Other examples of sequences that must be defined by words are the daily balances in your checking account or the rolls of a random die.
TWO WELL-KNOWN SEQUENCES
Sequence 1: The even numbers: 2, 4, 6, 8, …
Formula: an = 2n
Recursive definition: a1 = 2, an = an−1 + 2
Sequence 2: The squares: 1, 4, 9, 16, …
Formula: an = n2
Recursive definition: a1 = 1, an = an−1 + 2n − 1
One of the parts of a standard intelligence test is to predict the next number in the sequence. This is a case where too much knowledge is a dangerous thing, at least from the standpoint of scoring well on the intelligence test! For instance, suppose you were asked to predict the next number in a sequence whose first three terms were 1, 2, and 4. If you decide that the sequence is recursively defined by a1 = 1, an = 2an−1, then the next term of the sequence would be 8. If you decide instead that the sequence is recursively defined by a1 = 1, an = an−1 + n − 1, then the next term of the sequence is 7!
Sequences can consist of terms other than numbers. This book, for instance, is a sequence of letters, numbers, and symbols (and all the terms after about five hundred thousand are the blank symbol).
ARITHMETIC PROGRESSIONS
Look at the sequence 2, 5, 8, 11, …, which we define recursively by a1 = 2, an = an−1 + 3. Notice that the difference between any two consecutive terms is 3, which we can see simply by taking the recursion formula an = an−1 + 3, and subtracting an−1 from each side to obtain an − an−1 = 3. This is an example of an arithmetic sequence, more commonly called an arithmetic progression, which is a sequence in which the difference between any two consecutive terms is a constant. This difference is called the common difference. In an arithmetic sequence, one can find the common difference by subtracting any term from the term immediately following it. In the sequence 2, 5, 8, 11, …, 3 = 5 − 2 = 8 − 5 = 11 − 8, etc.
If we are given the first term 2 and the common difference 3 of an arithmetic progression, we can immediately write down the recursive definition: a1 = 2, an = an−1 + 3. We can also use the pattern to write down the first few terms of the sequence
a1 = 2 = 2 + (0 × 3)
a2 = 2 + 3 = 5 = 2 + (1 × 3)
a3 = (2 + 3) + 3 = 8 = 2 + (2 × 3)
a4 = (2 + 3 + 3) + 3 = 11 = 2 + (3 × 3)
From this pattern, we conclude that
an = 2 + (n − 1) × 3
Since we could have done exactly the same thing if we had had an arithmetic progression with first term a1 and common difference d, we have the following two formulas for arithmetic progressions.
FORMULAS FOR ARITHMETIC PROGRESSIONS
An arithmetic progression with first term f and common difference d can be defined recursively by
a1 = f, an = an−1 + d
or by means of the formula
an = f + (n − 1)d
Arithmetic progressions frequently occur in daily life in the total amount paid by a consumer making installment payments.
Example 1: Susan makes a down payment on a car of $2,000 and monthly payments of $180. Describe her equity (the total of the payments she has made) as an arithmetic progression. If she must make 48 monthly payments, how much will she have paid to buy the car?
Solution: The total amount Susan paid is an arithmetic progression, with a1 = $2,000 and common difference $180. Therefore, an = $2,000 + $180(n − 1). After she has made 48 payments, the total paid will be a49 = $10,640. ■
Nature, too, has many examples of arithmetic progressions at its disposal. Many natural phenomena, such as the intervals between eclipses or the closest approaches of planets to the sun (which can be used to mark the changing seasons), represent examples of arithmetic progressions.
Example 2: The great American author Mark Twain (the same one who, as Freddy remarked, found it easy to give up smoking!) died in 1910, the year that marked the fourth official sighting of Halley’s Comet, which appears every seventy-six years. When was Halley’s Comet first officially sighted? When did it last appear? When will it next appear?
Solution: We know that a4 = 1910 = f + (3 × 76) = f + 228, so f = a1 = 1682. (The comet had reappeared many times before then, but this was the year that Halley first recognized it and predicted when it would next reappear.) The last sighting was a5 = 1682 + (4 × 76) = 1986, and its next appearance will be in a6 = 1682 + (5 × 76) = 2062.
Incidentally, not only did Mark Twain die during a4, but he was born during a3! ■
SUMS OF ARITHMETIC PROGRESSIONS
(Number of cigarettes needed continued from p. 35)
The basic use of multiplication is for repeated addition of the same number—3 × 4 is a shorthand for 4 + 4 + 4. However, there are certain attractive situations where there are formulas in which multiplication can be used to find the sum of different numbers. An example of such a situation was discovered by one of the greatest mathematicians of all time, Carl Friedrich Gauss.
One day while Gauss was attending school, his teacher excused himself from the classroom for a short period and asked the students to add the numbers from 1 through 100 in his absence. Most people, when confronted with this problem, proceed in the obvious fashion. Faced with finding the sum 1 + 2 + 3 + … + 100, they compute the sum of 1 + 2, getting 3. To this result they add 3, getting 6. To 6, they then add 4, getting 10. And so on.
Gauss, however, took a different approach. Letting S denote the sum 1 + 2 + 3 + … + 100, he noticed that S could also be written 100 + 99 + 98 + … + 1. Writing these two expressions under each other, we have
Adding both these equations gives
S + S = (1 + 100) + (2 + 99) + … + (99 + 2) + (100 + 1)
The first number in each parenthesis comes from equation [4.1] and the second number from equation [4.2]. But each sum in parentheses adds up to 101, and there are obviously 100 such pairs. So
2S = S + S = 100 × 101 = 10,100
and therefore S = 5,050.
Gauss was about eight years old when he discovered this phenomenon. In his honor, mathematicians now refer to it as the Gauss trick. It would be the highlight of many a mathematician’s career to come up with as cute a trick.
There is obviously nothing special about the number 100. It could have been 1,000, or 8 million, or anything at all. Let’s suppose, therefore, that we want to add all the numbers from 1 through N. Letting S denote the sum 1 + 2 + … + N, we write down the sum forward and backward.
Adding equations [4.3] and [4.4], we get
2S = S + S = (1 + N) + (2 + (N − 1)) + … + ((N − 1) + 2) + (N + 1)
As before, the first term in each pair of parentheses comes from equation [4.3], and the second term from equation [4.4]. Each parenthetical sum is N + 1, and there are obviously N pairs of parentheses. Therefore,
2S = S + S = N × (N + 1)
and so S = (N × (N + 1))/2.
Even in an age of ultrahigh-speed computers, it is still extremely important to discover short formulas for sums. After all, why should one make even a high-speed computer add a thousand numbers, when with a little work we can derive a formula that only requires a few additions and multiplications? Sophisticated problems often require billions or even trillions of computations, and shortcut formulas can sometimes eliminate more than 99% of the calculations.
Example 3: In the story, Freddy wished to cut down his cigarettes by one a day. How many cigarettes would he have had to buy if he had wanted to cut down by two a day?
Solution: It’s easy to see that 40 + 38 + … + 2 = 2(20 + 19 + … + 1) = 2 × ((20 × 21)/2) = 420. So Freddy would have to buy two cartons and one pack (a carton contains ten packs, and a pack contains twenty cigarettes). Remember that he smokes two packs or forty cigarettes a day. ■
You can find the sum of any arithmetic progression using a similar idea.
APPENDIX 5
ALGEBRA, THE LANGUAGE OF QUANTITATIVE RELATIONSHIPS, IN “THE ACCIDENTAL GUEST”
This material focuses primarily on one of th
e portions of algebra that causes a lot of trouble—setting up and solving standard story problems. Maybe this describes you. Or, if you are a parent whose child is floundering in this area, read it yourself and impress your child with your skills. If your son or daughter is taking algebra, he or she is almost certainly a teenager, and this will be your last chance to impress him or her for another decade or so.
Algebra is as much a language as it is a subject. Unfortunately, the way algebra is sometimes taught in high school makes it very difficult to see this. In high school algebra, one learns how to simplify algebraic expressions, how to factor polynomials, how to solve quadratic equations, and so on ad infinitum (and, unfortunately, often ad nauseam). What is learned is an assortment of techniques.
Most people who use algebra a lot regard it as a language that describes relationships between quantities. One of the more important aspects of a language is that it enables questions to be raised and answered. In this sense, arithmetic is a language, too—the questions that it raises concern the relationships among numbers. An example of an arithmetic question is, “What is two plus two?” It has only one correct answer and is a question about numbers. As we have mentioned, questions about numbers are the concern of arithmetic.
The extremely funny—and somewhat morbid—cartoonist Gahan Wilson probably spoke for many algebra students in his classic cartoon “Hell’s Library.” We see a grinning devil with pitchfork, and flames surround the cartoon, leaving no doubt that we are, indeed, in the nether regions. In the background we see a bookshelf. All the books have titles like The Big Book of Story Problems and Even More Story Problems.
We can’t remove all the torments from story problems, but the following two principles provide a great place to start, as they apply to many story problems. Sadly, not all—but many.
THE TOTALS PRINCIPLE AND THE RATE PRINCIPLE
Totals Principle: Totals are sums of subtotals. The Totals Principle is based on addition.
Rate Principle: The cost of a number of objects, each of which sells for the same price, is the product of the price times the number of objects. The Rate Principle is based on multiplication and is applicable in any situation involving a rate (not just when the rate is a monetary one).
(Algebra continued from p. 48)
Pete uses both of these principles when he expresses the total cost of the car bill in dollars (before taxes) as $60.00 + $0.15 M, where M is the number of miles that the car was driven. The Totals Principle shows that the total cost is the sum of two subtotals; the weekend charge and the mileage cost. The weekend charge is $60, and the mileage cost is given by the Rate Principle as $0.15 M; you’re buying M miles at a rate of $0.15 per mile.
All that is left is to solve the equation $60 + $0.15 M = $120.30, and most simple equations yield to repeated applications of the Golden Rule of Equations; Do unto the right side of the equation as ye do unto the left side. In this case,
$120.30 = $60.00 + $0.15 M
$60.30 = $0.15 M (subtract $60 from both sides)
402 = M (divide both sides by $0.15)
Of course, the $ sign isn’t really necessary here; you could work with 120.30 = 60 + 0.15 M as long as you realized that the eventual answer is to be expressed in miles.
One step up in difficulty level from single linear equations (the word linear is used because the graph of y = 60 + 0.15 x is a straight line) is a problem with two linear equations in two unknowns.
Example 1: After attending the movies on Wednesday, everyone adjourns to the local pizza palace for pizzas and pitchers. Last Wednesday, they had five pizzas and three pitchers, and the bill came to $34. This Wednesday, they had six pizzas and four pitchers, for a total of $42. What is the cost of a pizza? What is the cost of a pitcher?
This problem is different from the ones we have encountered earlier because we are seeking not one, but two items of information. Suppose we let P denote the cost of a pizza, and let B denote the cost of a pitcher (the B is the leading letter of the liquid most frequently filling the pitcher). Last Wednesday, the five pizzas cost 5P and the three pitchers cost 3B, so the total of $34 results in the equation
Similarly, the six pizzas cost 6P, the four pitchers cost 4B, and since the bill for this came to $42, we obtain the equation
To solve this system of two equations, there is a standard step-by-step procedure. We pause the solution of this problem to outline it.
PROCEDURE FOR SOLVING TWO EQUATIONS IN TWO UNKNOWNS
Step 1: Solve one of the equations for one of the unknowns in terms of the other.
Step 2: Substitute the result of step 1 into the other equation.
Step 3: Solve the resulting equation, which only contains one unknown.
Step 4: Substitute the result of step 3 into the result of step 1 to obtain the value of the other unknown.
We apply step 1 to equation [5.1] in the pizza–pitcher problem. The result is
Substituting this into equation [5.2], we get
6(6.8 − 0.6 B) + 4B = 42
40.8 − 3.6B + 4B = 42
0.4B = 1.2
So B = 1.2/0.4 = 3. Substituting this value into equation [5.3] yields
P = 6.8 − 0.6(3) = 6.8 − 1.8 = 5
Although checking is not a formal part of the solution procedure, it should always be performed to be sure that an arithmetic or algebraic foul-up has not occurred. Assuming that a pizza costs $5 and a pitcher $3, last Wednesday, the five pizzas cost $25 and the three pitchers $9, for a total of $34. This Wednesday, the six pizzas cost $30 and the four pitchers $12, for a total of $42.
Notice that we did not check the problem by seeing whether the values B = 3 and P = 5 satisfied equations [5.1] and [5.2]; we checked the problem by seeing whether the answers we obtained satisfied the original conditions of the problem. Although it may seem like these two procedures are the same, they are not—it is possible to derive incorrect equations from the problem but solve the equations correctly without solving the problem correctly!
The preceding technique for solving two equations in two unknowns, known as elimination (the elimination takes place in step 3), always works for real-world problems with consistent answers. If, for instance, the price of a pizza was increased from last Wednesday to this Wednesday, the technique will still give answers, but these answers just won’t represent actual prices.
The following example uses the Totals Principle and the Rate Principle in a different environment.
Example 2: A triathlete trains by running for two hours and biking for three hours on Tuesday, covering a total of 110 miles. The next day he runs one hour and bicycles four hours, covering 130 miles. How fast does he run? How fast does he bike?
Solution: Notice that we must assume he always runs at the same rate, which we shall denote R, and always bikes at the same rate, which we shall denote B. Using the Totals Principle, we see that the total distance traveled is the sum of the distance run and the distance biked. Using the distance = rate × time formula (the Rate Principle), on Tuesday he ran 2R miles and biked 3B miles, so 2R + 3B = 110. On Wednesday, he ran R miles and biked 4B miles, so R + 4B = 130. The solution to these two equations is R = 10 miles per hour and B = 30 miles per hour. ■
The technique of elimination can be extended to more than two linear equations in more than two unknowns. Many reasonably priced pocket calculators can handle the solution of up to thirty equations in thirty unknowns, and computers can deal with thousands of equations in thousands of unknowns. However, as yet no computer, other than the human brain, can handle the much more important and difficult task of translating a real-world problem into the language of algebra. Artificial intelligence experts are working on this, but it’s probably many years in the future.
APPENDIX 6
MATHEMATICS OF FINANCE IN “MESSAGE FROM A CORPSE”
If you are a typical American, during your lifetime you will be involved with a substantial number of financial transactions. As a result, your ability to understand the mathematics of borr
owing and lending is probably worth a minimum of tens of thousands of dollars to you, and possibly even more.
It is astounding how a person who carefully evaluates which of two types of TV to buy, where the cost of the two TVs may differ by at most a couple of hundred dollars, immediately accepts the first loan shoved under his or her nose by anyone who agrees to finance his or her purchase of a car or a house. Just as there are “best buys” in TVs, there are “best buys” in credit, for when you shop for a loan, or decide what to do with your investment capital, you are making a purchase of money.
This chapter can save you substantial amounts of money during your lifetime—far more than you might imagine—just by enabling you to see what borrowing money costs you. Since this chapter deals extensively with calculations, you would do well to own a calculator. A simple one, costing only a few dollars, that just has the arithmetic functions (+, −, ×, and ÷) will suffice, but life will be easier if you have one with exponentiation (this key is usually denoted yx or xy). It will only cost a few dollars more, and there’s almost certainly one on your computer under “Accessories.”
SIMPLE AND COMPOUND INTEREST
Many jobs pay wages that are computed by multiplying the length of time worked by the pay rate. If you are paid $8 per hour and work for 10 hours, you are paid $80. Simple interest can be thought of as the wages earned by money, and it is computed exactly the same way, by multiplying the length of time the money is working by the interest rate.
Example 1: Suppose that Sue loans John $500 for three years at an interest rate of 6% per year. How much does the money that Sue loans earn? How much does John have to pay back? Assume that simple interest is being charged.
Solution: Because 6% of $500 is 0.06 × $500 = $30, each year John has the money costs $30. Since he borrows it for three years, the money “earns” 3 × $30 = $90 during that period. John will have to repay the $500 he borrowed, plus the $90 interest, for a total of $590. ■
L.A. Math: Romance, Crime, and Mathematics in the City of Angels Page 18