Lesson 2: Perpetuities, Geometric Progressions, and ...gregorybard.com/finite/S14_geometric_progression.pdf · rather important and very easy to understand are “The Arithmetic Progression,”

Lesson 4.2 Page 439 of 915.

Lesson 2: Perpetuities, Geometric Progressions, and Reserve Ratios

In this section, we’re going to visit the uses & purposes of geometric progressions. For usthis mostly pertains to calculating the value of perpetuities, a rare but fascinating kind ofinvestment. Perpetuities, in turn, will become a tool for calculating things about mortgagesas well as retirement savings. Two other applications of geometric progressions include themoney multiplier and reserve ratios (which are intimately related). We will also explorewhy 0.999 = 1.

At one time, there was a lottery ticket for sale in New York City that, if you won, wouldo↵er you $ 1000 per week, for the remainder of your life. An interesting question is to lookat the cash value of such a promise. If I were to o↵er you $ 1000 a week for all time, whatwould that really be worth?

Believe it or not, a fixed amount of money, delivered at regular intervals, over the courseof eternity is actually equivalent to one lump sum of cash! Personally, I was not willing toaccept this so easily when it was explained to me as an undergrad. To my intuition thesewere two very di↵erent situations.

Let us suppose that the prevailing interest rate were 5.2% on readily available money-marketaccounts. I’m going to take $ 1,000,000, and deposit it into such an account. Accounts ofthis type are usually made to compound monthly, or quarterly, but let us suppose that theaccount I’ve chosen is to compounded annually, because this is the lesson’s first exampleand we’d like to keep it simple. How much interest would I earn in the first year? Well, thatwould be (0.052)⇥ (1, 000, 000) = 52, 000 per year. Yet if I, in the first year, get $ 52,000 ininterest, then it is very easy for me to give you $ 1000 per week, because there are 52 weeksin the year.

This is not exactly precise, as I’d have to pay you $ 1000 right away, and $ 1000 after aweek, another $ 1000 after two weeks, and so on, and so I should be using the Time Valueof Money Formula, not to mention that I’d only get the first interest payment after oneyear. It turns out that we can be more accurate. In any case, the next year, I would have$ 1,000,000 again, and the entire process repeats, year after year, forever.

Roughly speaking, if the interest rate were 5.2% per year, then the value of my promiseto you of giving you $ 1000 per week for all time, would be worth “something in theneighborhood of” $ 1,000,000. Now let’s get more precise!

COPYRIGHT NOTICE: This is a work in-progress by Prof. Gregory V. Bard, which is intended to be eventually released under the Creative

Commons License (specifically agreement # 3 “attribution and non-commercial.”) Until such time as the document is completed, however, the

author reserves all rights, to ensure that imperfect copies are not widely circulated.


We will now try to make the previous two boxes and their computations more precise. Onceagain, I am promising you $ 1,000 a week, for the remainder of your life. The first week’spayment is definitely worth $ 1,000, no one can doubt that. However, the next week’spayment is worth just a hair less.

Let r be the prevailing interest rate per year (the prevailing nominal rate). Then leti = r/52 signify the prevailing interest rate per week. Using the Time-Value of MoneyFormula we know the second payment is worth 1000/(1 + i), and the third one is worth1000/(1 + i)2, and also the fourth one is worth 1000/(1 + i)3. This is because the second,third, and fourth payment are due after one week, two weeks, and three weeks, into thefuture.

So if I wanted to value the whole set of payments I’d have to consider the followinginfinite sum:

1000

1+

1000

(1 + i)+

1000

(1 + i)2+

1000

(1 + i)3+

1000

(1 + i)4+

1000

(1 + i)5+

1000

(1 + i)6+ · · ·

But maybe I don’t quite remember all the rules for infinite sums!

Mathematicians have many uses for infinite sums. They are vitally important, particularlythe field known as Scientific Computation (which is a favorite topic of mine) but that’snot something which a↵ects us in this course. Two types of sums, however, which arerather important and very easy to understand are “The Arithmetic Progression,” and “TheGeometric Progression.”

The sum in the box before the last box is an example of a geometric progression—eventhough there’s no geometry involved. Any easy example of a geometric progression is thefollowing example:

3 + 1 +1

3+

1

9+

1

27+

1

81+

1

273+

1

729+ · · ·

where each member of the sum is 1/3⇥ the size of the member to its left, and 3⇥ the sizeof the member to its right. We would say that this is a geometric progression of “commonratio equal to 1/3,” because going deeper and deeper into the progression means repeatedlymultiplying consecutive terms by 1/3.

A geometric progression is a sequence of numbers added together where each number is crtimes as large as the one before it, and 1/cr times as large as the one after it. The numbercr is called the common ratio of the geometric progression.

Other names for geometric progressions include geometric sums or geometric series.The latter is particularly used for sums that are infinitely long.





Try summing the first ten terms in the sum

3 + 1 +1

3+

1

9+

1

27+

1

81+

1

273+

1

729+ · · ·

• What do you get for the sum? [Answer: 4.49947 · · · .]

• Where do you think this is leading? [Answer: perhaps 4.5?]

• Of course, the 4.5 is merely a guess, but we’ll learn how to solve this without guessingmomentarily.

How about the first ten terms of

4 + 1 + 1/4 + 1/16 + 1/64 + 1/256 + 1/1024 + 1/4096 + · · ·

• What do you get for the sum? [Answer: 5.33332 · · · .]

• Where do you think this is leading? [Answer: perhaps 16/3?]

• Again, 16/3 is merely a guess, but we’ll learn momentarily how to solve that withoutguessing.

I now present you with two options. First, you could simply accept the formula in the next box as a gift. On the otherhand, if you are curious, I’m going to explain where it comes from, in the box after that.

As noted earlier, a geometric progression of common ratio cr is an infinite sum, where eachentry is cr times as big as the one before it. In this book, we restrict ourselves only to crbetween 0 and 1. We can write either

S = a+ acr + ac2r + ac3r + ac4r + ac5r + ac6r + · · ·

or, if you like big-Sigma notation, then

S =1X

i=0

acir

The large Greek letter that looks like a spring, or sideways “M,” is the capital Greekletter “Sigma.” Most students do not like big-Sigma notation, so I will not use it in thisbook.

Given a geometric progression of common ratio cr, with 0 < cr < 1, and whose firstterm is a, then the value of the sum is

S =a

1� cr

The formula we just learned is pretty amazing from a philosophical point of view. By it, we can calculate the value ofthe sum of infinitely many numbers, but with a very finite amount of work—in fact, almost no work at all.





Now we’re going to try to explain where the formula in the previous box comes from. Sincefew of us like big-Sigma notation, we will use ordinary notation. Let’s begin

S = a+ acr + ac2r + ac3r + ac4r + ac5r + ac6r + · · ·

and then let us suppose we multiply the whole entire thing by cr to obtain

crS = acr + ac2r + ac3r + ac4r + ac5r + ac6r + ac7r + · · ·

and then subtract the two equations

S � crS = a� acr + acr � ac2r + ac2r � ac3r + ac3r � ac4r + ac4r + · · ·

or equivalently

S � crS = a+ (�acr + acr) +�

�ac2r + ac2r�

+�

�ac3r + ac3r�

+�

�ac4r + ac4r�

+ · · ·

Now look at what happened to the right-hand side: every term except the lone a has beencancelled out! For example, (�acr + acr) can clearly become just 0. This means we are leftwith

S � crS = a ) S(1� cr) = a ) S =a

1� cr

which is what we wanted!

In calculus, there are four main objects: the derivative, the limit, the integral and theinfinite sum. Because the infinite sum often comes last in the curriculum, it is the leastspoken of, and least remembered.

We required that 0 < cr < 1. Perhaps you might be curious about the other cases. Thecase of cr = 0 is pretty obvious, as all terms but the first are equal to zero.

One of things you will learn in calculus is that if cr � 1, which we forbade earlier, thenthe total usually does not come to some nice real number, but rather usually approaches±1. When this is the case, if a = 0 then the sum would be 0, because every term wouldbe 0. On the other hand, if a > 0 then the sum is 1, and if a < 0 then the sum is �1.

To convince yourself of this, try calculating the first 20 terms for cr = 5 and see whathappens with various values of a.

The case of a negative common ratio, in other words cr < 0, is somewhat more compli-cated, and you’ll have to wait until a course in calculus to learn how to deal with these.

Gaining a better understanding of infinity is one of the many benefits of learning a smallbit of calculus.

The sum we had earlier was

3 + 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/273 + 1/729 + · · ·

which we can see has first term 3 and common ratio 1/3. Then the value of the sum shouldbe

3

1� 1/3=

3

2/3=

9

2= 4.5





Likewise, consider

4 + 2 + 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + · · ·

and this results in4

1� 1/2=

4

1/2=

8

1= 8

Consider the sum1 + 1/10 + 1/100 + 1/1000 + 1/10, 000 + · · ·

which could have been written as

1 + 0.1 + 0.01 + 0.001 + 0.0001 + · · ·

What do these add up to? [Answer: 10/9]

Consider the sum

1000/1 + 1000/(1.1) + 1000/(1.1)2 + 1000/(1.1)3 + 10, 000/(1.14) + · · ·

where we o↵er you the hint that each member of the sum is 1/1.1 times as big as the memberbefore it.

• What is a? [Answer: 1000.]

• What is cr? [Answer: 1/1.1 = 0.9090.]

• What does it all add up to? [Answer: 11,000]

In a past example, we had $ 1000 a week for the rest of our lives. If the nominal availableinterest rate is 5.2%, then that’s 0.1% per week. So, the first payment was worth 1000,then the next one 1000/(1.001), and the next one after that 1000/(1.001)2, followed by1000/(1.001)3, forever. This would make the total come to

S =1000

1+

1000

(1.001)+

1000

(1.001)2+

1000

(1.001)3+

1000

(1.001)4+ · · ·

and while it is not obvious, that means each term is 1/1.001 times as big as the one beforeit. So the common ratio is cr = 1/1.001. Then the value is

S =1000

1� 1/1.001=

1000

1� 0.999000 · · · =1000

0.000999000 · · · = 1, 000, 999.99 · · ·

which is basically $ 1,001,000.00. This type of investment is called a perpetuity .





• Try the previous example, with 10.4% prevailing interest (which would be very highindeed, but it is just an example). [Answer: $ 501,000].

• Try the previous example, with 2.6% prevailing interest (which would be kind of low,but in Spring of 2010, that was what was happening). [Answer: $ 2,001,000].

The previous three examples show a common pattern, so perhaps we can construct a shortcutformula that will be faster. Let a be the amount of each payment, and let i = r/m be theinterest rate adjusted to the time scale of the payments. In other words, for weekly payments,i = r/52 and for monthly payments i = r/12, but for quarterly payments, i = r/4, and soon.

Then a/(1+ i) is the second payment, a/(1+ i)2 is the third payment, a/(1+ i)3 is thefourth payment, and so on. The total value would be

S =a

1+

a

(1 + i)+

a

(1 + i)2+

a

(1 + i)3+

a

(1 + i)4+ · · ·

which is a geometric progression of common ratio cr = 1/(1 + i), and first term a.Then the sum comes to

S =a

1� cr=

a

1� 1

1+i

=a

1+i1+i �

1

1+i

=ai

1+i

=a(1 + i)

i=

a+ ai

i=

a

i+ a

allowing us to conclude that S = (a/i) + a is the value of this sequence of payments.

The value V of an infinite sequence of payments, each of a dollars is equal to

V =a

i+ a

where i = r/m, and r is the “prevailing interest rate.” Such an arrangement is called aperpetuity .

Note, the m represents whether the payments are to be given weekly (m = 52), monthly(m = 12), quarterly (m = 4), et cetera.

Personal injury law suits often have the following calculation: Suppose a factory worker iskilled in an industrial accident, and that he used to make $ 70,000 per year. To compensatethe widow for the earning power of her husband, the lawyers for the widow will ask for aperpetuity of $ 70,000 per year. Suppose the prevailing interest rate is 7%, which meansone can readily go out and find investments which are paying 7% interest.

The value of the perpetuity is given by

70, 000

0.07+ 70, 000 = 1, 070, 000

and so the lawyers will ask for 1.07 million dollars. The widow would then be expectedto approach reputable companies to purchase a perpetuity. Typically they are sold bylife-insurance companies.





Suppose the widow takes the $ 1,070,000 settlement, and uses the first $ 70,000 to live o↵.Suppose the widow then deposits the million remaining in some investment, making 7%compounded annually, and takes the interest to live o↵, in the sense of siphoning describedon Page 212. We can then use simple interest to find out what she gets each year. Theinterest for any year would be

I = Prt = (1, 000, 000)(0.07)(1) = 70, 000

as desired. The steps in this box are a good way to check your work.

Now suppose that the prevailing interest rate were 5% and that the salary were $ 60,000.What would the perpetuity have to be? [Answer: $ 1,260,000.]

We’ve glossed over one detail. Imagine the prevailing interest rate is 6%. Suppose you gointo a company that sells annuities on May 1st, and say that you want a payment of $ 1000on the last day of the month, every month forever and ever. Then there is a payment onMay 31st, June 30th, July 31st, August 31st, September 30th, October 31st, November 30th,December 31st, January 31st, and so on, forever. This is called a perpetuity-due, but we’llcall it “Plan A.”

Alternatively, if you wanted the payment on the first of the month (including May), thenyou’d get one on May 1st, June 1st, July 1st, August 1st, September 1st, and so on, forever.Because you get some money immediately (May 1st), this is called a perpetuity-immediate.If you think about it, other than the payment at the start of May, this is basically thesame thing as the perpetuity in the previous paragraph. That’s because there is not muchdi↵erence between May 31st and June 1st, nor between June 30th and July 1st, and so forth.We’ll call this “Plan B.” Note that Plan B is the one we already know how to calculate thecost of. In the next box, we’ll learn how to calculate the cost of Plan A.

To make everything extra clear, Plan A and Plan B are nearly identical. The di↵erenceis a payment at the end of the month, or at the start of the month, as well as a payment“right now” which Plan B includes, but Plan A excludes.

How can we figure out how much a perpetuity-due should cost?The June 1st payment of Plan B is equivalent to the May 31st payment of Plan A;

the July 1st payment of Plan B is equivalent to the June 30th payment of Plan A; and theAugust 1st payment of Plan B is equivalent to the July 31st payment of Plan A, et cetera.

The formula for the price of this kind of perpetuity, namely a perpetuity-immediate (aswe learned before) is V = a/i+a, and so we need to find i. Of course, 6% per year is 0.50%per month. Thus we have

V = a/i+ a = 1000/0.005 + 1000 = 200, 000 + 1000 = 201, 000

What about the perpetuity-due? The only di↵erence is that we do not receive the May1st payment. In no other way are the two schemes di↵erent. Thus, the perpetuity-dueshould be $ 1,000 less, because it is missing the first payment. It would be $ 200,000 incost. Or in general, the cost is of a perpetuity-due is V = a/i.





• The formula for the value of a perpetuity-due is V = a/i, and it gives fixed paymentsof a dollars at the end of each interest period—the interest rate being i.

• The formula for the value of a perpetuity-immediate is V = a/i+ a, and it gives fixedpayments of a dollars at the start of each interest period, including a single paymentthe moment you buy it.

• A perpetuity-immediate is sometimes called an ordinary perpetuity .

A young fellow forms a band, and after many nights of playing only for tips at cheapbars, he finally gets a record contract. He signs for ten million dollars. After recording abunch of songs, he marries his high-school sweetheart and they have a huge wedding andhoneymoon costing one million dollars. His wife then takes the nine million dollars left over,believing that he’ll likely never produce another album again, and buys a perpetuity-due.The prevailing interest rate is 4.75%. How much will they get per year?

Well, we know that V = a/i for a perpetuity-due. So we have 9, 000, 000 = a/0.0475and thus a = (9, 000, 000) ⇥ (0.0475) = 427, 500. Thus, for his wife’s good prudence, he’llreceive the yearly salary equivalent to an excellent lawyer or an above-average doctor, forthe remainder of his life, without doing any work. If only more rock stars were this prudent!

What if they spent two million on a house, instead of one million on the wedding andhoneymoon, and the prevailing rate was 6%. What would the annual payment be? [Answer:$ 480,000.]

What if, instead of a perpetuity-due, it were a perpetuity-immediate in the previous box?Well, we simply use the formula V = a/i+ a, with i = 0.06 and V = 8, 000, 000. Then

we have

8, 000, 000 = a/0.06 + a

0.06(8, 000, 000) = a+ 0.06a

480, 000 = (1.06)a

480, 000/1.06 = a

452, 830.18 · · · = a

or just slightly less than a perpetuity-due.

Repeat problem in the previous box for the interest rates 4%, 5%, and 7%. What do youobtain? [Answer: $ 307,692.30, $ 380,952.38, and $ 523,364.48.] As you can see, perpetuitiesare very sensitive to the interest rate!





We’ve learned that the musician and his wife will earn between $ 307,692.30 and $ 523,364.48for their investment, every year, forever. However, when they are aged 70, perhaps 45years into the future, then inflation will have shrunk the apparent worth of that six-digitpayment. Assuming a 4.5% inflation rate, which is a bit high, what would the $ 307,692.30and $ 523,364.48 be worth 45 years from now, in today’s dollars? Hint, see Page 357 if you’veforgotten how to do this. [Answer: $ 42,450.57 and $ 72,205.64, that is much smaller!] Thisis why it is important to always take into account the force of inflation, when calculatingthe value of future payments.

You might wonder if it is possible to have a perpetuity that makes payments which will grow with inflation. The answeris yes, and we will cover this on Page 544. One common financial instrument that almost does this, except for the factthat the payments are a bit irregular and unpredictable, is preferred stock.

The reason you’re learning about perpetuities is because they are a financial model; we’regoing to use them to derive formulas and explain concepts. Rarely do they come into actualplay in finance, except in certain types of lawsuits. However, there is an exception!

Starting in 1751, the British government issued “consols,” or “consol bonds,” whichare perpetuities. Even though 259 years have elapsed between 1751 and 2010, there arestill people collecting the payments on these bonds. Unlike most bonds, they do not paysemi-annually, but they pay quarterly instead. The idea is that a wealthy Duke or Duchesscould provide a quarterly income to his or her descendants, forever, by purchasing one ofthese financial instruments.

Let’s play with this concept and see what we can discover about it. London is crowdedand transportation is expensive. Getting from Point A to Point B takes time, and can beunpleasant—but not nearly as unpleasant as smaller cities in the USA that don’t even havea subway! Let us assume that the person who owns a consol must travel to some o�ceacross town to pick up their quarterly payment. If the payment were tiny, then they mightnot bother with the hassle. So if a person owns one of these consol bonds first issued in1751, then we have to assume that the payment that they are traveling to go pick up hasto be worth at least $ 75 of our dollars—just to be worth the hassle of going to get it, tosay nothing of transportation costs. Maybe the correct value should be $ 50 or $ 100, butlet’s just settle on an arbitrary value of $ 75. This is 50.67 pounds sterling, at the exchangerate of 1.48:1 which was in place in Spring of 2010.

These consol bonds pay quarterly (forever), and so the yearly payments would be 4 ⇥50.67 = 202.68 pounds sterling. The rate of interest in 1752 was 3.5%, and this meansthat the face value for a perpetuity-due would have been 202.68/0.035 = 5790.85 poundssterling. In 1752, this would have been a phenomenal sum of money.

On the other hand, over 258 years, this consol bond has made

258⇥ 4⇥ 50.67 = 52, 291.44

pounds sterling in total payments—which is pretty impressive, for an investment that onlycost a bit less than 5800 pounds.





Using the English inflation calculator from Page 913, we learn that 5790.85 pounds (they’dhave called it 5790 pounds and 17 shillings) in 1752 would have had a modern value in 2008of 759,000 or 8,450,000 pounds sterling, depending which formula you use. In dollars, thatis $ 1,123,320 to $ 12,506,000—not the sum of money that a common person would haveunder their mattress. And remember, this was a rough minimum.

Obviously $ 1.1 million and $ 12.5 million are a bit far apart. The former is calculatedusing currency inflation, and the latter via the prevailing wage. These two methods tendto diverge over long periods of time, and 258 years definitely qualifies as a long time. Sincewe don’t have time machines at our disposal, we have no choice but to accept that level ofuncertainty. Inflation is not an easy force to measure over centuries. With that much timegone by, it is only natural that the uncertainty be large, but it is a shockingly huge sum ofmoney, in either case.

Let’s say that Alfred, who is 10 years old, has a wealthy great-uncle that dies. The uncle’swill says that Alfred shall receive $ 10,000 per year, on his birthday, but starting with his25th birthday, for his entire life. What is the value of that perpetuity? The problem with thisperpetuity is that it doesn’t start now, nor does it start at the end of this month, but rather,it starts 15 years in the future. This is called a perpetuity-forborne, where forborne meansto withhold from doing something. The perpetuity “forbears” to give Alfred a payment forthe first 15 years. “Forborne” is a good Shakespearean-era word to know.

In the meantime, let us estimate the prevailing rate at 7% compounded annually. Thenwe can agree that on the day that Alfred turns 25, that this perpetuity is equivalent, at thatmoment, to a perpetuity-immediate. Then, by our previous formula, it should be worth

V = a/i+ a = (10, 000/0.07) + 10, 000 = 142, 857. · · ·+ 10, 000 = 152, 857. · · ·

at that moment, 15 years from now. Using the time-value of money, we can covert that FVto a present-day PV . And then we have

PV =FV

(1 + i)n=

152, 857. · · ·(1 + 0.07)15

=152, 857. · · ·2.75903 · · · = 55, 402.46

and so we learn that this gift, right now, is worth $ 55,402.46.

Alfred’s older brother is 12 and his younger sister is 7. They both got the same deal—namely, $ 10,000 per year for life, starting on the 25th birthday. What is the value of theperpetuity-forborne in each case? Just as we assumed Alfred was exactly 10 years old inthe previous box, you may assume that the brother is exactly 12 years old and the sisteris exactly 7 years old. [Answer: $ 45,224.91 for the younger sister and $ 63,430.27 for theolder brother.]

A perpetuity-forborne is a perpetuity that begins n units of time after the purchase date.The formula for this is

PV =a/i+ a

(1 + i)n

which can be algebraically manipulated to become

PV =a/i+ a

(1 + i)n=

a+ ai

i(1 + i)n=

a(1 + i)

i(1 + i)n=

a

i(1 + i)n�1

where a is the repeated payment, and i is the interest rate per unit of time.





I think that a perpetuity-forborne is a rare enough situation that it is not useful to memorize the previous formula.Just think of a perpetuity-immediate and use the Time-Value-of-Money idea.

• (Fairly Easy) A perpetuity-due is functionally a perpetuity-forborne with n = 1. Plugn = 1 into the above formula and see that you do indeed get PV = a/i.

• (Slightly Harder) A perpetuity-immediate is functionally a perpetuity-forborne withn = 0. Plug n = 0 into the above formula and show that you do indeed get PV =a/i+ a. This might require some algebraic manipulations.

Why should we want to know the present value of a perpetuity? There are many possiblereasons. First, in writing the will, the author of the will needs to know how much fundsare required for each type of gift, otherwise he might endow more than what he can a↵ord,and then the heirs will be stuck in probate court for years. Second, the IRS will definitelywant to know this detail for inheritance tax purposes. Third, the estate lawyer needsto negotiate with providers of annuities and perpetuities (usually these are sold by life-insurance companies) to actually arrange this gift, and that, in turn, requires knowingthe fair market value—otherwise how can you negotiate?! While it is very unlikely thatyou’ll ever need this calculation ever in your life, we’re going to use the present-value of aperpetuity as a step in deriving a formula for mortgages on Page 470. Our true use willbe using a perpetuity-forborne as a mathematical instrument, rather than as a financialinstrument.

A Pause for Reflection. . .A perpetuity, as we’ve seen described here, is an infinite sequence of payments that continuesforever, with the payments occurring on a regular schedule. Because this is an infinitelylong sequence of payments, many people imagine that the value of such an arrangementis infinite, or nearly so. I, as a math professor, have heard many people say “if you addinfinitely many numbers together, how could you possibly get anything other than 1?!”

Innumeracy can be very depressing; throughout this lesson you will see infinitely longsums that come to a finite numerical total. Why do you think many mathematically unini-tiated people have trouble with this?

In physics, there is the notion of the conservation of matter-energy. Prior to 1905, we couldsay that “energy is never created nor destroyed.” This means that you can measure theenergy of various objects in a physical system, and then using various physics formulas, youcan find out what happens later to those objects. The main mathematical idea is that thetotal energy always adds up to the same number, all the time. This is a really convenientand useful fact, and it remains useful across many di↵erent types of science courses.

Then in 1905, Albert Einstein published some important papers. There, and in whatfollowed and in what came before, he explains how matter and energy can sometimes convertinto each other. One example of that is an atomic explosion when a nuclear weapon isdetonated. Since most physics lab classes at most universities do not involve the use ofnuclear weapons, usually this point can be ignored, but instead of saying “energy is nevercreated nor destroyed” we say “matter-energy is never created nor destroyed.”





A physics professor, who shall remain nameless, attempted to construct a ‘new theory ofeconomics’ based on the idea that money cannot be created nor destroyed. However, whilehis work was quite rigorous, after building up from that embryonic idea—it was wastede↵ort because money can be created and destroyed. We are now going to learn about the“money multiplier,” which is how that can occur.

In order to introduce “the money multiplier,” the following discussion is an extended description of a situation wherea creation phenomenon occurs.

1. Suppose you hire a plumber (call him Luigi) and pay him $ 300. You write him a check, he deposits it, and hehas now $ 300 in his checking account. What does the bank do with his money when he deposits it? Of course,they lend it out to others, to charge them interest. This is how they pay you interest, and pay their expensesand employees. The law requires banks to keep 10% of their deposits on hand, in case customers come and makea withdrawal, and this 10% is called the reserves ratio. The remaining 90% get invested—mostly via loans tocustomers. Thus, 0.9(300) = 270 gets invested elsewhere, and $ 30 sits in the vault. Perhaps (bundled with manyother amounts) it becomes part of someone’s home equity loan. Let’s call the homeowner Fred.

2. Fred takes out a home equity loan from the bank, and it turns out that $ 270 of the dollars he borrows are fromthe $ 270 out of $ 300 that Luigi deposited. Fred takes the check that the bank gave him, and deposits it in hischecking account. Fred has $ 270 on deposit, ready to spend on home improvement, and Luigi still has $ 300 inhis bank account.

Note: From the original $ 300 there is now 300+ 270 = 570 on deposit. Both people (Fred and Liugi) imagine they canwithdraw their funds any time that they wish, so we have to add it together.

3. The bank takes the $ 270 which Fred deposited, and decides to issue car loans. The bank gets $ 270 from Fred’sdeposit, and according to the law, the bank can lend out 0.9(270) = 243 dollars, keeping the remaining $ 27 inthe vault. Someone takes out a car loan that day, call him Bob, and it turns out that $ 243 of the dollars thatBob receives from the bank are the $ 243 from Fred’s deposit. The money from the car loan is given to the cardealership.

Note: From the original $ 300 there is now 300 + 270 + 243 = 813 on deposit, representing Luigi, Fred, and the cardealership.

4. The car dealership receives the money from Bob’s car, and they deposit it in their operating expenses checkingaccount. The money is much more than $ 243 of course, but $ 243 of it came from Fred. The bank can loan out0.9(243) = 218.70 of this and $ 24.30 stays in the vault.

5. Meanwhile, it turns out that Luigi is getting engaged and so he goes and buys a ring. Sadly, he did not quitehave enough saved up for the ring that he wanted, and he had to get a signature loan to cover the di↵erence.(Clearly Luigi hasn’t read Page 221 of this book.) In any case, the signature loan happens to include $ 218.70from the dealership’s deposit. This money is then given to the jeweler.

Note: From the original $ 300 there is now 300+ 270+ 243+ 218.70 = 1031.70 on deposit, representing Luigi, Fred, thecar dealership, and the jeweler.

6. The jeweler deposits the check from Luigi. This means that $ 218.70 is on deposit in the jeweler’s account. Thebank can loan out 218.70(0.9) = 196.83, perhaps to Luigi’s bride Maria who can buy a dress, and then 21.87 muststay in the vault. Maria writes a check to the bridal shop, which deposits the $ 196.83.

Note: From the original $ 300 there is now 300 + 270 + 243 + 218.70 + 196.83 = 1228.53 on deposit, representing Luigi,Fred, the car dealership, the jeweler, and the bridal shop.





That previous box was a bit complex, but read it again, and try to verify that the followingtable represents what happend:Intake Intake On Outgo OutgoSource Quantity Reserve Quantity DestinationLuigi’s deposit 300.00 = 30.00 + 270.00 Fred’s home equityFred’s deposit 270.00 = 27.00 + 243.00 Bob’s car loanDealership’s deposit 243.00 = 24.30 + 218.70 Luigi’s signature loanJeweler’s deposit 218.70 = 21.87 + 196.83 Maria’s signature loan

Take a moment to verify that

300.00 = 300⇥ 0.90

270.00 = 300⇥ 0.91

243.00 = 300⇥ 0.92

218.70 = 300⇥ 0.93

196.83 = 300⇥ 0.94

That was tedious. In order to really take the process to its natural conclusion, we’d have tokeep continuing the story until the dollar values are so small that we really don’t care anymore. Since the dollar values are dropping very slowly (five steps brought us from $ 300 to$ 196.83), that might be a very long problem. Ironically, it is easier to proceed with algebra.

Suppose someone deposits D dollars in Bank Alpha. If the reserve ratio is 10%, thenBank Alpha must keep 0.1D in its reserves, and 0.9D is invested in various ways. Someof it will go to consumers who will spend it, and the recipient will deposit it. Some ofit will go into investment banks. In any case, let us represent this by putting the 0.9Ddollars into Bank Beta. Then Bank Beta must keep 0.1 ⇥ 0.9D = 0.09D in reserve, and0.9 ⇥ 0.9D = 0.81D can be loaned out. Suppose all of it is deposited with a centralbank, Bank Gamma. Then Bank Gamma must keep 0.1 ⇥ 0.81D = 0.081D on reserve,and 0.9 ⇥ 0.81D = 0.729D is invested. Perhaps it goes to a hedge fund operated byBank Delta. Then Bank Delta will keep 0.1 ⇥ 0.729D = 0.0729D on reserve, and caninvest 0.9⇥ 0.729D = 0.6561D of it. This cycle continues forever.

The next box will analyze the total amount of money in circulation.

Continuing with the previous box, now we ask ourselves: what is the total on deposit amongall banks? What is on reserve, among all banks? The invested part is

D + 0.9D + 0.81D + 0.729D + 0.6561D + · · ·

which can be rewritten as

D + 0.9D + 0.92D + 0.93D + 0.94D + · · ·

That is obviously an infinite geometric progression with first term D and common ratio 0.9.Thus the value of this infinite sum is going to be

a

1� cr=

D

1� 0.9=

D

0.1= 10D

and the 10 in that 10D is the money multiplier, as we will define shortly.





Continuing with the previous two boxes, we will now consider how much is on reserve at allthe banks in total. The amount on reserve is

0.1D + 0.09D + 0.081D + 0.0729D + · · ·

and that can be rewritten

0.1D + (0.9)(0.1D) + (0.9)2(0.1D) + (0.9)3(0.1D) + · · ·

which is clearly also an infinite geometric progression with first term 0.1D and commonratio 0.9. Thus the value of the infinite sum is going to be

a

1� cr=

0.1D

1� 0.9=

0.1D

0.1= D

Last but not least, note that “in the vault” is a figure of speech. It is probably an electronicdeposit, with a central bank. A central bank is a bank that ordinary banks use for the sametasks that you yourself use an ordinary bank; banks deposit wealth, make withdrawals anduse credit at central banks.

Until the advent of electronic banking in the 20th century, reserves would indeed becash in the vault. Thus it makes sense that the sum of all the amounts on reserve must addup to D for a deposit of D. Consider the Middle Ages. If you give D gold coins to a banker,and it gets re-invested repeatedly in the above process, then people can imagine what theylike about what is on their balance sheet, but there are still only D coins. Thus the reservestotal had better add up to D. If it added to less than D, then that would mean some of thecoins have mysteriously vanished—barring theft, that is nonsense. If it added up to morethan D then that would indicate some of the coins had managed to appear in two di↵erentplaces at once—which is equally absurd!

So whatever value D happens to take, the sum of all the reserves of all the banks hasto come to D.

As we saw two boxes ago, the total on deposit in the checking accounts at each bank addsup to 10D when the reserve ratio is 10%. In general, this total will be D/r, where r is thereserve ratio. In the United States, in Spring 2010, it was the law that 10% of the checkingaccount deposits must remain on reserve (though probably most of it in electronic form).Thus r = 0.1, and a deposit of D dollars represents an increase of 10D in the entire moneysupply . That is because

money multiplier =1

r=

1

0.1= 10

The money supply is the total that everyone imagines that they have available to spendwithout having to use credit. This factor of 10 is called the money multiplier , and is alwaysequal to 1/r. In Europe, the reserve ratios are often around 2%, and so the money multiplieris 50 there. The reserve ratio varies from government to government.





• Suppose that we are looking at a country with reserve ratio 5%. What is the moneymultiplier? [Answer: 20.]

• And if it were 3%? What is the money multiplier? [Answer: 33.33.]

• And if it were 2%? What is the money multiplier? [Answer: 50.]

Suppose that I have 700 dollars, and I deposit it in a bank in the USA where the reserveratio is 10%. Some of it must remain as reserves, but all of the rest gets deposited in anotherbank, and in another, and another, and so forth.

• What is the total (among all banks) held in reserve? [Answer: $ 700 on reserve (thinkof the gold coins).]

• What is the total money (among all banks) on deposit? [Answer: $ 7000 on deposit,because the money multiplier is 10.]

• What if, in the previous box, the reserve ratio had been 8%?

– What is on deposit? [Answer: $ 8750.]

– What is on reserve? [Answer: $ 700.]

• And if it were 12%?

– What is on deposit? [Answer: $ 5833.33.]

– What is on reserve? [Answer: $ 700.]

A Pause for Reflection. . .Imagine your local bank. Do you think most people realize that, if every customer of thebank were to simultaneously show up on Monday morning and demand their money back,that the bank would be unable to pay? Do you think most people know where the interestthey receive comes from? Do you think most people could understand the money multiplier,if it were well-explained? Do you think we are better o↵ with the masses being ignorant orknowledgeable?

For certain types of accounts in the USA, the reserve ratio is r = 0. I don’t like to thinkabout what that means for the money multiplier: If you try to calculate 1/r to get themoney multiplier, you discover that you are asked to divide by zero, which is impossible.If you use 10�6 or 10�9 as approximate values of r, what does that mean for the moneymultiplier? [Answer: The money multiplier would be a million or a billion.]





Suppose you have a geometric progression that is not an infinite sum, but that “stops early.”How would we go about calculating the value of that sum, without going through the tediumof adding each member of the sum? An example of this would be

S = 3+1+1/3+1/9+1/27+1/81+1/35+1/36+1/37+1/38+ · · ·+1/398+1/399+1/3100

and if we were to multiply all that by 3, we would get

3S = 9+3+1+1/3+ 1/9+ 1/27+ 1/81+ 1/35 +1/36 +1/37 + · · ·+1/397 +1/398 +1/399

yet observe that every term in the lower sum exists in the upper sum and vice versa, except9 and 1/3100. Therefore, we can subtract the two equations

3S � S = 9� 1/3100

2S = 9� 3�100

S =9� 3�100

2

and there is our answer.

Examining the previous box, of course, 3�100 is very tiny. Therefore, that value of S isapproximately the same thing as S = 9/2. The infinite sum would have come to 9/2 exactly(as you surely remember from Page 442 and so we see that the infinite sum is a good modelfor a finite but a very long sum.

Now let’s convert the above technique into a general-purpose formula. We start with somesort of geometric progression with common ratio cr and first term equal to a. This wouldbe

S = a+ acr + acr2 + acr

3 + acr4 + · · ·+ acr

n�3 + acrn�2 + acr

n�1

and as you can see, that sum has exactly n terms in it. Now, let’s multiply the entire thingby cr which yields

crS = acr + acr2 + acr

3 + acr4 + acr

5 + · · ·+ acrn�2 + acr

n�1 + acrn

and again you can see that every member of the lower sum is a member of the upper sum,except for a and acrn.

Now, we can subtract them and get

S � crS = a� acrn ) S(1� cr) = a� acr

n ) S =a� acrn

1� cr

That is our general-purpose formula for geometric progressions that stop early. Amaz-ingly, this formula will help us with mortgages on Page 468.





Suppose one has the sum:

512 + 128 + 32 + 8 + 2 +1

2+

1

8+

1

32+

1

128

and one wants to use the shortcut formula for geometric progressions that stop early. Thenwe have cr = 1/4 and a = 512, as well as n = 9. This comes to

S =512� 512(1/4)9

1� 1/4=

512� 512(3.81469 · · ·⇥ 10�6)0.75

=512� 0.00195312 · · ·

0.75= 682.664 · · ·

which makes sense, because the infinite sum would come to

S =512

1� 1/4=

512

3/4=

2048

3= 682.666 · · ·

It is really hard to believe that the two sums would actually be all that di↵erent, sinceall the “missing” terms are quite tiny—the next two would be 1/512 and 1/2048, small incomparison to 682.666 · · · . That’s why they di↵er only in the sixth significant figure.

Do you see the 512(1/4)9 in the middle formula of this box? That is the next (butunseen) term in the progression, i.e. 1/512. This is not a coincidence. That acrn term willalways be the next member of the sum that you would have written down if the progressiondid not stop early.

If you have a geometric progression of n terms, with first term a and a common ratio of cr,then the sum of that progression is

S =a� acrn

1� cr

or alternatively

S =a� zcr1� cr

where z is the last term in your progression.

Given the sum:

512 + 256 + 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 + 1/2 + 1/4

find the total using the formula for a geometric progression stopping early.[Answer = 1023.75].

Consider the geometric progression given by

2000

(1.005)+

2000

(1.005)2+

2000

(1.005)3+

2000

(1.005)4+

2000

(1.005)5+

2000

(1.005)6+ · · ·+ 2000

(1.005)360

which stops early. What is the value of the sum? [Hint: cr = 1/1.005].[Answer = 333, 583. · · · ].





The rest of this lesson contains advanced material, and some instructors might want toinclude it, while other instructors might not want to include it.

Suppose a company has the following profit history:1996 $ 541,368 1999 $ 781,0771997 $ 611,746 2000 $ 886,5221998 $ 685,155 2001 $ 1,010,636

which, using an exponential regression, can be modeled by

p(t) = (541, 368)e(0.124847··· )(t�1996)

You will learn how to construct exponential regressions yourself in a later lesson, onPage ??. If you wanted to forecast some number of years into the future, you could simplyuse p(t) to find the profit for those years (e.g. 2002, 2003, 2004, 2005, . . . ) and add themup to find the total profit.

Alternatively, you could recognize that it is a geometric progression. Each year ise0.124847··· times as profitable as the previous. (In plain English, if I increase t by one, thenI get another e0.124847···. So we have a geometric progression, with common ratio equal toe0.124847···. The computation is finished in the next box.

If you wanted to know the total projected profit for the years 2001 to 2011, then you could realize that this is ageometric progression, stopping early, with 11 members, and first term 1,010,636. Then the total profit is

S =a� acnr1� cr

=1, 010, 636� 1, 010, 636(e0.124847···)11

1� e0.124847···

=1, 010, 636� (1, 010, 636)(3.94842 · · · )

1� 1.13297 · · ·=�2, 979, 779. · · ·�0.13297 · · ·

= 22, 408, 597.59

Remember that those last few digits are highly subject to rounding errors. The pennies there would be the ninthand tenth significant figures, and no initial data could possibly be trusted to that level of accuracy.

Suppose now you wished to check the model. Calculate the total profit for the years 1996to 2001, as a geometric progression with common ratio e0.124847··· and first term 541,368.Then actually add up the stated profit figures 1996 to 2001. What does the model predict?What was the stated profit? What is the absolute and relative error? [Answer: The modelestimates $ 4,539,620. The actual is $ 4,516,504. The absolute error is $ 23,116; and therelative error is 0.51%, not bad at all!]





Suppose a company, perhaps a “dot-com,” has the following profit history.1996 $ 4,876,381 1999 $ 10,961,9091997 $ 6,388,059 2000 $ 14,250,4821998 $ 8,304,476 2001 $ 18,383,122

which, using an exponential regression, one can calculate is modeled by

p(t) = (4, 876, 381)e(0.265406··· )(t�1996)

Now find the predicted total profit for 1996 to 2001 via a geometric progression of6 terms, common ratio e(0.265406··· ) and first term 4,876,381. Alternatively, find the trueprofit. Finally, what is the absolute error and relative error? What does this predict willbe the total profit for 2002–2010? [Answer: The model estimates $ 62,819,027. The actualis $ 63,164,429. The absolute error is $ 345,402; and the relative error is 0.54%. For the2002–2010 period, the prediction is $ 780,638,516.70, which is very hard to believe.]

“Past performance is no guarantee of future results.” This phrase appears very frequently ininvestment literature, in particular, because the growth of the companies is unpredicatble.It might be exponential over some limited time frame but that does not mean that itwill continue to do that over many more years. In fact, eventually the trend must end,otherwise the small company would be making a profit greater than the amount of moneyin circulation in the entire world—which is absurd. After all, look at the previous box.We had an excellent model for the data (with only 0.54% error), but then the prediction,because we looked far into the future, gave an absurd answer. Thus, it should be notedthat such models should only be trusted for short-term predictions. This is much like theE. Coli problem on Page 379.

Suppose a newspaper estimates that 5% of their readers will switch from the print-versionto the online-version each quarter, and this pattern will continue for the forseeable future.They have a print readership of 100,000. How many people will they lose in the next fouryears?

First, it is obvious that they’ll lose 0.05(100, 000) = 5000 readers this quarter. Second,you should observe that next quarter, the loss will be 5% less than that, or 0.95(5000) =4750. After that, 0.95(4750) but that can be written as 0.95(0.95)5000. Clearly, this is ageometric progression, with first term 5000 and common ratio 0.95.

To sum up the losses over the next four years, we should realize that 16 quarters arecovered by four years. Just as the second member of the sum was 0.951(5000) and the thirdone was 0.952(5000), the sixteenth will be 0.9515(5000). So, for our sum of 16 quarters wehave z = 0.9515(5000) and a = 5000.

The total is pretty easy to find

S =a� zcr1� cr

=5000� 0.9515(5000)0.95

1� 0.95=

5000� 2200.63 · · ·0.05

= 55, 987.3 · · ·

and we must recall that the “point three,” in other words, the fact that the answer is notan integer, is due to the fact that this model cannot possibly be exact.





Now the question that could be put to you is to ask when will the newspaper in the previousbox be down to 20,000 readers? This means, of course, that there is 80,000 in total lostreaders. All we must do is replace the 15 in the previous formula with q� 1, where q� 1 isthe number of elapsed quarters. Then we have

S =a� zcr1� cr

=5000� 0.95q�1(5000)0.95

1� 0.95

and we can set that S equal to 80,000 and solve as follows:

S = 80, 000 =5000� 0.95q�1(5000)0.95

1� 0.95

80, 000 = 5000⇥ 1� 0.95q

1� 0.95

16 =1� 0.95q

0.050.8 = 1� 0.95q

0.2 = 0.95q

log 0.2 = log 0.95q

log 0.2 = q log 0.95log 0.2

log 0.95= q

31.3771 · · · = q

This means that in the 31st quarter they’ll have more than 20,000 readers (since it isn’t31.3771 . . . yet) and in the 32nd quarter they’ll have less than 20,000 readers (since it is past31.3771 . . . at that point).

We actually derived a useful fact about geometric progressions in the previous box, withoutstating the general formula. Here it is now:

The formula for the nth term of a geometric progression with common ratio cr andinitial term a is given by

acn�1

r

It is always important to check our work. Let’s check the newspaper readership problem.In this case, we’ll find out what the 31st quarter and 32nd quarter cumulative losses are,and make sure that they straddle 80,000. Since a = 5000 and cr = 0.95, we have

S = a�zcr

1�cr

S = a�zcr

1�cr

S = 5000�5000(0.95)300.951�0.95 S = 5000�5000(0.95)310.95

1�0.95

S = 5000�1019.53···0.05 S = 5000�968.557···

0.05S = 5000�1019.53···

0.05 S = 4031.44···0.05

S = 79, 609.3 · · · S = 80, 628.8 · · ·





A large computer programming firm is discovering that about 2% of their employees permonth are leaving to go work for internet start-ups. Unfortunately, there is not enoughmoney for raises available and the company has to absorb the losses. They have 56,000employees at the present moment.

• How many will be recruited away after 2 years? [Answer: 21, 516.3 · · · .]

• How many will be recruited away after 3 years? [Answer: 28, 940.0 · · · .]

• When will their total losses be 5000 employees? [Answer: between the 4th and 5th

months.]

• What are the total losses when the 4th month has ended? [Answer: 4347.38 · · · .]

• What are the total losses when the 5th month has ended? [Answer: 5380.43 · · · .]

Let’s try some science fiction. A space-exploration mission realizes that some strange virushas infected their crew. About 3% are dying each week. At the moment, the vessel has5680 crewmen remaining.

• What will the total losses be after 8 weeks? [Answer: 1228.33 · · · dead.]

• In which week will they lose roughly 151 people? [Answer: the fifth week.]

• When will they be down to 1000 people? [Answer: between the 57th and 58th week.]

• How many people will they have lost after 57 weeks? [Answer: 4679.22 · · · .]

• How many people will they have lost after 58 weeks? [Answer: 4709.24 · · · .]

Suppose a Petri dish is 8 cm across, and has 900 bacteria per square centimeter, as measuredby a lab assistant using a microscope. A new antibiotic soap is added to the dish, and thebacteria die at a rate of 8% per second.

• What is the area of the dish? [Answer: 50.2654 · · · cm2.]

• How many bacteria are in the dish? [Answer: Roughly 45, 238.9 · · · .]

• What will the total bacteria killed be after 10 seconds? [Answer: 25, 587.6 · · · dead.]

• At what time will 80% of the bacteria be dead? [Answer: 19.3020 · · · seconds.]

• How many bacteria will they have killed after 18 seconds? As a percentage of theoriginal count? [Answer: 35, 153.5 · · · , or 77.7063 · · ·%.]







Our previous four boxes are examples of “plague problems.” These are when populations decay, with the number ofpeople dying o↵ forming a geometric progression. They were quite common in epidemiology books and demographybooks of about 100 years ago, but luckily catastrophic pandemics have become extremely uncommon. On the otherhand, the plague phenomenon can occur in business situations, such as the newspaper problem and the computerprogramming firm losing sta↵ to start-up ventures.

I’m now going to share with you a very old mathematics problem. Nearly every mathematicstextbook that covers geometric progressions has this problem, so this book should as well.

As I was going to St. Ives,I met a man with seven wives,Each wife had seven sacks,Each sack had seven cats,Each cat had seven kits,Kits, cats, sacks, wives,How many were going to St. Ives?

Now we’re going to solve the St. Ives problem. As you can see, there are 7 wives, and 72

sacks. Then that means there are 73 cats and 74 kits [kittens]. Since only kits, cats, sacks,and wives are the domain of the question, the answer is a geometric progression that stopsearly. To be precise,

S = 7 + 72 + 73 + 74

As you can see, the first term is 7 and the common ratio is 7. The last term is 74. Sowe obtain:

S =a� zcr1� cr

=7� 74(7)

1� 7=�16800�6 = 2800

On the other hand, some include the man, and a = 1 in that case:

S =a� zcr1� cr

=1� 74(7)

1� 7=�16806�6 = 2801

Continue the above calculation, if each kitten had 7 fleas, and each flea had 7 gnats. Howmany, gnats, fleas, kittens, cats, sacks, wives, including the man, are going to St. Ives?[Answer: 137,257.]





Manuscripts with that St. Ives problem, with the rhyme as given, date from roughly the18th century. However, an important document called “the Rhind Mathematical Papyrus”has precisely this calculation, and is dated to around 1650 bc. The Rhind MathematicalPapyrus is a collect of math problems from ancient Egypt which were discovered around 150years ago during an 1858 archeological dig, by the Scottish archeologist Alexander Rhind(1833–1863).

The phrasing is slightly di↵erent:

There are seven houses;In each house there are seven cats;Each cat catches seven mice;Each mouse would have eaten seven ears of corn;If sown, each ear of corn would have produced seven hekat[?] of grain.How many things are mentioned altogether?

It is neat to imagine how many generations of students have thought about this com-putation. However, we have no reason to believe that its presence in math books has beencontinuous over those 36.5 centuries, particularly during the European Dark Ages in theEarly Medieval Period, and the crisis between the Bronze Age and the Iron Age.

Let’s look critically, one last time, at the St. Ives problem. Do we really need the geometricseries formula? Of course, you could just add up the numbers

1 + 7 + 72 + 73 + 74 = 2801

but performing the calculations by hand, without electronic aid, would require a good dealof personal time.

In comparison, the geometric progression formula saves you a lot of e↵ort. Consider ifeach gnat has seven Yersinia Pestis bacterium, for example, and we’d have

1 + 7 + 72 + 73 + 74 + 75 + 76 + 77 =1� 78

1� 7=

1� 5, 764, 801

�6 = 960, 800

You can find 78, relatively easily. Start with 7. Then one squaring tells you that 72 = 49,and another squaring says that 49⇥ 49 = 74 = 2801. Finally, a third squaring gives you

28012 = 494 = 78 = 5, 764, 801

Thus we’ve found 78 without finding 77 or 76, to say nothing of 75 or 73. That’s whythis shortcut was important to the Egyptians.

The next four boxes are a particularly theoretical excursion. They can be skipped by those who are not interested inexploring something more advanced.





This gives us another way to derive the formula for a geometric progression stopping early.Rather than re-derive with symbols, let’s suppose we are calculating

1 + 7 + 72 + 73 + 74 + 75 + 76

We can think of that geometric progression, which starts with 1 goes on forever

1 + 7 + 72 + 73 + 74 + 75 + 76| {z }

part we want

+77 + 78 + 79 + 710 + · · ·| {z }

part we don’t want| {z }

whole thing

but where we want only one part of that progression, and not the other.How can we break them apart? We will find out in the next box!

Continuing with the previous box, while it seems to make no sense, we could observe thelarger progression (labelled “whole thing”) has first term 1 and common ratio 7, and there-fore a sum equal to

S =a

1� cr=

1

1� 7=

1

�6 = �1/6

while the smaller part has first term 77 and common ratio 7, and therefore a sum equal to

S =a

1� cr=

77

1� 7=

77

�6 = �77/6

and so the part that we’d actually want is

�16� �7

7

6=�1 + 77

6=�1 + 823, 543

6=

823, 542

6= 137, 257

This is funny because we have a sum for two geometric progressions that are clearly goingto infinity. Furthermore, it is a sum of positive numbers, and the total comes out negative.We seem to have “broken” the S = a/(1� cr) formula. Why did this occur? This occurredbecause we assumed 0 < cr < 1 in the derivation of the S = a/(1 � cr) formula. Bychoosing to work with a progression that has cr = 7, we have violated an assumption of thederivation. Therefore, the formula has every right to give us a wrong answer.

Now, I could repeat the previous theory box with cr being a variable, instead of lockedat 7, and a equal to a variable, instead of 1. Yet this would be a dishonest parlor trick: Itturns out to work, but it passes through the land of nonsense to get there.

Just to be ultra-clear, the objection mentioned in the previous box is to sums of positive numbers coming out negative.Namely,

1 + 7 + 72 + 73 + 74 + 75 + · · · = �16

STRANGE

and also

77 + 78 + 79 + 710 + 711 + 712 + · · · = �77

6 STRANGE

which seem to be nonsense at first glance.





The previous three boxes were written for two reasons. The minor reason is to teach youthat if the common ratio is bigger than one, i.e. the progression is growing, then you cannotsum the infinite geometric sum, even if you can sum those that stop early.

The major reason is that you must become aware that formulas are derived with as-sumptions. If you have a problem which violates one of the assumptions, you will get awrong answer if you use the formula. This might be the first time in your mathematicalcareer where you’ve seen that occur, where a formula will lie to your face—but it won’t bethe last time.

When mathematicians state the assumptions at play when a theorem is derived, it isn’tto amuse themselves; tragically, most students ignore those very necessary sentences, andthey do so at their peril.

In the next five boxes, not including this one, I will explore the applications of the geometric progression to pure math,in particular, to the question of the value of repeating decimals. Many instructors will skip this, so check with yourinstructor.

We’re going to now present you with two di↵erent ways of seeing that 0.99 = 1. First, thetraditional proof:

Suppose x = 0.99. Then 10x = 9.99. Next, what is 10x� x? Observe,

10x = 9.99999 · · ·x = 0.99999 · · ·

9x = 9.00000 · · ·

As you can see, each of the nines to the right of the decimal point was cancelled out! Soif 9x = 9 then x = 1, and we learn that 1 = 0.99. Most students find this proof unsatisfying.

What does it mean when we write 2.84? Of course, that means 2 + 8

10

+ 4

100

, and likewise,what does it mean for us to write 12.3456? Of course, that means

12 +3

10+

4

100+

5

1000+

6

10, 000

as we all know and understand.Therefore

0.99 = 0.999999 · · · = 9

10+

9

100+

9

1000+

9

10, 000+

9

100, 000+

9

1, 000, 000+ · · ·

but what is that? That is merely an infinite geometric progression with first term 9/10, andthen common ratio 1/10. Then the value of the sum would be

9/10

1� 1/10=

9/10

9/10= 1

and thus 0.99 = 1. Many students find this proof more convincing.





Let’s try that again for 3.3636. We can write this as

3.3636 = 3 +3

10+

6

100+

3

1000+

6

10, 000+

3

100, 000+

6

1, 000, 000+ · · ·

which is not a geometric progression. Yet we can regroup as follows

3.3636 = 3 +

✓

3

10+

6

100

◆

+

✓

3

1000+

6

10, 000

◆

+

✓

3

100, 000+

6

1, 000, 000

◆

+ · · ·�

and that becomes

3.3636 = 3 +

✓

36

100

◆

+

✓

36

10, 000

◆

+

✓

36

1, 000, 000

◆

+ · · ·�

You can see that what is inside the “[” and “]” is an infinite geometric progression withfirst term 36/100, and common ratio 1/100. Thus the sum of the geometric progression is

36/100

1� 1/100=

36/100

99/100=

36

99=

12

33=

4

11

and thus we learn that 3.3636 = 3 + 4/11 = 37/11.

Use a similar argument to the above to prove that 2.99 = 3.

1. Write 2.99 as an infinite sum of fractions.

2. Identify the first term and the common ratio of this geometric progression.

3. Write the formula for the sum of an infinite geometric progression.

4. Calculate the value of the sum in this case.

5. Check your answer, which can be found on Page 465.

Now try that to prove that

• 0.33 = 1/3.

• 0.22 = 2/9.

• 0.4545 = 5/11.

• 0.123123 = 41/333.





Here are the solutions to the chessboard boxes on Page 464.

• First we observe

2.99 = 2 +910

+9

100+

91000

+9

10, 000+ · · ·

which is 2 plus an infinite geometric progression with first term 9/10 and common ratio1/10. Then we have

2.99 = 2 +9/10

1� 1/10= 2 +

9/109/10

= 2 + 1 = 3

• Now we observe

0.33 = 0 +310

+3

100+

31000

+3

10, 000+ · · ·


0.33 = 0 +3/10

1� 1/10= 0 +

3/109/10

= 0 + 1/3 = 1/3

• Again we observe

0.22 = 0 +210

+2

100+

21000

+2

10, 000+ · · ·


0.22 = 0 +2/10

1� 1/10= 0 +

2/109/10

= 0 + 2/9 = 2/9

• This next one is a bit di↵erent

0.4545 = 0 +410

+5

100+

41000

+5

10, 000+

4100, 000

+5

1, 000, 000+ · · ·

which is not a geometric progression at all—but, we can regroup!

0.4545 = 0 +45100

+45

10, 000+

451, 000, 000

+ · · ·


0.4545 = 0 +45/100

1� 1/100= 0 +

45/10099/100

= 0 + 45/99 = (5⇥ 9)/(11⇥ 9) = 5/11

• This last one is a bit harder

0.123123 = 0 +110

+2

100+

31000

+1

10, 000+

2100, 000

+3

1, 000, 000+ · · ·

which is not a geometric progression at all, as we saw before. However, as before, we canregroup!

0.123123 = 0 +1231000

+123

1, 000, 000+ · · ·


0.123123 = 0 +123/10001� 1/1000

= 0 +123/1000999/1000

= 0 + 123/999 = (41⇥ 3)/(333⇥ 3) = 41/333





We have learned the following skills in this lesson:

• To identify a perpetuity.

• To examine a geometric progressions, including those that stop early, and note thefirst term, the common ratio, and final term (if any).

• To calculate the sums of those geometric progressions.

• To distinguish between a perpetuity-due and a perpetuity-immediate.

• To calculate the value of a perpetuity, whether it is a perpetuity-due or perpetuity-immediate.

• To calculate the impact of inflation on a perpetuity.

• To apply to concept of a perpetuity to problems involving sudden unexpected wealth,estate planning, or accidental dismemberment.

• To calculate the relationship between the reserves ratio and the money multiplier.

• To find the value of perpetuities-forborne.

• To work with and calculate the value of businesses using a geometric progression astheir model of profit, but also, to reject the possibility that such models will hold trueforever.

• To use geometric progressions to model a disease or the reduction of the size a set ofpeople (i.e. plague problems), including if the population is the number of program-mers at a company or the number of customers of a newspaper.

• To use geometric progressions to evaluate repeating decimals.

• As well as the vocabulary terms: common ratio, geometric progression, geometric se-ries, geometric sum, money multiplier, money supply, ordinary perpetuity, perpetuity,perpetuity-due, perpetuity-forborne, perpetuity-immediate, reserves ratio.

Coming soon!




Documents

Lesson 2: Perpetuities, Geometric Progressions, and ...gregorybard.com/finite/S14_geometric_progression.pdf · rather important and very easy to understand are “The Arithmetic Progression,”