.09 eV; on Ni of 0.23 and 0.46 eV; and Pt of 0.04 and 0.67 eV; on Pd, a barrier of 1.03 eV was found for the diffusion of , but is not formed on Pd(111). Table 5. The diffusion barriers for the movement of CH 2 and CH 3 adsorbed on M(111) surfaces from initial to final position in aqueous medium. \| Adsorbates on M(111) \| Diffusion Barrier (E a) (eV) \| :---: \| \| Co(111) \| Ni(111) \| Pd(111) \| Pt(111) \| \| \| 0.48 \| 0.46 \| 1.03 \| 0.67 \| \| \| 0.04 \| 0.50 \| 0.28 \| 0.65 \| Open in a new tab The diffusion of methyl radicals () on the surface presents a minimal barrier on Co (0.04 eV), while the barrier is higher for Ni (0.50 eV), Pd (0.28 eV), and even higher for Pt (0.65 eV). These barriers are lower than the relevant barriers of reactions 7, 11, and 12; therefore, they are not expected to affect these reactions. The rate constants of the relevant processes are given in Table 6. The structures of the initial state (IS), TS, and final state (FS) for Co(111), Ni(111), Pd(111), and Pt(111) are presented in Tables S15, S16, S17 and S18, respectively. Table 6. The rate constants (k) for the diffusion of CH 2 and CH 3 adsorbed, reactions 7, 11, and 12 on M(111) surfaces in aqueous medium. \| Processes \| Rate Constants (k) (M−1 s−1) \| :---: \| \| Co(111) \| Ni(111) \| Pd(111) \| Pt(111) \| \| Diffusion of \| 6.19 × 10 3 \| 1.68 × 10 4 \| 3.15 × 10−6 \| 4.18 \| \| Diffusion of \| 9.65 × 10 11 \| 3.07 × 10 3 \| 2.25 × 10 7 \| 1.26 × 10 1 \| \| Reaction 7 () \| 1.06 × 10−9 \| 2.18 × 10−18 \| 9.38 × 10−3 \| NC \| \| Reaction 7 (5) \| 1.69 × 10 3 \| NC \| 1.03 × 10 1 \| 1.06 × 10−11 \| \| Reaction 7 (6) \| NC \| NC \| 1.02 × 10 5 \| 4.76 × 10−11 \| \| Reaction 11 () \| NC \| NC \| NC \| 9.65 × 10−14 \| \| Reaction 12 () \| NC \| NC \| NC \| 1.06 × 10−11 \| Open in a new tab NC: not calculated. In Table 6, the rate constants for the diffusion of and and for the formation of ethane, according to the LH mechanism (reaction 7) and RE mechanism (reaction 8) are displayed. The rate constants are calculated according to the barriers. The diffusion rate constants of and are higher than the rate constant for the reactions given in Table 6; therefore, the diffusion of these species does not affect the rate of these reactions. 3. Computational Methods We used the Vienna ab-initio Simulation Package (VASP) [58,59], version 6.3.2, to conduct a non-spin polarized first-principles computation within the framework of density functional theory (DFT) . To deal with electron–ion–core focused interfaces, the projector augmented wave (PAW) has been used . We have operated the Perdew–Burke–Ernzerhof (PBE) [62,63] function to manage electron exchange and correlation in addition to the PAW technique. The Grimme’s DFT-D2 dispersion correction [64,65] was used to describe the long-range van der Waals (vdW) interactions. The solvent effect was considered using an implicit self-consistent electrolyte solvation model, VASPsol . Explicit water molecules were not used, as no significant interactions are expected between the methyl radicals and the solvent (H 2 O). An energy cut-off of 500 eV was employed for each slab. A Monkhorst–Pack mesh of 6 × 6 × 1 k-point was used to sample the Brillouin zone . The convergence accuracy criteria for all our calculations were 10−3 eVÅ−1 and 10−5 eV for forces and energy. The transition states (TS) were located using the Climbing Image Nudged Elastic Band method (Cl-NEB) by considering five images for every state. We investigated the p- and d-band center values by using VASPKIT Standard Edition 1.5.1 (27 January 2024) . All M(111) (M = Co, Ni, Pd, Pt) surfaces were represented by a six-layer slab; every layer is made up of eight metal atoms, and a 16 Å vacuum space is included between the slabs in the z-direction to prevent undesired interactions. The vacuum space is substituted with an aqueous medium utilizing VASPsol. The harmonic oscillator method was employed to conduct phonon calculations for every optimized structure, utilizing a step width of 0.015 Å to derive the zero-point vibration energies (ZPVE) of the system. These computations were employed to confirm that the optimized configurations represent real minimum or transition states (one extra imaginary frequency). The ZPVE values were utilized to determine the Gibbs free energy. The VESTA code [70,71,72] was used to illustrate the stable structures and the TS. The adsorption energy (E ads) of adsorbate on the M(111) surfaces is calculated using Equation (17): (17) (18) E, T, and S are electronic energy, room temperature (298.15 K), and entropy. is the free energy of the surface with the adsorbate, is the free energy of the M(111) surfaces, and is the free energy of the aqueous adsorbate. Negative values mean that adsorption is favored, and vice versa. The free energy () of a reaction was calculated using the equation: (19) The amount of charge transfer (CT) of the adsorbate on M(111) is determined through Bader charge analysis [73,74]. Positive CT values indicate CT from the surface towards the adsorbate, whereas negative values signify CT from the adsorbate towards the surface. We determined the rate constants (k) through the application of the Arrhenius equation , (20) where is the activation energy barrier, T is the absolute room temperature (298.15 K), is the Boltzmann constant, and A is the pre-exponential factor. In this context, we have utilized the pre-exponential factor (A) value of 10 12 to determine the rate constant. In this research, pre-adsorbed water molecules are not considered. 4. Concluding Remarks In summary, results of an in-depth investigation of the properties of methyl radicals at the M(111) (M = Co, Ni, Pd, Pt) surfaces, using the density functional approach, were obtained. The adsorption of methyl radicals on M(111) surfaces, consisting of eight atoms, was explored, revealing that up to five methyl radicals can be adsorbed on Co(111), while on Pd(111) and Pt(111) surfaces up to six, and on Ni(111), only four radicals can be adsorbed. The best adsorption sites are fcc (Co and Ni) and atop (Pd and Pt). Production of ethane via the LH mechanism is endergonic on all metals for two adsorbed methyl radicals. The production of ethane is exergonic for high coverage ratios. Kinetically, it is produced at room temperature on Co at a high coverage ratio (5/5) with a barrier of 0.52 eV, and on Pd at a coverage ratio of 4/6, 5/6, and 6/6 with barriers of 0.83, 0.65, and 0.41 eV, respectively. Ethane can also be produced via the RE mechanism, one radical that is moving randomly in the aqueous solution hits an adsorbed CH 3 radical on the surface. In this scenario, the products are either ethane or and . On Pd(111), such a collision produces only . On Ni(111) and Co(111) surfaces only production of + is observed without a barrier, while on Pt(111), both processes are observed; neither has a barrier, and, therefore, both are plausible. The reaction of two radicals to produce is exergonic on Pt(111), but is very slow due to its large barrier (1.48 eV). In addition, the reaction of with to produce is exergonic on Pt(111) with a similar barrier (1.36 eV). The diffusion of the on the M(111) surfaces has a moderate barrier on Co, Ni, and Pt (0.48 eV, 0.46 eV, 0.67 eV, respectively), the barrier for Pd (1.03 eV) is significantly higher, but is not formed on Pd(111) via the reaction of methyl radicals. The diffusion of the methyl radicals on the surface is almost barrierless for Co (0.04 eV), and moderate for the other surfaces, 0.50, 0.28, and 0.65 eV for Ni(111), Pd(111), and Pt(111), respectively. These barriers are lower than the relevant barriers of reactions 7, 11, and 12; the diffusion is faster than the reaction of these intermediates on the surface, therefore, they are not expected to affect these reactions. and were detected experimentally on Pt(111) . The formation of on Pt(111) occurs via the RE mechanism in a two-step mechanism, is adsorbed on the Pt(111) surface, and then it reacts with to produce . This reaction is plausible due to the high adsorption energy of (−3.18 eV). is produced by the de-hydrogenation of in an exergonic reaction (ΔG 0 = −0.43 eV) with a barrier of 0.65 eV on Pt(111) surface. These products that are produced on Pt(111) were not found on Ag(111), Au(111), and Cu(111) surfaces . Acknowledgments The authors acknowledge the Ariel HPC Center at Ariel University for providing computing resources that have contributed to the research results reported in this paper. P.K. appreciates the fellowship provided by Ariel University, Israel. Supplementary Materials The following supporting information can be downloaded at: molecules-30-03065-s001.zip (1.4MB, zip) Author Contributions Conceptualization, D.M. and H.K.; methodology, P.K. and H.K.; software, P.K. and H.K.; validation, P.K., H.K. and D.M.; formal analysis, P.K.; investigation, P.K.; resources, P.K. and H.K.; data curation, P.K., H.K. and D.M.; writing—original draft preparation, P.K., H.K. and D.M.; writing—review and editing, H.K. and D.M.; visualization, P.K.; supervision, H.K. and D.M.; project administration, H.K. and D.M.; funding acquisition, A.M. and D.M. All authors have read and agreed to the published version of the manuscript. Institutional Review Board Statement Not applicable. Informed Consent Statement Not applicable. Data Availability Statement The original contributions presented in this study are included in the article and Supplementary Materials. Further inquiries can be directed to the corresponding author(s). 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Phys. 1935;3:107–115. doi: 10.1063/1.1749604. [DOI] [Google Scholar] Associated Data This section collects any data citations, data availability statements, or supplementary materials included in this article. Supplementary Materials molecules-30-03065-s001.zip (1.4MB, zip) Data Availability Statement The original contributions presented in this study are included in the article and Supplementary Materials. Further inquiries can be directed to the corresponding author(s). Articles from Molecules are provided here courtesy of Multidisciplinary Digital Publishing Institute (MDPI) ACTIONS View on publisher site PDF (4.7 MB) Cite Collections Permalink PERMALINK Copy RESOURCES Similar articles Cited by other articles Links to NCBI Databases On this page Abstract 1. Introduction 2. Result Analysis 3. Computational Methods 4. 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15776	https://www.elledecor.com/design-decorate/trends/g30123056/chair-types-styles/	Skip to Content Every item on this page was chosen by an ELLE Decor editor. We may earn commission on some of the items you choose to buy. Courtesy of Herman Miller These Are the 23 Most Iconic Chairs of All Time These seats stand the test of time. By Rachel Silva 1 Gaetano Pesce’s Up Chair Indianapolis Museum of Art at Newfields // Getty Images Year: 1969 Designer: Gaetano Pesce Provenance: Italy The late Italian designer Gaetano Pesce was in the shower in 1968 when the idea for the Up chair came to him. “I had the sponge in my hand,” he told Architectural Digest in 2017. “When I pressed the sponge, it shrank, and when I released it, it returned to its original volume.” This visual inspired his design of the pop-up polyurethane chair—a series of “Up” chairs that feature a rounded, sculptural form and are made with a single piece of polyurethane foam. Nearly 50 years after its creation, versions of it grace the halls of design museums around the world, including the Museum of Modern Art in New York City. Like all statement pieces, the “Up” chair—also known as La Mamma for its sensual rounded contours—has been the subject of controversy among some. In 2019, an Italian feminist group staged a protest around a 26-foot installation depicting the famous Up chair in Milan’s Piazza del Duomo during Milan Design Week, calling it a “further violence against women, represented as helpless and formless bodies, where man is not called into question.” Pesce responded by saying, “I believe these feminists have not read nor understood the meaning of my work in Milan." GET THE LOOK $6,511 at B&B Italia Photography courtesy of Knoll Archive 2 Queen Anne Side Chair Year: 1983 Designer: Robert Venturi and Denise Scott Brown, designed for Knoll Provenance: New York City American architect Robert Venturi is considered one of the founders of postmodernism—and his Queen Anne side chair (created in collaboration with his wife and creative partner, Denise Scott Brown) is a significant contribution to this movement. Venturi and Scott Brown transformed the elegant shape of an 18th-century Queen Anne chair to a flattened silhouette in bent plywood. The “Grandmother” pattern covering the piece is based on a mass-produced floral tablecloth that was owned by the grandmother of one of their employees. The chair, which—in typical PoMo style blends historic and modern styles with references to high and low culture—is a cheeky comment on consumer society’s desire to be linked with history. Advertisement - Continue Reading Below Courtesy Emeco 3 Emeco Navy Chair Year: 1944 Designer: Wilton C. Dinges Provenance: Hanover, Pennsylvania The Emeco 1006, also known as the Navy chair, is an aluminum chair developed by Emeco founder Wilton C. Dinges in 1944 in collaboration with the Aluminum Company of America (ALCOA). It was originally designed for the U.S. Navy warships, which needed a chair for the deck of battleships that could survive water, salt, sea air, sailors, and torpedo blasts. It’s the result of a 77-step (!) process in which welders melt recycled aluminum to form the sturdy, lightweight (just seven pounds) frame that far exceeded the navy’s specifications. Dinges reportedly demonstrated the chair’s durability when competing for the navy contract by throwing it out of an eighth floor window of a Chicago hotel. It bounced but did not bend or break. The chair has since become ubiquitous, showing up in many a restaurant scene and inspiring many a dupe. In the 1990s, the company began creating designer versions of the chair, called the 111 Navy chair, made from 111 plastic bottles, 65 percent recycled PET plastic and 35 percent glass fiber and pigment. Since its 2010 launch, the chair has diverted more than 20 million bottles out of landfills. GET THE LOOK $740, Emeco 4 Bentwood Thonet 209 Armchair Mark Kauffman // Getty Images Year: 1859 Designer: Michael Thonet Provenance: Austria The bentwood chair, also known as the Thonet 209, not only deserves a place in the history books for its popularity as the quintessential restaurant chair, but also because it emerged from one of the most significant innovations in the timeline of the modern chair. Created by Prussian-born cabinetmaker Michael Thonet, who was experimenting with new methods of bending solid wood with steam and molding it to shape with mechanical presses, this chair was originally designed for the Daum coffeehouse in Vienna. Thonet set up a furniture manufacturing company in Vienna with his five sons, where the chair is still in production today. “It is the biggest selling chair model of all time, with 50 million units sold by 1930,” Fiell explains. With its near-circular seat frame and its hoop splat back, the Thonet 209 armchair is the icon for modernists. Even Le Corbusier himself considered this the chair that best shares the aesthetic of his modern architecture, placing it in nearly all of his buildings. GET THE LOOK $486, Design Within Reach Advertisement - Continue Reading Below 5 Wassily B3 and Cesca Chair Keystone-France // Getty Images Year: 1925 and 1928 Designer: Marcel Breuer Provenance: Italy Behold the Wassily B3 Chair, the first successful cantilever chair that defied the four-legged standard. Fascinated by bicycle handlebars, Hungarian architect Marcel Breuer introduced the first chair made from tubular steel in 1925. While his cohorts, the Dutch architect Mart Stam and German architect Ludwig Mies van der Rohe, were experimenting with the same materials and showed their designs in public first at the Die Wohnung exhibition, it was Breuer’s version—which had better proportions and was more hard-wearing and comfortable—that ultimately revolutionized furniture. The sculptural, abstract chair was a milestone in the history of modern furniture, offering a sitting experience in which one is suspended over the base, seemingly floating on air, with just two legs for support (and a comfortable bounce). Three years later, Breuer would introduce through Knoll the Cesca, a simplified version of the chair that features a cane seat. While the Wassily and Cesca chairs have become the holy grail for modern designers, their design is as surprising now as it was in its first years. The chairs can be found in the permanent collections of museums like MoMA and the V&A. GET THE LOOK $3,633, Knoll GET THE LOOK $1,216, Knoll 6 Eileen Gray Transat Chair Courtesy B2H Year: 1927 Designer: Eileen Gray Provenance: France Used as a deck chair at Eileen Gray’s famous villa E-1027 in Roquebrune-Cap-Martin, France, the Transat chair has become one of the architect’s most enduring pieces. It gets its name from “transatlantic,” referring to deck chairs commonly used on steamships. This style would be a recurring theme in Gray’s buildings, which often resembled houseboats ready to sail. Flaunting a sleek frame with complicated joinery and an adjustable headrest, the Transat chair evokes the effortless pairing of comfort and class—and emerged as a Gray icon when it was published in Badovici’s L’Architecture Vivante in 1929. A hand-laquered specimen saw a spot in the homes of Yves Saint Laurent and Pierre Bergé, and sold at auction in 2009 for a staggering $27.8 million—making it still the most expensive chair ever sold in the world. GET THE LOOK $5,173, B2H Advertisement - Continue Reading Below Courtesy Design Within Reach 7 Grand Confort LC2 Year: 1928 Designer: Charlotte Perriand, Le Corbusier, and Pierre Jeanneret Provenance: France Now this is a chair that isn’t afraid to show its true self. Inspired by the boxy-looking gentleman’s club chairs of the 1920s, Charlotte Perriand and Pierre Jeanneret designed this chair as part of an architectural commission for Le Corbusier’s design practice in Paris. What made it special? The LC2 wears its bones on the exterior, flaunting a tubular steel exoskeleton that evokes the rationalized geometric forms and machine-age aesthetic of Le Corbusier’s architecture. Dubbed by Le Corbusier as the “cushion basket,” the Grand Confort LC2 is a clever way to contain cushions in an open frame—and an enduring icon of modernist style. GET THE LOOK $6,555, Design Within Reach 8 Barcelona MR90 Chair Courtesy Design Within Reach Year: 1929 Designer: Ludwig Mies van der Rohe Provenance: U.S. Some chairs fall short of their intended design statement. This one does not. A chair truly fit for royalty, the MR90 was designed by German-American architect Mies van der Rohe and his partner, tubular design pioneer Lilly Reich, to furnish the German pavilion that the King of Spain Alfonso XIII and his queen would visit for the 1929 International Exhibition in Barcelona (hence the chair’s nickname, the Barcelona). While van der Rohe and Reich subscribed fully to the modern “less is more” approach, they looked to historic models when designing the MR90. Its scissor-shaped X-frame design was adopted from the luxurious chairs of Imperial Rome and ancient Egypt. Their own interpretation featured chrome-plated steel and leather upholstery with button stitching. While it was never intended to be mass-produced, Knoll began manufacturing the chair in 1953 (Fiell tells us first-edition examples are incredibly rare), enhancing the original prototypes by reconfiguring the frame with seamless welded joints. The MR90 has since become one of the most sought-after chairs in the seating world and would launch more than a thousand copies and continue to appear in interiors today. GET THE LOOK $8,134, Knoll Advertisement - Continue Reading Below 9 Eames LCW Chair Courtesy Herman Miller Year: 1945 Designer: Charles and Ray Eames Provenance: U.S. American designer couple Charles and Ray Eames spent years experimenting with new materials and furniture-making processes, ultimately developing a machine called the “Kazam! Machine” that could mold plywood sheets into compound forms that could then be bent. The famed LCW plywood chair is the result of this process, introduced in 1946 after a series of award-winning chairs designed for the Museum of Modern Art’s 1940 “Organic Design in Home Furnishings” exhibition. Curiously, the seat was inspired by the gentle curve of a potato chip, with its soft cradling form shaped to host a sitter of any size—a design feature that was way ahead of anything that was being produced in Europe in the 1940s, according to the late chair historian Judith Miller. “Ingeniously, the Eameses [also] used rubber discs called shock mounts to fix the back and seat to the lumbar support and base, making the chair more flexible as well as more comfortable—and creating one of the first examples of a responsive backrest in the history of furniture,” she writes in her book Chairs. Today, this elegant, stackable chair remains easy to mass-produce (and therefore affordable) and a longstanding testament to the Eameses’ ingenuity. GET THE LOOK $1,116, Herman Miller 10 Eames Lounge Chair and Ottoman Courtesy Herman Miller Year: 1956 Designer: Charles and Ray Eames Provenance: U.S. If the LCW is the workaday sedan of design, the Eames lounge chair and matching ottoman, No. 670 and 671, are the Rolls-Royce of 20th-century chairs. “Unlike most other chairs by Charles and Ray Eames, this one was aimed squarely at the luxury end of the market,” Fiell says. “It was and remains very expensive.” It’s one of the most instantly recognizable designs by the Eames duo—and for good reason. “Comfortable and puffy like an old-fashioned club sofa, it is the mid-20th-century answer to the Victorian daybed,” Miller writes. Best known for their efforts to create inexpensive furniture pieces that could be mass-produced easily, the Eames set is the husband-and-wife team’s take on luxury. “The couple’s genius lies in the fact that the chair and ottoman look equally appropriate in the executive office, a home study, or the family den,” Miller adds. It’s also the best example of the evolution of the lounge chair typology, Vitra’s Liv Buur explains. “Charles and Ray Eames studied the chesterfield and the English club chair, in an aim to design a chair with similar comfort but a more easy and contemporary appearance. They succeeded—big time—both in design and longevity.” The seat was released by Herman Miller in 1956. Vitra produces and distributes the chair and ottoman in Europe and the Middle East and Herman Miller does so for the rest of the world. The first chairs introduced in the 1950s still exist today and continue to be one of the most collectible models of sexy (not to mention comfortable) midcentury design. GET THE LOOK $5,916, Herman Miller Advertisement - Continue Reading Below 11 DAR Chair Courtesy Design Within Reach Year: 1948–50 Designer: Charles and Ray Eames Provenance: U.S. Quite possibly the most famous of all the chairs (Fiell says he considers these the most important series of chairs ever designed), Charles and Ray Eames’s DAR seat is an example of survival of the fittest. “The DAR is part of a greater group of chairs, the Eames Shell chairs, which were the first chair with a seat and backrest formed from a single plastic shell,” Liv Buur says. “It was a ground-breaking innovation developed by the Eameses to meet the needs of a changing society.” The DAR chair, which stands for Dining Armchair with Rod base, merges Paris-chic with classic ’50s style, incorporating a base that resembles the Eiffel Tower and a gracefully ergonomic sculptural shell. The piece was the first to succeed the idea of the chair as part of a greater system, conceived to offer a variety of interchangeable bases including a rocker with a rod base on wooden runners and a four-legged base in tubular steel. The design has been in production by Herman Miller since the early 1950s and has spurred countless copies. GET THE LOOK $600, Design Within Reach 12 Womb Chair Courtesy Design Within Reach Year: 1947-48 Designer: Eero Saarinen Provenance: U.S. Fed up with seating options that offered comfort at the expense of style, pioneering midcentury modern architect and furniture designer Florence Knoll issued a challenge in 1948 to a family friend, rising design star Eero Saarinen, to create a cozy chair that she could curl up in. The Womb chair was the Finnish-American architect’s answer. The first piece of mass-produced furniture with an integrated seat shell made of fiber-reinforced plastic (thanks to the help of a boat builder in New Jersey who was experimenting with fiberglass and resin), this all-enveloping piece offers endless posture options and extra elbow room for a comfortable and stylish spot to snuggle in. GET THE LOOK $5,268, Knoll Advertisement - Continue Reading Below 13 Wishbone Chair Courtesy Design Within Reach Year: 1949 Designer: Hans Wegner Provenance: Denmark Inspired by a painting of Danish merchants in Ming dynasty chairs in 1944, the Wishbone chair was drawn up exclusively for Carl Hansen & Søn in 1949. Also known as the Y chair, this specimen requires over 100 manufacturing steps to create the steam-bent top and minimal, wishbone-shaped back, including a handwoven seat that takes an hour and approximately 400 feet of paper cord to create. It’s worth it. Its elegant simplicity makes it a masterpiece of Scandinavian design and a mainstay in dining rooms and offices alike. GET THE LOOK $790, Design Within Reach 14 Ant Chair Courtesy Design Within Reach Year: 1951–52 Designer: Arne Jacobsen Provenance: Denmark Arne Jacobsen was already an esteemed architect when he designed this lightweight stacking chair for the factory canteen of the Danish pharmaceutical firm Novo Nordisk. It was the first chair to be made from a single piece of plywood, featuring a slightly curved seat that was designed with the body’s needs in mind. But in fact, the original three-legged piece was not an instant hit. In fact, it almost ended up being another one of Jacobsen’s prototypes as Novo ordered only 300 chairs. It was only in 1980 after Jacobsen’s death (he was adamant about the three legs) that a fourth leg was added and a variety of colors was offered. It flew off the shelves. This landmark design, produced by Fritz Hansen, became the basis for many other designs including the similarly constructed and massive selling 3107 chair from his Series 7 seating chair. GET THE LOOK $8,436, Design Within Reach Advertisement - Continue Reading Below 15 Egg Chair Courtesy Design Within Reach Year: 1957-58 Designer: Arne Jacobsen Provenance: Denmark Like so many innovations before it, the Egg chair was the result of a series of trials and errors. Jacobsen experimented with wire and plastic in his garage before he landed on this now-legendary piece of functional art. Originally produced in an apple red upholstery, the Egg chair was designed for the SAS Royal Hotel in Copenhagen, part of a hotel that was designed down to every detail by Jacobsen and has become an enduring symbol of Danish modern style at its finest—and the Egg chair is its crown jewel. Composed of a steel frame that is covered in an upholstered foam shell, this large and voluptuous chair is adored for its ability to embrace the sitter in a swaddling embrace. GET THE LOOK $8,436, Design Within Reach 16 Swan Chair Courtesy Design Within Reach Year: 1957-58 Designer: Arne Jacobsen Provenance: Denmark Also crafted for the Radisson SAS Royal Hotel in Copenhagen by Arne Jacobsen in 1958, this pedestal lounge chair has also endured alongside its sister chair (the Egg chair). A similarly cocooning piece made of molded synthetic material on a steel-and-aluminum base that lets its sitters swivel on a 360-degree rotation, the Swan chair was a technological innovation in furniture design. And important to note, it had no straight lines. A sitter that easily adapts to lobbies, lounges, and homes across the world, the Swan chair is still in production today at Fritz Hansen. GET THE LOOK $5,456, Design Within Reach Advertisement - Continue Reading Below 17 Papa Bear Chair San Francisco Chronicle/Hearst Newspapers via Getty Images // Getty Images Year: 1954 Designer: Hans Wegner Provenance: Denmark A new AP19 chair, creatively dubbed by a journalist the Papa Bear chair, by Hans J. Wegner retails for approximately $18,400. It’s no wonder, with its being a definite favorite among the nearly 500 chairs Wegner designed in his lifetime. With its tall back and side panels that extend for each arm (believed to have been created in England in the 1600s to help shield drafts), the wingback chair is the ultimate, gloriously comfy, lounge seat. Be right back, you can find us in this bear of a chair with a book in hand. GET THE LOOK $3,972, Design Within Reach Courtesy Design Within Reach 18 Tulip Armchair Year: 1955–56 Designer: Eero Saarinen Provenance: U.S. In 1957, Eero Saarinen unveiled the now-famous Pedestal Collection, a series of chairs that are a solution to the “ugly, confusing, unrestful world” underneath tables and chairs. Among them, the Tulip chair stood out as a piece that eschewed the jumble of traditional chair legs with one sleek, tuliplike base. It was a chic, space-age model that looked good from every angle. “A supremely elegant design, the Tulip armchair was one of the first serious attempts at creating a single-material, single-form chair,” Fiell explains. “It couldn’t be done, and thus resulted in the marriage of a cast aluminum base with a plastic seat shell. While the design wasn’t structurally integrated, it was visually unified.” Knoll has produced the Tulip chair since 1957, making it a gigantic commercial success. GET THE LOOK $3,903, Knoll Advertisement - Continue Reading Below 19 Panton Chair Courtesy Design Within Reach Year: 1959-60 Designer: Verner Panton Provenance: Denmark Technological advances in the postwar years facilitated the production of what would become one of the largest industries in the world: plastic. “Verner Panton was captivated by the potential of plastic, a novel material in the 1950s,” Liv Buur explains. “His aim was to create a comfortable chair made in one piece that could be used anywhere.” Enter the Panton chair. This stackable seat—the first all-plastic chair made in one piece with a cantilever design—took over 10 years of research and development, as well as several years of searching for a manufacturer, before making its debut via Vitra in 1967. “The groundbreaking innovation was hailed as a sensation and received numerous prizes,” Liv Buur adds. “Today the recognizable organic shape of the Panton chair is an icon.” GET THE LOOK $500, Design Within Reach Courtesy Design Within Reach 20 Platner Arm Chair Year: 1966 Designer: Warren Platner, designed for Knoll Provenance: U.S. The Platner chair's seemingly effortless curvature never fooled anyone so well. Each chair requires the welds of up to 1,000 nickel-plated steel rods. A clever reinterpretation of the period Louis XV–style chair, the Platner chair, designed by American designer Warren Platner, is an elegant silhouette of a chair that doesn’t have to shout to be noticed. Bonus: It’s now available in an 18K-gold-plated finish that takes a glamorous icon to the next level—if at all possible. GET THE LOOK $6,727, Knoll Rachel Silva Associate Digital Editor Rachel Silva is the associate digital editor at ELLE DECOR, where she covers all things design, architecture, and lifestyle. She also oversees the publication’s feature article coverage, and is, at any moment, knee-deep in an investigation on everything from the best spa gifts to the best faux florals on the internet right now. She has more than 16 years of experience in editorial, working as a photo assignment editor at Time and acting as the president of Women in Media in NYC. She went to Columbia Journalism School, and her work has been nominated for awards from ASME, the Society of Publication Designers, and World Press Photo. Read full bio Watch Next Advertisement - Continue Reading Below How Wellness Conquered the Great Outdoors Everything to Know About Building a Garden Room 5 Summer Design Trends, According to Designers Chunky Sofas and Velvet? They're Back Advertisement - Continue Reading Below From ELLE Decor for Walmart Find the Top Summer Interior Trends at Walmart These Are The New Scandi Color Trends Carrie Just Made This Wall Trend a Must-Have The Coolest Trend in Design? Going Back in Time 12 Design Trends From London's Chicest Showhouse Nothing Says Summer Like Serving In Silver When Did Beige Become Evil? Is a Candle Actually a Good Gift? Advertisement - Continue Reading Below We use technologies that provide information about your interactions with this site to others for functionality, analytics, targeted advertising, and other uses. Learn more in our Privacy Notice.
15777	https://people.math.sc.edu/filaseta/gradcourses/Math788M/ComputationalMaterialOnPolynomials.pdf	Chapter 4 Irreducible Polynomials over Finite Fields §4.1 Construction of Finite Fields As we will see, modular arithmetic aids in testing the irreducibility of poly-nomials and even in completely factoring polynomials in Z[x]. If we expect a polynomial f(x) is irreducible, for example, it is not unreasonable to try to ﬁnd a prime p such that f(x) is irreducible modulo p. If we can ﬁnd such a prime p and p does not divide the leading coeﬃcient of f(x), then f(x) is irreducible in Z[x] (see the example on page 4). It is the case that there exist polynomials which are irreducible in Z[x] but are reducible modulo every prime (see Exer-cise (4.1)), but as it turns out, one can show that such polynomials are rare and verifying that a polynomial f(x) is irreducible by trying to ﬁnd a prime p for which f(x) is irreducible modulo p will almost always work rather quickly (see Chapter 5). This is already strong motivation for looking into the idea of using modular arithmetic, but in this chapter, we plan to explore other aspects of modular arithmetic as well. We begin with a deﬁnition. Let p be a prime, and let f(x) ∈Z[x]. Suppose further that f(x) ̸≡0 (mod p). We say that u(x) ≡v(x) (mod p, f(x)), where u(x) and v(x) are in Z[x], if there exist g(x) and h(x) in Z[x] such that u(x) = v(x) + f(x)g(x) + ph(x). (In other words, u(x) ≡v(x) (mod p, f(x)) if u(x) − v(x) is in the ideal generated by p and f(x) in the ring Z[x].) One easily checks that if u(x) ≡v(x) (mod p, f(x)) and v(x) ≡w(x) (mod p, f(x)), then u(x) ≡w(x) (mod p, f(x)). Suppose that u1(x) ≡v1(x) (mod p, f(x)) and u2(x) ≡v2(x) (mod p, f(x)). Then u1(x) ± u2(x) ≡v1(x) ± v2(x) (mod p, f(x)). 61 62 Also, using u1(x)u2(x) −v1(x)v2(x) = u1(x) (u2(x) −v2(x)) + v2(x) (u1(x) −v1(x)) , we easily see that u1(x)u2(x) ≡v1(x)v2(x) (mod p, f(x)). We note that if u(x) ≡v(x) (mod p), then u(x) ≡v(x) (mod p, f(x)) (by taking g(x) ≡0), and if u(x) ≡v(x) (mod f(x)), then u(x) ≡v(x) (mod p, f(x)) (by taking h(x) ≡ 0). A further important and useful observation is that u(x) ≡0 (mod p, f(x)) if and only if f(x) is a factor of u(x) modulo p. Let f(x) be monic. If u(x) ≡v(x) (mod p, f(x)) where u(x) and v(x) are in Z[x], then there are polynomials g0(x) and h0(x) in Z[x] such that u(x) − v(x) = f(x)g0(x) + ph0(x). Recall (see Exercise (1.14) (b)) that when dividing a polynomial in Z[x] by a monic polynomial in Z[x], the quotient and remainder will be in Z[x]. It follows that there are polynomials q(x) and r(x) in Z[x] with r(x) ≡0 or deg r < deg f such that h0(x) = f(x)q(x) + r(x). Taking g(x) = g0(x) + pq(x) and h(x) = r(x), we deduce that if u(x) ≡v(x) (mod p, f(x)), then there are polynomials g(x) and h(x) in Z[x] with h(x) ≡0 or deg h < deg f such that u(x) −v(x) = f(x)g(x) + ph(x). A simple argument shows further that such a g(x) and h(x) are unique (given u(x), v(x), f(x), and p). We will also make use of the following convention. Let p be a prime, and suppose f(x) = Pn j=0 ajxj ∈Z[x] (with f(x) ̸≡0 (mod p)). Then we refer to the degree of f(x) modulo p as the largest positive integer k ≤n for which p does not divide ak. Thus, for example, 2x3 + 3x2 + 4 is a polynomial of degree 2 modulo 2. With the added condition that an = 1, we easily see that any g(x) ∈Z[x] is congruent (mod p, f(x)) to one of the pn polynomials of degree ≤n −1 with coeﬃcients from {0, 1, . . . , p −1}. Also, each of these pn poly-nomials are incongruent (mod p, f(x)). In other words, we can view these pn polynomials as representatives of the pn distinct residue classes (mod p, f(x)). Consider now the possibility that an ̸= 1, and let k denote the degree of f(x) modulo p. Exercise (4.6) implies that arithmetic (mod p, f(x)) is the same as arithmetic (mod p, f1(x)) where f1(x) ≡f(x) (mod p) and deg f1(x) = k. Exercise (4.5) further implies that arithmetic (mod p, f1(x)) is the same as arithmetic (mod p, f2(x)) where f2(x) is an appropriate monic polynomial with deg f2(x) = k. It follows that there are precisely pk distinct residue classes (mod p, f(x)) with representatives given by the polynomials of degree ≤k −1 with coeﬃcients from {0, 1, . . . , p −1}. Theorem 4.1.1. Let p be a prime. If f(x) ∈Z[x] is of degree n modulo p and f(x) is irreducible modulo p, then xpn ≡x (mod p, f(x)). We clarify that in Theorem 4.1.1, as is usual, xpn = x(pn). Before we prove this theorem, we consider an example. We show that f(x) = xp −x−1 is irreducible modulo p (and hence irreducible over Z). Consider u(x) Chapter 11 Computational Considerations §11.1 Berlekamp’s Method and Hensel Lifting We begin this chapter by discussing some important and classical methods for factoring polynomials in Z[x]. From this book’s point of view, we are mainly interested in knowing whether a given polynomial in Z[x] is irreducible over Q. If it is reducible over Q, factoring methods will allow us to ﬁnd a non-trivial factorization of it in Z[x]. A key to factoring techniques for polynomials in Z[x] is to make use of a fac-toring algorithm in Fp[x], where Fp as before is the ﬁeld of arithmetic modulo p. We describe an approach due to Berlekamp (1984). This algorithm determines the factorization of a polynomial f(x) in Fp[x] where p is a prime (or more gen-erally over ﬁnite ﬁelds). For simplicity, we suppose f(x) is monic and squarefree in Fp[x]. Let n = deg f(x). For w(x) ∈Z[x], deﬁne w(x) modd (p, f(x)) as the unique g(x) ∈Z[x] satisfying deg g ≤n, each coeﬃcient of g(x) is in the set {0, 1, . . . , p −1} and g(x) ≡w(x) (mod p, f(x)). Observe that we can view w(x) modd (p, f(x)) as also being in Fp[x]. Let A be the matrix with jth column derived from the coeﬃcients of x(j−1)p modd (p, f(x)). Speciﬁcally, write x(j−1)p modd (p, f(x)) = n X i=1 aijxi−1 for 1 ≤j ≤n. 255 256 Then we set A = (aij)n×n. Note that the ﬁrst column consists of a one followed by n−1 zeroes. In particular, ⟨1, 0, 0, . . . , 0⟩will be an eigenvector for A associ-ated with the eigenvalue 1. We are interested in determining the complete set of eigenvectors associated with the eigenvalue 1. In other words, we would like to know the null space of B = A −I where I represents the n × n identity matrix. It will be spanned by k = n −rank(B) linearly independent vectors which can be determined by performing row operations on B. Suppose ⃗ v = ⟨b1, b2, . . . , bn⟩ is one of these vectors, and set g(x) = Pn j=1 bjxj−1. Observe that g(xp) ≡g(x) (mod p, f(x)). Moreover, the g(x) with this property are precisely the g(x) with coeﬃcients obtained from the components of vectors ⃗ v in the null space of B. Our ﬁrst result in this chapter connects the factorization of f(x) in Fp[x] with the computations of greatest common divisors of f(x) and polynomials g(x) −s where s ∈{0, 1, . . . , p −1}. These greatest common divisors must be computed in Fp[x], and we recall the discussion following Deﬁnition 1.3.1. For clariﬁcation in what follows, if u(x) and v(x) are in Z[x] or Fp[x], then we use the notation gcd p u(x), v(x) to denote the greatest common divisor of u(x) and v(x) computed over the ﬁeld Fp. Theorem 11.1.1. Let f(x) be a monic polynomial in Z[x]. Suppose f(x) is squarefree in Fp[x]. Let g(x) be a polynomial with coeﬃcients obtained from a vector in the null space of B = A −I as described above. Then f(x) ≡ p−1 Y s=0 gcd p g(x) −s, f(x) (mod p). Proof. Observe that g(x)p −g(x) ≡ p−1 Y s=0 (g(x) −s) (mod p). Since g(x)p ≡g(xp) (mod p), we deduce that f(x) divides the product on the right in Fp. Since the factors g(x) −s, for s ∈{0, 1, . . . , p −1}, are pairwise relatively prime in Fp, we deduce that each monic irreducible factor of f(x) divides exactly one of the expressions gcd p g(x) −s, f(x) appearing on the right. The result follows. Observe that if g(x) is not a constant, then 1 ≤deg(g(x) −s) < deg f(x) for each s so the above claim implies we get a non-trivial factorization of f(x) in Fp[x]. On the other hand, f(x) will not necessarily be completely factored. One can completely factor f(x) by repeating the above procedure for each factor obtained from the claim; but it is simpler to use (and not diﬃcult to show) that if one takes the product of the greatest common divisors of each factor of 257 f(x) obtained above with h(x) −s (with 0 ≤s ≤p −1) where h(x) is obtained from another of the k vectors spanning the null space of B, then one will obtain a new non-trivial factor of f(x) in Fp[x]. Continuing to use all k vectors will produce a complete factorization of f(x) in Fp[x]. As an example, we factor x7 + x4 + x3 + x + 1 in F2[x]. The matrices A and B are A =                 1 0 0 0 0 1 0 0 0 0 0 1 1 1 0 1 0 0 1 0 0 0 0 0 0 0 0 1 0 0 1 0 1 0 1 0 0 0 0 1 0 1 0 0 0 1 0 1 0                 and B =                 0 0 0 0 0 1 0 0 1 0 0 1 1 1 0 1 1 0 1 0 0 0 0 0 1 0 0 1 0 0 1 0 0 0 1 0 0 0 0 1 1 1 0 0 0 1 0 1 1                 Performing elementary row operations in F2 on the matrix B, one can obtain the matrix                 0 1 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 1 0 0 1 0 0 0 0 1 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0                 . Consequently, the dimension of the null space is 2, and it is spanned by the two vectors ⟨1, 0, 0, 0, 0, 0, 0⟩and ⟨0, 0, 1, 1, 1, 0, 1⟩. The ﬁrst of these vectors corre-sponds to polynomial g(x) = 1. Since this is a constant, Theorem 11.1.1 only leads to the trivial factorization for this choice of g(x). The second eigenvector gives g(x) = x6 + x4 + x3 + x2. As g(x), in this case, is not constant, we know that Theorem 11.1.1 must lead to a non-trivial factorization of f(x). In fact, we get gcd 2(f(x), g(x)) = x3+x2+1 and gcd 2(f(x), g(x)−1) = x4+x3+x2+x+1, so that we can deduce f(x) ≡(x3 + x2 + 1)(x4 + x3 + x2 + x + 1) (mod 2). Next, we describe Hensel Lifting, which is a procedure for using the factor-ization of f(x) in Fp[x] (p a prime) to produce a factorization of f(x) modulo 258 pk for an arbitrary positive integer k. Suppose that u(x) and v(x) are relatively prime polynomials in Fp[x] for which f(x) ≡u(x)v(x) (mod p). We continue to view f(x) as being monic, for simplicity, so we take u(x) and v(x) also to be monic. Then Hensel Lifting will produce, for any positive integer k, monic polynomials uk(x) and vk(x) in Z[x] satisfying uk(x) ≡u(x) (mod p), vk(x) ≡v(x) (mod p), and f(x) ≡uk(x)vk(x) (mod pk). When k = 1, it is clear how to choose uk(x) and vk(x). For k ≥1, we determine values of uk+1(x) and vk+1(x) from the values of uk(x) and vk(x) as follows. We compute wk(x) ≡1 pk (f(x) −uk(x)vk(x)) (mod p). Observe that deg wk(x) < deg f(x) and pkwk(x) ≡f(x) −uk(x)vk(x) (mod pk+1). Since u(x) and v(x) are relatively prime in Fp[x], we can ﬁnd a(x) and b(x) in Fp[x] (depending on k) such that a(x)u(x) + b(x)v(x) ≡wk(x) (mod p). By choosing t(x) ∈Fp[x] appropriately and replacing a(x) with a(x) −t(x)v(x) and b(x) with b(x) + t(x)u(x), we see that we may suppose deg a(x) < deg v(x). The above equation then implies further that we may take deg b(x) < deg u(x). Setting uk+1(x) = uk(x) + b(x)pk and vk+1(x) = vk(x) + a(x)pk, we see that uk+1(x) and vk+1(x) are monic and uk+1(x)vk+1(x) ≡ uk(x) + b(x)pkvk(x) + a(x)pk ≡uk(x)vk(x) + pka(x)u(x) + b(x)v(x) ≡uk(x)vk(x) + pkwk(x) ≡uk(x)vk(x) + f(x) −uk(x)vk(x) ≡f(x) (mod pk+1). A complete factorization of f(x) modulo pk can be obtained from a com-plete factorization of f(x) modulo p by modifying this idea. We do not elab-orate on the best approach here, but note that such a factorization can be 259 achieved easily as follows. If f(x) is a product of r monic irreducible polyno-mials g1(x), g2(x), . . . , gr(x) modulo p, then one can factor f(x) modulo pk by taking u(x) = g1(x) and v(x) = g2(x)g3(x) · · · gr(x) above. This will produce a factor uk(x) modulo pk that is congruent to u(x) modulo p and another factor vk(x) congruent to v(x) modulo p. One can then replace the role of f(x) with vk(x) which is g2(x)g3(x) · · · gr(x) modulo p and repeat the process, factoring vk(x) modulo pk as a polynomial which is congruent to g2(x) modulo p and a polynomial congruent to g3(x)g4(x) · · · gr(x) modulo p. Continuing in this man-ner, one gets a complete factorization of f(x) into a product of monic irreducible polynomials modulo pk. By way of example, setting f(x) = x7 + x4 + x3 + x + 1, u(x) = x3 + x2 + 1 and v(x) = x4 + x3 + x2 + x + 1, we recall that, in our previous example, we showed f(x) ≡u(x)v(x) (mod 2). Then w1(x) ≡1 2 (f(x) −u(x)v(x)) ≡x6 + x5 + x4 + x3 + x2 (mod 2). Taking a(x) = 0 and b(x) = x2, we see that a(x)u(x) + b(x)v(x) ≡w1(x) (mod 2). Hence, we take u2(x) = u(x) + 2x2 = x3 + 3x2 + 1 and v2(x) = v(x) = x4 + x3 + x2 + x + 1. We deduce that f(x) ≡(x3 + 3x2 + 1)(x4 + x3 + x2 + x + 1) (mod 4). Continuing in this manner, we obtain f(x) = (x3 + 7 x2 + 1)(x4 + x3 + x2 + x + 1) (mod 8) f(x) = (x3 + 15 x2 + 1)(x4 + x3 + x2 + x + 1) (mod 16) f(x) = (x3 + 31 x2 + 1)(x4 + x3 + x2 + x + 1) (mod 32) f(x) = (x3 + 63 x2 + 1)(x4 + x3 + x2 + x + 1) (mod 64) Perhaps the above is enough for the reader to guess how f(x) factors in Z[x]. We return to this at the end of the next section. 260 §11.2 An Inequality of Landau and an Approach of Zassenhaus Landau’s inequality gives an upper bound on the “size” of the factors of a given polynomial in Z[x]. For (11.2.1) f(x) = n X j=0 ajxj = an n Y j=1 (x −αj), we recall the notations ∥f∥= n X j=0 a2 j 1/2 and M(f) = \|an\| n Y j=1 max{1, \|αj\|}, the latter being the Mahler measure of the polynomial f(x). We make use of following two easily established properties of Mahler measure: (i) If g(x) and h(x) are in C[x], then M(gh) = M(g)M(h). (ii) If g(x) is in Z[x], then M(g) ≥1. For a ﬁxed f(x) ∈Z[x], we want an upper bound on ∥g∥where g(x) is a factor of f(x) in Z[x]. Theorem 11.2.1. If f(x), g(x), and h(x) in Z[x] are such that f(x) = g(x)h(x), then ∥g∥≤2deg g∥f∥. Proof. We begin by proving that for f(x) ∈R[x], (11.2.2) M(f) ≤∥f∥≤2deg fM(f). For w(x) ∈C[x], we use the reciprocal polynomial e w(x) = xdeg ww(1/x). The coeﬃcient of xdeg w in the expanded product w(x) e w(x) is ∥w∥2. For f(x) as in (11.2.1), we consider w(x) = an Y 1≤j≤n \|αj\|>1 (x −αj) Y 1≤j≤n \|αj\|≤1 (αjx −1). Observe that e w(x) = an Y 1≤j≤n \|αj\|>1 (1 −αjx) Y 1≤j≤n \|αj\|≤1 (αj −x). Hence, w(x) e w(x) = a2 n n Y j=1 (x −αj) n Y j=1 (1 −αjx) = f(x) e f(x). 261 By comparing coeﬃcients of xn, we deduce that ∥w∥= ∥f∥. Also, the deﬁnition of w(x) implies \|w(0)\| = M(f). Thus, writing w(x) = Pn j=0 cjxj, we obtain M(f) = \|c0\| ≤(c2 0 + c2 1 + · · · + c2 n)1/2 = ∥w∥= ∥f∥, establishing the ﬁrst inequality in (11.2.2). For the second inequality, observe that for any k ∈{1, 2, . . . , n}, the product of any k of the αj has absolute value ≤M(f)/\|an\|. It follows that \|an−k\|/\|an\|, which is the sum of the products of the roots taken k at a time, is ≤ n k × M(f)/\|an\|. Hence, \|an−k\| ≤ n k M(f) = n n −k M(f). The second inequality in (11.2.2) now follows from ∥f∥= n X j=0 a2 j 1/2 ≤ n X j=0 \|aj\| ≤ n X j=0 n j M(f) = 2nM(f). Now, we make use of (11.2.2) and properties (i) and (ii) of Mahler measure above to deduce ∥g∥≤2deg gM(g) ≤2deg gM(g)M(h) = 2deg gM(gh) = 2deg gM(f) ≤2deg g∥f∥. This establishes the theorem. We explain a method for factoring a given f(x) ∈Z[x] with the added assumptions that f(x) is monic and squarefree. This approach has its origins in a paper by Zassenhaus (1969). The latter we can test by computing gcd(f, f ′), which will give us a nontrivial factor of f(x) if f(x) is not squarefree. If f(x) is not monic, then one needs to add a little more to the ideas below (but not much). Let B = 2⌊(deg f)/2⌋∥f∥. Then if f(x) has a nontrivial factor g(x) in Z[x], it has such a factor of degree ≤⌊(deg f)/2⌋so that by Theorem 11.2.1, we can use B as a bound on ∥g∥. Next, we ﬁnd a prime p for which f is squarefree modulo p. There are a variety of ways this can be done. There are only a ﬁnite number of primes for which f is not squarefree modulo p. These primes divide the resultant R(f, f ′). So one can compute R(f, f ′) and avoid primes which divide R(f, f ′). Alternatively, one can compute gcd p(f(x), f ′(x)) modulo p or simply using Berlekamp’s factoring algorithm until a squarefree factorization occurs. We choose a positive integer r as small as possible such that pr > 2B. Then we factor f(x) modulo p by Berlekamp’s algorithm and use Hensel lifting to 262 obtain the factorization of f(x) modulo pr. Given our conditions on f(x), we can suppose all irreducible factors are monic and do so. Next, we can determine if f(x) = g(x)h(x) for some monic g(x) and h(x) in Z[x] with ∥g∥≤B as follows. We observe that the coeﬃcients of g(x) are in [−B, B]. We use a residue system modulo pr that includes this interval, namely (−pr/2, pr/2], and consider each factorization of f(x) modulo pr with coeﬃcients in this residue system as a product of two monic polynomials u(x) and v(x). Since f(x) = g(x)h(x), there must be some factorization where g(x) ≡u(x) (mod pr) and h(x) ≡v (mod pr). On the other hand, the coeﬃcients of g(x) and u(x) are all in (−pr/2, pr/2] so that the coeﬃcients of g(x) −u(x) are each divisible by pr and are each < pr in absolute value. This implies g(x) = u(x). Thus, we can determine if a factor g(x) exists as above by simply checking each monic factor of f(x) modulo pr with coeﬃcients in (−pr/2, pr/2]. Recall we factored f(x) = x7 +x4 +x3 +x+1 modulo various powers of 2 in the previous section. Using the above approach, we can deduce a factorization of f(x) in Z[x]. In the notation above B = 23∥f∥= 8 √ 5 < 20. Since 26 = 64 > 2B, we can use the factorization of f(x) that we obtained modulo 64, namely f(x) = (x3 + 63 x2 + 1)(x4 + x3 + x2 + x + 1) (mod 64). We see that if f(x) is reducible, then it must have two factors, one congruent to x3 + 63 x2 + 1 modulo 64 and one congruent to x4 + x3 + x2 + x + 1 modulo 64. The ﬁrst of these is of particular interest to us as its degree is < (deg f)/2. If g(x) ∈Z[x] divides f(x) and g(x) ≡x3 + 63 x2 + 1 (mod 64), then the arguments above imply that g(x) must equal the polynomial obtained by taking the coeﬃcients on the right to be in the interval [−32, 32]. In other words, from x3 + 63 x2 + 1 ≡x3 −x2 + 1 (mod 64), we deduce g(x) = x3 −x2 + 1. To clarify, this means that if f(x) is reducible over Z, then g(x) will be a factor of f(x). To establish that f(x) in fact has g(x) as a factor, we are left with checking if g(x) divides f(x). In fact, we have the (perhaps not unexpected) factorization f(x) = (x3 −x2 + 1)(x4 + x3 + x2 + x + 1). 263 §11.3 Swinnerton-Dyer’s Example The algorithm just described above for factoring a polynomial f(x) ∈Z[x] of degree n can take time that is exponential in n for some, albeit rare, f(x). This has been illustrated by a nice example due to Swinnerton-Dyer. We formulate his example as follows. Let a1, a2, . . . , am be arbitrary squarefree pairwise rela-tively prime integers > 1. Let Sm be the set of 2m diﬀerent m-tuples (ε1, . . . , εm) where each εj ∈{1, −1}. We justify that the polynomial (11.3.1) f(x) = Y (ε1,...,εm)∈Sm x −(ε1 √a1 + · · · + εm √am) has the properties: (i) The polynomial f(x) is in Z[x]. (ii) It is irreducible over the rationals. (iii) It factors as a product of linear and quadratic polynomials modulo every prime p. Some of the arguments below (in particular, the argument for (ii)) makes some use of Galois theory. We elaborate on the details but note in advance that some background is needed in this direction for the presentation given here. One can deduce (i) by observing that the coeﬃcients of f(x) are symmetric polynomials with integer coeﬃcients in the roots of (x2 −a1)(x2 −a2) · · · (x2 −am). Alternatively, an easy induction argument on m can be done as follows. For any squarefree positive integer a1, we have x −√a1 x + √a1 = x2 −a1. Suppose f(x), as above, is in Z[x] whenever m = t where t is a positive integer. Let a1, a2, . . . , at+1 be arbitrary squarefree pairwise relatively prime integers > 1. The induction hypothesis implies ft(x) = Y (ε1,...,εt)∈St x −(ε1 √a1 + · · · + εt √at) ∈Z[x]. By elementary symmetric functions associated with the two roots of the quadratic x2 −at+1, we see that ft+1(x) = Y (ε1,...,εt+1)∈St+1 x −(ε1 √a1 + · · · + εt+1 √at+1) = ft x + √at+1 ft x −√at+1 ∈Z[x]. 264 The use of elementary symmetric functions can be avoided by observing that ft x + √at+1 = u(x) + v(x)√at+1 for some u(x) and v(x) in Z[x] and, conse-quently, ft x−√at+1 = u(x)−v(x)√at+1. Hence, the product of ft x+√at+1 and ft x −√at+1 is u(x)2 −at+1v(x)2 ∈Z[x]. Thus, (i) holds. To establish (ii), it suﬃces to show that the minimal polynomial for αm = √a1 + √a2 + · · · + √am, that is the monic irreducible polynomial in Q[x] that has αm as a root, has all 2m numbers of the form ε1 √a1 + · · · + εm √am, where (ε1, . . . , εm) ∈Sm, as roots and that these 2m numbers are distinct. We begin, however, ﬁrst by looking at the algebraic number ﬁeld Km = Q √a1, √a2, . . . , √am . Observe that Km is the splitting ﬁeld for the polynomial (x2 −a1)(x2 −a2) · · · (x2 −am), and therefore forms a Galois extension over Q. We prove (∗) The number ﬁeld Km has degree 2m over Q, and the 2m elements of the Galois group Gal(Km/Q) of Km over Q are given by the mappings σ √aj = εj√aj, for 1 ≤j ≤m, where (ε1, . . . , εm) varies over the 2m elements of Sm. We proceed by induction on m to establish that (∗) holds for each positive integer m and for all choices of a1, a2, . . . , am squarefree pairwise relatively prime integers > 1. We begin with m = 0 and m = 1. For m = 0, we have K0 = Q which has degree 1 over Q and implies Gal(K0/Q) only consists of the identity element. Thus, (∗) holds for m = 0. Suppose m = 1. Given a1 is squarefree and > 1, the quadratic x2 −a1 is irreducible over Q. Therefore, the number ﬁeld K1 = Q √a1 has degree 2 over Q. We deduce Gal(K1/Q) has exactly one non-identity element σ. As a root of x2 −a1 must be mapped to a root of x2 −a1 by σ, we see that σ √a1 = ±√a1. Since σ is not the identity mapping, we deduce σ √a1 = −√a1. This establishes what we want to start our induction, namely (∗) for m = 1. Suppose now that (∗) holds for m ≤t, where t is a positive integer, and let a1, a2, . . . , at+1 be arbitrary squarefree pairwise relatively prime integers > 1. With the already established notation above, we have that the degree of Kt over Q is 2t and that there is a σ ∈Gal(Kt/Q) that satisﬁes σ √a1 = −√a1 265 and σ √aj = √aj for 2 ≤j ≤t. We argue next that √at+1 ̸∈Kt. Assume otherwise. Let S = t Y j=2 √aj ηj : ηj ∈{0, 1} for each j . Thus, S consists of 2t−1 elements. Every element of Kt can be expressed uniquely as a linear combination of the elements of S and the elements of S times √a1 with coeﬃcients from Q. In other words, each element of Kt can be written as X s∈S b(s)s + X s∈S c(s)s√a1, for exactly one choice of b(s) and c(s) in Q. Fix such b(s) and c(s) so that the above sum represents √at+1. We make use of the automorphism σ deﬁned above. Observe that σ X s∈S b(s)s + X s∈S c(s)s√a1 = X s∈S b(s)s − X s∈S c(s)s√a1. Since σ √at+1 2 = σ(at+1) = at+1, we deduce that σ √at+1 is either √at+1 or −√at+1. In the former case, we use that 2√at+1 = √at+1 + σ √at+1 = 2 X s∈S b(s)s ∈Q √a2, . . . , √at . In the latter case, we use that 2√at+1 = √at+1 −σ √at+1 = 2√a1 X s∈S c(s)s which implies √a1at+1 ∈Q √a2, . . . , √at . On the other hand, by the induction hypothesis, Q √a2, . . . , √at has degree 2t−1 over Q and both the ﬁelds Q √a2, . . . , √at, √at+1 and Q √a2, . . . , √at, √a1at+1 have degree 2t over Q. This is a contradiction. Thus, √at+1 ̸∈Kt. In other words, x2 −at+1 is irreducible over Kt. Note that Kt+1 is formed by adjoining a root of x2 −at+1 to Kt. Since the degree of Kt over Q is 2t, we deduce that the degree of Kt+1 over Q is 2t+1. Observe that if σ is an automorphism of Kt+1 that ﬁxes Q, then its action on the elments of Kt+1 are determined by 266 the values of √aj for j ∈{1, 2, . . . , t + 1}. Also, for each j ∈{1, 2, . . . , t}, the element √aj in Kt+1 must be mapped by σ to one of √aj or −√aj. Since the degree of Kt+1 over Q is 2t+1, there are 2t+1 elements of the Galois group of Kt+1 over Q, and we see that they are precisely the 2t+1 mappings given in (∗) with m = t + 1. Thus, (∗) holds for m = t + 1, and the induction argument in complete. We return to establishing (ii). We show next that (11.3.2) ε1 √a1 + · · · + εm √am ̸= ε′ 1 √a1 + · · · + ε′ m √am, where (ε1, . . . , εm) and (ε′ 1, . . . , ε′ m) are distinct elements of Sm. Taking j max-imal such that εj ̸= ε′ j, we see that if (11.3.2) does not hold, then √aj ∈Q √a1, . . . , √aj−1 . The argument above showing that √at+1 ̸∈Kt implies a contradiction. Thus, the 2m numbers ε1 √a1 + · · · + εm √am, with (ε1, . . . , εm) ∈Sm are distinct. Observe that αm ∈Km. If w(x) is the minimal polynomial for αm, then each σ ∈Gal(Km/Q) must map αm to a root of w(x). Fix (ε1, . . . , εm) ∈Sm. From (∗), we know that there is a σ ∈ Gal(Km/Q) satisfying σ(αm) = ε1 √a1 + · · · + εm √am. Therefore, each root of f(x), given by (11.3.1), is a root of w(x). We deduce f(x) = w(x) and, hence, is irreducible. This completes the proof of (ii). To see (iii), we again use induction on m. We give two such induction argu-ments, the ﬁrst more direct for someone accustomed to working in extensions with factorizations modulo a prime and the second more self-contained given the contents of this book. The case that m = 1 is clear, since in this case f(x), as deﬁned in (11.3.1), is a quadratic. Suppose (iii) holds for m = t where t is a positive integer. Fix a prime p. We make use of the notation in (11.3.1) with m = t + 1 and write (11.3.3) f(x) = gj x + √aj gj x −√aj , where j ∈{1, 2, . . . , t + 1} and gj(x) is deﬁned by gj(x) = Y (ε1,...,εt)∈St x −(ε1 √a1 + · · · + εj−1√aj−1 + εj√aj+1 + · · · + εt √at+1) . The induction hypothesis implies that each gj(x) factors modulo p as a product of linear and quadratic polynomials. If some aj is a square modulo p, then there is some integer b such that b2 ≡aj (mod p) and, hence, f(x) ≡gj(x+b)gj(x−b) (mod p). Since each of gj(x+b) and gj(x−b) factors as a product of linear and quadratic polynomials modulo p, we are through in this case. Now, suppose 267 no aj is a square modulo p. Fix (ε1, . . . , εm) ∈Sm and observe that when the product x + (ε1 √a1 + · · · + εm √am) x −(ε1 √a1 + · · · + εm √am) is expanded the result is an expression with each radicand the product of two of the aj. Since no aj is a square modulo p, each such product of two of the aj will be (since the product of two non-quadratic residues is a quadratic residue). This means that the above product can be expressed as a quadratic polynomial modulo p with coeﬃcients from {0, 1, . . . , p −1}. Pairing then the linear factors of f(x) appropriately leads to the desired factorization modulo p. For the second argument, we make use of Exercise (4.2)(b) (also, see Exercise (8.4)). We prove by induction that f(x), as deﬁned in (11.3.1), is a product modulo each prime p of linear and quadratic monic polynomials, with the latter of the form x2 + 2bx + c for some integers b and c (so that modulo 2 the middle coeﬃcient is necessarily 0). We start the induction the same, noting the case m = 1 holds and supposing we know that such a factorization holds for m = t where t is a positive integer and for each prime p. We now ﬁx a prime p and make use of the notation in (11.3.1) and in (11.3.3) with m = j = t+1. The induction hypothesis implies there are monic polynomials u1(x), u2(x), . . . , ur(x) in Z[x], each of degree 1 or 2, and a polynomial v(x) ∈Z[x] such that gt+1(x) = u1(x)u2(x) · · · ur(x) + pv(x). Hence, f(x) = gt+1 x + √at+1 gt+1 x −√at+1 = u1 x + √at+1 u1 x −√at+1 · · · ur x + √at+1 ur x −√at+1 + pw(x), where w(x) is a polynomial with each coeﬃcient a symmetric polynomial in √at+1 and −√at+1 with coeﬃcients in Z. We deduce then that in fact w(x) ∈ Z[x]. By the induction hypothesis, we further may take each uj(x) of one of the forms uj(x) = x + b and uj(x) = x2 + 2bx + c where b and c are integers. In the ﬁrst case, we observe that uj x + √at+1 uj x −√at+1 = (x + b)2 −at+1. Since this is a monic quadratic with even middle term, this is a factor of f(x) of a form we want. In the case that uj(x) = x2 + 2bx + c, we set d = c −b2 and write uj(x) = (x + b)2 + d. We deduce then that uj x + √at+1 uj x −√at+1 = (x + b + √at+1)2 + d (x + b −√at+1)2 + d = (x + b)2 −at+1 2 + 2d (x + b)2 + at+1 + d2. Observe that the above is of the form h (x+b)2 where h(x) is a monic quadratic with an even coeﬃcient for x and a constant term equal to a2 t+1 + 2dat+1 + d2 = (at+1 + d)2. 268 Modulo 2, we see that h(x2) and, hence, h (x + b)2 factors as a product of two monic quadratics with the coeﬃcient of x equal to 0 for each quadratic. For odd primes p, Exercise (4.2)(b) implies h(x2) and, hence, h (x + b)2 is reducible modulo p. Note that k is a root of h(x2) modulo p if and only if −k is a root of h(x2) modulo p. Furthermore, 0 is a root of h(x2) if and only if x2 is a factor of h(x2). Thus, h (x + b)2 factors as a product of linear and quadratic monic polynomials modulo p. Since any integer is congruent to twice an integer modulo an odd prime p, we can take the coeﬃcient of x appearing in any quadratic to be even. The induction argument is therefore complete. §11.4 The Lattice Base Reduction Algorithm Lenstra, Lenstra, and Lovasz (1982) showed that it is possible to factor a poly-nomial f(x) = Pn j=0 ajxj ∈Z[x] in polynomial time. If n is the degree of f(x) (so an ̸= 0) and H is the height of f(x), that is the maximum of \|aj\| for 0 ≤j ≤n, then the quantity n(log2 H + log2 n + 2) can be viewed as an upper bound on the length of the input polynomial f(x). A polynomial time algorithm for factoring f(x) corresponds to an algorithm that runs in time that is polynomial in n and log H. The previous factoring algorithm we described is not polynomial as was seen from the example of Swinnerton-Dyer. The main problem there (which is notably atypical) is that the polynomial f(x) can factor into many small irreducible factors modulo every prime p causing us to have to consider exponentially many possibilities for the mod p reduction of any non-trivial factor of f(x). The algorithm of Lenstra, Lenstra and Lov´ asz, called the lattice base reduction algorithm or the LLL-algorithm, is an approach for get-ting around having to consider all such mod p reductions and thereby provides a polynomial time algorithm for factoring f(x) over the rationals. To describe the lattice base reduction algorithm, we turn now to some back-ground on lattices. Let Qn denote the set of vectors ⟨a1, a2, . . . , an⟩with aj ∈Q. For ⃗ b = ⟨a1, a2, . . . , an⟩∈Q and ⃗ b′ = ⟨a′ 1, a′ 2, . . . , a′ n⟩∈Q, we deﬁne the usual dot product ⃗ b ·⃗ b′ by ⃗ b ·⃗ b′ = a1a′ 1 + a2a′ 2 + · · · + ana′ n. Also, we set ∥⃗ b∥= q a2 1 + a2 2 + · · · + a2 n. Further, we use AT to denote the transpose of a matrix A, so the rows and columns of A are the same as the columns and rows of AT , respectively. Let ⃗ b1, . . . ,⃗ bn ∈Qn, and let A = ⃗ b1, . . . ,⃗ bn be the n × n matrix with column vectors ⃗ b1, . . . ,⃗ bn. The lattice L generated by ⃗ b1, . . . ,⃗ bn is L = L(A) = ⃗ b1Z + · · · +⃗ bnZ. 269 We will be interested mainly in the case that ⃗ b1, . . . ,⃗ bn are linearly independent; in this case, ⃗ b1, . . . ,⃗ bn is called a basis for L. Observe that given L, the value of \| det A\| is the same regardless of the basis ⃗ b1, . . . ,⃗ bn that is used to describe L. To see this, observe that if ⃗ b′ 1, . . . ,⃗ b′ n is another basis for L, there are matrices A and B with integer entries such that ⃗ b1, . . . ,⃗ bn AB = ⃗ b′ 1, . . . ,⃗ b′ n B = ⃗ b1, . . . ,⃗ bn . Given that ⃗ b1, . . . ,⃗ bn is a basis for Rn, it follows that AB is the identity matrix and det B = ±1. The second equation above then implies \| det ⃗ b′ 1, . . . ,⃗ b′ n \| = \| det ⃗ b1, . . . ,⃗ bn \|. We set det L to be this common value. Next, we describe the Gram-Schmidt orthogonalization process. Deﬁne re-cursively ⃗ b∗ i = ⃗ bi − i−1 X j=1 µij⃗ b∗ j, for 1 ≤i ≤n, where µij = µi,j = ⃗ bi ·⃗ b∗ j ⃗ b∗ j ·⃗ b∗ j , for 1 ≤j < i ≤n. Then for each i ∈{1, . . . , n}, the vectors ⃗ b∗ 1, . . . ,⃗ b∗ i span the same subspace of Rn as ⃗ b1, . . . ,⃗ bi. In other words, a1⃗ b∗ 1 + · · · + ai⃗ b∗ i : aj ∈R for 1 ≤j ≤i = a1⃗ b1 + · · · + ai⃗ bi : aj ∈R for 1 ≤j ≤i . Furthermore, the vectors ⃗ b∗ 1, . . . ,⃗ b∗ n are linearly independent (hence, non-zero) and pairwise orthogonal (i.e., for distinct i and j, we have ⃗ b∗ i ·⃗ b∗ j = 0). We leave veriﬁcation of these facts as exercises. We turn next to Hadamard’s inequality. The value of det L can be viewed as the volume of the polyhedron with edges parallel to and the same length as ⃗ b1, . . . ,⃗ bn. As indicated by the above remarks, this volume is independent of the basis. Geometrically, it is apparent that det L ≤∥⃗ b1∥∥⃗ b2∥· · · ∥⃗ bn∥ (where “apparent” is limited somewhat to the dimensions we can think in). This is Hadamard’s inequality. One can also use the vectors ⃗ b∗ j to provide a proof in any dimensions as follows. One checks that det ⃗ b1, . . . ,⃗ bn = det ⃗ b∗ 1,⃗ b∗ 2 + µ21⃗ b∗ 1, . . . ,⃗ b∗ n + n−1 X j=1 µnj⃗ b∗ j = det ⃗ b∗ 1, . . . ,⃗ b∗ n . 270 Since ⃗ b1, . . . ,⃗ bn is a basis for L, we deduce that (det L)2 = det (⃗ b∗ 1, . . . ,⃗ b∗ n)T (⃗ b∗ 1, . . . ,⃗ b∗ n) = det    ∥⃗ b∗ 1∥2 0 . . . 0 . . . ... . . . 0 . . . 0 ∥⃗ b∗ n∥2    = n Y i=1 ∥⃗ b∗ i ∥ 2 . Thus, det L = Qn i=1 ∥⃗ b∗ i ∥. So it suﬃces to show ∥⃗ b∗ i ∥≤∥⃗ bi∥. The orthogonality of the ⃗ b∗ i ’s implies ∥⃗ bi∥2 = ⃗ b∗ i + i−1 X j=1 µij⃗ b∗ j 2 = ∥⃗ b∗ i ∥2 + i−1 X j=1 µ2 ij∥⃗ b∗ j∥2. The sum on the right above is clearly positive so that ∥⃗ b∗ i ∥≤∥⃗ bi∥follows. The Hadamard inequality provides an upper bound on the value of det L. Hermite proved that there is a constant cn (depending only on n) such that for some basis ⃗ b1, . . . ,⃗ bn of L, we have ∥⃗ b1∥∥⃗ b2∥· · · ∥⃗ bn∥≤cn det L. It is known that cn ≤nn. To clarify a point, Minkowski has shown that there exist n linearly independent vectors ⃗ b′ 1, . . . ,⃗ b′ n in L such that ∥⃗ b′ 1∥∥⃗ b′ 2∥· · · ∥⃗ b′ n∥≤nn/2 det L, but ⃗ b′ 1, . . . ,⃗ b′ n is not necessarily a basis for L. Further, we note that the problem of ﬁnding a basis ⃗ b1, . . . ,⃗ bn of L for which ∥⃗ b1∥· · · ∥⃗ bn∥is minimal is known to be NP-hard. The problem of ﬁnding a vector ⃗ b ∈L with ∥⃗ b∥minimal is not known to be NP-complete, but it may well be. In any case, no one knows a polynomial time algorithm for this problem. We note that Lagarias (1985) has, however, proved that the problem of ﬁnding a vector⃗ b ∈L which minimizes the maximal absolute value of a component is NP-hard. Observe that Hermite’s result mentioned above implies that there is a constant c′ n, depending only on n, such that ∥⃗ b∥≤ c′ n n √ det L. It is possible for a lattice L to contain a vector that is much shorter than this, but it is known that the best constant c′ n for all lattices L satisﬁes p n/(2eπ) ≤c′ n ≤ p n/(eπ). The vectors ⃗ b∗ j obtained from the Gram-Schmidt orthogonalization process can be used to obtain a lower bound for the shortest vector in a lattice L. More 271 precisely, we have (11.4.1) ⃗ b ∈L, ⃗ b ̸= 0 = ⇒ ∥⃗ b∥≥min{∥⃗ b∗ 1∥, ∥⃗ b∗ 2∥, . . . , ∥⃗ b∗ n∥}. To see this, express ⃗ b in the form ⃗ b = u1⃗ b1 + · · · + uk⃗ bk, where each uj ∈Z and uk ̸= 0. Observe that the deﬁnition of the ⃗ b∗ j imply then that ⃗ b = v1⃗ b∗ 1 + · · · + vk⃗ b∗ k, for some vj ∈Q with vk = uk. In particular, vk is a non-zero integer. We deduce that ∥⃗ b∥2 = v1⃗ b∗ 1 + · · · + vk⃗ b∗ k · v1⃗ b∗ 1 + · · · + vk⃗ b∗ k = v2 1∥⃗ b∗ 1∥2 + · · · + v2 k∥⃗ b∗ k∥2 ≥∥⃗ b∗ k∥2, from which (11.4.1) follows. §11.5 Reduced Bases and Factoring with LLL We will want the following important deﬁntion. Deﬁnition 11.5.1. Let ⃗ b1, . . . ,⃗ bn be a basis for a lattice L and ⃗ b∗ 1, . . . ,⃗ b∗ n the corresponding basis for Rn obtained from the Gram-Schmidt orthogonalization process, with µij as deﬁned before. Then ⃗ b1, . . . ,⃗ bn is said to be reduced if both of the following hold (i) ∥µij∥≤1 2 for 1 ≤j < i ≤n (ii) ∥⃗ b∗ i + µi,i−1⃗ b∗ i−1∥2 ≥3 4 ∥⃗ b∗ i−1∥2 for 1 < i ≤n. The main work of Lenstra, Lenstra, and Lovasz (1982) establishes an algorithm that runs in polynomial time which constructs a reduced basis of L from an arbitrary basis ⃗ b1, . . . ,⃗ bn of L. Our main goal below is to explain how such a reduced basis can be used to factor a polynomial f(x) in polynomial time. We will need to describe the related lattice and an initial basis for it. We begin, however, with some properties of reduced bases. Let⃗ b1, . . . ,⃗ bn be a reduced basis for a lattice L and⃗ b∗ 1, . . . ,⃗ b∗ n the correspond-ing basis for Rn obtained from the Gram-Schmidt orthogonalization process with µij as before. The argument for (11.4.1) can be modiﬁed to show that (11.5.1) ⃗ b ∈L, ⃗ b ̸= 0 = ⇒ ∥⃗ b1∥≤2(n−1)/2∥⃗ b∥. 272 In particular, the above inequality holds for the shortest vector ⃗ b ∈L. To prove (11.5.1), observe that (i) and (ii) imply ∥⃗ b∗ i ∥2 + 1 4∥⃗ b∗ i−1∥2 ≥∥⃗ b∗ i ∥2 + µ2 i,i−1∥⃗ b∗ i−1∥2 = ∥⃗ b∗ i + µi,i−1⃗ b∗ i−1∥2 ≥3 4 ∥⃗ b∗ i−1∥2. Hence, ∥⃗ b∗ i ∥2 ≥(1/2)∥⃗ b∗ i−1∥2. We deduce that (11.5.2) ∥⃗ b∗ i ∥2 ≥ 1 2i−j ∥⃗ b∗ j∥2 for 1 ≤j < i ≤n. Deﬁning k as in the proof of (11.4.1) and following the argument there, we obtain ∥⃗ b∥2 ≥∥⃗ b∗ k∥2. Hence, ∥⃗ b∥2 ≥∥⃗ b∗ k∥2 ≥ 1 2k−1 ∥⃗ b∗ 1∥2 ≥ 1 2n−1 ∥⃗ b∗ 1∥2 = 1 2n−1 ∥⃗ b1∥2, where the last equation makes use of ⃗ b∗ 1 = ⃗ b1. Thus, (11.5.1) follows. Recall that ∥⃗ bi∥2 = ∥⃗ b∗ i ∥2 + i−1 X j=1 µ2 ij∥⃗ b∗ j∥2. From (i) and (11.5.2), we obtain ∥⃗ bi∥2 ≤∥⃗ b∗ i ∥2 + 1 4 i−1 X j=1 ∥⃗ b∗ j∥2 ≤∥⃗ b∗ i ∥2 + 1 4 i−1 X j=1 2i−j∥⃗ b∗ i ∥2 = 1 + 1 4(2i −2) ∥⃗ b∗ i ∥2 ≤2i−1∥⃗ b∗ i ∥2. Using (11.5.2) again, we deduce (11.5.3) ∥⃗ bj∥2 ≤2j−1∥⃗ b∗ j∥2 ≤2i−1∥⃗ b∗ i ∥2 for 1 ≤j ≤i ≤n. We show now the following improvement of (11.5.1). Let ⃗ x1, ⃗ x2, . . . , ⃗ xt be t linearly independent vectors in L. Then (11.5.4) ∥⃗ bj∥≤2(n−1)/2 max{∥⃗ x1∥2, ∥⃗ x2∥2, . . . , ∥⃗ xt∥2} for 1 ≤j ≤t. For each 1 ≤j ≤t, deﬁne a positive integer m(j) and integers uji by ⃗ xj = m(j) X i=1 uji⃗ bi, ujm(j) ̸= 0. By reordering the ⃗ xj, we may suppose further that m(1) ≤m(2) ≤· · · ≤m(t). The linear independence of the ⃗ xj implies that m(j) ≥j for 1 ≤j ≤t. The proof of (11.4.1) implies here that ∥⃗ xj∥≥∥⃗ b∗ m(j)∥ for 1 ≤j ≤t. 273 From (11.5.3), we deduce ∥⃗ bj∥2 ≤2m(j)−1∥⃗ b∗ m(j)∥2 ≤2m(j)−1∥⃗ xj∥2 ≤2n−1∥⃗ xj∥2 for 1 ≤j ≤t. The inequality in (11.5.4) now follows. Recall that det L = Qn i=1 ∥⃗ b∗ i ∥. We obtain from (11.5.3) that n Y i=1 ∥⃗ bi∥2 ≤ n Y i=1 2i−1∥⃗ b∗ i ∥2 ≤2n(n−1)/2 n Y i=1 ∥⃗ b∗ i ∥2 = 2n(n−1)/2(det L)2. Thus, from Hadamard’s inequality, we obtain 2−n(n−1)/4∥⃗ b1∥∥⃗ b2∥· · · ∥⃗ bn∥≤det L ≤∥⃗ b′ 1∥∥⃗ b′ 2∥· · · ∥⃗ b′ n∥ for any basis ⃗ b′ 1, . . . ,⃗ b′ n of L. Recall that ﬁnding a basis ⃗ b′ 1, . . . ,⃗ b′ n for which the product on the right is minimal is NP-hard. The above implies that a reduced basis is close to being such a basis. We also note that Hermite’s inequality mentioned earlier is a consequence of the above inequality. Suppose now that we want to factor a non-zero polynomial f(x) ∈Z[x]. Let p be a prime, and consider a monic irreducible factor h(x) of f(x) modulo pk (obtained say through Berlekamp’s algorithm and Hensel lifting). Now, let h0(x) denote an irreducible factor of f(x) in Z[x] such that h0(x) is divisible by h(x) modulo pk. Note that h0(x) being irreducible in Z[x] implies that the content of h0(x) (the greatest common divisor of its coeﬃcients) is 1. Our goal here is to show how one can determine h0(x) without worrying about other factors of f(x) modulo pk (to avoid the diﬃculty suggested by Swinnerton-Dyer’s example). We describe a lattice for this approach. Let ℓ= deg h. We need only consider the case that ℓ< n. Fix an integer m ∈{ℓ, ℓ+1, . . . , n−1}. We will successively consider such m beginning with ℓand working our way up until we ﬁnd h0(x). In the end, m will correspond to the degree of h0(x); and if no such h0(x) is found for ℓ≤m ≤n −1, then we can deduce that f(x) is irreducible. We associate with each polynomial w(x) = amxm + · · · + a1x + a0 ∈Z[x], a vector ⃗ b = ⟨a0, a1, . . . , am⟩∈Zm+1. Observe that ∥⃗ b∥= ∥w(x)∥. Let L be the lattice in Zm+1 spanned by the vectors associated with wj(x) = ( pkxj−1 for 1 ≤j ≤ℓ h(x)xj−ℓ−1 for ℓ+ 1 ≤j ≤m + 1. It is not diﬃcult to see that these vectors form a basis. Furthermore, the poly-nomials associated with the vectors in L correspond precisely to the polynomials in Z[x] of degree ≤m that are divisible by h(x) modulo pk. In particular, if m ≥deg h0, the vector, say ⃗ b0, associated with h0(x) is in L. Observe that if k is large enough and deg h0 > ℓ, the coeﬃcients of h(x) are “presumably” 274 large and the value of ∥⃗ b0∥is “seemingly” small. We will show that in fact if k is large enough and m = deg h0 and ⃗ b1, . . . ,⃗ bm+1 is a reduced basis for L, then ⃗ b# 1 = ⃗ b0, where ⃗ b# 1 corresponds to the vector obtained by dividing the components of ⃗ b1 by the greatest common divisor of these components (i.e., the polynomial associated with ⃗ b# 1 is the polynomial associated with ⃗ b1 with its content removed). The lattice L seemingly has little to do with f(x) as its deﬁnition only depends on h(x). Fix h0(x) as above. We show that if k is large enough, then h0(x) is the only irreducible polynomial in Z[x] which is associated with a short vector in L. For this purpose, suppose g0(x) is an irreducible polynomial in Z[x] divisible by h(x) but diﬀerent from h0(x) and that R is the resultant of h0(x) and g0(x). Note that since h0(x) and g0(x) are irreducible in Z[x], we have R ̸= 0. The deﬁnition of the resultant implies that if R is large, then ∥g0(x)∥ must be large (since we are viewing h0(x) as ﬁxed). So suppose R is not large. There are polynomials u(x) and v(x) in Z[x] such that h0(x)u(x) + g0(x)v(x) = R. We wish to take advantage now of the fact that the left-hand side above is divisible by h(x) modulo pk, but at the same time we want to keep in mind that unique factorization does not exist modulo pk. Since h(x) is monic of degree ℓ≥1, the left-hand side is of the form h(x)w(x) modulo pk where we can now easily deduce that every coeﬃcient of w(x) is divisible by pk. This implies pk\|R. Hence, given k is large, we deduce R is large, giving us the desired conclusion that ∥g0(x)∥is large. The above argument does more. If m = deg h0(x) and ⃗ b ∈L, then viewing g0(x) ∈L as the polynomial associated with ⃗ b, we deduce from the above that either ∥g0(x)∥is large or R = 0. In the latter case, since h0(x) is irreducible and deg g0 ≤m = deg h0, we obtain that ⃗ b# = ⃗ b0. How large is large? We take ⃗ b = ⃗ b1 above (i.e., g(x) is the polynomial associated with the ﬁrst vector in a reduced basis). Recall that Theorem 11.2.1 gives ∥h0(x)∥≤2m∥f(x)∥. On the other hand, by considering the vectors associated with g0(x) (that is, ⃗ b1) and h0(x) in L ⊆Zm+1, we deduce from (11.5.1) that ∥g0(x)∥≤2m/2∥h0(x)∥. Thus, ∥g0(x)∥≤23m/2∥f(x)∥. We want this bound on ∥g0(x)∥to assure that R = 0 so that ⃗ b# = ⃗ b0. To see how large pk needs to be, we recall the Sylvester form of the resultant given by (2.2.1). We are interested in the resultant R of g0(x) and h0(x) where 275 we may suppose that deg g0 ≤m and deg h0 ≤m. From Hadamard’s inequality and Theorem 11.2.1, we deduce \|R\| ≤∥g0(x)∥m∥h0(x)∥m ≤∥g0(x)∥m2m∥f(x)∥ m = 2m2∥g0(x)∥m∥f(x)∥m. Our upper bound on ∥g0(x)∥now implies \|R\| ≤25m2/2∥f(x)∥2m. Hence, we see that if pk > 25m2/2∥f(x)∥2m, then the vector ⃗ b1 in a reduced basis for the lattice L, where m = deg h0(x), is such that the polynomial corresponding to ⃗ b# 1 is h0(x). §11.6 Sparse Polynomial Computations In the previous sections of this chapter, we explored algorithms for factoring polynomials with an interest in them as irreducibility tests for polynomials in Z[x]. We began with an approach due to Zassenhaus (1969). An example due to Swinnerton-Dyer shows that this algorithm can take time that is exponential in the degree of the input polynomial f(x). In general, this algorithm is nev-ertheless quite practical as examples of f(x) for which the running time of the algorithm is exponential in deg f are rare. We then explored an algorithm due to Lenstra, Lenstra, and Lovasz (1982) which can be shown to have running time that is polynomial in deg f as well as the height of f(x) (the maximum of the absolute values of the coeﬃcients of f(x)). In this section, we describe an algorithm given by Filaseta, Granville, and Schinzel (2008) for determining whether a non-reciprocal polynomial f(x) is irreducible. Take particular note that this algorithm requires the input polynomial f(x) to be non-reciprocal. This algorithm is signiﬁcant when a polynomial is sparse. The algorithm has running time that is almost linear in log deg f, but its dependence on the num-ber of terms of f(x) and the height of f(x) is not so good. In particular, the dependence on the number of terms of f(x) is worse than exponential. On the other hand, if one considers all polynomials in Z[x] with a ﬁxed bound on the number of terms and a ﬁxed bound on the height of f(x), then the running time of the algorithm will just depend on the deg f and will be close to linear in log deg f. More precisely, we have the following result of Filaseta, Granville, and Schinzel (2008). Theorem 11.6.1. Let f(x) = Pr j=0 ajxdj ∈Z[x] with each aj ̸= 0 and with 0 = d0 < d1 < · · · < dr−1 < dr = n. Suppose r ≥1 and n ≥16. Let H = H(f) denote the height of f(x), so H = max0≤j≤r{\|aj\|}. Then there is a constant c1 = c1(r, H) such that an algorithm 276 exists for determining whether a given non-reciprocal polynomial f(x) ∈Z[x] as above is irreducible and that runs in time O c1 log n (log log n)2 log log log n . The algorithm for Theorem 11.6.1 also provides some information on the factorization of f(x) in the case that f(x) is reducible (with the same running time). Speciﬁcally, we have the following: (i) If f(x) has a cyclotomic factor, then the algorithm will detect this and output an m ∈Z+ such that the cyclotomic polynomial Φm(x) divides f(x). (ii) If f(x) does not have a cyclotomic factor but has a non-constant reciprocal factor, then the algorithm will produce such a factor. In fact, the algorithm will produce a reciprocal factor of f(x) of maximal degree. (iii) Otherwise, if f(x) is reducible, then the algorithm outputs a complete factorization of f(x) as a product of irreducible polynomials over Q. These can in fact be viewed as basic parts of the algorithm. First, we will check if f(x) has a cyclotomic factor by making use of the algorithm for Theorem 6.9.1. If it does, the algorithm will produce m as in (i) and stop. If it does not, then the algorithm will check if f(x) has a non-cyclotomic non-constant reciprocal factor. If it does, then the algorithm will produce such a factor as in (ii) and stop. If it does not, then the algorithm will output a complete factorization of f(x) as indicated in (iii). For (ii), we will make use of another algorithm for computing the greatest common divisor of two polynomials in Z[x], that is for computing gcdZ(f(x), g(x)) as deﬁned in Section 10.2. Here, we will have the added condition that at least one of the two polynomials is not divisible by a cyclotomic polynomial, which in general can be checked by making use of the algorithm in Theorem 6.9.1. This result, also due to Filaseta, Granville, and Schinzel (2008), is as follows. Theorem 11.6.2. There is an algorithm which takes as input two polynomials f(x) and g(x) in Z[x], each of degree ≤n and height ≤H and having ≤r + 1 nonzero terms, with at least one of f(x) and g(x) free of cyclotomic factors, and outputs the value of gcdZ(f(x), g(x)) and runs in time O c2 log n for some constant c2 = c2(r, H). In discussing the above theorems in this section, we describe the algorithms which lead to the proofs but do not address details of the running times. The reader should consult the work of Filaseta, Granville, and Schinzel (2008) for further details on the running time estimates. The proof of Theorem 11.6.2 result relies heavily on a result Bombieri and Zannier described in an appendix by the latter in Schinzel (2000b). Alterna-tively, one can use the later work of Bombieri, Masser, and Zannier (2007). As a consequence of their either of these, we have the following. 277 Theorem 11.6.3. Let F(x1, . . . , xk), G(x1, . . . , xk) ∈Q[x1, . . . , xk] be two coprime polynomials. There exists an eﬀectively computable number B(F, G) with the following property. If − → u = ⟨u1, . . . , uk⟩∈Zk, ξ ̸= 0 is al-gebraic and F(ξu1, . . . , ξuk) = G(ξu1, . . . , ξuk) = 0, then either ξ is a root of unity or there exists a nonzero vector − → v ∈Zk having components bounded in absolute value by B(F, G) and orthogonal to − → u . Our proof of Theorem 11.6.2 has similarities to an application of Theo-rem 11.6.3 by Schinzel (1999c, 2000b). In particular, we make use of the follow-ing lemma which is Corollary 6 in Appendix E of Schinzel (2000b). A proof is given there. Lemma 11.6.4. Let ℓbe a positive integer and − → v ∈Zℓwith − → v nonzero. The lattice of vectors − → u ∈Zℓorthogonal to − → v has a basis − → v1′, − → v2′, . . . , − − → vℓ−1′ such that the maximum absolute value of a component of any vector − → vj ′ is bounded by ℓ/2 times the maximum absolute value of a component of − → v . We make use of the notation Or,H α(n) to denote a function which has absolute value bounded by Cα(n) for some constant C > 0 and for n suﬃciently large. Another result of use to us is the following. Lemma 11.6.5. There is an algorithm with the following property. Given an r × t integral matrix M = (mij) of rank t ≤r and max{\|mij\|} = Or,H(1) and given an integral vector − → d = ⟨d1, . . . , dr⟩with max{\|dj\|} = Or,H(n), the algorithm determines whether there is an integral vector − → v = ⟨v1, . . . , vt⟩for which    d1 . . . dr   = M    v1 . . . vt    holds, and if such a − → v exists, the algorithm outputs the solution vector − → v . Furthermore, max{\|vj\|} = Or,H(n) and the algorithm runs in time Or,H(log n). There are a variety of ways we can determine if − → d = M− → v has a solution and to determine the solution if there is one within the required time Or,H(log n). We use Gaussian elimination to explain an approach. Again, we refer to reader to Filaseta, Granville, and Schinzel (2008) for details on the running time. Per-forming elementary row operations on M and multiplying by entries from the matrix as one proceeds to use only integer arithmetic allows us to rewrite M in the form of an r × t matrix M ′ = (m′ ij) with each m′ ij ∈Z and the ﬁrst t rows of M ′ forming a t × t diagonal matrix with nonzero integers along the diagonal. We perform the analogous row operations and integer multiplications on the vector − → d = ⟨d1, d2, . . . , dr⟩to solve − → d = M− → v for − → v . We are thus left 278 with an equation of the form − → d′ = M ′− → v where the entries of M ′ are integers that are Or,H(1) and the components of − → d′ = ⟨d′ 1, d′ 2, . . . , d′ r⟩are integers that are Or,H(n). For each j ∈{1, 2, . . . , t}, we check if d′ j ≡0 (mod m′ jj). If for some j ∈{1, 2, . . . , t} we have d′ j ̸≡0 (mod m′ jj), then a solution to the original equation − → d = M− → v , if it exists, must be such that vj ̸∈Z. In this case, an integral vector − → v does not exist. Now, suppose instead that d′ j ≡0 (mod m′ jj) for every j ∈{1, 2, . . . , t}. Then we divide d′ j by m′ jj to determine the vector − → v . This vector may or may not be a solution to the equation − → d = M− → v . We check whether it is by a direct computation. If it is not a solution to the equation − → d = M− → v , then there are no solutions to the equation. Otherwise, − → v is an integral vector satisfying − → d = M− → v , and we output the vector − → v . This ﬁnishes our explanation for Lemma 11.6.5. We also make use of the following notation. For a polynomial F x1, . . . , xr, x−1 1 , . . . , x−1 r , in the variables x1, . . . , xr and their reciprocals x−1 1 , . . . , x−1 r , we deﬁne J F = xu1 1 · · · xur r F x1, . . . , xr, x−1 1 , . . . , x−1 r , where each uj is an integer chosen as small as possible so that J F is a polynomial in x1, . . . , xr. In the way of examples, if F = x2 + 4x−1y + y3 and G = 2xyw −x2z−3w −12w, then J F = x3 + 4y + xy3 and J G = 2xyz3 −x2 −12z3. In particular, note that although w is a variable in G, the polynomial J G does not involve w. We call a multi-variable polynomial F(x1, . . . , xr) ∈Q[x1, . . . , xr] reciprocal if J F x−1 1 , . . . , x−1 r = ±F(x1, . . . , xr). For example, x1x2 −x1 −x2 + 1 and x1x2 −x3x4 are reciprocal. Note that this is consistent with our deﬁnition of a reciprocal polynomial f(x) ∈Z[x]. For our proof of Theorem 11.6.2, we can suppose that f(x) does not have a cyclotomic factor and do so. We consider only the case that f(0)g(0) ̸= 0 as computing gcdZ(f(x), g(x)) can easily be reduced to this case by initially removing an appropriate power of x from each of f(x) and g(x). This would need to be followed up by possibly multiplying by a power of x after our gcd computation. We furthermore only consider the case that the content of f(x), that is the greatest common divisor of its coeﬃcients, and the content of g(x) are 1. Otherwise, we simply divide by the contents before proceeding and then multiply the ﬁnal result by the greatest common divisor of the two contents. 279 We express our two polynomials in the form f(x) = k X j=0 ajxdj and g(x) = k X j=0 bjxdj, where above we have possibly extended the lists of exponents and coeﬃcients describing f(x) and g(x) so that the exponent lists are identical and the co-eﬃcient lists are allowed to include coeﬃcients which are 0. We do this in such a way that \|aj\| + \|bj\| ̸= 0 for each j ∈{0, 1, . . . , k}. Also, we take 0 = d0 < d1 < · · · < dk−1 < dk. Thus, d0 = 0, a0b0 ̸= 0 and k ≤2r. Let w(x) denote gcdZ(f(x), g(x)). We will apply Theorem 11.6.3 to construct two ﬁnite sequences of polynomials in several variables Fu and Gu with integer coeﬃcients and a corresponding ﬁnite sequence of vectors − → d (u) that will enable us to determine a polynomial in Z[x] that has the common zeros, to the correct multiplicity, of f(x) and g(x). This then will allow us to compute w(x). Let ξ be a zero of w(x), if it exists. Observe that ξ ̸= 0, and since ξ is a zero of f(x) which has no cyclotomic factors, we have ξ is not a root of unity. Since ξ is a common zero of f(x) and g(x), we have k X j=0 ajξdj = k X j=0 bjξdj = 0. We recursively construct Fu, Gu and − → d (u), for 0 ≤u ≤s, where s is to be determined, beginning with F0 = F0(x1, . . . , xk) = a0 + k X j=1 ajxj, G0 = G0(x1, . . . , xk) = b0 + k X j=1 bjxj, (11.6.1) and − → d (0) = ⟨d1, d2, . . . , dk⟩. As u increases, the number of variables deﬁning Fu and Gu will decrease. The value of s then will be ≤k. Observe that F0(xd1, . . . , xdk) = f(x) and G0(xd1, . . . , xdk) = g(x). We deduce that F0 and G0, being linear, are coprime in Q[x1, . . . , xk] and that (11.6.2) F0(ξd1, . . . , ξdk) = G0(ξd1, . . . , ξdk) = 0. Now, suppose for some u ≥0 that nonzero polynomials Fu and Gu in Z[x1, . . . , xku] and a vector − → d (u) = ⟨d(u) 1 , . . . , d(u) ku ⟩∈Zku have been determined, where ku < ku−1 < · · · < k0 = k. Furthermore, suppose that Fu and Gu are coprime in Q[x1, . . . , xku] and that we have at least one zero ξ of w(x) such that (11.6.3) Fu ξd(u) 1 , . . . , ξd(u) ku = Gu ξd(u) 1 , . . . , ξd(u) ku = 0. 280 In particular, ξ ̸= 0 and ξ is not a root of unity. Note that the d(u) j may be negative. We will require (11.6.4) J Fu xd(u) 1 , . . . , xd(u) ku \| f(x) and J Gu xd(u) 1 , . . . , xd(u) ku \| g(x). Observe that J Fu xd(u) 1 , . . . , xd(u) ku and f(x) are in Z[x]. We take (11.6.4) to mean that there is a polynomial h(x) ∈Z[x] such that f(x) = h(x) · J Fu xd(u) 1 , . . . , xd(u) ku with an analogous equation holding for g(x) and J Gu xd(u) 1 , . . . , xd(u) ku . In par-ticular, we want J Fu xd(u) 1 , . . . , xd(u) ku and J Gu xd(u) 1 , . . . , xd(u) ku to be nonzero. Note that these conditions which are being imposed on Fu and Gu are satisﬁed for u = 0 provided w(x) is not constant. For 0 ≤u < s, we describe next how to recursively construct Fu+1 and Gu+1 having analogous properties. There is a computable bound B(Fu, Gu) as described in Theorem 11.6.3. We deduce that there is a nonzero vector − → v = ⟨v1, v2, . . . , vku⟩∈Zku such that each \|vi\| ≤B(Fu, Gu) and − → v is orthogonal to − → d (u). From Lemma 11.6.4, there is a ku × (ku −1) matrix M with each entry of M having absolute value ≤kuB(Fu, Gu)/2 and such that − → d (u) = M− → v (u) for some − → v (u) ∈Zku−1, where we view the vectors as column vectors. We deﬁne integers mij (written also mi,j) and v(u) j , depending on u, by the conditions M =    m11 · · · m1,ku−1 . . . ... . . . mku1 · · · mku,ku−1    and − → v (u) = ⟨v(u) 1 , . . . , v(u) ku−1⟩. The relations xi = ymi1 1 · · · ymi,ku−1 ku−1 , for 1 ≤i ≤ku, transform the polynomials Fu(x1, . . . , xku) and Gu(x1, . . . , xku) into polynomi-als in some, possibly all, of the variables y1, . . . , yku−1. These new polynomials we call Fu and Gu, respectively. More precisely, we deﬁne (11.6.5) Fu(y1, . . . , yku−1) = J Fu ym11 1 · · · ym1,ku−1 ku−1 , . . . , ymku1 1 · · · ymku,ku−1 ku−1 and (11.6.6) Gu(y1, . . . , yku−1) = J Gu ym11 1 · · · ym1,ku−1 ku−1 , . . . , ymku1 1 · · · ymku,ku−1 ku−1 . The polynomials Fu and Gu will depend on the matrix M so that there may be many choices for Fu and Gu for each Fu and Gu. We need only consider one such Fu and Gu and do so. Note that this still may require considering various M until we ﬁnd one for which − → d (u) = M− → v (u) is satisﬁed for some − → v (u) ∈Zku−1. 281 The equation − → d (u) = M− → v (u) implies that for some integers ef(u) and eg(u) we have (11.6.7) Fu xv(u) 1 , . . . , xv(u) ku−1 = xef (u)Fu xd(u) 1 , . . . , xd(u) ku and (11.6.8) Gu xv(u) 1 , . . . , xv(u) ku−1 = xeg(u)Gu xd(u) 1 , . . . , xd(u) ku . In particular, Fu and Gu are nonzero. Also, (11.6.9) J Fu xv(u) 1 , . . . , xv(u) ku−1 \| f(x) and J Gu xv(u) 1 , . . . , xv(u) ku−1 \| g(x). Furthermore, with ξ as in (11.6.3), we have Fu ξv(u) 1 , . . . , ξv(u) ku−1 = Gu ξv(u) 1 , . . . , ξv(u) ku−1 = 0. The idea is to suppress the variables, if they exist, which do not occur in Fu and Gu and the corresponding components of − → v (u) to obtain the polynomials Fu+1 and Gu+1 and the vector − → d (u+1) for our recursive construction. However, there is one other matter to consider. The polynomials Fu and Gu may not be coprime, and we require Fu+1 and Gu+1 to be coprime. Hence, we adjust this idea slightly. Let (11.6.10) Du = Du(y1, . . . , yku−1) = gcdZ(Fu, Gu) ∈Z[y1, . . . , yku−1]. Recall that f(0)g(0) ̸= 0. Hence, (11.6.7), (11.6.8) and (11.6.9) imply that J Du xv(u) 1 , . . . , xv(u) ku−1 divides gcdZ(f, g) in Z[x]. We deﬁne (11.6.11) Fu+1 = Fu(y1, . . . , yku−1) Du(y1, . . . , yku−1) and Gu+1 = Gu(y1, . . . , yku−1) Du(y1, . . . , yku−1), and set ku+1 ≤ku−1 to be the total number of variables y1, . . . , yku−1 appearing in Fu+1 and Gu+1. Note that Fu+1 and Gu+1 are coprime and that (11.6.4) holds with u replaced by u + 1 and the appropriate change of variables. We describe next how the recursive construction will end. Suppose we have just constructed Fu, Gu and − → d (u) and proceed as above to the next step of constructing Fu+1, Gu+1 and − → d (u+1). At this point, Du−1 will have been de-ﬁned but not Du. We want to ﬁnd M and a − → v (u) such that − → d (u) = M− → v (u) where M is a ku × (ku −1) matrix with entries bounded in absolute value by kuB(Fu, Gu)/2. So we compute B(Fu, Gu) and the bound kuB(Fu, Gu)/2 on the absolute values of the entries of M. We consider such M and apply the algorithm of Lemma 11.6.5 to see if there is an integral vector − → v (u) for which − → d (u) = M− → v (u). Once such an M and − → v (u) are found, we can proceed with the construction of Fu+1 and Gu+1 given above. On the other hand, it is possible 282 that no such M and − → v (u) will be found. Given Theorem 11.6.3, this will be the case only if the supposition that (11.6.3) holds for some zero ξ of w(x) is incorrect. In particular, (11.6.3) does not hold for some zero ξ of w(x) if Fu and Gu are coprime polynomials in < 2 variables (i.e., ku ≤1), but it is also possible that (11.6.3) does not hold for some u with Fu and Gu polynomials in ≥2 variables (i.e., ku ≥2). Given that M is a ku ×(ku −1) matrix, we consider it to be vacuously true that no M and − → v (u) exist satisfying − → d (u) = M− → v (u) in the case that ku ≤1. If no such M and − → v (u) exist, we consider the recursive construction of the polynomials Fu and Gu complete and set s = u. We will want the values of Du for every 1 ≤u ≤s −1, so we save these as we proceed. The motivation discussed above can be summarized into a procedure for describing the algorithm associated with Theorem 11.6.2. Beginning with F0 and G0 as in (11.6.1) and − → d (0) = ⟨d1, . . . , dk⟩, we construct the multi-variable polynomials Fu and Gu and vectors − → d (u) = ⟨d(u) 1 , . . . , d(u) ku ⟩∈Zku recursively. Given Fu, Gu and − → d (u), we compute B(Fu, Gu) and search for a ku × (ku − 1) matrix M with integer entries having absolute value ≤kuB(Fu, Gu)/2 for which − → d (u) = M− → v (u) is solvable with − → v (u) = ⟨v(u) 1 , . . . , v(u) ku−1⟩∈Zku−1. We check for solvability and determine the solution − → v (u) if it exists by using the algorithm in Lemma 11.6.5. If no such M and − → v (u) exist, then we set s = u and stop our construction. Otherwise, once such an M = (mij) and − → v (u) are determined, we deﬁne Fu+1 and Gu+1 using (11.6.5), (11.6.6), (11.6.10) and (11.6.11). After using (11.6.11) to construct Fu+1 and Gu+1, we determine the variables y1, . . . , yku−1 which occur in Fu+1 and Gu+1 and deﬁne − → d (u+1) as the vector with corresponding components from v(u) 1 , . . . , v(u) ku−1; in other words, if yj is the ith variable occurring in Fu+1 and Gu+1, then v(u) j is the ith component of − → d (u+1). One important aspect of the construction is that when we divide by Du to construct Fu+1 and Gu+1, we obtain not simply that J Du xv(u) 1 , . . . , xv(u) ku−1 divides gcdZ(f, g) in Z[x] but also (11.6.12) u Y j=0 J Dj xv(j) 1 , . . . , x v(j) kj −1 divides gcdZ(f, g) in Z[x]. This can be seen inductively by observing that (11.6.13) J Fu xv(u) 1 , . . . , xv(u) ku−1 = f(x) u−1 Y j=0 J Dj xv(j) 1 , . . . , x v(j) kj −1 and (11.6.14) J Gu xv(u) 1 , . . . , xv(u) ku−1 = g(x) u−1 Y j=0 J Dj xv(j) 1 , . . . , x v(j) kj −1 . 283 The algorithm for Theorem 11.6.2 ends by making use of the identity (11.6.15) gcdZ f(x), g(x) = s−1 Y u=0 J Du xv(u) 1 , . . . , xv(u) ku−1 . We justify (11.6.15). Recall that we have denoted the left side by w(x). Observe that (11.6.12) implies that the expression on the right of (11.6.15) divides w(x). By the deﬁnition of s, when we arrive at u = s in our recursive construction, (11.6.3) fails to hold for every zero ξ of w(x). Therefore, taking u = s −1 in (11.6.11), (11.6.13) and (11.6.14)) implies that the right-hand side of (11.6.15) vanishes at all the zeros of w(x) and to the same multiplicity. As noted earlier, we are considering the case that the contents of f(x) and g(x) are 1. We deduce that (11.6.15) holds. We are now left with considering the case that f(x) does not have any non-constant reciprocal factor. For this part, we make use of the a weakened form of a result of Schinzel (1970a). Before stating this result, we give some background. Similar to the approach just used for computing the greatest common divisor of two sparse polynomials, we set F(x1, . . . , xr) = arxr + · · · + a1x1 + a0 ∈Z[x1, . . . , xr], with the plan of connecting the factorization of f(x) = F(xd1, xd2, . . . , xdr) with the factorization of a multi-variable polynomial of the form J F ym11 1 · · · ym1t t , . . . , ymr1 1 · · · ymrt t , where the number of variables t is ≤r and mij ∈Z for 1 ≤i ≤r and 1 ≤j ≤t. The above multi-variable polynomial can be expressed as yu1 1 · · · yut t F(ym11 1 · · · ym1t t , . . . , ymr1 1 · · · ymrt t ), where uj = −min{m1j, m2j, . . . , mrj, 0} for 1 ≤j ≤t. To make the connection with the factorization of f(x), we want the matrix M = (mij) to be such that (11.6.16)    d1 . . . dr   = M    v1 . . . vt    for some integers v1, v2, . . . , vt. In this way, the substitution yj = xvj for 1 ≤ j ≤t takes any factorization yu1 1 · · · yut t F(ym11 1 · · · ym1t t , . . . , ymr1 1 · · · ymrt t ) = F1(y1, . . . , yt) · · · Fs(y1, . . . , yt) (11.6.17) 284 in Z[y1, . . . , yt] into the form xu1v1+···+utvtF(xd1, xd2, . . . , xdr) = F1(xv1, . . . , xvt) · · · Fs(xv1, . . . , xvt). (11.6.18) We restrict our attention to factorizations in (11.6.17) where the Fi(y1, . . . , yt) are non-constant. We will be interested in the case that s is maximal; in other words, we will want the right-hand side of (11.6.17) to be a complete factoriza-tion of the left-hand side of (11.6.17) into irreducibles over Q. For achieving the results in this paper, we want some algorithm for obtaining such a complete fac-torization of multi-variable polynomials; among the various sources for this, we note the work by Lenstra (1987) provides such an algorithm. For the moment, though, we need not take s maximal. Since f(x) = F(xd1, xd2, . . . , xdr), the above describes a factorization of f(x), except that we need to take some caution as some vj may be negative so the expressions Fi(xv1, . . . , xvt) may not be polynomials in x. For 1 ≤i ≤s, deﬁne wi as the integer satisfying (11.6.19) J Fi(xv1, . . . , xvt) = xwiFi(xv1, . . . , xvt). We obtain from (11.6.18) that xu1v1+···+utvt+w1+···+wsf(x) = s Y i=1 xwiFi(xv1, . . . , xvt). The deﬁnition of wi implies that this product is over polynomials in Z[x] that are not divisible by x. The conditions a0 ̸= 0 and d0 = 0 imposed on f(x) in the introduction imply that f(x) is not divisible by x. Hence, the exponent of x appearing on the left must be 0, and we obtain the factorization (11.6.20) f(x) = s Y i=1 xwiFi(xv1, . . . , xvt) = s Y i=1 J Fi(xv1, . . . , xvt). The factorization given in (11.6.20) is crucial to our algorithm. As we are interested in the case that f(x) has no non-constant reciprocal factor, we re-strict our attention to this case. From (11.6.20), we see that the polynomials xwiFi(xv1, . . . , xvt) cannot have a non-constant reciprocal factor. There are, however, still two possibilities that we need to consider for each i ∈{1, 2, . . . , s}: (i′) Fi(y1, . . . , yt) is reciprocal. (ii′) J Fi(xv1, . . . , xvt) ∈Z. Although we will not need to know a connection between (i′) and (ii′), we show here that if (i′) holds for some i, then (ii′) does as well. We consider then the possibility that (11.6.21) J Fi y−1 1 , . . . , y−1 t = ±Fi(y1, . . . , yt). 285 In other words, suppose that (11.6.22) ye1 1 · · · yet t Fi y−1 1 , . . . , y−1 t = ±Fi(y1, . . . , yt), where ej = ej(i) is the degree of Fi(y1, . . . , yt) as a polynomial in yj. Substi-tuting yj = xvj into (11.6.22), we obtain (11.6.23) xwi+e1v1+···+etvtFi x−v1, . . . , x−vt = ±xwiFi xv1, . . . , xvt . By the deﬁnition of wi, the polynomial on the right does not vanish at 0. Assume (ii′) does not hold. Let α be a zero of this polynomial. Then substituting x = 1/α into (11.6.23) shows that 1/α is also a zero. On the other hand, we have already demonstrated in (11.6.20) that the right-hand side of (11.6.23) is a factor of f(x). This contradicts that f(x) has no non-constant reciprocal factor. Hence, (ii′) holds. As noted earlier, we make use of a special case of a result due to Schinzel (1970a). In particular, the more general result implies that the above idea can in fact always be used to factor f(x) if f(x) has two nonreciprocal irreducible factors. In other words, there exist a matrix M and vj satisfying (11.6.16) and a factorization of the form (11.6.17) that leads to a non-trivial factorization of f(x), if it exists, through the substitution yj = xvj. We are interested in the case that f(x) has no non-constant reciprocal factor. In this case, we can obtain a complete factorization of f(x) into irreducibles. Theorem 11.6.6. Fix F = F(x1, . . . , xr) = arxr + · · · + a1x1 + a0, where the aj are nonzero integers. There exists a ﬁnite computable set of ma-trices S with integer entries, depending only on F, with the following property: Suppose the vector − → d = ⟨d1, d2, . . . , dr⟩is in Zr with dr > dr−1 > · · · > d1 > 0 and such that f(x) = F(xd1, xd2, . . . , xdr) has no non-constant reciprocal fac-tor. Then there is an r × t matrix M = (mij) ∈S of rank t ≤r and a vector − → v = ⟨v1, v2, . . . , vt⟩in Zt such that (11.6.16) holds and the factorization given by (11.6.17) in Z[y1, . . . , yt] of a polynomial in t variables y1, y2, . . . , yt as a product of s irreducible polynomials over Q implies the factorization of f(x) given by (11.6.20) as a product of polynomials in Z[x] each of which is either irreducible over Q or a constant. We are ready now to apply the above to ﬁnish describing the algorithm for establishing Theorem 11.6.1, that is to describe how (iii) can be done. As before, the reader can look at the work of Filaseta, Granville, and Schinzel (2008) to obtain details for estimating the running time. As suggested by the statement of Theorem 11.6.6, we take the coeﬃcients aj of f(x) and consider the multi-variable polynomial F = F(x1, . . . , xr). We compute the set S. Since f(x) = F(xd1, . . . , xdr) has no non-constant reciprocal factors, there is a matrix M = (mij) ∈S of rank t ≤r and a vector − → v in Zt as in Theorem 11.6.6. 286 We go through each of the Or,H(1) matrices M in S and solve for the vectors − → v = ⟨v1, v2, . . . , vt⟩in Zt satisfying − → d = M− → v , where t is the number of columns in M and we interpret − → d and − → v as column vectors. From the deﬁnition of S, we have that the rank of M is t and t ≤r. Hence, there can be at most one such vector − → v for each M ∈S. However, for each − → d , there may be many M ∈S and − → v for which − → d = M− → v , and we will consider all of them. The algorithm for Lemma 11.6.5 is performed for each of the Or,H(1) matri-ces M in S. This leads to Or,H(1) factorizations of the form given in (11.6.17) into irreducibles, each having a potentially diﬀerent value for s. For each of these, we compute the values of Fi xv1, . . . , xvt and determine wi as in (11.6.19). We produce then Or,H(1) factorizations of f(x) as in (11.6.20). As we obtain these factorizations, we keep track of the number of non-constant polyno-mials xwiFi xv1, . . . , xvt appearing in (11.6.20). We choose a factorization for which this number is maximal. Recalling that (11.6.20) follows from − → d = M− → v and (11.6.17), we deduce from Theorem 11.6.6 that the factorization of f(x) we have chosen provides a factorization of f(x) with each xwiFi xv1, . . . , xvt either irreducible or constant. For a factorization of f(x) into irreducibles over Q, we multiply together the constants appearing on the right of (11.6.20) and one of the irreducible polynomials J Fi xv1, . . . , xvt . This completes the proof of Theorem 11.6.1. 287 §11.7 Exercises (11.1) Use Berlekamp’s algorithm to factor f(x) = x6 +x3 +x2 +x+1 modulo 2. You should obtain two polynomials u(x) and v(x) of degrees < 6 such that f(x) ≡u(x)v(x) (mod 2). (11.2) Use the previous problem and Hensel lifting to factor f(x) = x6 + x3 + x2 + x + 1 modulo 32. To help, let u(x) and v(x) be as in the previous problem with deg u > deg v. Then you can take advantage of the following: u(x) + x3v(x) ≡x5 + 1 (mod 2) (x + 1)u(x) + (x2 + 1)v(x) ≡x5 + x4 (mod 2) xu(x) + (x2 + x)v(x) ≡x5 (mod 2). (11.3) In each part, apply the Gram-Schmidt orthogonalization process to the given subset S of V to obtain an orthogonal basis for V . (a) S = {⟨1, 1, 1⟩, ⟨1, 0, 1⟩, ⟨2, 1, 1⟩}, V = Q3 (b) S = {⟨1, 0, 0, 1⟩, ⟨1, 0, 2, 2⟩, ⟨1, 1, 0, 1⟩, ⟨2, 1, 1, 0⟩}, V = Q4 (11.4) Let β1 = ⟨1, 0, −5⟩, β2 = ⟨11, 0, −3⟩and β3 = ⟨41, 2, −23⟩. (a) Verify that {β1, β2, β3} form a basis for R3. In other words, show that if a1, a2 and a3 are real numbers, then there exist real numbers b1, b2 and b3 such that b1β1 + b2β2 + b3β3 = ⟨a1, a2, a3⟩. (b) Show that {β1, β2, β3} is not an orthogonal basis for R3. (c) Calculate the orthogonal basis {α1, α2, α3} for R3 obtained by the Gram-Schmidt orthogonalization process. (d) Write ⟨1, 2, 3⟩∈R3 as a linear combination of α1, α2 and α3 by considering projections of vectors rather than solving by linear equations. (11.5) Let ⃗ b1, . . . ,⃗ bn be in Qn, and suppose they are linearly independent over R. Let ⃗ b∗ 1, . . . ,⃗ b∗ n be the vectors obtained from ⃗ b1, . . . ,⃗ bn by the Gram-Schmidt orthogonalization process. (a) For each i ∈{1, . . . , n}, show that the vectors ⃗ b∗ 1, . . . ,⃗ b∗ i span the same subspace of Rn as ⃗ b1, . . . ,⃗ bi. (b) Show that ⃗ b∗ 1, . . . ,⃗ b∗ n are linearly independent over R. (c) Show that ⃗ b∗ 1, . . . ,⃗ b∗ n are pairwise orthogonal (i.e., for distinct i and j, we have ⃗ b∗ i ·⃗ b∗ j = 0). 288 (11.6) (a) Let α1, α2, . . . , αn ∈C with \|α1\|, \|α2\|, . . . , \|αm\| each ≥1 and \|αm+1\|, \|αm+2\|, . . . , \|αn\| each < 1. Set M = α1α2 · · · αm. For k ∈{1, 2, . . . , n}, let σk = X 1≤j1<j2<··· 1. Discuss what conditions are really needed on the aj’s. In particular, consider the possibility that some of the aj’s are negative. (b) We have shown that x4 +1 is reducible modulo every prime. Show that this follows from part (a), possibly redoing part (a) so that it does. (11.11) (a) Let a and b be relatively prime positive integers. Prove that gcd xab −1, (xa −1)(xb −1) = (xa −1)(xb −1) x −1 . (b) Explain how (a) implies that Theorem 11.6.2 is not true if we omit the condition that at least one of f(x) and g(x) is free of cyclotomic factors.
15778	https://forvo.com/word/thriving/	How to pronounce thriving in English - Definition and synonyms of thriving in English Forvo - Do Not Process My Personal Information If you wish to opt-out of the sale, sharing to third parties, or processing of your personal or sensitive information for targeted advertising by us, please use the below opt-out section to confirm your selection. Please note that after your opt-out request is processed you may continue seeing interest-based ads based on personal information utilized by us or personal information disclosed to third parties prior to your opt-out. You may separately opt-out of the further disclosure of your personal information by third parties on the IAB’s list of downstream participants. This information may also be disclosed by us to third parties on the IAB’s List of Downstream Participants that may further disclose it to other third parties. Personal Data Processing Opt Outs CONFIRM Data DeletionData AccessPrivacy Policy Menu Search Pronounce English Deutsch Español Français Italiano 日本語 Nederlands Polski Português Русский Türkçe 汉语 العربية Български Bosanski Català Čeština Dansk Ελληνικά Euskara پارسی Suomi 客家语 עברית हिन्दी Hrvatski Magyar Հայերեն Bahasa Indonesia 한국어 Kurdî / كوردی Latviešu Norsk ਪੰਜਾਬੀ Română Slovenčina Српски / Srpski Svenska ไทย Татар теле Українська Tiếng Việt 粵文 Log in Languages Guides Categories Events Users Blog Search for a word Languages Pronunciation Search [x] Filter language and accent: [x] English 2 [x] American 2 Filter(1) Not satisfied? Request a new pronunciation Pronunciation in: Is there anything wrong with this word/phrase? How to pronounce thriving Listened to:5.0K times Filter language and accent(1) _thriving pronunciation in English[en] Phonetic spelling:ˈθraɪvɪŋ_ Accent:American thriving pronunciation Pronunciation by SorenV (Male from United States)Male from United States Pronunciation by SorenV 5 votes Good Bad Add to favorites Download MP3 Report thriving pronunciation Pronunciation by tipit (Male from United States)Male from United States Pronunciation by tipit User information Follow 0 votes Good Bad Add to favorites Download MP3 Report Phrases thriving example in a phrase There is a thriving digital start-up community in this part of town There is a thriving digital start-up community in this part of town pronunciation Pronunciation by stlowery(Male from United States) When he bought the business it was a thriving commercial enterprise When he bought the business it was a thriving commercial enterprise pronunciation Pronunciation by stlowery(Male from United States) Translation Definition - Synonyms Definition of thriving very lively and profitable Synonyms of thriving sprouting pronunciation sprouting[en] budding pronunciation budding[en] flourishing pronunciation flourishing[en] blooming pronunciation blooming[en] lush pronunciation lush[en] luxuriant pronunciation luxuriant[en] prolific pronunciation prolific[en] prosperous pronunciation prosperous[en] wealthy pronunciation wealthy[en] affluent pronunciation affluent[en] Can you pronounce it better? Or with a different accent? Pronounce thriving in English Share the pronunciation of thriving in English: Facebook Twitter Email Accents & languages on maps +- Leaflet \| Map data © OpenStreetMap contributors, CC-BY-SA _Do you know how to pronounce thriving? thriving is pending pronunciation in:_ Record pronunciation for thriving thriving[en - uk] Record pronunciation for thriving thriving[en - other] Random words: water,antidisestablishmentarianism,hello,tomato,caramel How to pronounce Gyro Gif Nguyen Acai Quinoa Google Pho Meme Qatar Worcestershire Gnocchi Niche Saoirse Siobhan Salmon Turmeric Cthulhu Cache Guide Nietzsche Caramel Hijab Balayage Porsche Gal Gadot Dachshund Charcuterie Tinnitus General Tso Uranus Curacao Pecan Niger Nevada Samhain Macabre Chipotle La Croix Phonetic Caribbean Asus Received Data Ng Goethe Sikh Saoirse Ronan Key Caprese Peridot Melee Doge Daesh Chopin Cacao Lychee Homage Bruschetta Dictionary Choose your language: Deutsch English Español Français Italiano 日本語 Nederlands Polski Português Русский Türkçe 汉语 and even more languages العربية Български Bosanski Català Čeština Dansk Ελληνικά Euskara پارسی Suomi 客家语 עברית हिन्दी Hrvatski Magyar Հայերեն Bahasa Indonesia 한국어 Kurdî / كوردی Latviešu Norsk ਪੰਜਾਬੀ Română Slovenčina Српски / Srpski Svenska ไทย Татар теле Українська Tiếng Việt 粵文 Blog App Forvo Kids Videos API Terms and conditions Privacy About Forvo Contact us FAQ Certificates Donate
15779	https://pmc.ncbi.nlm.nih.gov/articles/PMC4888901/	Molecular and Cellular Mechanisms of Cardiovascular Disorders in Diabetes - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Journal List User Guide View on publisher site Download PDF Add to Collections Cite Permalink PERMALINK Copy As a library, NLM provides access to scientific literature. 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Published in final edited form as: Circ Res. 2016 May 27;118(11):1808–1829. doi: 10.1161/CIRCRESAHA.116.306923 Search in PMC Search in PubMed View in NLM Catalog Add to search Molecular and Cellular Mechanisms of Cardiovascular Disorders in Diabetes Manasi S Shah Manasi S Shah, M.D. 1 Diabetes Research Center, Albert Einstein College of Medicine, Bronx, New York 2 Department of Medicine, Albert Einstein College of Medicine, Bronx, New York Find articles by Manasi S Shah 1,2, Michael Brownlee Michael Brownlee, M.D. 1 Diabetes Research Center, Albert Einstein College of Medicine, Bronx, New York 2 Department of Medicine, Albert Einstein College of Medicine, Bronx, New York 3 Department of Pathology, Albert Einstein College of Medicine, Bronx, New York Find articles by Michael Brownlee 1,2,3, Author information Copyright and License information 1 Diabetes Research Center, Albert Einstein College of Medicine, Bronx, New York 2 Department of Medicine, Albert Einstein College of Medicine, Bronx, New York 3 Department of Pathology, Albert Einstein College of Medicine, Bronx, New York Address correspondence to: Dr. Michael Brownlee, Diabetes Research Center, Albert Einstein College of Medicine, 1300 Morris Park Avenue, F-530, Bronx, New York 10461-1602, Phone: +1 718 430-3636; Fax: 1 718 430 8570, michael.brownlee@einstein.yu.edu PMC Copyright notice PMCID: PMC4888901 NIHMSID: NIHMS782186 PMID: 27230643 The publisher's version of this article is available at Circ Res Abstract The clinical correlations linking diabetes with accelerated atherosclerosis, cardiomyopathy, and increased post-MI fatality rates are increasingly understood in mechanistic terms. The multiple mechanisms discussed in this review seem to share a common element: prolonged increases in ROS production in diabetic cardiovascular cells. Intracellular hyperglycemia causes excessive ROS production. This activates nuclear poly(ADP ribose) polymerase (PARP), which inhibits GAPDH, shunting early glycolytic intermediates into pathogenic signaling pathways. ROS and PARP also reduce sirtuin, PGC1α, and AMPK activity. These changes cause decreased mitochondrial biogenesis, increased ROS production, and disturbed circadian clock synchronization of glucose and lipid metabolism. Excessive ROS production also facilitates nuclear transport of pro-atherogenic transcription factors, increases transcription of the neutrophil enzyme initiating NETosis, PAD4, and activates the NRLP3 inflammasome. Insulin resistance causes excessive cardiomyocyte ROS production by increasing fatty acid flux and oxidation. This stimulates overexpression of the nuclear receptor PPARα and nuclear translocation of FOXO1, which cause cardiomyopathy. ROS also shift the balance between mitochondrial fusion and fission in favor of increased fission, reducing the metabolic capacity and efficiency of the mitochondrial electron transport chain and ATP synthesis. Mitochondrial oxidative stress also plays a central role in angiotensin II-induced gap junction remodeling and arrhythmogenesis. ROS contribute to sudden death in diabetics after MI by increasing post-translational protein modifications which cause increased ryanodine receptor phosphorylation and downregulation of SERCA2a transcription. Increased ROS also depress autonomic ganglion synaptic transmission by oxidizing the nAch receptor α3 subunit, potentially contributing to the increased risk of fatal cardiac arrhythmias associated with diabetic cardiac autonomic neuropathy. Keywords: Diabetes mellitus, insulin resistance, atherosclerosis, heart failure, reactive oxygen species Subject Terms: Obesity; Cardiovascular Disease; Diabetes, type 2; Vascular Biology; Myocardial Biology; Pathophysiology; Metabolism; Oxidant Stress INTRODUCTION We are in the midst of a global diabetes epidemic. Since 1985, the number of people with diabetes has increased from 30 million to nearly 400 million, and the number of people with diabetes is increasing rapidly in every country. Predictably, as the number of people with diabetes has increased, a world-wide epidemic of diabetic complications has followed. Cardiovascular disorders are by far the leading cause of death in people with diabetes, reducing the median life expectancy for diabetic adults aged 55–64 years by 8 years1. Unlike microvascular complications, which are unique to diabetes, diabetic cardiovascular disorders are clinically similar to cardiovascular diseases in people without diabetes. However, there are important differences with major clinical implications. First, diabetes causes accelerated atherosclerosis, with greater inflammatory infiltrate (macrophages and T lymphocytes), larger necrotic core size, and more diffuse atherosclerosis in the coronary arteries2. In the general U.S. population, deaths due to coronary artery disease have declined substantially over the past decades in the general population. In contrast, in people with diabetes the reduction in deaths due to CAD has been much less dramatic3. Second, diabetes increases both diastolic heart failure with preserved ejection fraction (HFpEF) and systolic heart failure with reduced ejection fraction (HFrEF). Even after adjustment for standard risk factors, diabetes increases heart failure risk four-fold. Contributing factors include diabetes-induced cardiomyocyte dysfunction (cardiomyopathy), impaired microvascular perfusion due to defective endothelial function, increased collagen deposition with fibrosis, and maladaptive remodeling after myocardial infarction, leading to both diastolic and systolic heart failure4. Third, diabetes increases both early (30 days) and late (1 yr) post-MI fatality rates5. The fatality rate for people with diabetes is nearly twice the rate for people without diabetes at both time points. A major cause of post-MI mortality is ventricular arrhythmia. Diabetes-induced downregulation of SERCA2a transcription and increased phosphorylation of the ryanodine receptor by activated CaMKII6, 7 increase intracellular Ca++, contributing to potentially fatal arrhythmias such as premature ventricular complexes and delayed afterdepolarizations, and activation of a mitochondrial/oxidized-CaMKII pathway8 contributes to increased sudden death in diabetics after myocardial infarction. Cardiac autonomic neuropathy, present in almost 50% of patients with diabetes and CAD, is associated with a significantly increased risk for fatal cardiac arrhythmias9. BIOCHEMICAL, MOLECULAR, AND CELLULAR MECHANISMS Conceptual Overview Virtually all the data to be discussed in this section comes from studies of murine models and of cultured cells. Murine models are valuable tools for defining the pathogenesis of diabetic cardiovascular disorders. They have significant limitations, however, and it is important to recognize some of their limits. First, no diabetic animal model, regardless of genetic background, recapitulates the structural and functional alterations of human diabetic cardiovascular disease. Rodent models of diabetes do not develop coronary atherosclerosis with complex plaque formation and plaque rupture, nor do they develop the degree of fibrosis seen in human hearts with late stage diabetes-associated heart failure. Second, most of the mechanistic data currently available comes from studies of the earliest stages of each complication. Murine atherosclerosis, for example, is limited to foam cell accumulation and fatty streaks in a very restricted anatomical distribution. Mechanisms dominant in the pathogenesis of fibrous plaques, complicated lesions, and plaque rupture are likely quite different from those dominant in fatty streak formation. Cultured cell experiments also have important limitations. Cells from species that do not develop diabetic cardiovascular disease are unlikely to respond to hormonal and metabolic perturbations in the same way as human cells. Immortalized human cell lines metabolize glucose and other relevant substrates quite differently from primary human cells. Even primary human cell studies are limited by the lack of important interactions with other cell types within and between different tissues that occur in human cardiovascular disease. For example, the development of human atherosclerosis involves cross-talk among endothelial, smooth muscle, inflammatory and phagocytic cells in the arterial wall, as well as interactions with cells whose functions are altered by diabetes in the bone marrow, visceral fat, and liver. Despite these limitations, however, there has been enormous recent progress in understanding the biochemical, molecular, and cellular mechanisms involved in the pathogenesis of diabetic cardiovascular disorders. Many of these are described in several excellent recent reviews10–12. Hyperglycemia and insulin resistance are the two major consequences of diabetes responsible for cardiovascular disorders in patients with diabetes, and the mechanisms associated with each are presented in the following sections. The multiple mechanisms discussed in this review seem to share a common element: prolonged increases in ROS production in diabetic cardiovascular cells. Physiologic levels of ROS (H 2 O 2) are signaling molecules essential for normal cardiovascular cell homeostasis. However, ROS production at too high a level, for too long, or at an inappropriate subcellular location, leads to impaired cellular function and cardiovascular pathology. Mechanisms of Hyperglycemia-Induced Cardiovascular Damage Increased aldose reductase (AKR1B1) substrate conversion Aldose reductase (AKR1B1), a member of the large aldo-keto reductase superfamily, catalyzes the reduction of a wide variety of hydrophobic and hydrophilic carbonyl-containing compounds– including glucose and several glycolytic intermediates– to their corresponding alcohols. This enzyme is cytosolic, and requires NADPH as a co-factor. In some cell types, where glucose is converted to the sugar alcohol sorbitol, sorbitol is then converted to fructose by another enzyme, sorbitol dehydrogenase, using NAD+ as a cofactor. This series of reactions, termed the polyol pathway, has been implicated in the pathogenesis of several diabetic complications including diabetic cardiovascular disease13. In diabetic apolipoprotein E knockout mice, overexpression of human aldose reductase (hAR) accelerated atherosclerosis and pharmacological inhibition of the enzyme prevented this14. In the same model, diabetes caused sustained activation of the C2H2-type zinc-finger transcription factor Egr-1, with subsequent induction of its downstream targets, the pro-thrombotic protein tissue factor (TF) and the pro-inflammatory vascular cell adhesion molecule-1 (VCAM-1)15. In contrast, in mice expressing physiologic levels of aldose reductase, knockout or pharmacological inhibition of the enzyme unexpectedly caused increased early atherosclerotic lesion size in both diabetic and non-diabetic mice16. Differences in total enzyme activity, cofactor levels, and levels of alternative intracellular substrates, coupled with known differences in enzyme kinetics for different substrates, likely explain these seemingly paradoxical observations. Increased intracellular formation of the major AGE-precursor methylglyoxal The post-translational modifications of proteins called Advanced Glycation Endproducts (AGEs) are formed by glucose-derived dicarbonyls reacting with amino groups of unprotonated lysine and arginine residues of proteins. Methylglyoxal, formed by the non-enzymatic fragmentation of the glycolytic intermediate triose-phosphate, accounts for the majority of hyperglycemia-induced increase in AGE adducts in diabetic tissues17. Intracellular methylglyoxal is detoxified by the glyoxalase system18. The enzyme glyoxalase I, together with glyoxalase II and a catalytic amount of glutathione, reduces this highly reactive α-oxoaldehyde to D-lactate. In cells, methylglyoxal reacts with unprotonated arginine residues to form the major methylglyoxal-derived epitope MG-H1. Intracellular production of AGE precursors damages target cells by three general mechanisms. First, AGE modification of intracellular proteins changes their function. Second, AGE-modification of extracellular matrix components alters their interaction with other matrix components and with integrin matrix receptors. Third, intracellular methylglyoxal increases expression of both the pattern recognition receptor RAGE (Receptor for AGEs) and its major endogenous ligands, the proinflammatory S100 calgranulins19. Ligation of these ligands with RAGE causes cooperative interaction with the innate immune system signaling molecule toll-like receptor 4 (TLR4)20. Expressions of RAGE, S100A8, S100A12, and HMGB1 are all increased by high levels of glucose in cell culture and in diabetic animals. This hyperglycemia-induced overexpression is mediated by ROS-induced increases in methylglyoxal, which increase binding of the transcription factors NFκB and activator protein 1 (AP-1) to the promoters of RAGE and of RAGE ligands, respectively19. Recent work has identified increased methylglyoxal as an important element in the pathogenesis of both diabetic atherosclerosis and diabetic cardiomyopathy. In non-diabetic Apo E null mice, increasing plasma methylglyoxal levels to diabetic levels using a Glo1 inhibitor caused endothelial inflammation and atherogenesis similar to that induced by diabetes21. In human atherosclerotic plaques, MG-H1 levels were associated with rupture-prone plaques having increased levels of the inflammatory mediators IL-8 and MCP-1, and higher MMP-9 activity. MG-H1 was primarily found in lesion macrophages surrounding the necrotic core, and co-localized with cleaved caspase-322. In the diabetic heart, methylglyoxal preferentially reacts with both ryanodine receptor 2, the major myocardial intracellular mediator of calcium-induced calcium release, and with sarco-endoplasmic reticulum Ca++-ATPase (SERCA2a), which is responsible for the synchronized reuptake of released intracellular calcium23. This coordinated process of calcium cycling is critical for efficient cardiac contractions, and diabetes-induced defects caused by increased methylglyoxal adduct formation and increased O-GlcNAcylation likely contribute to impaired systolic function. Increased methylglyoxal production also appears to be responsible for poor cardiac stem cell-mediated repair and angiogenic capacity24. Cardiac stem cells from biopsies of hearts from human diabetics were less able to repair post-infarction damage in immunodeficient mice than cardiac stem cells from non-diabetic patients, and conditioned medium from these cells had less angiogenic capacity. Culture of non-diabetic murine cardiac stem cells in high glucose induced the same cardiac repair and angiogenesis defects seen in human diabetic cells. In both human and mouse cells, overexpression of glyoxalase 1 restored the angiogenic defects24. In diabetic mice with defective post-ischemia hind limb revascularization, overexpression of the methylglyoxal-metabolizing enzyme glyoxalase-1 (GLO1) exclusively in bone marrow cells (BMCs) was sufficient to restore BMC function and neovascularization of ischemic tissue in diabetes25. Increased methylglyoxal also activates the unfolded protein response (UPR) in cardiomyocytes26. While transient activation of the UPR relieves ER stress, prolonged activation of the UPR in CVD triggers apoptosis, mediated by the downstream effector CHOP (C/EBP-homologous protein). CHOP plays a critical role in macrophage apoptosis, a process involved in plaque necrosis in advanced atheromata. In Chop−/−Apoe −/− mice, lesion area plaque necrosis was reduced by 50%. In high fat fed Apo E −/− and LDL receptor −/− mice, CHOP promoted plaque growth, apoptosis, and plaque necrosis27. In cardiomyocytes, methylglyoxal also induces apoptosis via CHOP26. Infusion of methylglyoxal in non-diabetic mice induced cardiomyocyte apoptosis, inflammation, and a significant reduction in LV fractional shortening and LV ejection fraction. Each of these adverse effects was prevented in CHOP−/− mice. In the hearts of diabetic mice, overexpression of GLO1 in the vasculature (Vuleseveic, B. et al., Diabetes 2016, in press) prevented diabetes-induced reduction of myocardial capillary density, increased apoptosis, and loss of cardiac function. Neuregulin production, which transduces signals between the heart’s microvasculature and cardiomyocytes28, eNOS dimerization, and Bcl-2 expression were also maintained in the diabetic GLO1 transgenic hearts. Activation of protein kinase C β, δ, and θ Protein kinase C is a family of protein kinase enzymes with 15 isoforms that are involved in the regulation of protein function. Nine of these 15 PKC isoforms are activated by a lipid second-messenger, diacylglycerol (DAG). Elevated intracellular glucose levels increase DAG levels in a variety of diabetic target tissues, including arterial smooth muscle cells and cardiomyocytes29–31 by de novo synthesis. Hyperglycemia primarily activates the β and δ isoforms of PKC, but increases in activity of several other isoforms have also been found. These PKC isoforms can also be activated by intracellular ROS in the absence of DAG or Ca++. The regulatory domain of these PKC isoforms contains two pairs of zinc fingers with six cysteine residues and two zinc atoms, which can be oxidized by intracellular ROS. Oxidation alters zinc finger conformation and activates PKC32. Many cellular abnormalities involved in diabetic cardiovascular disease have been linked to PKC activation. These include endothelial dysfunction, increased vascular permeability, impaired angiogenesis and increased apoptosis. Molecular mechanisms affected by diabetes-induced PKC activation include alterations in functionally significant enzymatic activities such as mitogen-activated protein kinase (MAPK), cytosolic phospholipase A2, and Na+–K+–ATPase, and alterations in several transcription factors33. Hyperglycemia-induced activation of PKCβ promotes vascular inflammation and acceleration of atherosclerosis in diabetic ApoE null mice by augmenting expression of inflammatory mediators. In addition, it increased macrophage expression of CD11c (integrin, alpha X complement component 3 receptor 4 subunit), chemokines (C-C motif) ligand 2), and interleukin-1β via increased extracellular signal-regulated kinase 1 and 2 (ERK1/2) and Jun-N-terminus kinase-mitogen-activated kinase (JNK kinase)34. In this same diabetic model, PKCβ activation increased transcription of the proinflammatory cytokine IL-18 and inhibited transcription of IL-18-binding protein in the aorta. Diabetic mice showed increased plaque formation, cholesteryl ester content and macrophage infiltration. Treatment with a PKCβ inhibitor prevented these35. PKC-β2 in endothelial cells from transgenic ApoE null mice overexpressing PKCβ decreased insulin-stimulated Akt/eNOS activation and increased basal and angiotensin-induced expression of the vasoconstrictor endothelin-1. These dual effects increased endothelial dysfunction and accelerated atherosclerosis in this model compared with apoE−/− mice36. PKC activity has also been linked to myocardial dysfunction, causing cardiomyopathy and cardiac failure. Ruboxistaurin, a PKC inhibitor, improved the metabolic gene profile and reduced PKC activity in diabetic hearts without altering levels of circulating metabolites. Selective overexpression of PKC-β2 in the myocardium of diabetic mice increased expression of connective tissue growth factor (CTGF) and TGF-β1, cardiomyopathy and cardiac fibrosis37. More recently, activation of PKCα/β in diabetic hearts has been shown to mediate reactivation of fetal splicing programs in diabetic hearts by phosphorylation and upregulation of the RNA binding proteins CELF1 and Rbfox238. Chronic activation of PKC isozymes α, β, and δ promotes diastolic and systolic dysfunction, fibrosis, cardiomyocyte hypertrophy, and apoptosis39. Another PKC isoform, PKC θ, plays crucial roles in the proliferation, differentiation and activation of mature T-cells via activation of several transcription factors in the nuclei of T-cells, including NFAT, c-Jun, c-Fos and AP-1. Diabetes-induced cardiac interstitial fibrosis, reduced contractility, reduced expression of the tight junction maintaining protein ZO-1, and T-cell infiltration were all improved by treatment with an isoform-specific PKC θ inhibitor. Increased protein modification by O-GlcNAc The hexosamine pathway causes reversible post-translational modification of intracellular protein serine and threonine residues by N-acetylglucosamine. In cells damaged by hyperglycemia, excess intracellular glucose provides increased fructose-6-phosphate (Fruc-6-P) which is converted to glucosamine 6-phosphate (Glc-6-P) by the rate limiting enzyme glutamine:fructose-6-phosphate amidotransferase (GFAT). Glucosamine-6-phosphate (Glc-6-P) is further converted to N-acetylglucosamine-6-phosphate (GlcNAc-6-P) and finally to UDP-N-acetylglucoseamine (UDP-GlcNAc). The enzyme O-GlcNAc transferase (OGT) uses UDP-GlcNAc to transfer N-acetylglucosamine to a variety of proteins, resulting in increased protein modification by N-acetylglucosamine. Another enzyme, N-acetylglucosaminidase (O-GlcNAcase or OGA), removes this protein modification. Alternative splicing of the genes encoding the O-linked GlcNAc cycling enzymes OGT and OGA yields isoforms targeted to discrete sites in the nucleus, cytoplasm, and mitochondria. O-GlcNAc serves as a nutrient/stress sensor regulating cellular homeostasis by altering signaling, transcription, metabolism, organelle biogenesis, cytoskeletal dynamics and apoptosis40, 41. The role of the hexosamine pathway in CVD has been reviewed recently42. Studies have linked chronically elevated O-GlcNAc levels to diabetic cardiovascular complications. Adverse cardiac effects of chronically increased O-GlcNAcylation include decreased mitochondrial function, decreased autophagic signaling, and decreased contractile function. In mouse coronary endothelial cells (MCECs) isolated from diabetic mice, O-GlcNAcase protein expression was significantly decreased compared with control MCECs. In contrast, OGT protein expression was markedly increased43. The resultant increased protein modification by O-GlcNAc was responsible for decreased endothelium-dependent relaxation of the coronary arteries and reduced capillary density in the left ventricle. Both of these defects were restored by overexpression of O-GlcNAcase. Decreased endothelium-dependent relaxation of the coronary arteries and reduced capillary density both reflect inhibition of endothelial nitric oxide synthase (eNOS), which is required for endothelium-dependent arterial relaxation and for mobilization of stem and progenitor cells from the bone marrow compartment44. In human arterial endothelial cells, activation of eNOS by phosphorylation at Serine 1177 is inhibited directly by hyperglycemia-induced O-GlcNAcylation at this site45, and indirectly by reduced insulin-stimulated phosphatidylinositol 3-kinase (PI3-K) and Akt activity by O-GlcNAc modification. eNOS activity is also affected by several other post-translational modifications, but the effect of diabetes on these has not yet been determined46. Carotid plaques from diabetic patients have a marked increase of O-GlcNAcylation in both cytoplasm and nuclear compartments of endothelial cells compared with non-diabetic subjects47. Increased O-GlcNAcylation may also contribute to diabetic accelerated atherosclerosis by increasing ubiquitination and proteasomal degradation of the anti-inflammatory NF-κB inhibitory protein A20 in coronary endothelial and smooth muscle cells48. Chronically elevated O-GlcNAc levels also adversely affect myocardial function. Ventricular contraction and relaxation are controlled mainly by release and uptake of Ca2+ by the sarcoplasmic reticulum Ca++-ATPase (SERCA2) pump. In hypertrophied and failing myocardium, SERCA2 protein level and its Ca2+ uptake function are depressed. Overexpression of O-GlcNAc-transferase (OGT) significantly reduced transcription of SERCA2, causing decreased calcium re-uptake and impaired diastolic relaxation49. High glucose also increased O-GlcNAc modification of the calcium/calmodulin-dependent protein kinase IIδ (CaMKIIδ), an enzyme critical for Ca2+homeostasis and reuptake in cardiomyocytes. O-GlcNAc-modified CaMKII at Ser 279 is increased in the heart of diabetic humans and rats7, causing autonomous activation of CaMKII. Thus, CaMKII remains activated even after intracellular Ca++ declines. This contributes to decreased cardiac contractility and potentially fatal arrhythmias such as premature ventricular complexes and delayed afterdepolarizations (Figure 1). Delayed afterdepolarizations are associated with the initiation of long QT-interval arrhythmias such as torsade de pointes. Overexpression of GlcNAcase or inhibition of GlcNAc modification increased expression of SERCA2a, ablated sarcoplasmic reticulum Ca++ leak, improved cardiac contractility, and reduced arrhythmic events. Increased levels of ROS also cause autonomous activation of CaMKII by oxidation of adjacent methionine residues in its regulatory domain50. Activation of this mitochondrial ROS-oxidized CaMKII pathway increased mortality after myocardial infarction in diabetic mouse models8. Figure 1. Hyperglycemia-induced myocardial protein modification by O-GlcNAc causes increased intracellular Ca++ and delayed afterpolarizations. Open in a new tab Increased intracellular glucose flux provides more substrate for the enzyme O-GlcNAc-transferase (OGT). This increases O-GlcNAc modification of calcium/calmodulin-dependent protein kinase IIδ (CaMKII), causing autonomous CaMKII activation. CaMKII increases intracellular Ca++ by phosphorylating ryanodine receptor 2 (RyR). OGT also modifies transcription complex factors regulating expression of sarcoplasmic reticulum Ca2+-ATPase (SERCA2), reducing SERCA2A expression and contributing toincreased intracellular Ca++. Increased O-GlcNAc modification of these proteins causes delayed afterdepolarizations in cardiomyocytes. PLB, phospholamban. Mitochondrial O-GlcNAc transferase (OGT) is increased in diabetic cardiac mitochondria, while O-GlcNAcase (OGA) is reduced, causing increased O-GlcNAcylation of cardiac mitochondrial proteins. Inhibition of OGA and the resulting increased mitochondrial protein modification by O-GlcNAc increases oxygen consumption and reduces reserve capacity6. Reduced bioenergenic reserve capacity makes cells more sensitive to stress and cell death. Different hyperglycemia-induced pathogenic mechanisms reflect a single upstream process: overproduction of ROS A single upstream hyperglycemia-induced process — overproduction of superoxide by the mitochondrial electron transport chain—activates all four mechanisms described in the previous sections51, 52 (Figure 2). Enhanced intracellular glucose transport and oxidation leads to mitochondrial overproduction of superoxide52–54. This can, in turn, activate other superoxide production pathways that may amplify the original damaging effect of hyperglycemia55. Examples of amplification mechanisms include ROS-mediated uncoupling of eNOS dimers to eNOS monomers in endothelial cells, activation of various NADPH oxidase isoforms in cardiovascular cells, and increased mitochondrial fission mediated by the rho-associated protein kinase ROCK156, 57. The initiating role of mitochondrial ROS is suggested by the observation that cells lacking mitochondrial electron transport chain function (ρ 0 cells)53 fail to increase ROS production in response to high glucose. Figure 2. Four hyperglycemia-induced pathogenic mechanisms are activated by overproduction of ROS. Open in a new tab Increased intracellular glucose flux causes mitochondrial overproduction of reactive oxygen species (ROS), which can further amplify ROS production by activating NADPH oxidases and uncoupling eNOS. Stable ROS species diffuse into the nucleus, where they cause DNA damage and activation of poly(ADP ribose) polymerase (PARP). PolyADP-ribosylation of glyceraldehyde-3-dehydrogenase (GAPDH) by PARP reduces GAPDH activity, which causes upstream accumulation of early glycolytic intermediates which are diverted into four pathogenic signaling pathways. AKR1B1, aldose reductase; Gln, glucosamine; GFAT, glutamine fructose-6-phosphate amidotransferase; UDP-GlcNAc, uridine diphosphate N-acetylglucosamine; DHAP, dihydroxyacetone phosphate; DAG, diacylglycerol; PKC, protein kinase C; NF-κB, nuclear factor-κB; AGEs, advanced glycation end-products; RAGE, Receptor for Advanced Glycation Endproducts. In mitochondria, increased superoxide causes the release of Fe2+ from ferritin and iron sulfur cluster–containing proteins. Interaction of this released free iron with diffused mitochondrial superoxide-derived hydrogen peroxide forms hydroxyl radicals, the only ROS species capable of cleaving bonds in macromolecules58. This results in ROS-mediated DNA double strand breaks in the nucleus which activate DNA repair mechanisms, including the enzyme poly(ADP ribose) polymerase 1 (PARP1). Activation of PARP1 inhibits the key glycolytic enzyme glyceraldehyde-3-phosphate dehydrogenase (GAPDH) by polyADP-ribosylation, and depletes intracellular NAD+ by degrading it to ADP-ribose and nicotinamide. Inhibition of GAPDH activity causes upstream accumulation of early glycolytic intermediates which are diverted into the four pathogenic signaling pathways52, 59. Diversion of glucose increases polyol pathway flux, while diversion of fructose-6-phosphate increases hexosamine pathway activity. Diversion of glyceraldehyde-3-phosphate to α-glycerol phosphate activates PKC, and reduced activity of GAPDH diverts glyceraldehyde-3-phosphate to the highly reactive α-dicarbonyl methylglyoxal, which increases expression of the receptor for advanced glycation end-products (RAGE) and its activating ligand S100A8/9. Together, these diversions and pathway activations lead to cellular dysfunction, inflammation, apoptosis, and fibrosis in cells exposed to excessive glucose flux. The central importance of ROS in initiating each of these processes is illustrated by the fact that each can be prevented when hyperglycemia-mediated ROS generation is curtailed by transgenic expression of the enzyme superoxide dismutase52, 59. In arteries from patients undergoing coronary artery bypass surgery, the levels and activity of NAD(P)H oxidase protein subunits (p22phox, p67phox, and p47phox) were significantly increased. The endothelium was an additional net source of superoxide production because of dysfunctional endothelial NO synthase60. Overexpression of the mitochondrial isoform of SOD also prevents hyperglycemia-induced inhibition of the antiatherogenic enzyme prostacyclin synthase53. In diabetes, inhibition of prostacyclin synthase causes the common prostacyclin and thromboxane precursor prostaglandin H 2 to be shunted towards thromboxane synthesis. Activation of thromboxane receptors triggers vasoconstriction, platelet aggregation, increased expression of leukocyte adhesion molecules, and apoptosis61. In mice overexpressing the mitochondrial isoform of the superoxide scavenging enzyme catalase (mCAT) in macrophages, lesional macrophage accumulation was successfully suppressed, causing a significant reduction in lesional area. The mCAT lesions had fewer monocyte-derived cells, fewer Ly6c(hi) monocyte infiltration into lesions, and lower levels of monocyte chemotactic protein-1(MCP-1). The decrease in lesional MCP-1 was associated with the suppression of other markers of inflammation and with decreased phosphorylation of RelA (NF-κB p65), indicating decreased activation of the proinflammatory NF-κB pathway. Thus, mitochondrial overproduction of ROS in lesional macrophages amplifies atherosclerotic lesion development by promoting NF-κB-mediated entry of monocytes and other inflammatory processes62. Nox4 expression and activity are also increased in cardiomyocytes exposed to high glucose63 and in the heart of diabetic mouse models64. Transgenic overexpression of the antioxidant enzymes Mn-SOD and catalase reduced ROS, and prevented diabetes-induced abnormalities in cardiac contractility in an animal model of diabetic cardiomyopathy65, 66. Physiologic ROS production is essential for normal intracellular signaling In normal cardiovascular physiology, ROS production is coupled to circadian clocks and metabolic networks, and ROS species (H 2 O 2) function as signaling molecules essential for normal cellular homeostasis67, 68. Physiologic ROS (H 2 O 2) signaling is essential for normal intracellular communication, cell differentiation, autophagy, response to insulin and growth factor stimulation, and the generation of physiologic inflammatory responses67. Enhanced production of H 2 O 2 from a mitochondrial source of superoxide is observed when flow rate is increased in human coronary resistance vessels. Hydrogen peroxide (H 2 O 2) hyperpolarizes and dilates human coronary arterioles through opening of Ca 2+-activated K+ channels. Catalase, a scavenger of H 2 O 2, greatly inhibited this flow-induced dilation69. Similarly, H 2 O 2 exerted a beneficial effect on vasodilator function and reduced blood pressure in transgenic mice with endothelium-targeted Nox4 overexpression70. In the heart, low levels of hydrogen peroxide induce proliferation of mouse embryonic stem (ES) cells as well as neonatal cardiomyocytes, and ROS induce expression of cardiac-specific genes, transcription factors and growth factors in ES cells. These effects are dampened by free radical scavengers. ROS also act as transducers of mechanical strain-induced cardiovascular differentiation of embryonic stem cells (ES)71. ROS-dependent activation of integrins and subsequent induction of PI3K/Akt signaling are also involved in cyclic strain-mediated cardiomyogenesis72. Localized ROS production appears to play a role in stretch-induced augmentation of cardiac contractile activity as well73. Physiologic excitation-contraction coupling in heart muscle may also involve ROS signaling. Prosser et al. showed that physiologic stretch rapidly activates NOX2 located on sarcolemmal and t-tubule membranes in cardiomyocytes. The local ROS produced sensitizes ryanodine receptors (RyRs) in the sarcoplasmic reticulum (SR). This triggers a burst of Ca2+ sparks, thereby increasing Ca2+ signaling sensitivity in healthy cardiomyocytes. Thus, ROS production at an inappropriate place or time, for too long, at too high a level or of inappropriate forms lead to impaired cellular function and pathological gain of function, while ROS production at the right time, place, level, and duration plays a crucial role in physiological homeostasis74. Decreased Nrf2 activity Mechanisms which can damage the cardiovascular system are normally counterbalanced by protective mechanisms that maintain homeostasis. In diabetes, however, the hyperglycemia, quantitative and qualitative changes in lipids, and insulin resistance which together promote tissue injury often inhibit these protective mechanisms39. The number of recognized ROS-regulating enzymes has increased in recent years74. Examples include superoxide dismutases, catalases, glutathione peroxidases, glutathione reductase, thioredoxins, thioredoxin reductases, methionine sulfoxide reductases, and peroxiredoxins74. The activity of these enzymes is largely determined by ROS-induced changes in their transcription. Increased transcription of many of these antioxidant enzymes is mediated by the transcription factor Nrf2 (nuclear erythroid–related factor 2), a member of the cap ‘n’ collar (CNC) subfamily of basic region leucine zipper (bZip) transcription factors75. By regulating oxidant levels and oxidant signaling, Nrf2 participates in the control of inflammasome signaling, the unfolded protein response, apoptosis, mitochondrial biogenesis and stem cell regulation. Nrf2 also increases transcription of glyoxalase 1 (Glo1), the rate-limiting enzyme of the glyoxalase system which prevents posttranslational modification of proteins by methylglyoxal, the major AGE precursor76. It also increases transcription of the rate-controlling enzyme in the non-oxidative branch of the pentose phosphate pathway, transketolase. Activation of transketolase by the lipid-soluble thiamine derivative benfotiamine inhibits three of the major hyperglycemia-driven pathways implicated in the pathogenesis of vascular diabetic vascular damage (the DAG-protein kinase C (PKC) pathway, the methylglyoxal-advanced glycation end product (AGE) formation pathway, and the hexosamine pathway) and inhibits hyperglycemia-induced NFκB activation77. Transketolase activation by benfotiamine also prevented high glucose-induced arterial endothelial cell death78, improved diastolic and systolic function, prevented left ventricular end-diastolic pressure increase and chamber dilatation, improved cardiac perfusion, and reduced cardiomyocyte apoptosis and interstitial fibrosis79. Nrf2 is expressed constitutively, and its intranuclear levels are controlled post-translationally. In the absence of inducers, Nrf2 associates with the redox-sensitive protein Keap1 (Kelch-like erythroid cell-derived protein with CNC homology-associated protein 1), where it is rapidly polyubiquinated by Keap1-associated cullin-3 (Cul3)–RING E2 ubiquitin ligase proteins and degraded by proteasomes. ROS oxidation of critical cysteine thiols of Keap1 or the reaction of these thiols with ROS-generated electrophiles such as methylgloxal from glycolysis-derived triose phosphate and the lipid peroxidation product 4-hydroxynonenal causes the release of bound Nrf2 protein. Phosphorylation of Nrf2 by protein kinases such as CK2 may help target Nrf2 to the nucleus. After forming heterodimers with small Maf proteins, Nrf2 binds to the antioxidant response element (ARE) to induce transcription of its target genes. Export of Nrf2 from the nucleus is controlled by phosphorylation75. Src family members such as Fyn phosphorylate Nrf2 at Tyr568, causing export from the nucleus and degradation80. Reduction of Nrf2 protein in the cytosolic compartment is mediated by β-transducin repeat-containing protein (β-TrCP), a substrate adaptor for the S-phase kinase-associated protein-1 (Skp1)–Cul1–F-box protein (SCF) E3 ubiquitin ligase which targets Nrf2 phosphorylated byGSK3β to the proteosome81 (Figure 3). Figure 3. Diabetes reduces Nrf2 protein in diabetic heart. Open in a new tab Nuclear erythroid–related factor 2 (Nr2), the master regulator of antioxidant gene expression, associates with the redox-sensitive protein Kelch-like erythroid cell-derived protein with CNC homology-associated protein 1(Keap1), where it is rapidly polyubiquinated by Keap1-associated cullin-3 (Cul3)–RING E2 ubiquitin ligase proteins and degraded by proteasomes. ROS oxidation of critical cysteine thiols of causes release of bound Nrf2 protein. Phosphorylation of Nrf2 by protein kinases such as CK2 help target Nrf2 to the nucleus. Nrf2 forms heterodimers with small Maf proteins which bind to the antioxidant response element (ARE) in its target gene promoters. After export of Nrf2 from the nucleus, cytosolic Nrf2 is phosphorylated by GSK3β. This phosphorylated Nrf2 recognized by β-transducin repeat-containing protein (β-TrCP), a substrate adaptor for the S-phase kinase-associated protein-1 (Skp1)–Cul1–F-box protein (SCF) E3 ubiquitin ligase which targets Nrf2 phosphorylated by GSK3β to the proteosome. Modification of critical Keap1 cysteine thiols by the dietary isothiocyanate sulforaphane also releases Nrf282. In endothelial cells, sulforaphane prevented hyperglycemia-induced activation of the hexosamine and PKC pathways and prevented increased cellular accumulation and excretion of the major AGE precursor methylglyoxal. In the aortae of diabetic mice, sulforaphane treatment restored aortic levels of Nrf2 and Nrf2-dependent antioxidant gene expression, preventing diabetes-induced increases in wall thickness, fibrosis, inflammation (TNFα and VCAM-1 expression), apoptosis, and increased cell proliferation83. Diabetic cardiomyopathy was also prevented in mouse models by sulforaphane treatment84. Restoration of Nrf2 activity prevented these, as well as preventing diabetes-associated inflammation, fibrosis, and cardiac lipid accumulation. In addition, decreased autophagy was restored. In hearts from diabetic patients, Nrf2 protein is dramatically reduced85. In mice, cardiac Nrf2 protein was similarly reduced after 5 months of diabetes85. What mechanisms are responsible for this long-term decrease in Nrf2 protein content and function in diabetic heart? Since hyperglycemia-induced mitochondrial overproduction of ROS has been shown to increase GSK3β activity by inhibiting Akt1-dependent phosphorylation of GSK3β at Serine 958, increased proteasomal degradation of cytosolic Nrf2 phosphorylated by GSK3β mediated by the β-transducin repeat-containing protein (β-TrCP)/Skp1−Cul1−F-box protein (SCF) E3 ubiquitin ligase is currently the only mechanism supported by experimental data81. NFAT activation The transcription factor Nuclear Factor of Activated T Cells (NFAT) has been implicated in the development of diabetic cardiovascular complications. Many of the cell types involved in diabetic cardiovascular disease express one or more of the four calcium-dependent NFAT isoforms, NFATc1-NFATc4. In resting cells, NFAT proteins are phosphorylated and are located in the cytoplasm. In diabetes, intracellular calcium is increased by increased ROS. Increased intracellular calcium then activates NFAT by increasing NFAT dephosphoryation by the Ca 2+/calmodulin-dependent serine/threonine phosphatase calcineurin, which facilitates NFAT translocation into the nucleus. Once in the nucleus, NFAT interacts with coregulators to achieve optimal NFAT activation86. In the nucleus, ADP-ribosylation mediated by PARP-1 acts as a molecular switch to positively regulate NFAT-dependent cytokine gene transcription87. In diabetes, nuclear PARP-1 is activated by intracellular hyperglycemia via ROS-induced DNA strand breaks (Figure 4). Figure 4. Increased mitochondrial oxidation of glucose or fatty acids activates NFAT-mediated transcription of genes promoting diabetic atherosclerosis and heart failure. Open in a new tab Mitochondrial overproduction of ROS causes increased intracellular Ca++, which activates the calcium-activated neutral cysteine protease calpain. Calpain then activates the Ca 2+/calmodulin-dependent /CaM) (Ca2+)serine/threonine phosphatase calcineurin. Dephosphorylation facilitates nuclear translocation of the transcription factor Nuclear Factor of Activated T Cells (NFAT). In the nucleus, NFAT interacts with polyADP-ribose polymerase (PARP), which increases NFAT transcriptional activity via NFAT polyADP ribosylation. MCP-1, monocyte chemoattractant protein-1; ICAM-1, Intercellular Adhesion Molecule 1; IL-6, Interleukin-6. In diabetic mice, activated NFATc3 plays a role in accelerated atherosclerosis. Activated NFATc3 induces arterial cell expression of the pro-inflammatory matrix protein osteopontin (OPN), a cytokine that promotes atherosclerosis and diabetic vascular disease. NFAT inhibition effectively reduced osteopontin, IL-6, monocyte chemotactic protein 1, intercellular adhesion molecule 1, CD68 and tissue factor expression in the arterial wall, and lowered plasma IL-6 in diabetic mice88. In diabetic ApoE−/− mice, inhibition of NFAT-signaling completely suppressed a 2.2-fold increase in atherosclerotic plaque area. Inhibition of NFAT also reduced lipid content in the plaque of diabetic mice independent of plasma glucose and lipid levels89. NFATc3 activated by increased mitochondrial ROS production also increases arterial vasoconstrictor reactivity to endothelin-190. NFAT activation also appears to play a role in cardiac hypertrophy, fibrosis, and cardiomyocyte apoptosis91. In the diabetic heart, NFAT is activated by the calcium-activated neutral cysteine protease calpain, which in turn activates calcineurin. In cardiomyocytes, high glucose increases calpain activity, which activates NFAT-dependent cardiac hypertrophy and heart failure92. In two mouse models of diabetes, cardiac-specific deletion of calpain reduced myocardial hypertrophy and fibrosis, leading to the improvement of myocardial function. Calpain activation correlated with increased activity of NFAT and NFκB, consistent with calpain’s role in activation of calcineurin and degradation of the cytosolic NF-κB inhibitor, NF-κB inhibitor alpha (IκBα)93, 94. Increased PAD4 and NETosis activation Atherosclerotic lesions begin with the deposition of cholesterol-rich lipoproteins in the artery wall, followed by the entry of inflammatory leukocytes into lesions. Early lesions are characterized by infiltration of neutrophils and lipid-filled monocyte-derived macrophages. In CAD, inflammation becomes chronic, and lesions progress rather than resolve. During lesion progression, neutrophils and macrophages continue to accumulate, and recruit pro-inflammatory IL-17-producing T-cells. Smooth muscle cell proliferation and altered matrix production occur. Advanced lesions containing a variety of inflammatory cell types accumulate a necrotic core, which contains dead macrophage foam cells, prothrombotic molecules, and matrix proteases. A large necrotic core predisposes to release of thrombogenic material due to protease erosion or rupture of the plaque95. Recently, Warnatsch et al96 showed that neutrophils prime macrophages for pro-inflammatory responses in atherosclerotic plaques. This priming is mediated by neutrophil extracellular traps (NETs), extracellular webs of DNA bound to cytotoxic histones which are released by activated neutrophils97. This process, called NETosis, appears to follow a coordinated multi-step process: histone citrullination, chromatin decondensation, migration of elastase and other granule enzymes into the nucleus, disintegration of the nuclear membrane and release of DNA, histones and granule proteins into the extracellular space98. The release of NETs prime macrophages to produce pro-IL-1β, which is cleaved to mature pro-inflammatory IL-1β by caspase 1. Caspase 1, in turn, is secreted by macrophages in response to activation of the NLRP3 (NOD-like receptor family, pyrin domain-containing 3) inflammasome (discussed in the following section)96 (Figure 5). Figure 5. Diabetes increases neutrophil extracellular traps (NETs), priming macrophages for inflammation. Open in a new tab Increased ROS increase transcription and activation of peptidylarginine deiminase 4 (PAD4), the enzyme which initiates formation and release of NETs by citrullination of histones. Released NETs prime macrophages to produce pro-IL-1β, which is cleaved to mature pro-inflammatory IL-1β by caspase 1 secreted by macrophages in response to NLRP3 inflammasome activation. Cit, citrullination; αH4R, αhistone 4 Arginine; αH4cit, αhistone 4 with arginine residues converted to citrulline. Apoe−/− mice with deletions of two serine proteases that localize to NETs, neutrophil elastase and proteinase-3 (Apoe−/−/Ela2−/−/Prtn3−/−), developed dramatically smaller atherosclerotic lesions compared to Apoe−/− control animals, despite similar lipid concentrations and leukocyte counts in blood. Triple mutant mice had no NETs, lower systemic IL-1β concentration, and fewer lesional IL-17 producing T cells99. NETs are also pro-thrombotic96. Neutrophils from type 1 and type 2 diabetic humans as well as mice are primed to produce NETs, and have increased expression of peptidylarginine deiminase 4 (PAD4), the enzyme critical for histone citrullination-mediated chromatin decondensation and NET formation100, 101. Increased PAD4 transcription is driven by NFκB102, which is chronically activated in diabetes103. Hyperglycemia-induced ROS likely activate PAD4 as well, by increasing intracellular Ca++ concentration. Other consequences of increased intracellular ROS such as PKC activation increase levels of the NETosis-priming cytokine tumor necrosis factor-α 7. Normal resolution of inflammation caused by infiltration of neutrophils and macrophages involves a switch from synthesis of arachidonic acid–derived prostaglandins and leukotrienes to synthesis of lipoxins, which stop neutropil recruitment. At the same time, increased production of resolvins and protectins from omega-3 polyunsaturated fatty acids induce neutrophil apoptosis. Phagocytosis of these apoptotic neutrophils by macrophages causes a switch to an anti-inflammatory macrophage phenotype, which secretes anti-inflammatory and reparative cytokines104. The mechanism responsible for defective inflammatory lesion resolution in atherosclerosis and diabetes is not known. NLRP3 inflammasome activation Increased expression of the NRLP3 inflammasome components Nlrp3 and ASC was found in monocytes from new, untreated patients with type 2 diabetes105. Along with increased expression, there was increased inflammasome activation. Consistent with this, the drug-naïve type 2 diabetic patients had significantly higher serum levels of the pro-inflammatory cytokines IL-1β and IL-18 than did healthy subjects105. In a Type 2 diabetic rat model, excessive activation of NLRP3 was associated with cardiac inflammation, cell death, disorganized ultrastructure, and fibrosis. NLRP3 gene silencing ameliorated cardiac inflammation, apoptosis, fibrosis, and left ventricular cardiac dysfunction106. The NLRP3 inflammasome is formed by oligomerization of inactive NLRP3, associated with apoptosis-associated speck-like protein (ASC), and procaspase-1. This complex, in turn, catalyzes the conversion of procaspase-1 to caspase-1, which contributes to the production and secretion of mature pro-inflammatory IL-1β and IL-18107. The transcription factor NF-κB, which is chronically active in mononuclear cells from diabetic patients and in vascular endothelial cells of diabetic rats103 promotes transcription of NLRP3, proIL-1β, and proIL-18108. These proteins remain in the cytoplasm in inactive forms. A second signal activates the NLRP3 inflammasome by facilitating the oligomerization of inactive NLRP3, apoptosis-associated speck-like protein (ASC), and procaspase-1. Many, but not all reported activators of the NRLP3 inflammasome converge on excessive production of ROS (Figure 6). The ROS-induced reduction of intracellular NAD+ levels, discussed previously, reduces the activity of SIRT2, causing accumulation of acetylated α-tubulin109. Acetylated α-tubulin regulates the transport of mitochondria and helps form an efficient interaction between the adaptor protein ASC and NLRP3. ROS also activate NLRP3 by opening the cell membrane calcium channel TRPM2, increasing Ca2+ influx. The mitochondrial membrane phospholipid cardiolipin, discussed in the following section, also activates the NLRP3 inflammasome after translocation to the outer mitochondrial membrane, where it binds to NLRP3110. It is currently not known what causes cardolipin to move to the outer mitochondrial membrane, but ROS-induced cardiolipin remodeling may be one explanation111, 112. Activation of NLRP3 in diabetic heart and artery may also reflect the increased cell surface CD 36 induced by insulin-resistance (discussed in the following section), which facilitates internalization of oxidized LDL (ox-LDL) and intracellular conversion of ox-LDL to cholesterol crystals. Figure 6. NLRP3 inflammasome activation in diabetic atherosclerosis. Open in a new tab Intracellular hyperglycemia-induced ROS in monocytes and vascular endothelial cells increases RAGE expression, which heterodimerizes with TLR4. Signaling from this complex causes NFκB-mediated transcription of inactive NOD-like receptor family, pyrin domain-containing 3 (NRLP3), pro-IL-1β, and pro-IL-18. Increased intracellular Ca++ triggers oligomerization of inactive NLRP3, associated with apoptosis-associated speck-like protein (ASC), and pro-caspase-1. This activated inflammsome complex catalyzes the conversion of procaspase-1 to caspase-1, and of pro-IL-1β and pro-IL-18 to mature IL-1β and IL-18. S100A8/12, calgranulin A/B heterodimer ligand for RAGE. Consequences of oxidative miRNA modification and ROS-induced downreguation of specific miRNAs ROS oxidatively modify certain microRNAs. ROS can hydroxylate guanine to produce 8-oxo-7,8,-dihydroguanosine (8OHG)113. Oxidized miR-184 was shown to mismatch with the 3′ UTRs of the anti-apoptotic proteins Bcl-xL and Bcl-w, thereby sensitizing cardiomyocytes to apoptosis. Administration of oxidatively modified miR-184 increased infarct size in an ischemia/reperfusion model. Increased expression of several miRNAs which may contribute to diabetic atherosclerosis is correlated with insulin resistance, but virtually nothing is known about the mechanisms regulating expression of these miRNAs. A recent study of miRNA-QTL using liver tissue from 424 morbidly obese, insulin-resistant subjects identified an association of miR-128-1 and miR-148a expression with SNPs linked to abnormal human blood lipid levels. In vivo studies in the Apoe−/− mouse verified that miR-128-1 and miR-148a do regulate cholesterol/lipid and energy homeostasis114. MiR-128-1 also regulates expression of SIRT1, the NAD+-dependent lysine deacetylase, and these miRNAs also affect expression of several components of the heterotrimeric AMP-activated protein kinase (AMPK). Diabetes-induced ROS have been implicated in the downregulation of miR-499, -133a, and -373 in diabetic cardiomyocytes115. Although most individual miRNAs target hundreds of specific mRNAs, thereby coordinately regulating complex gene networks, transgenic mice overexpressing miR-499 displayed pathological cardiac remodeling accompanied by cardiac dysfunction116. Downregulation of MiR-133a in a normal adult genetic background was sufficient to induce cardiac hypertrophy, and its downregulation is a prerequisite for the development of apoptosis, fibrosis, and prolongation of the QT interval in animal models117. Insulin-resistance associated down-regulation of miR-128-1 and perhaps also miR-34a may also play a critical role in diabetic atherosclerosis. MiR-34a directly targets and decreases SIRT1 expression. SIRT1 activity would be reduced further by miR34a targeting of the rate-limiting enzyme in the salvage pathway for NAD+ biosynthesis, NAMPT118. In a well-established porcine model of diabetic atherosclerosis which develops complex atherosclerotic plaques resembling human complex plaques, the expression and activity of the NAD+-dependent deacetylase SIRT1 were markedly reduced119. As mentioned earlier, ROS-activated PARP depletes NAD+ in the process of synthesizing ADP-ribose by cleaving NAD+ into ADP-ribose and nicotinamide. This depletion of NAD+ would also inhibit SIRT1’s enzymatic activity. In the porcine diabetic atherosclerosis model, the activity of AMPK was also reduced by 40%, due to decreased phosphorylation at Thr-172. Since SIRT1-mediated deacetylation is necessary for full function of the AMPK-activating kinase LKB1, AMPK-dependent inhibition of fatty acid synthesis would be reduced by diabetes-induced increased production of mitochondrial ROS120. In addition, overexpression of miR-451 in diabetes reduces AMPK activating phosphorylation of LKB1 by reducing levels of the LKB1 scaffold protein Calcium-binding protein 39 (Cab39)120. Normally, SIRT1 activity and AMPK activity suppress the activating cleavage processing of sterol regulatory element binding proteins (SREBP-1 and -2). In the porcine model of diabetic atherosclerosis, the cleavage processing of SREBP-1 and -2 and expression of their target genes were increased in endothelial cells and infiltrating macrophages of both fatty streaks and of advanced lesions with fibrous caps, necrotic cores, and cholesterol cores. SIRT1 and AMPK activity were also reduced119. Increased SREBP-1a in macrophages also directly upregulates transcription of NLRP3 inflammasomes121, and in porcine diabetic atherosclerosis, increased NLRP3 was found both in macrophages of advanced lesions, and in endothelial cells and smooth muscle cells. The changes found in porcine diabetic atherosclerosis were also present in coronary atherosclerosis samples from diabetic humans. Mechanisms of Insulin Resistance-Induced Cardiovascular Damage Insulin resistance is operationally defined as an impaired ability of different cell types to respond normally to insulin. In people with type 2 diabetes, insulin resistance is caused by underlying heritable factors and is exacerbated by environmental factors such as obesity. Insulin signaling is initiated by its binding to tyrosine kinase insulin receptors (IR). Insulin receptor activation of the phosphatidylinositol 3-kinase- (PI3K)–Akt pathway is responsible for most of the metabolic actions of insulin. The IR has two splice isoforms, both of which can phosphorylate at least six known insulin receptor substrate proteins (IRS). These IRSs are capable of interacting with eight known forms of the PI3K regulatory subunit. PI3K regulatory subunits in turn can associate with three forms of the PI3K catalytic subunit, and the product of PI3K activity can then activate three isoforms of Akt. The combinatorial possibilities of the IR-IRS-PI3K-Akt part of the insulin signaling pathway alone exceed 1,000. When differential compartmentalization, stoichiometry, and kinetics of the various downstream signaling components are included, this number increases dramatically122. Many of these steps are negatively regulated by action of phosphatases or inhibitory proteins. The complexity of this signaling system is essential to selectively mediate the large variety of known responses to insulin123. Because of this complexity, the molecular pathogenesis of insulin resistance is still incompletely understood. It has been suggested that insulin resistance is a physiologic mechanism that protects the cardiovascular system from nutrient-induced injury, and that therapies attempting to override it with intensive insulin therapy in an effort to lower plasma glucose levels could therefore be harmful124. Since insulin resistance in critical tissues appears to be pathway-specific for glucose metabolism, attempts to treat insulin resistance with intensive insulin therapy may well increase deleterious effects of insulin on lipid and lipoprotein metabolism125. However, physiologic hyperinsulinemia in response to pathway-specific insulin resistance in liver without exogenous insulin is responsible for the flooding of the heart with triglyceride-derived fatty acids125. Insulin resistance itself also appears to be harmful due to associated metabolic inflexibility, in which nutrient overload and heightened substrate competition result in mitochondrial dysfunction, impaired fuel switching, and energy dysregulation. Metabolic inflexibility occurs early in the course of glucose intolerance, and obesity-induced perturbations in substrate switching persist in isolated muscle mitochondria. Current evidence suggests that myocardial cells (and those of striated muscle and liver) function optimally when they retain their capacity to switch freely between oxidative substrates in response to nutritional physiologic cues126. Increased endothelial fatty acid oxidation, increased mitochondrial ROS, and diabetic atherosclerosis Insulin resistance increases fatty acid oxidation in arterial endothelial cells. In two insulin- resistant nondiabetic animal models, inhibition of either free fatty acid (FFA) release from adipocytes or FFA oxidation in arterial endothelium prevented the increased production of ROS and its damaging effects127. In arterial endothelial cells, this FFA-induced increase in ROS activates the same damaging pathways seen with high glucose: AGEs, PKC, the hexosamine pathway (GlcNAc), and NFκB. FFA-induced overproduction of superoxide also activates a variety of proinflammatory signals previously implicated in hyperglycemia-induced vascular damage, and inactivates two important antiatherogenic enzymes, prostacyclin synthase and eNOS127. King first proposed that in diabetes, a pathway-specific insulin resistance in diabetic vascular cells reduces the antiatherogenic actions of insulin, while leaving insulin’s proatherogenic actions unaffected128. Subsequently, this group showed in apoE null mice that specific overexpression of PKCβ2 in endothelial cells caused pathway-specific insulin resistance by inhibiting downstream PI3K-Akt signaling, thereby inhibiting eNOS activation36. NO released from endothelial cells is a potent inhibitor of platelet aggregation and adhesion to the vascular wall. Endothelial NO also controls the expression of genes involved in atherogenesis. It decreases expression of MCP-1, and of surface adhesion molecules such as CD11/CD18, P-selectin, VCAM-1 and ICAM-1. Since PI3K-Akt signaling also causes the sequestration of FOXO transcription factors in the cytoplasm, thereby preventing transcription of a variety of pro-atherogenic genes129, inhibition of PI3K-Akt signaling by PKCβ2 would be expected to increase transcription of these pro-atherogenic genes. Surprisingly, in human arterial endothelial cells, cells from diabetics showed higher levels of the activating Ser 1177 phosphorylation of eNOS, despite increased PKCβ expression, reduced PI3K-Akt signaling, and increased levels of oxidative stress. PKCβ activity was associated with lower flow-mediated dilation which was reversible by PKC inhibition130. These results are consistent with hyperglycemia- and fatty acid-induced ROS overproduction. Increased ROS activates PKC, which like PI3K-Akt, phosphorylates eNOS at Ser 1177. However, increased ROS also reduce eNOS activity by two mechanisms: oxidation of tetrahydrobiopterin (BH4), the essential cofactor of endothelial nitric oxide synthase (eNOS), and uncoupling of dimeric eNOS to monomeric eNOS. These changes convert the nitric oxide-producing dimeric eNOS to superoxide-producing monomeric eNOS131. In contrast to IR inhibition of PI3K-Akt signaling, insulin signaling through MAPK and ERK is unaffected by insulin resistance. In arterial endothelial cells, this resulted in increased expression of pro-atherogenic proteins like endothelin-1. In apoE null mice with specific overexpression of PKCβ2 in endothelial cells, the severity of atherosclerotic lesions was increased by 70%36. Increased myocardial fatty acid oxidation, increased mitochondrial ROS, and diabetic cardiomyopathy The normal adult heart consumes a large amount of energy and uses a variety of substrates to produce ATP. 70% of total energy is derived from mitochondrial oxidative phosphorylation of fatty acids, with most of the rest derived from glucose oxidation132. Fatty acid uptake in the heart is mediated by the cluster of differentiation 36 (CD36), a scavenger receptor class B type I, and fatty acid translocase (FAT), while glucose uptake is mediated by insulin-stimulated translocation of glucose transporter 4 (GLUT4). In diabetes, myocardial insulin resistance with impaired insulin signaling decreases GLUT4 translocation to the cell surface, while increasing cell-surface CD36133. At the same time, insulin resistance in adipose tissue and liver causes an increased delivery of triglycerides and triglyceride-rich VLDL to the heart, where heart-specific LpL releases fatty acids. In diabetes, heart-specific LPL activity is upregulated134, 135. Thus, insulin resistance causes a major increase in fatty acid flux into the myocardium. Similarly, murine models of insulin-dependent and non-insulin-dependent diabetes had serum triglyceride levels 2.6- and 4.2-fold higher, respectively, than normal mice, and 7- and 3.5-fold higher levels of heart microsomal CD36, respectively, than control mice136. Thus, there is a dramatic shift away from glucose utilization and an overreliance on fatty acids as the energy source in the diabetic heart. Even in TIDM, where substrate-induced insulin resistance is much less severe than the intrinsic insulin resistance of T2D, studies have shown a dramatic impairment in glucose uptake in cardiomyocytes because of diminished insulin-induced transcription and translocation of GLUT-4137, 138. In streptozotocin-induced TIDM animals, levels of GLUT-4 were significantly reduced, forcing the cardiomyocytes to rely on fatty acids139. Both increased fatty acid uptake and increased fatty acid oxidation in the diabetic heart are mediated in part by increased PPARα activity140, 141. Like the other two members of this nuclear receptor family, PPARβ/δ and PPAR γ, PPARα forms heterodimers with the retinoid X receptors (RXRs), which bind to PPAR-responsive elements (PPREs). PPARα activation by binding of its endogenous ligand (1-palmitoyl-2-oleoyl-sn-glycerol-3-phosphocholine (16:0/18:1-GPC)142 causes bound transcriptional repressors (such as SMRT and N-CoR) to be exchanged for transcriptional co-activators—primarily PGC-1α and PGC-1β—which form active transcriptional complexes with CBP/p300, SRC-1, and a number of other proteins. PPARα transcriptional activity is further increased by PKCβ2-mediated Ser phosphorylation143. Since increased ROS activate PKCβ2 and several other PKC isoforms (discussed previously), this may contribute to the upregulation of PPARα-target genes in diabetic heart. PPARα (−/−) mice were protected from the development of diabetes-induced cardiac hypertrophy, while the combination of diabetes and the MHC-PPARα genotype resulted in a more severe cardiomyopathic phenotype than either did alone. Cardiomyopathy in diabetic MHC-PPARα mice was accompanied by myocardial long-chain triglyceride accumulation. Reactive oxygen intermediates were identified as candidate mediators of cardiomyopathic effects in MHC-PPARα mice140. Recently, the muscle ring finger-3 (MuRF3) ubiquitin ligase was shown to stabilize PPARα activity in vivo144. Both diabetes and treatment with the cardiotoxic ROS-producing chemotherapeutic agent doxorubicin increase MuRF3 expression. MuRF3 expression was increased in hearts from insulin resistant high fat fed mice, and was shown to mono-ubiquitinate cardiac PPARα. Cardiomyopathy resulting from this increased fatty acid flux is called “lipotoxicity”. Two general mechanisms have been proposed to explain lipotoxicity. One mechanism proposes that in diabetic cardiomyocytes, increased fatty acid flux results in increased synthesis of diacylglycerols, diglycerides, and ceramide145. Aberrant accumulation of these signaling intermediates, particularly ceramide, appears to be cardiotoxic146. In transgenic mice with heart-specific LpL overexpression, more lipid accumulated in hearts expressing the transgene, and myocytes were enlarged and exhibited abnormal architecture. Hearts of transgenic mice were dilated, and left ventricular systolic function was impaired135. Inhibition of de novo ceramide biosynthesis reduced fatty acid oxidation and increased glucose oxidation in isolated perfused LpL hearts, and improved systolic function and prolonged survival rates of cardiac specific LpL overexpressing mice146. The second general mechanism proposes that increased fatty acid oxidation results in increased ROS production, which causes cardiomyopathy by a variety of downstream actions. Although the oxidation of fatty acids normally yields significantly more energy per carbon atom than does the oxidation of glucose, oxidative phosphorylation capacity is impaired in hearts from insulin-resistant db/db mice. However, H 2 O 2 production is increased147. In human myocardium from insulin resistant diabetic patients, mitochondrial H 2 O 2 production was also increased during oxidation of lipid-based substrates compared with carbohydrate-based substrates148. The explanation for this apparent paradox is that the major site of electron leakage from increased fatty acid oxidation is electron transfer flavoprotein (ETF), which receives electrons from the FADH 2 formed during the first oxidation step of β-oxidation54 (Figure 7). Increased myocardial ROS production in the diabetic heart occurs very early, before accumulation of TGs are evident, due to increased fatty acid oxidation and resultant oxidative damage to mitochondrial cardiolipin (CL)149, the specific phospholipid of mitochondrial membranes. Cardiolipin is important for efficient electron flux, ATP synthesis, and reduced ROS formation. In addition, cardiolipin is involved in mitochondrial-mediated apoptosis, and it plays a critical role in regulating mitochondrial fission and fusion150–152. In non-diabetic heart, the major species of cardiolipin contains four linoleic acids (tetra 18:2 cardiolipin). This unique acyl composition is not derived from de novo synthesis of CL, but rather from a remodeling process that involves phospholipases and acyltransferase-transacylases. In diabetic myocardium from murine models with insulin resistance (ob/ob, db/db, high-fat diet) and in models of severe insulin-deficient type 1 diabetes (STZ), the fatty acyl content of the more saturated 18:2 cardiolipin is dramatically reduced, while the content of longer chain, more unsaturated fatty acyl cardiolipin is substantially increased112. Because of this increase in highly unsaturated side chains, diabetic heart cardiolipin is more vulnerable to oxidative damage. Cardiac overexpression of cardiolipin synthase increases tetra 18:2 cardiolipin in diabetic mice and prevents diabetes-induced changes in cardiolipin lipid remodeling. The cardiolipin deficiency and profound remodeling caused by diabetes and by diet-induced obesity is caused by ROS-induced transcription of Acyl-CoA:lysocardiolipin acyltransferase 1 (ALCAT1)(Figure 7). ALCAT1 catalyzes the transfer of linoleoyl-CoA onto mono- or dilysocardiolipin. Overexpression of ALCAT1 caused cardiolipin deficiency and fatty acid compositional changes similar to diabetes and obesity, with increased production of ROS, while ALCAT1 deficiency increased levels of tetra 18:2 cardiolipin in mouse heart and reduced ROS production111. Figure 7. Increased cardiac fatty acid oxidation, ROS formation and cardiolipin remodeling. Open in a new tab Insulin resistance-induced increased cardiac β oxidation of free fatty acids (FFA) causes greater H 2 O 2 production than increased glucose oxidation because of increased electron leakage from the electron transfer flavoprotein (ETF) complex. These ROS activate Acyl-CoA:lysocardiolipin acyltransferase 1 (ALCAT1). ALCAT1, located in the mitochondrial associated membrane (MAM) of the ER, which causes pathologic remodeling of cardiolipin from tetra 18:2 cardiolipin to cardiolipin with highly unsaturated fatty acid side chains and cardiolipin deficiency due to oxidative damage. This reduces ETC electron flux, ATP synthesis, and further increases ROS. High levels of ROS, through JNK signaling, are also a major proximal activator of the forkhead box O (FOXO) family of transcription factors. Increased ROS stimulate FOXO translocation from the cytosol into the nucleus by increasing FOXO GlcNAcylation, direct cysteine oxidation, and CaMKII activation153, 154. Numerous other signals and post-translational modifications can also regulate FOXOs, and FOXOs regulate a multitude of diverse processes by interacting with many different transcription factors and other nuclear proteins. In the hearts of diabetic mice and of mice with high fat diet-induced insulin resistance, FOXO proteins were persistently activated155. This persistent activation was associated with downregulation of insulin receptor substrate 1 (IRS1), reduced activity of IRS1 and its downstream target Akt, and the development of cardiomyopathy. In cardiomyocyte-specific FOXO1 knockout mice fed a high fat diet, neither insulin resistance nor cardiomyopathy occurred. Altered mitochondrial dynamics Mitochondria dynamics are the continuous processes of mitochondrial fusion, fission, biogenesis and mitophagy which maintain optimal cellular bioenergetics and ROS homeostasis. The topic of altered mitochondrial dynamics and cardiovascular disease has been reviewed quite recently156, and the reader is referred to that comprehensive review for more extensive discussion. In the context of insulin resistance and diabetic cardiomyopathy, the balance between mitochondrial fusion and fission is altered in favor of increased fission and decreased fusion. This increased ratio of fission to fusion causes increased production of ROS and decreased mtDNA, reducing the metabolic capacity and efficiency of the mitochondrial electron transport chain and ATP synthesis. The primary regulators of mitochondrial fusion are dynamin-related GTPases termed mitofusins 1 and 2 (MFN1 and MFN2) and optic atrophy protein 1 (OPA1) located in the inner mitochondrial membrane. In humans with Type 2 diabetes, expression of the fusion protein MFN1 is decreased in myocardium157. The primary regulator of mammalian mitochondrial fission is the GTPase dynamin-related protein 1 (DRP1), which is recruited from the cytosol to the mitochondrial outer membrane where it binds to four DRP1 receptors- — mitochondrial fission factor (MFF), Mitochondrial dynamics protein of 49 kDa (MID49) and 51kDa (MID51), and FIS1158, 159. Oligomerization of Drp1 is believed to provide the mechanical force to constrict mitochondrial membranes and to fragment the organelle. Drp1 activity is regulated by the Ca 2+/calmodulin-dependent serine/threonine phosphatase calcineurin. Dephosphorylation of Drp1 at Ser 656 by calcineurin activates DRP1 and increases mitochondrial fission. Calcineurin itself is activated by the calcium-activated neutral cysteine protease calpain. Since increased ROS activate calpain and calcineurin in cardiomyocytes160, 161, the increased mitochondrial ROS production from increased fatty acid oxidation described earlier could explain, in part, the observed increased fission in hearts of insulin resistant diabetics (Figure 8). Increased mitochondrial fragmentation also occurs in mitochondria from mouse coronary endothelial cells (MCECs), where the level of the fusion protein OPA1 is decreased, and the level of DRP1 is increased162. Insulin resistance itself may also contribute to the increased ratio of fission to fusion. In non-diabetic non-insulin resistant cardiomyocytes, insulin increased OPA-1 levels and mitochondrial fusion, and reduction of either OPA-1 or MFN2 with siRNA prevented this163. 52 Figure 8. Diabetes-induces increased mitochondrial fission. Open in a new tab Increased ROS from either excess glucose in vascular cells or fatty acids in cardiomyocytes increases intracellular Ca++. Increased Ca++ activates the calcium-activated neutral cysteine protease calpain, which activates the Ca 2+/calmodulin-dependent/CaM) (Ca2+)serine/threonine phosphatase calcineurin. Calcineurin dephosphorylates the GTPase dynamin-related protein 1 (DRP1), which is then recruited from the cytosol to the mitochondrial outer membrane where it binds to four DRP1 receptors- — mitochondrial fission factor (MFF), Mitochondrial dynamics protein of 49 kDa (MID49) and 51 kDa (MID51), and FIS1. Drp1 oligimerization provides the mechanical force to constrict mitochondrial membranes and fragment the organelle (mitochondrial fission). Increased fission causes further increases in ROS production and mitochondrial dysfunction. FIS1, Mitochondrial fission 1 protein. Circadian clocks, Metabolic Networks, and Diabetic Cardiovascular Disease Cell autonomous circadian clocks encoded by a transcription-translation feedback loop exist in most peripheral tissues including liver, fat, muscle, and pancreatic β cells. In the liver, they promote lipid catabolism, gluconeogenesis, and mitochondrial biogenesis during sleep/fasting, and lipogenesis, glycogen synthesis, and cholesterol and bile acid synthesis in the awake/feeding state164. In adipose tissue, they promote lipid catabolism during sleep/fasting, and lipogenesis when awake and feeding. In muscle, they promote oxidative metabolism during sleep/fasting, and fatty acid update during feeding. Different tissues exhibit distinct clock-controlled properties164. The core transcriptional components of the mammalian circadian clock are the transcriptional activators CLOCK and BMAL1, which co-activate transcription at E-box-containing gene promoters, including the period (Per) and cryptochrome (Cry) genes. PER and CRY proteins form a complex that is imported into the nucleus and inhibits their own transcription. During the night/fasting, these proteins are degraded, resulting in reactivation of their transcription in the early morning. A short feedback loop consisting of the orphan nuclear receptor REV-ERBα and the retinoic acid orphan receptors (ROR) α and β activate and repress, respectively, Bmal1 transcription. REV-ERBα controls oscillation, abundance and activation of SREBPs through modulation of INSIG2. PGC-1alpha, a transcriptional coactivator that regulates mitochondrial biogenesis and energy metabolism, is rhythmically expressed in the liver and skeletal muscle of mice. PGC-1alpha stimulates the expression of clock genes including Bmal1 and Rev-Erbα through co-activation of the retinoic acid orphan receptors (ROR) family of nuclear receptors. Mice lacking PGC-1alpha show abnormal diurnal rhythms metabolic rate. The disruption of physiological rhythms in these animals is correlated with aberrant expression of clock genes and those involved in energy metabolism. Analyses of PGC-1α-deficient fibroblasts and mice with liver-specific knockdown of PGC-1α indicate that it is required for cell-autonomous clock function165. Metabolic networks influenced by oscillation of clock genes and downstream transcription factors also reciprocally influence clock function164, 166. Major metabolic coupling signals include NAD+, SIRT1, AMPK, and ROS164, 166, 167. NAD+ activates SIRT1, which deacetylates and thereby inhibits the CLOCK:BMAL1 complex. PARP-1 also ADP-ribosylates the CLOCK protein. AMPK controls proteolytic degradation of PER and CRY. CLOCK mutant mice develop hypertriglyceridemia, and hepatic steatosis168, while ablation of Bmal1 increases arterial superoxide, uncoupling of eNOS169, and arterial wall lesion extent after endothelial injury170. Circadian transcription factors also regulate levels of the pro-thrombotic plasminogen activator inhibitor type 1 (PAI-1)171, 172. Circadian rhythmicity of the clock-dependent oscillator krupple-like factor 15 (Klf15) controls cardiac expression of Kv channel-interacting protein 2 (KChIP2), a critical subunit required for generating the transient outward potassium current and QT-interval duration. Deficiency or excess of Klf15 causes loss of rhythmic QT variation, abnormal repolarization and enhanced susceptibility to ventricular arrhythmias173. In the diabetic cardiovascular system, however, continuously increased ROS production caused by hyperglycemia and insulin resistance depletes NAD+ by activating PARP1, which cleaves NAD+ into ADP-ribose and nicotinamide in the process of synthesizing ADP-ribose. This depletion of NAD+ inhibits SIRT1’s enzymatic activity, which normally deacetylates and activates both PGC1α and LKB1, the kinase that activates AMPKα2. Decreased SIRT1 activity would decrease activity of LKB1, PGC1α, and AMPK174, causing decreased mitochondrial biogenesis, increased ROS production, and a profoundly disturbed clock synchronization of glucose and lipid metabolism. Recently, the flavone nobiletin was discovered to improve the amplitude of the clock repressor PERIOD 2, acting through RORs. Nobiletin treatment of wild type mice on a high fat diet prevented weight gain and visceral adiposity, while treatment had no effect in Clock homozygous mutant mice. Nobiletin had potent insulin-sensitizing actions, consistent with the observed reduction in hepatic and serum triglycerides, and reversed high fat diet-induced alterations in clock gene expression in liver and fat164, 175. Conclusion Knowledge about the biochemical, molecular, and cellular mechanisms responsible for cardiovascular disorders in diabetes has increased enormously in recent years. As a consequence, the clinical correlations linking hyperglycemia and insulin resistance with accelerated atherosclerosis, heart failure (both HFpEF and HFrEF), and increased post-MI fatality rates are increasingly understood in mechanistic terms. For each of these clinical entities, the multiple mechanisms discussed in this review seem to share a common element: prolonged increases in ROS production in diabetic cardiovascular cells. In normal cardiovascular physiology, ROS production is coupled to circadian clocks and metabolic networks, and ROS species (H 2 O 2) function as signaling molecules essential for normal cellular homeostasis67, 68. In contrast, ROS production at too high a level, for too long, or at an inappropriate location, leads to impaired cellular function and cardiovascular pathology74. Intracellular hyperglycemia causes excessive ROS production by increasing electron leak from the mitochondrial electron transport chain. This can be amplified by ROS-induced uncoupling of nitric oxide synthase (eNOS) and activation of NADPH oxidases. Increased mitochondrial ROS cause DNA double strand breaks damage by releasing free iron and H 2 O 2, which diffuse into the nucleus. These DNA strand breaks activate latent nuclear poly(ADP ribose) polymerase (PARP). PolyADP-ribosylation of glyceraldehyde-3-dehydrogenase (GAPDH) by PARP leads to partial inhibition of this key glycolytic enzyme. As a result, early glycolytic intermediates accumulate and are then diverted into pathogenic signaling pathways. These pathways include increased substrate conversion by the enzyme AKR1B1, increased formation of methylglyoxal, the major advanced glycation product precursor, activation of protein kinase C isoforms β,δ, and θ, and increased protein modification by O-GlcNAc. Activation of GS3Kβ by excessive ROS production reduces intracellular Nrf2, the transcription factor that normally increases expression of ROS-degrading enzymes. Nrf2 also regulates expression of the rate-limiting enzyme of the glyoxalase system, GLO1, which degrades the major AGE precursor methylglyoxal. PARP activation also degrades NAD+, reducing activity of the NAD+-dependent sirtuin deacetylases. SIRT1 normally deacetylates and activates both PGC1α and LKB1, the kinase that activates AMPKα2. Reduced SIRT1 deacetylation decreases activities of LKB1, PGC1α, and AMPKα2174, causing decreased mitochondrial biogenesis, increased ROS production, and a profoundly disturbed circadian clock synchronization of glucose and lipid metabolism. Excessive ROS production also facilitates nuclear transport of NFAT transcription factors by activating the Ca 2+/calmodulin-dependent serine/threonine phosphatase calcineurin. Excessive ROS also increase transcription of Pad4, thee neutrophil enzyme initiating NETosis, which primes macrophages for pro-inflammatory responses in atherosclerotic plaques. Many reported activators of the NRLP3 inflammasome also converge on excessive production of ROS. Insulin resistance causes excessive ROS production by cardiomyocytes because it increases fatty acid uptake and oxidation. Thus, there is a dramatic shift away from glucose utilization and an overreliance on fatty acids as the energy source in the diabetic heart. Although oxidation of fatty acids normally yields significantly more energy per carbon atom than does the oxidation of glucose, oxidative phosphorylation capacity is impaired in hearts from insulin-resistant animal models and human diabetics. Despite a reduced oxidative phosphorylation capacity mitochondrial H 2 O 2 production is increased during oxidation of lipid-based substrates compared with carbohydrate-based substrates. This reflects the fact that the major site of electron leakage from increased fatty acid oxidation is electron transfer flavoprotein (ETF), which receives electrons from β oxidation-generated FADH 2, independent of the electron transport chain complexes. Increased myocardial ROS production in the diabetic heart occurs very early, before accumulation of TGs are evident, due to increased fatty acid oxidation and resultant ROS-induced cardiolipin remodeling by ALCAT1. Finally, in the context of insulin resistance and diabetic cardiomyopathy, the balance between mitochondrial fusion and fission is altered in favor of increased fission and decreased fusion. This increased ratio of fission to fusion causes increased production of ROS and decreased mtDNA, reducing the metabolic capacity and efficiency of the mitochondrial electron transport chain and ATP synthesis. Reactive oxygen intermediates have also been identified as candidate mediators of the cardiomyopathic effects of cardiac overexpression of the nuclear receptor PPARα140. High levels of ROS are also a major proximal activator of the forkhead box O (FOXO) family of transcription factors. The increased ROS associated with insulin resistance stimulate translocation of FOXO from the cytosol into the nucleus, and persistent myocardial FOXO1 activation in insulin resistant diabetic mice causes cardiomyopathy. In contrast to the mechanisms underlying diabetic accelerated atherosclerosis and diabetic cardiomyopathy, much less is known about the mechanisms responsible for diabetes-associated increases in both early and post-MI fatality rates. A major cause of post-MI mortality is ventricular arrhythmia. Mitochondrial ROS play a central role in gap junction remodeling and fatal arrhythmia generation176. ROS also increase myocardial protein modification by O-GlcNAc, which downregulates SERCA2a transcription and increases phosphorylation of the ryanodine receptor by autonomously activating CaMKII6, 7. This contributes to potentially fatal arrhythmias such as premature ventricular complexes and delayed afterdepolarizations. Overexpression of GlcNAcase normalized CaMKII activity, restored SERCA2a transcription, and reduced arrhythmic events. In addition, ROS can activate CaMKII directly through a mitochondrial/oxidized-CaMKII pathway8. Activation of this mitochondrial ROS-oxidized CaMKII pathway increased mortality after myocardial infarction in diabetic mouse models8. Deficiency or excess of the clock-dependent oscillator krupple-like factor 15(Klf15) causes loss of rhythmic QT variation, abnormal repolarization and enhanced susceptibility to ventricular arrhythmias in mice173. While human repolarization occurs through a complex interaction of multiple repolarizing ionic currents, rather than the outward potassium current which is central in mice, altered klf15 expression may also contribute to human arrhythmogenesis by the ROS-dependent mechanisms described above. klf15 controls rhythmic expression of key enzymes involved in normal glucose, lipid, and nitrogen homeostasis. Disruption of branched chain amino acid (BCAA) catabolism caused by reduced klf15 suppresses cardiac mitochondrial respiration and induces superoxide production177. In human diabetic hearts, regional myocardial autonomic denervation may also predispose patients to malignant arrhythmias178. Much work remains to be done in this area. However, increased ROS have been shown to depress sympathetic ganglion synaptic transmission by oxidizing residue Cys 239 of the nAch receptor α3 subunit179. This subunit is also present in parasympathetic neurons. 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15780	https://www.talkshop.com.au/what-speech-sounds-should-my-child-be-able-to-say-at-their-age/	02 7209 3838 Call Us What Speech Sounds Should My Child Be Able to Say at Their Age? For Parents What to do if you think your child is falling behind? Did you know there are 44 speech sounds in the English language? 20 Vowel Sounds and 24 Consonant Sounds.1 As children develop their speech sounds, they tend to follow a pattern or sequence2. At first baby vocalisations are reflexive, such as crying, coughing and burping. Between 8 – 16 weeks babies will typically start producing their first vowel sounds “ah” and “oo”. These are the ‘cute’ coo’s, goo’s and gah’s we all enjoy. As children listen to the language being spoken around them and to them, their early speech sounds develop. Vocal play typically takes place from 16 – 26 weeks. Babies will start to experiment with making sounds of different pitches and volumes. From 6 months babies start babbling and producing their first consonant sounds, most commonly “ma ma ma” “da da da” and “bah bah bah”. Their babbling should show the beginnings of conversation from 10 months. A child’s speech sound development continues for approximately 7 years until the muscles of their lips, teeth and tongue have developed and learnt the sophisticated movements necessary to produce all 44 sounds. Not sure if your child needs speech therapy? Book a free consultation. Book a Free Consultation Speech Sound Development Chart \| \| \| --- \| \| 3 months \| Making cooing sounds. E.g. ahhh \| \| 5 months \| Child is laughing and making playful sounds. \| \| 6 months \| Speech-sounding babbling sounds start, e.g. me, pah, bah, da \| \| 1 year \| Babbling begins to sound like real words as they string more sounds together. E.g. upup, babada, mama \| \| 2-3 years \| Able to say the sounds: p, b, m, d, n, h, t, k, g, w, ng, f, y in words. By 3 years old, familiar people should be able to understand your child’s speech. \| \| 4 years \| Able to say the sounds: l, j, ch, s, v, sh, z in words. Your child may still be making mistakes on: s, sh, ch, j, ng, th, z, l, and r. Most people should understand your child’s speech at this age. \| \| 5 years \| Able to say r, zh (e.g. ‘measure’), th (voiced – e.g. in ‘that’ and ‘there’) \| \| 6 years \| Able to say a voiceless th (e.g. in ‘thunder’ and ‘thick’) \| Consonant vs Vowel Sound development Looking at the chart above you can be forgiven for thinking “what about the 20 vowel sounds?” The chart above is almost exclusively about consonants. For most English language learners the most common speech sound errors are consonant sounds. It is much less common for a child to struggle with vowel sounds. In fact, vowel sound errors are often indicative of a more complex condition called Childhood Apraxia of Speech (CAS)3. Learn more about CAS here. Consonant Sound Development Children usually follow a similar sequence when developing the 24 consonant sounds. This progression is categorised into the “early 8”, “middle 8 and “late 8”. \| Early 8 Sounds \| Middle 8 Sounds \| Late 8 Sounds \| --- \| Emerging between ages 1-3. Consistent production ~ age 3 \| Emerging between ages 3-6 ½. Consistent production ~ age 5 ½ \| Emerging between ages 5 -7 ½ Consistent production ~ age 7 ½ \| \| /m/ as in “mama” /b/ as in “baby” /y/ as in “you” /n/ as in “no” /w/ as in “we” /d/ as in “daddy” /p/ as in “pop” /h/ as in “hi” \| /t/ as in “two” /ng/ as in “running” /k/ as in “cup” /g/ as in “go” /f/ as in “fish” /v/ as in “van” /ch/ as in “chew” “j” as in “jump” \| /sh/ as in “sheep” /s/ as in “see” /th/ as in “think” /th/ as in “that” /r/ as in “red” /z/ as in “zoo” /l/ as in “like” /zh/ as in “measure” \| Common Speech Sound Errors. While a child is learning a speech sound it is normal for them to make mistakes. It is helpful to be able to recognise these mistakes for a couple of reasons. First, if you recognise one of these speech sound errors it is very helpful to correct them in the moment. Doing so helps your child hear the correction and practice correcting it. Second is to notice if they get stuck. Muscle movements depend on neural pathways. If a child creates strong neural pathways that create a speech sound incorrectly there is an increasing chance this will become a speech disorder and they will need speech therapy to help reprogram new neural pathways for the correct sound. Below is a list of common speech sound errors broken into age categories where the sound error is developmentally normal. If your child continues to make these errors into older ages it may be necessary to get help from a speech pathologist. 1-2 years old: When your child replaces a quiet (voiceless) sound to a noisy (voiced) one, this is called voicing. For example, if ‘pea’ becomes ‘bee’. A long sound with airflow (e.g. s, sh, f) becomes a short sound with no airflow (e.g. t, d). This is called stopping. For example, if ‘sea’ becomes ‘tea’ If your child leaves off the end of a word. This is called final consonant deletion. For example ‘cat’ becomes ‘ca’ When back sounds (e.g. k, g) are said at the front of the mouth and turn into t and d. This is called velar fronting. For example, when ‘car’ becomes ‘tar’ The tongue moves forward in the mouth and turns sh into s. This is called palatal fronting. For example ‘shed’ becomes ‘said’ When a child leaves out a non-stressed syllable in a word, this is called weak syllable deletion. For example, tomato becomes mato. When two consonants in a word and reduced to only one consonant (for example ‘tree’ becomes ‘tee’) this is called cluster reduction When r and l sounds become w and y (for example ‘rabbit’ becomes ‘wabbit’), this is called gliding. When th becomes f (e.g. ‘thunder’ becomes ‘funder’) and v (‘them’ becomes ‘vem’) 2-3 years old: We still expect to see lots of the same processes in 1-2 year olds like: voicing, stopping, final consonant deletion, velar fronting, palatal fronting, weak syllable deletion, cluster reduction, gliding and ‘th’ becomes ‘v’ or ‘f’ (see explanation of these processes under the 1-2 years old category) 3-4 years old: Your child should start fixing many of the above processes. It is still common for them to have: weak syllable deletion, cluster reduction, gliding and ‘th’ becomes ‘v’ or ‘f’ (see explanation of these processes under the 1-2 years old category) 4-6 years old: Now your child should begin mastering many consonant sounds and be easily understood in conversation. We may still expect to see some gliding and difficulty with the ‘th’ sound Ideally, we would start working on ‘r’ and ‘th’ sounds before your child starts school, so that they can have the best start possible for their reading and writing skills. 6-7 years old: The only difficult sounds for children this age may be ‘th’, all other processes should be CORRECTED 7-8 years old: NONE! Your child should be speaking fluently and be able to pronounce all consonants accurately. How can I support my child’s development of these speech sounds? Nurturing and responsive environments in which children are frequently exposed to lots of language can dramatically improve a child’s speech development6. Here’s 5 helpful tips you can follow to help your child develop to their potential7. Imitate and play with their early speech sounds. Treat every communication attempt as meaningful. Try to respond with affection and attention each time. Don’t be afraid to sound silly as you repeat back to the child their sounds and early words Enrich your day to day interactions with language. Use every opportunity to comment and narrate what is going on in the world round you and your child. Make it fun! Stimulate interaction with games like peek-a-boo. Sing nursery rhymes with your child. Follow your child’s interests and play preferences or make up nonsense words can help them learn more about different sounds and rhyming words. Model, model, model! Repeat clear and exaggerated sounds as an opportunity for your child to imitate you. If your child mispronounces a sound, simply model the correct pronunciation and DO NOT focus on the mistake – make this a positive experience. E.g. If your child says ‘I heard funder’ instead of ‘thunder’, you could respond with interest by saying ‘Oh, you heard thunder? I heard it too. The thunder was so loud!’ Practice sounds face-to-face with your child or side by side in the mirror. This will allow them to see your mouth movements as you say each sound. Get them to copy your mouth movements. This way you can easily see what they are doing with their lips, tongue and teeth. You can add some silly faces in there to make it extra fun! What can you do if you suspect your child’s speech sounds aren’t following the typical developmental pattern? If you notice that your child continues to have speech sound errors past the developmentally appropriate age then early intervention is best. Start by including as many of the tips above in day-to-day life and book in to get assistance from a speech pathology service. Speech Pathologists are specifically trained to be able to identify speech sounds disorders and are uniquely skilled in therapy approaches to target a speech sound error. If your child is heading towards school age and is often difficult to understand or showing mispronunciation of words, we recommend bringing them in for a speech sound assessment. Feel free to book an initial consultation on our website or give us a call on (02) 7209 3838 if you have any questions! Reference Rao, C. S. (2015). The intelligibility of English sounds: A study of phonetics. English for Specific Purposes World, ISSN, 1682-3257. Berk, L. E. (2013). Child Development (9th Edition). Upper Saddle River, New Jersey: Pearson Education. Bloom, L., & Lahey, M. (1978). Language development and language disorders. New York, NY, US: John Wiley & Sons. American Speech-Language-Hearing Association. (2007). Childhood apraxia of speech [Technical Report]. Shriberg, L. (1993). Four new speech and voice-prosody measures for genetics research and other studies in developmental phonological disorders. Journal of Speech, Language, and Hearing Research, 36, 105–140. ASHA. n.d. Speech Sound Disorders. [online] Harrison, L. J., & McLeod, S. (2010). Journal of Speech Language and Hearing Research, 53(2), 508. Risk and protective factors associated with speech and language impairment in a nationally representative sample of 4- to 5-year-old children. Levy, A., 2018. speech sounds – What to Expect and How to Help your Child Learn Sounds \| Clear Communicators Speech Therapy. [online] Clear Communicators Speech Therapy. word, phonology, language, speech, brain, early 8 sounds, consonant, sound, pathology, language development, syllable, voice, english language, tongue, reading, babbling, mouth, research, phoneme, vowel, lip, literacy, phonics, email, phonological development, learning, auditory processing disorder, lisp, education, spoken language, teacher, signs and symptoms, audiology, apraxia of speech, consonant cluster, developmental verbal dyspraxia, cluster reduction, phonological awareness, spelling, fluency, therapy, linguistics, risk, phone, pdf, medicine, communication disorder, disability, language acquisition, speech delay, otitis media, intelligibility, phonetics, airflow, infant, speech production, health, language disorder, speech perception, development, stress, classroom, attention, pronunciation, fricative, disease, apraxia, memory, vocabulary, evidence, complexity, skill, ear, perception, library, conversation, experiment, terms of service, autism spectrum, data, noise, genetics, palate, tone, grammar, dialect, gesture, phonemic awareness, information, evaluation, community, hearing, individualized education program, sentence, health care, phrase, pinterest, human voice, developmental disability, advertising, mirror, knowledge, speech acquisition, reading comprehension, discrimination, developmental psychology, faq, muscle, auditory system, dysarthria, parent, united states, affricate, frequency, symbol, learning disability, feedback, nonverbal communication, child development, communication, articulation, preschool, chart, rhyme, thumb, developmental speech sound chart, behavior, cognition, adult, experience, life expectancy, caregiver, middle age, psychosocial, sleep, generation, identity, erik erikson, puberty, dementia, europe, society, autonomy, human, young adult, psychoanalysis, culture, pubmed, guilt, infection, intimate relationship, psychology, cognitive development, european union, patient, theory, mind, sigmund freud, failure, shame, prenatal development, black sea, dna, book, early childhood, neolithic, arabs, france, ukraine, western europe, tool, science, psychosexual development, toilet training, leadership, identity crisis, caucasus, germany, food, anxiety, iberian peninsula, steppe, virtue, attachment theory, protein, mood, peer review, identity formation, personal development, emotion, cell, obesity, pediatrics, internet, balkans, art, music, pubmed central, pathogen, sequence, bronze age, activities of daily living, danube, confidence, mediterranean sea, prehistory, wisdom, philosophy, understanding, gender, concept, clothing, creative commons, jean piaget, cure, agriculture, health professional, sex, england, old age, mass, iron, africa, volga, weight, evolution, mathematics, civilization, learning theory, depression, plos one, risk factor, wales, innovation, hypothesis, yersinia pestis, confusion, aggression, copper, pneumonia, genotype, lecanemab, archaeology, youth, trust, john bowlby, encyclopædia britannica, population, genetic history of europe, dream, eurasian steppe, movement disorder, tooth, quiz, feeling, intelligence, lewy body, fidelity, length, nature, physician, pannonian basin, middle east, nutrition, time, organization, new york city, asia, mental health, social emotional development, anatolia, romania, transgender, russia, denmark, ratio, central europe, signs, family, orient, medication, mixture, fear, psychologist, javascript, hormone, levant, quality of life, horse, citation, cancer, doubt, wildfire, sexual identity, nursing home, curiosity, policy, stroke, national institutes of health, malnutrition, manuscript, tissue, metal, immune system, epidemic, southern ukraine, mosaic Frequently Asked Questions How do infants progress in vocalization stages? Infants progress in vocalization stages by going through cooing, babbling, and eventually forming words. Cooing starts at around 2 months, babbling around 6 months, and words typically start forming around the 12-month mark, showcasing their language development. What cues promote sound development in toddlers? Promoting sound development in toddlers can be facilitated by providing a nurturing environment with engaging activities, ample social interaction, consistent routines, healthy nutrition, regular exercise, and age-appropriate toys that stimulate sensory and cognitive skills. Encouraging exploration, offering positive reinforcement, and ensuring sufficient rest and sleep are also crucial for toddlers' overall wellbeing and development. Which phonemes emerge by specific ages in children? By the age of 3, children typically develop sounds like /p, b, m, n, t, d, k, g, h, w/. By 6, they usually acquire /f, v, sh, ch, j, l, s, z, y/. By 7-8, they should master all English phonemes. Can infants distinguish phonetic contrasts by month eight? Infants can distinguish phonetic contrasts by month eight, demonstrating cognitive development in language processing and auditory discrimination skills early in their developmental stages. These abilities are vital for language acquisition and communication proficiency in the future. What milestones indicate typical phonological development? Typical phonological development milestones include babbling, first words around 12 months, development of consonant sounds, expanding vocabulary, correct sound productions by age 3-4, and mastering more complex sounds like blends and clusters. Phonological awareness and speech intelligibility continue to improve as a child grows. How does babbling evolve into meaningful speech? Babbling evolves into meaningful speech through the development of motor skills and cognitive abilities. Initially, infants experiment with sounds, progressing to imitate familiar words and eventually forming coherent speech patterns through language exposure and practice. When do children typically start combining syllables? Children typically start combining syllables around 9 to 18 months of age, as they transition from single-word utterances to more complex forms of speech. This stage marks an important milestone in language development, showcasing their evolving ability to form more sophisticated vocalizations and express themselves with increasing fluency. What sound patterns do babies exhibit at different stages? Babies exhibit various sound patterns at different stages of development. Initially, they produce cooing and babbling sounds, later progressing to more complex sound combinations. By 12 months, most babies can say simple words like "mama" or "dada." This gradual development showcases their growing language skills. When should a child's babbling turn into words? Children's babbling typically transitions into recognizable words around 10-14 months of age. Consistent exposure to language, responsive interactions, and encouragement help facilitate this developmental milestone. Encouraging attempts at communication and mimicking sounds can support a smooth progression from babbling to spoken words. How do nursery rhymes impact speech sound mastery? Nursery rhymes play a vital role in developing speech sound mastery in children. Repetition in rhymes aids in phonological awareness, enhancing speech clarity. Rhymes also introduce new vocabulary and promote language skills crucial for communication and literacy development. The rhythmic patterns and playful nature of nursery rhymes engage children, making the learning process fun and memorable. By reciting and singing along to nursery rhymes, children can improve their pronunciation, fluency, and overall speech sound proficiency. What techniques improve sound recognition in toddlers? To enhance sound recognition in toddlers, consider employing the following techniques: Engage in music and singing activities Utilize sound-based toys and games Encourage listening to various sounds in the environment Repetition and reinforcement of sounds Modeling sound imitation Creating sound-rich environments When should consonant clusters become clear in speech? Consonant clusters typically become clearer in speech as language skills develop, usually by age 4-5. Mastery varies among individuals based on exposure and practice. Pronunciation accuracy improves with age and linguistic competence, supporting clear consonant cluster production. At what stage do initial word approximations appear? Initial word approximations typically appear during the early stages of language development, usually around the age of 12-18 months. Children start approximating and attempting to imitate words they hear, marking the beginning of their vocabulary acquisition journey. How can caregivers track speech development over time? Caregivers can track speech development over time by regularly noting new words, sounds, and phrases used by the child, monitoring progress in language milestones, and seeking professional assessment if delays are suspected. Recording observations and engaging in interactive activities can further aid in monitoring progress effectively. What strategies encourage vocal play in babies? To encourage vocal play in babies, consider the following strategies:Engage in interactive conversations with varied tones and sounds.Imitate your baby's vocalizations to encourage reciprocation.Use musical toys to stimulate auditory responses. When do first words typically replace babbling? First words typically replace babbling in a child's development around the age of 12-18 months. During this period, infants transition from producing nonspecific sounds to uttering recognizable words, signaling the onset of language acquisition and communication skills. How do early vocalizations predict later speech fluency? Early vocalizations, such as babbling and cooing, serve as indicators for future speech development. These initial sounds help in building the necessary oral motor skills and vocal control, laying the foundation for later speech fluency and language acquisition. Are there critical periods for specific phoneme acquisition? Specific phoneme acquisition in language learning may have critical periods. Research suggests that early exposure aids in accurate acquisition, but flexibility exists later in life. Understanding these timeframes can enhance language development and proficiency. What role do caregivers play in sound development? Caregivers play a crucial role in fostering sound development in individuals. They provide emotional support, create a safe environment, offer guidance, and facilitate learning opportunities crucial for cognitive, emotional, and social development. Caregivers shape individual behaviors, values, and relationships through their consistent presence and nurturing interactions. Their role is vital in laying the foundation for healthy growth and well-being. How do children's speech patterns change with age? Children's speech patterns evolve as they grow older. Initially, they may use simple sentences and vocabulary. As they age, syntax and grammar become more complex. Pronunciation and fluency improve, along with the ability to express ideas logically. Social interactions impact language development. What activities enhance articulation in early childhood? Activities like storytelling, singing, tongue twisters, and playing with age-appropriate puzzles can enhance articulation in early childhood. Encouraging children to mimic sounds, practice rhymes, and engage in interactive games can also support speech development. Additionally, activities that promote oral motor skills, such as blowing bubbles or using straws, can further enhance articulation abilities in young children. Can delayed babbling signal developmental concerns? Delayed babbling can sometimes indicate potential developmental issues, such as speech delays. If a child is not babbling within the typical timeframe, it could be beneficial to consult with a pediatrician or speech therapist for further evaluation and support. Be attentive to developmental milestones and seek early intervention if needed. When do most children achieve adult-like speech? Most children achieve adult-like speech by the age of 8 years old. Language development progresses through various stages, with children gradually improving pronunciation, vocabulary, and grammar skills. Exposure to language, parental interaction, and individual learning pace significantly influence this milestone. How does bilingualism affect early sound differentiation? Bilingualism can enhance early sound differentiation in children, as exposure to multiple languages can improve phonetic awareness and sound discrimination skills. Research suggests that bilingual children may develop a more nuanced ability to recognize and distinguish between various sounds compared to monolingual peers. When do speech sounds align with adult norms? Speech sounds align with adult norms typically by the age of 8, marking the completion of speech sound development in most children. However, variations exist among individuals, influenced by factors like language exposure, motor skills, and phonological awareness. How should parents interpret varied babbling stages? Parents should understand that varied babbling stages indicate normal speech development in infants. Initially, sounds are random, progressing to canonical babbling (consonant-vowel combinations). Final stages include jargon and meaningful words. Monitor progress and consult professionals if concerned. What indicates atypical speech sound development? Atypical speech sound development may be indicated by errors in articulation, inconsistent errors, difficulty with speech intelligibility, restricted sound repertoire, or lack of progress in mastering age-appropriate speech sounds. Observing these signs can help identify potential speech development concerns in children. How do different environments impact sound acquisition? Various environments affect sound acquisition differently. Factors like background noise, room acoustics, and distance from the sound source influence how sounds are perceived. Noise pollution, physical barriers, and reverberation can distort or absorb sound, hindering its clarity and reach. In contrast, quiet, well-designed spaces enhance sound reception, enabling better communication and understanding. Does sound imitation accuracy predict language proficiency? Sound imitation accuracy correlates with language proficiency, as it showcases phonetic abilities. However, it is just one aspect; overall language skills, including vocabulary and grammar, play significant roles in proficiency assessment. Other factors like culture and environment also influence language development. At what age should children master complex clusters? Children typically master complex clusters in speech between the ages of 4 and 7, varying individually. 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15781	https://medium.com/mlearning-ai/linalg-dot-product-101f548b23c2	Sitemap Open in app Sign in Sign in LINALG — Dot product Jaehoon Jang 4 min readJul 15, 2022 Dot product and duality — 3Blue1Brown Dot product The dot product is a way of multiplying vectors. So how do we compute the dot product? In numerical view, if you have two vectors in the same dimension, you pair up all the coordinates, multiply them, and add them altogether. In geometrical view, the dot product of two vectors v and w can be said as the length of the projected w on to v multiplied by the length of v. Here, order does not matter. Dot product and vector directions There are three cases when computing dot product: when the dot product has positive value, negative value, and zero value. Let’s see this in geometrical point of view. When two vectors have similar directions like the example above, the dot product of the two vectors become positive. When one vector has opposite direction like the example above, we put a minus sign in front of the value. So it can be said that if the dot product of the two vectors has negative value, the two vectors have opposite directions. When one vector is perpendicular to the other, that is, vector w projected on to vector v becomes a zero vector, the dot product would be zero. Connecting the two views We looked at the two views regarding the dot product. One was the numerical view that paired up the coordinates, multiply them and add them altogether, and the other was the geometrical view, which defined the dot product as the multiplication of a projection of a certain vector to the other and the other vector. When looking at the dot product with the numerical point of view, it uses two vectors and puts out a certain value. In the previous post, we saw something that takes in a vector, and returns a number(+decrease in dimension) The 1 x 2 matrix(linear transformation) was the one. Get Jaehoon Jang’s stories in your inbox Join Medium for free to get updates from this writer. We are going to prove that the computation of the dot product is the same as vector projection by 5 steps. Place a line diagonally while keeping the origin. Imagine that there is a unit vector u. We can think of those orange dots as vectors that have their roots at the origin. We can project the dots to the diagonal line, which has vector u as the basis vector(1d). This means that we transform a 2d plane into a 1d line. This can be represented as a 1 x 2 matrix. The mystery matrix represents the landing place of the i-hat and j-hat. As we saw in matrix multiplication, we can see the transformation as following the transformed i-hat and j-hat, and scaling them. So in order to find out where the transformed vector goes, we must follow where i-hat and j-hat goes. Let’s draw a line of symmetry in purple. So what would the transformation look like? Where would i-hat and j-hat land when projected to the vector u? Because we drew a line of symmetry, i-hat projected to vector u would have the same value as vector u projected to i-hat. Similarly, j-hat projected to vector u would have the same value as vector u projected to j-hat. So the i-hat lands on the x coordinate of the vector u, and the j-hat lands on the y coordinates of the vector u. So the mystery transformation would be: Conclusion So, the dot product is a useful way to find out about the directions of vectors, and understand about projections. However, if we dive deeper, the lesson is that anytime you have one of those linear transformations that has the output space of a number line, that transformation would have a corresponding vector. Also applying that transformation would be the same as taking the dot product with that corresponding vector. ## Mlearning.ai Submission Suggestions How to become a writer on Mlearning.ai medium.com 3blue1brown Linear Algebra Machine Learning Data Science Ml So Good ## Written by Jaehoon Jang 72 followers ·2 following Masters in Artificial Intelligence, Korea University, South Korea No responses yet Write a response What are your thoughts? More from Jaehoon Jang Jaehoon Jang ## ML-How to choose Lambda About the lambda in regularization term-Andrew Ng Feb 3, 2022 2 Maximilian Vogel ## The ChatGPT list of lists: A collection of 3000+ prompts, GPTs, use-cases, tools, APIs, extensions… Updated Aug-8, 2025. Added New Models, Prompts, Lists and Tools Feb 7, 2023 13.8K 197 Musstafa ## Optimizers in Deep Learning What is an optimizer? Mar 27, 2021 445 1 Jaehoon Jang ## LINALG — Linear combinations, span and basis vectors Basic vector concepts— 3Blue1Brown Jun 28, 2022 13 1 See all from Jaehoon Jang Recommended from Medium In Long. Sweet. Valuable. by Ossai Chinedum ## I’ll Instantly Know You Used Chat Gpt If I See This Trust me you’re not as slick as you think May 16 24K 1440 Jordan Gibbs ## ChatGPT Is Poisoning Your Brain… Here‘s How to Stop It Before It’s Too Late. Apr 29 26K 1287 AI & Tech by Nidhika, PhD ## Integrate success and differentiate Pains. In layman terms #calculus Note: The original is on authors webpage. Jun 19 Abhinav ## Docker Is Dead — And It’s About Time Docker changed the game when it launched in 2013, making containers accessible and turning “Dockerize it” into a developer catchphrase. Jun 8 6.6K 182 Venkatkumar (VK) ## Physics-Informed Neural Networks (PINNs): Learning from My First Experience💡 Introduction: Why PINNs? 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15782	https://www.teacherspayteachers.com/browse/activities/activities/math?search=seating%20arrangements	Log InSign Up Cart is empty Total: View Wish ListView Cart Seating Arrangements 40+results Relevance Rating Rating Count Price (Ascending) Price (Descending) Most Recent Relevance Rating Rating Count Price (Ascending) Price (Descending) Most Recent Filters Grade Elementary Preschool Kindergarten 1st grade 2nd grade 3rd grade 4th grade 5th grade Middle school 6th grade 7th grade 8th grade High school 9th grade 10th grade 11th grade 12th grade Higher education Adult education Not grade specific Subject Art Art history Visual arts Other (arts) English language arts Balanced literacy Close reading ELA test prep Grammar Handwriting Informational text Literature Phonics & phonological awareness Poetry Reading Reading strategies Short stories Spelling Vocabulary Writing Other (ELA) Math Algebra Algebra 2 Applied math Basic operations Fractions Geometry Graphing Math test prep Measurement Mental math Numbers Place value Statistics Other (math) Performing arts Drama Instrumental music Music Music composition Vocal music Other (performing arts) Science Anatomy Basic principles Biology Chemistry Computer science - technology Earth sciences Engineering Family consumer sciences General science Instructional technology Physical science Physics Other (science) Social emotional Character education Classroom community School counseling School psychology Social emotional learning Social studies Ancient history Asian studies Australian history Canadian history Civics Economics Elections - voting European history Geography Government Religion U.S. history World history Other (social studies) Speaking & listening World languages French Spanish Other (world language) For all subjects Not subject specific Price Format Easel Easel Activities Google Apps Image Interactive whiteboards SMART Notebook Microsoft Microsoft PowerPoint Microsoft Word PDF Resource type Classroom decor Bulletin board ideas Posters Forms Classroom forms Professional documents Teacher tools Lessons Classroom management Teacher manuals Thematic unit plans Tools for common core Unit plans Printables Clip art Hands-on activities Activities Centers Projects Internet activities Cultural activities Games Project-based learning Scripts Instruction Handouts Interactive notebooks Scaffolded notes Student assessment Assessment Critical thinking and problem solving Study skills Test preparation Student practice Independent work packet Worksheets Flash cards Graphic organizers Homework Task cards Workbooks Standard Theme Seasonal Back to school Spring Holiday Christmas-Chanukah-Kwanzaa St. Patrick's Day Thanksgiving Audience TPT sellers Homeschool Supports Special education Ordinal Numbers and Position Words Activity Task Cards \| Seating Arrangement Created by Instill Skills and Drills Looking for a fun theme on Ordinal Numbers and Position Words? Grab this 2 in 1 spiral review on Ordinal Numbers and Position Words. This resource has 15+ problem based riddles. Each riddle is numbered and presented on individual slides. You can choose to flash the riddles on your screen... or you can also turn them into task cards and get your students to rotate and take turns to do them within a time frame. Each riddle tests on different set of skills and outcomes, serving you the perfect K- 2nd Mental Math, Numbers, Other (Math) Also included in: K-1st Grade Math All Year Round Morning Work & Spiral Review Worksheets BUNDLE $2.50 Original Price $2.50 Math Seatfinder: A Seating Chart Management Activity {Freebie} Created by Math Snippets Are you looking for a way to mix up how you manage your seating arrangements? Let the students earn their seat! With this seating chart management activity, students pick a task card and complete the math problem shown. The answer to their problem determines their seat number. Included in this resource: -28 task cards covering a variety of fifth grade math standards -4 blank task cards so that you can adapt the activity to suit your students' needs -Answer Key 5th - 8th Basic Operations, Math FREE Rated 4.95 out of 5, based on 24 reviews 5.0 (24) Partner Mats: Directions, Operations, Time Created by Hear the Hummingbird These partner table/desk mats can be used as a curriculum connection throughout the year. There are mats for different seating arrangements of 6, 5, 4, 3, or 2 students. There is also a set of teacher visuals that can be printed for the board or digitally added to Powerpoint/Slides. Curriculum ConnectionsHorizontal, Vertical, DiagonalNorth, South, East, West, NW, NEOperations +,-, x, / Shoulder, Face, AcrossTimeSeating Arrangements6, 5, 4, 3, 2 Extra ResourcesColor/ Black and White VersionsTea Not Grade Specific Geography, Math $4.50 Original Price $4.50 Rated 5 out of 5, based on 1 reviews 5.0 (1) Area and Perimeter Spaghetti and Meatballs for All Activity CCSS 4.MD.3 Created by Clowning Around in Third Grade This is a pdf file where students use the book Spaghetti and Meatballs for all to draw all the different seating arrangements to figure out the perimeter and area. This is a great introduction to area and perimeter. It can be completed in partners or used as a math center activity. I also have this lesson as a Smart Board file to be used as whole class. Please check out that item in my store. All items intended for single classroom use only! Thanks! 3rd - 5th Measurement $2.00 Original Price $2.00 Rated 4.9 out of 5, based on 10 reviews 4.9 (10) Random Grouping Cards \| Groovy Retro Theme Created by Middle Math Resources - Taylor Schurig Make visibly random groups of students using these grouping cards! Quickly put students in teams for activities, small groups, projects, games, or seating arrangements. Includes 15 Groovy/Retro themed categories (groups) and can be used to form groups of 2, 3, 4, or more. Low prep: Simply print, cut, and hand out to students! Use for:Improve classroom management and streamline transitions.Great for teachers utilizing Building Thinking Classrooms (BTC).Promote positive classroom community and Not Grade Specific Classroom Community, Math $3.00 Original Price $3.00 Area & Perimeter PBL: Spaghetti and Meatballs for All Created by Melanie Zasadil In this engaging hands-on math lesson students listen to the story "Spaghetti and Meatballs for All!" by Marilyn Burns and create the different seating arrangements in the story using square tiles. (paper or plastic) Students are then asked to find the area and perimeter of each of the seating arrangements. This file includes a graphic organizer for them to record the family members that arrive, the characters solution to the seating, and perimeter and area. This lesson gives students a real lif 3rd - 4th Math CCSS 3.MD.C.5 , 3.MD.C.5a , 3.MD.C.5b +13 FREE Rated 4.75 out of 5, based on 4 reviews 4.8 (4) Venn Diagram Get to Know You Back to School Activity Created by Math Stop Great Back to School activity. This is an ideal activity to use to help new groups get to know each other (i.e., at the beginning of the year, or right after a new seating arrangement). It is an ideal tool to reinforce students’ understanding of the important display of data, Venn Diagram. Students brainstorm ways they are alike and different and include this info in the appropriate area on the diagram. This activity will help them: -- get to know each other a bit better (an important compone 4th - 8th Math, Other (Math) $3.00 Original Price $3.00 Rated 5 out of 5, based on 5 reviews 5.0 (5) 3-DIGIT addition with regroup SCAVENGER HUNT WORD PROBLEM pdf & digital activity Created by Orange Square Feet 3-Digit Addition with regrouping INTERCHANGEABLE SCAVENGER HUNT! SELF-CHECKING! PDF and DIGITAL included!!! So much fun! Practice 3-digit addition with (or without) word problems -- your choice! Students are up out of their seats and LOVE it! This is a 12 problem scavenger hunt. Students can start their journey at ANY letter! The answer will lead them to the next card/letter. sum < 1,000 The answer key has a “string” of letters. This will be the same string of letters each studen 2nd - 4th Math, Mental Math, Other (Math) CCSS 3.NBT.A.2 , 4.NBT.B.4 Also included in: 3-DIGIT ADDITION W & WO REGROUPING INTERCHANGEABLE SCAVENGER HUNT BUNDLE LOOK! $4.00 Original Price $4.00 3-DIGIT SUBTRACTION WITH regroup SCAVENGER HUNT WORD PROBLEM pdf & digital Created by Orange Square Feet 3-Digit SUBTRACTION with regrouping INTERCHANGEABLE SCAVENGER HUNT! SELF-CHECKING! PDF and DIGITAL included!!! So much fun! Practice 3-digit subtraction with (or without) word problems -- your choice! Students are up out of their seats and LOVE it! This is a 12 problem scavenger hunt. Students can start their journey at ANY letter! The answer will lead them to the next card/letter. sum < 1,000 The answer key has a “string” of letters. This will be the same string of letters each 2nd - 4th Math, Mental Math, Other (Math) CCSS 3.NBT.A.2 , 4.NBT.B.4 Also included in: 3-DIGIT SUBTRACTION W & WO REGROUPING INTERCHANGEABLE SCAVENGER HUNT BUNDLE LOOK $4.00 Original Price $4.00 3-DIGIT SUBTRACTION without regroup SCAVENGER HUNT WORD PROBLEM pdf & digital Created by Orange Square Feet 3-Digit SUBTRACTION without regrouping INTERCHANGEABLE SCAVENGER HUNT! SELF-CHECKING! PDF and DIGITAL included!!! So much fun! Practice 3-digit subtraction with (or without) word problems -- your choice! Students are up out of their seats and LOVE it! This is a 12 problem scavenger hunt. Students can start their journey at ANY letter! The answer will lead them to the next card/letter. sum < 1,000 The answer key has a “string” of letters. This will be the same string of letters ea 2nd - 4th Math, Mental Math, Other (Math) CCSS 3.NBT.A.2 , 4.NBT.B.4 Also included in: 3-DIGIT SUBTRACTION W & WO REGROUPING INTERCHANGEABLE SCAVENGER HUNT BUNDLE LOOK $4.00 Original Price $4.00 3-DIGIT addition no regroup SCAVENGER HUNT WORD PROBLEMS pdf & digital activity Created by Orange Square Feet 3-Digit Addition without regrouping INTERCHANGEABLE SCAVENGER HUNT! SELF-CHECKING! PDF and DIGITAL included!!! So much fun! Practice 3-digit addition with (or without) word problems -- your choice! Students are up out of their seats and LOVE it! This is a 12 problem scavenger hunt. Students can start their journey at ANY letter! The answer will lead them to the next card/letter. sum < 1,000 The answer key has a “string” of letters. This will be the same string of letters each stu 2nd - 4th Math, Mental Math, Other (Math) CCSS 3.NBT.A.2 , 4.NBT.B.4 Also included in: 3-DIGIT ADDITION W & WO REGROUPING INTERCHANGEABLE SCAVENGER HUNT BUNDLE LOOK! $4.00 Original Price $4.00 Operations with Radical Expressions ( +, - , ) - Speed Dating Activity Created by RR Resources and Rooms This activity is centered around simplifying radicals using the operations of addition, subtraction, and multiplication with radical expressions. 12 questions in total Work Page (back and front)Answer Key with worked out solutionsPrep: Line your desks/seating arrangement into two lines where students are sitting across from each other with their desks touching. I did 12 desks in each row to match the number of questions. Depending on your class size, you can adjust how many questions to hand 6th - 12th Algebra, Algebra 2, Math $3.50 Original Price $3.50 Rated 5 out of 5, based on 1 reviews 5.0 (1) Real World Math Using Squares, Factors, Multiples, & Divisibility Rules Created by Sandra Matthews Teacher Place This is Mrs. Matthews' real world example of "why we need to learn this stuff." Using the scenario of helping the principal organize the fifth grade moving-up ceremony, students use their knowledge of factors, multiples, square numbers, arrays, and divisibility rules to determine seating arrangements and the ordering of products for the fifth grade party. Teachers can use this activity as a review of number and operations concepts or as an assessment for the unit. Please take a minute to chec 5th Applied Math, Math, Other (Math) $2.00 Original Price $2.00 Rated 5 out of 5, based on 7 reviews 5.0 (7) Simplifying Radicals - Rationalizing the Denominator - Speed Dating Activity Created by RR Resources and Rooms This activity is centered around simplifying radicals by rationalizing the denominator. Included: 11 questions in total ( includes radical in the denominator, radical in the denominator and numerator, and perfect squares)Work Page (back and front)Answer Key with worked out solutionsPrep: Line your desks/seating arrangement into two lines where students are sitting across from each other with their desks touching. I did 11 desks in each row to match the number of questions. Depending on your 6th - 12th Algebra, Algebra 2, Math $3.50 Original Price $3.50 Logic Puzzles Worksheet 4: Think Outside the Box (Grades 6-9) Created by TYSK Learning Hub Product Description: Logic Puzzles Worksheet 4 - Think Outside the BoxThis engaging worksheet is designed for middle schoolers (grades 6–9) to practice and enhance their logical reasoning, analytical thinking, and pattern recognition skills. It features a variety of 10 puzzles, including word scrambles, pattern completion, logical seating arrangements, math challenges, and critical thinking problems that encourage students to think creatively and "outside the box." How This Resource Can Be Helpf 6th - 9th Math Also included in: Logic Puzzles Mega Bundle for Middle School $2.50Original Price $2.50 $1.75 Price $1.75 Area and Perimeter Spaghetti and Meatballs for All Lesson CCSS 4.MD.3 Created by Clowning Around in Third Grade This is a smart board activity that teaches area and perimeter while using the children's literature story Spaghetti and Meatballs for All. It is a good introduction to area and perimeter. If you do not have a smart board, I have this for sale as a pdf and students can work together to draw the seating arrangements and figure out the area and perimeter. Check for this item in my store. All purchases intended for single classroom use only! Thanks! 3rd - 5th Measurement $2.00 Original Price $2.00 Rated 4.93 out of 5, based on 14 reviews 4.9 (14) Deductive Reasoning Puzzles Worksheet Bundle (Book 9) Created by TYSK Learning Hub Deductive Reasoning Puzzles Worksheet challenges students to think logically and solve problems using given clues. The activities include mystery scenarios, number patterns, problem-solving situations, and strategic challenges. Students will analyze information, make logical connections, and explain their answers clearly. The puzzles involve real-life situations such as solving a classroom mystery, figuring out a seating arrangement, and cracking a number sequence. Each section encourages studen 6th - 9th Math Also included in: Deductive Reasoning WorkSheet Mega Bundle $1.50Original Price $1.50 $1.05 Price $1.05 The Thrills of Roller Coasters: Using Data to Make Recommendations Created by allthingsdata This lesson builds on research of how real world data investigations can be used to engage students to learn key concepts in data science and statistics. The lesson uses a free online data visualization tool (CODAP) and a dataset of 635 roller coasters from US amusement parks in operation in Spring of 2020. Variables like top speed, material used, seating arrangement, length of track, and height quickly engage students in making sense of the data and using their real world and scientific underst 7th - 10th Engineering, Physical Science, Statistics FREE Thanksgiving Dinner Math Project Created by You Betcha Ill Teach Ya Have your students pretend they are hosting Thanksgiving Dinner for up to 16 guests of their choice! Students will: - create their guest list and menu - calculate how much food they will need to purchase and the cost of it all - use elapsed time to figure out when to start preparing the food on their menu - use area and perimeter to design the seating arrangement - calculate guests' length of stay with elapsed time - graph the distance each guest travelled using Google Maps Use all parts of the 3rd - 5th Geometry, Math CCSS 3.MD.A.1 , 3.MD.B.3 , 3.MD.C.5 +7 $7.00 Original Price $7.00 Rated 5 out of 5, based on 1 reviews 5.0 (1) Famous Feast Math Project Created by You Betcha Ill Teach Ya Have your students pretend they are hosting a feast for famous people (alive, dead, real, or fantasty)! Students will: - create their guest list and menu - calculate how much food they will need to purchase and the cost of it all - use elapsed time to figure out when to start preparing the food on their menu - use area and perimeter to design the seating arrangement - calculate guests' length of stay with elapsed time - graph the distance each guest travelled using Google Maps Use all parts of 3rd - 5th Graphing, Math, Measurement CCSS 3.MD.A.1 , 3.MD.B.3 , 3.MD.C.5 +7 $7.00 Original Price $7.00 Fraction Center Activities Created by Tiarra's Teaching Techniques A perfect combination of Six No Prep Problem-Solving Activities ALL focusing on Fractions. Use in your MATH centers, small groups, partners or even as a Sub Activity. Students will practice with all four operations using fractions that require problem solving skills. The first page is great as a whole group instruction to asses pre-requisite skills using whole numbers and leads into break out activities. Many possibilities! Included:• Seating Arrangement (whole number multiplication and sums) 6th - 7th Basic Operations, Fractions, Math CCSS , 7.NS.A.1 , 7.NS.A.2 $4.00 Original Price $4.00 Rated 5 out of 5, based on 3 reviews 5.0 (3) Slope from Equations: Entrance Cards Created by SR Math When I first started teaching, I would place a problem or two on the board for the students to complete when they entered the room. However, what happened more often than not, the students came in the room, sat down in their seat, and proceeded to ignore the problems on the board. I began every single class frustrated that the students wouldn’t take the initiative to start the problems without me. At the beginning of this school year, in an effort to get to know my students, I tried a differen 7th - 10th Algebra, Math Also included in: Algebra 1 Entrance Cards Bundle Original Price $1.25 Rated 5 out of 5, based on 3 reviews 5.0 (3) Real World Equivalent Ratios Thanksgiving Scavenger Hunt- Digital and Print Created by The Middle School Math Maniac Engage your students in a festive fall math adventure with our "Real-World Equivalent Ratios Thanksgiving Scavenger Hunt - Digital and Print!" This Thanksgiving-themed activity has your class solving real-world ratio problems in a fun scavenger hunt format, adding a touch of holiday excitement to math practice. Perfect for the fall season, this resource combines the warmth of Thanksgiving with essential skills in finding equivalent ratios, keeping students entertained and engaged. This resource 6th - 7th Math, Other (Math) CCSS , 6.RP.A.3a Original Price $3.25 End of unit/block student survey Created by Manisha Arora Are you looking for an effective way to gather student feedback at the end of each unit or block? The End of Block Survey is designed to help teachers gain valuable insights into their students' experiences throughout the course. This easy-to-use survey provides teachers with a clear understanding of what worked well, what could be improved, and how students feel about various aspects of the class such as: Class content and assignments: Learn which types of assignments resonate most with studen 9th - 12th, Adult Education, Higher Education English Language Arts, Math, Social Studies $2.00 Original Price $2.00 Showing 1-24of 40+ results TPT is the largest marketplace for PreK-12 resources, powered by a community of educators. 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15783	https://cmjcollege.ac.in/wp-content/uploads/2020/05/Mr.-Vikas-Kumar-10-04-2020.pdf	VIKAS KUMAR C.M.J.College,Khutauna Mathematics System of Circles 1. A set of circles is said to be a system of circles if it contains atleast two circles. 2. Two circles S = 0 and S′ = 0 are said to touch each other if they have if they have a unique point P in common. The common point P is called point of contact of the circles S = 0 and S′ = 0. 3. If two circles touch each other then there exists only one tangent at the point of contact of the two circles. 4. Let S = 0, S′ = 0 be two circles with centres c1, c2 and radii r1 , r2 respectively. 5. If C1C2 > r1 + r2 then each circle lies completely outside the other circle. 6. If C1C2 = r1 + r2 then the two circles touch each other externally. The point of contact divides C1C2 in the ratio r1 : r2 internally. 7. If \|r1 – r2\| < C1C2 < r1+ r2 then the two circles intersect at two points P and Q. The chord PQ is called common chord of the circles. 8. If C1C2 = \|r1 – r2\| then the two circles touch each other internally. The point of contact divides C1C2 in the ratio r1: r2 externally. 9. If C1C2 < \|r1 – r2\| then one circle lies completely inside the other circle. 10. If two circles S = 0 and S′ = 0 intersect at P, then the angle between the tangents of the two circles at P is called the angle between the circles at P. 11. If d is the distance between the centres of two intersecting circles with radii r1, r2 and θ is the angle between the circles then cosθ = 1 2 2 2 2 1 2 2r r d − r − r 12. If θ is the angle between the circles S = x2 + y2 + 2gx + 2fy + c= 0, S′=x2 + y2 + 2g′x + 2f′y+c′ =0 then cosθ = ( ) 2 g f c g f c c c 2 gg ff 2 + 2 − ′2 + ′2 − ′ + ′ − ′ + ′ 13. Two intersecting circles are said to cut each other orthogonally if the angle between the circles is a right angle. 14. Let d be the distance between the centres of two intersecting circles with radii r1, r2. The two circles cut orthogonally iff d2 = 2 2 2 r1 + r . 15. The condition that the two circles S ≡ x2+y2 + 2gx + 2fy + c = 0, S′ ≡ x2 + y2 + 2g′x +2f′y + c′= 0 may cut each other orthogonally is 2gg′ +2ff′ = c + c′. 16. A common tangent L = 0 of the circles S =0, S′ = 0 is said to be a direct common tangent of the circles if the two circles S = 0. S′ = 0 lie on the same side of L = 0. System of Circles 17. A common tangent L = 0 of the circle S = 0, S′ =0 is said to be a transverse common tangent of the circles if the two circles S = 0, S′ = 0 lie on the opposite (either) sides of L = 0. 18. Let S = 0, S′ = 0 be two circles with centres C1 , C2 and radii r1, r2 respectively and n be the number of common tangents. 19. If C1C2 > r1 + r2 then n = 4 20. If C1C2 = r1 + r2 then n = 3 21. If \|r1 – r2\| < C1 C2 < r1 + r2 then n = 2 22. If C1C2 = \| r1 – r2\| then n = 1 23. If C1C2 < \|r1 – r2\| then n = 0 24. Let S = 0, S′ = 0 be two circles. (i) The point of intersection of direct common tangents of S = 0, S′ = 0 is called external centre of similitude. (ii) The point of intersection of transverse common tangents S = 0, S′ =0 is called internal centre of similitude. 25. Let S = 0, S′= 0 be two circles with centres C1, C2 and radii r1, r2 respectively. If A1 and A2 are respectively the internal and external centres of similitude of the circles S = 0, S′= 0 then i) A1 divides C1C2 in the ratio r1 : r2 internally ii) A2 divides C1C2 in the ratio r1 : r2 externally 26. If the radii of two circles are equal then the external centre of similitude does not exist. 27. The locus of a point, for which the powers with respect of two given nonconcentric circles are equal, is a straight line, called the radical axis of the given circles. 28. The equation of the radical axis of the circles S= 0, S′ = 0 is S - S′= 0. 29. The lengths of tangent from a point on the radical axis of two circles are equal, if exists. 30. The radical axis of two circles bisects all common tangents of the two circles. 31. The radical axis of two circles is perpendicular to their line of centres. 32. If two circles intersect, then the radical axis is their common chord. 33. If two circles touch each other, then the radical axis is their common tangent at the point of contact. 34. Any point on the radical axis of two circles S=0, S′= 0 lies externally or lies internally or lies on both the circles simultaneously. 35. The radical axes of three circles, whose centres are noncollinear, taken in pairs, are concurrent. System of Circles 36. The point of concurrence of the radical axes of three circles, whose centres are noncollinear, taken in pairs, is called the radical centre of the circles. 37. The powers of the radical centre of three circles with respect to each of the three circles are equal. 38. The centre of a circle cutting two circles orthogonally lies on the radical axis of the two circles. 39. The centre of the circle cutting three circles orthogonally is the radical centre of the three circles. The radius of the circle cutting three circles orthogonally is the length of tangent from the radical centre to any of the three circles. 40. If P is the radical centre of three circles and r is the length of tangent from P to any of the circles then the circle with centre P and radius r cuts the three circles orthogonally. 41. A system of circles is said to be a system of coaxal circles or coaxal system of circles if every pair of circles has the same radical axis. 42. Since the radical axis is perpendicular to the line of centres, it follows that the centres of circles in a coaxal system are collinear. 43. If S=0, S′ = 0 are two circles then λ1S + λ2S′ = 0 where λ1, λ2 are parameters such that λ1+λ2 ≠ 0, represents the coaxal system of all circles containing S = 0, S′ = 0. 44. If S =0, S′ = 0 are two circles then the coaxal system λ1S + λ2S′ = 0 is called the coaxal system determined by the circles S = 0, S′= 0. 45. If two circles intersect, then the radical axis is the common chord and hence λ1S + λ2S′= 0 represents a coaxal system of circles passing through the points of intersection of the circles S = 0, S′ = 0. 46. If S = 0 is a circle and L = 0 is a line then S + λL =0 where λ is a parameter, represents the coaxal system of all circles of which S= 0 is a member and L =0 is the radical axis of the system. 47. The coaxal system S + λL = 0 is called the coaxal system determined by the circle S = 0 and the line L = 0 as the radical axis. 48. If S = 0 is a circle in the coaxal system having radical axis L = 0 then every circle in the system is of the form S + λL = 0 for some constant λ. 49. If S = 0, S′ = 0 be two circles then every circle in the coaxal system λ1S + λ2S′ = 0 except S′=0 can be taken as S + λL = 0 for some constant λ where L = S - S′. 50. The coaxal system of circles is said to be in the simplest form it its line of centres is x-axis and the radical axis is y –axis. 51. The equation to the system of coaxal circles in the simplest form is x2 + y2 + 2λx +c =0 where λ is a parameter and c is a fixed constant. 52. Let x2 + y2 + 2λx + c = 0, λ is a parameter, c is a fixed constant, be a coaxal system of circles. Then System of Circles i) If c < 0 then the system of circles is an intersecting coaxal system. ii) If c = 0 the then system of circles is a touching coaxal system. iii) If c > 0 then the system of circles is a nonintersecting coaxal system. 53. The point circles in a coaxal system are called the limiting points of the coaxal system. 54. The limiting points of the coaxal system x2 + y2 + 2λx + c = 0 are ( ∓ c ,0). 55. A nonintersecting coaxal system has two limiting points. 56. A touching coaxal system has only one limiting point which is the point of contact of the circles of the system. 57. An intersecting coaxal system has no limiting point. 58. If P and Q are the two limiting points of a coaxal system then Q is the image of P with respect to the radical axis. 59. If P, Q are the two limiting points of a coaxal system then P, Q are inverse points with respect to the coaxal system. 60. The limiting points of a coaxal system are conjugate points with respect to every circle in the coaxal system. 61. The coaxal system of circle ρ = 0 is said to be orthogonal to a coaxal system of circles ρ′ = 0 if every circle in ρ = 0 is orthogonal to every circle in ρ′ = 0. 62. The equation of the coaxal system of circles orthogonal to the coaxal system x2 + y2 + 2λx + c = 0 is x2 + y2 + 2μ y – c = 0 where μ is a parameter. 63. The coaxal systems ρ ≡ x2 + y2 + 2λx + c = 0, ρ′ = x2 + y2 + 2μy – c = 0 are called conjugate coaxal systems. 64. If ρ = x2 + y2 + 2λx + c = 0 is a nonintersecting coaxal system of circles then every circle in the coaxal system ρ′ = x2 + y2 + 2μy – c = 0 passes through the limiting points of ρ = 0. 65. If r1, r2(r1 > r2) are the radii of two circles and d is the distance between the centres of the circles then i) the length of the direct common tangent = 2 1 2 d2 − (r − r ) . ii) the length of the transverse common tangent = 2 1 2 d2 − (r + r ) . 66. The line joining the midpoints of the pair of tangents drawn from P to the circle S = 0 is the radical axes of the point circle of P and S = 0. System of Circles 67. If origin is a limiting point of the coaxal system containing the circle x2 +y2 + 2gx +2fy + c= 0 then the other limiting point is ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + − + − 2 2 g2 f 2 , fc g f gc . 68. A common tangent drawn to any two circles of a coaxial system subtends angles at the limiting points which are equal to π/2. 69. A circle passing through the limiting points of a given co-axal system cuts any member of the system at an angle of 90° 70. Length of direct common tangent is 2 1 2 2 (c1c2 ) − (r − r ) . Where c1, c2 are centres and r1 , r2 are radii of given circles. 71. Length of Transverse common tangent is 2 1 2 2 (c1c2 ) − (r + r ) . 72. Two circles of radii r1 and r2 intersect at angle θ. The length of their common chord is + − θ θ r r 2r r cos 2r r sin 1 2 2 2 2 1 1 2 . 73. Two circles with centres C1 and C2 and radius a cut each other orthogonally. Then a = 2 C1C2
15784	https://www.khanacademy.org/science/ap-college-physics-1/xf557a762645cccc5:force-and-translational-dynamics/xf557a762645cccc5:forces-and-systems/e/free-body-diagrams-ap-physics-1	Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer. More information Manage Consent Preferences Strictly Necessary Cookies Always Active Certain cookies and other technologies are essential in order to enable our Service to provide the features you have requested, such as making it possible for you to access our product and information related to your account. For example, each time you log into our Service, a Strictly Necessary Cookie authenticates that it is you logging in and allows you to use the Service without having to re-enter your password when you visit a new page or new unit during your browsing session. Functional Cookies These cookies provide you with a more tailored experience and allow you to make certain selections on our Service. For example, these cookies store information such as your preferred language and website preferences. Targeting Cookies These cookies are used on a limited basis, only on pages directed to adults (teachers, donors, or parents). We use these cookies to inform our own digital marketing and help us connect with people who are interested in our Service and our mission. We do not use cookies to serve third party ads on our Service. Performance Cookies These cookies and other technologies allow us to understand how you interact with our Service (e.g., how often you use our Service, where you are accessing the Service from and the content that you’re interacting with). Analytic cookies enable us to support and improve how our Service operates. For example, we use Google Analytics cookies to help us measure traffic and usage trends for the Service, and to understand more about the demographics of our users. We also may use web beacons to gauge the effectiveness of certain communications and the effectiveness of our marketing campaigns via HTML emails. Cookie List Consent Leg.Interest label label label
15785	https://www.gauthmath.com/solution/1985457153921412/What-are-some-of-God-s-basic-attributes-orange-round-bouncy-Loving-merciful-crea	Solved: What are some of God's basic attributes? orange, round, bouncy Loving, merciful, creator b [Others] Drag Image or Click Here to upload Command+to paste Upgrade Sign in Homework Homework Assignment Solver Assignment Calculator Calculator Resources Resources Blog Blog App App Gauth Unlimited answers Gauth AI Pro Start Free Trial Homework Helper Study Resources Others Questions Question What are some of God's basic attributes? orange, round, bouncy Loving, merciful, creator blue, fast, new crunchy, juicy, green Show transcript Expert Verified Solution 100%(20 rated) Answer The correct answer is Loving, merciful, creator. Explanation Attributes are qualities or characteristics that describe someone or something. God's attributes are the characteristics that define His nature. Loving, merciful, and creator are attributes that describe God's nature and actions. These are qualities often associated with God in various religious traditions. Here are further explanations. Option 1: orange, round, bouncy. These are physical characteristics that describe objects, not attributes of God. Option 3: blue, fast, new. These are also descriptive words, but they relate to physical properties or conditions, not divine attributes. Option 4: crunchy, juicy, green. These are sensory qualities, typically used to describe food or other tangible items, and are not applicable to describing God's attributes. Helpful Not Helpful Explain Simplify this solution Gauth AI Pro Back-to-School 3 Day Free Trial Limited offer! Enjoy unlimited answers for free. Join Gauth PLUS for $0 Previous questionNext question Related What kind of long-term memories shows the greatest age-related declines? Semantic Episodic Explicit Implicit 100% (6 rated) Planning Appealing Meals: Match the term on the left with the correct description on the right. Ease of eating [ Choose ] [ Choose ] Texture Foods that are crunchy, creamy, or juicy. Foods that are cut into slices, wedges, or sticks. Foods that are cold, warm, or hot. Temprature Foods that are red, orange, yellow, green, or blue. Foods that require less chewing, swallowing, and handling. Color [ Choose ] Shapes [ Choose ] 100% (4 rated) Universal Intellectual Standards should be applied to thinking whenever you are interested in evaluating your quality of reasoning about a problem, issue, or situation. True False 100% (2 rated) Apples and oranges are both fruits, but they have many differences. Apples are round and come in many different colors like red, green, and yellow. They have a smooth skin and a crunchy texture. Oranges, on the other hand, are more round and come in the color orange. They have a bumpy skin and a juicy texture. What text structure is the above paragraph? sequence problem/solution compare/contrast cause/effect 100% (3 rated) Layla, a 15-year-old, has an older sister and a younger sister. According to Adler, which of the following is most applicable to Layla in the context of birth order? Multiple Choice She is likely to be the center of attention in comparison to her siblings. She is most likely of the three to suffer from a psychological disorder. She is most likely of the three to engage in thefts and other petty crimes. She is most likely to be inspired by her older sister to strive for superiority. 60% (5 rated) Choose the option that best answers the question. One of the greatest_ to the endangered status of the giant panda is the systematic destruction of its habitat. Making its home in bamboo forests, this gentle giant is known for its slow, Which choice completes the text with the most logical constant munching of the plant. Chinese and international demand for and precise word? bamboo, though, has dramatically reduced possible habitat area. contributions providers contributors factors 72% (14 rated) 60+ Heartfelt Easter Texts for Girlfriend Writing Examples Mastering Your Howard University Personal Statement: A Comprehensive Guide Writing Examples Understanding 15 Most Commonly Used Tones When Writing an Essay Blog How to Write a Literary Analysis Essay: 4 Tips, 3 Common Mistakes, and Outlining Blog Gauth it, Ace it! contact@gauthmath.com Company About UsExpertsWriting Examples Legal Honor CodePrivacy PolicyTerms of Service Download App
15786	https://www.chegg.com/homework-help/questions-and-answers/-mixing-ideal-gases-spontaneous-gibbs-energy-change-mixing-process-positive-ii-gibbs-energ-q99475314	Your solution’s ready to go! Our expert help has broken down your problem into an easy-to-learn solution you can count on. Question: A. Mixing of ideal gases is spontaneous because: i) The Gibbs energy change of the mixing process is positive. ii) The Gibbs energy change of the mixing process is negative. iii) The enthalpy change of the mixing process is negative. iv) The enthalpy change of the mixing process is positive. B. At a phase boundary for a pure substance: i) The enthalpy of one A)For an ideal solution, the Gibbs free energy of mixing is always negative, meaning that mixing of ideal solutions is always spontaneous.... In ideal mixtures the enth… Not the question you’re looking for? Post any question and get expert help quickly. Chegg Products & ServicesChegg Products & Services Chegg Products & Services CompanyCompany Company Chegg NetworkChegg Network Chegg Network Customer ServiceCustomer Service Customer Service EducatorsEducators Educators
15787	https://physics.stackexchange.com/questions/566360/radioactivity-and-half-life	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Radioactivity and half life Ask Question Asked Modified 5 years, 2 months ago Viewed 932 times 0 $\begingroup$ Does the process of decaying in radioactive elements occur every second? The equation consists of $\rm e$, so it must mean that decay due to radiation must occur every second, right? radiation radioactivity half-life Share Improve this question edited Jul 18, 2020 at 14:40 user258881 asked Jul 17, 2020 at 15:17 user794763user794763 10511 gold badge11 silver badge33 bronze badges $\endgroup$ 5 $\begingroup$ Certainly many isotopes have half-lives much longer than a second, so it is not clear what you mean. Please clarify. $\endgroup$ Jon Custer – Jon Custer 2020-07-17 15:28:10 +00:00 Commented Jul 17, 2020 at 15:28 $\begingroup$ @JonCuster I am not talking about half lives ,if we take a radioactive element say strontium 90 ,it takes about 28 years to lose half of its quantity, that's ok but how does it do so,after 28.79 yrs approx does the half of it disappear into thin air(or) does it lose it's quantity every instant and after 28.79 years it happens to have half of its initial mass? $\endgroup$ user794763 – user794763 2020-07-17 15:33:48 +00:00 Commented Jul 17, 2020 at 15:33 $\begingroup$ Decay is an ongoing statistical process. $\endgroup$ Jon Custer – Jon Custer 2020-07-17 15:34:42 +00:00 Commented Jul 17, 2020 at 15:34 $\begingroup$ @JonCuster pls elaborate $\endgroup$ user794763 – user794763 2020-07-17 15:36:19 +00:00 Commented Jul 17, 2020 at 15:36 2 $\begingroup$ Re, "every second." Nature doesn't know anything about seconds. Nor does it know anything about years or months or any other time scale we might choose to talk about radioactive half lives. Whatever is happening, it happens continuously. $\endgroup$ Solomon Slow – Solomon Slow 2020-07-17 16:38:27 +00:00 Commented Jul 17, 2020 at 16:38 Add a comment \| 7 Answers 7 Reset to default 2 $\begingroup$ The process of radioactive decay occurs at every instant in which th substance exists. The rate at which the decay happens is proportional to the number of active particles at the given instant. It isn't stroboscopic. But if you solve the differential equation, you will get a fair approximation of how many radioactive particles remain after some amount of time. Share Improve this answer answered Jul 17, 2020 at 15:48 SK DashSK Dash 1,88077 silver badges2121 bronze badges $\endgroup$ Add a comment \| 2 $\begingroup$ does the half of it disappear into thin air...? Note! It doesn't disappear. It becomes something else. You gave, as an example, 90Sr. When at atom of 90Sr decays, it emits a beta particle, and it becomes an atom of 90Y (Yttrium). Beta emission happens in a nucleus that has too many neutrons to be stable. One of the neutrons spontaneously becomes; a proton, an energetic electron (a.k.a., "beta particle"), and an antineutrino. Because the nucleus now has one more proton than previously, the atomic number goes up by one. 90Y has a half-life of only a few days, before it decays (again by beta emission) and it becomes a stable 90Zr (Zirconium) atom. Share Improve this answer edited Jul 17, 2020 at 17:07 answered Jul 17, 2020 at 16:51 Solomon SlowSolomon Slow 17.3k11 gold badge3939 silver badges5656 bronze badges $\endgroup$ Add a comment \| 1 $\begingroup$ No it doesn't. It decays at a rate depending on its half-life. As @Jon Custer said its a statistical process and the nuclei will randomly decay in some period. The decay rate, decay per unit time is given by A=Î»N where A is decay rate, Î» = decay constant and N is number of Nuclei in sample. Share Improve this answer answered Jul 17, 2020 at 15:47 NatsfanNatsfan 2,69822 gold badges1111 silver badges1212 bronze badges $\endgroup$ Add a comment \| 1 $\begingroup$ When we talk about half-time we deal with a collection of many atoms (macroscopic quantity, i.e. of the order of Avogadro number). Every atom has a probability $p=\frac{1}{\tau}$ to decay per unit time, whereas the average number of atoms that have not decayed is given by $N_0e^{-t/\tau}$, where $N_0$ is the initial quantity of isotopes. Share Improve this answer edited Jul 17, 2020 at 17:21 answered Jul 17, 2020 at 15:46 Roger V.Roger V. 70.8k77 gold badges7878 silver badges251251 bronze badges $\endgroup$ 0 Add a comment \| 1 $\begingroup$ What really happens It really depends. If you have a moderate value of decay rate ($\sim 10^{10} \:\rm dps$), then the decay of multiple particles every second is inevitable. However, if the value of decay rate is small enough ($\sim 2 \: \rm dps$), then it's quite possible that you might find a particular second where none of the particles decayed. In the cases where the decays rate $¤1\:\rm dps$, it's highly likely to find such time intervals. Why does it happen? The reason why this is the way it is, is because of the quantum mechanical nature of nuclear decay. You can't really predict whether a certain particle will have definitely decayed by a certain time $t$, because of the quantum mechanical nature of the process of nuclear decay. However, the probability of particle being decayed by a time $t$ increases as the time $t$ increases, so eventually it will be highly probable that the particle would decay. However since we are dealing with probabilities, we can never be sure that a particle decayed. If you're into quantum mechanics, you might also know that the state of a nucleus is a superposition of decayed and undecayed. And you can't know whether it decayed until you collapse its wavefunction by making a measurement, and checking if it decayed. (see the last section for the continuation of this line of thought) How come we write a continuous fuction describing decay? The function that we write can be regarded as "empirical", or more appropriately, an approximation of the reality rather than the absolute truth. Thus if you were to perform a nuclear decay experiment, it's highly likely that you would find measurements which are quite close (though not exact) to the mathematical equation of nuclear decay: $$N(t)=N_0\mathrm e^{-\lambda t}\tag{1}$$ Now, no matter how many times you repeat the experiment, or how precise you make your experimental procedure, you'd always get results which are closer to equation $(1)$, but you'd never be able to do the experiment and get the exactly same results as predicted by the equation. And no, it's not because of the experimental errors that might have crept in, it's because of the uncertain and probabilistic nature of nuclear decay. Many-Worlds interpretation This is a bit of an off-shoot section, so it isn't strictly related to the question, but it's worth reading :-) Now if you believe in the many-worlds interpretation of quantum mechanics, then you'd be fascinated to hear what it predicts during nuclear decay. It predicts that our universe branches multiple (gazillion times) during the process of nuclear decay, with every universe being different. In other words, whenever there are two possibilities for a nucleus, decay or not decay, our universe branches into two other universes, where in one of the universes, the nucleus has decayed while in the other one, it hasn't. And this is true for all the nuclei. This line of logic implies an extremely mind-boggling result. It implies that there might be a universe where none of the particles have decayed in the first second of nuclear decay (despite the nuclear decay having a rate of $10^{10}\:\rm dps$). At first, this sounds to be unreal, thus thinking us into doubting the many-worlds interpretation. However, if we think further, then we can notice that the probability of us being in such a universe is extremely low (I mean extremely extremely extremely low). Why? Because everytime a universe branches into two new universes, we have almost equal probability of being in one or the other. But to reach to a universe where no particles have decayed in the first second, you'd alwyas have to end up in the "undecayed" version of the universe, every single time, for $10^{10}$ branchings. And this is extremely unlikely, thus we never find ourselves conducting such an experiment. But do note that if we were to perform our experiment $2^{10^{10}}$ times, we might have a chance to witness that "special" decay where nothing decays in the first second. Share Improve this answer edited Jul 18, 2020 at 15:56 answered Jul 18, 2020 at 14:39 user258881user258881 $\endgroup$ Add a comment \| 1 $\begingroup$ The poster may be asking if radioactive decay is a continuous process or a discrete process. It is a continuous process. For any given increment in time, there is a probability that a certain amount of material will decay. This probability is called the "decay constant" and usually denoted with the greek symbol "$\lambda$". In the form of a differential equation, the "rate of change" of the nuclide concentration is proportional to the decay constant times the concentration itself. $$ \frac{dN}{dt}=-\lambda \, N $$ The solution of the differential equation makes it clear that it is a continuous process $$ N(t)=N(0) \, \exp(-\lambda t) $$ The decay constant isn't a very intuitive variable since the units are "per time". Therefore, the decay constant usually converted to a "half-life", which is more intuitive to understand because it has units of time. You can use the previous equations to show that the half-life is related to the decay constant by the expression $$ t_{h} = \frac{\ln(2)}{\lambda} $$ The confusion may be that the half-life has units of time (e.g. seconds), but it is still refers to a continuous process. Share Improve this answer answered Jul 22, 2020 at 13:54 NuclearFissionNuclearFission 89644 silver badges77 bronze badges $\endgroup$ Add a comment \| 1 $\begingroup$ In order to understand radioactive decay, we have to see what is in the nucleus of an atom and how its parts interact. As you may know, all nuclei, no matter what kind of atom (element), consist of protons and neutrons (nucleons). A proton is a positively charged heavy body, and a neutron is just slightly heavier than a proton and has no charge. The neutron can be thought of as a proton with an electron (or, more accurately, beta particle) bound to it by the weak nuclear force. The protons of a nucleus repel one another very strongly through the electric force of repulsion between like-charged bodies. However, the attractive strong nuclear force is much stronger than the electric force at such small distances, and so the strong nuclear force overcomes the electrostatic (Coulomb) repulsion and holds the protons and neutrons together. This union results in a ball of protons and neutrons (to a certain approximation) shaking back and forth violently within the nucleus, but held together with the strong nuclear force. Sometimes, the nucleus' (or, isotope's) configuration (shape) is always energetically "stable," and will never break apart no matter how much time passes - like an intact water balloon (so long as something strong enough doesn't come along and break it). Other nuclei (of a different isotope), every so often in this violent vibration, assume a shape that cannot be held by the volumetric and surface tension. These are called "unstable" nuclei, or "radioactive" nuclei. When this happens, a piece of the nucleus, a particle, breaks off. This is called "decay," or "radioactive decay." When the nucleus decays, it does not disappear. It simply breaks into multiple pieces. We cannot say, for any given "unstable" nucleus, exactly when it will assume a configuration leading to decay. But, if we have a large number of nuclei (say, $N = 10^{23}$; around the number of atoms when amassed you can see with your naked eye) we can say approximately that the number of decaying nuclei at time t, dN(t), must be proportional to the number of nuclei present at time t, N(t). Additionally, we can say that dN(t) should be proportional to the time that passes over a small enough duration, dt. Note that in these proportionalities N is considered continuous rather than discrete. This is an approximation - or, really, an error - since we cannot really have a fraction of a radioactive particle, by definition. Continuing, nonetheless, we have said $dN \propto N dt$. Rearranging, and introducing a proportionality constant, $\lambda$ (the "decay constant"), we see $\frac{dN(t)}{dt} = - \lambda N(t)$ The constant $\lambda$ is considered positive by definition, so the negative is introduced to capture that the change in the number of nuclei is negative over time period dt. Rearranging the equation gives $\frac{dN(t)}{N(t)} = -\lambda dt$ The solution to this equation through basic calculus is $N(t) = N(t=0) e^{-\lambda t}$ This is where the "e" comes from. Now, your question is "does decay happen every second?" The problem is, the question assumes there is a yes or no answer. Also, it is useful to explain the half-life. The half-life, by definition, is the time during which half of the nuclei in the sample should have decayed (become something else... not disappeared). We can calculate this time by using the equation above as $\frac{N(t_{1/2})}{N(t=0)} \equiv 1/2 = e^{-\lambda t_{1/2}}$ where $t_{1/2} = \text{half-life}$. Solving the right side, we get $-ln(2) = -\lambda t_{1/2}$ or $t_{1/2} = ln(2)/\lambda$ This shows the relationship between the half-life and the decay constant. Over the time $t_{1/2}$, half of the original nuclei will have decayed, leaving N(t=0)/2 nuclei in their original state. After another half-life, half of the remaining undecayed nuclei will have decayed, with N(t=0)/4 remaining originals. Generally, after H half-lives, $N(t=0)/2^{H}$ nuclei will remain undecayed. The problem with the above derivation, as mentioned, is that it calculates an average behavior of a large number of nuclei (or, more accurately, a proportion of total nuclei) that, as far as the equation is concerned, is continuous. This is called the "classical" approach. In order to derive the actual decay of a number of nuclei N, we should start with the statistical representation, which correctly treats the number of nuclei as discrete, but - since we only know the probability that any given nucleus will decay in a certain duration - gives a probability distribution of final outcomes instead of a deterministic result. Therefore, the answer to your question is: "In each second, there is a probability P that decay will happen, and there is a probability (1-P) that decay will not happen." Of course, once all nuclei have decayed, the probability P is zero. It is possible, albeit exceedingly improbable for large numbers of nuclei, that in a sample of radioactive nuclei, all of them will decay in the same short time dt. We can call this outcome #1. There is only one way that this outcome can be achieved. If the probability that a nucleus decays in time dt is p, then the probability it does not decay in the same time is (1 - p). The probability of outcome #1 is $p^{N}$. There are N ways in which outcome #2 occurs, where all but one nuclei decay in time dt. This means the probability of outcome #2 is $(\frac{N!}{(N-1)!})(1-p)p^{N-1} = N(1-p)p^{N-1}$. Outcome #3 is that all but 2 nuclei decay in time dt. The probability of outcome #3 is $\frac{N!}{(N-2)!2!}(1-p)^{2}p^{N-2}$. In general, Outcome #k, that all but k-1 nuclei decay in time dt has probability $\frac{N!}{(N-(k-1))!(k-1)!}(1-p)^{k-1}p^{N-(k-1)}$. One of all of these N+1 outcomes must be fulfilled at the end of time dt, so the summation of all these probabilities is one. It is useful for the purpose of answering your question to mention that Outcome #N+1 is that no decay occurs over the time period under consideration, and the probability of this event is $(1-p)^{N}$. The irony here is that I've slipped a "classical" model of the nucleus itself into the discussion - we could call it the "liquid drop" model - when, in fact, the nucleus itself is just a superposition of a great number of possible configurations or accessible states based on its internal energy, and we can, in principle, count these number of possible states to arrive at the statistical, and correct, model of the nucleus as well. We can also relate the two approaches. If we set dt = 1, then in the first equation this corresponds to t=1 (at this point, it doesn't matter what we choose for units, but we do know whatever units we choose are consistent with $\lambda$), leading to equation $\frac{N(1)}{N(0)} = e^{-\lambda} = \frac{\text{Number of nuclei not decayed at time 1}}{\text{Number of nuclei at time 0}} = 1 - p$ where, admittedly there was a bit of a sleight of hand since p was for a single atom while N corresponds to a population of atoms; but, since p is derived from a population of atoms, it's okay. Share Improve this answer edited Jul 29, 2020 at 17:40 answered Jul 17, 2020 at 22:18 jpfjpf 54033 silver badges1313 bronze badges $\endgroup$ Add a comment \| Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions radiation radioactivity half-life See similar questions with these tags. 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15788	https://probability.oer.math.uconn.edu/wp-content/uploads/sites/2187/2018/01/prob3160ch1.pdf	CHAPTER 1 Combinatorics 1.1. Basic counting principle and combinatorics 1.1.1. Basic counting principle. The rst basic counting principle is to multiply. Namely, if there are n possible outcomes of doing something and m outcomes of doing another thing, then there are m · n possible outcomes of performing both actions. Basic counting principle Suppose that two experiments are to be performed. Then if experiment 1 can result in any one of m possible outcomes, and if for each outcome of experiment 1 there are n possible outcomes of experiment 2, then there are m·n possible outcomes of the two experiments together. Example 1.1. Suppose we have 4 shirts of 4 dierent colors and 3 pants of dierent colors. How many dierent outts are there? For each shirt there are 3 dierent colors of pants, so altogether there are 4 × 3 = 12 possibilities. Example 1.2. How many dierent license plate numbers with 3 letters followed by 3 numbers are possible? Solution: (26)3(10)3. Indeed, the English alphabet has 26 dierent letters, therefore there are 26 possibilities for the rst place, 26 for the second, 26 for the third, 10 for the fourth, 10 for the fth, and 10 for the sixth. We multiply. 1.1.2. Permutations. How many ways can one arrange letters a, b, c? We can list all possibilities, namely, abc acb bac bca cab cba. There are 3 possibilities for the rst position. Once we have chosen the letter in the rst position, there are 2 possibilities for the second position, and once we have chosen the rst two letters, there is only 1 choice left for the third. So there are 3 × 2 × 1 = 6 = 3! arrangements. In general, if there are n distinct letters, there are n! dierent arrangements of these letters. Example 1.3. What is the number of possible batting orders (in baseball) with 9 players? © Copyright 2013 Richard F. Bass, 2020 Masha Gordina Typesetting date: March 31, 2020 3 4 1. COMBINATORICS Solution: 9! = 362880. Example 1.4. How many ways can one arrange 4 math books, 3 chemistry books, 2 physics books, and 1 biology book on a bookshelf so that all the math books are together, all the chemistry books are together, and all the physics books are together? Solution: 4!·(4!·3!·2!·1!) = 6912. We can arrange the math books in 4! ways, the chemistry books in 3! ways, the physics books in 2! ways, and the biology book in 1! = 1 way. But we also have to decide which set of books go on the left, which next, and so on. That is the same as the number of ways of arranging of four objects (such as the letters M, C, P, B), and there are 4! ways of doing that. In permutations the order does matter as is illustrated by the next example. Example 1.5. How many ways can one arrange the letters a, a, b, c? Let us label them rst as A, a, b, c. There are 4! = 24 ways to arrange these letters. But we have repeats: we could have Aa or aA which are the same. So we have a repeat for each possibility, and so the answer should be 4!/2! = 12. If there were 3 as, 4 bs, and 2 cs, we would have 9! 3!4!2! = 1260. What we just did is called nding the number of permutations. These are permutations of a given set of objects (elements) unlike the example with the licence plate numbers where we could choose the same letter as many times as we wished. Permutations The number of permutations of n objects is equal to n! := 1 · ... · n, with the usual convention 0! = 1. 1.1.3. Combinations. Now let us look at what are known as combinations. Example 1.6. How many ways can we choose 3 letters out of 5? If the letters are a, b, c, d, e and order matters, then there would be 5 choices for the rst position, 4 for the second, and 3 for the third, for a total of 5 × 4 × 3. Suppose now the letters selected were a, b, c. If order does not matter, in our counting we will have the letters a, b, c six times, because there are 3! ways of arranging three letters. The same is true for any choice of three letters. So we should have 5 × 4 × 3/3!. We can rewrite this as 5 · 4 · 3 3! = 5! 3!2! = 10 1.1. BASIC COUNTING PRINCIPLE AND COMBINATORICS 5 This is often written as 5 3 , read 5 choose 3. Sometimes this is written C5,3 or 5C3. Combinations (binomial coecients) The number of dierent groups of k objects chosen from a total of n objects is equal to n k = n! k! (n −k)!. Note that this is true when the order of selection is irrelevant, and if the order of selection is relevant, then there are n · (n −1) · ... · (n −k + 1) = n! (n −k)! ways of choosing k objects out of n. Example 1.7. How many ways can one choose a committee of 3 out of 10 people? Solution: 10 3 = 120. Example 1.8. Suppose there are 8 men and 8 women. How many ways can we choose a committee that has 2 men and 2 women? Solution: we can choose 2 men in 8 2 ways and 2 women in 8 2 ways. The number of possible committees is then the product 8 2 · 8 2 = 28 · 28 = 784. Example 1.9. Suppose one has 9 people and one wants to divide them into one committee of 3, one committee of 4, and the last one of 2. There are 9 3 ways of choosing the rst committee. Once that is done, there are 6 people left and there are 6 4 ways of choosing the second committee. Once that is done, the remainder must go in the third committee. So the answer is 9! 3!6! 6! 4!2! = 9! 3!4!2!. Example 1.10. For any k ⩽n we have n k = n n−k . Indeed, the left-hand side gives the number of dierent groups of k objects chosen from a total of n objects which is the same to choose n −k objects not to be in the group of k objects which is the number on the right-hand side. 6 1. COMBINATORICS Combinations (multinomial coecients) The number of ways to divide n objects into one group of n1 objects, one group of n2, . . ., and a kth group of nk objects, where n = n1 + · · · + nk, is equal to n n1,...,nk = n! n1!n2! · · · nk!. Example 1.11. Suppose we have 4 Americans and 6 Canadians. (a) How many ways can we arrange them in a line? (b) How many ways if all the Americans have to stand together? (c) How many ways if not all the Americans are together? (d) Suppose you want to choose a committee of 3, which will be all Americans or all Cana-dians. How many ways can this be done? (e) How many ways for a committee of 3 that is not all Americans or all Canadians? Solution: (a) This is just the number of arrangements of 10 elements, that is, 10! (b) Consider the Americans as one group (element) and each Canadian as a distinct group (6 elements); this gives 7 distinct groups (elements) to be arranged, which can be done in 7! ways. Once we have these seven groups arranged, we can arrange the Americans within their group in 4! ways, so we get 4!7! by the basic counting principle. (c) This is the answer to (a) minus the answer to (b): 10! −4!7! (d) We can choose a committee of 3 Americans in 4 3 ways and a committee of 3 Canadians in 6 3 ways, so the answer is 4 3 + 6 3 . (e) We can choose a committee of 3 out of 10 in 10 3 ways, so the answer is 10 3 − 4 3 − 6 3 . Finally, we consider three interrelated examples. Example 1.12. First, suppose one has 8 copies of o and two copies of \|. How many ways can one arrange these symbols in order? There are 10 spots, and we want to select 8 of them in which we place the os. So we have 10 8 . Example 1.13. Next, suppose one has 8 indistinguishable balls. How many ways can one put them in 3 boxes? Let us use sequences of os and \| s to represent an arrangement of balls in these 3 boxes; any such sequence that has \| at each side, 2 other \| s, and 8 os represents a way of arranging balls into boxes. For example, if one has \| o o \| o o o \| o o o \|, this would represent 2 balls in the rst box, 3 in the second, and 3 in the third. Altogether there are 8 + 4 symbols, the rst is a \| as is the last, so there are 10 symbols that can be 1.1. BASIC COUNTING PRINCIPLE AND COMBINATORICS 7 either \| or o. Also, 8 of them must be o. How many ways out of 10 spaces can one pick 8 of them into which to put a o? We just did that, so the answer is 10 8 . Example 1.14. Now, to nish, suppose we have $8,000 to invest in 3 mutual funds. Each mutual fund required you to make investments in increments of $1,000. How many ways can we do this? This is the same as putting 8 indistinguishable balls in 3 boxes, and we know the answer is 10 8 . 8 1. COMBINATORICS 1.2. Further examples and explanations 1.2.1. Generalized counting principle. Here we expand on the basic counting prin-ciple formulated in Section 1.1.1. One can visualize this principle by using the box method below. Suppose we have two experiments to be performed, namely, one experiment can result in n outcomes, and the second experiment can result in m outcomes. Each box represents the number of possible outcomes in that experiment. Experiment 1 Experiment 2 = Experiment 1 and 2 together m n = mn Example 1.15. There are 20 teachers and 100 students in a school. How many ways can we pick a teacher and student of the year? Solution: using the box method we get 20 × 100 = 2000. Generalized counting principle Suppose that k experiments are to be performed, with the number of possible outcomes being ni for the ith experiment. Then there are n1 · ... · nk possible outcomes of all k experiments together. Example 1.16. A college planning committee consists of 3 freshmen, 4 sophomores, 5 juniors, and 2 seniors. A subcommittee of 4 consists of 1 person from each class. How many choices are possible? The counting principle or the box method gives 3 × 4 × 5 × 2 = 120. Example 1.17 (Example 1.2 revisited). Recall that for 6-place license plates, with the rst three places occupied by letters and the last three by numbers, we have 26·26·26·10·10·10 choices. What if no repetition is allowed? Solution: the counting principle or the box method 26 · 25 · 24 · 10 · 9 · 8. Example 1.18. How many functions dened on k points are possible if each function can take values as either 0 or 1. Solution: the counting principle or the box method on the 1, . . . , k points gives us 2k possible functions. This is the generalized counting principle with n1 = n2 = ... = nk = 2. 1.2.2. Permutations. Now we give more examples on permutations, and we start with a more general results on the number of possible permutations. © Copyright 2017 Phanuel Mariano, 2020 Masha Gordina 1.2. FURTHER EXAMPLES AND EXPLANATIONS 9 Permutations revisited The number of dierent permutations of n objects of which n1 are alike, n2 are alike, ..., n2 are alike is equal to n! n1! · · · nr!. Example 1.19. How many ways can one arrange 5 math books, 6 chemistry books, 7 physics books, and 8 biology books on a bookshelf so that all the math books are together, all the chemistry books are together, and all the physics books are together. Solution: We can arrange the math books in 5! ways, the chemistry in 6! ways, the physics in 7! ways, and biology books in 8! ways. We also have to decide which set of books go on the left, which next, and so on. That is the same as the number of ways of arranging the letters M,C,P, and B, and there are 4! ways of doing that. So the total is 4! · (5! · 6! · 7! · 8!) ways. Now consider a couple of examples with repetitions. Example 1.20. How many ways can one arrange the letters a, a, b, b, c, c? Solution: let us rst re-label the letters by A, a, B, b, C, c. Then there are 6! = 720 ways to arrange these letters. But we have repeats (for example, Aa or aA) which produce the same arrangement for the original letters. So dividing by the number of repeats for A, a, B, b and C, c, so the answer is 6! (2!)3 = 90. Example 1.21. How many dierent letter arrangements can be formed from the word PEPPER? Solution: There are three copies of P and two copies of E, and one of R. So the answer is 6! 3!2!1! = 60. Example 1.22. Suppose there are 4 Czech tennis players, 4 U.S. players, and 3 Russian players, in how many ways could they be arranged, if we do not distinguish players from the same country? Solution: 11! 4!4!3!. 1.2.3. Combinations. Below are more examples on combintations. 10 1. COMBINATORICS Example 1.23. Suppose there are 9 men and 8 women. How many ways can we choose a committee that has 2 men and 3 women? Solution: We can choose 2 men in 9 2 ways and 3 women in 8 3 ways. The number of committees is then the product 9 2 · 8 3 . Example 1.24. Suppose somebody has n friends, of whom k are to be invited to a meeting. (1) How many choices do exist for such a meeting if two of the friends will not attend together? (2) How many choices do exist if 2 of the friends will only attend together? Solution: (1) We can divide all possible groups into two (disjoint) parts: one is for groups of friends none of which are these two, and another which includes exactly one of these two friends. There are n−2 k groups in the rst part, and n−2 k−1 in the second. For the latter we also need to account for a choice of one out of these two incompatible friends. So altogether we have n −2 k + 2 1 · n −2 k −1 (2) Again, we split all possible groups into two parts: one for groups which have none of the two inseparable friends, and the other for groups which include both of these two friends. Then n −2 k + 1 · 1 · n −2 k −2 . Theorem 1.1 The binomial theorem (x + y)n = n X k=0 n k xkyn−k. Proof. We give two proofs. First proof: let us expand the left-hand side (x+y)·...·(x+y). This is the sum of 2n terms, and each term has n factors. For now we keep each product in the order we expanded the left-hand side, therefore we have all possible (nite) sequences of variables x and y, with the total power being n. We would like to collect all the terms having the same number of xs and ys. Counting all the terms having k copies of x and n −k copies of n is the same as asking in a sequence of n positions, how many ways can one choose k of them in which to put x. The 1.2. FURTHER EXAMPLES AND EXPLANATIONS 11 answer is n k which gives the coecient for xkyn−k. To illustrate it we take k = 2 and n = 3, then all possible terms are x · x · y x · y · x y · x · x Second proof: we will use (mathematical) induction on n. For n = 1 we have that the left-hand side is x + y, and the right-hand side 1 X k=0 1 k xky1−k = 1 0 x0y1−0 + 1 1 x1y1−1 = y + x = x + y, so the statement holds for n = 1. Suppose now that the statement holds for n = N, we would like to show it for n = N + 1. (x + y)N+1 = (x + y) (x + y)N = (x + y) N X k=0 N k xkyN−k = x N X k=0 N k xkyN−k + y N X k=0 N k xkyN−k = N X k=0 N k xk+1yN−k + N X k=0 N k xkyN−k+1 = N+1 X k=1 N k −1 xkyN−k+1 + N X k=0 N k xkyN−k+1, where we replaced k by k −1 in the rst sum. Then we see that (x + y)N+1 = N N xN+1y0 + N X k=1 N k −1 + N k xkyN−k+1 + N 0 x0yN+1 = xN+1 + N X k=1 N k −1 + N k xkyN−k+1 + yN+1 = N+1 X k=0 N + 1 k . Here we used Example 1.26. □ Example 1.25. We can use combinatorics to show that 10 4 = 9 3 + 9 4 without evaluating these expressions explicitly. Solution: the left-hand side represents the number of committees consisting of 4 people out of the group of 10 people. Now we would like to represent the right-hand side. Let's say Tom Brady is one these ten people, and he might be in one of these committees and he is 12 1. COMBINATORICS 0 0 1 0 1 1 2 0 2 1 2 2 3 0 3 1 3 2 3 3 4 0 4 1 4 2 4 3 4 4 5 0 5 1 5 2 5 3 5 4 5 5 6 0 6 1 6 2 6 3 6 4 6 5 6 6 7 0 7 1 7 2 7 3 7 4 7 5 7 6 7 7 Pascal's triangle special, so we want to know when he will be there or not. When he is in the committee of 4, then there are 1 · 9 3 number of ways of having a committee with Tom Brady as a member, while 9 4 is the number of committees that do not have Tom Brady as a member. Adding it up gives us the number of committees of 4 people chosen out of the 10. Example 1.26. The more general identity is n k = n−1 k−1 + n−1 k which can be proven either using the same argument or a formula for binomial coecients. Example 1.27. Expand (x + y)3. Solution: (x + y)3 = y3 + 3xy2 + 3x2y + x3. 1.2.4. Multinomial Coecients. Example 1.28. Suppose we are to assign 10 police ocers: 6 patrols, 2 in station, 2 in schools. Then there are 10! 6!2!2! dierent assignments. Example 1.29. We have 10 ags: 5 of them are blue, 3 are red, and 2 are yellow. These ags are indistinguishable, except for their color. How many dierent ways can we order them on a ag pole? Solution: 10! 5!3!2!. 1.2. FURTHER EXAMPLES AND EXPLANATIONS 13 Example 1.30 (Exercise 1.13 revisited). Suppose one has n indistinguishable balls. How many ways can one put them in k boxes, assuming n > k? Solution: as in Exercise 1.13 we use sequences of os and \| s to represent each an arrangement of balls in boxes; any such sequence that has \| at each side, k −1 copies of \| s, and n copies of os. How many dierent ways can we arrange this, if we have to start with \| and end with \|? Between these, we are only arranging n + k −1 symbols, of which only n are os. So the question can be re-formulated as this: how many ways out of n+k −1 spaces can one pick n of them into which to put an o? This gives n+k−1 n . Note that this counts all possible ways including the ones when some of the boxes can be empty. Suppose now we want to distribute n balls in k boxes so that none of the boxes are empty. Then we can line up n balls represented by os, instead of putting them in boxes we can place \| s in spaces between them. Note that we should have a \| on each side, as all balls have to be put to a box. So we are left with k −1 copies of \| s to be placed among n balls. This means that we have n −1 places, and we need to pick k −1 out of these to place \| s. So we can reformulate the problem as choose k −1 places out of n −1, and so the answer is n−1 k−1 . We can check that for n = 3 and k = 2 we indeed have 4 ways of distributing three balls in two boxes, and only two ways if every box has to have at least one ball. 14 1. COMBINATORICS 1.3. Exercises Exercise 1.1. Suppose a license plate must consist of 7 numbers or letters. How many license plates are there if (A) there can only be letters? (B) the rst three places are numbers and the last four are letters? (C) the rst three places are numbers and the last four are letters, but there can not be any repetitions in the same license plate? Exercise 1.2. A school of 50 students has awards for the top math, English, history and science student in the school (A) How many ways can these awards be given if each student can only win one award? (B) How many ways can these awards be given if students can win multiple awards? Exercise 1.3. A password can be made up of any 4 digit combination. (A) How many dierent passwords are possible? (B) How many are possible if all the digits are odd? (C) How many can be made in which all digits are dierent or all digits are the same? Exercise 1.4. There is a school class of 25 people made up of 11 guys and 14 girls. (A) How many ways are there to make a committee of 5 people? (B) How many ways are there to pick a committee of all girls? (C) How many ways are there to pick a committee of 3 girls and 2 guys? Exercise 1.5. If a student council contains 10 people, how many ways are there to elect a president, a vice president, and a 3 person prom committee from the group of 10 students? Exercise 1.6. Suppose you are organizing your textbooks on a book shelf. You have three chemistry books, 5 math books, 2 history books and 3 English books. (A) How many ways can you order the textbooks if you must have math books rst, English books second, chemistry third, and history fourth? (B) How many ways can you order the books if each subject must be ordered together? Exercise 1.7. If you buy a Powerball lottery ticket, you can choose 5 numbers between 1 and 59 (picked on white balls) and one number between 1 and 35 (picked on a red ball). How many ways can you (A) win the jackpot (guess all the numbers correctly)? (B) match all the white balls but not the red ball? (C) match exactly 3 white balls and the red ball? (D) match at least 3 white balls and the red ball? 1.3. EXERCISES 15 Exercise 1.8. A couple wants to invite their friends to be in their wedding party. The groom has 8 possible groomsmen and the bride has 11 possible bridesmaids. The wedding party will consist of 5 groomsmen and 5 bridesmaids. (A) How many wedding party's are possible? (B) Suppose that two of the possible groomsmen are feuding and will only accept an invi-tation if the other one is not going. How many wedding parties are possible? (C) Suppose that two of the possible bridesmaids are feuding and will only accept an invi-tation if the other one is not going. How many wedding parties are possible? (D) Suppose that one possible groomsman and one possible bridesmaid refuse to serve to-gether. How many wedding parties are possible? Exercise 1.9. There are 52 cards in a standard deck of playing cards. The poker hand consists of ve cards. How many poker hands are there? Exercise 1.10. There are 30 people in a communications class. Each student must inter-view one another for a class project. How many total interviews will there be? Exercise 1.11. Suppose a college basketball tournament consists of 64 teams playing head to head in a knockout style tournament. There are 6 rounds, the round of 64, round of 32, round of 16, round of 8, the nal four teams, and the nals. Suppose you are lling out a bracket, such as this, which species which teams will win each game in each round. How many possible brackets can you make? Exercise 1.12. We need to choose a group of 3 women and 3 men out of 5 women and 6 men. In how many ways can we do it if 2 of the men refuse to be chosen together? Exercise 1.13. Find the coecient in front of x4 in the expansion of (2x2 + 3y)4. Exercise 1.14. In how many ways can you choose 2 or less (maybe none!) toppings for your ice-cream sundae if 6 dierent toppings are available? (You can use combinations here, but you do not have to. Next, try to nd a general formula to compute in how many ways you can choose k or less toppings if n dierent toppings are available 16 1. COMBINATORICS Exercise∗1.1. Use the binomial theorem to show that n X k=0 n k = 2n, n X k=0 (−1)k n k = 0. Exercise∗1.2. Prove the multinomial theorem (x1 + ... + xk)n = X (n1,··· ,nk) n1+···+nk=n n n1,...,nk xn1 1 · ... · xnk k . Exercise∗1.3. Show that there are n−1 k−1 distinct positive integer-valued vectors (x1, ..., xk) satisfying x1 + ... + xk = n, xi > 0 for all i = 1, ..., k. Exercise∗1.4. Show that there are n+k−1 k−1 distinct non-positive integer-valued vectors (x1, ..., xk) satisfying x1 + ... + xk = n, xi ⩾0 for all i = 1, ..., k. Exercise∗1.5. Consider a smooth function of n variables. How many dierent partial derivatives of order k does f possess? 1.4. SELECTED SOLUTIONS 17 1.4. Selected solutions Solution to Exercise 1.1(A): in each of the seven places we can put any of the 26 letters giving 267 possible letter combinations. Solution to Exercise 1.1(B): in each of the rst three places we can place any of the 10 digits, and in each of the last four places we can put any of the 26 letters giving a total of 103 · 264. Solution to Exercise 1.1(C): if we can not repeat a letter or a number on a license plate, then the number of license plates becomes (10 · 9 · 8) · (·26 · 25 · 24 · 23) . Solution to Exercise 1.2(A): 50 · 49 · 48 · 47 Solution to Exercise 1.2(B): 504 Solution to Exercise 1.3(A): 104 Solution to Exercise 1.3(B): 54 Solution to Exercise 1.3(C): 10 · 9 · 8 · 7 + 10 Solution to Exercise 1.4(A): 25 5 Solution to Exercise 1.4(B): 14 5 Solution to Exercise 1.4(C): 14 3 · 11 2 Solution to Exercise 1.5: 10 · 9 · 8 3 Solution to Exercise 1.6(A): 5!3!3!2! Solution to Exercise 1.6(B): 4! (5!3!3!2!) Solution to Exercise 1.7(A): 1 18 1. COMBINATORICS Solution to Exercise 1.7(B): 1 · 34 Solution to Exercise 1.7(C): 5 3 · 54 2 · 1 1 Solution to Exercise 1.7(D): 5 3 · 54 2 · 1 1 + 5 4 · 54 1 · 1 1 + 1 Solution to Exercise 1.8(A): 8 5 · 11 5 Solution to Exercise 1.8(B): 6 5 · 11 5 + 2 1 · 6 4 · 11 5 Solution to Exercise 1.8(C): 8 5 · 9 5 + 8 5 · 2 1 · 9 4 Solution to Exercise 1.8(D): 7 5 · 10 5 + 1 · 7 4 · 10 5 + 7 5 · 1 · 10 4 Solution to Exercise 1.9: 52 5 Solution to Exercise 1.10: 30 2 Solution to Exercise 1.11: First notice that the 64 teams play 63 total games: 32 games in the rst round, 16 in the second round, 8 in the 3rd round, 4 in the regional nals, 2 in the nal four, and then the national championship game. That is, 32+16+8+4+2+1= 63. Since there are 63 games to be played, and you have two choices at each stage in your bracket, there are 263 dierent ways to ll out the bracket. That is, 263 = 9, 223, 372, 036, 854, 775, 808. Solution to Exercise∗1.1: use the binomial formula (x + y)n = n X k=0 n k xkyn−k with x = y = 1 to see 2n = (1 + 1)n = n X k=0 n k · 1k · 1n−k = n X k=0 n k , 1.4. SELECTED SOLUTIONS 19 and with x = −1, y = 1 0 = (−1 + 1)n = n X k=0 n k · (−1)k · (1)n−k = n X k=0 n k (−1)k . Solution to Exercise∗1.2: we can prove the statement using mathematical induction on k. For k = 1 we have (x1)n = X n1=n n n1 x1 = xn 1, which is true; for k = 2 we have (x1 + x2)n = X (n1,n2) n1+n2=n n n1,n2 xn1 1 · xn2 2 = n X n1=0 n n1 xn1 1 · xn−n1 2 , which is the binomial formula itself. Now suppose the multinomial formula holds for k = K (induction hypothesis), that is, (x1 + ... + xK)n = X (n1,··· ,nK) n1+···+nK=n n n1,...,nK · xn1 1 · ... · xnK K , and we need to show (x1 + ... + xK+1)n = X (n1,··· ,nK+1) n1+···+nK+1=n n n1,...,nK+1 · xn1 1 · ... · xnK+1 K+1 . Denote y1 = x1, .., yK−1 := xK−1, yK := xK + xK+1, then by the induction hypothesis (x1 + ... + xK+1)n = (y1 + ... + yK)n = X (n1,··· ,nK) n1+···+nK=n n n1,...,nK · yn1 1 · ... · ynK−1 K−1 · ynK K = X (n1,··· ,nK) n1+···+nK=n n n1,...,nK · xn1 1 · ... · xnK−1 K−1 · (xK + xK+1)nK . By the binomial formula (xK + xK+1)nK = nK X m=1 nK m · xm K · xnK−m K+1 , therefore 20 1. COMBINATORICS (x1 + ... + xK+1)n = X (n1,··· ,nK) n1+···+nK=n n n1,...,nK · xn1 1 · ... · xnK−1 K−1 · nK X m=1 nK m · xm K · xnK−m K+1 . It is easy to see (using the denition of multinomial coecients) that n n1,...,nK nK m = n n1,...,nK,m , n1 + ... + nK + m = n. Indeed, n n1,...,nK nK m = n! n1!n2! · ... · nK−1! · nK! nK! m! (nK −m)! = n! n1!n2! · ... · nK−1! · m! (nK −m)! = n n1,...,nK,m . Thus (x1 + ... + xK+1)n = X (n1,··· ,nK) n1+···+nK=n nK X m=1 n n1,...,nK,m · xn1 1 · ... · xnK−1 K−1 · ·xm K · xnK−m K+1 . Note that nK = m+(nK −m), so if we denote m1 := n1, m2 := n2, ..., mK−1 := nK−1, mK := m, mK+1 := nK −m then we see that (x1 + ... + xK+1)n = X (m1,··· ,mK,mK+1) m1+···+mK+1=n n m1,...,mK,mK+1 · xm1 1 · ... · xmK−1 K−1 · xmK K · xmK+1 K+1 which is what we wanted to show. Solution to Exercise∗1.3: this is the same problem as dividing n indistinguishable balls into k boxes in such a way that each box has at least one ball. To do so, you can select k −1 of the n −1 spaces between the objects. There are n−1 k−1 possible selections that is equal to the number of possible positive integer solutions to the equation. Solution to Exercise∗1.4: dene yi := xi + 1 and apply the previous problem. Solution to Exercise∗1.5: the same answer as in the previous problem.
15789	https://electricala2z.com/electrical-circuits/electrical-units-and-metric-prefixes-examples/	Skip to content Electrical Units and Metric Prefixes by Ahmed Faizan The article covers the use of electrical units such as amperes, volts, ohms, and watts, as well as the application of metric prefixes to express large or small electrical values. It also explains the importance of converting between base units and their multiples or submultiples for accurate measurements in electrical circuits. Metric Prefixes Most often measurements made on electric circuits are that of current, voltage, resistance, and power. The base units—ampere, volt, ohm, and watts—are the values most commonly used to measure them. Table 1 lists these basic electrical quantities and the symbols that identify them. Table 1: Electrical Units, Symbols, and Definition In certain circuit applications the basic electrical units—volt, ampere, ohm, and watt—are either too small or too big to express conveniently. In such cases metric prefixes are often used. Recognizing the meaning of a prefix reduces the possibility of confusion in interpreting data. Common metric prefixes are shown in Table 2. Table 2: Common Metric Prefixes and their Symbols A metric prefix precedes a unit of measure or its symbol to form decimal multiples or submultiples. Prefixes are used to reduce the quantity of zeros in numerical equivalencies. For example, in an electrical system, the signal from a sensor may have strength of 0.00125 V, while the voltage applied to the input of a distribution transformer may be in the 27,000-V range. With prefixes, these values would be expressed as 1.25 mV (millivolts) and 27 kV (kilovolts), respectively. Figure 1 shows examples of prefixes used in the rating of electric components. Figure 1 Prefixes used in the rating of electric components. Knowing how to convert metric prefixes back to base units is needed when reading digital multimeters or using electric circuit formulas. Figure 2 and the following examples illustrate how many positions the decimal point is moved to get from a base unit to a multiple or a submultiple of the base unit. Figure 2 Movement of the decimal point to and from base units. EXAMPLE 1 To convert amperes (A) to milliamperes (mA), it is necessary to move the decimal point three places to the right (this is the same as multiplying the number by 1,000). EXAMPLE 2 To convert milliamperes (mA) to amperes (A), it is necessary to move the decimal point three places to the left (this is the same as multiplying by 0.001). EXAMPLE 3 To convert volts (V) to kilovolts (kV), it is necessary to move the decimal point three places to the left. EXAMPLE 4 To convert from megohms (MΩ) to ohms (Ω), it is necessary to move the decimal point six places to the right. EXAMPLE 5 To convert from microamperes (μA) to amperes (A), it is necessary to move the decimal point six places to the left. Review Questions What is the base unit and symbol used for electric: Current Voltage Resistance Power Write the metric prefix and symbol used to represent each of the following: One-thousandth One million One millionth One thousand Convert each of the following: 2,500 Ω to kilohms 120 kΩ to ohms 1,500,000 Ω to megohms 2.03 MΩ to ohms 0.000466 A to micro-amps 0.000466 A to milliamps 378 mV to volts 475 Ω to kilohms 28 μA to amps 5 kΩ + 850 Ω to kilohms 40,000 kV to megavolts 4,600,000 μA to amps 2.2 kΩ to ohms Review Questions – Answers (a) Ampere A, (b) Volt V, (c) Ohm Ω, (d) Watt W (a) Milli m, (b) Mega M, (c) Micro μ, (d) Kilo k (a) 2.5 kΩ, (b) 120,000 Ω, (c) 1.5 MΩ, (d) 2,030,000 Ω, (e) 466 μA, (f) 0.466 mA, (g) 0.378 V, (h) 0.475 kΩ, (i) 0.000028 A, (j) 5.85 kΩ, (k) 40 MV, (l) 4.6 A, (m) 2,200 Ω Electrical Units Key Takeaways Understanding electrical units and the use of metric prefixes is essential for accurate measurements in electrical circuits. These concepts help simplify the representation of large or small values, making data easier to interpret and reducing the risk of confusion. The ability to convert between base units and their multiples or submultiples is crucial for effectively analyzing and troubleshooting electrical systems, ensuring precise calculations and efficient system design. Electrical A2Z is an internal initiative of AVO Engineering, a collaborative team of engineers, designers, developers, and technical content writers, primarily engaged in collaborating with electronic manufacturers and distributors. Our motivation behind the inception of Electrical A2Z stemmed from our recognition that individuals have diverse learning preferences. Consequently, we aspired to offer another, ideally high-quality, learning resource for individuals seeking to gain proficiency in electronics. About Us All contents are Copyright © 2025 by electricala2z.com
15790	https://books.google.com.cu/books?id=lpUwmgEACAAJ&source=gbs_book_other_versions_r&cad=1	Physics for Scientists and Engineers with Modern Physics - Douglas C. Giancoli - Google Libros Sign in Hidden fields Libros Add to my library Mi biblioteca Ayuda Búsqueda avanzada de libros Mi biblioteca Mi historial Physics for Scientists and Engineers with Modern Physics Douglas C. Giancoli Pearson Education, 2008 - 950 páginas Key Message:This book aims to explain physics in a readable and interesting manner that is accessible and clear, and to teach readers by anticipating their needs and difficulties without oversimplifying. Physics is a description of reality, and thus each topic begins with concrete observations and experiences that readers can directly relate to. We then move on to the generalizations and more formal treatment of the topic. Not only does this make the material more interesting and easier to understand, but it is closer to the way physics is actually practiced. Key Topics: ELECTRIC CHARGE AND ELECTRIC FIELD, GAUSS'S LAW, ELECTRIC POTENTIAL, CAPACITANCE, DIELECTRICS, ELECTRIC ENERGY STORAGE, ELECTRIC CURRENTS AND RESISTANCE, DC CIRCUITS, MAGNETISM, SOURCES OF MAGNETIC FIELD, ELECTROMAGNETIC INDUCTION AND FARADAY'S LAW, INDUCTANCE, ELECTROMAGNETIC OSCILLATIONS, AND AC CIRCUITS, MAXWELL'S EQUATIONS AND ELECTROMAGNETIC WAVES, LIGHT: REFLECTION AND REFRACTION, LENSES AND OPTICAL INSTRUMENTS, THE WAVE NATURE OF LIGHT; INTERFERENCE, DIFFRACTION AND POLARIZATION, Market Description:This book is written for readers interested in learning the basics of physics. More » Otras ediciones - Ver todas ‹ 2008 Limited preview 2004 No preview 2007 M07 27 No preview 2008 No preview › Acerca del autor(2008) Douglas C. Giancoli obtained his BA in physics (summa cum laude) from UC Berkeley, his MS in physics at MIT, and his PhD in elementary particle physics back at the UC Berkeley. He spent 2 years as a post-doctoral fellow at UC Berkeley's Virus lab developing skills in molecular biology and biophysics. His mentors include Nobel winners Emilio Segr and Donald Glaser. He has taught a wide range of undergraduate courses, traditional as well as innovative ones, and continues to update his textbooks meticulously, seeking ways to better provide an understanding of physics for students. Doug's favorite spare-time activity is the outdoors, especially climbing peaks. He says climbing peaks is like learning physics: it takes effort and the rewards are great. Información bibliográfica Título Physics for Scientists and Engineers with Modern Physics Página 2 de Physics for Scientists & Engineers with Modern Physics, Douglas C. Giancoli, ISBN 0321542142, 9780321542144 AutorDouglas C. Giancoli Edición 4, ilustrada Editor Pearson Education, 2008 ISBN 0132273594, 9780132273596 Largo 950 páginas Exportar citaBiBTeXEndNoteRefMan Acerca de Google Libros - Política de Privacidad - Condiciones de uso - Información para editores - Informar un problema - Ayuda - Página principal de Google
15791	https://www3.nd.edu/~dgalvin1/pdf/mindegree.pdf	Counting independent sets of a ﬁxed size in graphs with a given minimum degree John Engbers David Galvin April 24, 2012∗ Abstract Galvin showed that for all ﬁxed δ and suﬃciently large n, the n-vertex graph with minimum degree δ that admits the most independent sets is the complete bipartite graph Kδ,n−δ. He conjectured that except perhaps for some small values of t, the same graph yields the maximum count of independent sets of size t for each possible t. Evidence for this conjecture was recently provided by Alexander, Cutler, and Mink, who showed that for all triples (n, δ, t) with t ≥3, no n-vertex bipartite graph with minimum degree δ admits more independent sets of size t than Kδ,n−δ. Here we make further progress. We show that for all triples (n, δ, t) with δ ≤3 and t ≥3, no n-vertex graph with minimum degree δ admits more independent sets of size t than Kδ,n−δ, and we obtain the same conclusion for δ > 3 and t ≥2δ + 1. Our proofs lead us naturally to the study of an interesting family of critical graphs, namely those of minimum degree δ whose minimum degree drops on deletion of an edge or a vertex. 1 Introduction and statement of results An independent set (a.k.a. stable set) in a graph is a set of vertices spanning no edges. For a simple, loopless ﬁnite graph G = (V, E), denote by i(G) the number of independent sets in G. In this quantity is referred to as the Fibonacci number of G, motivated by the fact that for the path graph Pn its value is a Fibonacci number. It has also been studied in the ﬁeld of molecular chemistry, where it is referred to as the Merriﬁeld-Simmons index of G . A natural extremal enumerative question is the following: as G ranges over some family G, what is the maximum value attained by i(G), and which graphs achieve this ∗{jengbers, dgalvin1}@nd.edu; Department of Mathematics, University of Notre Dame, Notre Dame IN 46556. Galvin in part supported by National Security Agency grant H98230-10-1-0364. 1 maximum? This question has been addressed for numerous families. Prodinger and Tichy considered the family of n-vertex trees, and showed that the maximum is uniquely attained by the star K1,n−1. Granville, motivated by a question in combinatorial group theory, raised the question for the family of n-vertex, d-regular graphs (see for more details). An approximate answer – i(G) ≤2n/2(1+o(1)) for all such G, where o(1) →0 as d →∞– was given by Alon in , and he speculated a more exact result, that the maximizing graph, at least in the case 2d\|n, is the disjoint union of n/2d copies of Kd,d. This speculation was conﬁrmed for bipartite G by Kahn , and for general regular G by Zhao . The family of n-vertex, m-edge graphs was considered by Cutler and Radcliﬀe in , and they observed that it is a corollary of the Kruskal-Katona theorem that the lex graph L(n, m) (on vertex set {1, . . . , n}, with edges being the ﬁrst m pairs in lexicographic order) maximizes i(G) in this class. Zykov considered the family of graphs with a ﬁxed number of vertices and ﬁxed independence number, and showed that the maximum is attained by the complement of a certain Tur´ an graph. (Zykov was actually considering cliques in a graph with given clique number, but by complementation this is equivalent to considering independent sets in a graph with given independence number.) Other papers addressing questions of this kind include , , and . Having resolved the question of maximizing i(G) for G in a particular family, it is natural to ask which graph maximizes it(G), the number of independent sets of size t in G, for each possible t. For many families, it turns out that the graph which maximizes i(G) also maximizes it(G) for all t. Wingard showed this for trees, Zykov showed this for graphs with a given independence number (see for a short proof), and Cutler and Radcliﬀe showed this for graphs on a ﬁxed number of edges (again, as a corollary of Kruskal-Katona). In , Kahn conjectured that for all 2d\|n and all t, no n-vertex, d-regular graph admits more independent sets of size t than the disjoint union of n/2d copies of Kd,d; this conjecture remains open, although asymptotic evidence appears in . The focus of this paper is the family G(n, δ) of n-vertex graphs with minimum degree δ. One might imagine that since removing edges increases the count of independent sets, the graph in G(n, δ) that maximizes the count of independent sets would be δ-regular (or close to), but this turns out not to be the case, even for δ = 1. The following result is from . Theorem 1.1. For n ≥2 and G ∈G(n, 1), we have i(G) ≤i(K1,n−1). For δ ≥2, n ≥4δ2 and G ∈G(n, δ), we have i(G) ≤i(Kδ,n−δ). What about maximizing it(G) for each t? The family G(n, δ) is an example of a family for which the maximizer of the total count is not the maximizer for each individual t. Indeed, consider the case t = 2. Maximizing the number of independent sets of size two is the same as minimizing the number of edges, and it is easy to see that for all ﬁxed δ and suﬃciently large n, there are n-vertex graphs with minimum 2 degree at least δ which have fewer edges than Kδ,n−δ (consider for example a δ-regular graph, or one which has one vertex of degree δ +1 and the rest of degree δ). However, we expect that anomalies like this occur for very few values of t. Indeed, the following conjecture is made in . Conjecture 1.2. For each δ ≥1 there is a C(δ) such that for all t ≥C(δ), n ≥2δ and G ∈G(n, δ), we have it(G) ≤it(Kδ,n−δ) = n −δ t + δ t . The case δ = 1 of Conjecture 1.2 is proved in , with C(1) as small as it possible can be, namely C(1) = 3. In , Alexander, Cutler and Mink looked at the subfamily Gbip(n, δ) of bipartite graphs in G(n, δ), and resolved the conjecture in the strongest possible way for this family. Theorem 1.3. For δ ≥1, n ≥2δ, t ≥3 and G ∈Gbip(n, δ), we have it(G) ≤ it(Kδ,n−δ). This provides good evidence for the truth of the strongest possible form of Conjecture 1.2, namely that we may take C(δ) = 3. The purpose of this paper is to make signiﬁcant progress towards this strongest possible conjecture. We completely resolve the cases δ = 2 and 3, and for larger δ we deal with all but a small fraction of cases. Theorem 1.4. 1. For δ = 2, t ≥3 and G ∈G(n, 2), we have it(G) ≤it(K2,n−2). For n ≥5 and 3 ≤t ≤n −2 we have equality iﬀG = K2,n−2 or G is obtained from K2,n−2 by joining the two vertices in the partite set of size 2. 2. For δ = 3, t ≥3 and G ∈G(n, 3), we have it(G) ≤it(K3,n−3). For n ≥6 and t = 3 we have equality iﬀG = K3,n−3; for n ≥7 and 4 ≤t ≤n −3 we have equality iﬀG = K3,n−3 or G is obtained from K3,n−3 by adding some edges inside the partite set of size 3. 3. For δ ≥3, t ≥2δ + 1 and G ∈G(n, δ), we have it(G) ≤it(Kδ,n−δ). For n ≥3δ + 1 and 2δ + 1 ≤t ≤n −δ we have equality iﬀG = Kδ,n−δ or G is obtained from Kδ,n−δ by adding some edges inside the partite set of size δ. (Note that there is some overlap between parts 2 and 3 above.) Part 1 above provides an alternate proof of the δ = 2 case of the total count of independent sets, originally proved in . Corollary 1.5. For n ≥4 and G ∈G(n, 2), we have i(G) ≤i(K2,n−2). For n = 4 and n ≥6 there is equality iﬀG = K2,n−2. 3 Proof. The result is trivial for n = 4. For n = 5, it is easily veriﬁed by inspection, and we ﬁnd that both C5 and K2,3 have the same total number of independent sets. So we may assume n ≥6. We clearly have i(K′ 2,n−2) < i(K2,n−2), where K′ 2,n−2 is the graph obtained from K2,n−2 by joining the two vertices in the partite set of size 2. For all G ∈G(n, 2) diﬀerent from both K2,n−2 and K′ 2,n−2, Theorem 1.4 part 1 tells us that it(G) ≤ it(K2,n−2) −1 for 3 ≤t ≤n −2. For t = 0, 1, n −1 and n we have it(G) = it(K2,n−2) (with the values being 1, n, 0 and 0 respectively). We have i2(G) ≤ n 2 −n (this is the number of non-edges in a 2-regular graph), and so i2(G) ≤i2(K2,n−2) + n 2 −n − n −2 2 −1 = i2(K2,n−2) + n −4. (1) Putting all this together we get i(G) ≤i(K2,n−2). If G is not 2-regular then we have strict inequality in (1) and so i(G) < i(K2,n−2). If G is 2-regular, then (as we will show presently) we have i3(G) < i3(K2,n−2) −1 and so again i(G) < i(K2,n−2). To see the inequality concerning independent sets of size 3 note that in any 2-regular graph the number of independent sets of size 3 that include a ﬁxed vertex v is the number of non-edges in the graph induced by the n −3 vertices V \ {v, x, y} (where x and y are the neighbors of v), which is at most n−3 2 −(n −4). It follows that i3(G) ≤1 3 n n −3 2 −(n −4) < n −2 3 −1. The paper is laid out as follows. In Section 2 we make some easy preliminary observations that will be used throughout the rest of the paper, and we introduce the important ideas of ordered independent sets and critical graphs for a given minimum degree. In Section 3 we deal with the case δ = 2 (part 1 of Theorem 1.4). We begin Section 4 with the proof of part 3 of Theorem 1.4, and then explain how the argument can be improved (within the class of critical graphs). This improvement will be an important ingredient in the δ = 3 case (part 2 of Theorem 1.4) whose proof we present in section 5. Finally we present some concluding remarks and conjectures in Section 6. Notation: Throughout the paper we use N(v) for the set of vertices adjacent to v, and d(v) for \|N(v)\|. We write u ∼v to indicate that u and v are adjacent (and u ≁v to indicate that they are not). We use G[Y ] to denote the subgraph induced by a subset Y of the vertices, and E(Y ) for the edge set of this subgraph. Finally, for t ∈N we use xt to indicate the falling power x(x −1) . . . (x −(t −1)). 4 2 Preliminary remarks For integers n, δ and t, let P(n, δ, t) denote the statement that for every G ∈G(n, δ), we have it(G) ≤it(Kδ,n−δ). An important observation is that if we prove P(n, δ, t) for some triple (n, δ, t) with t ≥δ + 1, we automatically have P(n, δ, t + 1). The proof introduces the important idea of considering ordered independent sets, that is, independent sets in which an order is placed on the vertices. Lemma 2.1. For δ ≥2 and t ≥δ + 1, if G ∈G(n, δ) satisﬁes it(G) ≤it(Kδ,n−δ) then it+1(G) ≤it+1(Kδ,n−δ). Moreover, if t < n −δ and it(G) < it(Kδ,n−δ) then it+1(G) < it+1(Kδ,n−δ). Corollary 2.2. For δ ≥2 and t ≥δ + 1, P(n, δ, t) ⇒P(n, δ, t + 1). Proof. Fix G ∈G(n, δ). By hypothesis, the number of ordered independent sets in G of size t is at most (n −δ)t. For each ordered independent set of size t in G there are at most n−(t+δ) vertices that can be added to it to form an ordered independent set of size t + 1 (no vertex of the independent set can be chosen, nor can any neighbor of any particular vertex in the independent set). This leads to a bound on the number of ordered independent sets in G of size t + 1 of (n −δ)t(n −(t + δ)) = (n −δ)t+1. Dividing by (t + 1)!, we ﬁnd that it+1(G) ≤ n−δ t+1 = it+1(Kδ,n−δ). If we have it(G) < n−δ t then we have strict inequality in the count of ordered independent sets of size t, and so also as long as n −(δ + t) > 0 we have strict inequality in the count for t + 1, and so it+1(G) < n−δ t+1 . Given Corollary 2.2, in order to prove P(n, δ, t) for n ≥n(δ) and t ≥t(δ) it will be enough to prove P(n, δ, t(δ)). Many of our proofs will be by induction on n, and will be considerably aided by the following simple observation. Lemma 2.3. Fix t ≥3. Suppose we know P(m, δ, t) for all m < n. Let G ∈G(n, δ) be such that there is v ∈V (G) with G −v ∈G(n −1, δ) (that is, G −v has minimum degree δ). Then it(G) ≤it(Kδ,n−δ). Equality can only occur if all of 1) it(G −v) = it(Kδ,n−1−δ), 2) G −v −N(v) is empty (has no edges), and 3) d(v) = δ hold. Proof. Counting ﬁrst the independent sets of size t in G that do not include v and then those that do, and bounding the former by our hypothesis on P(m, δ, t) and the latter by the number of subsets of size t −1 in G −v −N(v), we have (with Ek 5 denoting the empty graph on k vertices) it(G) = it(G −v) + it−1(G −v −N(v)) ≤ it(Kδ,n−1−δ) + it−1(En−1−d(v)) = n −1 −δ t + δ t + n −1 −δ t −1 = n −δ t + δ t = it(Kδ,n−δ). The statement concerning equality is evident. Lemma 2.3 allows us to focus on graphs with the property that each vertex has a neighbor of degree δ. Another simple lemma further restricts the graphs that must be considered. Lemma 2.4. If G′ is obtained from G by deleting edges, then for each t we have it(G) ≤it(G′). This leads to the following deﬁnition. Deﬁnition 2.5. Fix δ ≥1. A graph G with minimum degree δ is edge-critical if for any edge e in G, the minimum degree of G −e is δ −1. It is vertex-critical if for any vertex v in G, the minimum degree of G −v is δ −1. If it is both edge- and vertex-critical, we say that G is critical. Lemmas 2.3 and 2.4 allow us to concentrate mostly on critical graphs. In Section 3 (speciﬁcally Lemma 3.2) we obtain structural information about critical graphs in the case δ = 2, while much of Section 5 is concerned with the same problem for δ = 3. An easy upper bound on the number of independent sets of size t ≥1 in a graph with minimum degree δ is it(G) ≤n(n −(δ + 1))(n −(δ + 2)) · · · (n −(δ + (t −1))) t! . (2) This bound assumes that each vertex has degree δ, and moreover that all vertices share the same neighborhood. We will obtain better upper bounds by considering more carefully when these two conditions actually hold, as having many vertices which share the same neighborhood forces those vertices in the neighborhood to have large degree. To begin this process, it will be helpful to distinguish between vertices with degree δ and those with degree larger than δ. Set V=δ = {v ∈V (G) : d(v) = δ} 6 and V>δ = {v ∈V (G) : d(v) > δ}. Most of the proofs proceed by realizing that a critical graph must have at least one of a small list of diﬀerent structures in it, and we exploit the presence of a structure to signiﬁcantly dampen the easy upper bound. 3 Proof of Theorem 1.4, part 1 (δ = 2) Recall that we want to show that for δ = 2, t ≥3 and G ∈G(n, 2), we have it(G) ≤it(K2,n−2), and that for n ≥5 and 3 ≤t ≤n −2 we have equality iﬀ G = K2,n−2 or K′ 2,n−2 (obtained from G by joining the two vertices in the partite set of size 2). We concern ourselves initially with the inequality, and discuss the cases of equality at the end. By Corollary 2.2, it is enough to consider t = 3, and we will prove this case by induction on n, the base cases n ≤5 being trivial. So from here on we assume that n > 5 and that P(m, 2, 3) has been established for all m < n, and let G ∈G(n, 2) be given. By Lemmas 2.3 and 2.4 we may assume that G is critical. We begin with two lemmas, the ﬁrst of which is well-known (see e.g. ), and the second of which gives structural information about critical graphs (in the case δ = 2). Lemma 3.1. Let k ≥1 and 0 ≤t ≤k + 1. In the k-path Pk we have it(Pk) = k + 1 −t t . Let k ≥3 and 0 ≤t ≤k −1. In the k-cycle Ck we have it(Ck) = k −t t + k −t −1 t −1 . Lemma 3.2. Fix δ = 2. Let G be a connected n-vertex critical graph. Either 1. G is a cycle or 2. V (G) can be partitioned as Y1 ∪Y2 with 2 ≤\|Y1\| ≤n −3 in such a way that Y1 induces a path, Y2 induces a graph with minimum degree 2, each endvertex of the path induced by Y1 has exactly one edge to Y2, the endpoints of these two edges in Y2 are either the same or non-adjacent, and there are no other edges from Y1 to Y2. Proof. If G is not a cycle, then it has some vertices of degree greater than 2. If there is exactly one such vertex, say v, then by parity considerations d(v) is even and at least 4. Since all degrees are even, the edge set may be partitioned into cycles. Take 7 any cycle through v and remove v from it to get a path whose vertex set can be taken to be Y1. There remains the case when G has at least two vertices with degree larger than 2. Since G is edge-critical, V>δ forms an independent set and so there is a path on at least 3 vertices joining distinct vertices v1, v2 ∈V>δ, all of whose internal vertices u1, . . . , uk have degree 2 (the shortest path joining two vertices in V>δ would work). Since G is vertex-critical we must in fact have k ≥2, since otherwise u1 would be a vertex whose deletion leaves a graph with minimum degree 2. We may now take Y1 = {u1, . . . , uk}. Note that the Y2 endpoints (v1 and v2) of the edges from u1 and uk to Y2 are both in V>δ and so are non-adjacent. Armed with Lemmas 3.1 and 3.2 we now show that for critical G we have i3(G) < i3(K2,n−2) = n −2 3 . If G is the n-cycle, then we are done by Lemma 3.1. If G is a disjoint union of cycles, then choose one such, of length k, and call its vertex set Y1, and set Y2 = V \Y1. We will count the number of independent sets of size 3 in G by considering how the independent set splits across Y1 and Y2. For k ≥4 Lemma 3.1 gives a count of k−3 3 + k−4 2 for the number of independent sets of size 3 in Y1, and this is still a valid upper bound when k = 3. By induction there are at most n−k−2 3 independent sets of size 3 in Y2. There are k−1 2 −1 (n −k) independent sets with two vertices in Y1 and one in Y2 (the ﬁrst factor here simply counting the number of non-edges in a k-cycle). Finally, there are k n−k−1 2 −1 independent sets with one vertex in Y1 and two in Y2 (the second factor counting the number of non-edges in a 2-regular graph on n −k vertices). The sum of these bounds is easily seen to be n−2 3 −k, so strictly smaller than n−2 3 . We may now assume that G has a component that is not 2-regular. Choose one such component. Let Y1 be as constructed in Lemma 3.2 and let Y2 be augmented by including the vertex sets of all other components. Denote by v1, v2 the neighbors in Y2 of the endpoints of the path. Note that it is possible that v1 = v2, but if not then by Lemma 3.2 we have v1 ≁v2. We will again upper bound i3(G) by considering the possible splitting of independent sets across Y1 and Y2. By Lemma 3.1, there are k−2 3 independent sets of size 3 in Y1, and by induction there are at most n−k−2 3 independent sets of size 3 in Y2. The number of independent sets of size 3 in G that have two vertices in Y1 and one in Y2 is at most k −3 2 (n −k) + k −1 2 − k −3 2 (n −k −1). The ﬁrst term above counts those independent sets in which neither endpoint of the k-path is among the two vertices from Y1, and uses Lemma 3.1. The second term 8 upper bounds the number of independent sets in which at least one endpoint of the k-path is among the two vertices from Y1, and again uses Lemma 3.1. (Note that when k = 2 the application of Lemma 3.1 is not valid, since when we remove the endvertices we are dealing with a path of length 0, outside the range of validity of the lemma; however, the above bound is valid for k = 2, since it equals 1 in this case.) Finally, the number of independent sets of size 3 in G that have one vertex in Y1 and two in Y2 is at most (k −2) n −k 2 −\|E(Y2)\| + 2 X i=1 n −k −1 2 −\|E(Y2)\| + dY2(vi) . The ﬁrst term here counts the number of independent sets in which the one vertex from Y1 is not an endvertex, the second factor being simply the number of non-edges in G[Y2]. The second term counts those with the vertex from Y1 being the neighbor of vi, the second factor being the number of non-edges in G[Y2] −vi. The sum of all of these bounds, when subtracted from n−2 3 , simpliﬁes to −(k −1)n + k2 + k −3 + k\|E(Y2)\| −dY2(v1) −dY2(v2), (3) a quantity which we wish to show is strictly positive. Suppose ﬁrst that Y1 can be chosen so that v1 ̸= v2. Recall that in this case v1 ≁v2, so dY2(v1) + dY2(v2) ≤\|E(Y2)\|. Combining this with \|E(Y2)\| ≥n −k we get that (3) is at most 2k −3, which is indeed strictly positive for k ≥2. If v1 = v2 = v, then we ﬁrst note that \|E(Y2)\| = 1 2 X w∈Y2 dY2(w) ≥dY2(v) 2 + (n −k −1) (since G[Y2] has minimum degree 2). Inserting into (3) we ﬁnd that (3) is at most n −3 + k 2 −2 dY2(v). (4) This is clearly strictly positive for k ≥4, and for k = 3 strict positivity follows from dY2(v) < 2(n −3), which is true since in fact dY2(v) < n −3 in this case. If k = 2, then (4) is strictly positive unless dY2 = n −3 (the largest possible value it can take in this case). There is just one critical graph G with the property that for all possible choices of Y1 satisfying the conclusions of Lemma 3.2 we have \|Y1\| = 2, v1 = v2 = v and dY2(v) = n −3; this is the windmill graph (see Figure 1) consisting of (n −1)/2 triangles with a single vertex in common to all the triangles, and otherwise no overlap between the vertex sets (note that the degree condition on v forces G to be connected). A direct count gives (n −1)(n −3)(n −5)/6 < n−2 3 independent sets of size 3 in this particular graph. 9 · · · Figure 1: The windmill graph. This completes the proof that it(G) ≤it(K2,n−2) for all t ≥3 and G ∈G(n, 2). We now turn to considering the cases where equality holds in the range n ≥5 and 3 ≤t ≤n −2. For t = 3 and n = 5, by inspection we see that we have equality iﬀ G = K2,3 or K′ 2,3 (obtained from K2,3 by adding an edge inside the partite set of size 2). For larger n, we prove by induction that equality can be achieved only for these two graphs. If a graph G is not edge-critical, we delete edges until we obtain a graph G′ which is edge-critical, using Lemma 2.4 to get it(G) ≤it(G′). If G′ is critical, then the discussion in this section shows that we cannot achieve equality. If G′ is not vertex-critical, Lemma 2.3 and our induction hypothesis shows that we only achieve equality for G′ if there is v ∈V (G′) with G′ −v = K2,n−3 or K′ 2,n−3, G′ −v −N(v) empty, and d(v) = 2. First, notice that G′ −v = K′ 2,n−3 implies that G′ is not edge-critical, so equality can only occur when G′ −v = K2,n−3. If G′ −v = K2,n−3, the second and third conditions tell us that N(v) is exactly the partite set of size 2 in K2,n−3, that is, that G′ = K2,n−2. From here it is evident that equality can only occur for G = K2,n−2 or K′ 2,n−2. Now for each ﬁxed n ≥5, we conclude from Lemma 2.1 that for 3 ≤t ≤n −2 we cannot have equality unless G = K2,n−2 or K′ 2,n−2; and since the equality is trivial for these two cases, the proof is complete. 4 Proof of Theorem 1.4, part 3 (δ ≥3) Throughout this section we set h = \|V>δ\| and ℓ= \|V=δ\|; note that h + ℓ= n. We begin this section with the proof of Theorem 1.4 part 3; we then show how the method used may be improved to obtain a stronger result within the class of critical graphs (Lemma 4.1 below), a result which will play a role in the proof of Theorem 1.4, part 2 (δ = 3) that will be given in Section 5. Recall that we are trying to show that for δ ≥3, t ≥2δ + 1 and G ∈G(n, δ), we have it(G) ≤it(Kδ,n−δ), and that for n ≥3δ + 1 and 2δ + 1 ≤t ≤n −δ there is 10 equality iﬀG is obtained from Kδ,n−δ by adding some edges inside the partite set of size δ. As with Theorem 1.4 part 1 we begin with the inequality and discuss cases of equality at the end. By Corollary 2.2 it is enough to consider t = 2δ + 1. We prove P(n, δ, 2δ + 1) by induction on n. For n < 3δ + 1 the result is trivial, since in this range all G ∈G(n, δ) have it(G) = 0. It is also trivial for n = 3δ + 1, since the only graphs G in G(n, δ) with it(G) > 0 in this case are those that are obtained from Kδ,n−δ by the addition of some edges inside the partite set of size δ, and all such G have it(G) = 1 = it(Kδ,n−δ). So from now on we assume n ≥3δ + 2 and that P(m, δ, 2δ + 1) is true for all m < n, and we seek to establish P(n, δ, 2δ + 1). By Lemmas 2.3 and 2.4 we may restrict attention to G which are critical (for minimum degree δ). To allow the induction to proceed, we need to show that the number of ordered independent sets of size 2δ + 1 in G is at most (n −δ)2δ+1. We partition ordered independent sets according to whether the ﬁrst vertex is in V>δ or in V=δ. In the ﬁrst case (ﬁrst vertex in V>δ) there are at most h(n −(δ + 2))(n −(δ + 3)) · · · (n −(3δ + 1)) = h n n(n −(δ + 2))2δ < h n(n −δ)2δ+1 (5) ordered independent sets of size 2δ + 1, since once the ﬁrst vertex has been chosen there are at most n −(δ + 2) choices for the second vertex, then at most n −(δ + 3) choices for the third, and so on. In the second case (ﬁrst vertex in V=δ) there are at most ℓ(n −(δ + 1))(n −(δ + 2)) · · · (n −2δ) ways to choose the ﬁrst δ + 1 vertices in the ordered independent set. The key observation now is that since G is vertex-critical there can be at most δ −1 vertices distinct from v with the same neighborhood as v, where v is the ﬁrst vertex of the ordered independent set. It follows that one of choices 2 through δ has a neighbor w outside of N(v). Since w cannot be included in the independent set, there are at most (n −(2δ + 2))(n −(2δ + 3)) · · · (n −(3δ + 1)) choices for the ﬁnal δ vertices. Combining these bounds, there are at most ℓ n n(n −(δ + 1))2δ+1 n −(2δ + 1) < ℓ n(n −δ)2δ+1 ordered independent sets of size 2δ +1 that begin with a vertex from V=δ. Combining with (5) we get i2δ+1(G) < (n −δ)2δ+1/(2δ + 1)!, as required. This completes the proof that it(G) ≤it(Kδ,n−δ) for all t ≥2δ+1 and G ∈G(n, δ). We now turn to considering the cases where equality holds in the range n ≥3δ + 1 11 and 2δ + 1 ≤t ≤n −δ. For t = 2δ + 1 and n = 3δ + 1, we clearly have equality iﬀ G is obtained from Kδ,2δ+1 by adding some edges inside the partite set of size δ. For larger n, we prove by induction that equality can be achieved only for a graph of this form. If a graph G is not edge-critical, we delete edges until we obtain a graph G′ which is edge-critical, using Lemma 2.4 to get it(G) ≤it(G′). If G′ is critical, then the discussion in this section shows that we cannot achieve equality. If G′ is not vertex-critical, Lemma 2.3 and our induction hypothesis shows that we only achieve equality for G′ if there is v ∈V (G′) with G′ −v obtained from Kδ,n−δ−1 by adding some edges inside the partite set of size δ, G′ −v −N(v) empty, and d(v) = δ. First, notice that the cases where G′ −v ̸= Kδ,n−δ−1 imply that G′ is not edge-critical, so in fact equality can only occur when G′ −v = Kδ,n−δ−1. Since d(v) = δ the neighborhood of v cannot include all of the partite set of size n −1 −δ. If it fails to include a vertex of the partite set of size δ, there must be an edge in G −v −N(v); so in fact, N(v) is exactly the partite set of size δ and G′ = Kδ,n−δ. From here it is evident that equality can only occur for G obtained from Kδ,n−δ by adding some edges inside the partite set of size δ. Now for each ﬁxed n ≥3δ + 1, we conclude from Lemma 2.1 that for 2δ + 1 ≤ t ≤n −δ we cannot have equality unless G is obtained from Kδ,n−δ by adding some edges inside the partite set of size δ; and since the equality is trivial in these cases, the proof is complete. The ideas introduced here to bound the number of ordered independent sets in a critical graph can be modiﬁed to give a result that covers a slightly larger range of t, at the expense of requiring n to be a little larger. Speciﬁcally we have the following: Lemma 4.1. For all δ ≥3, t ≥δ + 1, n ≥3.2δ and vertex-critical G ∈G(n, δ), we have it(G) < it(Kδ,n−δ). For δ = 3 and t = 4 we get the same conclusion for vertex-critical G ∈G(n, 3) with n ≥8. Remark. The constant 3.2 has not been optimized here, but rather chosen for convenience. Proof. By Lemma 2.1 it is enough to consider t = δ + 1. The argument breaks into two cases, depending on whether G has at most δ −2 vertices with degree larger than m (a parameter to be speciﬁed later), or at least δ −1. The intuition is that in the former case, after an initial vertex v has been chosen for an ordered independent set, many choices for the second vertex should have at least two neighbors outside of N(v), which reduces subsequent options, whereas in the latter case, an initial choice of one of the at least δ −1 vertices with large degree should lead to few ordered independent sets. First suppose that G has at most δ −2 vertices with degree larger than m. Just as in (5), a simple upper bound on the number of ordered independent sets of size t 12 whose ﬁrst vertex is in V>δ is h n (n(n −(δ + 2))(n −(δ + 3)) · · · (n −(2δ + 1))) < h n(n −δ)δ+1. (6) There are ℓchoices for the ﬁrst vertex v of an ordered independent set that begins with a vertex from V=δ. For each such v, we consider the number of extensions to an ordered independent set of size δ + 1. This is at most x(n −(δ + 2))δ−1 + y(n −(δ + 3))δ−1 + z(n −(δ + 4))δ−1 (7) where x is the number of vertices in V (G) \ ({v} ∪N(v)) that have no neighbors outside N(v), y is the number with one neighbor outside N(v), and z is the number with at least 2 neighbors outside N(v). Note that x + y + z = n −δ −1, and that by vertex-criticality x ≤δ −1. Let u1 and u2 be the two lowest degree neighbors of v. By vertex-criticality and our assumption on the number of vertices with degree greater than m, the sum of the degrees of u1 and u2 is at most δ + m. Each vertex counted by y is adjacent to either u1 or u2, so counting edges out of u1 and u2 there are at most m + δ −2x −2 such vertices. For ﬁxed x we obtain an upper bound on (7) by taking y as large as possible, so we should take y = m + δ −2x −2 and z = n −m −2δ + x + 1. With these choices of y and z, a little calculus shows us that we obtain an upper bound by taking x as large as possible, that is, x = δ −1. This leads to an upper bound on the number of ordered independent sets of size t whose ﬁrst vertex is in V=δ of ℓ   (δ −1)(n −(δ + 2))δ−1+ (m −δ)(n −(δ + 3))δ−1+ (n −m −δ)(n −(δ + 4))δ−1  . Combining with (6) we see that are done (for the case G has at most δ −2 vertices with degree larger than m) as long as we can show that the expression above is strictly less than ℓ(n −δ)δ+1/n, or equivalently that n   (δ −1)(n −(δ + 2))(n −(δ + 3))+ (m −δ)(n −(δ + 3))(n −(2δ + 1))+ (n −m −δ)(n −(2δ + 1))(n −(2δ + 2))  < (n −δ)4 . (8) We will return to this presently; but ﬁrst we consider the case where G has at least δ −1 vertices with degree larger than m. An ordered independent set of size δ + 1 in this case either begins with one of δ −1 vertices of largest degree, in which case there are strictly fewer than (n −m −1)δ extensions, or it begins with one of the remaining n −δ + 1 vertices. For each such vertex v in this second case, the second vertex chosen is either one of the k = k(v) ≤δ −1 vertices that have the 13 same neighborhood as v, in which case there are at most (n −(δ + 2))δ−1 extensions, or it is one of the n−d(v)−1−k vertices that have a neighbor that is not a neighbor of v, in which case there are at most (n −(δ + 3))δ−1 extensions. We get an upper bound on the total number of extensions in this second case (starting with a vertex not among the δ −1 of largest degree) by taking k as large as possible and d(v) as small as possible; this leads to a strict upper bound on the number of ordered independent sets of size δ + 1 in the case G has at least δ −1 vertices with degree larger than m of (δ −1)(n −m −1)δ + (n −δ + 1) (δ −1)(n −(δ + 2))δ−1+ (n −2δ)(n −(δ + 3))δ−1 . We wish to show that this is at most (n −δ)δ+1. As long as m ≥δ we have n −m −i ≤n −δ −i, and so what we want is implied by   (δ −1)(n −m −1)(n −m −2)+ (n −δ + 1)(δ −1)(n −(δ + 2))+ (n −δ + 1)(n −2δ)(n −(2δ + 1))  ≤(n −δ)3. (9) Setting m = n/2, we ﬁnd that for δ ≥3, both (8) and (9) hold for all n ≥3.2δ. Indeed, in both cases at n = 3.2δ the right-hand side minus the left-hand side is a polynomial in δ (a quartic in the ﬁrst case and a cubic in the second) that is easily seen to be positive for all δ ≥3; and in both cases we can check that for each ﬁxed δ ≥3, when viewed as a function of n the right-hand side minus the left-hand side has positive derivative for all n ≥3.2δ. This completes the proof of the ﬁrst statement. It is an easy check that both (8) and (9) hold for all n ≥8 in the case δ = 3, completing the proof of the lemma. 5 Proof of Theorem 1.4, part 2 (δ = 3) Recall that we are trying to show that for δ = 3, t ≥3 and G ∈G(n, 3), we have it(G) ≤it(K3,n−3), and that for n ≥6 and t = 3 we have equality iﬀG = K3,n−3, while for n ≥7 and 4 ≤t ≤n −3 we have equality iﬀG is obtained from K3,n−3 by adding some edges inside the partite set of size 3. For t = 4 and n ≥7 we prove the result (including the characterization of uniqueness) by induction on n, with the base case n = 7 trivial. For n ≥8, Lemma 4.1 gives strict inequality for all vertex-critical G, so we may assume that we are working with a G which is non-vertex-critical. Lemma 2.3 now gives the inequality i4(G) ≤i4(K3,n−3), and the characterization of cases of inequality goes through exactly as it did for Theorem 1.4 parts 1 and 3. The result for larger t (including the characterization of uniqueness) now follows from Lemma 2.1. For t = 3, we also argue by induction on n, with the base case n = 6 trivial. For n ≥7, if G is not vertex-critical then the inequality i3(G) ≤i3(K3,n−3) follows 14 from Lemma 2.3, and the fact that there is equality in this case only for G = K3,n−3 follows exactly as it did in the proofs of Theorem 1.4 parts 1 and 3. So we may assume that G is vertex-critical. We will also assume that G is edge-critical (this assumption is justiﬁed because in what follows we will show i3(G) < i3(K3,n−3), and restoring the edges removed to achieve edge-criticality maintains the strictness of the inequality). Our study of critical 3-regular graphs will be based on a case analysis that adds ever more structure to the G under consideration. A useful preliminary observation is the following. Lemma 5.1. Fix δ = 3. If a critical graph G has a vertex w of degree n −3 or greater, then i3(G) < i3(K3,n−3). Proof. If d(w) > n −3 then there are no independent sets of size 3 containing w, and by Theorem 1.4 part 1 the number of independent sets of size 3 in G −w (a graph of minimum degree 2) is at most n−3 3 < i3(K3,n−3). If d(w) = n −3 and the two non-neighbors of w are adjacent, then we get the same bound. If they are not adjacent (so there is one independent set of size 3 containing w) and G −w is not extremal among minimum degree 2 graphs for the count of independent sets of size 3, then we also get the same bound, since now i3(G −w) ≤ n−3 3 −1. If G −w is extremal it is either K2,n−3 or K′ 2,n−3, and in either case w must be adjacent to everything in the partite set of size n −3 (to ensure that G has minimum degree 3), and then, since the non-neighbors of w are non-adjacent, it must be that G = K3,n−3, a contradiction since we are assuming that G is critical. 5.1 Regular G If G is 3-regular then we have i3(G) < n−3 3 + 1. We see this by considering ordered independent sets of size 3. Given an initial vertex v, we extend to an ordered independent set of size 3 by adding ordered non-edges from V \ (N(v) ∪{v}). Since G is 3-regular there are 3n ordered edges in total, with at most 18 of them adjacent either to v or to something in N(v). This means that the number of ordered independent sets of size 3 in G is at most n((n −4)(n −5) −(3n −18)) < (n −3)(n −4)(n −5) + 6 with the inequality valid as long as n ≥7. So from here on we may assume that G is not 3-regular, or equivalently that V>3 ̸= ∅. Remark. The argument above generalizes to show that δ-regular graphs have at most n−δ 3 + δ 3 independent sets of size 3, with equality only possible when n = 2δ. Let v ∈V (G) have a neighbor in V>δ. By criticality d(v) = 3. Let w1, w2, and w3 be the neighbors of v, listed in decreasing order of degree, so d(w1) = d, d(w2) = x and d(w3) = 3 satisfy 3 ≤x ≤d ≤n −4, the last inequality by Lemma 5.1 as well as d > 3 (see Figure 2). 15 v w1 with degree d > 3 w2 with degree 3 ≤x ≤d w3 with degree 3 Figure 2: The generic situation from the end of Section 5.1 on. 5.2 No edge between w3 and w2 We now precede by a case analysis that depends on the value of x as well as on the set of edges present among the wi’s. The ﬁrst case we consider is w3 ≁w2. In this case we give upper bounds on the number of independent sets of size 3 which contain v and the number which do not. There are n−4 2 −\|E(Y )\| independent sets of size 3 which include v, where Y = V \ (N(v) ∪{v}). We lower bound \|E(Y )\| by lower bounding the sum of the degrees in Y and then subtracting oﬀthe number of edges from Y to {v} ∪N(v). This gives \|E(Y )\| ≥3(n −4) −2 −(d −1) −(x −1) 2 = 3(n −4) −x −d 2 . (10) To bound the number of independent sets of size 3 which don’t include v, we begin by forming G′ from G by deleting v and (to restore minimum degree 3) adding an edge between w3 and w2 (we will later account for independent sets in G that contain both w2 and w3). The number of independent sets of size 3 in G′ is, by induction, at most i3(K3,n−4). But in fact, we may assume that the count is strictly smaller than this. To see this, note that if we get exactly i3(K3,n−4) then by induction G′ = K3,n−4. For n = 7 this forces G to have a vertex of degree 4 and so i3(G) < i3(K3,4) by Lemma 5.1. For n > 7, w3 must be in the partite set of size n −4 in G′ (to have degree 3) so since w2 ∼w3 (in G′), w2 must be in the partite set of size 3. To avoid creating a vertex of degree n −3 in G, w1 must be in the partite set of size n −4. But then all other vertices in the partite set of size n −4 only have neighbors of degree n −4 (in G), contradicting criticality. So we may now assume that the number of independent sets of size 3 in G which do not include v is at most n −4 3 + (n −x −2), (11) the extra n −x −2 being an upper bound on the number of independent sets of size 3 in G that include both w3 and w2. Combining (10) and (11) we ﬁnd that in this 16 case i3(G) ≤ n −4 2 −3(n −4) −x −d 2 + n −4 3 + (n −x −2). (12) As long as d < n + x −6 this is strictly smaller that i3(K3,n−3). Since x ≥3 and d < n −3, this completes the case w3 ≁w2. 5.3 Edge between w3 and w2, no edge between w3 and w1, degree of w2 large The next case we consider is w3 ∼w2, w3 ≁w1, and x > 3. In this case we can run an almost identical the argument to that of Section 5.2, this time adding the edge from w1 to w3 when counting the number of independent sets of size 3 that don’t include v. We add 1 to the right-hand side of (10) (to account for the fact that there is now only one edge from w3 to Y instead of 2, and only x −2 edges from w2 to Y instead of x −1) and replace (11) with n−4 3 + 1 + (n −d −2) (the 1 since in this case we do not need strict inequality in the induction step). Upper bounding −d in this latter expression by −x, we get the same inequality as (12). 5.4 Edge between w3 and w2, edge between w3 and w1, degree of w2 large Next we consider the case w3 ∼w2, w3 ∼w1, and x > 3. Here we must have w1 ≁w2, since otherwise G would not be edge-critical. The situation is illustrated in Figure 3. To bound i3(G), we consider v and w3. Arguing as in Section 5.2 (around w3 w2 v w1 Figure 3: The situation in Section 5.4. (10)), the number of independent sets including one of v, w3 is at most 2 n −4 2 −3(n −4) −(d −2) −(x −2) 2 17 To obtain an upper bound on the number of independent sets including neither v nor w3, we delete both vertices, add an edge from w1 to w2 (to restore minimum degree 3) and use induction to get a bound of n −5 3 + 1 + (n −d −2) (where the n −d −2 bounds the number of independent sets containing both w1 and w2). Since x ≤n −2 the sum of these two bound is strictly smaller than i3(K3,n−3). 5.5 None of the above If there is no v of degree 3 that puts us into one of the previous cases, then every v of degree 3 that has a neighbor w1 of degree strictly greater than 3 may be assumed to have two others of degree 3, w2 and w3 say, with vw2w3 a triangle (see Figure 4). v w1 w2 w3 Figure 4: The situation in Section 5.5. Since every neighbor of a vertex of degree greater than 3 has degree exactly 3 (by criticality) it follows that for every w1 of degree greater than 3, every neighbor of w1 is a vertex of a triangle all of whose vertices have degree 3. We claim that two of these triangles must be vertex disjoint. Indeed, if w1 has two neighbors a and b with a ∼b then the triangles associated with a and b must be the same, and by considering degrees we see that the triangle associated with any other neighbor of w1 must be vertex disjoint from it. If a and b are not adjacent and their associated triangles have no vertex in common, then we are done; but if they have a vertex in common then (again by considering degrees) they must have two vertices in common, and the triangle associated with any other neighbor of w1 must be vertex disjoint from both. By suitable relabeling, we may therefore assume that G has distinct vertices w1 (of degree greater than 3) and x, y2, y3, v, w2 and w3 (all of degree 3), with x and v adjacent to w1, and with xy2y3 and vw2w3 forming triangles (see Figure 5). By 18 considering degrees, we may also assume that the wi’s and yi’s are ordered so that wi ≁yi for i = 1, 2. v w1 w2 w3 x y2 y3 Figure 5: The forced structure in Section 5.5, before modiﬁcation. From G we create G′ by removing the edges w2w3 and y2y3, and adding the edges w2y2 and w3y3 (see Figure 6). We will argue that i3(G) ≤i3(G′); but then by the argument of Section 5.2 we have i3(G′) < i3(K3,n−3), and the proof will be complete. v w1 w2 w3 x y2 y3 Figure 6: The forced structure in Section 5.5, after modiﬁcation (i.e. in G′). Independent sets of size 3 in G partition into Iw2y2 (those containing both w2 and y2, and so neither of y3, w3), Iw3y3 (containing both w3 and y3), and Irest, the 19 rest. Independent sets of size 3 in G′ partition into I′ w2w3, I′ y2y3, and I′ rest. We have \|Irest\| = \|I′ rest\| (in fact Irest = I′ rest). We will show i3(G) ≤i3(G′) by exhibiting an injection from Iw2y2 into I′ w2w3 and one from Iw3y3 into I′ y2y3. If it happens that for every independent set {w2, y2, a} in G, the set {w2, w3, a} is also an independent set in G′, then we have a simple injection from Iw2y2 into I′ w2w3. There is only one way it can happen that {w2, y2, a′} is an independent set in G but {w2, w3, a′} is not one in G′; this is when a′ is the neighbor of w3 that is not v or w2. If {w2, y2, a′} is indeed an independent set in G in this case, then letting b′ be the neighbor of y2 that is not x or y3, we ﬁnd that {w2, w3, b′} is an independent set in G′, but {w2, y2, b′} is not one in G. So in this case we get an injection from Iw2y2 into I′ w2w3 by sending {w2, y2, a} to {w2, w3, a} for all a ̸= a′, and sending {w2, y2, a′} to {w2, w3, b′}. The injection from Iw3y3 into I′ y2y3 is almost identical and we omit the details. 6 Concluding remarks There now seems to be ample evidence to extend Conjecture 1.2 as follows. Conjecture 6.1. For each δ ≥1, n ≥2δ, t ≥3 and G ∈G(n, δ), we have it(G) ≤it(Kδ,n−δ). Throughout we have considered n ≥2δ, that is, δ small compared to n. It is natural to ask what happens in the complementary range δ > n/2. In the range n ≥2δ we (conjecturally) maximize the count of independent sets by extracting as large an independent set as possible (one of size n −δ). In the range δ > n/2 this is still the largest independent set size, but now it is possible to have many disjoint independent sets of this size. The following conjecture seems quite reasonable. Conjecture 6.2. For δ ≥1, n ≥δ + 1 and G ∈G(n, δ), we have i(G) ≤ i(Kn−δ,n−δ,...,n−δ,x), where 0 ≤x < n −δ satisﬁes n ≡x (mod n −δ). Question 6.3. For δ ≥1, n ≥δ + 1 and t ≥3, which G ∈G(n, δ) maximizes it(G)? When n −δ divides n (that is, x = 0), both Conjecture 6.2 and Question 6.3 turn out to be easy; in this case (2) gives that for all 1 ≤t ≤n −δ and all G ∈G(n, δ) we have it(G) ≤it(Kn−δ,n−δ,...,n−δ) and so also i(G) ≤i(Kn−δ,n−δ,...,n−δ) (the case n = 2δ was observed in ). The problem seems considerably more delicate when x ̸= 0. Lemmas 2.3 and 2.4 allow us in the present paper to focus attention on the class of edge- and vertex-critical graphs. Lemma 3.2 gives us a good understanding of critical graphs in the case δ = 2, and the bulk of Section 5 concerns structural properties of critical graphs for δ = 3. It is clear that approaching even the case δ = 4 by similar arguments would be considerable work. Any answer to the following question would help signiﬁcantly. 20 Question 6.4. For δ ≥4, what can be said about the structure of edge- and vertex-critical graphs? References J. Alexander, J. Cutler, and T. Mink, Independent sets in graphs with given minimum degree, manuscript. N. Alon, Independent sets in regular graphs and sum-free subsets of ﬁnite groups, Israel Journal of Mathematics 73 (1991), 247-256. T. Carroll, D. Galvin, and P. Tetali, Matchings and independent sets of a ﬁxed size in regular graphs, Journal of Combinatorial Theory Series A 116 (2009), 1219-1227. J. Cutler and A. J. Radcliﬀe, Extremal problems for independent set enumeration, Electronic Journal of Combinatorics 18(1) (2011), #P169. , Extremal graphs for homomorphisms, Journal of Graph Theory 67 (2011), 261-284. D. Galvin, Two problems on independent sets in graphs, Discrete Mathematics 311 (2011), 2105-2112. G. Hopkins and W. Staton, Some identities arising from the Fibonacci numbers of certain graphs, Fibonacci Quarterly 22 (1984), 255-258. H. Hua, A sharp upper bound for the number of stable sets in graphs with given number of cut edges, Applied Mathematics Letters 22 (2009), 1380-1385. J. Kahn, An entropy approach to the hard-core model on bipartite graphs, Combinatorics, Probability & Computing 10 (2001), 219-237. C. Lin and S. Lin, Trees and forests with large and small independent indices, Chinese Journal of Mathematics 23 (1995), 199-210. R. Merriﬁeld and H. Simmons, Topological Methods in Chemistry, Wiley, New York, 1989. A. Pedersen and P. Vestergaard, Bounds on the number of vertex independent sets in a graph, Taiwanese Journal of Mathematics 10 (2006), 1575-1587. H. Prodinger and R. Tichy, Fibonacci numbers of graphs, The Fibonacci Quarterly 20 (1982), 16-21. A.A. Sapozhenko, On the number of independent sets in bipartite graphs with large minimum degree, DIMACS Technical Report (2000), no. 2000-25. G. Wingard, Properties and applications of the Fibonacci polynomial of a graph, Ph.D. thesis, University of Mississippi, May 1995. Y. Zhao, The Number of Independent Sets in a Regular Graph, Combinatorics, Probability & Computing 19 (2010), 315-320. A.A. Zykov, On some properties of linear complexes, Mat. Sbornik N.S. 24(66) (1949), 163-188. 21
15792	https://entokey.com/third-nerve-palsies/	Third Nerve Palsies \| Ento Key Ento Key Fastest Otolaryngology & Ophthalmology Insight Engine Home Log In Register Categories » HEAD AND NECK SURGERY GENERAL OPHTHALMOLOGY OTOLARYNGOLOGY About Gold Membership Contact Menu Home Log In Register Categories » HEAD AND NECK SURGERY GENERAL OPHTHALMOLOGY OTOLARYNGOLOGY More References » Abdominal Key Anesthesia Key Basicmedical Key Otolaryngology & Ophthalmology Musculoskeletal Key Neupsy Key Nurse Key Obstetric, Gynecology and Pediatric Oncology & Hematology Plastic Surgery & Dermatology Clinical Dentistry Radiology Key Thoracic Key Veterinary Medicine About Gold Membership Contact Showing most revelant items. Click here or hit Enter for more. Third Nerve Palsies Third Nerve Palsies What Are the Clinical Features of a Third Cranial Nerve Palsy? The oculomotor nerve (third cranial nerve) supplies four extraocular muscles (medial, superior and inferior recti, and inferior oblique) as well as the levator of the lid, and contains parasympathetic fibers that supply the sphincter of the pupil and the ciliary body A complete peripheral third nerve palsy (TNP) thus causes ptosis, a fixed and dilated pupil, and a down (hypotropic) and out (exotropic) resting eye position. Partial TNPs may cause (in combination or isolation) variable ptosis; variable paresis of eye adduction, elevation, and depression; and variable pupillary involvement. In this section, we discuss the localization of TNPs associated with other neurologic signs (nonisolated TNPs) and TNPs without other associated neurologic or neuro-ophthalmologic deficits (isolated TNPs) (Lee, 1999). Is the TNP Isolated or Nonisolated? Can the TNP Be Localized? We classify TNPs as either nonisolated or isolated. The isolated TNPs were defined as TNPs without associated neurologic findings (e.g., headache, other cranial neuropathies). Patients with evidence for myasthenia gravis (e.g., variability, fatigue, Cogan’s lid twitch sign, enhancement of ptosis) are not included in the isolated TNP group. We define six types of TNP inTable 11–1. The localization of TNP is outlined in Table 11–2. Etiologies of TNPs by localization are outlined in Table 11–3. Table 11–1. Definitions of the Six Types of Third Nerve Palsy (TNP)Type 1: nonisolated TNP is considered nonisolated if it has the following features: Orbital disease (e.g., chemosis, proptosis, lid swelling, injection, and positive forced ductions) Evidence to suggest myasthenia gravis (e.g., fatigability of the motility defect, Cogan’s lid twitch sign, orbicularis oculi weakness) Multiple cranial nerve palsies (including bilateral TNP) or radiculopathy Brainstem signs (e.g., hemiplegia, cerebellar signs, other cranial nerve deficits) Systemic, infectious, or inflammatory risk factors for TNP (e.g., history of previous malignancy, giant cell arteritis, collagen vascular disease) Severe headache Type 2: traumatic Isolated unilateral TNP, which has a clearly established temporal relationship to significant previous head trauma and does not progress, is considered traumatic in origin; patients with minor head trauma are not included Type 3: congenital Patient born with an isolated TNP Type 4: acquired, nontraumatic isolated Type 4A: TNP with a normal pupillary sphincter with completely palsied extraocular muscles Type 4B: TNP with normal pupillary sphincter and incomplete palsied extraocular muscles Type 4C: TNP with subnormal pupillary sphincter dysfunction and partial or complete extraocular muscle palsies Type 5: progressive or unresolved Patients with TNP that worsens after the acute stage (more than 2 weeks) or who develop new neurologic findings are considered to have progressive TNP; patients without resolution of TNP after 12 to 16 weeks are considered unresolved Type 6: signs of aberrant regeneration Is the TNP Due to a Nuclear Lesion? Lesions of the third nerve nucleus are rare and often associated with other signs of mesencephalic involvement, especially vertical gaze impairment (Bengel, 1994; Bogousslavsky, 1994; Chee, 1990; Gaymard, 1990; Nakao, 1998; Saeki, 2000b). Nuclear lesions may be due to infarction, hemorrhage, tumor, infection, or trauma and, thus, should be investigated by magnetic resonance imaging (MRI). Paresis of an isolated muscle innervated by the oculomotor nerve almost always results from a lesion in the orbit or from disease of the muscle or neuromuscular junction. For example, isolated inferior rectus paresis may develop with trauma, myasthenia gravis, or vascular disease and may also occur on a congenital or idiopathic basis (von Noorden, 1991). Lesions of the inferior rectus subnucleus, however, may also give rise to isolated weakness of the inferior rectus muscle (Chou, 1998; Lee, 2000b; Tezer, 2000). Isolated inferior rectus paresis may also occur on a supranuclear basis with a lesion selectively interrupting fibers descending from the right medial longitudinal fasciculus (MLF) to the inferior rectus subnucleus (Tezer, 2000). The levator palpebrae superioris muscles, the superior recti, and the constrictors of the pupils are affected bilaterally with nuclear lesions. Because medial rectus neurons probably lie at three different locations within the oculomotor nucleus, it is unlikely that a medial rectus paralysis (unilateral or bilateral) would be the sole manifestation of a nuclear lesion (Umapathi, 2000). Most characteristicof oculomotor nuclear involvement is unilateral TNP, weakness of the ipsilateral and contralateral superior rectus, and bilateral incomplete ptosis (Pratt, 1995). On rare occasions the ipsilateral superior rectus is spared while the contralateral superior rectus is paretic. Bilateral TNPs with sparing of the lid levators may also be caused by nuclear lesions (the central caudal levator subnucleus is spared) (Bryan, 1992). Isolated bilateral ptosis with sparing of the extraocular muscles and pupils may occur with lesions involving the levator subnucleus and sparing more rostral oculomotor subnuclei (Martin, 1996). After surgery for a fourth ventricle ependymoma, bilateral nuclear oculomotor palsies affecting only the levator and superior recti subnuclei have been described, resulting in third nerve paresis affecting only the levators and superior recti bilaterally (Sanli, 1995). Bilateral total ophthalmoplegia, bilateral complete ptosis, and large, unreactive pupils have been described with midbrain hematoma (Worthington, 1996). This constellation of findings was thought due to bilateral third nerve nuclear or fascicular damage or both, bilateral involvement of the interstitial nucleus of Cajal and the rostral nucleus of the MLF, and involvement of bilateral horizontal saccadic and smooth pursuit pathways. Table 11–2. The Localization of TNP (Structure Involved: Clinical Manifestation)Lesions affecting the third nerve nucleus Oculomotor nucleus: ipsilateral complete cranial nerve (CN) III palsy; contralateral ptosis and superior rectus paresis Oculomotor subnucleus: isolated muscle palsy (e.g., inferior rectus) Isolated levator subnucleus: isolated bilateral ptosis Lesions affecting the third nerve fasciculus Isolated fascicle: partial or complete isolated CN III palsy with or without pupil involvement Paramedian mesencephalon: plus-minus syndrome (ipsilateral ptosis and contralateral eyelid retraction) Fascicle, red nucleus/cerebellar peduncle: ipsilateral CN III palsy with contralateral ataxia and tremor (Claude) Fascicle and cerebral peduncle: ipsilateral CN III palsy with contralateral hemiparesis (Weber) Fascicle and red nucleus/substantia nigra: ipsilateral CN III palsy with contralateral choreiform movements (Benedikt) Lesions affecting the third nerve in the subarachnoid space Oculomotor nerve: complete CN III palsy with or without other cranial nerve involvement; superior or inferior division palsy Lesions affecting the third nerve in the cavernous sinus Cavernous sinus lesion: painful or painless CN III palsy; with or without palsies of CN IV, VI, and VI; CN III palsy with small pupil (Horner syndrome); primary aberrant CN III regeneration Lesions affecting the third nerve in the superior orbital fissure Superior orbital fissure lesion: CN III palsy with or without palsies of CN IV, VI, and VI; often with proptosis Lesion affecting the third nerve in the orbit Oculomotor nerve: CN III palsy; superior or inferior CN III branch palsy CN III and optic nerve or other orbital structures: visual loss; proptosis; swelling of lids; Chemosis Source: Modified from Brazis, 2001, with permission from Lippincott Williams & Wilkins.Table 11–3. Etiologies of Third Nerve Palsy (TNP) by Topographical LocalizationNuclear TNP Infarction or hemorrhage (Bengel, 1994; Bogousslavsky, 1994; Bryan, 1992; Chee, 1990; Gaymard, 1990; Saeki, 2000a; Tezer, 2000; Worthington, 1996) Tumor (Chou, 1998; Nakao, 1998; Sanli, 1995) Infection Trauma Multiple sclerosis (Lee, 2000b) Fascicular TNP Infarction or hemorrhage (Breen, 1991; Castro, 1990; Gaymard, 1990, 2000; Getenet, 1994; Guy, 1989a; Hopf, 1990; Kim, 1993; Ksiazek, 1994; Liu, 1991; Messe, 2001; Oishi, 1997; Onozu, 1998; Saeki, 1996, 2000a,b; Thömke, 1995; Umapathi, 2000) Tumor (Andreo, 1994; Barbas, 1995; Eggenberger, 1993; Ishikawa, 1997; Landolfi, 1998; Vetrugno, 1997) Multiple sclerosis (Newman, 1990; Onozu, 1998; Thömke, 1997) Stereotactic surgery (Borras, 1997) Subarachnoid space Aneurysms of the internal carotid–posterior communicating, superior cerebellar, basilar, or posterior cerebral arteries (Birchall, 1999; Branley, 1992; DiMario, 1992; Friedman, 2001; Giombini, 1991; Good, 1990; Greenspan, 1990; Griffiths, 1994; Horikoshi, 1999; Keane, 1996; McFadzean, 1998; Mudgil, 1999; Park-Matsumoto, 1997; Ranganadham, 1992; Renowden, 1993; Richards, 1992; Striph, 1993; Teasdale 1990; Tomsak, 1991; Tummala, 2001; Walter, 1994; Weinberg, 1996; Wolin, 1992; Zimmer, 1991) Ectatic vessels (Hashimoto, 1998b; Nakagawa, 1991; Zingale, 1993) Tumors, especially meningiomas, chordomas, metastases, or primary tumors of the third nerve (Cullom, 1993; Egan, 2001; Hardenack, 1994; Ide, 1997; Jacobson, 2001; Kadota, 1993; Kajiya, 1995; Kawasaki, 1999; Kawase, 1996; Kaye-Wilson, 1994; Kodsi, 1992; Mehta, 1990; Norman, 2001; Ogilvy, 1993; Reifenberger, 1996; Robertson, 1998; Sanchez Dalmau, 1998; Schultheiss, 1993; Takano, 1990; Winterkorn, 2001) Infectious or inflammatory processes of the meninges (e.g., sarcoidosis and Wegener’s) and carcinomatous or lymphomatous meningitis (Balm, 1996; Galetta, 1992; Guarino, 1995; Ing, 1992; Ishibashi, 1998; Jacobson, 2001; Keane, 1993; Mark, 1992; McFadzean, 1998; Newman, 1995; Renowden, 1993; Straube, 1993; Ueyama, 1997) Ophthalmoplegic migraine (O’Hara, 2001) Subarachnoid hemorrhage with leukemia (Papke, 1993) Pseudotumor cerebri Spontaneous intracranial hypotension (Ferrante, 1998) Trauma, especially during neurosurgical procedures (Balcar, 1996; Hedges, 1993; Horikoshi, 1999; Kudo, 1990; Lepore, 1995) Nerve infarction from diabetes, atherosclerosis, giant cell arteritis, or systemic lupus erythematosus (nerve infarction may also occur in the cavernous sinus or anywhere along the course of nerve) (Berlit, 1991; Bondenson, 1997; Capo, 1992; Cullom, 1995; Davies, 1994; Jacobson, 1994, 1995, 1998a, 2001; Naghmi, 1990; Renowden, 1993; Richards, 1992) Uncal herniation Hydrocephalus Cavernous sinus/superior orbital fissure Aneurysm of the internal carotid or posterior communicating artery (Hahn, 2000; Ikeda, 2001; Jacobson, 2001; Keane, 1996; Lanzino, 1993; Silva, 1999; Zingale, 1997) Dural carotid cavernous sinus fistula (Acierno, 1995; Brazis, 1994; Keane, 1996; Lee, 1996; Miyachi, 1993; Perez Sempere, 1991; Uehara, 1998; Yen, 1998) Ballon test occlusion of the cervical internal carotid artery (Lopes, 1998) Cavernous sinus thrombosis or infection (e.g., tuberculoma); superior ophthalmic vein thrombosis (Bikhazi, 1998; Grayeli, 1998; Holland, 1998; Polito, 1996) Tumors, including pituitary adenoma, meningioma, esthesioneuroblastoma, arachnoid cyst, neurinoma, nasopharyngeal carcinoma, myeloma, lymphoma, Hodgkin’s disease, and metastases (Barr, 1999; Cullom, 1993; Ing, 1997; Kasner, 1996; Keane, 1996; Kurokawa, 1992; Lee, 2000c; Liu, 1993; Manabe, 2000; Moster, 1996; North, 1993; Shen, 1993; Tao, 1992; Wake, 1993) Pituitary infarction or hemorrhage (pituitary apoplexy) (Lee, 2000c; Robinson, 1990; Rossitch, 1992; Seyer, 1992) Gammopathy Intraneural hemorrhage (Miyao, 1993) Mucocele of the sphenoid sinus (Ashwin, 2001) Sphenoid sinusitis (Chotmongkol, 1999) Tolosa-Hunt syndrome, Wegener’s granulomatosis, or other granulomatous diseases (Herman, 1999; Jacobson, 2001; Keane, 1996) Orbit Infections, inflammations, and granulomatous processes (e.g., orbital pseudotumor) (Kondoh, 1998; Ohtsuka, 1997; Stefanis, 1993) Sphenoid sinus mucocele (Sethi, 1997) Tumors (Goldberg, 1990a,b) Dural arteriovenous malformation (Gray, 1999) Trauma Unknown localization Congenital (Good, 1991; Hamed, 1991; Ing, 1992; Mudgil, 1999; Parmeggiani 1992; Patel, 1993; Pratt, 1995; Schumacher-Feero, 1999; Tsaloumas, 1997; White, 1992) Migraine (Mark, 1998; O’Halloran, 1999; Prats, 1999) Viral infections (including herpes zoster ophthalmicus or Ramsay Hunt syndrome) and immunizations (Capoferri, 1997; Chang-Godinich, 1997; Mansour, 1997; Saeki, 2000c; Sood, 1999; Zurevinsky, 1993) Lyme disease (Savas, 1997) Diffuse neuropathic processes (e.g., Fisher’s syndrome and chronic inflammatory polyradiculo-neuropathy (CIDP) (Arroyo, 1995; Nagaoka, 1996) Cervical carotid artery dissection, stenosis, or occlusion (Balcar, 1997; Hollinger, 1999; Koennecke, 1998; Mokri, 1996; Schievink, 1993) Subdural hematomas (Okuchi, 1999; Phookan, 1994) Glioblastoma multiforme (Al-Yamany, 1999) Unintentional subdural catheter (Haughton, 1999) Submucosal diathermy to the inferior turbinates to improve the nasal airway (Green, 2000) Toxic effects of drugs (Pacifici, 1993; Soysal, 1993) Cocaine (Migita, 1998) Sildenafil citrate (Viagra) (Donahue, 1998) Internal carotid cisplatin infusion (inferolateral trunk carotid artery neurovascular toxicity) (Alderson, 1996; Wu, 1997) Dental anesthesia Radiation therapy (Ebner, 1995) Partial TNP associated with elevated anti-galactocerebroside and anti-GM 1 antibodies (Go, 2000)Is the TNP Due to a Fascicular Lesion? Lesions of the third nerve fascicle often accompany nuclear lesions because infarction is a common cause of a nuclear TNP, and the paramedian branches near the top of the basilar artery often feed both structures. For example, infarction of the dorsal paramedian midbrain may cause bilateral ptosis associated with unilateral paresis of all other muscles innervated by the oculomotor nerve (pupil spared) with sparing of the contralateral superior rectus muscle (Liu, 1991). These unique findings suggest a lesion of the proximal third nerve fascicles and the central caudal subnucleus. Third nerve fascicular lesions are often caused by infarction, hemorrhage, or demyelination. Pure fascicular lesions cause a unilateral peripheral type of oculomotor palsy. Involvement of brainstem structures other than the fascicle of the third nerve identifies the mesencephalic location of the lesion (Liu, 1992). Concomitant damage of the red nucleus and superior cerebellar peduncle causes contralateral ataxia and outflow tract cerebellar tremor (Claude’s syndrome), whereas a more anterior lesion, affecting the peduncle, gives rise to oculomotor palsy with contralateral hemiparesis (Weber’s syndrome). The TNP with Weber’s syndrome may affect or spare the pupillary fibers (Saeki, 1996). Larger lesions that affect the oculomotor fascicle and the red nucleus/substantia nigra region may produce TNP with contralateral choreiform movements or tremor (Benedikt’s syndrome) (Borras, 1997), sometimes associated with contralateral hemiparesis if the cerebral peduncle is also involved. A pupil-sparing TNP associated with binocular ocular torsion to the contralateral side—thereby indicating a left-sided midbrain lesion that included the fascicle of the third nerve and the supranuclear integration centers for torsional eye movements, the interstitial nucleus of Cajal, and the rostral interstitial nucleus of the MLF—has been described with a paramedian rostral midbrain infarction in a diabetic with giant cell arteritis (Dichgans, 1995). Ipsilateral TNP and contralateral downbeat nystagmus may be caused by unilateral paramedian thalamopeduncular infarction (Oishi, 1997). Rarely, a unilateral or bilateral fascicular third nerve lesion may occur in isolation without other ocular motor or neurologic signs or symptoms (see below) (Andreo, 1994; Barbas, 1995; Getenet, 1994; Kim, 1993; Newman, 1990; Thömke, 1995). Fascicular lesions, even when bilateral, may occasionally spare the pupil(s). Bilateral preganglionic internal ophthalmoplegia has been described with bilateral partial oculomotor fascicular lesions (Hashimoto, 1998a). Because of the intraaxial topographic arrangement of fibers, fascicular lesions may cause TNP limited to specific oculomotor-innervated muscles (Ksiazek, 1994). Fascicular lesions have resulted in the following: 1.Isolated inferior oblique paresis (Castro, 1990) 2.Unilateral fixed, dilated pupil unassociated with other neurologic dysfunction (Shuaib, 1989) 3.Paresis of the superior rectus and inferior oblique without other evidence of oculomotor nerve involvement (Gauntt, 1995) 4.Paresis of the superior and medial rectus (Saeki, 2000a) 5.Paresis of the levator muscle, superior rectus, and medial rectus (Onozu, 1998) 6.Paresis of the inferior oblique, superior rectus, medial rectus, and levator muscle with sparing of the inferior rectus muscle and pupil (Naudea, 1983; Schwartz, 1995; Shuaib, 1987) 7.Paresis of the inferior oblique, superior rectus, medial rectus, levator, and inferior rectus with pupillary sparing (Breen, 1991; Naudea, 1983)8.Paresis of the left inferior rectus, left pupil, right superior rectus, convergence, and left medial rectus (Umapathi, 2000). Based on these clinical studies, it has been proposed that individual third nerve fascicles in the ventral mesencephalon are arranged topographically from lateral to medial as follows: inferior oblique, superior rectus, medial rectus and levator palpebrae, inferior rectus, and pupillary fibers (Castro, 1990). A rostral-caudal topographic arrangement has also been suggested with pupillary fibers most superior, followed by fibers to the inferior rectus, inferior oblique, medial rectus, superior rectus, and levator, in that order (Saeki, 2000a; Schwartz, 1995). This model also accounts for the description of superior and inferior division oculomotor palsies. The superior division paresis involves the superior rectus and levator muscles without involvement of other groups (Guy, 1989a; Hriso, 1990; Ksiazek, 1989). The inferior division oculomotor palsies cause paresis of inferior rectus, inferior oblique, medial rectus, and pupillary fibers with sparing of the superior rectus and levator (Abdollah, 1990; Eggenberger, 1993; Ksiazek, 1989). Both divisional palsies may be associated with intraaxial midbrain lesions. Thus, although superior and inferior divisional TNP have classically been localized to anterior cavernous sinus or posterior orbital lesions, a divisional TNP may occur from damage at any location along the course of the oculomotor nerve, from the fascicle to the orbit (Ksiazek, 1989). Fascicular TNP may occasionally be associated with ipsilateral ptosis and contralateral eyelid retraction (plus-minus lid syndrome) (Gaymard, 1992; Vetrugno, 1997). This syndrome occurs with a small lesion located in the paramedian mesencephalon. There is involvement of the ipsilateral levator palpebrae fascicles as they emerge from the central caudal nucleus (the central caudal nucleus is spared), and the inhibitory pathways projecting on the levator palpebrae motor neurons immediately before their entrance in the central caudal nucleus. The plus-minus syndrome has been described with bilateral glioma extending to the paramedian midbrain and thalamic-mesencephalic infarction; it also may occur with peripheral processes such as peripheral TNP, myasthenia gravis, orbital myositis, congenital ptosis, or orbital trauma. Is the TNP Due to a Subarachnoid Lesion? An isolated pupil spared peripheral TNP is most often related to an ischemic neuropathy or a lesion affecting its subarachnoid portion. Subarachnoid lesions may distort or injure the brainstem, and diffuse processes will show signs of meningeal irritation. Etiologies of TNP due to a subarachnoid lesion are outlined inTable 11–3. Third nerve schwannoma may cause a painful relapsing-remitting TNP mimicking the clinical syndrome of ophthalmoplegic migraine (Kawasaki, 1999). Monocular elevator paresis from isolated superior rectus and/or inferior oblique dysfunction may occur in neurofibromatosis type 2-related schwannoma (Egan, 2001). The third nerve is also susceptible to trauma in the subarachnoid space, especially during neurosurgical procedures (Hedges, 1993; Horikoshi, 1999; Kudo, 1990). Closed head trauma may cause TNP due to shearing injury resulting in distal fascicular damage or partial root avulsion (Balcar, 1996). Walter et al described two patients with TNP precipitated by minor head trauma with negative brain computed tomography (CT) scans who were subsequently discovered to have ipsilateral posterior communicating artery aneurysms (Walter, 1994). Park-Matsumoto and Tazawa described a similar case (Park-Matsumoto, 1997). Compression of the third nerve by an aneurysm characteristically causes dilatation and unresponsiveness of the pupil. Compressive subarachnoid lesions may occasionally spare the pupil, however. Two explanations have been proposed: (1) compression may be evenly distributed and the relatively pressure-resistant, smaller-caliber pupillomotor fibers escape injury; or (2) the lesion compresses only the inferior portion of the nerve and spares the dorsally situated pupillomotor fibers. The TNP due to an aneurysm may be incomplete with at least one element of nerve dysfunction (i.e., ptosis, mydriasis, or extraocular muscle weakness) being absent. Ptosis has been described in isolation as the sole manifestation of third nerve compression by a posterior communicating artery aneurysm (Good, 1990). Rarely, aneurysmal TNP may even be transient and clear spontaneously (Greenspan, 1990). A normal pupil in the setting of a complete somatic oculomotor paresis, however, essentially excludes a diagnosis of aneurysm (see below). A single patient has been described in whom a painless, pupil-sparing but otherwise complete oculomotor paresis was the only sign of an aneurysm arising from the basilar artery (Lustbader, 1988). Conversely, an isolated pupillary paralysis without ptosis or ophthalmoparesis is rarely caused by an aneurysm or other subarachnoid lesion (Kaye-Wilson, 1994; Wilhelm, 1995). Koennecke and Seyfert reported a patient with a common carotid artery dissection from intraoperative trauma whose mydriasis preceded a complete TNP by 12 hours (Koennecke, 1998). Is the TNP Due to a Cavernous Sinus Lesion? Lesions of the third nerve in the cavernous sinus often also involve the other ocular motor nerves, the ophthalmic branch of the trigeminal nerve, and sympathetic fibers. Sensory fibers from the ophthalmic division of the fifth cranial nerve join the oculomotor nerve within the lateral wall of the cavernous sinus. The frontal-orbital pain experienced by patients with enlarging aneurysms could thus be caused by direct irritation of the third nerve (Lanzino, 1993). Compressive cavernous sinus lesions may also spare the pupil because they often preferentially involve only the superior division of the oculomotor nerve that carries no pupillomotor fibers (Silva, 1999) or the superior aspect of the nerve anterior to the point where the pupillomotor fibers descend in their course near the inferior oblique muscle. The pupillary “sparing” with anterior cavernous sinus lesions may be more apparent than real, resulting from simultaneous injury of nerve fibers to both the pupillary sphincter and dilator, causing a mid-position fixed pupil or from aberrant regeneration (see below). Ikeda et al described a patient with a painful, “severe” TNP with normal pupils due to a cavernous sinus aneurysm (Ikeda, 2001). Lesions in the neighborhood of the posterior clinoid process may for some time affect only the third nerve as it pierces the dura (e.g., breast and prostatic carcinoma) (Cullom, 1993). Medial lesions in the cavernous sinus, such as a carotid artery aneurysm, may affect only the ocular motor nerves but spare the more laterally located ophthalmic branch of the trigeminal nerve, resulting in painless ophthalmoplegia. Lesions that begin laterally present with retro-orbital pain first, and only later does ophthalmoparesis supervene. Lesions located in the cavernous sinus causing TNP are outlined inTable 11–3. The clinical findings and etiologies for processes located in the superior orbital fissure are similar to those of the cavernous sinus syndrome. Is the TNP Due to an Orbital Lesion? Lesions within the orbit that produce third nerve dysfunction usually produce other ocular motor dysfunction as well as optic neuropathy and proptosis (Goldberg, 1990a,b). Lesions may extend from the cavernous sinus to the orbital apex and vice versa so that a clear distinction between the two syndromes may be impossible. Isolated involvement of the muscles innervated by either the superior or the inferior oculomotor branch has classically been localized to an orbital process: often trauma, tumor, or infection, or a sphenocavernous lesion (Stefanis, 1993). However, as we noted, the functional division of the third nerve is present probably even at the fascicular level, and a divisional pattern may occur from damage anywhere along the course of the nerve. Superior division or inferior division third nerve paresis may occur with subarachnoid lesions (Guy, 1985), and isolated superior division paresis has been described with a superior cerebellar-posterior cerebral artery junction aneurysm that compressed and flattened the interpeduncular third nerve from below (Guy, 1989b). Superior branch palsy has also been described with basilar artery aneurysm, intracavernous carotid aneurysm, migraine, diabetes, lymphoma, sphenoidal abscess, sphenoid sinusitis, frontal sinus mucocele, viral illness, meningitis, and after craniotomy (Chotmongkol, 1992, 1999; Ehrenpries, 1995; Guy, 1989b; Manabe, 2000; O’Halloran, 1999; Saeki, 2000c; Silva, 1999; Stefanis, 1993). Even ophthalmoplegic migraine may cause recurrent paroxysmal superior division oculomotor palsy. Isolated superior division-like paresis may be mimicked by myasthenia gravis (Dehaene, 1995). Isolated inferior division involvement has occurred with trauma, mesencephalic infarction and tumor (Abdollah, 1990; Eggenberger, 1993; Ksiazek, 1989), basilar artery aneurysm (Kardon, 1991), parasellar tumors (e.g., meningioma, schwannoma) (Carlow, 1990), viral illness, orbital dural arteriovenous malformation (Gray, 1999), as part of a more generalized vasculitic or demyelinating neuropathy (Cunningham, 1994), and in association with elevated antigalactocerebroside and anti-GM1 antibodies (Go, 2000). Inferior division involvement with tumors may be pupil-sparing, perhaps because of insidious tumor growth sparing pressure-resistant pupillomotor fibers. Partial or complete TNP may rarely follow dental anesthesia, presumably due to inadvertent injection of an anesthetic agent into the inferior dental artery or superior alveolar artery with subsequent retrograde flow into the maxillary, middle meningeal, and finally the lacrimal branch of the ophthalmic artery. What Is the Evaluation of Nonisolated TNP? Nonisolated TNP should undergo neuroimaging, with attention to areas suggested topographically by the associated neurologic signs and symptoms. Appropriate investigations and neuroimaging studies are directed at the precise area of interest, and this area is determined by the associated localizing features with the TNP (Brazis, 1991; Lee, 1999). In general, MRI with and without gadolinium enhancement is the neuroimaging modality of choice for all these processes (Renowden, 1993). Contrast-enhanced CTscanning with narrow (2-mm) collimation is reserved for those patients who cannot tolerate MRI or in whom MRI is contraindicated (e.g., pacemaker, claustrophobia, metallic clips in head, etc.) (Renowden, 1993; Teasdale, 1990). CT scanning is also the appropriate first-choice neuroimaging study in patients being evaluated for acute head trauma or acute vascular events (infarction or hemorrhage). If there are clinical signs of a meningeal process, lumbar puncture should be performed. The evaluation of a patient with TNP is summarized inFigure 11–1. Is the TNP Due to Trauma? Traumatic isolated TNP (type 2) should undergo CT scanning to search for associated central nervous system damage (e.g., subdural or intracerebral hematoma) as indicated by associated neurologic signs and symptoms (Balcar, 1996; Hedges, 1993; Kudo, 1990; Lepore, 1995; Phookan, 1994). TNP after mild head trauma have been observed in association with otherwise asymptomatic lesions (e.g., cerebral aneurysm) (Park-Matsumoto, 1997; Walter, 1994). Although uncommon, neuroimaging may be warranted in patients with TNP after minimal or trivial trauma to exclude mass lesions or cerebral aneurysms (class III–IV, level C). Is the TNP Congential? Congenital isolated TNP (type 3) is rare, usually unilateral, and may occur in isolation or in association with other neurologic and systemic abnormalities, including congenital facial nerve palsies or other cranial neuropathies, facial capillary hemangioma, cerebellar hypoplasia, gaze palsy, ipsilateral nevus sebaceous of Jadassohn, mental retardation, and digital anomalies (Good, 1991; Hamed, 1991; Ing, 1992; Parmeggiani, 1992; Patel, 1993; Pratt, 1995; Shumacher-Feero, 1999; White, 1992). All patients have some degree of ptosis and ophthalmoplegia, and nearly all have pupillary involvement. In most cases, the pupil is miotic rather than dilated, probably because of aberrant third nerve regeneration, and usually trace reactive or nonreactive to light. Rarely the pupil may be spared. Amblyopia is common (Schumacher-Feero, 1999). Most cases are spontaneous, but familial cases have been described. We recommend MRI in all patients with congenital TNPs, mainly to search for associated structural abnormalities of the brain (class III–IV, level C). Is the Isolated TNP Acquired and Nontraumatic? Acquired, nontraumatic isolated TNP (type 4) may occur with lesions localized anywhere along the course of the third nerve from the fascicle to the orbit (Renowden 1993). For clinical purposes, isolated TNP may be divided into three types (types 4A–4C) (Jacobson, 1999; Lee, 1999; Trobe 1985) (Table 11–1). Figure 11–1. Evaluation of third nerve palsy (TNP).Does the Patient Have an Acquired Isolated TNP with a Normal Pupillary Sphincter with Completely Palsied Extraocular Muscles (Type 4A TNP)? TNP with a normal pupillary sphincter and completely palsied extraocular muscles is almost never due to an intracranial aneurysm. However, a single patient has been described in whom a painless, pupil-sparing, but otherwise complete TNP was the only sign of an aneurysm arising from the basilar artery (Lustbader 1988). A similar painful TNP palsy has been described with an aneurysm in the cavernous sinus (Ikeda, 2001), and pupillary sparing may rarely occur with pituitary adenoma. This type of TNP is most commonly caused by ischemia, especially associated with diabetes mellitus. In a retrospective review of 34 consecutive cases of isolated atraumatic TNP, diabetes mellitus was the most common etiology accounting for 46% of the cases (Renowden 1993). Ischemic TNP may also occur with giant cell arteritis (Berlit, 1991; Bondenson, 1997; Da vies, 1994; Renowden, 1993; Richards, 1992) and systemic lupus erythematosus. Pupil-sparing TNP has also been reported with sildenafil citrate (Viagra) (Donahue, 1998) and cocaine use (Migita, 1998). Significant risk factors for ischemic oculomotor nerve palsies include diabetes, left ventricular hypertrophy, and elevated hematocrit (Jacobson, 1994). Obesity, hypertension, and smoking are also probable risk factors. Ischemic damage to the trigeminal fibers in the oculomotor nerve may be the source of pain in ischemic-diabetic TNPs (Bortolami, 1993). Only gold members can continue reading.Log In or Register to continue Share this: Click to share on Twitter (Opens in new window) Click to share on Facebook (Opens in new window) Related Related posts: Papilledema The Diagnosis of Optic Neuropathies Transient Visual Loss Thyroid Eye Disease: Graves’ Ophthalmopathy Stay updated, free articles. Join our Telegram channel ------------------------------------------------------ Join Tags: Clinical Pathways in Neuro-Ophthalmology Jun 4, 2016 \| Posted by admin in OPHTHALMOLOGY \| Comments Off on Third Nerve Palsies Full access? 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Home 2. Campus Bookshelves 3. Widener University 4. CHEM 176: General Chemistry II (Fischer-Drowos) 5. 3: Kinetics 6. 3.2: Chemical Reaction Rates Expand/collapse global location CHEM 176: General Chemistry II (Fischer-Drowos) Front Matter 1: Composition of Substances and Solutions 2: Solutions and Colloids 3: Kinetics 4: Fundamental Equilibrium Concepts 5: Acid-Base Equilibria 6: Equilibria of Other Reaction Classes 7: Gases 8: Thermodynamics 9: Electrochemistry 10: Polymers 11: Radioactivity and Nuclear Chemistry 12: Electronic Structure and Periodic Properties 13: Chemical Bonding and Molecular Geometry 14: Advanced Theories of Covalent Bonding Back Matter 3.2: Chemical Reaction Rates Last updated Aug 20, 2023 Save as PDF 3.1: Introduction 3.3: Factors Affecting Reaction Rates Page ID 454833 OpenStax OpenStax ( \newcommand{\kernel}{\mathrm{null}\,}) Table of contents 1. Learning Objectives 2. Chemistry in Everyday Life: Reaction Rates in Analysis: Test Strips for Urinalysis 3. Relative Rates of Reaction 1. Example 3.2.1: Expressions for Relative Reaction Rates 1. Solution 2. Exercise 3.2.1/03:_Kinetics/3.02:_Chemical_Reaction_Rates#Exercise_.5C(.5CPageIndex.7B1.7D.5C)) 3. Example 3.2.2: Reaction Rate Expressions for Decomposition of H2O2/03:_Kinetics/3.02:_Chemical_Reaction_Rates#Example_.5C(.5CPageIndex.7B2.7D.5C):_Reaction_Rate_Expressions_for_Decomposition_of_H2O2) 1. Solution/03:_Kinetics/3.02:_Chemical_Reaction_Rates#Solution_2) 4. Exercise 3.2.2/03:_Kinetics/3.02:_Chemical_Reaction_Rates#Exercise_.5C(.5CPageIndex.7B2.7D.5C)) Learning Objectives By the end of this section, you will be able to: Define chemical reaction rate Derive rate expressions from the balanced equation for a given chemical reaction Calculate reaction rates from experimental data A rate is a measure of how some property varies with time. Speed is a familiar rate that expresses the distance traveled by an object in a given amount of time. Wage is a rate that represents the amount of money earned by a person working for a given amount of time. Likewise, the rate of a chemical reaction is a measure of how much reactant is consumed, or how much product is produced, by the reaction in a given amount of time. The rate of reaction is the change in the amount of a reactant or product per unit time. Reaction rates are therefore determined by measuring the time dependence of some property that can be related to reactant or product amounts. Rates of reactions that consume or produce gaseous substances, for example, are conveniently determined by measuring changes in volume or pressure. For reactions involving one or more colored substances, rates may be monitored via measurements of light absorption. For reactions involving aqueous electrolytes, rates may be measured via changes in a solution’s conductivity. For reactants and products in solution, their relative amounts (concentrations) are conveniently used for purposes of expressing reaction rates. For example, the concentration of hydrogen peroxide, H 2 O 2, in an aqueous solution changes slowly over time as it decomposes according to the equation: 2⁢H 2⁡O 2⁡(a⁢q)⟶2⁢H 2⁡O⁡(l)+O 2⁡(g) The rate at which the hydrogen peroxide decomposes can be expressed in terms of the rate of change of its concentration, as shown here: \text { rate of decomposition of } \begin{align} H_2O_2 & =-\dfrac{\text { change in concentration of reactant }}{\text { time interval }} \[4pt] & =-\dfrac{\left[ H_2 O_2\right]{t_2}-\left[ H_2 O_2\right]{t_1}}{t_2-t_1} \[4pt] & =-\dfrac{\Delta\left[ H_2 O_2\right]}{\Delta t} \end{align} This mathematical representation of the change in species concentration over time is the rate expression for the reaction. The brackets indicate molar concentrations, and the symbol delta (Δ) indicates “change in.” Thus,[H⁡A 2⁢O⁢A 2]t 1 represents the molar concentration of hydrogen peroxide at some time t 1; likewise, [H⁡A 2⁢O⁢A 2]t 2 represents the molar concentration of hydrogen peroxide at a later time t2; and Δ[H2O2] represents the change in molar concentration of hydrogen peroxide during the time interval Δ⁢t (that is, t 2−t 1). Since the reactant concentration decreases as the reaction proceeds, Δ⁡[H⁡A 2⁢O⁢A 2] is a negative quantity. Reaction rates are, by convention, positive quantities, and so this negative change in concentration is multiplied by −1. Figure 3.2.1 provides an example of data collected during the decomposition of H⁡A 2⁢O⁢A 2. Figure 3.2.1: The rate of decomposition of H 2 O 2 in an aqueous solution decreases as the concentration of H 2 O 2 decreases. To obtain the tabulated results for this decomposition, the concentration of hydrogen peroxide was measured every 6 hours over the course of a day at a constant temperature of 40 °C. Reaction rates were computed for each time interval by dividing the change in concentration by the corresponding time increment, as shown here for the first 6-hour period: −Δ⁡[H 2⁡O 2]Δ⁡t=−(0.500⁢m⁢o⁢l/L−1.000⁢m⁢o⁢l/L)(6.00⁢h−0.00⁢h)=0.0833⁢m⁢o⁢l⁢L−1⁢h−1 Notice that the reaction rates vary with time, decreasing as the reaction proceeds. Results for the last 6-hour period yield a reaction rate of: −Δ⁡[H 2⁡O 2]Δ⁡t=−(0.0625⁢m⁢o⁢l/L−0.125⁢m⁢o⁢l/L)(24.00⁢h−18.00⁢h)=0.010⁢m⁢o⁢l⁢L−1⁢h−1 This behavior indicates the reaction continually slows with time. Using the concentrations at the beginning and end of a time period over which the reaction rate is changing results in the calculation of an average rate for the reaction over this time interval. At any specific time, the rate at which a reaction is proceeding is known as its instantaneous rate. The instantaneous rate of a reaction at “time zero,” when the reaction commences, is its initial rate. Consider the analogy of a car slowing down as it approaches a stop sign. The vehicle’s initial rate—analogous to the beginning of a chemical reaction—would be the speedometer reading at the moment the driver begins pressing the brakes (t 0). A few moments later, the instantaneous rate at a specific moment—call it t 1—would be somewhat slower, as indicated by the speedometer reading at that point in time. As time passes, the instantaneous rate will continue to fall until it reaches zero, when the car (or reaction) stops. Unlike instantaneous speed, the car’s average speed is not indicated by the speedometer; but it can be calculated as the ratio of the distance traveled to the time required to bring the vehicle to a complete stop (Δ t). Like the decelerating car, the average rate of a chemical reaction will fall somewhere between its initial and final rates. The instantaneous rate of a reaction may be determined one of two ways. If experimental conditions permit the measurement of concentration changes over very short time intervals, then average rates computed as described earlier provide reasonably good approximations of instantaneous rates. Alternatively, a graphical procedure may be used that, in effect, yields the results that would be obtained if short time interval measurements were possible. In a plot of the concentration of hydrogen peroxide against time, the instantaneous rate of decomposition of H⁡A 2⁢O⁢A 2 at any time t is given by the slope of a straight line that is tangent to the curve at that time (Figure 3.2.2). These tangent line slopes may be evaluated using calculus, but the procedure for doing so is beyond the scope of this chapter. Figure 3.2.2: This graph shows a plot of concentration versus time for a 1.000 M solution of H 2 O 2. The rate at any time is equal to the negative of the slope of a line tangent to the curve at that time. Tangents are shown at t = 0 h (“initial rate”) and at t = 12 h (“instantaneous rate” at 12 h). Chemistry in Everyday Life: Reaction Rates in Analysis: Test Strips for Urinalysis Physicians often use disposable test strips to measure the amounts of various substances in a patient’s urine (Figure 3.2.3). These test strips contain various chemical reagents, embedded in small pads at various locations along the strip, which undergo changes in color upon exposure to sufficient concentrations of specific substances. The usage instructions for test strips often stress that proper read time is critical for optimal results. This emphasis on read time suggests that kinetic aspects of the chemical reactions occurring on the test strip are important considerations. The test for urinary glucose relies on a two-step process represented by the chemical equations shown here: C 6⁢H 12⁡O 6+O 2⟶catalyst C 6⁢H 10⁡O 6+H 2⁡O 2 2⁢H 2⁡O 2+2⁢I−⟶catalyst I 2+2⁢H 2⁡O+O 2 The first equation depicts the oxidation of glucose in the urine to yield glucolactone and hydrogen peroxide. The hydrogen peroxide produced subsequently oxidizes colorless iodide ion to yield brown iodine, which may be visually detected. Some strips include an additional substance that reacts with iodine to produce a more distinct color change. The two test reactions shown above are inherently very slow, but their rates are increased by special enzymes embedded in the test strip pad. This is an example of catalysis, a topic discussed later in this chapter. A typical glucose test strip for use with urine requires approximately 30 seconds for completion of the color-forming reactions. Reading the result too soon might lead one to conclude that the glucose concentration of the urine sample is lower than it actually is (a false-negative result). Waiting too long to assess the color change can lead to a false positive due to the slower (not catalyzed) oxidation of iodide ion by other substances found in urine. Figure 3.2.3: Test strips are commonly used to detect the presence of specific substances in a person’s urine. Many test strips have several pads containing various reagents to permit the detection of multiple substances on a single strip. (credit: Iqbal Osman) Relative Rates of Reaction The rate of a reaction may be expressed as the change in concentration of any reactant or product. For any given reaction, these rate expressions are all related simply to one another according to the reaction stoichiometry. The rate of the general reaction (3.2.1)A⟶B can be expressed in terms of the decrease in the concentration of A or the increase in the concentration of B. These two rate expressions are related by the stoichiometry of the reaction: rate=−(1 a)⁢(Δ⁢A Δ⁢t)=(1 b)⁢(Δ⁢B Δ⁢t) Consider the reaction represented by the following equation: 2⁢N⁡H 3⁡(g)⟶N 2⁡(g)+3⁢H 2⁡(g) The relation between the reaction rates expressed in terms of nitrogen production and ammonia consumption, for example, is: −Δ⁢m⁢o⁢l N⁢H 3 Δ⁢t×1⁢m⁢o⁢l N 2 2⁢m⁢o⁢l N⁢H 3=Δ⁢m⁢o⁢l N 2 Δ⁢t This may be represented in an abbreviated format by omitting the units of the stoichiometric factor: −1 2⁢Δ⁢m⁢o⁢l N⁢H 3 Δ⁢t=Δ⁢m⁢o⁢l N 2 Δ⁢t Note that a negative sign has been included as a factor to account for the opposite signs of the two amount changes (the reactant amount is decreasing while the product amount is increasing). For homogeneous reactions, both the reactants and products are present in the same solution and thus occupy the same volume, so the molar amounts may be replaced with molar concentrations: −1 2⁢Δ⁡[N⁢H 3]Δ⁡t=Δ⁡[N 2]Δ⁡t Similarly, the rate of formation of H 2 is three times the rate of formation of N 2 because three moles of H 2 are produced for each mole of N 2 produced. 1 3⁢Δ⁡[H 2]Δ⁡t=Δ⁡[N 2]Δ⁡t Figure 3.2.4: illustrates the change in concentrations over time for the decomposition of ammonia into nitrogen and hydrogen at 1100 °C. Slopes of the tangent lines at t = 500 s show that the instantaneous rates derived from all three species involved in the reaction are related by their stoichiometric factors. The rate of hydrogen production, for example, is observed to be three times greater than that for nitrogen production: 2.91×10−6⁢M/s 9.70×10−7⁢M/s≈3 Figure 3.2.4: Changes in concentrations of the reactant and products for the reaction 2 NH⁢A 3N⁢A 2+3⁢H⁡A 2. The rates of change of the three concentrations are related by the reaction stoichiometry, as shown by the different slopes of the tangents at t = 500 s. Example 3.2.1: Expressions for Relative Reaction Rates The first step in the production of nitric acid is the combustion of ammonia: (3.2.2)4⁢N⁢H 3⁡(g)+5⁢O 2⁡(g)⟶4⁢N⁢O⁡(g)+6⁢H 2⁡O⁡(g) Write the equations that relate the rates of consumption of the reactants and the rates of formation of the products. Solution Considering the stoichiometry of this homogeneous reaction, the rates for the consumption of reactants and formation of products are: −1 4⁢Δ⁡[N⁢H 3]Δ⁡t=−1 5⁢Δ⁡[O 2]Δ⁡t=1 4⁢Δ⁡[N⁢O]Δ⁡t=1 6⁢Δ⁡[H 2⁡O]Δ⁡t Exercise 3.2.1 The rate of formation of Br 2 is 6.0 10−6 mol/L/s in a reaction described by the following net ionic equation: 5⁢B⁢r−+B⁢r⁢O 3−+6⁢H+⟶3⁢B⁢r 2+3⁢H 2⁡O Write the equations that relate the rates of consumption of the reactants and the rates of formation of the products. Answer −1 5⁢Δ⁡[B⁢r−]Δ⁡t=−Δ⁡[B⁢r⁢O 3−]Δ⁡t=−1 6⁢Δ⁡[H+]Δ⁡t=1 3⁢Δ⁡[B⁢r 2]Δ⁡t=1 3⁢Δ⁡[H 2⁡O]Δ⁡t Example 3.2.2: Reaction Rate Expressions for Decomposition of H 2 O 2 The graph in Figure 3.2.2: shows the rate of the decomposition of H 2 O 2 over time: 2⁢H 2⁡O 2⟶2⁢H 2⁡O+O 2 Based on these data, the instantaneous rate of decomposition of H 2 O 2 at t = 11.1 h is determined to be 3.20 10−2 mol/L/h, that is: (3.2.3)−1 2⁢Δ⁡[H 2⁡O 2]Δ⁡t=1 2⁢Δ⁡[H 2⁡O]Δ⁡t=Δ⁡[O 2]Δ⁡t What is the instantaneous rate of production of H⁡A 2⁢O and O⁢A 2? Solution The reaction stoichiometry shows that (3.2.4)−1 2⁢Δ⁡[H 2⁡O 2]Δ⁡t=1 2⁢Δ⁡[H 2⁡O]Δ⁡t=Δ⁡[O 2]Δ⁡t Therefore: 1 2×3.20×10−2⁢m⁢o⁢l⁢L−1⁢h−1=Δ⁡[O 2]Δ⁡t and Δ⁡[O 2]Δ⁡t=1.60×10−2⁢m⁢o⁢l⁢L−1⁢h−1 Exercise 3.2.2 If the rate of decomposition of ammonia, NH 3, at 1150 K is 2.10 10−6 mol/L/s, what is the rate of production of nitrogen and hydrogen? Answer 1.05 10−6 mol/L/s, N 2 and 3.15 10−6 mol/L/s, H 2. This page titled 3.2: Chemical Reaction Rates is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform. Back to top 3.1: Introduction 3.3: Factors Affecting Reaction Rates Was this article helpful? Yes No Recommended articles 3: KineticsChemical kinetics is the study of rates of chemical processes and includes investigations of how different experimental conditions can influence the s... Article typeSection or PageAuthorOpenStaxLicenseCC BYLicense Version4.0OER program or PublisherOpenStaxShow Page TOCno on pageTranscludedyes Tags Author tag:OpenStax average rate initial rate instantaneous rate rate expression rate of reaction source-chem-38258 source-chem-414682 source@ © Copyright 2025 Chemistry LibreTexts Powered by CXone Expert ® ? 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15794	https://www.youtube.com/watch?v=oyyFpAwyt6w	Probability Must-Knows for Machine Learning \| Math for ML (Part 1) Kylie Ying 83700 subscribers 1081 likes Description 24265 views Posted: 30 Jun 2024 Want to start learning machine learning? Follow this series and supplement your learning with Brilliant. To try everything Brilliant has to offer—free—for a full 30 days, visit . You’ll also get 20% off an annual premium subscription. This video provides an introductory crash course to probability, with the intention of teaching concepts that are foundational to machine learning. Timestamps 00:00 Introduction 00:44 Defining Probability 06:31 Conditions of Probability 10:39 Joint Probability 14:37 Conditional Probability 18:25 Independence 20:12 Sum Rule 22:40 Law of Total Probability 25:30 Law of Total Probability Example 28:40 Bayes' Rule 32:17 Bayes' Rule Example Feel free to leave any questions. Please consider subscribing if you liked this video: Thanks for watching everyone! ~~~~~~~~~~~~~~~~~~~~~~~~ Follow me on Instagram: Follow me on Twitter: Check out my website: 44 comments Transcript: Introduction in this video I'll be teaching basic probability for machine learning so probability it's all about How likely something is to happen and assigning a number to that the reason why that's important in machine learning is because machine learning is all about predicting some sort of outcome and often what a machine learning model is doing is it's actually outputting the most probable answer to whatever input so today we're going to go over just a crash course in probability and we're going to look at different types of probability such as joint probability conditional probability and then finally we're going to go over something called base rule let's first start with the definition of probability so if I have some you know Defining Probability box over here so this box is something that I'm going to call the sample space or S and what the is is it represents all the possible outcomes so all possible outcomes of something of some experiment you could say now inside of this box I have different points and each of these points is actually an event so this is an event and an event is a possible outcome of this trial so it's a possible outcome and I'm just going to say that this that an event I'm just going to name it a so you know this specific event here is known as a and what probability is well so I'm going to use P to denote probability I'm just taking the probability of a so I'm just taking the probability of some event and I'm assigning this to some number so actually let's take a look at an example so how about a six-sided dice which a lot of people are familiar with but it looks something like this and each side so it's a cube and each side has a number on it so that might be 1 2 three four five six so that means what are all the possible outcomes when we roll This dice I'm going to use these curly brackets so to note a set but the possible outcomes are 1 2 3 4 5 and 6 so in this case s equals to the set of numbers 1 2 3 4 5 6 because every single time I roll a dice that's my experiment here it's rolling the dice these numbers those are my possible outputs okay so now each of these events in the in the sample space I'm going to say okay 1 is a 1 2 is A2 3 is A3 and you get the point so if I came back to this diagram the sample space diagram up here then I could add some more points so this could be representing A1 A2 A3 A4 A5 and A6 so there's six possible points so if I go down here and I say what is the probability of A1 well A1 means a my event is equal to one so the probability of rolling this dice to be equal to one um what is that value well we know that this is a fair dice right and which means that every single side has an equal probability of coming up so this probability of A1 actually equals 16 and the probability of A2 also equals 16 and all the probabilities so this probability of A6 also equals 16 they all equal 16 so these values over here these are the probabilities that we assign to some event so let's put this in English the probability oops probability of event A1 is 1 six so that's what this means it means the probability of event A1 is 16 that's great so we're getting an introduction to probability and that's what this course is all about but there's so much more left to learn in probability and that brings me to today's sponsor brilliant brilliant is where you can learn by doing with thousands of interactive lessons in math data analysis programming and AI each lesson on Brilliance platform is filled with handson problem solving that actually lets you play with Concepts which is a method that is extremely effective for learning because you're not just memorizing but you're actually practicing those critical thinking skills and you can build up real knowledge by just doing a few minutes every single day a great way to supplement these video courses is actually through practice problems on brilliant's platform currently brilliant has just put out a lot of new datar related content so there's introduction to probability predicting with probability it's perfect for Learners of really any level to get started myself included what's really awesome about brilliant is that you actually get hands on practice with real data sets such as ones from X previously Twitter and Spotify and best of all you can actually try brilliant completely free for 30 days just visit brilliant.org Kylie Ying or click on the link in the description below and if you do decide to subscribe this will give you 20% off an annual premium sub subscription in order to Conditions of Probability qualify as a probability this P there's three conditions that it has to satisfy so the first condition is that for every single event a so for every a the probability of a has to be greater than or equal to zero so that basically just means that every single probability has to be non negative because what does a negative probability mean I we don't really have a meaning for that the second is that the probability of the entire sample space so all of this is equal to one that means that the probability of any of the events happening well that has to equal one and then finally if we have many events so A1 A2 a etc up to a n so just n events so if these are disjoint the sum the probability of any of the events happening so that's A1 + A2 plus you know all the way to a n is equal to actually the sum of all the probabilities so all the individual probabilities so probability of A1 Plus probability of A2 plus probability dot dot dot uh of a n so that's just summing up all the probabilities now let's go back one step what does disjoint mean that basically just means that they don't overlap so that means that another word for disjoint is mutually exclusive and this means that A1 cannot overlap with A2 so in here this is zero okay so that means that they're disjoint when two events cannot overlap with one another um so in our dice example can we roll a one and a two at the same time on the same roll the answer is no we can roll a one and then on the next roll we can roll a two but on the same roll it has to be a one or a two or a three or you know whatever up to six but we can't have two numbers at the same time so that is an example of something that's disjoint and now this here means that the probability that any of these events happen so event A1 Plus A2 would be the probability that these two happen happen A1 + A2 + A3 would be the probability of these three so all of them would be this which is actually just the entire sample space s but we don't have to go all the way up to S so what this is saying is that if these events are mutually exclusive then the probability of a1+ A2 plus whatever number of events is actually equal to the probability of all the events themselves added up so in our dice example the probability of rolling a 1 or a two or a three is actually equal to the probability of these individually so probability of A1 Plus probability of A2 plus probability of A3 which is equal to 16 + 16 + 16 or 1 half now that we understand basic probability let's discuss joint and conditional probability so the first one is joint Joint Probability probability and Joint probability is just what is the probability of two different events happening so the probability of some event X and the probability of some Event Y so X and Y they're the same has a in this example up here they're just two different events so event X and Event Y um and this can be represented in mathematical terms with this funky U upside down U thing or a comma so all of these just mean the probability of X and Y now I'm going to use a 52 card deck for a lot of my examples and in a 52 card deck we have two colors so we have black cards and we have red cards and under black we have Spades we have clubs and under red we have hearts and we have have diamonds so these are the four different Suits now under each suit we have 13 different cards we have Ace one oh nope there's no one 2 3 4 5 6 7 8 9 10 Jack queen king right so that's 13 cards and under clubes we have the exact same thing so I'm not going to rewrite all of that but Ace through King so that's 13 cards same with hearts Ace through King 13 cards and under diamonds we have again Ace through King 13 cards okay so this is a 52 card deck let's use this 52 card deck to show what we mean by probabilities so I'm going to draw a random card okay so the probability that the card that I draw let's ask what is the probability that that random card is a black card so there are 13 + 13 black cards right so that's 26 / 52 total cards which equals 1/2 well now let's say what is the probability of drawing a card and that card being a four well there's a four here there's a four somewhere in here somewhere here and somewhere here so there's four four fours in this deck out of 52 and that equals 13th okay but Kylie where where does joint probability come in so now let's ask what is the probability that we draw a four and it's a black card so how many cards are there in this whole deck that are that equal four and and are black so again we have the four some four here some four here and some four here well these two over here those are red so we don't actually count those which means that we have one two cards in our deck that are both four and black so this is now equal to two out of we still have 52 cards in the deck so two out of 52 which equals 126 so that is what joint probability means it means the probability of both of these things now let's move on to conditional Conditional Probability probability conditional probability looks something like this we have the probability of some event x with this vertical bar Y and now what this bar here means it means given so translating this this means the probability of X of event x given the pro given Event Y so that means that y has already happened okay we're already saying y has happened given that y has happened what is the probability of X so let's go back to our 52 card deck and let's first ask let's let first look at an example so what is the probability of uh let's say Hearts right um given that we chose a red card so how many hearts are there there's 13 well now we're only looking at this section over here because we're saying that we chosen a red card we're giving the reader some information we've chosen a red card so that means instead of out of 52 we're only eliminating this to 26 cards so the probability that we chose a heart given that we already chose a given that we that this card is red is 13 out of 26 or 1/2 now let me ask you what is the probability that we draw a heart given that we drew a black card okay well if we're looking over here none of these cards are red so let me erase that cuz we're not focused on this part anymore none of these cards sorry none of these cards are Hearts so that means how many hearts are in that box there's zero out of 26 so the probability that we draw a heart given that we like we already knew we drew a black C that's zero and now finally what is the probability of a four given that we drew a black card let's go back to our example okay we know we're still in this box because we we drew a black card and now how many fours are in that box there's again uh one two so two over now how many black cards are there 13 + 13 so 26 so there's a 1 and 13 chance that we drew a four given that we drew a black card and so this you might actually notice that this is equal to the probability of four of the card being four and the card being black divided by the probability of the card being a black card so that actually brings me to the mathematical formula for conditional probability the probability of x given Y is equal to the probability of X and Y divided by the probability of Y so yeah again the probability of x given that y happened is equal to the probability of X and Y happening divided by the probability that y happened this leads me to a really important Concept Independence in probability called Independence Independence means are these two events related or not does the does one event happening affect the other event happening so for example um let's say that we're drawing cards and uh you know I'm rolling a dice does the probability that I draw a four out of the deck change if my dice you know rolls of one or rolls a six no these are two different independent events so to formally put this something is independent two events are independent if the probability of X is equal to the probability of x given y so this means that X and Y are independent why because the probability of x well here we're saying okay y already happened but the probability is not changing the probability is still equal so that means that there's no way that y actually affects X because it doesn't matter if y happened or not the chances of X happening are the same so another way to write this would actually be the probability of X and Y if that equals the probability of x times the probability of Y then these two are also independent but this is a little bit harder to conceptualize why now that we understand joint and conditional probabilities let's move on to explain a few other Concepts in probability so let's say I have my sample space s right and let's say that Sum Rule there's a few different partitions of this sample space so I'm just going to say okay over here I have this blue partition I'm going to call this C1 and then I have you know this red part and I'm going to call that C2 and then I have this uh purple part that I'm going to call C3 and so let's say that you know my event a is going to be represented by this chunk of the sample space so this is a okay okay so first I want to talk about something called the sum Rule now let's say we want to find the probability of a we don't know that number but we do know a few different numbers so let's say we don't know you know what this total black space is but let's say that we are given this chunk and this chunk and this chunk that's in yellow that's you can barely see but what it's saying is that the probability of a is equal to summing over these chunks and taking the probability of a and that chunk so in other words in our example this equals the probability of a and C1 plus the probability of a and C2 Plus the probability of a and C3 so that's effectively just summing up this green portion over here sorry let me change a marker okay let's use orange okay so that means that we're summing up this section over here and then this middle section here where the orange barely shows up and then this section over here and that's just saying that when we sum up these three sections that equals the probability of this event space a which looks really obvious from this diagram but now we're just putting it into a mathematical formula then we have something called The Law of Law of Total Probability total probability earlier we actually said that the probability of x given y so that's a conditional probability is equal to the probability of x and Y ID by the probability of Y so we can actually rewrite this so let's multiply by the probability of Y on both sides now the probability of x given y times the probability of Y is equal to the probability of X and Y and what we're going to do in order to derive the law of total probability we're just going to plug this thing into this portion of that formula so the law of total probability says the probability of a sorry to change the letters on you again but I'm going back to this probability of a here is equal to the sum over all of those chunks of you know the sample space c i and then the probability of a given that chunk times the probability of that chunk all right so this here this is the law of total probability and let's try to visualize that okay so the probability of a given some CI so what that means is what is this chunk here when we're looking at this section so that this value here that gives us the probability of a given some C1 and then the probability of C1 is actually equal to this green portion out of the total event space so that's this green portion here including that middle part so the pro the probability of C1 is equal to this portion here divided by all of that so in this case it's approximately 1/3 a little bit less than that and now if you take that over this section and this section then we can actually find the probability of a okay again it seems like this is a really roundabout way of doing that so why would we even even bother the reason why we do this is because sometimes this is the information that we're given for Law of Total Probability Example example let's say that um you know we're trying to measure the probability of doing our homework so probability of doing homework and now let's say that the probability of doing homework changes whether or not it rains or it's sunny outside so if it's raining then I'm more likely to stay inside and the probability that I do my homework is 95% now if it's sunny that probability drops to 30% okay and we also know that let's say we live in California and the probability of rain in California is equal to 15% of the time on a random day what is the probability that I am doing homework so probability of homework well I don't know the total probability of homework but what I do know is I know the probabilities given if it's raining or if it's sunny and I know the probability that it will rain right and so what this actually ends up equaling is the probability that I do my homework give given the rain times the probability that it rains plus the probability that I do my homework given that it's sunny times the probability of sun and now this sun we actually know is equal to 1 minus the probability of rain because either it's rainy or it's sunny there's no in between so that means that this is equal to 85% and we know that probability of rain is 15% the probability that I do homework given that it's raining okay well that's 95% the probability that I do homework given it sunny 30% so now you know I can all of a sudden solve for the probability that I'm doing homework on a given day and that equals 0.95 0.15 plus 0.3 0.85 which if you plug that into a calculator this comes out to 0. 3975 or approximately a 39. 75% chance so 40% chance that I'm doing my homework on a given day so cool would you have guessed that from these numbers probably not but now with the law of total probability we have a way to solve for that and that's why that's why this complicated equation over here is all of a sudden so useful sometimes now given the law of total probability I want to focus on something Bayes' Rule called B rule so base rule is really important probably one of the most important things about this specific lesson so base rule is another way to solve for a conditional probability it's another way to solve for probability of a given some probability B and we know that this is the probability of A and B over the probability of B right we've already we've already talked about that and we also have a probability for A and B right and this can be the probability of a given B times the probability of B which is not really that helpful for us because if we use that then we cancel these two B's and we're just left with probability of a given b equals probability of a given B okay not very helpful so let's get rid of that but the probability of A and B is the same as the probability of b and a so this is actually interchangeable and we can write this as the probability of B given a times the probability of a and now we can say okay the probability of a given B can actually be expressed in terms of probability of B given a and all of a sudden that's really powerful because sometimes we don't know a given B and we want to solve for it and we're given B given a so this here is base rule okay so what we can do is actually rewrite this and then I'm going to name a few terms so let's rewrite this probability of a given B is equal to the probability of a times the probability of B given a / by the probability of B this here is what we call a prior okay so before we get some evidence B the probability of a that's that's what we have so so that's what we know that's our prior probability and now this here probability of B this is called our marginal probability and that means the probability of some evidence be the probability that this evidence is occurring this probability of B given a this is our likelihood so How likely is it that you know this evidence is occurring given that this a this event a this belief is true and then finally over here we have our posterior so this is our updated belief this is our updated probability of a after some evidence of B in machine learning what actually happens is we're constantly updating this prior and we're getting a new distribution out of it every single time we see new evidence and that's how probability ties into machine learning is we're constantly updating what we think is true and the probability well we're constantly updating the probability of something to be true and finally before you know we finish this lesson let's just do a quick example with Bas rule let's talk about covid Bayes' Rule Example testing okay when you have covid you can test positive or you can test negative right so there's two possibilities of covid so if somebody actually has covid and someone who does not have covid now if I actually have covid and I take a covid test I can test positive right and that would make sense because I actually have Co but I can actually also test negative right if I didn't wait long enough to take the test then um then my symptoms are whatever and the test come back comes back negative even though I'm actually positive so this is something called a false negative now somebody who doesn't have covid can also test positive and they can also test negative and when somebody doesn't have covid so they're actually negative but they test positive this is called a false positive and in this specific case like test makers actually know what these probabilities are in the case where if we think about it what is worse testing positive when you don't actually covid or testing negative when you actually do have covid well if you test negative and you actually have covid and then you go around somebody you know maybe your grandma well that's bad because the test said you don't have covid and now you may have just exposed somebody else to covid maybe Co is like not that bad anymore but let's say that it was some new mysterious disease that um has a 50% death rate well it's really really bad to actually test negative because like because you don't want to be spreading that around right whereas if you don't actually have it and you test positive you just play it a little bit more safe it's not the end of the world so in this specific case when test makers are making these test they actually prefer like they they think it's okay to have a higher false positive rate than a higher false negative so let's say that these tests when you actually have covid uh they are 95% of the time true positive so so you actually have Co and you test positive and 5% of the time okay fine you test false negative well when you don't have covid they are okay with that maybe being a 90% negative test and then when you test positive that's maybe 10% okay and then now at any given point let's say we also have approximate estimate of how how much of the population has covid how prevalent is co and let's say that's 2% so 98% of people actually don't have covid okay so suppose we test positive what is the actual probability that we have covid and this is not knowing anything else this is not knowing oh you know I have a cough and a fever and symptoms like covid no right now we just know that we just tested positive um what is the probability that we have Co okay in mathematical terms what we're trying to solve is the probability that we have covid given that we tested positive right and now we can use base rule on this so This equals the probability that we get a positive test given covid times the probability that you know Co is actually a thing divided by the probability of a positive test okay so let's look at our tree andall for each of these the probability that we test positive given that we have Co okay so given that we actually have Co the probability that we test positive here is 95% so this here is 0.95 and then the probability that we have covid okay well what is the probability that we actually have Co here so that's 2% 0.02 because we have to divide that by 100 2ide 100 0.02 and now divided by the probability of testing positive okay so the probability of testing positive we actually have to use what we learned up here the law of total probability because we don't actually know straight up what is the probability testing positive we have it over here and over here let me get rid of a few of these circles first right so we can test positive here and we can test positive here so the probability of testing positive here is equal to the probability of a positive test given that you know we do have that covid that given that we actually have covid times the probability that we actually have covid plus the probability that we're positive given no covid times the probability of not having covid so the probability that we test positive given Co I mean we already have that value up here we already actually have this top part up here that's our numerator so I'm just going to copy that over 0.95 0.02 plus now this denominator here is just the other Branch so the probability that we test positive given that we don't have covid is 10% so 0.1 time the probability of no covid is 98% so times 098 and let me just move this part over so we don't uh it doesn't intersect oops okay so this here equals this and now if we actually do the math here and do these multiplications out we get 0.19 in the numerator plus 0.19 over 0.19 plus 0.1 0.98 so 0 uh 098 okay and this actually ends up giving us 0 around 66 so now given that these probabilities are true which you know in the real world they probably are not this is just an example problem but given that these are the numbers if we actually if we test positive and we don't know that you know we have symptoms or anything the probability that we actually have covid is only 66% so yeah that's base Rule and that's what a lot of probability is based on base rule is often used for things such as statistical inference and modeling and also it's used in different machine learning algorithms such as naive Baye or expectation maximization so it's really critical to have a solid foundation of probability and so hopefully this video today has given you enough of a foundation to understand some basic probability Concepts so that you can at least get started in machine learning e
15795	https://ntrs.nasa.gov/api/citations/20220005627/downloads/Simmons_eVTOL_Prop_Modeling_Aviation22_v04232022.pdf	Efficient Variable-Pitch Propeller Aerodynamic Model Development for Vectored-Thrust eVTOL Aircraft Benjamin M. Simmons∗ NASA Langley Research Center, Hampton, Virginia 23681 This paper describes a variable-pitch propeller testing and aerodynamic modeling method-ology for electric vertical takeoff and landing (eVTOL) aircraft using wind tunnel testing. Propellers used for eVTOL aircraft propulsion systems experience a wide range of operating conditions resulting in significant variation of axial thrust and torque, as well as off-axis forces and moments, across a typical flight envelope. An experimental design is developed to collect informative wind tunnel data in an efficient manner using response surface methods. System identification methods are then applied to isolated propeller wind tunnel data to develop a mathematical model of the propeller valid throughout the transition envelope of a vectored-thrust eVTOL aircraft. Multiple explanatory variable definitions are postulated and compared using modeling and prediction performance metrics. Modeling results validated against data withheld from the modeling process indicate good predictive capability and agree with theoretical expectations. All identified propeller models are provided, which allows the models to be used in future eVTOL aircraft flight dynamics simulations. Nomenclature 𝐶𝑇𝑥, 𝐶𝑇𝑦, 𝐶𝑇𝑧 = propeller force coefficients 𝐶𝑄𝑥, 𝐶𝑄𝑦, 𝐶𝑄𝑧 = propeller moment coefficients 𝐷 = propeller diameter, ft 𝑖𝑝 = propeller incidence angle, rad 𝐽 = advance ratio 𝐽𝑥 = normal advance ratio 𝐽𝑧 = edgewise advance ratio 𝑛 = propeller and motor rotational speed, revolutions/s 𝑄𝑥, 𝑄𝑦, 𝑄𝑧 = propeller moments, ft·lbf ¯ 𝑞 = freestream dynamic pressure, lbf/ft2 𝑇𝑥, 𝑇𝑦, 𝑇𝑧 = propeller forces, lbf 𝑉 = freestream velocity, ft/s 𝛿𝑐 = collective pitch (blade root) angle, rad 𝜂𝑚 = motor pulse width modulation command, 𝜇s 𝜂𝑐 = collective pitch servo-actuator pulse width modulation command, 𝜇s 𝜌 = air density, slug/ft3 I. Introduction E lectric vertical takeoff and landing (eVTOL) aircraft are currently drawing significant interest in the aerospace industry as an enabling technology for future Urban Air Mobility (UAM) transportation missions. Mission requirements for eVTOL aircraft necessitate operations in a wide variety of flight conditions spanning hover, transition, and forward flight. Consequently, eVTOL aircraft propellers experience aerodynamic conditions that significantly differ from conventional propeller and rotor operation in a typical flight profile. Propeller aerodynamics are conventionally modeled in the axial airflow condition where data tables or functional representations of axial thrust and torque coefficients are sufficient to model the propeller aerodynamics. At high incidence angles, axial thrust and torque deviate from nominal axial airflow values and off-axis propulsive forces and moments become significant. ∗Research Engineer, Flight Dynamics Branch, MS 308, Member AIAA. 1 Several works have employed methods for theoretically and computationally predicting propeller aerodynamics at incidence [1–10]; however, experimental techniques provide the most accuracy in revealing the highly complex and nonlinear behavior of high incidence angle propeller aerodynamics. Previous experimental studies of propellers operating at high incidence angles include Refs. [11–19]. In particular, the work described in this paper builds on experimental design and modeling techniques for fixed-pitch, eVTOL aircraft propellers described in Refs. [16–19]. The significant new contributions of this work include characterization of a variable-pitch eVTOL aircraft propeller, an extended experimental design methodology, a comparison of modeling results using different explanatory variable definitions, and distribution of variable-pitch propeller aerodynamics models valid through a significant portion of an eVTOL aircraft flight envelope. The objective of this work is to identify a model of the propeller aerodynamics enabling accurate flight dynamics simulation development for a new vectored-thrust eVTOL vehicle. Vectored thrust in the context of eVTOL aircraft includes both tilt-wing and tilt-rotor configurations. The intended aircraft is similar in scale and utility to the Langley Aerodrome No. 8 (LA-8) tandem tilt-wing, eVTOL, subscale wind-tunnel/flight test (SWFT) aircraft pictured in Fig. 1 . Design of experiments (DOE) and response surface methodology (RSM) techniques are leveraged to develop an efficient, statistically-rigorous wind tunnel experiment [21, 22]. Response surface models are then developed using propeller aerodynamic model structure insight and system identification algorithms . The proposed propeller model identification strategies can be applied for many tilt-wing and tilt-rotor eVTOL vehicles currently under development∗and future eVTOL vehicles. This work is complemented by recent eVTOL aircraft modeling research leveraging isolated propeller models for flight dynamics simulation development [24, 25]. Fig. 1 LA-8 mounted in the NASA Langley 12-Foot Low-Speed Tunnel. This paper is organized as follows. Section II provides background information on propeller aerodynamics. Section III describes the wind tunnel experimental setup. The experimental design leveraging DOE/RSM techniques is developed in Sec. IV. Section V outlines the overall propeller aerodynamic modeling approach, which is followed by an overview of the employed model identification methods given in Sec. VI. Propeller aerodynamic modeling results are presented in Sec. VII. Overall conclusions are summarized in Sec. VIII. II. Background Propeller aerodynamics are well-defined and well-researched for nominal operating conditions in axial flow where aerodynamic predictions can be made analytically and/or experimentally. Theoretical techniques include momentum theory, blade element methods, and vortex theories . Experimental techniques typically consist of developing data tables or functional representations from wind tunnel data. For a propeller operating in airflow normal to the propeller disk, only a net thrust force and a net aerodynamic torque acting along the axis of rotation are generated . The individual propeller blades can be thought of as rotating wings which each produce a lift force perpendicular to the relative flow direction and a drag force parallel to the relative flow direction . The summed lift forces produced by the propeller blades is the propeller thrust 𝑇𝑥. The summed drag forces result in a net moment about the propeller shaft opposite to the direction of rotation, which is the propeller aerodynamic torque 𝑄𝑥. ∗Information available online at [retrieved 29 October 2021] 2 Propeller data are generally nondimensionalized and presented in terms of thrust coefficient 𝐶𝑇𝑥and torque coefficient 𝐶𝑄𝑥(or equivalently by power coefficient 𝐶𝑃= 2𝜋𝐶𝑄𝑥where 𝑃= 2𝜋𝑛𝑄𝑥). The thrust and torque coefficients are defined as: 𝐶𝑇𝑥= 𝑇𝑥 𝜌𝑛2𝐷4 (1) 𝐶𝑄𝑥= 𝑄𝑥 𝜌𝑛2𝐷5 (2) The thrust and torque coefficients can be shown through dimensional analysis to have a functional dependence on advance ratio 𝐽, propeller blade Reynolds number 𝑅𝑒, and propeller tip Mach number 𝑀tip for a given fixed-pitch propeller design [27, 28]. For propellers with variable-pitch hubs, thrust and torque also vary with blade pitch angle. Advance ratio 𝐽, which relates to the linear distance traveled by the propeller in one revolution, is defined as: 𝐽= 𝑉 𝑛𝐷 (3) Advance ratio generally has the largest effect on fixed-pitch propeller aerodynamics, and accordingly, thrust and torque coefficient representations are commonly expressed as only a function of advance ratio. Representing propeller aerodynamics only as a function of advance ratio for a fixed blade pitch angle requires that airflow is parallel to the propeller axis of rotation as well as the assumptions that viscous and compressibility effects are negligible . Reynolds number is a dimensionless quantity which corresponds to the ratio of inertial to viscous forces acting on a body. For large propellers, the propeller blade Reynolds number effects are minimal and can generally be neglected. For small propellers, the Reynolds number is lower, indicating that the viscous forces become important and results in propeller performance degradation [29, 30]. Following the definition given in Ref. , the propeller blade Reynolds number 𝑅𝑒is 𝑅𝑒= 𝜌𝑉𝑝𝑐 𝜇 (4) where 𝑐is the propeller chord at 75% blade length, 𝜇is the dynamic viscosity, and 𝑉𝑝= 0.75𝜋𝑛𝐷is the propeller blade linear speed at 75% blade length. Mach number is the ratio of flow speed to the speed of sound 𝑎, which physically represents the ratio of inertial forces to forces related to compressibility of the fluid . The propeller tip Mach number, which quantifies the averaged compressibility effects, is defined as : 𝑀tip = 𝜋𝑛𝐷 𝑎 (5) For the propeller studied in this paper, the propeller blade Reynolds number and tip Mach number effects were small compared to advance ratio and blade pitch angle effects; however, the influence of propeller blade Reynolds number and/or tip Mach number, which are both proportional to rotational speed 𝑛, was found to be beneficial to consider for model development. This will be discussed further in Sec. VII. When the airflow relative to a propeller is not parallel to the axis of rotation, the propeller will produce auxiliary forces and moments other than the axial thrust and torque . In this condition, periodic variation in propeller blade local angle of attack results in a non-uniform load distribution on the propeller disk. Thus, in a general case of arbitrary flow direction relative to the propeller disk, propeller forces and moments will also be dependent on angle between the freestream velocity and propeller axis of rotation, in addition to advance ratio, propeller blade pitch angle, propeller blade Reynolds number, tip Mach number, and the propeller design. This angle between the freestream airflow and propeller rotation axis is the propeller incidence angle, 𝑖𝑝, shown in Fig. 2. The value of 𝑖𝑝is zero when airflow is normal to the propeller disk, opposing the direction of axial thrust. Previous work has demonstrated the benefit of using the normal and edgewise (tangential) component of advance ratio 𝐽𝑥= 𝑉cos𝑖𝑝 𝑛𝐷 = 𝐽cos𝑖𝑝 (6) 𝐽𝑧= 𝑉sin𝑖𝑝 𝑛𝐷 = 𝐽sin𝑖𝑝 (7) as explanatory variables for modeling propeller aerodynamics at high incidence angles . Similar advance ratio component definitions are used for rotorcraft [31, 32]; this representation of advance ratio was also used in a previous propeller modeling effort which developed lookup tables for propulsive forces and moments for a quadrotor vehicle . 3 𝑉 𝑧 𝑖𝑝 𝑥 Fig. 2 Propeller incidence angle definition and coordinate system. The propeller side force𝑇𝑦, normal force𝑇𝑧, pitching moment𝑄𝑦, and yawing moment𝑄𝑧can be non-dimensionalized in a manner similar to the thrust 𝑇𝑥and torque 𝑄𝑥. The propeller normal force coefficient 𝐶𝑇𝑧, side force coefficient 𝐶𝑇𝑦, pitching moment coefficient 𝐶𝑄𝑦, and yawing moment coefficients 𝐶𝑄𝑧, are defined as: 𝐶𝑇𝑦= 𝑇𝑦 𝜌𝑛2𝐷4 , 𝐶𝑇𝑧= 𝑇𝑧 𝜌𝑛2𝐷4 , 𝐶𝑄𝑦= 𝑄𝑦 𝜌𝑛2𝐷5 , 𝐶𝑄𝑧= 𝑄𝑧 𝜌𝑛2𝐷5 (8) The propeller force and moment sign convention used in this work follows the right-handed propeller coordinate system shown in Fig. 2, where the 𝑦-axis is pointed into the page. III. Wind Tunnel Test Setup The test article for this study was a variable-pitch, three-bladed, 19.5-inch diameter, clockwise rotating propeller. Assuming that all clockwise and counterclockwise rotating propellers on an aircraft have perfectly mirrored designs, only one orientation needs to be tested. The propeller data and models would be identical, except that the signs of the lateral propeller force and moment components, 𝑄𝑥, 𝑇𝑦, and 𝑄𝑧, would be reversed for the counterclockwise rotating propeller. Figure 3 shows the assembled propeller and variable-pitch hub on the wind tunnel sting. The propeller was powered by a 220 KV electric motor and 100 amp electronic speed control (ESC). A variable-pitch mechanism was designed and fabricated to equip the flight vehicle with high-bandwidth variable-pitch control for aircraft stabilization. The ability to leverage variable propeller pitch as a control effector becomes important for certain full-scale eVTOL aircraft configurations because motor/propeller rotational speed control becomes slower as propeller diameter and inertia increase. As configured for testing, the custom variable-pitch mechanism allowed setting the collective pitch angle 𝛿𝑐 (blade root pitch angle) between −9.6 and +7.2 deg. A multi-exposure image showing propeller blade collective pitch movement is shown in Fig. 4. The isolated propeller wind tunnel experiment was performed in the NASA Langley Research Center 12-Foot Low-Speed Tunnel.† The test was designed to cover a significant portion of the operational flight envelope of a future eVTOL SWFT aircraft. For operational convenience and test facility implementation consideration, test factors selected to control during wind tunnel testing were freestream velocity 𝑉(or dynamic pressure ¯ 𝑞), incidence angle 𝑖𝑝, motor pulse width modulation (PWM) command 𝜂𝑚, and collective pitch servo-actuator PWM command 𝜂𝑐. The factor settings to test were determined using DOE/RSM experimental design techniques that will be discussed in Sec. IV. At each test point, the six force and moment components were measured using a strain gauge balance and a propeller rotational speed measurement was provided by the ESC. A polynomial calibration model between the collective servo-actuator PWM command 𝜂𝑐and the collective pitch angle 𝛿𝑐was used to determine the tested collective pitch angle. Each individual wind tunnel run was executed by automatically commanding incidence angle, motor PWM command, and collective PWM command at each test point throughout the run. Incidence angle variation is depicted in the multi-exposure image displayed in Fig. 5a. The propeller test coordinate system and incidence angle measurement are shown in Fig. 5b. The wind tunnel dynamic pressure setting had to be manually changed from the control room and was slower to change compared to the other test factors, but could be adjusted during the course of a run without †Information available online at [retrieved 29 October 2021] 4 Fig. 3 Experimental propeller mounted on the wind tunnel sting. Fig. 4 Multi-exposure image showing propeller blade pitch movement. shutting off tunnel airflow. In accordance with the ease of changing the settings for each test factor, the incidence angle, motor PWM command, and collective servo-actuator PWM command were designated as easy-to-change (ETC) factors; freestream velocity (dynamic pressure) was designated as a hard-to-change (HTC) factor. These considerations informed the experimental design discussed in the next section. 𝑉 Sting Rotation (a) Multi-exposure image showing incidence angle variation 𝑖𝑝 𝑥 𝑦𝑧 𝑉 (b) Propeller test coordinate system Fig. 5 Experimental propeller mounted in the NASA Langley 12-Foot Low-Speed Tunnel (viewed from above). Each of the test factors were varied to encompass the anticipated transition envelope for the future eVTOL vehicle. Although the test point conditions were specified using motor PWM command and collective pitch servo-actuator PWM command, these settings were used to indirectly sweep variables more pertinent to propeller aerodynamics: propeller rotational speed and collective pitch angle. Freestream velocity and propeller rotational speed were then further reduced to calculate propeller advance ratio. Changes in flow velocity drive changes in propeller advance ratio. Changes in motor PWM command change the propeller rotational speed, which has the primary effect of changing the propeller 5 blade Reynolds number and tip Mach number, but also contributes to changes in the propeller advance ratio. IV. Experimental Design As mentioned previously, the propeller test matrix was designed using four variable factors: freestream velocity 𝑉(HTC), incidence angle 𝑖𝑝(ETC), motor command 𝜂𝑚(ETC), and collective servo-actuator command 𝜂𝑐(ETC). For the wind tunnel used for this study, changing 𝑉required more time and effort than to change 𝑖𝑝, 𝜂𝑚, and 𝜂𝑐. Consequently, for testing efficiency, 𝑉was treated as a HTC factor held at a constant value for several consecutive test points, rather than being varied between each test point. Four different test regions were designed and tested: 1) a 2-factor Hover Region varying 𝜂𝑚and 𝜂𝑐with no airflow, 2) a 4-factor Low Incidence Region with a higher airspeed range (high-speed transition), 3) a 4-factor High Incidence Region with a lower airspeed range (low-speed transition), and 4) a 3-factor Descent Region surrounding the vortex ring state (VRS) condition at 𝑖𝑝= 180 deg. The factor ranges tested for each region are listed in Table 1, with a visual overview shown in Fig. 6.‡ These test ranges were selected based on the expected transition flight envelope, test hardware restrictions, and insight from a previous study [16, 17], where it was expected that a cubic response surface model would be able to adequately describe the propeller force and moment coefficient variation in each region. Note that a lower collective command setting 𝜂𝑐 corresponds to a higher collective pitch angle 𝛿𝑐, and vice versa. Table 1 Factor ranges for each propeller test region Factor Units Hover Region Low Incidence Region High Incidence Region Descent Region 𝑉(HTC) ft/s 0 (fixed) 14.5 to 71.1 14.5 to 54.3 14.5 to 45.9 𝑖𝑝(ETC) deg N/A (fixed at 0) 0 to 65 50 to 100 180 (fixed) 𝜂𝑚(ETC) 𝜇s 1500 to 1960 1500 to 1960 1500 to 1960 1500 to 1960 𝜂𝑐(ETC) 𝜇s 1000 to 2000 1000 to 1500 1200 to 1600 1200 to 1600 Fig. 6 Freestream airspeed and incidence angle range for each propeller test region. Modeling test matrix design was accomplished with the aid of Design-Expert®, a commercially available statistical software package.§ The test factor settings in each region were determined using I-Optimal response surface designs. I-optimal designs minimize the integrated prediction variance for a predefined model order over the range of test factors, which reduces prediction error for the identified models [21, 22]. The primary goal for this study was to develop propeller aerodynamic models with low prediction error over the region of operability, so an I-optimal design is a good response surface design choice. The Hover Region was run as a completely randomized design, meaning that each test factor was varied at each new test point. The Low Incidence Region, High Incidence Region, and Descent Region employed split-plot designs [34, 35], or designs with restricted randomization for HTC factor(s). Split-plot designs are used to conduct efficient experiments with HTC factors, while still executing a statistically-rigorous designed experiment. The HTC factors (whole-plot factors) are held at a constant value for several consecutive test points while ‡Propeller incidence angle 𝑖𝑝is undefined in hover (𝑉= 0). For Hover Region testing, the sting orientation angle was fixed at 0 deg, as indicated in Table 1. The Hover Region is plotted from 0 to 180 deg in Fig. 6 to reflect that hover can be approached from all incidence angles at low speeds. §Information available online at [retrieved 29 October 2021] 6 the ETC factors (sub-plot factors) are varied between each test point. The set of consecutive test points where the settings for HTC factor(s) are held constant is referred to as a group. The different amount of randomization for HTC factors and ETC factors results in different whole-plot and sub-plot variance components, which must be considered for experiment design and data analysis . All split-plot designs used for this work were ordinary least-squares equivalent [22, 36–38], meaning that the ordinary least-squares parameter estimates are equal to parameter estimates obtained using generalized least-squares with sub-plot/whole-plot variance component estimates. In other words, the design performance and estimates of model parameters are decoupled from the sub-plot and whole-plot variance components. A set of completely randomized and split-plot I-optimal experiment designs were developed with different design model complexities to investigate the adequacy of fitting a cubic model. Assessment of the prediction variance of a response surface design for a given model structure provides insight into its precision of prediction and allows comparison of different response surface designs. For a completely randomized design, the variance of the predicted response is Var[ ˆ 𝑦(𝒙0)] = 𝜎2𝒙𝑇 0 𝑿𝑇𝑿 −1 𝒙0 (9) where ˆ 𝑦(𝒙0) is the predicted response evaluated at the design space location 𝒙0 expanded to the form of the model structure, 𝑿is a matrix composed of the designed test points in the form of the model structure, and 𝜎2 is the measurement error variance . As can be seen by examining Eq. (9), the prediction variance is a function of the experimental design, the model structure, the location in the design space, and the measurement facility error variance. The scaled prediction variance (SPV) and unscaled prediction variance (UPV), which remove the dependence on 𝜎2, are used to compare experimental designs prior to experimentation. SPV is defined as SPV = 𝑁Var[ ˆ 𝑦(𝒙0)] 𝜎2 = 𝑁𝒙𝑇 0 𝑿𝑇𝑿 −1 𝒙0 (10) where the number of test points 𝑁penalizes a larger design size . The SPV considers the prediction accuracy as well as the expense of test points when comparing designs. The UPV, defined as UPV = Var[ ˆ 𝑦(𝒙0)] 𝜎2 = 𝒙𝑇 0 𝑿𝑇𝑿 −1 𝒙0 (11) provides an assessment of the prediction precision independent from the size of the experimental design. For split-plot designs, restricted randomization for HTC factors and the consequent compound error structure results in a different prediction variance computation . The prediction variance for split-plot designs is calculated as: Var[ ˆ 𝑦(𝒙0)] = 𝒙𝑇 0 𝑿𝑇𝚺−1𝑿 −1 𝒙0 (12) The SPV and UPV for split-plot designs are: SPV = 𝑁Var[ ˆ 𝑦(𝒙0)] 𝜎2 sp + 𝜎2 wp = 𝑁𝒙𝑇 0 𝑿𝑇𝚺−1𝑿−1 𝒙0 𝜎2 sp + 𝜎2 wp (13) UPV = Var[ ˆ 𝑦(𝒙0)] 𝜎2 sp + 𝜎2 wp = 𝒙𝑇 0 𝑿𝑇𝚺−1𝑿−1 𝒙0 𝜎2 sp + 𝜎2 wp (14) Here, 𝜎2 sp is the sub-plot variance and 𝜎2 wp is the whole-plot variance, with the measurement error variance of a single data point being 𝜎2 sp + 𝜎2 wp. For 𝑛wp total whole-plots (groups), the covariance matrix 𝚺is 𝚺=         𝚺1 . . . 0 . . . ... . . . 0 . . . 𝚺𝑛wp         with the covariance matrix for each 𝑗th whole plot given as : 𝚺𝑗=         𝜎2 sp + 𝜎2 wp . . . 𝜎2 wp . . . ... . . . 𝜎2 wp . . . 𝜎2 sp + 𝜎2 wp         , 𝑗= 1, 2, ..., 𝑛wp 7 Graphical presentation of the distribution of prediction variance throughout the design space is an effective way to assess experimental designs. Each candidate design used for this work was compared using fraction of design space (FDS) plots . FDS graphs depict the prediction variance distribution over the design space in a concise manner, where the prediction variance metrics are plotted against the FDS encompassing a prediction variance less than or equal to a particular value. It is also useful to consider the FDS including a particular model precision, quantified by the confidence interval half-width 𝛿[41–43]. The model precision 𝛿normalized by the response standard deviation 𝜎 plotted against FDS provides further insight into the prediction capability of the model developed from a particular experiment design, prior to conducting the experiment. For this study, a design was deemed to be adequate for fitting a particular model if 𝛿/𝜎was less than two for greater than 95% of the design space. The prediction variance threshold PV∗used to determine the FDS within a given model precision level is PV∗= 𝛿/𝜎 𝑡𝛼/2,𝑁−𝑛𝑝 2 (15) where 𝑁is the number of test points, 𝑛𝑝is the number of parameters in the model, and 𝛼is the significance level (chosen as 𝛼= 0.05 for this study). A. Hover Region: 2-Factor Completely Randomized Design In the Hover Region, only the motor PWM command and the collective PWM command were varied with no tunnel airflow and sting incidence angle fixed. A comparison of candidate 2-factor completely randomized I-optimal designs using a cubic, quartic, quintic, and sixth-order design model complexity is shown in Table 2 and Fig. 7. Each candidate design contains five model points beyond the minimum number of test points needed to fit a full model of the specified order, as well as one center point. Table 2 lists the number of test points, as well as the FDS with 𝛿/𝜎≤2 for a cubic and quartic evaluation model. Due to the small number of factors, increasing the design model complexity only marginally increases the number of test points. An adequate FDS (FDS ≥0.95) for a normalized model precision 𝛿/𝜎≤2 is obtained with a design order one power larger than the evaluation model order. Figure 7 shows the UPV, SPV, and 𝛿/𝜎threshold values against FDS for a cubic evaluation model. The UPV and 𝛿/𝜎threshold curves decrease in value and become more uniform (flat) as the design order increases. The SPV curve is similar for each design order. Table 2 Comparison of candidate Hover Region experimental designs Cubic Model Quartic Model Design Order Points FDS with 𝛿/𝜎≤2 FDS with 𝛿/𝜎≤2 Cubic Design 16 0.921 0.000 Quartic Design 21 0.998 0.865 Quintic Design 27 1.000 0.998 Sixth-Order Design 34 1.000 0.999 Fig. 7 Hover Region FDS plots for a cubic evaluation model. 8 Although the quartic design meets the FDS criteria for a cubic evaluation model, the quintic design was selected to use for the experiment due to the increased data density and the relatively low expense of executing the additional test points for the higher order design. Prior to running the experiment, three additional center points were added and spaced throughout the test matrix (resulting in four total center points). The center points aid in further reduction of the prediction variance within the experimental region and allow estimation of pure error . Six additional test points selected using a random number generator were added throughout the design to use as validation data withheld from model identification. The commanded Hover Region motor and collective PWM factor settings are shown in Fig. 8a. The observed values of the propeller rotational speed and collective pitch angle are shown in Fig. 8b, where it can be seen that a higher collective pitch angle setting results in a lower propeller speed. This is due to increased loading on the propeller at higher collective pitch settings and the open-loop nature of the motor commands used for the experiment. Although propeller speed could be commanded directly and regulated using a feedback controller, using motor PWM settings as the commanded test factor is more operationally convenient from an experimental design and wind tunnel implementation perspective, with minimal impact on the modeling results . (a) Motor command against collective command (b) Observed rotational speed against collective pitch angle Fig. 8 Designed and observed test variable values for the Hover Region. B. Low and High Incidence Regions: 4-Factor Split-Plot Design In the Low Incidence Region and High Incidence Region, all four test factors were varied to characterize the propeller aerodynamics across the transition flight envelope for the intended eVTOL vehicle. Each region used the same experiment design in coded units, but translated to the factor settings in engineering units shown in Table 1. A comparison of candidate 4-factor split-plot I-optimal designs using a cubic, quartic, quintic, and sixth-order design model complexity is shown in Table 3 and Fig. 9. Each candidate design contains five model points beyond the minimum number of test points and two groups beyond the minimum number of groups needed to fit a full model of the specified order. Table 3 lists the number of test points and number of groups, as well as the FDS with 𝛿/𝜎≤2 for a cubic and quartic evaluation model. An adequate FDS (FDS ≥0.95) for a normalized model precision 𝛿/𝜎≤2 is obtained with a design order two powers larger than the evaluation model order. Figure 9 shows the UPV, SPV, and 𝛿/𝜎threshold values against FDS for a cubic evaluation model. The UPV and 𝛿/𝜎threshold curves decrease in value and become more uniform as the design order increases. The SPV curve increases in value as the design order increases. 9 Table 3 Comparison of candidate Low/High Incidence Region experimental designs Cubic Model Quartic Model Design Order Points Groups FDS with 𝛿/𝜎≤2 FDS with 𝛿/𝜎≤2 Cubic Design 40 6 0.000 0.000 Quartic Design 75 7 0.922 0.000 Quintic Design 131 8 0.988 0.785 Sixth-Order Design 215 9 0.999 0.982 Fig. 9 Low and High Incidence Region FDS plots for a cubic evaluation model. Balancing design quality to support identifying a cubic model and test efficiency, the quintic split-plot design was selected for the Low Incidence and High Incidence Region experiments. The base quintic design was augmented with three center groups containing three center points each (resulting in nine total center points) near the start, middle, and end of the test matrix to allow estimation of pure error and further reduce prediction variance. The ETC factor settings were reset between center points within the center point groups by commanding randomized changes in the ETC factors, which also functioned as validation test points. Furthermore, two test points selected using a random number generator were appended to each non-center point group to use as additional validation data withheld from model identification. These validation data were found to provide a good estimate of prediction error while remaining a modest number of test points for the present application. The experimental test points for the Low Incidence Region are shown in Fig. 10. Figure 10a shows the commanded freestream velocity (HTC factor) against test point number. Figure 10b shows the incidence angle against freestream velocity. The commanded motor and collective factor settings are shown in Fig. 10c. The observed values of the propeller rotational speed and collective pitch angle are shown in Fig. 10d. The coded test matrix for the high incidence angle region was identical, and thus, followed the same test matrix as shown in Figs. 10a-10c with different values of test factors in engineering units. 10 (a) Freestream velocity (HTC factor) against test point number (b) Incidence angle against freestream velocity (c) Motor command against collective command (d) Observed rotational speed against collective pitch angle Fig. 10 Designed and observed test variable values for the Low Incidence Region. C. Descent Region: 3-Factor Split-Plot Design In the Descent Region, the freestream velocity, motor PWM command, and collective PWM command were varied at a fixed incidence angle of 𝑖𝑝= 180 deg to characterize the location and average thrust losses associated with the VRS condition. A comparison of candidate 3-factor split-plot I-optimal designs using a cubic, quartic, quintic, and sixth-order design model complexity is shown in Table 4 and Fig. 11. Each candidate design contains five model points beyond the minimum number of test points and two groups beyond the minimum number of groups needed to fit a full model of the specified order. The candidate design comparison interpretation is similar to that described in Sec. IV.B. Table 4 Comparison of candidate Descent Region experimental designs Cubic Model Quartic Model Design Order Points Groups FDS with 𝛿/𝜎≤2 FDS with 𝛿/𝜎≤2 Cubic Design 25 6 0.000 0.000 Quartic Design 40 7 0.837 0.000 Quintic Design 61 8 0.965 0.492 Sixth-Order Design 89 9 0.997 0.956 11 Fig. 11 Descent Region FDS plots for a cubic evaluation model. Due to the possibility of increased thrust sensitivity near the VRS condition, the sixth-order design was selected to run for the experiment for the increased data density throughout the experimental region. Center points and validation points were added throughout the test matrix following the same approach and justification as discussed in Sec. IV.B The commanded and observed test variable values for the Descent region are shown in Fig. 12. (a) Freestream velocity (HTC factor) against test point number (b) Motor command against freestream velocity (c) Motor command against collective command (d) Observed rotational speed against collective pitch angle Fig. 12 Designed and observed test variable values for the Descent Region. 12 D. Final Design Summary The final design qualities for each propeller test region are summarized in Table 5 and Fig. 13 following the same format as the design comparisons presented in Secs. IV.A-IV.C. The validation points are included in the total number of points but are withheld from the FDS calculations. The prediction variance distribution is relatively uniform and has a low overall value for each design. The model precision is also adequate for fitting at least a full cubic model. Table 5 Final experimental design properties for each test region Cubic Model Quartic Model Design Order Points Groups FDS with 𝛿/𝜎≤2 FDS with 𝛿/𝜎≤2 Hover Region 36 N/A 1.000 0.999 Low/High Incidence Region 162 11 0.993 0.864 Descent Region 122 12 0.997 0.963 Fig. 13 Final design FDS plots for a cubic evaluation model in each test region. Figure 14 shows a four-dimensional representation of the observed freestream velocity, incidence angle, propeller rotational speed, and collective pitch angle for each experimental region with non-zero freestream velocity. Figure 15 shows the 𝐽𝑥and 𝐽𝑧coverage for each experimental region. Table 6 shows the observed ranges of several propeller test variables in each experimental region. Excluding experimental setup time, execution of all test points described in this paper took approximately 3.5 hours. This test strategy leveraging DOE/RSM experimental design techniques is significantly faster compared to a similar propeller characterization study conducted using a one-factor-at-a-time (OFAT) test matrix [16, 17]. Additionally, the randomized test point execution makes the designs more robust to time-varying systematic instrumentation errors and extraneous variables . These errors are transferred to the parameter variance as opposed to corrupting the parameter estimates. 13 Fig. 14 Observed propeller test variable values for the Low Incidence, High Incidence, and Descent Regions. Fig. 15 Observed normal and edgewise advance ratio values in each experimental region. Table 6 Observed ranges of propeller test variables Variable Hover Region Low Incidence Region High Incidence Region Descent Region 𝑛, rpm [2240, 5450] [2230, 5640] [2220, 5350] [2290, 5190] 𝛿𝑐, deg [−9.56, 7.23] [−0.387, 7.23] [−2.33, 4.9] [−2.3, 4.77] 𝐽 0 [0.11, 1.19] [0.103, 0.905] [0.123, 0.661] 𝐽𝑥 0 [0.0476, 0.697] [−0.157, 0.438] [−0.661, −0.123] 𝐽𝑧 0 [0, 1.08] [0.0974, 0.891] 0 𝑅𝑒 [98300, 239000] [96300, 246000] [95300, 231000] [98700, 224000] 𝑀tip [0.17, 0.415] [0.169, 0.429] [0.168, 0.405] [0.174, 0.393] 14 V. Propeller Aerodynamic Modeling Approach The main goal of this effort was to develop a propeller aerodynamic model relating the propeller states to output propulsive forces and moments. The response variables were the dimensionless propeller force and moment coefficients, 𝐶𝑇𝑥, 𝐶𝑇𝑦, 𝐶𝑇𝑧, 𝐶𝑄𝑥, 𝐶𝑄𝑦, and 𝐶𝑄𝑧, defined previously in Eqs. (1), (2) and (8). Response surface models developed from designed experiments conventionally evaluate the factors under test directly as candidates for explanatory variables. For this work, analysis was instead performed by redefining certain explanatory variables for modeling due to the unique characteristics of propeller aerodynamics at incidence and test facility integration considerations. As mentioned in Secs. III-IV, wind tunnel testing was performed by varying motor PWM command 𝜂𝑚and collective PWM command 𝜂𝑐directly; however, modeling was performed using the measured propeller rotational speed 𝑛and collective pitch angle 𝛿𝑐, which are more directly related to propeller aerodynamics. Three different explanatory variable definitions were considered and compared for this work: 1) Explanatory Variable Definition I (EV-I): Using ˆ 𝑉, ˆ 𝑖𝑝, ˆ 𝑛, and ˆ 𝛿𝑐as explanatory variables, which is the closest explanatory variable representation to the factors under test 2) Explanatory Variable Definition II (EV-II): Using ˆ 𝐽𝑥, ˆ 𝐽𝑧, and ˆ 𝛿𝑐as explanatory variables, which was hypothesized to better represent propeller aerodynamics when 𝑅𝑒and 𝑀tip effects are negligible 3) Explanatory Variable Definition III (EV-III): Using ˆ 𝐽𝑥, ˆ 𝐽𝑧, ˆ 𝑛, and ˆ 𝛿𝑐as explanatory variables, which was hypothesized to better represent propeller aerodynamics when 𝑅𝑒and/or 𝑀tip effects are significant. The hat (ˆ) notation is used to indicate that the variables are centered on the median value within the modeling data. It is important to perform modeling with explanatory variables expressed in coded units, or in engineering units with the explanatory variables centered on a reference value, to maintain low correlation among candidate regressors. Note that the variables not directly varied in the Hover and Descent Regions are omitted for their respective modeling analyses. The utility of each of these three parameterizations are compared in Sec. VII. Since the transformations between commanded test factors and the explanatory variables used for modeling were not linear, data collinearity assessment was performed to assess the impact on model identification. Data collinearity is defined as a correlation between regressors high enough to cause corrupted model identification . Data collinearity will cause difficulty in both model structure determination and parameter estimation because the effects of certain regressors on the response cannot be distinguished. Collinearity assessment is useful for confirming that a choice of modeling candidate regressors from a given experiment design or data set are sufficiently decorrelated for model identification. Correlation between two candidate regressors can be assessed using the pairwise correlation coefficient 𝑟𝑖𝑗= (𝝃𝑖−¯ 𝜉𝑖)𝑇(𝝃𝑗−¯ 𝜉𝑗) √︁ (𝝃𝑖−¯ 𝜉𝑖)𝑇(𝝃𝑖−¯ 𝜉𝑖) √︃ (𝝃𝑗−¯ 𝜉𝑗)𝑇(𝝃𝑗−¯ 𝜉𝑗) (16) where 𝝃𝑖and 𝝃𝑗are two regressor measurement histories, with means denoted ¯ 𝜉𝑖and ¯ 𝜉𝑗, respectively. A correlation coefficient value of zero means the signals are uncorrelated, or orthogonal, and an absolute correlation coefficient of one indicates that the signals are completely correlated. A correlation coefficient between regressors with magnitude greater than 0.9 indicates that data collinearity problems may be encountered . Figure 16 shows the pairwise correlation coefficient values between all candidate regressors sorted in ascending order within a full cubic model structure in the Low Incidence Region. The candidate regressors are assembled using the coded test factors as well as the three explanatory variable definitions centered on each respective median value. The highest correlation for each case is between the linear and cubic candidate regressors for each respective explanatory variable, which are challenging to decorrelate for any modeling problem. As would be expected, the overall lowest correlation is observed for the candidate regressors assembled from the test factors, or the designed test matrix. Low candidate regressor correlation, however, is still sufficiently maintained for a large majority of the candidate regressors for each of the explanatory variable definitions (EV-I, EV-II, and EV-III). This supports the validity of using the transformed explanatory variable definitions for model identification. Additionally, the model structure selection method used for this work, discussed next, automatically avoids adding highly correlated regressors into the model, thus, providing additional protection against data collinearity . 15 Fig. 16 Pairwise correlation between candidate regressors for a cubic model in the Low Incidence Region. VI. Model Identification Methodology As highlighted previously, most of the data used for model development in this paper were obtained from split-plot experiments [34, 35], or experiments with restriction to randomization for certain test factor(s). This results in a compound error structure with different whole-plot and sub-plot variance estimates. Generally, restricted maximum likelihood (REML) analysis is used to estimate the variance components and generalized least-squares is used for parameter estimation when developing models for split-plot experiments . However, because the split-plot experimental designs used for this work were ordinary least-squares equivalent, ordinary least-squares regression can be used for parameter estimation and provides the same solution as generalized least-squares regression, independent of the variance component estimates [22, 36–38]. Model structure determination, however, is still complicated by the compound error structure . Statistical software packages generally use REML analysis to estimate variance components and significance testing leveraging methods described in Ref. to refine the model structure. For this work, an alternative model structure determination approach was sought that does not rely on significance testing or the variance component estimation. An approach meeting these requirements allows more model development flexibility for when the explanatory variables used for modeling are transformed versions of the designed test factors, which was necessary for this study for the reasons explained in Sec. V. For example, for explanatory variable definitions EV-II and EV-III, the whole-plot and sub-plot factors are combined to calculate the advance ratio components which precludes using conventional model structure development techniques used for split-plot experiments. This approach represents an example of an engineering solution used to develop an improved, aerodynamically-informed model taking priority over statistical convention. Although it is not a statistically-based choice, using explanatory variable definitions different from the test factors was necessary to accommodate the practical considerations associated with this experimental study and yield the most useful mathematical model for the intended engineering application. Multivariate orthogonal function (MOF) modeling, described in Refs. [23, 46], meets the aforementioned model structure identification requirements—it does not rely on significance testing or variance component estimation to determine the model structure. MOF modeling instead selects regressors to include in the model based on their independent ability to improve characterization of the response variable. MOF modeling has also been successfully used in previous modeling studies using wind tunnel testing to characterize fixed-wing aircraft [47, 48] and propeller aerodynamics at incidence . Consequently, MOF modeling was selected as the model structure determination technique for this work. The MOF modeling approach [23, 46] is initiated by orthogonalizing a predefined set of candidate regressors using an algorithm such as Gram-Schmidt orthogonalization or QR decomposition. Orthogonal regressors are convenient for model structure development because of the ability to independently assess the potential of the orthogonalized candidate regressors to model the response variable—this facilitates selecting only the model terms that significantly contribute to model effectiveness. For model selection, the orthogonal regressors are ranked from highest to lowest decrease in the mean squared fit error (MSFE), MSFE = 1 𝑁(𝒛−ˆ 𝒚)𝑇(𝒛−ˆ 𝒚) (17) where ˆ 𝒚is the length 𝑁model response vector and 𝒛is the length 𝑁measured response vector. In other words, this ranks the regressors from highest to lowest ability to improve the model. Candidate regressors are brought into the model structure in this order. Note that this procedure prevents data collinearity because if there is high correlation between candidate regressors, after the first candidate regressor is orthogonalized, any other highly correlated candidate 16 regressors will be close to zero after passing through the orthogonalization process, which will prevent the latter model terms from being included in the model structure . Deciding which terms to include in the final model can be done using one or more statistical metrics. A common threshold for MOF modeling is to minimize the predicted squared error (PSE) [23, 49]. The PSE is the sum of the MSFE for a model and a complexity penalty term related to the number of terms included in the model PSE = MSFE + 2𝜎2 max 𝑛𝑝 𝑁 (18) where 𝑛𝑝is the number of terms in the current model structure and 𝜎2 max is an estimate of the upper-bound of mean squared error for the model prediction of data not used to develop the model. The model complexity penalty term contains a factor of 2 because the PSE metric is being employed for model structure identification, where inadequate model forms are being evaluated [47, 48]. In wind tunnel testing, the quantity 𝜎2 max can be estimated using the variance of measured response between repeated data points or from the variance between the measured response 𝒛and mean measured response ¯ 𝒛. The quantity 𝜎2 max is estimated for this work using the measurement error variance between repeated data points ˆ 𝜎2 as [23, 47, 48]: 𝜎2 max = 25 ˆ 𝜎2 (19) Another statistical metric that has been used as a stopping criterion for MOF modeling is the coefficient of determination 𝑅2 [50, 51]. The 𝑅2 metric, calculated as 𝑅2 = ˆ 𝒚𝑇𝒛−𝑁¯ 𝒛2 𝒛𝑇𝒛−𝑁¯ 𝒛2 (20) quantifies the amount of variation in the response variable about its mean value that is described by the model. Because 𝑅2 always increases with the addition of more model terms, it is important that each model term added on the basis of the 𝑅2 metric significantly increases its value. A common 𝑅2 increase constituting a significant increase with the addition of a new model term is Δ𝑅2 = 0.5% . This means that the model term describes a minimum of 0.5% of the total variation about the mean response. Both PSE and 𝑅2 were used as a cutoff threshold for candidate model terms to include in the final model structure. After the orthogonal regressors were ranked by their ability to reduce the MSFE, the cutoff for model term addition was chosen to be either the candidate model term that minimized the PSE or the last term to increase 𝑅2 by 0.5%, whichever admitted more terms into the model. This selection was made because in certain circumstances, PSE minimization was found to admit too few terms into the model due to having a rough estimated value of 𝜎2 max. After determining the model terms to include in the model structure, the final parameter values were estimated using ordinary least-squares regression in ordinary regressor space as: ˆ 𝜽= 𝑿𝑇𝑿 −1 𝑿𝑇𝒛 (21) Here, ˆ 𝜽is a vector of 𝑛𝑝estimated parameters, 𝑿is a 𝑁× 𝑛𝑝matrix consisting of column vectors of regressors, and 𝒛is the measured response variable. The software implementing the MOF modeling and least-squares regression algorithm was from the System IDentification Programs for AirCraft (SIDPAC) software toolbox.¶ After identifying the model structure and parameter estimates, model adequacy was examined using data withheld from the model development process. Regression methods minimize the summation of squared modeling residuals between modeled and measured response, so inspection of modeling fit metrics and modeling residuals alone does not provide information about the model predictive capability. Assessment of model performance using validation data not used for modeling provides a more reliable estimate of model prediction accuracy. Validation assessment can be performed by analyzing the residuals between the measured response 𝒛and predicted response ˆ 𝒚for the same explanatory variable inputs. Comparison of modeling and prediction residuals is useful because a significant increase in the spread of prediction residuals compared to modeling residuals is a way of diagnosing an improper model fit. For this work, modeling and prediction residuals are compared using the normalized root-mean-square modeling error (NRMSE) metric: NRMSE = 1 range(𝒛) √︄ (𝒛−ˆ 𝒚)𝑇(𝒛−ˆ 𝒚) 𝑁 (22) ¶Information available online at [retrieved 29 October 2021] 17 Range normalization, as opposed to other normalization metrics such as the mean or maximum absolute value of the response variable, provides the fairest comparison between prediction error metrics for this work because axial propeller responses are generally biased above or below zero and off-axis propeller responses are centered about zero . A prediction error metric defined using critical binomial analysis of validation residuals is also useful as a quantitative measure of the model adequacy. For this analysis, each validation data point is considered to either pass or fail relative to a prediction error threshold. Failed trials can indicate model inadequacy or measurement error. The binomial test provides a threshold to determine when the number of failures is statistically significant. For this metric, the prediction error level associated with the number of successful trials being equal to the critical binomial number quantifies the model prediction capability. The process of computing critical binomial analysis of residuals prediction error metric (𝑒∗ 𝑐𝑣) is shown in Ref. and further explanation of critical binomial analysis of residuals and justification for using this metric to assess prediction error are given in Ref. . VII. Modeling Results Separate propeller aerodynamic models were identified for each test region using each of the three explanatory variable definitions [EV-I ( ˆ 𝑉, ˆ 𝑖𝑝, ˆ 𝑛, ˆ 𝛿𝑐), EV-II ( ˆ 𝐽𝑥, ˆ 𝐽𝑧, ˆ 𝛿𝑐), and EV-III ( ˆ 𝐽𝑥, ˆ 𝐽𝑧, ˆ 𝑛, ˆ 𝛿𝑐)] postulated in Sec. V. To analyze the adherence to regression modeling assumptions and model fit adequacy, sample residual plots for 𝐶𝑇𝑥in the Low Incidence Region are shown in Fig. 17 for the EV-I model and Fig. 18 for the EV-III model. Figures 17a and 18a (a) Residual history (b) Normal probability plot (c) Residuals against predicted response Fig. 17 Externally studentized residual diagnostics for the EV-I 𝑪𝑻𝒙model in the Low Incidence Region. (a) Residual history (b) Normal probability plot (c) Residuals against predicted response Fig. 18 Externally studentized residual diagnostics for the EV-III 𝑪𝑻𝒙model in the Low Incidence Region. show the externally studentized modeling residuals, 𝑡𝑖, against data point number [22, 53]. Figures 17b and 18b show normal probability plots for the externally studentized modeling residuals. Figures 17c and 18c show the externally studentized modeling residuals against the predicted response. These plots show that the residuals for each model definition are reasonably independent, normally distributed, and have constant variance, satisfying the regression modeling assumptions and suggesting that the identified model parameters are meaningful. There is a slight trend observed in Figs. 17a and 18a, but this was deemed to be small enough as to not significantly impact the modeling 18 results—parameter estimation results are robust to time-varying instrumentation errors due to the randomization in the experimental design. Residuals with similar character were obtained for the other response variables, test regions, and explanatory variable definitions. Figures 19-23 compare modeling and prediction error metrics for each modeling region and explanatory variable definition. Tables 7-11 at the end of this paper list the numerical values shown in the figures. These metrics are used to assess the quality of the modeling results and compare the utility of the different explanatory variable combinations. Fig. 19 Coefficient of determination, 𝑹2, for each local model. Fig. 20 Number of model parameters identified for each local model. Fig. 21 Modeling data NRMSE for each local model response. Figure 19 (Table 7) shows the coefficient of determination (𝑅2) value for each local model. The thrust coefficient 𝐶𝑇𝑥models have an 𝑅2 value above 98% indicating that a large majority of the variation in the response is described by the models. The 𝐶𝑇𝑥models for EV-III generally have the highest 𝑅2 values, although the difference in value is small compared to the EV-I and EV-II models. For the other propeller force and moment coefficients, it is observed that EV-II most often has the lowest 𝑅2 value compared to EV-I and EV-III. This suggests that Reynolds number and/or tip Mach 19 Fig. 22 Validation data NRMSE for each local model response. Fig. 23 Binomial analysis of residuals prediction error metric, 𝒆∗ 𝒄𝒗, for each local model. number effects become significant for characterizing the variation of the torque coefficient 𝐶𝑄𝑥and the off-axis force and moment coefficients. The 𝐶𝑇𝑧and 𝐶𝑄𝑥models for EV-I and EV-III have 𝑅2 values above 97%. The 𝐶𝑄𝑦and 𝐶𝑄𝑧 models for EV-I and EV-III have 𝑅2 values above 82%, with models in the High Incidence Region having 𝑅2 values above 94%. The models for 𝐶𝑇𝑦have the lowest 𝑅2 values, but this is expected since 𝐶𝑇𝑦is the weakest response for propellers at incidence . Figure 20 (Table 8) shows the number of model terms identified in each local model. Generally, the models corresponding to EV-I have the largest number of model parameters for the stronger response variables compared to respective EV-II and EV-III models. The models for EV-II are seen to generally have the fewest number of model terms, but as explained above, they omit variables that characterize propeller blade Reynolds number and tip Mach number effects leading to a worse model fit. This suggests that propeller blade Reynolds number effects or tip Mach number effects are important to consider for the propeller used in this study. Figures 21-22 (Tables 9-10) show the NRMSE for modeling data (NRMSE𝑚) and validation data (NRMSE𝑣), respectively. Although 𝑅2 and NRMSE𝑚reflect the model fit quality, model prediction metrics calculated using validation data not used in the modeling process are generally considered a more reliable indicator of modeling success. The NRMSE values calculated using modeling and validation data for each respective local model are similar and low-valued indicating that a quality model has been identified. The binomial analysis of residuals prediction error metric (𝑒∗ 𝑐𝑣) value for each model is shown in Figure 23 (Table 11). In the Hover, Low Incidence, and High Incidence regions, the prediction error for thrust coefficient is less than 3%; in the Descent region prediction error for thrust coefficient is around 5% or less. The models for 𝐶𝑇𝑧and 𝐶𝑄𝑥in all respective regions, as well as 𝐶𝑄𝑦and 𝐶𝑄𝑧in the High Incidence region, have prediction error values of approximately 10% or less. This indicates that high-quality propeller aerodynamic models have been developed. The 𝐶𝑇𝑦models and the 𝐶𝑄𝑦, 𝐶𝑄𝑧models in the Low Incidence region have prediction error values of roughly 20% or less, which are reasonable values because these models characterize relatively weak responses with a lower signal-to-noise ratio. After investigating the three different explanatory variable definitions, one definition needed to be selected for the final model. Since the EV-II models were suspected to lack the parameterization to best describe the aerodynamics for the propeller described in this work (i.e., EV-II models cannot characterize propeller blade Reynolds number or 20 tip Mach number effects), EV-II was not selected as the final model explanatory variable type. Comparing EV-I and EV-III in general, and especially for stronger propeller force and moment coefficients, using EV-III ( ˆ 𝐽𝑥, ˆ 𝐽𝑧, ˆ 𝑛, and ˆ 𝛿𝑐as explanatory variables) results in a better modeling fit and lower prediction error, while also having a lower number of modeling terms. Additionally, using dimensionless propeller variables ˆ 𝐽𝑥, ˆ 𝐽𝑧, as opposed to 𝑉and 𝑖𝑝extends the extrapolation capabilities of the model. For these reasons, the models developed with ˆ 𝐽𝑥, ˆ 𝐽𝑧, ˆ 𝑛, and ˆ 𝛿𝑐as explanatory variables (EV-III) were selected as the final models. The EV-III model parameter estimates ˆ 𝜃and parameter standard errors 𝑠( ˆ 𝜃) for each propeller force and moment coefficient in each test region are shown in Tables 12-27. The model parameters are presented in the order in which they were added to the model, so terms appearing first are most significant to the model. As an example of how the polynomial models would appear for usage, the 𝐶𝑇𝑥model in the Low Incidence Region in Table 14 is expressed in polynomial equation form as: 𝐶𝑇𝑥= 𝐶𝑇𝑥𝑜+ 𝐶𝑇𝑥ˆ 𝐽𝑥 ˆ 𝐽𝑥+ 𝐶𝑇𝑥ˆ 𝛿𝑐 ˆ 𝛿𝑐+ 𝐶𝑇𝑥ˆ 𝐽2 𝑥 ˆ 𝐽2 𝑥+ 𝐶𝑇𝑥ˆ 𝐽𝑧 ˆ 𝐽𝑧+ 𝐶𝑇𝑥ˆ 𝐽𝑥ˆ 𝛿𝑐 ˆ 𝐽𝑥ˆ 𝛿𝑐 Recall that the explanatory variables are all centered on a reference value in the model equations (see Sec. V). The median values used to center each explanatory variable are given in Table 28. As an example for 𝐽𝑥, the centered explanatory variable is ˆ 𝐽𝑥= 𝐽𝑥−𝐽𝑥𝑜 where the value of 𝐽𝑥𝑜, as well as the respective reference values for the other explanatory variables, are listed in Table 28. Figures 24-26 show the modeled response for axial thrust coefficient 𝐶𝑇𝑥in each test region compared to modeling and validation wind tunnel data. For the Hover Region (Fig. 24), the 𝐶𝑇𝑥model and data are plotted against rotational Fig. 24 Hover Region 𝑪𝑻𝒙response surface model compared to measured data. Fig. 25 Low and High Incidence Region 𝑪𝑻𝒙models compared to measured data. speed 𝑛and collective pitch angle 𝛿𝑐, which were the only two explanatory variables varied in the experiment. For the Low and High Incidence Regions (Fig. 25), the 𝐶𝑇𝑥model is plotted against normal advance ratio 𝐽𝑥and edgewise advance ratio 𝐽𝑧at a collective pitch angle of 𝛿𝑐= 4 deg and a propeller rotational speed of 𝑛= 3500 rpm. The modeling and validation test data shown in the figure have a collective pitch angle between 3 deg and 5 deg (i.e., near the nominal value used to display the modeled response). All tested propeller rotational speed values are shown due to the relatively weak influence of propeller speed on the 𝐶𝑇𝑥response compared to the other explanatory variables. As can be seen in Table 14 and Table 20, the 𝐶𝑇𝑥model in the Low Incidence Region is not dependent on 𝑛and the 𝐶𝑇𝑥model in the High Incidence Region only has one relatively weak model term dependent on 𝑛. The 𝑛effects, attributed to propeller blade Reynolds number and/or tip Mach effects, were more influential in other propeller force and moment coefficients. Small differences seen between the model and data are partially attributed to the collective pitch angle and propeller rotational speed settings for the displayed test data not perfectly matching the displayed modeled response surface value. For the Descent Region (Fig. 25), the 𝐶𝑇𝑥model and data are plotted against normal advance ratio 𝐽𝑥and collective pitch angle 21 Fig. 26 Descent Region 𝑪𝑻𝒙response surface model compared to measured data. Fig. 27 Modeled 𝑪𝑻𝒙response in the Descent Region at different collective pitch angle settings. of 𝛿𝑐. As can be seen in Table 26, the 𝐶𝑇𝑥model in the Descent Region is independent of direct 𝑛effects (propeller blade Reynolds number and/or tip Mach effects). The modeled response in each region shows overall good agreement with the modeling and validation data points indicating that the models are suitable for their intended application of flight dynamics simulation development. Note that models identified over different ranges of certain variables require blending methods to eliminate discontinuities for simulation, where it is desirable to have continuous, differentiable transition between modeling regions. Reference presents an approach to smoothly blend propeller models for this purpose, in addition to other practical aspects of using the identified propeller models in a flight dynamics simulation. Figure 27 shows the modeled 𝐶𝑇𝑥response in the Descent Region against normal advance ratio at several different collective pitch angle settings shown in the legend. The minimum 𝐶𝑇𝑥for each collective pitch angle on the plot is the approximate location of the VRS at each collective pitch angle setting. It can be seen that the VRS location occurs at a more negative 𝐽𝑥value (greater total advance ratio 𝐽) as the collective pitch angle increases. VIII. Conclusions A method for variable-pitch propeller aerodynamic model development suitable for use in flight dynamics simulations for vectored-thrust eVTOL aircraft has been presented. Wind tunnel data were collected using designed experiments allowing identification of accurate cubic response surface propeller aerodynamic models in four regions of flight: hover, high-incidence/low-speed transition, low-incidence/high-speed transition, and descent. Application of I-optimal completely randomized and split-plot experiment designs enabled efficient collection of high-quality wind tunnel data for model identification. The propeller model identification approach leveraged multivariate orthogonal function modeling to identity local polynomial models, which was a suitable and effective choice for the selected experimental designs and modeling problem definitions. Three different explanatory variable formulations were compared and the results indicated that the variables derived from propeller aerodynamics theory provided the best models. The final models describe the variation of the dimensionless force and moment coefficients in each test region as a function of normal advance ratio, edgewise advance ratio, propeller rotational speed, and collective pitch angle. Collectively, the models identified in the four different test regions describe propeller aerodynamics over a wide range of flight conditions seen in operational flight for eVTOL vehicles. Model validation assessment indicated that the models are high quality and sufficient for the purpose of supporting flight dynamics model development for a future eVTOL aircraft. Because eVTOL vehicle dynamics are highly dependent on propulsive effects, accurate propeller aerodynamic modeling is essential for high-fidelity simulator development. This paper demonstrated several novel propeller modeling techniques useful for modeling future eVTOL vehicles and provides progress in this new area of aerodynamic modeling research. 22 Acknowledgments This research was funded by the NASA Aeronautics Research Mission Directorate (ARMD) Transformational Tools and Technologies (TTT) project. Wind tunnel testing support was provided by Gregory Howland, Ronald Busan, Wes O’Neal, George Altamirano, Rose Weinstein, and Clinton Duncan. Wes O’Neal and David Hatke added a new capability to the 12-Foot Low-Speed Tunnel software allowing dynamic pressure to be changed in an automated data collection run. The automatic propeller variable-pitch mechanism was developed by Gregory Howland. Photography support was provided by Lee Pollard. Additional team members contributing to the wind tunnel testing effort included Earl Harris, Stephen Riddick, Sue Grafton, Jacob Cook, David North, Brian Duvall, Matthew Gray, Steven Geuther, Jason Welstead, and Siena Whiteside. Modeling and Validation Metric Tables Table 7 Coefficient of determination, 𝑹2, for each local model (expressed as a percentage) Region Explanatory Variables 𝐶𝑇𝑥 𝐶𝑇𝑦 𝐶𝑇𝑧 𝐶𝑄𝑥 𝐶𝑄𝑦 𝐶𝑄𝑧 Hover Region EV-I: ˆ 𝑛, ˆ 𝛿𝑐 99.97 — — 99.91 — — Hover Region EV-II: ˆ 𝛿𝑐 99.97 — — 99.80 — — Hover Region EV-III: ˆ 𝑛, ˆ 𝛿𝑐 99.97 — — 99.91 — — Low Incidence Region EV-I: ˆ 𝑉, ˆ 𝑖𝑝, ˆ 𝑛, ˆ 𝛿𝑐 99.21 84.21 98.52 97.40 82.93 86.08 Low Incidence Region EV-II: ˆ 𝐽𝑥, ˆ 𝐽𝑧, ˆ 𝛿𝑐 99.40 75.95 97.02 92.78 60.57 71.68 Low Incidence Region EV-III: ˆ 𝐽𝑥, ˆ 𝐽𝑧, ˆ 𝑛, ˆ 𝛿𝑐 99.40 86.13 98.27 98.12 84.92 83.26 High Incidence Region EV-I: ˆ 𝑉, ˆ 𝑖𝑝, ˆ 𝑛, ˆ 𝛿𝑐 99.75 73.55 97.65 97.76 94.62 95.88 High Incidence Region EV-II: ˆ 𝐽𝑥, ˆ 𝐽𝑧, ˆ 𝛿𝑐 99.84 79.10 97.76 95.21 86.82 93.30 High Incidence Region EV-III: ˆ 𝐽𝑥, ˆ 𝐽𝑧, ˆ 𝑛, ˆ 𝛿𝑐 99.87 82.68 97.76 98.09 95.70 96.39 Descent Region EV-I: ˆ 𝑉, ˆ 𝑛, ˆ 𝛿𝑐 98.41 — — 98.54 — — Descent Region EV-II: ˆ 𝐽𝑥, ˆ 𝛿𝑐 98.43 — — 98.18 — — Descent Region EV-III: ˆ 𝐽𝑥, ˆ 𝑛, ˆ 𝛿𝑐 98.43 — — 98.18 — — Table 8 Number of model parameters identified for each local model Region Explanatory Variables 𝐶𝑇𝑥 𝐶𝑇𝑦 𝐶𝑇𝑧 𝐶𝑄𝑥 𝐶𝑄𝑦 𝐶𝑄𝑧 Hover Region EV-I: ˆ 𝑛, ˆ 𝛿𝑐 4 — — 4 — — Hover Region EV-II: ˆ 𝛿𝑐 4 — — 3 — — Hover Region EV-III: ˆ 𝑛, ˆ 𝛿𝑐 4 — — 4 — — Low Incidence Region EV-I: ˆ 𝑉, ˆ 𝑖𝑝, ˆ 𝑛, ˆ 𝛿𝑐 15 11 20 11 17 14 Low Incidence Region EV-II: ˆ 𝐽𝑥, ˆ 𝐽𝑧, ˆ 𝛿𝑐 6 13 8 6 7 7 Low Incidence Region EV-III: ˆ 𝐽𝑥, ˆ 𝐽𝑧, ˆ 𝑛, ˆ 𝛿𝑐 6 15 10 7 18 9 High Incidence Region EV-I: ˆ 𝑉, ˆ 𝑖𝑝, ˆ 𝑛, ˆ 𝛿𝑐 15 17 13 16 9 15 High Incidence Region EV-II: ˆ 𝐽𝑥, ˆ 𝐽𝑧, ˆ 𝛿𝑐 11 16 6 10 9 6 High Incidence Region EV-III: ˆ 𝐽𝑥, ˆ 𝐽𝑧, ˆ 𝑛, ˆ 𝛿𝑐 13 20 6 11 10 9 Descent Region EV-I: ˆ 𝑉, ˆ 𝑛, ˆ 𝛿𝑐 9 — — 8 — — Descent Region EV-II: ˆ 𝐽𝑥, ˆ 𝛿𝑐 6 — — 4 — — Descent Region EV-III: ˆ 𝐽𝑥, ˆ 𝑛, ˆ 𝛿𝑐 6 — — 4 — — 23 Table 9 Modeling data NRMSE for each local model response (expressed as a percentage) Region Explanatory Variables 𝐶𝑇𝑥 𝐶𝑇𝑦 𝐶𝑇𝑧 𝐶𝑄𝑥 𝐶𝑄𝑦 𝐶𝑄𝑧 Hover Region EV-I: ˆ 𝑛, ˆ 𝛿𝑐 0.64 — — 1.02 — — Hover Region EV-II: ˆ 𝛿𝑐 0.64 — — 1.52 — — Hover Region EV-III: ˆ 𝑛, ˆ 𝛿𝑐 0.64 — — 1.02 — — Low Incidence Region EV-I: ˆ 𝑉, ˆ 𝑖𝑝, ˆ 𝑛, ˆ 𝛿𝑐 2.24 7.82 1.88 3.77 7.43 4.93 Low Incidence Region EV-II: ˆ 𝐽𝑥, ˆ 𝐽𝑧, ˆ 𝛿𝑐 1.97 9.66 2.67 6.29 11.30 7.03 Low Incidence Region EV-III: ˆ 𝐽𝑥, ˆ 𝐽𝑧, ˆ 𝑛, ˆ 𝛿𝑐 1.97 7.33 2.03 3.21 6.99 5.41 High Incidence Region EV-I: ˆ 𝑉, ˆ 𝑖𝑝, ˆ 𝑛, ˆ 𝛿𝑐 0.81 9.01 2.99 3.23 5.02 2.98 High Incidence Region EV-II: ˆ 𝐽𝑥, ˆ 𝐽𝑧, ˆ 𝛿𝑐 0.64 8.01 2.92 4.73 7.86 3.80 High Incidence Region EV-III: ˆ 𝐽𝑥, ˆ 𝐽𝑧, ˆ 𝑛, ˆ 𝛿𝑐 0.58 7.29 2.92 2.99 4.49 2.79 Descent Region EV-I: ˆ 𝑉, ˆ 𝑛, ˆ 𝛿𝑐 2.59 — — 3.40 — — Descent Region EV-II: ˆ 𝐽𝑥, ˆ 𝛿𝑐 2.57 — — 3.79 — — Descent Region EV-III: ˆ 𝐽𝑥, ˆ 𝑛, ˆ 𝛿𝑐 2.57 — — 3.79 — — Table 10 Validation data NRMSE for each local model response (expressed as a percentage) Region Explanatory Variables 𝐶𝑇𝑥 𝐶𝑇𝑦 𝐶𝑇𝑧 𝐶𝑄𝑥 𝐶𝑄𝑦 𝐶𝑄𝑧 Hover Region EV-I: ˆ 𝑛, ˆ 𝛿𝑐 1.02 — — 0.82 — — Hover Region EV-II: ˆ 𝛿𝑐 1.02 — — 1.67 — — Hover Region EV-III: ˆ 𝑛, ˆ 𝛿𝑐 1.02 — — 0.82 — — Low Incidence Region EV-I: ˆ 𝑉, ˆ 𝑖𝑝, ˆ 𝑛, ˆ 𝛿𝑐 2.45 12.61 4.55 6.46 12.11 13.06 Low Incidence Region EV-II: ˆ 𝐽𝑥, ˆ 𝐽𝑧, ˆ 𝛿𝑐 2.10 11.91 4.94 7.03 15.79 15.11 Low Incidence Region EV-III: ˆ 𝐽𝑥, ˆ 𝐽𝑧, ˆ 𝑛, ˆ 𝛿𝑐 2.10 11.93 4.03 4.85 12.40 14.33 High Incidence Region EV-I: ˆ 𝑉, ˆ 𝑖𝑝, ˆ 𝑛, ˆ 𝛿𝑐 2.20 16.95 3.40 5.70 8.09 6.78 High Incidence Region EV-II: ˆ 𝐽𝑥, ˆ 𝐽𝑧, ˆ 𝛿𝑐 1.83 13.33 3.09 6.23 7.94 7.14 High Incidence Region EV-III: ˆ 𝐽𝑥, ˆ 𝐽𝑧, ˆ 𝑛, ˆ 𝛿𝑐 1.68 17.46 3.09 6.02 6.37 5.31 Descent Region EV-I: ˆ 𝑉, ˆ 𝑛, ˆ 𝛿𝑐 3.73 — — 3.33 — — Descent Region EV-II: ˆ 𝐽𝑥, ˆ 𝛿𝑐 3.53 — — 3.47 — — Descent Region EV-III: ˆ 𝐽𝑥, ˆ 𝑛, ˆ 𝛿𝑐 3.53 — — 3.47 — — 24 Table 11 Binomial analysis of residuals prediction error metric, 𝒆∗ 𝒄𝒗, for each local model (expressed as a percentage) Region Explanatory Variables 𝐶𝑇𝑥 𝐶𝑇𝑦 𝐶𝑇𝑧 𝐶𝑄𝑥 𝐶𝑄𝑦 𝐶𝑄𝑧 Hover Region EV-I: ˆ 𝑛, ˆ 𝛿𝑐 0.99 — — 0.25 — — Hover Region EV-II: ˆ 𝛿𝑐 0.99 — — 1.59 — — Hover Region EV-III: ˆ 𝑛, ˆ 𝛿𝑐 0.99 — — 0.25 — — Low Incidence Region EV-I: ˆ 𝑉, ˆ 𝑖𝑝, ˆ 𝑛, ˆ 𝛿𝑐 2.98 13.99 5.77 6.37 11.51 18.66 Low Incidence Region EV-II: ˆ 𝐽𝑥, ˆ 𝐽𝑧, ˆ 𝛿𝑐 2.64 13.24 5.85 10.04 15.44 17.89 Low Incidence Region EV-III: ˆ 𝐽𝑥, ˆ 𝐽𝑧, ˆ 𝑛, ˆ 𝛿𝑐 2.64 15.36 4.83 6.45 13.62 18.73 High Incidence Region EV-I: ˆ 𝑉, ˆ 𝑖𝑝, ˆ 𝑛, ˆ 𝛿𝑐 2.15 23.87 4.03 9.06 9.68 7.79 High Incidence Region EV-II: ˆ 𝐽𝑥, ˆ 𝐽𝑧, ˆ 𝛿𝑐 1.96 15.15 3.23 8.00 11.14 10.38 High Incidence Region EV-III: ˆ 𝐽𝑥, ˆ 𝐽𝑧, ˆ 𝑛, ˆ 𝛿𝑐 1.79 19.51 3.23 6.96 9.21 6.62 Descent Region EV-I: ˆ 𝑉, ˆ 𝑛, ˆ 𝛿𝑐 4.34 — — 3.87 — — Descent Region EV-II: ˆ 𝐽𝑥, ˆ 𝛿𝑐 5.02 — — 3.51 — — Descent Region EV-III: ˆ 𝐽𝑥, ˆ 𝑛, ˆ 𝛿𝑐 5.02 — — 3.51 — — EV-III Parameter Estimate Tables Table 12 Parameter estimates for 𝑪𝑻𝒙in the Hover Region (𝑹2 = 99.97%) Parameter ˆ 𝜃 𝑠( ˆ 𝜃) 𝐶𝑇𝑥𝑜 +7.054E−02 2.643E−04 𝐶𝑇𝑥ˆ 𝛿𝑐 +4.877E−01 5.187E−03 𝐶𝑇𝑥ˆ 𝛿2 𝑐 +2.844E−01 2.273E−02 𝐶𝑇𝑥ˆ 𝛿3 𝑐 −1.696E+00 2.918E−01 Table 13 Parameter estimates for 𝑪𝑸𝒙in the Hover Region (𝑹2 = 99.91%) Parameter ˆ 𝜃 𝑠( ˆ 𝜃) 𝐶𝑄𝑥𝑜 −3.706E−03 2.478E−05 𝐶𝑄𝑥ˆ 𝛿𝑐 −2.757E−02 1.761E−04 𝐶𝑄𝑥ˆ 𝛿2 𝑐 −8.434E−02 1.872E−03 𝐶𝑄𝑥ˆ 𝑛 +6.931E−06 1.232E−06 Table 14 Parameter estimates for 𝑪𝑻𝒙in the Low Inci-dence Region (𝑹2 = 99.40%) Parameter ˆ 𝜃 𝑠( ˆ 𝜃) 𝐶𝑇𝑥𝑜 +5.102E−02 4.433E−04 𝐶𝑇𝑥ˆ 𝐽𝑥 −2.317E−01 1.832E−03 𝐶𝑇𝑥ˆ 𝛿𝑐 +5.686E−01 6.237E−03 𝐶𝑇𝑥ˆ 𝐽2 𝑥 −2.301E−01 1.242E−02 𝐶𝑇𝑥ˆ 𝐽𝑧 +1.927E−02 1.414E−03 𝐶𝑇𝑥ˆ 𝐽𝑥ˆ 𝛿𝑐 +3.828E−01 4.092E−02 25 Table 15 Parameter estimates for 𝑪𝑻𝒚in the Low Inci-dence Region (𝑹2 = 86.13%) Parameter ˆ 𝜃 𝑠( ˆ 𝜃) 𝐶𝑇𝑦𝑜 −1.894E−03 1.572E−04 𝐶𝑇𝑦ˆ 𝐽𝑧 −8.849E−03 6.870E−04 𝐶𝑇𝑦ˆ 𝐽𝑥 +8.063E−03 5.899E−04 𝐶𝑇𝑦ˆ 𝑛 +8.497E−05 1.053E−05 𝐶𝑇𝑦ˆ 𝐽𝑧ˆ 𝛿𝑐 −4.461E−02 8.483E−03 𝐶𝑇𝑦ˆ 𝐽𝑥ˆ 𝐽𝑧 +2.554E−02 4.590E−03 𝐶𝑇𝑦ˆ 𝐽𝑥ˆ 𝐽2 𝑧 −6.945E−02 1.630E−02 𝐶𝑇𝑦ˆ 𝛿3 𝑐 −1.378E+00 4.614E−01 𝐶𝑇𝑦ˆ 𝛿2 𝑐ˆ 𝑛 −6.985E−03 3.081E−03 𝐶𝑇𝑦ˆ 𝐽2 𝑥ˆ 𝐽𝑧 +1.253E−01 2.599E−02 𝐶𝑇𝑦ˆ 𝐽2 𝑧ˆ 𝑛 −3.445E−04 9.993E−05 𝐶𝑇𝑦ˆ 𝑛2 −1.376E−06 4.495E−07 𝐶𝑇𝑦ˆ 𝐽𝑥ˆ 𝐽𝑧ˆ 𝑛 +7.399E−04 3.290E−04 𝐶𝑇𝑦ˆ 𝐽𝑥ˆ 𝛿𝑐 +3.368E−02 1.154E−02 𝐶𝑇𝑦ˆ 𝐽2 𝑥 −1.032E−02 3.554E−03 Table 16 Parameter estimates for 𝑪𝑻𝒛in the Low Inci-dence Region (𝑹2 = 98.27%) Parameter ˆ 𝜃 𝑠( ˆ 𝜃) 𝐶𝑇𝑧ˆ 𝐽𝑧 −3.929E−02 1.094E−03 𝐶𝑇𝑧𝑜 −6.643E−03 2.001E−04 𝐶𝑇𝑧ˆ 𝐽𝑧ˆ 𝛿𝑐 −1.883E−01 2.088E−02 𝐶𝑇𝑧ˆ 𝐽𝑥 +1.137E−02 9.038E−04 𝐶𝑇𝑧ˆ 𝛿𝑐 −2.922E−02 3.387E−03 𝐶𝑇𝑧ˆ 𝑛 +7.735E−05 1.025E−05 𝐶𝑇𝑧ˆ 𝐽2 𝑧 −1.551E−02 3.114E−03 𝐶𝑇𝑧ˆ 𝑛2 −2.945E−06 6.773E−07 𝐶𝑇𝑧ˆ 𝐽𝑥ˆ 𝐽2 𝑧 −9.038E−02 1.668E−02 𝐶𝑇𝑧ˆ 𝐽2 𝑧ˆ 𝛿𝑐 +1.919E−01 4.802E−02 Table 17 Parameter estimates for 𝑪𝑸𝒙in the Low In-cidence Region (𝑹2 = 98.12%) Parameter ˆ 𝜃 𝑠( ˆ 𝜃) 𝐶𝑄𝑥𝑜 −5.958E−03 4.606E−05 𝐶𝑄𝑥ˆ 𝛿𝑐 −3.964E−02 6.993E−04 𝐶𝑄𝑥ˆ 𝐽𝑥 +1.064E−02 2.020E−04 𝐶𝑄𝑥ˆ 𝑛 +4.634E−05 2.593E−06 𝐶𝑄𝑥ˆ 𝐽2 𝑥 +2.048E−02 1.218E−03 𝐶𝑄𝑥ˆ 𝛿𝑐ˆ 𝑛 +3.002E−04 4.551E−05 𝐶𝑄𝑥ˆ 𝐽2 𝑧ˆ 𝑛 +9.493E−05 1.538E−05 Table 18 Parameter estimates for 𝑪𝑸𝒚in the Low In-cidence Region (𝑹2 = 84.92%) Parameter ˆ 𝜃 𝑠( ˆ 𝜃) 𝐶𝑄𝑦ˆ 𝐽𝑥 −2.664E−03 1.221E−03 𝐶𝑄𝑦ˆ 𝐽𝑧ˆ 𝑛 +9.627E−05 3.513E−05 𝐶𝑄𝑦ˆ 𝑛 +5.446E−05 7.954E−06 𝐶𝑄𝑦𝑜 −7.585E−04 1.221E−04 𝐶𝑄𝑦ˆ 𝛿𝑐 +6.799E−03 2.169E−03 𝐶𝑄𝑦ˆ 𝐽𝑥ˆ 𝐽𝑧 −1.915E−02 3.680E−03 𝐶𝑄𝑦ˆ 𝐽2 𝑥 +1.586E−02 3.095E−03 𝐶𝑄𝑦ˆ 𝐽𝑥ˆ 𝑛 +5.184E−05 3.617E−05 𝐶𝑄𝑦ˆ 𝐽3 𝑧 +7.998E−03 2.447E−03 𝐶𝑄𝑦ˆ 𝐽𝑧ˆ 𝑛2 −7.827E−06 1.701E−06 𝐶𝑄𝑦ˆ 𝑛2 −1.540E−06 3.586E−07 𝐶𝑄𝑦ˆ 𝐽2 𝑧ˆ 𝛿𝑐 +5.352E−02 1.550E−02 𝐶𝑄𝑦ˆ 𝐽2 𝑥ˆ 𝑛 −9.327E−04 2.494E−04 𝐶𝑄𝑦ˆ 𝛿𝑐ˆ 𝑛2 −1.389E−05 7.449E−06 𝐶𝑄𝑦ˆ 𝐽𝑥ˆ 𝑛2 −8.027E−06 2.750E−06 𝐶𝑄𝑦ˆ 𝐽𝑥ˆ 𝐽𝑧ˆ 𝑛 −8.403E−04 2.783E−04 𝐶𝑄𝑦ˆ 𝐽3 𝑥 −4.838E−02 1.733E−02 𝐶𝑄𝑦ˆ 𝐽𝑥ˆ 𝐽2 𝑧 −3.084E−02 1.477E−02 26 Table 19 Parameter estimates for 𝑪𝑸𝒛in the Low In-cidence Region (𝑹2 = 83.26%) Parameter ˆ 𝜃 𝑠( ˆ 𝜃) 𝐶𝑄𝑧ˆ 𝐽𝑧 −1.266E−02 1.142E−03 𝐶𝑄𝑧ˆ 𝐽𝑥ˆ 𝑛2 −9.130E−06 4.602E−06 𝐶𝑄𝑧ˆ 𝑛 −9.283E−05 1.064E−05 𝐶𝑄𝑧ˆ 𝑛2 +2.317E−06 5.943E−07 𝐶𝑄𝑧ˆ 𝐽2 𝑧ˆ 𝛿𝑐 −1.583E−01 2.820E−02 𝐶𝑄𝑧ˆ 𝐽3 𝑥 −8.719E−02 1.859E−02 𝐶𝑄𝑧ˆ 𝐽2 𝑥 +1.611E−02 4.730E−03 𝐶𝑄𝑧ˆ 𝐽2 𝑧 −8.568E−03 2.738E−03 𝐶𝑄𝑧ˆ 𝐽𝑥ˆ 𝐽𝑧ˆ 𝑛 −1.241E−03 4.331E−04 Table 20 Parameter estimates for 𝑪𝑻𝒙in the High Incidence Region (𝑹2 = 99.87%) Parameter ˆ 𝜃 𝑠( ˆ 𝜃) 𝐶𝑇𝑥𝑜 +8.638E−02 2.227E−04 𝐶𝑇𝑥ˆ 𝐽𝑥 −1.667E−01 1.936E−03 𝐶𝑇𝑥ˆ 𝛿𝑐 +4.844E−01 2.783E−03 𝐶𝑇𝑥ˆ 𝐽𝑧 +6.300E−02 1.394E−03 𝐶𝑇𝑥ˆ 𝐽𝑥ˆ 𝐽𝑧 −3.751E−01 1.330E−02 𝐶𝑇𝑥ˆ 𝐽2 𝑧 +1.544E−01 6.887E−03 𝐶𝑇𝑥ˆ 𝐽𝑥ˆ 𝐽2 𝑧 +3.234E−01 3.279E−02 𝐶𝑇𝑥ˆ 𝐽𝑥ˆ 𝛿𝑐 +3.205E−01 2.174E−02 𝐶𝑇𝑥ˆ 𝐽3 𝑧 −2.401E−01 1.832E−02 𝐶𝑇𝑥ˆ 𝐽3 𝑥 −4.661E−01 5.525E−02 𝐶𝑇𝑥ˆ 𝐽2 𝑥ˆ 𝐽𝑧 +2.554E−01 5.030E−02 𝐶𝑇𝑥ˆ 𝑛2 −2.828E−06 6.887E−07 𝐶𝑇𝑥ˆ 𝐽𝑧ˆ 𝛿𝑐 +6.817E−02 1.876E−02 Table 21 Parameter estimates for 𝑪𝑻𝒚in the High Incidence Region (𝑹2 = 82.68%) Parameter ˆ 𝜃 𝑠( ˆ 𝜃) 𝐶𝑇𝑦𝑜 −3.768E−03 9.358E−05 𝐶𝑇𝑦ˆ 𝛿𝑐 −2.000E−02 1.746E−03 𝐶𝑇𝑦ˆ 𝐽𝑥ˆ 𝐽𝑧ˆ 𝛿𝑐 −3.965E−01 6.011E−02 𝐶𝑇𝑦ˆ 𝐽𝑧 −1.692E−03 5.718E−04 𝐶𝑇𝑦ˆ 𝐽𝑧ˆ 𝑛 −6.493E−06 3.799E−05 𝐶𝑇𝑦ˆ 𝐽𝑧ˆ 𝛿𝑐ˆ 𝑛 −9.258E−04 7.534E−04 𝐶𝑇𝑦ˆ 𝐽3 𝑧 −6.420E−02 7.830E−03 𝐶𝑇𝑦ˆ 𝐽2 𝑧 +2.734E−02 3.522E−03 𝐶𝑇𝑦ˆ 𝐽2 𝑧ˆ 𝛿𝑐 +2.527E−01 5.708E−02 𝐶𝑇𝑦ˆ 𝐽2 𝑥 +1.803E−02 3.241E−03 𝐶𝑇𝑦ˆ 𝐽𝑥ˆ 𝐽𝑧 −3.029E−02 4.665E−03 𝐶𝑇𝑦ˆ 𝐽𝑥ˆ 𝐽2 𝑧 +1.321E−01 2.068E−02 𝐶𝑇𝑦ˆ 𝐽𝑥ˆ 𝐽𝑧ˆ 𝑛 +1.166E−03 3.400E−04 𝐶𝑇𝑦ˆ 𝐽𝑧ˆ 𝛿𝑐 −3.342E−02 8.854E−03 𝐶𝑇𝑦ˆ 𝑛 +2.014E−05 5.943E−06 𝐶𝑇𝑦ˆ 𝐽2 𝑥ˆ 𝛿𝑐 +1.519E−01 6.317E−02 𝐶𝑇𝑦ˆ 𝛿2 𝑐 +6.441E−02 2.982E−02 𝐶𝑇𝑦ˆ 𝐽𝑥 +2.395E−03 7.666E−04 𝐶𝑇𝑦ˆ 𝐽𝑥ˆ 𝛿2 𝑐 −6.251E−01 2.434E−01 𝐶𝑇𝑦ˆ 𝛿2 𝑐ˆ 𝑛 −4.123E−03 2.078E−03 Table 22 Parameter estimates for 𝑪𝑻𝒛in the High In-cidence Region (𝑹2 = 97.76%) Parameter ˆ 𝜃 𝑠( ˆ 𝜃) 𝐶𝑇𝑧𝑜 −1.132E−02 8.805E−05 𝐶𝑇𝑧ˆ 𝐽𝑧 −3.037E−02 7.763E−04 𝐶𝑇𝑧ˆ 𝛿𝑐 −3.567E−02 1.866E−03 𝐶𝑇𝑧ˆ 𝐽𝑥ˆ 𝐽𝑧 −5.633E−02 4.439E−03 𝐶𝑇𝑧ˆ 𝐽3 𝑧 −5.915E−02 6.191E−03 𝐶𝑇𝑧ˆ 𝐽𝑥 −4.666E−03 8.625E−04 27 Table 23 Parameter estimates for 𝑪𝑸𝒙in the High Incidence Region (𝑹2 = 98.09%) Parameter ˆ 𝜃 𝑠( ˆ 𝜃) 𝐶𝑄𝑥𝑜 −4.968E−03 4.436E−05 𝐶𝑄𝑥ˆ 𝛿𝑐 −3.383E−02 7.032E−04 𝐶𝑄𝑥ˆ 𝐽2 𝑧ˆ 𝑛 +3.540E−04 1.044E−04 𝐶𝑄𝑥ˆ 𝐽2 𝑥 +1.605E−02 1.064E−03 𝐶𝑄𝑥ˆ 𝛿2 𝑐 −1.240E−01 1.450E−02 𝐶𝑄𝑥ˆ 𝑛 +2.015E−05 2.127E−06 𝐶𝑄𝑥ˆ 𝐽𝑧ˆ 𝛿𝑐 −1.032E−02 4.055E−03 𝐶𝑄𝑥ˆ 𝐽𝑥ˆ 𝐽𝑧ˆ 𝛿𝑐 +2.035E−01 2.301E−02 𝐶𝑄𝑥ˆ 𝐽2 𝑥ˆ 𝛿𝑐 −1.635E−01 3.022E−02 𝐶𝑄𝑥ˆ 𝐽2 𝑧 −6.122E−03 1.125E−03 𝐶𝑄𝑥ˆ 𝐽3 𝑧 +2.022E−02 5.217E−03 Table 24 Parameter estimates for 𝑪𝑸𝒚in the High Incidence Region (𝑹2 = 95.70%) Parameter ˆ 𝜃 𝑠( ˆ 𝜃) 𝐶𝑄𝑦ˆ 𝐽𝑥 −2.780E−02 7.570E−04 𝐶𝑄𝑦𝑜 +2.879E−03 9.625E−05 𝐶𝑄𝑦ˆ 𝐽𝑧 +1.403E−02 7.421E−04 𝐶𝑄𝑦ˆ 𝐽2 𝑧ˆ 𝑛 +5.117E−04 1.395E−04 𝐶𝑄𝑦ˆ 𝑛 +8.761E−05 6.092E−06 𝐶𝑄𝑦ˆ 𝐽𝑧ˆ 𝑛 +2.827E−04 3.718E−05 𝐶𝑄𝑦ˆ 𝐽2 𝑥 +3.307E−02 4.561E−03 𝐶𝑄𝑦ˆ 𝛿𝑐 +6.980E−03 1.543E−03 𝐶𝑄𝑦ˆ 𝐽𝑥ˆ 𝐽2 𝑧 +1.328E−01 1.812E−02 𝐶𝑄𝑦ˆ 𝐽𝑥ˆ 𝐽𝑧 −4.272E−02 6.805E−03 Table 25 Parameter estimates for 𝑪𝑸𝒛in the High Incidence Region (𝑹2 = 96.39%) Parameter ˆ 𝜃 𝑠( ˆ 𝜃) 𝐶𝑄𝑧ˆ 𝐽𝑧 −1.915E−02 5.020E−04 𝐶𝑄𝑧𝑜 −2.770E−03 9.572E−05 𝐶𝑄𝑧ˆ 𝐽2 𝑧ˆ 𝛿𝑐 −5.641E−02 4.267E−02 𝐶𝑄𝑧ˆ 𝛿𝑐 −2.216E−02 1.776E−03 𝐶𝑄𝑧ˆ 𝐽𝑧ˆ 𝛿𝑐 −9.223E−02 1.009E−02 𝐶𝑄𝑧ˆ 𝑛 −3.757E−05 4.960E−06 𝐶𝑄𝑧ˆ 𝐽𝑥ˆ 𝐽𝑧ˆ 𝑛 −8.877E−04 1.376E−04 𝐶𝑄𝑧ˆ 𝑛2 +1.979E−06 3.624E−07 𝐶𝑄𝑧ˆ 𝐽𝑧ˆ 𝑛 +1.533E−04 3.218E−05 Table 26 Parameter estimates for 𝑪𝑻𝒙in the Descent Region (𝑹2 = 98.43%) Parameter ˆ 𝜃 𝑠( ˆ 𝜃) 𝐶𝑇𝑥𝑜 +8.906E−02 6.646E−04 𝐶𝑇𝑥ˆ 𝐽3 𝑥 +1.126E+00 1.815E−01 𝐶𝑇𝑥ˆ 𝛿𝑐 +4.380E−01 1.109E−02 𝐶𝑇𝑥ˆ 𝐽𝑥 −1.445E−01 6.936E−03 𝐶𝑇𝑥ˆ 𝐽2 𝑥 +8.904E−01 4.206E−02 𝐶𝑇𝑥ˆ 𝐽𝑥ˆ 𝛿𝑐 +6.459E−01 8.246E−02 Table 27 Parameter estimates for 𝑪𝑸𝒙in the Descent Region (𝑹2 = 98.18%) Parameter ˆ 𝜃 𝑠( ˆ 𝜃) 𝐶𝑄𝑥𝑜 −4.858E−03 3.650E−05 𝐶𝑄𝑥ˆ 𝛿𝑐 −3.438E−02 5.012E−04 𝐶𝑄𝑥ˆ 𝐽2 𝑥 +1.424E−02 8.473E−04 𝐶𝑄𝑥ˆ 𝛿2 𝑐 −1.066E−01 1.466E−02 28 Table 28 Median reference values used to center explanatory variables for each modeling region Variable Hover Region Low Incidence Region High Incidence Region Descent Region 𝐽𝑥 — +3.320E−01 +8.333E−02 −3.038E−01 𝐽𝑧 — +1.693E−01 +3.215E−01 — 𝑛, rev/s +6.249E+01 +6.345E+01 +6.155E+01 +6.144E+01 𝛿𝑐, rad −6.747E−03 +6.992E−02 +2.676E−02 +2.550E−02 References Ribner, H. 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15796	https://tentativeplantscientist.wordpress.com/2013/08/18/plant-divisions-ginkgophyta/	Tentative Plant Scientist A gardener's exploration into the world of botany Plant Divisions: Ginkgophyta Ginkgo biloba Ginkgophyta is a plant division of non-flowering trees originating over 250 million years ago, in which all plants except for one, Ginkgo biloba, have become extinct. Ginkgo bilobas are large, deciduous trees with unusual looking cones and distinctive leaves, they can live for up to a thousand years. A few hundred million years ago whole forests existed around the world filled with different species of Ginkgos, but now the one remaining species is native only to China. Ginkgophyta Family Tree Ginkgophyta Family Tree Leaves Ginkgo biloba leaf Ginkgo leaves are bi-lobed, tough and more resistant to decay than other leaves. Some leaves are borne on long stems and turn yellow, die back in winter, then reappear in spring, while others are on shorter stems that may survive the winter. Trunk and Vascular System The bark of Ginkgos is fissured and the trunks may reach to 4m in diameter. The vascular system of Ginkgos, and also conifers, are different to that of flowering plants. While flowering plants have a series of tube-like cells to conduct water, Ginkgos have connecting cells with tiny perforations, these are valves that close when water is in short supply so that turgidity is preserved. Reproduction and Survival Ginkgo biloba with male cones Cone on female Ginkgo Ovules Ginkgos are dioecious. The male cones grow from the shoot tip in clusters and release pollen. The female ovules (cones) appear in twos on the end of a stalk and do not look much like the cones of conifers. Each ovule has a drop of fluid, the pollination drop, that traps pollen to enable fertilisation. Ginkgo pseudofruit Ginkgo sperm cells are motile, swimming to the ovule using thousands of hairs. This is something that occurs in cycads too (see previous blog) and in ferns, but not conifers or flowering plants, so is a throwback to a more primitive form of reproduction. Once fertilized the ovule grows into something resembling a fruit containing the seed. Ginkgo seedling Ginkgo seeds contain two cotyledons (seed leaves), but these never expand or emerge, instead they remain embedded in the seed providing nutrition for the seedling. The first leaves to appear above ground are true leaves with the distinctive Ginkgo shape, this is called hypogeal germination. Ginkgos have a few clever ways of surviving and reproducing: Like cycads, Ginkgos have been known to change sex, so that the male trees start producing ‘fruits’ and seeds. This is an effective way of propagating when there are no females around. Ginkgos have a tendency to put out suckers from the ground that point upwards, but older trees sometimes also have odd downward growths, called Chichi, hanging from a single branch like stalactites. When these growths hit the ground they can start growing new roots and eventually form into a new tree, this is seems to be a form of reproduction for when the main tree is coming to the end of its life. Chichi on Ginkgo The brilliant photo above was taken by Rebecca Sweet and posted on Gossip in the Garden If Ginkgos are hacked right back to the bare trunk they can regrow, either growing from the damaged stem or by putting out new shoots from the ground. Ginkgos are also very resistant to pests, diseases, fires and pollution. Medicinal Properties Ginkgo biloba Ginkgo biloba contains Flavonoids and Terpenoids which are naturally occurring chemical groups found in plants. Flavonoids Use for the plant: pigmentation, assisting in nitrogen fixation and cellular function Use for humans: thought to have anti-allergic, anti-inflammatory, anti-microbial, anti-cancer and anti-diarrheal properties although this is not fully proved. Terpenoids Use for the plant: provide pigmentation and smell. They are thought to act as a deterrent to herbivorous insects and an attractant to insects that may eat herbivorous insects. They also are found in flowering plants and are used to attract pollinators. They may have antioxidant benefits for plants. Use for humans: they have been used in traditional medicines for many years, although their effectiveness is not proved, they may have antibacterial properties and they may also have antioxidant benefits. (note: I have been unable to ascertain exactly what Terpenoids and Flavonoids do in Ginkgo biloba specifically, so this information refers to their function in plants in general.) Why do plants have medicinal properties? We have enemies in common: plants have evolved chemicals that fight some of the same insects, fungi and bacteria that also plague humans. Poisons can also be cures: mammals are often problematic for plants and so they have evolved ways to fight them off, but these ways may also, in small amounts, be cures. For example, Digitalis affects heart rate and is fatal in large amounts, but in small amounts can regulate heart rate. While researching this question I have come across a common belief that plants evolved medicines in order to benefit humans, that by cultivating plants we made it beneficial for them to produce certain chemicals. However since plants first evolved 400 million years ago and evolved those chemical defenses millions of years ago, yet Homo sapiens only evolved a few hundred thousand years ago and only started cultivating plants 12,000 years ago, this isn’t really likely. Further information about Ginkgos: A very good website here, with clear pictures and video ( although the video is unfortunately difficult to hear): Share this: Related Post navigation Blog notes Categories Blog Stats Categories
15797	https://www.khanacademy.org/math/precalculus/x9e81a4f98389efdf:series/x9e81a4f98389efdf:arith-series/v/formula-for-arithmetic-series	Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer. More information Manage Consent Preferences Strictly Necessary Cookies Always Active Certain cookies and other technologies are essential in order to enable our Service to provide the features you have requested, such as making it possible for you to access our product and information related to your account. For example, each time you log into our Service, a Strictly Necessary Cookie authenticates that it is you logging in and allows you to use the Service without having to re-enter your password when you visit a new page or new unit during your browsing session. Functional Cookies These cookies provide you with a more tailored experience and allow you to make certain selections on our Service. For example, these cookies store information such as your preferred language and website preferences. Targeting Cookies These cookies are used on a limited basis, only on pages directed to adults (teachers, donors, or parents). We use these cookies to inform our own digital marketing and help us connect with people who are interested in our Service and our mission. We do not use cookies to serve third party ads on our Service. Performance Cookies These cookies and other technologies allow us to understand how you interact with our Service (e.g., how often you use our Service, where you are accessing the Service from and the content that you’re interacting with). Analytic cookies enable us to support and improve how our Service operates. For example, we use Google Analytics cookies to help us measure traffic and usage trends for the Service, and to understand more about the demographics of our users. We also may use web beacons to gauge the effectiveness of certain communications and the effectiveness of our marketing campaigns via HTML emails.
15798	https://user.eng.umd.edu/~abarg/620/GF.pdf	Lecture 4: Generating functions 1 of 15 Course: Introduction to Stochastic Processes Term: Fall 2019 Instructor: Gordan Žitkovi´ c Lecture 4 Generating functions The path-counting method used in the previous lecture only works for ﬁnite-horizon walks, where the of the horizon T is given in advance. We will see later that most of the interesting questions do not fall into this category. For example, the distribution of the time it takes for the random walk to hit the level l ̸= 0 is like that. While we know that it will happen eventually, there is no way to give an a-priori upper bound on the number of steps it will take to get to l. To deal with a wider class of properties of random walks (and other processes), we need to develop some new mathematical tools. The central among those is the method of generating functions we describe in this lecture. 4.1 Deﬁnition and ﬁrst properties Generating functions provide a link between probability and analysis (cal-culus). The deﬁnition provided below is somewhat confusing at ﬁrst, as it initially lacks any intuition. To give the proper treatment to the “why” be-hind it, we would need to introduce a lot more mathematics than we can at the moment1. The best way to gain understanding without the advanced mathematical background is to learn how to work with generating functions and appreciate what they can do for us. Even though it can be used to deal with many other kinds of distributions, we will use generating functions to study distributions of random variables that take values in the set N0 of natural numbers (and zero). Such random variables often model random times, but can also be given other interpreta-tions. The distribution (table) of an N0-valued random variable looks like this 0 1 2 3 . . . p0 p1 p2 p3 . . . and so, the distribution is completely deﬁned by its probability-mass function (pmf) {pk}k∈N0. Each such pmf is a probability sequence, i.e., a sequence 1if you are curious, generating functions are a special case of the Fourier/Laplace transform and the proper setting for their understanding is within the subﬁeld of mathematics called harmonic analysis. Last Updated: September 25, 2019 Lecture 4: Generating functions 2 of 15 of numbers between 0 and 1 whose sum is 1. For example, the pmf of the random variable X which models a roll of a fair die is the sequence p0 = 0, p1 = 1/6, p2 = 1/6, . . . , p6 = 1/6, p7 = 0, p8 = 0, . . . Deﬁnition 4.1.1. The generating function of a probability sequence {pk}k∈N0 is the function P deﬁned by the following power series: P(s) = ∞ ∑ k=0 pksk. (4.1.1) Given an N0-valued random variable X, we deﬁne its generating func-tion simply as the generating function of its pmf, i.e., of the sequence {pk}k∈N0, with pk = P[X = k], k ∈N0. This generating function is denoted by PX. Since ∑k pk = 1 for each probability sequence the radius of convergence2 of {pk}k∈N0 is at least equal to 1. Therefore, the function P(s) given by (4.1.1) is well deﬁned for s ∈[−1, 1], and, perhaps, other values of s, too. Let us start by deriving expressions for the generating functions of some of the popular N0-valued distributions. Example 4.1.2. 1. Bernoulli (b(p)). Here p0 = q, p1 = p, and pk = 0, for k ≥2. Therefore, PX(s) = ps + q. 2. Binomial (b(n, p)). Since pk = (n k)pkqn−k, k = 0, . . . , n, we have PX(s) = n ∑ k=0 n k pkqn−ksk = (ps + q)n, by the binomial theorem. 3. Geometric (g(p)). For k ∈N0, pk = qkp, so that PX(s) = ∞ ∑ k=0 qkskp = p ∞ ∑ k=0 (qs)k = p 1 −qs. 4. Poisson (P(λ)). Given that pk = e−λ λk k! , k ∈N0, we have PX(s) = ∞ ∑ k=0 e−λ λk k! sk = e−λ ∞ ∑ k=0 (sλ)k k! = e−λesλ = eλ(s−1). 2Remember, that the radius of convergence of a power series ∑∞ k=0 akxk is the largest number R ∈[0, ∞] such that ∑∞ k=0 akxk converges absolutely whenever \|x\| < R. Last Updated: September 25, 2019 Lecture 4: Generating functions 3 of 15 Some of the most useful analytic properties of PX are listed in the follow-ing proposition Proposition 4.1.3. Let X be an N0-valued random variable, {pk}k∈N0 its pmf, and PX its generating function. Then 1. PX(s) = E[sX], s ∈[−1, 1], 2. PX(s) is convex and non-decreasing with 0 ≤PX(s) ≤1 for s ∈[0, 1] 3. PX(s) is inﬁnitely differentiable on (−1, 1) with dk dsk PX(s) = ∞ ∑ j=k j(j −1) . . . (j −k + 1)sj−kpj, k ∈N. (4.1.2) In particular, pk = 1 k! dk dsk PX(s) s=0 and so s 7→PX(s) uniquely determines the sequence {pk}k∈N0. Proof. Statement 1. follows directly from the formula E[g(X)] = ∞ ∑ k=0 g(k)pk, applied to g(x) = sx. As far as (3) is concerned, we only note that the expression (4.1.2) is exactly what you would get if you differentiated the expression (4.1.1) term by term. The rigorous proof of the fact this is allowed is beyond the scope of these notes. With 3. at our disposal, 2. follows by the fact that the ﬁrst two derivatives of the function PX are non-negative and that PX(1) = 1. Remark 4.1.4. 1. If you know about moment-generating functions, you will notice that PX(s) = MX(log(s)), for s ∈(0, 1), where MX(t) = E[exp(tX)] is the moment-generating function of X. 2. Generating functions can be used with sequences {ak}k∈N0 that are not necessarily pmf’s of random variables. The method is useful for any se-quence {ak}k∈N0 such that the power series ∑∞ k=0 aksk has a positive (non-zero) radius of convergence (see the problem about Fibbonacci numbers in the Problems section). 3. The name generating function comes from the last part of the property (3). The knowledge of PX implies the knowledge of the whole sequence {pk}k∈N0. Put differently, PX generates the whole distribution of X. Last Updated: September 25, 2019 Lecture 4: Generating functions 4 of 15 4.2 Convolution and moments The true power of generating functions comes from the fact that they behave very well under the usual operations in probability. Deﬁnition 4.2.1. Let {pk}k∈N0 and {qk}k∈N0 be two sequences. The convolution p ∗q of {pk}k∈N0 and {qk}k∈N0 is the sequence {rk}k∈N0, where rk = k ∑ j=0 pjqk−j, k ∈N0. This abstractly-deﬁned operation will become much clearer once we prove the following proposition: Proposition 4.2.2. Let X, Y be two independent N0-valued random vari-ables with pmfs {pk}k∈N0 and {qk}k∈N0. Then the sum Z = X + Y is also N0-valued and its pmf is the convolution of {pk}k∈N0 and {qk}k∈N0 in the sense of Deﬁnition 4.2.1. Proof. Clearly, Z is N0-valued. To obtain an expression for its pmf, we use the law of total probability: P[Z = k] = k ∑ j=0 P[X = j]P[Z = k\|X = j]. On the other hand, P[Z = k\|X = j] = P[X + Y = k\|X = j] = P[Y = k −j\|X = j] = P[Y = k −j], where the last equality follows from independence of X and Y. Therefore, P[Z = k] = k ∑ j=0 P[X = j]P[Y = k −j] = k ∑ j=0 pjqk−j. Corollary 4.2.3. Let {pk}k∈N0 and {pk}k∈N0 be two sequences. Then 1. p ∗q = q ∗p, i.e., convolution is commutative. 2. The convolution r = p ∗q of two probability sequences is a probability se-quence itself, i.e., rk ≥0, for all k ∈N0 and ∑∞ k=0 rk = 1. Corollary 4.2.4. Let {pk}k∈N0 and {qk}k∈N0 be two probability sequences, and let P(s) = ∞ ∑ k=0 pksk and Q(s) = ∞ ∑ k=0 qksk Last Updated: September 25, 2019 Lecture 4: Generating functions 5 of 15 be their generating functions. Then the generating function R(s) = ∑∞ k=0 rksk, of the convolution r = p ∗q is given by R(s) = P(s)Q(s). Equivalently, the generating function PX+Y of the sum of two independent N0-valued random variables is equal to the product PX+Y(s) = PX(s)PY(s), of the generating functions PX and PY of X and Y. Example 4.2.5. 1. The binomial b(n, p) distribution is a sum of n independent Ber-noullis b(p). Therefore, if we apply Corrolary 4.2.4 n times to the generating function (q + ps) of the Bernoulli b(p) distribution we immediately get that the generating function of the binomial is (q + ps) . . . (q + ps) = (q + ps)n. 2. More generally, we can show that the sum of m independent random variables with the b(n, p) distribution has a binomial distribution b(mn, p). If you try to sum binomials with different values of the parameter p you will not get a binomial. 3. What is even more interesting, the following statement can be shown: Suppose that the sum Z of two independent N0-valued random variables X and Y is binomially distributed with parame-ters n and p. Then both X and Y must be binomial with parameters nX, p and ny, p where nX + nY = n. In other words, the only way to get a binomial as a sum of independent random variables is if they are both binomial with the same p. Another useful thing about generating functions is that they make the computation of moments easier. Proposition 4.2.6. Let {pk}k∈N0 be the pmf of the N0-valued random vari-able X and let PX(s) be its generating function. For n ∈N the following two statements are equivalent 1. E[Xk] < ∞, 2. dkP(s) dsk s=1 exists (in the sense that the left limit lims↗1 dkP(s) dsk exists) Last Updated: September 25, 2019 Lecture 4: Generating functions 6 of 15 In either case, we have E[X(X −1)(X −2) . . . (X −k + 1)] = dk dsk P(s) s=1. The quantities E[X], E[X(X −1)], E[X(X −1)(X −2)], . . . are called factorial moments of the random variable X. You can get the classical moments from the factorial moments by solving a system of linear equations. It is very simple for the ﬁrst few: E[X] = E[X], E[X2] = E[X(X −1)] + E[X], E[X3] = E[X(X −1)(X −2)]] + 3E[X(X −1)] + E[X], . . . A useful identity which follows directly from the above results is the follow-ing: E[X] = P′(1) and Var[X] = P′′(1) + P′(1) −(P′(1))2, and is valid if all the involved derivatives exist. Example 4.2.7. Let X be a Poisson random variable with parameter λ. Its generating function is given by PX(s) = eλ(s−1). Therefore, dk dsk PX(1) = λk, and so, the sequence (E[X], E[X(X − 1)], E[X(X −1)(X −2)], . . . ) of factorial moments of X is just (λ, λ2, λ3, . . . ). It follows that E[X] = λ, E[X2] = λ2 + λ, Var[X] = λ E[X3] = λ3 + 3λ2 + λ, . . . 4.3 Random sums Our next application of generating functions in the theory of stochastic pro-cesses deals with so-called random sums. Let {ξk}k∈N be a sequence of ran-dom variables, and let N be a random time (we allow the value +∞here, too). We can deﬁne the random variable Y = N ∑ k=1 ξk Last Updated: September 25, 2019 Lecture 4: Generating functions 7 of 15 for ω ∈Ωby Y(ω) =      0, N(ω) = 0, ∑ N(ω) k=1 ξk(ω), 1 ≤N(ω) < ∞. ∑∞ k=1 ξk(ω), N(ω) = +∞. In the case when N does not take the value +∞, we can deﬁne this more generally: for an arbitrary stochastic process {Xk}k∈N0 we deﬁne the random variable XN by XN(ω) = XN(ω)(ω), for ω ∈Ω. When N is a constant (N = n), then XN is simply equal to Xn. In general, think of XN as a value of the stochastic process X taken at the time which is itself random. If Xn = ∑n k=1 ξk, then XN = ∑N k=1 ξk. Example 4.3.1. Let {ξk}k∈N be the increments of a symmetric simple random walk; we denoted these by {δk}k∈N when we talked about random walks. Let N be a random variable independent of all {ξk}k∈N with the following distribution N ∼ 0 1 2 1/3 1/3 1/3 Let us compute the distribution of Y = ∑N k=1 ξk in this case. The for-mula of total probability comes in very handy here: P[Y = m] = P[Y = m\|N = 0] P[N = 0] + P[Y = m\|N = 1] P[N = 1] + P[Y = m\|N = 2] P[N = 2] = P[ N ∑ k=0 ξk = m\|N = 0] P[N = 0] + P[ N ∑ k=0 ξk = m\|N = 1] P[N = 1] + P[ N ∑ k=0 ξk = m\|N = 2]P[N = 2] = 1 3 (P[0 = m] + P[ξ1 = m] + P[ξ1 + ξ2 = m]) . When m = 1 (for example), we get P[Y = 1] = 0 + 1 2 + 0 3 = 1/6. Perform the computation for some other values of m for yourself. What happens when N and {ξk}k∈N are dependent? This will usually Last Updated: September 25, 2019 Lecture 4: Generating functions 8 of 15 be the case in practice, as the value of the time N when we stop adding increments will typically depend on the behaviour of the sum itself. Example 4.3.2. Let {ξk}k∈N be as in Example 4.3.1 above - we can think of a situation where a gambler is repeatedly playing the same game in which a coin is tossed and the gambler wins a dollar if the outcome is heads and loses a dollar otherwise. A “smart” gambler enters the game and decides on the following tactic: Let’s see how the ﬁrst game goes. If I lose, I’ll play another 2 games and hopefully cover my losses, and If I win, I’ll quit then and there. The described strategy amounts to the choice of the random time N as follows: N = ( 1, ξ1 = 1, 3, ξ1 = −1. Then Y = ( 1, ξ1 = 1, −1 + ξ2 + ξ3, ξ1 = −1. Therefore, P[Y = 1] = P[Y = 1\|ξ1 = 1]P[ξ1 = 1] + P[Y = 1\|ξ1 = −1]P[ξ1 = −1] = 1 · P[ξ1 = 1] + P[ξ2 + ξ3 = 2]P[ξ1 = −1] = 1 2(1 + 1 4) = 5 8. Similarly, we get P[Y = −1] = 1 4 and P[Y = −3] = 1 8. The expectation E[Y] is equal to 1 · 5 8 + (−1) · 1 4 + (−3) · 1 8 = 0. This is not an accident. One of the ﬁrst powerful results of the beautiful martingale theory states that no matter how smart a strategy you employ, you cannot beat a fair gamble. We will return to the general (non-independent) case in the next lecture. Let us use generating functions to give a full description of the distribution of Y = ∑N k=1 ξk in this case. Proposition 4.3.3. Let {ξn}n∈N be a sequence of independent N0-valued random variables, all of which share the same distribution with pmf {pk}k∈N0 and generating function Pξ(s). Let N be a random time independent of {ξn}n∈N. Then the generating function PY of the random sum Y = ∑N k=1 ξk is given by PY(s) = PN(Pξ(s)). Proof. () We use the idea from Example 4.3.1 and condition on possible val-Last Updated: September 25, 2019 Lecture 4: Generating functions 9 of 15 ues of N. We also use the following fact (Tonelli’s theorem) without proof: If aij ≥0, for all i, j, then ∞ ∑ j=0 ∞ ∑ i=0 aij = ∞ ∑ i=0 ∞ ∑ j=0 aij. (4.3.1) PY(s) = ∞ ∑ k=0 skP[Y = k] = ∞ ∑ k=0 sk ∞ ∑ i=0 P[Y = k\|N = i]P[N = i] = ∞ ∑ k=0 sk ∞ ∑ i=0 P[ i ∑ j=0 ξj = k]P[N = i] (by independence) = ∞ ∑ i=0 ∞ ∑ k=0 skP[ i ∑ j=0 ξj = k]P[N = i] (by Tonelli) = ∞ ∑ i=0 P[N = i] ∞ ∑ k=0 skP[ i ∑ j=0 ξj = k] (by (4.3.1)) By (iteration of) Corollary 4.2.4, we know that the generating function of the random variable ∑i j=0 ξj - which is exactly what the second sum above repre-sents - is (Pξ(s))i. Therefore, the chain of equalities above can be continued as = ∞ ∑ i=0 PN = ii = PN(Pξ(s)). Corollary 4.3.4 (Wald’s Identity I). Let {ξn}n∈N and N be as in Proposition 4.3.3. Suppose, also, that E[N] < ∞and E[ξ1] < ∞. Then E[ N ∑ k=1 ξk] = E[N] E[ξ1]. Proof. We just apply the composition rule for derivatives to the equality PY = PN ◦Pξ to get P′ Y(s) = P′ N(Pξ(s))P′ ξ(s). After we let s ↗1, we get E[Y] = P′ Y(1) = P′ N(Pξ(1))P′ ξ(1) = P′ N(1)P′ ξ(1) = E[N] E[ξ1]. Example 4.3.5. Every time Springﬁeld Wildcats play in the Superbowl, their chance of winning is p ∈(0, 1). The number of years between two Superbowls they get to play in has the Poisson distribution P(λ), λ > Last Updated: September 25, 2019 Lecture 4: Generating functions 10 of 15 0. What is the expected number of years Y between two consecutive Superbowl wins? Let {ξk}k∈N be the sequence of independent P(λ)-random variables modeling the number of years between two consecutive Superbowl appearances by the Wildcats. Moreover, let ˜ N be a geometric g(p) random variable with success probability p, and let N = ˜ N + 1 be its “shifted-by-one” versiona. Every time the Wildcats lose the Superbowl, another ξ· years have to pass before they get another chance and the whole thing stops when they ﬁnally win. Therefore, Y = N ∑ k=1 ξk. To compute the expectation of Y we use Corollary 4.3.4 E[Y] = E[N] E[ξk] = λ p . athis is one of the examples where the version of the geometric that starts from 1, and not from 0, is better suited. Indeed, you cannot win two Superbowls in the same year. Problems Problem 4.3.1. If P(s) is the generating function of the random variable X, then the generating function of 2X + 1 is (a) P(2s + 1) (b) 2P(s) + 1 (c) P(s2 + 1) (d) sP(s2) (e) none of the above Problem 4.3.2. Let (p0, p1, p2, . . . ) be a sequence, and let P(s) be its generat-ing function. Then (1 −s)P(s2) is the generating function of the sequence: (a) (p0, p0 + p2, p0 + p2 + p4, p0 + p2 + p4 + p6, . . . ) (b) (p0, p0 + p1, p0 + p1 + p2, p0 + p1 + p2, . . . ) Last Updated: September 25, 2019 Lecture 4: Generating functions 11 of 15 (c) (p0, −p0, p1, −p1, p2, −p2, p3, −p3, . . . ) (d) (p0, p1 −p0, p2 −p1, p3 −p2, . . . ) (e) none of the above Problem 4.3.3. Let X be a random variable with the generating function PX. The generating function of the random variable X2 is (a) 2PX(s) (b) PX(2s) (c) PX(s2) (d) PX(s)2 (e) none of the above Problem 4.3.4. Let X be a random variable whose generating function PX is given by PX(s) = 1 2(1 + s)/(2 −s) Compute the following: 1. E[X], E[X2] and E[X3]. 2. P[X > 2]. 3. The generating function of the random variable Y = 3X + 2 4. The generating function of the random variable Z obtained as follows. A coin is tossed and the value of X is drawn (independently). If the outcome of the coin is H, we set Z = X. Otherwise, Z = 2X. Problem 4.3.5. 1. Use generating functions to compute the probability that the sum on two independent fair dice is 9. 2. Determine the distribution of the sum of two independent Poisson ran-dom variables with parameters λ1 > 0 and λ2. 3. Determine the distribution of the sum of two independent geometric ran-dom variables with (the same) parameter p > 0. Problem 4.3.6. Let X and Y be two N0-valued random variables, let PX(s) and PY(s) be their generating functions and let Z = X −Y, V = X + Y, W = XY. Then Last Updated: September 25, 2019 Lecture 4: Generating functions 12 of 15 (a) PX(s) = PZ(s)PY(s) (b) PX(s)PY(s) = PZ(s) (c) PW(s) = PX(PY(s)), (d) PZ(s)PV(s) = PX(s)PY(s) (e) none of the above. Problem 4.3.7. Let X be an N0-valued random variable and P(s) its generat-ing function. If Q(s) = P(s)/(1 −s), then (a) Q(s) is a generating function of a random variable, (b) Q(s) is a generating function of a non-decreasing sequence of non-negative numbers, (c) Q(s) is a concave function on (0, 1), (d) Q(0) = 1, (e) none of the above. Problem 4.3.8. The generating function of the N0-valued random variable X is given by PX(s) = s 1 + √ 1 −s2 . 1. Compute p0 = P[X = 0]. 2. Compute p1 = P[X = 1]. 3. Does E[X] exist? If so, ﬁnd its value; if not, explain why not. Problem 4.3.9. Let P(s) be the generating function of the sequence (p0, p1, . . . ) and Q(s) the generating function of the sequence (q0, q1, . . . ). If the sequence {rn}n∈N0 is deﬁned by rn = ( 0, n ≤1 ∑n−1 k=1 pkqn−1−k, n > 1, then its generating function is given by (Note: Careful! {rn}n∈N0 is not exactly the convolution of {pn}n∈N0 and {qn}n∈N0. ) (a) P(s)Q(s) −p0q0 Last Updated: September 25, 2019 Lecture 4: Generating functions 13 of 15 (b) (P(s) −p0)(Q(s) −q0) (c) 1 s (P(s) −p0)Q(s) (d) 1 s P(s)(Q(s) −q0) (e) s(P(s) −p0)Q(s) Problem 4.3.10. A fair coin and a fair 6-sided die are thrown repeatedly until the the ﬁrst time 6 appears on the die. Let X be the number of heads obtained (we are including the heads that may have occurred together with the ﬁrst 6) in the count. Find the generating function of X. Problem 4.3.11. Let N be geometrically distributed with parameter p = 1 2, and let {ξn}n∈N be iid with ξ1 ∼ 0 1 2 1 4 1 2 1 4 . Then the generating function of the random sum Y = ∑N k=0 ξk is (a) ( 7 4 −s 2 −s2 4 )−1 (b) 1 4 + s 2 + s2 4 (c) (s−3)2 4(s−2)2 (d) 1 4 + s 2 + s2 4 1 2 −1 2 s (e) none of the above Problem 4.3.12. A fair coin is tossed 100 times, and the number of H (heads) is denoted by N1. 1. After that, N1 fair coins are tossed and the number of H is denoted by N2. Compute P[N2 = 1]. 2. We continue by tossing N2 fair coins, count the number of heads obtained and denote the result by N3. Compute P[N3 = 1]. Last Updated: September 25, 2019 Lecture 4: Generating functions 14 of 15 Problem 4.3.13. Let N be a random time, independent of {ξk}k∈N, where {ξk}k∈N is a sequence of mutually independent Bernoulli ({0, 1}-valued) ran-dom variables with parameter pB ∈(0, 1). Suppose that N has a geometric distribution g(pg) with parameter pg ∈(0, 1). Compute the distribution of the random sum Y = N ∑ k=1 ξk, i.e., ﬁnd P[Y = k], for k ∈N0. (Note: You can think of Y as a binomial random variable b(n, p) with “random n”.) Problem 4.3.14. Six fair gold coins are tossed, and the total number of tails is recorded; let’s call this number N. Then, a set of three fair silver coins is tossed N times. Let X be the total number of times at least two heads are observed (among the N tosses of the set of silver coins). (Note: A typical outcome of such a procedure would be the following: out of the six gold coins 4 were tails and 2 were heads. Therefore N = 4 and the 4 tosses of the set of three silver coins may look something like {HHT, THT, TTT, HTH}, so that X = 2 in this state of the world. ) Find the generating function and the pmf of X. You don’t have to evaluate binomial coefﬁcients. Problem 4.3.15. Tony Soprano collects his cut from the local garbage man-agement companies. During a typical day he can visit a geometrically dis-tributed number of companies with parameter p = 0.1. According to many years’ worth of statistics gathered by his consigliere Silvio Dante, the amount he collects from the ith company is random with the following distribution Xi ∼ $1000 $2000 $3000 0.2 0.4 0.4 The amounts collected from different companies are independent of each other, and of the number of companies visited. 1. Find the (generating function of) the distribution of the amount of money S that Tony will collect on a given day. 2. Compute E[S] and P[S > 0]. Problem 4.3.16. () The Fibonacci sequence {Fn}n∈N0 is deﬁned recursively by F0 = F1 = 1, Fn+2 = Fn+1 + Fn, n ∈N0. 1. Find the generating function P(s) = ∑∞ k=0 Fksk of the sequence {Fn}n∈N0. (Hint: What is (1 −s −s2)P(s)?) Last Updated: September 25, 2019 Lecture 4: Generating functions 15 of 15 2. Use P to derive an explicit expression for Fn, n ∈N0. (Hint: use partial fractions ) Note: the purpose of this problem is to show that one can use generating functions to do other things, as well. Indeed {Fn}n∈N0 is not a probability distribution, but the generating function techniques still apply. Problem 4.3.17. () When two fair dice are thrown, and their sum is com-puted, different values come with different probabilities (getting a sum of 2 is less likely than getting the sum of 3, etc.). How about if we start with two loaded dice? Suppose that you can build a die with face probabilities of p1, . . . , p6, for any 6-tuple of positive numbers that sum to 1. Can you ﬁnd “loadings” p1, . . . , p6 for the ﬁrst die and q1, . . . , q6 for the second one, so that, when we toss them together, all possible sums (namely 2, 3, . . . 12) have the same probability (namely 1/11)? (Hint: Argue that no matter how you pick p1, . . . , p6 and q1, . . . , q6, the product of their generating functions must be of the form s2F(s), where F has a real zero.) Last Updated: September 25, 2019
15799	https://physics.stackexchange.com/questions/698572/calculating-coherence-length-and-spectral-width-given-an-interferogram	optics - Calculating Coherence Length and spectral width given an interferogram - Physics Stack Exchange Join Physics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Physics helpchat Physics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Calculating Coherence Length and spectral width given an interferogram Ask Question Asked 3 years, 6 months ago Modified1 year, 10 months ago Viewed 2k times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. I'm currently doing a lab in which we use a Michelson-Morley interferometer to analyse different light source. One of the mirrors in the interferometer is moved by a stepper motor. One of the light sources we use is a blue LED. Through a data collection programme, we build up an interferogram of the intensity of the light as it moves through the null point (with intensity on the y-axis and the amount of steps the motor has turned): If we take a fast-fourier transform of this we will get a wavelength spectrum which looks like this: We are then asked to calculate the coherence length and spectral width using the following formulae: Δ ν×L=c 2 π Δ ν×L=c 2 π and Δ ν=Δ λ c λ 2 Δ ν=Δ λ c λ 2 Not much context was given in the lab script for these formulae, but from what I could gather, I thought Δ ν Δ ν referred to the spectral width, L L referred to the coherence length (?), λ λ referred to the mean/peak wavelength and Δ λ Δ λ refers to the full width half maximum of the wavelength spectrum (?) Again, I'm not sure if I have interpreted that correctly, but assuming I did, the values I got for the coherence length and spectral width were as follows: L=1.12∗10−6 L=1.12∗10−6 and Δ ν=4.26∗10 13 Δ ν=4.26∗10 13 They seem to be on the correct order of magnitude, but the reason I am unsure is that when I look at different light sources (e.g. a white LED), the wavelength spectrum has more than one peak, so I'm not sure how I'm supposed to determine the "mean wavelength" and Δ λ Δ λ from those. optics visible-light data-analysis interferometry coherence Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Improve this question Follow Follow this question to receive notifications asked Mar 11, 2022 at 18:47 probablysidprobablysid 97 7 7 bronze badges Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 0 Save this answer. Show activity on this post. So, first, those seem to be in the right order. But I would mention that the Δ Δ's are typically 1/e 2 1/e 2 and not FWHM. Which answers then the question regarding broadband sources. You would in principle take the whole spectrum in the frequency domain and calculate its standard deviation to find the effective bandwidth of the source. The center of mass of the spectrum is its central frequency. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications answered Mar 12, 2022 at 12:58 José AndradeJosé Andrade 1,183 6 6 silver badges 16 16 bronze badges 2 When you say "frequency domain" are you still referring to a fourier transform of the interferogram? Since my interferogram was plotting position vs intensity, wouldnt the interferogram have to be plotting time vs intensity, then taking the fourier transform of that would give a frequency spectrum of frequency vs intensity?probablysid –probablysid 2022-03-12 15:10:58 +00:00 Commented Mar 12, 2022 at 15:10 Exactly, I meant ω ω, so rad/s. And yes, I was going to mention that I would have changed the distance axis to time delay and then proceed from there instead of doing it by going through wavelength. But the result is the same when you know how to go from wavelength to the frequency domain, which they provided the formula for you.José Andrade –José Andrade 2022-03-12 16:38:16 +00:00 Commented Mar 12, 2022 at 16:38 Add a comment\| Your Answer Thanks for contributing an answer to Physics Stack Exchange! Please be sure to answer the question. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. 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