and Exponentials Logarithms
Dawn Leigh Anderson University of Georgia Athens, Georgia Ann Willis Sebrell Godwin High School Richmond, Virginia Rebecca Berg Bowie State University Bowie, Maryland Donald W. Smith Albuquerque Academy Albuquerque, New Mexico
Historical Modules Project
Sponsored by the Mathematical Association of America Funded by the National Science Foundation
Table of Contents
Introduction Suggested courses for using module activities HISTORICAL AND CULTURAL OVERVIEW Timeline Visual timeline From the Top of the Mountain Transparencies John Napier Lesez Euler Portrait of Euler Frontispiece from Euler's Introductio in Analysin Infinitorum Contents of Euler's Introductio in Analysin Infinitorum Euler's thoughts on studying mathematics EXPONENTIAL FUNCTIONS Exponent Notation According to Newton Properties of Exponential Functions Properties of Exponents Compound Interest Activity: The Manhattan Purchase What Will Happen If ... Euler and Population Growth Logistic Growth Newton's Law of Cooling Euler and Exponential Functions: An Investigation Calculating ex Equations Using e 23 26 32 38 45 49 54 62 67 73 78 7 9 10 17 18 19 20 21 22 4 6
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LOGARITHMIC FUNCTIONS Introduction to Logarithms from Introductio in analysin infinitorum Development of Logarithms Using Sequences Calculation of Logarithms Using the Method of Napier and Briggs Constructing a Slide Rule Possible Influences on Logarithms Euler and Exponential Functions: An Investigation Revisited A Development of the Natural Base e Using Napier's Method An Application of Areas to Logarithms REFERENCES Bibliography Websites 135 137 82 92 102 109 113 118 121 127
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Introduction
The Principles and Standards for School Mathematics released in April 2000 by The National Council of Teachers of Mathematics, advise the reader that "the secondary school mathematics program must be both broad and deep" (p. 287). In light of this guidance, the study of Logarithmic and Exponential functions plays a very important role in secondary education, particularly in providing a way to model some of the real world problems that students meet outside of the classroom. Such problems are found in diverse places as medicine and finance, as well as, advanced mathematics and politics. More likely than not, if one asked those who studied mathematics in secondary school 25 years ago why they studied exponentials and logarithms, other than to learn about slide rule computation, many would say it was because the topic was a chapter in their textbook that had to be covered. One wonders why something that made a major impact on mathematics four centuries ago became so meaningless to the secondary students of the latter half of the 20th century, the "new math" era. Fortunately, textbooks which embody the Principles and Standards for School Mathematics today reward the student with the theory of exponents and logarithms along with the study of contemporary applications giving meaning to the study. In 1989, the Curriculum and Evaluation Standards for School Mathematics, which The Principles and Standards for School Mathematics were designed to build upon, were released. The document encouraged that attention be focused "on the need for student awareness of the interaction between mathematics and the historical situations from which it has developed and the impact that interaction has on or culture and our lives" (p.6). There is no doubt that the study of the history of exponents and logarithms is a prime example of how "theoretical mathematics (which) has burgeoned in its diversity and deepened in its complexity and abstraction...has become more ... vital to our technologically oriented society" (p.6). Goals of the Module: 1. To understand the historical background leading to the development of exponents, exponentials, and logarithms and to see how mathematical concepts evolve over periods of time. 2. To provide students with opportunities to apply their knowledge of exponents and logarithms to various concrete situations and problems in a historical context. 3. To show students the connection between exponential functions and logarithmic functions, that is the inverse relationship they share with one another. 4. To develop in students an appreciation of the history connected with the development of exponents and logarithms.
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How to use this module: The authors realize this module contains more material than most teachers can squeeze into an already overcrowded syllabus. It is recommended that the teacher review the material and pick and choose those topics that best fit teacher and students' needs. The material can be made to fit many different types of objectives, and it can be made to fit many different lesson plans. It may even be of interest to teachers of subjects other than mathematics. General historical material may be used for group or whole class discussion. Some quotations and biographical sketches may be more appropriate for introductory material while some are best used within activities. Students should be able to locate the places mentioned on a world map. The written assignments that are required in some of the activities could also include connections with units in social studies, science, economics, etc. If more comprehensive studies are required or desired, the authors have included a bibliography for students and teachers interested in the further study of content in the module. Time Schedule Time needed will depend on the activities and materials the individual teacher chooses to use and the level of the class. The activities may be used to introduce the topic and then further work may be accomplished using exercises found in student textbooks. The teacher is in no way obliged to ask students to perform all these activities; one is not constrained to follow the suggested sequence in which the activities are listed nor to follow the teacher hints. The authors believe that the history will enable students to understand the mathematics better. It is further suggested that the mathematics teacher team teach parts of this module with a history or social science teacher, if such an arrangement can be made. Materials Needed It is highly recommended that the teacher post a world map. Other materials which may be needed in these activities are: machine tape, straightedge, compass, ruler, lined paper, graphing paper, graphing calculator or graphing utility, computers. .Classroom Organization: When using the exploratory activities and projects of this module appears to be the appropriate time to have the class organized into cooperative groups. Using a constructivist approach, the lessons have been created with questions or hints that are designed to guide the student toward "discovering" the relevant results. The quotations of mathematicians may be best presented to the class as a whole on overhead transparencies; copies may be made for personal study by the students.
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Suggested Courses for Using Module Activities
Activity Exponent Notation According to Newton Properties of Exponential Functions Properties of Exponents Compound Interest Activity: The Manhattan Purchase What Will Happen If....? Euler and Population Growth Logistic Growth Newton's Law of Cooling Euler and Exponential Functions: An Investigation Calculating ex Equations using e Introduction to Logarithms from Introductio in analysin infinitorum Development of Logarithms Using Sequences Calculation of Logarithms Using the Method of Napier and Briggs Constructing a Slide Rule Possible Influences on Logarithms Euler and Exponential Functions: An Investigation Revisited A Development of the Natural Base e Using Napier's Method An Application of Areas to Logarithms Algebra 1 X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X Algebra 2 Precalculus/ Trigonometry
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Historical and Cultural Overview
A Time Line for Exponents and Logarithms
This is a brief listing of the contributions of various individuals to the development of exponents and logarithms. Diophantus (Greek living in Alexandria (?), c. 250) gave rules for multiplying and dividing exponents in Arithmetica Al-Karaji (Islamic, d. 1019) Formulated rules for exponents in al-Fakhri Al-Samaw'al (Islamic, c. 1125 1174) used negative exponents in al-Bahir (1144) Nicole Oresme (French, c. 1320 1382) discussed fractional and irrational exponents in Algorismus proportionum (c. 1360) Nicolas Chuquet (French, c. 1445 c. 1500) introduced exponential notation in Triparty (1484) Michael Stifel (German, 1487 1567) used the word `exponent' in Arithmetica integra (1544); correlated the terms of an arithmetic sequence with the terms of a geometric sequence; used negative exponents John Napier (Scottish, 1550 1617) began to work on the concept of logarithms about 1590; published the first description and table of logarithms in Mirifici logarithmorum canonis descriptio (1614); invented the word `logarithm' by combining the Greek words logos (ratio) and arithmos (number) Joost Brgi (Swiss, 1552 1630) independently of Napier also developed the equivalent concept of logarithms; published his table in 1620 in Progress tabulen Henry Briggs (British, 1561 1630) collaborated with Napier; wrote Arithmetica logarithmica (1624) giving values of logarithms base 10 for the integers from 1 to 20,000 and 90,000 to 100,000 to 14 decimal places William Oughtred (British, 1575 1660); using logarithms, designed the rectilinear slide rule around 1621, an early mechanical device used for computation Adriaan Vlacq (Dutch, 1600 c. 1666) wrote Het tweede deel van de Nieuwe telkonst (1627), giving the values of the logarithms of integers from 1 to 100,000 to 10 decimal places Ren Descartes (French, 1596 1650) gave the modern symbolization for exponents in La geometrie (1637)
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Alfonso Antonio de Sarasa (Belgian, 1618 1667) in Solutio problematis a Mersenno propositi (1649) gave the first identification of the logarithmic properties of the area bounded by the x-axis under the curve y = 1/x Nicolaus Mercator (Danish, c. 1619 1687) in Logarithmotechnia (1668) gave the first published method of computing logarithms using infinite series, based on the work of de Sarasa and Gregory of St. Vincent Isaac Newton (British, 1642 1727) in c. 1650 wrote the inherent calculus relationship between the natural logarithm and the area bounded by y = 1/x Gottfried Wilhelm Leibniz (German, 1646 1716) published Nova methodis (1684) stating the calculus relationship between the natural logarithm and the area bounded by y = 1/x Leonhard Euler (Swiss, 1707 1783) in Introductio in analysin infinitorum (1748) defined exponential functions and then logarithmic functions as their inverses
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Visual Timeline
0 200 400 600 800 1000 1200 1400 1500 1600
Diophantus (c. 250) rules for multiplication and division of numbers with exponents
Al-Karaji (c. 1000) rules for exponents Al-Samaw'al (c. 1144) negative exponents Oresme (c. 1350) fractional and irrational exponents Chuquet (1484) exponential notation Stifel (1544) negative exponents
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Napier (1614) table of logarithms Burgi (1620) table equivalent to logarithms Oughtred (1621) rectilinear slide rule Briggs (1624) Arithmetica logarithmica with Napier Descartes (1637) exponential symbolism de Sarasa (1649) logarithmic properties of the area bounded by the x-axis and the area under the curve y = 1/x Mercator (1668) computing logarithms using infinite series Newton (1670) calculus relationship between natural logarithms and de Sarasa's work Leibniz (1684) calculus relationship between natural logarithms and de Sarasa's work
Euler (1748) defined exponential functions and defined logarithmic functions as their inverses
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From the Top of the Mountain
It is an example of the truth that from the top of the mountain one can often see how the climb might have been made easier by deviations which to the climbers might well seem to be courting unnecessary difficulties. Lord Moulton, The Invention of Logarithms, Its Genesis and Growth (NTV, 21) The discovery of logarithms is attributed to the Scottish mathematician John Napier; in this section, you will be able to learn about Napier's life, how he formulated the first view of logarithms, and how they were quickly transformed during his own lifetime into something different. The temptation is severe to recast all of Napier's work through the lens of our present interpretation of logarithms, but the actual story is every bit as impressive. Moulton's quote above is useful to keep in mind when we consider the path which Napier took. John Napier was born into an influential and well-to-do family in 1550 in the family estate at Murchiston Castle near Edinburgh, Scotland as the eldest son of Archibald and Janet Napier. Before his death (also at Murchiston) in 1617, he was to become a theologian and an ardent supporter of the Protestant faith, an inventor of weaponry, and an exceptional mathematician. In mathematics, he was known for his work on spherical trigonometry, his invention of an early form of a mechanical calculator, and in particular, his discovery of and work with logarithms. It is also the case that Napier's name was spelled a variety of different ways, including by himself, among them being: Napeir, Nepeir, Neper, Nepper, Naper, Napare, and Naipper. We will adopt the spelling in ordinary usage at the present time. We really know quite little about Napier's early life. It seems he was a student at University of St. Andrews, but he never earned a degree there. He may have traveled abroad, but by 1571 he had returned to the family estates in Scotland where he remained for the rest of his life. He married Elizabeth Stirling in 1572, and after she died in 1579, he married Agnes Chisholm. He had a son and a daughter by his first wife and five sons and five daughters by his second. At this time in Scotland, religious strive was rampant. King James VI of Scotland had designs on the English throne, and there were rumors of the involvement of King Philip II of Spain in those plans. The issue of whether Scotland would become Catholic or Protestant became a widespread concern, and Napier came down very firmly on the side of the Protestants. Scotland suffered through a civil war in 1570-72, and in 1588 the English defeated the Spanish Armada. In 1593 Napier wrote the Plaine Discovery of the Whole Revelation of St. John; to gain a sense of this book, consider the following quote from a letter addressed to King James which formed the dedication: "...let it be your Majesty's continuall study...to reforme the universall enormities of your country, and first...to begin at your Majesty's owne house, familie and court, and purge the same of all suspicion of Papists and Atheists and Newtralls..." (NTV, 42) The book was widely read; between 1593 and 1645 there were 5 English editions, 2 Dutch editions, 5 French editions, and 3 in German. In it, among other things, Napier declared the Pope to be the Antichrist and predicted "that the day of God's judgement appears to fall betwixt the years of Christ 1688 and 1700."
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There is strong evidence that probably as early as 1594, Napier had begun to investigate certain properties of calculations which would make working with the trigonometric values of sines considerably easier to do in order to facilitate the computations involved in astronomy. The German mathematician Michael Stifel wrote the Arithmetica integra in 1544, and it contained a clear statement relating the arithmetic sequence of integers (1, 2, 3...) with the corresponding geometric sequence of 2 raised to those powers ( 2 1 , 2 2 , 2 3 , ...), and this may have suggested the possibility to Napier of using sums to compute products. Stifel even called the powers `exponents' and discussed what we would describe as "when you multiply the numbers, you add the exponents." Napier was also visited by Dr. John Craig, a member of the Scottish royal court, who met had the Danish astronomer Tycho Brahe in 1590 and seen some of Tycho's work with prosthaphaeresis. This was a long-known method of using "product to sum" formulas: namely, such trigonometric identities as 2 cos x cos y = cos(x + y) + cos(x y). Whatever the impetus, in 1614 Napier published the Mirifici Logarithmorum Canonis Descriptio (Description of the Wonderful Canon of Logarithms) and then posthumously in 1619 the Mirifici Logarithmorum Canonis Constructio (Construction of the Wonderful Canon of Logarithms) was published. It appears as if Napier wished his work to be as widely accessible as possible, and to this end he aided in Edward Wright's translation of the Descriptio into English in 1616. In this translation he explained that his intent with logarithms was to aid in the calculations involved in multiplication, division, and the taking of roots: Seeing there is nothing...that is so troublesome to mathematical practice, nor that doth more molest and hinder calculators, than the multiplication, division, square and cubical extractions of great numbers, which besides the tedious expense of time are for the most part subject to many slippery errors, I began therefore to consider in my mind by what certain and ready art I might remove those hindrances. Napier coined the word `logarithm' from the Greek words logos (ratio) and arithmos (number), and his original definition takes some getting used to, compared to our present interpretation of a logarithm. Keep in mind as well that he originally was working strictly with the sines of angles. In proposition 26 in the Constructio he states: "The logarithm of a given sine is that number which has increased arithmetically with the same velocity throughout as that with which radius began to decrease geometrically, and in the same time as radius has decreased to the given sine." (Napier, p. 19) T b g a S i
In the figure above, imagine that TS is a line segment of 10 7 units, and that bi is a ray from b. (Napier chose 10 7 for his radius so that he could use values of sines accurate to 7 decimal places but treat them as integers. It is also the case that Napier was one of the first mathematicians to widely use the decimal point as we now write it to separate the integer from the decimal fractional part of a number.) A point g starts at T and moves towards S at a rate
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equal to its distance from S; at the same time that g starts at T, point a starts at b and moves towards i at the same rate as g's initial velocity. The distance from b to a is then the logarithm of the length from g to S. There are some immediate conclusions that can be reached, stressing the difference from this original view of logarithms and our present perspective. Let Nap. log stand for Napier's definition. First, if g is still at T, then the point a is still at b; therefore, Nap. log 10 7 = 0 and Nap. log is a decreasing function. Second, it can be shown that Nap. log (xy) = Nap. log x + Nap. log y Nap. log 1. Recasting Napier's description in the language of differential equations and x solving, it is possible to demonstrate that Nap. log x = 10 7 log 1 7 ; however, it is critical to 10
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stress that Napier did not interpret his logarithms in anything approaching the modern sense of an exponent for some given base. It may be reassuring to us to solve a differential equation to obtain something more familiar, but Napier's vision clearly did not start there. From the top of the mountain, we can see that Napier's path and our more recent path of regarding logarithms as exponents clearly both get us to the top, but the paths themselves are very, very different. Napier regarded one of the essential properties of logarithms as "the logarithms of similarly proportioned sines are equidifferent." (Napier, p. 23) In other words, for a given k, log (kx) log x = log (ky) log y. This can clearly be considered, in light of the laws of logarithms ky kx which Napier himself knew, as equivalent to log = log . In another interesting side x y note, Napier's proposition 51 states: "All sines in the proportion of two to one have 6931469.22 for the difference of their logarithms." (Napier, p. 38) This statement is closely related to the fact that the natural logarithm of 2 is .693147180. Henry Briggs (1561 1630) was an English mathematician and professor of geometry at Gresham House, London at the time he met Napier; he went on to become the first Savilian professor of geometry at Merton College, Oxford. After having read the Descriptio, he went to Edinburgh in 1615 to meet with Napier. It is somewhat unusual in scientific history, but the two men immediately hit it off and in fact worked with each other until Napier's death in 1617. In 1617 with the publication of Logarithmorum chilias prima and in 1624 with Arithmetica logarithmica Briggs implemented the changes in Napier's logarithmic scheme that he and Napier had considered. Quoting from the preface to the Arithmetica logarithmica: That these logarithms differ from those which that illustrious man, the Baron of Merchiston, must not surprise you...concerning that matter I wrote immediately to the author himself...I journeyed to Edinburgh, where, being most hospitably received by him, I lingered for a whole month. But as we talked over the change in the logarithms he said that he had for some time been of the same opinion and had wished to accomplish it; he had however published those he had already prepared until he could construct more convenient ones if his affairs and his health would admit of it. But he was of the opinion that the change should be effected in this manner, that 0 should be the logarithm of unity and 10000000000
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that of the whole sine; which I could not but admit was by far the most convenient. (NTV, p. 126) It seems clear that Napier had already begun to consider a more accessible approach to his logarithms, and that Briggs' suggestions helped moved him in that direction. (As Napier himself observed at the end of his table of logarithms in the Constructio: "Nothing is perfect at birth." (Napier, p. 87.)) Since the logarithm of 1 becomes 0, this simplifies greatly the computational algorithms that Napier had already discovered, bringing them in line with our present theorems of logarithms. In the Constructio, an appendix was included "On the Construction of another and better kind of Logarithms, namely one in which the Logarithm of unity is 0." In this, Napier suggested but in fact did not do a revision of his own system, reflecting what he and Briggs had already determined would make computations with logarithms easier. Napier also observed in the Constructio that in this case, "if two numbers with known Logarithms be multiplied together, forming a third; the sum of their Logarithms will be the Logarithm of the third" (Napier, p. 50) (in other words, when you multiply, you add) and went on to state all the other familiar laws of logarithms which we use. He went ahead and then observed "whence the only difficulty is in finding the Logarithms of the prime numbers" (Napier, p. 51). Briggs went ahead with his own work after Napier's death, and the Logarithmorum chilias prima and the Arithmetica logarithmica contain what is for all intents and purposes the first table of common logarithms i.e., base 10 logarithms. One of the sections of this module discussed how to calculate logarithms somewhat in the method of Napier and Briggs. However, to get a sense of what they went through, consider the following. In constructing his tables, Napier started with the value of his radius 10,000,000 and began successive multiplications of that by 1-10 -7 ; in more modern terms, he generated the terms of a geometric sequence whose first term was 10 7 and whose ratio was 1-10 -7 . Moulton argues that in performing this process some 100 times, Napier became aware that decreasing a number by the same ratio each time was equivalent to repeatedly decreasing a line segment by the same proportional length, possibly giving rise to Napier's definition of a logarithm. It may have also been the case that through the sheer effort of this work, Napier eventually saw the utility of having (in modern terms) log 1 = 0 and log 10 = 1, but that he needed to go through the tremendous labor of the first construction to reach that conclusion. It is therefore tempting now to claim Napier must have therefore seen that he was constructing a system where the logarithm corresponded to the necessary power of 10 needed to obtain the given number, but that again is our view from the mountain, not Napier's. After working through the first 100 terms (his first table), Napier apparently became aware of the tremendous work left, so he changed to a geometric series whose first term was 10 7 and whose ratio was 1-10 -5 . He carried this out for 50 more terms, giving a second table. He then created two other series in a third table, one having ratio 1- 1/2000 and one having ratio 1 - 1/100. Using the data he had accumulated with his three tables, he was then able to calculate the logarithm values of numbers from 10 7 to almost exactly 5 x 10 6 . It is small wonder that Napier worked on his tables for 20 years before he published them! In the Logarithmorum chilias prima and the Arithmetica logarithmica, Briggs approached the problem differently. He began with log 10 = 1 and then began taking successive square roots, using well-known algorithms for the computations needed for the roots. (For
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example, if 10 = 3.162277, then log 3.162277 = .5.) Once Briggs obtained the logarithms of the primes, it was easy, using the laws of logarithms, to then generate the values of the logarithms of composite numbers. It may be somewhat astonishing to realize the work required by Briggs to obtain, in Arithmetica logarithmica, the logarithms of all natural numbers from 1 to 20,000 and 90,000 to 100,000, carried out to 14 places. (For example, in his calculations to obtain the logarithm of 2 he took 47 successive square roots.) Adriaan Vlacq a few years later completed Briggs' work and completed a table for all integers from 1 to 100,000 to 10 decimals places. (Until recently, these tables formed the basis of all modern logarithm tables.) Napier also used his idea to construct what were called "Napier's bones", a mechanical multiplication table. Within 7 years, William Oughtred used Napier's ideas to construct the first rectilinear slide rules, basically the first pocket calculator. Some two hundred years after Napier's publication, the French mathematician Pierre-Simon Laplace noted that "by shortening the labors, (Napier) doubled the life of the astronomer." Logarithms also present one of the instances when a given discovery must have been in the air; it now seems clear that roughly at the same time Napier was working on his own logarithms, the same essential concept occurred to the Swiss astronomer and watchmaker Joost Brgi (1552 - 1632). While the exact timing of Brgi's work is not entirely clear, he may have in fact worked on the concept before Napier did; however, it is the case that he did not publish his own tables (actually, his were tables of anti-logarithms) until 1620 in Arithmetische und geometrische Progress-Tabulen. Kepler's statement on the priority issue may be of some interest: "...(this) led the way for Justus Byrgius to the very same logarithms many years before Napier's edition, although this man, a procrastinator and guardian of his secrets, abandoned his child at birth and did not rear it for publicity." (Goldstine, p. 22) Brgi generated the values in his own table by starting with 10 8 and then multiplying by 1 + 10 -4 to obtain the successive terms. The ease with which the discovery of logarithms supported the massive calculations needed for astronomy and navigation generated intense interest among mathematicians. The work of Gregory of St. Vincent and his student A. A. de Sarasa by 1647 made the connection between the natural logarithm of a number a and the area bounded by the x-axis, the vertical a 1 lines x = 1 and x = a (for any a > 0) and the curve y = 1/x, what we would now call dx . x 1 (Natural logarithms are contrasted with common, or Briggsian logarithms by their base. If we use Euler's interpretation of a logarithm as the exponent of a given base, then natural logarithms use the number e = 2.718281828... as their base, while common logarithms use 10.) The Danish mathematician Nicolaus Mercator is usually credited with discovering the formula for the x2 x3 x4 expansion of the natural logarithm of x + 1 as the power series x - + - +... by 1668. 2 3 4 As early as 1664 Newton used this same idea to compute the value of the natural logarithm of 1.1 to 68 decimal places. The ground-breaking work of Newton and Leibniz in the calculus quickly suggested applications of logarithms very far indeed from Napier's original view of the technique.
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It turned out that the method that was developed to help with computational drudgery held a vital key in the new mathematics of calculus. The logarithm took on a life of its own, separate from the astonishing levels of sheer, brute arithmetic that it both replaced and which were required to establish the first logarithm tables. The logarithm continued to play a major role in computational sciences, but it had already begun to move away from its roots. As the 18th century progressed, mathematicians became increasingly aware of the necessary changes which calculus had brought to the techniques available to them, and also to the changes which were needed in their more basic structures in order to give the methods of calculus a concrete logical base. One of the more central concepts which evolved was that of a `function', namely the ability to regard an operation as the pairing of the number which was to be operated on with the number which resulted from the operation itself. In more abstract terms, it allowed mathematicians to regard the operation as an object itself, more divorced from the immediate numerical relationships. For example, while mathematicians had worked with the numerical relationships expressed in 2 2 = 4, 2 3 = 8, and so forth, they could now consider the relationship between the variables x and y in y = 2 x . This is an example of an `exponential' function, where the variable is the exponent of some positive base. Leonhard Euler, the enormously influential mathematician of the later 18th century, formulated the relationship between exponential functions and logarithmic functions when he defined log a b = c to be the case if and only if a c = b; therefore, y = a x was equivalent to the statement x = log a y. This was a long way indeed from Napier's original vision of a logarithm as involving the distances traveled by two moving points, yet it has proved to again be of critical use in many situations. It has provided a gateway into the solutions and analysis of many types of problems that would otherwise be extremely difficult to solve otherwise. An important observation therefore can be made that despite the view of many, both in the past and in present times, mathematics is indeed a changing science. A method that began as a strictly computational device had a significant impact on techniques of calculus, was featured in an important relationship of functions, and surfaced as an important problem solving technique, far from its beginnings. Until fairly recently, high school students versed in math and science were given a significant exposure to the computational methods which logarithms opened up and the mechanical devices which logarithms helped create. With the advent of inexpensive hand-held calculators, the computational roots of logarithms have become obsolete, yet the concepts which logarithms have led to continue to be of great utility. We need to be cautious whenever we decide a particular method or approach has reached a dead end; the history of logarithms provides an excellent example of how mathematics grows and develops new relationships for what may appear to be concepts and techniques which have been outgrown, but which can continue to provide important mathematics. It is often the case that when one finally does make it to the top of a mountain, after a long and difficult climb, that you find things up there very different from what you might have imagined. Vistas open up which are almost unimaginable until you reach the peak, and not only the path up but the ground below takes on a very different appearance. You can marvel at the difficulties of some paths and the ease of others, but it is difficult to not appreciate the sheer
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effort required along any path to make it to the top, and especially for those who pioneered a particular path. So too it is with the concepts and computations of logarithms; whether it be Napier, Brgi, Briggs, or any of the multitude who followed them, we must continue to remember and appreciate their contributions and exertions. Questions: 1. In Proposition 5 of the Constructio Napier writes "In numbers distinguished thus by a period in their midst, whatever is written after the period is a fraction, the denominator of which is unity with as many cyphers after it as there are figures after the period." (Napier, p. 8) What exactly does this appear to refer to? 2. In Proposition 52 Napier states "All sines in the proportion of ten to one will have 23025842.34 for the difference of their logarithms." (Napier, p. 38) What exactly is Napier saying in modern terms? 3. In an Appendix to the Constructio, Napier writes "Finally any common number being formed from other common numbers by multiplication, division, [raising to a power] or extraction [of a root]; its Logarithm is correspondingly formed from their Logarithms by addition, subtraction, multiplication, by 2, 3, & c. [or division by 2, 3, & c.]." (Napier, p. 51) Translate the meaning of this statement into more modern terms. 4. In 1617, Napier published the Rabdologi where he described the construction and use of what are now called `Napier's bones' or `Napier's rods'. Find out what these were and what they were used for.
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John Napier
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Lesez Euler, lesez Euler, c'est notre maitre a tous. Read Euler, read Euler! He is the master of us all.
Pierre-Simon Laplace 1749-1827
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Leonhard Euler
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Frontispiece from Euler's Introductio in Analysis Infinitorum
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Table of Contents from Euler's Introductio in Analysin Infinitorum
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I soon found an opportunity to be introduced to a famous professor Johann Bernoulli, whose good pleasure it was to advance me further in the mathematical sciences. True, he was very busy and so refused flatly to give me private lessons; but he gave me much more valuable advice to start reading more difficult mathematical books on my own and to study them as diligently as I could; if I cam across some obstacle of difficulty, I was given permission to visit him freely every Saturday afternoon and he kindly explained to me everything I could not understand, which happened with such greatly desired advantage that whenever he had obviated one difficulty for me, because of that ten others disappeared right away, and this undoubtedly, is the best method to succeed in mathematical subjects. Leonhard Euler Autobiography of 1767
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Exponent Notation According to Newton Teacher Notes
Level: This activity can be used in Algebra 1, and higher level courses Objective: This activity is designed to give an example of an original source of the introduction of exponents that Newton used at Cambridge University in England Materials: Historical Background for each student; Student Page for each student Time Frame: One class period How to Use: The teacher may want to distribute the Historical Background to the students the day before the activity is scheduled or present it as background on the day of the activity. The students may work in groups or individually. If the students need extra time to finish the problems, they could be assigned as homework. It is interesting to read the way that Newton introduced exponents and the terminology that he used. It is also interesting to note that this was part of a lecture Newton gave at Cambridge University - so your students are being exposed to the type of mathematics that was available to university students in the 1680's. An additional assignment would be to have the students write the complete passage in the background information using modern terminology. Bibliography: Newton, Isaac, Arithmetica Universalis, translated by D. T. Whiteside, Cambridge University Press, 1972.
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Exponent Notation According to Newton Solutions
Solutions to questions: 1. 2. If a is a quantity, "If a quantity should multiply itself," means a a. There will be various answers to this. Hopefully most will find it sufficient. The solution to a2b4 if a is 4 and b is 3 a2b4 = 42 x 34 = 4 4 3 3 x 3 3 = 1296 It means that the a is to multiplied together 3 times: a a a It means a a a. Since he is naming the powers, like a5 as cube-square, he is adding the powers. So a8 would be square-cube-cube or maybe square-square-square-square.
3. 4. 5.
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Exponent Notation According to Newton Student Page
Historical Background In Sir Isaac Newton's widely read Arithmetica Universalis, many beginning algebraic topics are included. One such topic is that of exponents. This activity comes directly from one of Newton's Arithmetica Universalis, which is a collection of the material covered in his lectures at Cambridge University in England in the 1680's. The following excerpts are from the 1972 translation by D. T. Whiteside from Newton's original Latin. t From page 61: If a quantity should multiply itself, for shortness the number of factors is usually appended. Thus in place of aaa we write a3, in place of aaaa we write a4, in place of aaaaa we write a5, and in place of aaabb we write a3bb or a3b2. If, say, a be 5 and b 2, then a3 will be 555 or 125, a4 will be 5555 or 625, and a3b2 will be 55522 or 500. Here note that a number written immediately between two variables relates always to the preceding one. Thus `3' in the quantity a3b2does not signify the b2 is to be taken 3 times, but denotes that a has twice to be multiplied into itself. Note also that these quantities are said to be of as many dimensions, powers or `dignities' as they comprise factors or quantities multiplying one another, and that the number appended is called the index of their power or dimension. Thus a2 is of two dimensions or powers, while a3, as its appended number `3' indicates is of three. Also a2 is said to be the square, a3 the cube, a4 the square-square (fourth power), a5 the square-cube (fifth power), a6 the cube-cube (sixth power), a7 the square-square-cube (seventh power) and so forth. Exercises 1. Explain what "If a quantity should multiply itself," means. 2. Is the explanation Sir Isaac Newton gave his students in the following enough for you to write terms using exponents? Give your reasons why or why not. Would you be able to evaluate a2b4 if a is 4 and b is 3? "Thus in place of aaa we write a3, in place of aaaa we write a4, in place of aaaaa we write a5, and in place of aaabb we write a3bb or a3b2. If, say, a be 5 and b 2, then a3 will be 555 or 125, a4 will be 5555 or 625, and a3b2 will be 55522 or 500." 3. According to Newton what does the `3' in the expression a3b2 mean? 4. What does the phrase "a has twice to be multiplied into itself" mean? 5. Using Newton's method of naming powers, how would he describe the eighth power of a, a8?
25
Properties of Exponential Functions Teacher Notes
This activity is a great way to give students the opportunity to graph and observe a variety of exponential functions and then summarize their findings. Students then compare and contrast their summaries with the one written by Leonhard Euler, the great eighteenth century mathematician. Level: Algebra 2 or Precalculus/Trigonometry Materials: Student Pages for each student; Calculator; Graphing calculator (opt.) Time Frame: One class period plus homework; 15 minutes the next class Objective: This activity is designed to enable students to develop most (if not all) of the properties of an exponential function and compare their list to the paragraph written by Euler. It is also designed to give students the opportunity of gaining the satisfaction of creating mathematical ideas similar to those of a well-known mathematician How to Use: The teacher may want to give out only the first two pages and do Exercise B with the students as a class exercise, then assign Exercise A and C-E to be completed as homework. The next day the teacher may hand out the third page and ask students to do Exercise F. Bibliography: Euler, Leonhard. Introduction to Analysis of the Infinite Book 1. Translated by John D. Blanton. Springer-Verlag, 1988.
26
Properties of Exponential Functions Solutions
A. z -3 -2 -1 0 1 2 2z 1/8 1/4 1/2 1 2 4
y
z -3 -2 -1 0 1 2
3z 1/27 1/9 1/3 1 3 9
y
z
z
B. Z
1 2 8 4 2 1 1/2 1/4
z
z
1 3
z
-3 -2 -1 0 1 2
-3 -2 -1 0 1 2 v
27 9 3 1 1/3 9 v
z
z
27
C. z -3 -2 -1 0 1 2
1z 1 1 1 1 1 1
y
z -3 -2 -1 0 1 2
1.1z .75 .83 .91 1 1.1 1.21
y
z
z
D. Answers will vary, but will be similar to the ones in A through C above. Students might choose negative or zero values for a. If they do, this is a good time to discuss why these values give very strange graphs, graphs which we will usually avoid. In general, in precalculus and calculus classes we generally want to talk just about continuous functions and functions whose inverses are also functions. E. Answers will vary, but will include some of the following observations. All the graphs intersect at the point (0,1). None of the graphs has negative y-values. If a > 0, as z increases, y increases. As a gets larger, the values of y increase faster as z increases. If a < 1, as z increases, y decreases. If a = 1, then y = 1; i.e. the function is constant. If a < 0, the graph jumps back and forth between positive and negative values for y as z increases. F. Answers will vary. Many students will find that their answers include some, if not all, of Euler's observations.
28
Properties of Exponential Functions Student Pages
Leonhard Euler wrote the first easy-to-read precalculus book in 1748. First, see how you might figure out properties of exponential functions; then compare your summary to his. Look at a general exponential function, y = a z , and see what happens to the graph as the value of a changes. Use the same scale for all graphs. Also, you may want to use the integers from -3 to 2 for the z-values. The equation uses z instead of x because that is what Euler used. A. Graph y = 2 z and y = 3z using a table of values.
z y z y
B. Also, graph y = (2-
z z y
1 z
)
1 1 1 z = and y = (3- ) = . 2 3
z y
z
z
29
C. Graph y = 1z and also y = (1.1)z.
z
y
z
y
D. Graph two other exponential equations of the form y = a z to enhance your knowledge of this type of graph. State the equations you are graphing. y=
z y
y=
z y
E. Compare the graphs in section A with those in B, C, and D. Summarize what you notice.
F. Compare what you have written with the description Euler writes concisely below.
30
Euler gives a magnificent and clear description of an exponential function in his precalculus book, Introduction to Analysis of the Infinite (1748). The values of the exponential a z depend primarily on the magnitude of the constant a . If a = 1 , then we always have a z = 1 , no matter what value is given z to z . If a > 1 , then a will have a greater value if the value of z is greater than it was originally and as z goes to infinity, so also a z increases to infinity. If z = 0 , then a z = 1 ; if z < 0 , then the values of a z become less than 1 and as z goes to z - , a goes to 0. On the other hand if a < 1 but still positive, then the values of a z decrease when z increases above 0. The exponential [ a z ] increases as z 1 increases in the negative direction. Since when a < 1 , we have < 1 , and if we a 1 -z z let = b , then a = b . For this reason we can examine the case when a < 1 a from the case when a > 1 .
31
Properties of Exponents Teacher Notes
This activity gives students the opportunity to look at how a master mathematician expressed several ideas about exponents in his great precalculus book, Introduction to Analysis of the Infinite, published in 1748. Level: Algebra 2 or Precalculus/Trigonometry, possibly honors Algebra 1 Objective: To afford students the chance to read a master mathematician's own words about exponents Materials: Student Pages for each student; Problem #8 if desired Time Frame: 20 minutes How to Use: The teacher may want to give this activity as an extension of what has been done in Algebra 1 or as an introduction to rational exponents or as a follow-up activity. The problem 8 is tough; many excellent students will find it a challenge and interesting. Bibliography: Euler, Leonhard. Introduction to Analysis of the Infinite Book 1. Translated by John D. Blanton. Springer-Verlag, 1988 N. Chuquet. in The History of Mathematics, a Reader, edited by John Fauvel and Jeremy Gray, The Open University, 1987, pp. 247-249 from Nicolas Chuquet, Renaissance Mathematician, Reidel, 1985, pp. 144, 151-153 tr. and ed. by H. G. Flagg, C. M. Hay, B. Moss. Newton, Isaac. Epistola Prior (1676) in The History of Mathematics, a Reader, edited by John Fauvel and Jeremy Gray, The Open University, 1987, p. 403 from H. W. Turnbull (ed.), The Mathematical Correspondence of Isaac Newton, II. Cambridge, 1960, pp. 332-333.
32
Properties of Exponents Solutions
1. 2.
3
a2 a
4
3. a
3
1 4
= (a 3 ) 4 = 4 a 3
1
1 2 4. a number between 4 and 4 , that is, between 4 and 16. 2 3 5. a number between 2 and 2 , that is, between 4 and 8. 0 1 6. a number between 32 and 32 , that is, between 1 and 32. 1 2 7. a number between 9 and 9 , that is, between 9 and 81. 8a. The topic is exponents. 8b. Proportional numbers are powers of the number. For example if the number is a, then the proportional numbers are a2, a3, a4, a5, etc. The denomination of a number is its exponent when it is expressed as a power of a. 8c. 1, 4, 16, 64, 256 since 40 = 1, etc. 8d. To Chuquet 42 means the number corresponding to denomination 2 in the list. To us it means 4 x 4 = 16. 8e. It tells how to multiply numbers with the same base, in this case 2, by adding their denominations (meaning our exponents). 8f. Chuquet's secret is the same as our answer to 8e.
33
Properties of Exponents Student Pages
What would happen if exponents were numbers other than natural numbers? Islamic mathematicians of the 11th century were the earliest people to use negative numbers as exponents. Nicole Oresme, a French mathematician associated with the University of Paris around 1350, had some ideas about fractional exponents and even tried to explain what it would mean to have an irrational number as an exponent. In 1655, John Wallis, an English mathematician, gave a clearer explanation of fractional exponents and a few years later, Isaac Newton was able to make use of them with ease. Here, however, we look at the explanation of exponents of various types by the prolific mathematician Leonhard Euler in 1748: Let the exponential to be considered be a where a is a constant and the exponent z is a variable. Since the exponent z stands for all determined numbers, it is clear at least that all positive integers can be substituted for z to give determined 1 2 3 4 5 6 values for a , a , a , a , a , a , etc. If for z we substitute the negative integers -1, -2, -3, etc. we obtain
z
1 1 1 1 1 1 , , , , , , etc. If z = 0, a1 a2 a 3 a4 a5 a6 then we have a0 = 1. If we substitute a fraction for z , for instance 1 1 2 1 3 , , , , , etc., we obtain 2 3 3 4 4
the values more values, since the extraction of roots gives several values. However, we will consider only their primary [now called principal] values, since
1 2 1 3 a, a 3 , a 3 , a 4 , a 4 , etc. These symbols can have two or
a a and 3 a ... In like manner we let z take irrational values, even though it is more 7 difficult to understand this concept. ... Thus a has a value which lies between 2 3 a and a ...
from Introduction to Analysis of the Infinite by Leonhard Euler Some statements in that selection probably look more familiar than other statements. For example, in the first sentence Euler talks about negative exponents. In the second sentence he
1 tells about square roots so that we understand a 2
z
5 2 lies between is to be thought of as a ... function. For this reason. a
2
= a
1 and by implication, a 3
= 3 a , etc.
We can use this to help us rewrite expressions with rational exponents as expressions with radicals instead. Complete the following using Euler's idea to rewrite the expressions using radicals.
34
2 1 1 2 3 = a 3 = a 2 3 = _________________ 1. a
( )
1 2. a 4
= _________________
3 3. a 4 =__________________= __________________=__________________
The sixth and seventh sentences refer to situations such as the fact that a given positive number has two different square roots, a positive one and a negative one. But since we want to define an exponential function, we will consider only the positive value.
5 lies between the two 2 7 successive integers 2 and 3. Finally, Euler also suggest that one can estimate the value of a when one knows that 7 lies between the two successive integers 2 and 3.
Use this method to estimate the following by stating the numbers between which the given number lies. Example. 4
5 2 3 2 = _______________________________________________________________ 4. 4 2 3 = a number between 4 and 4 , that is, between 16 and 64.
5 2 , since The next sentence suggests a method to estimate the value of a
5. 2 6. 32
5
= ______________________________________________________________
4 5 = ______________________________________________________________
3 7. 9 2 = ________________________________________________________________
Clearly this last observation by Euler does not give very precise estimates, but it is insightful!
8. Familiar, Yet Different, Yet Related. You are a mathematics historian; you even consider yourself a detective of sorts! As you read the selection below, think about these questions: What familiar concept is being discussed? The notation looks familiar, but wait, does it have the same meaning? Or is it a little different?
35
Nicolas Chuquet, a mathematician who taught in Lyons, France, wrote these words sometime during the fifteenth century. He invented the notation, but its meaning has evolved somewhat since then. To understand the reason why denomination of number is added to denomination, ..., it is necessary to set down several proportional numbers beginning with 1 and arranged in a continuous sequence, like 1, 2, 4, 8, 16, 32, etc. or 3, 9, 27, etc. Numbers 1 2 4 8 16 32 64 128 256 512 1024 2048 4096 8192 16384 32768 65536 131072 262144 524288 1048576 Denomination 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Now it is necessary to know that 1 represents and is in the place of numbers, whose denomination is 0, 2 represents [...] the first terms, whose denomination is 1. 4 holds the place of the second terms, whose denomination is 2. And 8 is in the place of the third terms, 16 holds the place of the fourth terms, 32 represents the fifth terms, and so for the others. Now whoever multiplies 1 by 1, it comes to 1, and because 1 multiplied by 1 does not change at all, neither does any other number when it is multiplied by 1 increase or diminish, and for this consideration, whoever multiplies a number by a number, it comes to a number, whose denomination is 0. And whoever adds 0 to 0 makes 0. Afterwards, whoever multiplies 2, which is the first number by 1, which is a number, the multiplication comes to 2; then afterwards, whoever adds their denominations, which are 0 and 1, it makes 1; thus the multiplication comes to 21. And from this it comes that when one multiplies numbers by first terms or vice versa, it comes to first terms. Also whoever multiplies 21 by 21, it comes to 4 which is a second number. Thus the multiplication amounts to 42. For 2 multiplied by 2 makes 4 when adding the
36
denominations, that is 1 with 1, makes 2. And from this it comes that whoever multiplies first terms by first terms, it comes to second terms. Likewise whoever multiplies 21 by 42, it comes to 83. For 2 multiplied by 4 and 1 added with 2 makes 83. And thus whoever multiplies first terms by second terms, it comes to third terms. Also, whoever multiplies 42 by 42, it comes to 16 which is a fourth number, and for this reason whoever multiplies second terms by second terms, it comes to fourth terms. Likewise whoever multiplies 4 which is a second number by 8 which is a third number makes 32 which is a fifth number. And thus whoever multiplies second terms by third terms or vice versa, it comes to fifth terms. And third terms by fourth terms comes to 7th terms, and fourth terms by fourth terms, it comes to 8th terms, and so for the others. In this discussion there is manifest a secret which is in the proportional numbers. It is that whoever multiplies a proportional number by itself, it comes to the number of the double of its denomination, as, whoever multiplies 8 which is a third number by itself, it comes to 64 which is a sixth. And 16 which is a fourth number multiplied by itself should come to 256, which is an eighth. And whoever multiplies 128 which is the 7th proportional by 512 which is the 9th, it should come to 65536 which is the 16th. from N. Chuquet, pp. 144, 151-153 a. Why is this essay included in this activity? In other words, what is the topic?
b. What does Chuquet mean by "proportional numbers?" "denomination of a number?"
c. What are the first five "proportional numbers" starting with 4?
d. Chuquet employs what we call exponent notation, but has it mean something different. What does 42 mean according to Chuquet?
What does 42 mean to us?
e. In the last sentence of this selection from Chuquet there is a description of a familiar property of exponents. State the property. f. What is Chuquet's "secret?"
37
Compound Interest Activity: The Manhattan Purchase Teacher Notes
Introduction: Compound interest is a useful application of exponential functions. Compound interest holds much promise for applying mathematical ideas to relatively important everyday applications, such as the amount of money an investor makes on a particular investment, a notion Americans are familiar with living in a capitalist society. The need to understand compound interest also arises in calculating the interest that accrues on credit card purchases, car, home, and boat loans, and financial investments. In the following activity, a historical example is used as a springboard to discuss compound interest. The historical context is centered on the European colonist's purchase of the island of Manhattan from the Native Americans. Level: Algebra I, Algebra II, Precalculus (Exposure to the compound interest formula is a prerequisite.) Objective: To investigate an application of exponential growth, compound interest, using an situation taken from the historical context of the Dutch "purchase" of the island of Manhattan from the Native Americans. Materials: Student Page; Calculator; Graphing calculator and Internet access (optional) Time Frame: Two 45-minute class periods; extended project would be done outside of class How to Use: To begin the discussion of exponential growth and compound interest embedded in a historical context, provide students with the historical opener on an overhead transparency (see Transparency #1). Have students read and reflect on this historical situation. You might ask students to develop their own questions that arise from the historical situation presented. Be prepared to consider social and historical questions as well as mathematical related questions. A second option would be to use the guiding questions in the student activity to develop the lesson. One way to begin the lesson is to look at question 1 involving compound interest. First, it might be helpful to review the general compound interest formula. You might also show students the derivation for the general compound interest formula. It might be helpful to derive the compound interest formula using an example first. Then show the students the formal algebraic derivation for the compound interest formula. Consider the following example involving compound interest. You have $100. You want to invest your money in a fund that guarantees 8% interest compounded quarterly over six years. You have the following given information: r = 8% = 0.08 annual interest rate t=6 time in years
38
n=4 number of compounding periods per year nt = 24 number of compounding periods P = $100 principal amount of money being invested S n = total amount of money at the end of n periods Because we are looking for the total amount of money at the end of 24 periods, we want to find S 24 . Guide the students through the derivation using the above given information. The total amount of money we have at the end of the first compounding period is 0.08 S1 = 100 + 100 = 100(1 + 0.02) . 4 The total amount of money we have at the end of the second compounding period is 0.08 2 S 2 = S1 + S1 = S1 (1 + 0.02) = 100(1 + 0.02)(1 + 0.02) = 100(1 + 0.02) . 4 The total amount of money we have at the end of the third compounding period is 0.08 2 3 S3 = S 2 + S2 = S 2 (1 + 0.02) = 100(1 + 0.02) (1 + 0.02) = 100(1 + 0.02) . 4 Explain to your students that we are looking for S 24 ; that is, the total amount of money we have accumulated at the end of the 24th period. Next guide students to the algebraic derivation of the general compound interest. We have the following given information: r = annual interest rate t = time in years n = number of compounding periods per year nt = total number of compounding periods P = principal amount of money being invested S n = total amount of money at the end of n periods r = interest rate per compounding period n nt = total number of compounding periods. The total amount of money we have at the end of the first compounding period is r r S 1 = P + P = P 1 + n n
39
The total amount of money we have at the end of the second compounding period is r r r r r r r S 2 = S1 + S1 = P1 + + P1 + = P1 + 1 + = P1 + . n n n n n n n The total amount of money we have at the end of the nt -th compounding period is S nt = P1 + r . n
nt 2
Using the derivation of the general compound interest formula, lead students to the exponential equation that describes the situation described in the historical opener. Let P = 24, r = 7%, n = 1 (compounded annually) and t = 372 (the number of years since 1626 this is being calculated for the year 1998) r y = S nt = P1 + n 0.07 y = 241 + 1
nt
(1)( 372 )
y = 24(1.07) 372 $2,046,334,096,410
Guide your students to discover that the initial formula for y represents an exponential growth function of the form, f ( x) = ab x where a is the initial amount of money, b is the growth factor, and x is the total number of compounding periods. Stress that because b > 1 in compound interest problems, then the function is an example of exponential growth. Guide students to change a and b in f(x) = abx to see the effect on the curve. Encourage students to examine the function for a > 0, a < 0, b >1, and 0 < b < 1 and to write a complete explanation of the results they found. Have students solve additional examples of exponential growth problems either using variations on the historical opener problem or problems from a textbook. Give students an opportunity to create their own problems and to explain them to the class.
40
Compound Interest Activity: The Manhattan Purchase Solutions
1. As indicated above, y = 24(1.07) 372 $2,046,334,096,410 2. The 3.8 billion dollars is less than the amount found in number 1, so perhaps, Minuit paid too much, or the interest rate is too high, or the tax assessment value is too low. 3. Minuit should have paid a great deal more. Land in the Old World was dear and certainly, an equivalent amount of land, in say Holland or England, would have been very expensive. But given that the amount found in number 1 is higher than the tax valuation in number 2, find a payment that would give the tax valuation. This time solve for the initial amount given that the final amount is 3.8 billion dollars and the interest rate is 7% with 372 years (this solution is being done in 1998) with annual compounding yields: 3.8 10 = a(1.07)
9 372
3.8 10 9 a= .044567 or about 4.5cents. (1.07) 372
4. Have students find the interest rate in 1626 via the Internet. The Federal Reserve Board holds past and present interest rates. 5. f ( x) = ab x where a is the initial amount of money, b is the growth factor, and x is the total number of compounding periods. 6. The students' explanation should include cases such as: a > 0, b > 1 exponential growth function a > 0, 0 < b < 1 exponential decay function a < 0, b > 1 exponential decay function a < 0, 0 < b < 1 exponential growth function
41
Compound Interest Activity: The Manhattan Purchase Student Page
Historical Opener In 1625, the Dutch Colony of New Netherland occupied sites in the present states of New York and New Jersey. Peter Minuit (1580-1638) was officially designated as the general director of the colony in 1626. At this time, Minuit was instructed from the West India Company to move the inhabitants, approximately 200 Native Americans, to the southern tip of the island of Manhattan. Minuit proceeded to strike a land deal with the Native Americans. Minuit negotiated with the local Native American sachems, the chief of a confederation of the Algonquian tribes of the north Atlantic coast, for the purchase of the island. After the final negotiations were agreed upon, Minuit purchased the island of Manhattan (New York City) by giving them trinkets valued at approximately 60 guilders (currency of the Netherlands), or $24. 1. Suppose that the Native Americans invested the money they received from the purchase of the land at 7% compounded annually. How much money would they have today? 2. Now consider that the current tax assessment value of Manhattan is 3.8 billion dollars. Discuss the implications of the land deal based on this information. 3. Based on question two, what might have been a fair price for Minuit to pay? 4. Find the actual interest rate at the time of the purchase in 1626. Using this interest rate, calculate the amount of money the Native Americans would have today based on that interest rate. 5. What is the general equation of an exponential function? 6. What effect do the variables have on the graph of the function?
42
Historical Opener The Manhattan Purchase In 1625, the Dutch Colony of New Netherland occupied sites in the present states of New York and New Jersey. Peter Minuit (1580-1638) was officially designated as the general director of the colony in 1626. At this time, Minuit was instructed from the West India Company to move the inhabitants, approximately 200 Native Americans, to the southern tip of the island of Manhattan. Minuit proceeded to strike a land deal with the Native Americans. Minuit negotiated with the local Native American sachems--the chief of a confederation of the Algonquian tribes of the north Atlantic coast--for the purchase of the island. After the final negotiations were agreed upon, Minuit purchased the island of Manhattan (New York City) by giving them trinkets valued at approximately 60 guilders (currency of the Netherlands), or $24.
Transparency #1
43
Suppose the natives invested the money they received from the purchase of the land at 7% compounded annually. How much money would they have today?
Transparency #2
44
What Will Happen If....? Teacher Notes
Introduction: This activity helps give students exposure to the number 2.71828 in a context with which they are already familiar, to make e more plausible to students later. Level: Algebra 2 or Precalculus/Trigonometry (prerequisite: the Compound Interest Activity from earlier in this module or work with the compound interest formula from a textbook) Objective: To relate 2.71828 to the compound interest formula thereby setting the stage for e to be discussed next Time Frame: 10 minutes in class, overnight to research the bonus question Materials: Student Pages for each student; calculator; graphing calculator (opt.) How to Use: The teacher may want to use this activity as an introduction to e . The bonus question should be discussed after the students have a chance to answer it if they choose to do so.
45
What Will Happen If...? Solutions
1. N 1 2 3 4 5 10 50 100 1000 10,000 100,000 1,000,000 10,000,000 100,000,000 2 2.25 2.37037 2.44241 2.48832 2.59374 2.69159 2.70481 2.71692 2.71815 2.71827 2.71828 2.71828 2.71828 S
2. Increases (but at an ever slower rate) 3. S seems to get closer and closer to 2.71828, rounded. [Note: some of your students may try x = 10,000,000,000,000, which yields S = 2.7606 or x = 100,000,000,000,000, which yields S = 1; these are aberrations of the calculator. For example, what is happening with x = 1014 is that the calculator evaluates the expression 1 1 + 1014
14
and rounds
1 1014
to zero first, then adds 1 and raises the sum to the 14th power.]
Bonus: The number approaches e as n increases without bound.
46
What Will Happen If...? Student Pages
What will happen if we use some very special values in the equation we developed for compound interest? r nt where S = amount of money after t years S = P 1 +
n
P = principal r = annual interest n = number of periods per year t = number of years Substitute interest rate = 100% = 1 time = 1 year principal = $1.
Then, S = 1 1 +
1 n1 = 1 + n
1 n . n
1. Create a table of values for n and the corresponding values of S. (Hint: You may use your calculator or graphing utility to find the values of S to five or six decimal places.) n 1 2 3 4 5 10 50 100 1000 10,000 100,000 1,000,000 S
47
2. As n increases, the value of S ________________________(increases or decreases). 3. What do you think will happen to the value of S as n approaches infinity?
Bonus: The answer to #3 is a very famous number. Research its name. [Hint: the first person to name it is Leonhard Euler.]
48
Euler and Population Growth Teacher Notes
Introduction: This activity highlights a problem about population growth, which was taken from Introductio in analysin infinitorum (Introduction to Analysis of the Infinite) (1748). This famous mathematics book was written by Leonhard Euler (1707-1783) in the middle of the eighteenth century. The fact that this problem appears in Euler's Introductio in analysin infinitorum shows that issues of population growth were of concern to the people living in the eighteenth century. Euler initially solved the problem using logarithms. However, this activity is designed to solve the problem using exponential functions first. Later, when logarithms and logarithmic functions are developed in the module, you can show students how to solve this problem using logarithms. Showing students how to solve this problem using both exponential and logarithmic functions provides a wonderful opportunity to reveal the inherent connection between exponential and logarithmic functions. Level: Algebra I, Algebra II, Precalculus (Prerequisites: Exposure to exponential functions and their graphs) Objectives: To develop students' understanding of exponential growth functions by using a problem of population growth taken from one of Euler's most famous works, Introductio in analysin infinitorum, which was written in the middle eighteenth century. To show students that mathematics was and is done in specific historical and social contexts. Materials: Student Pages; Calculator; Graphing calculator and Internet access (optional) Time Frame: One 45-minute class period. The extended project is to be completed outside of class. How to Use: Consider using cooperative groups to do this activity. At the very least, consider pairing up students to work on this activity. If students have not had much work with exponential functions, you might need to help them find the exponential function that represents the problem set forth by Euler. This activity can be completed without a graphing calculator. However, more exploration of the exponential function and its curve can occur if students have access to graphing calculators or a graphing utility. Bibliography: Euler, Leonhard. Introduction to Analysis of the Infinite, Book 1, translated. by John D. Blanton (New York: Springer-Verlag, 1988). pp. 75-91. Katz, Victor. A History of Mathematics: An Introduction (New York: HarperCollins, 1998).
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Euler and Population Growth Solutions
1.
y = f ( x) = ab x 1 y = 100,0001 + 30 2. Because we are dealing with people, stress to students that they need to round to the nearest whole number.
x 10 20 30 40 50 60 70 80 90 100 y = 100,000(1+1/30)x 138,805 192,668 267,432 371,208 515,255 715,198 992,729 1,377,955 1,912,668 2,654,874
x
3. a. The graph should represent an exponential growth function. b. If the initial population is increased or decreased, the population is still increasing by the one-thirtieth chosen by Euler. If the initial population is greater than the 100,000, the population after 100 years will be greater than 2,654,874; If the initial population is less than the 100,000, the population after 100 years will be less than 2,654,874. c. If the growth rate is larger than one-thirtieth a year, the population after 100 years will be greater than 2,654,874; if the growth rate is smaller than one-thirtieth a year, the population after 100 years will be less than 2,654,874. d. The relation will now be g(x) = 100,000(1 1/30)x. And the population after 100 years will be g(100) = 100,000(1 1/30)100 3370. Extended Project: One place to start is Euler's biography found at http://www.dcs.warwick.ac.uk/bshm/resources.html#biog
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Euler and Population Growth Student Pages
Introductio in analysin infinitorum (1748) Example II on page 85 Leonhard Euler (1707-1783) If the population in a certain region increases annually by one-thirtieth and at one time there were 100,000 inhabitants, we would like to know the population after 100 years.
The above population growth problem was taken from Introductio in analysin infinitorum (Introduction to Analysis of the Infinite) (1748), a famous mathematics book written by Leonhard Euler (1707-1783. The fact that this problem appears in Euler's Introductio in analysin infinitorum shows that issues of population growth were of concern to the people living in the eighteenth century. Can you think of reasons why population growth would have been an issue in the eighteenth century? Euler initially solved the problem using logarithms. However, you will solve the problem using exponential functions first. Later, when you learn about logarithms and logarithmic functions, you can solve this problem using logarithms. 1. Based on your knowledge of exponential functions, write the relation described in the above problem as an exponential function. 2. For the exponential function you found in problem 1, make a table of values for years 10, 20, 30, ...100. (If possible, use the table menu of a graphing calculator) x 10 20 30 40 50 60 70 80 90 100 f(x) = ?
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3. Graph the exponential function. Explore its curve.
4. Answer the following questions based on your exploration of the curve. a. What kind of exponential function does this function represent? b. What happens if the initial population is increased or decreased? c. What happens if the growth rate is increased or decreased? d. What happens if the population decreases by one-thirtieth each year? Extended Project: Discover where Euler was living in 1748. Determine the population of that area during the time he lived there. After you determine the population in 1748, determine today's population of that area based on the above exponential model. Compare with the actual number of people living there now. How close was your approximation?
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If the population in a certain region increases annually by one-thirtieth and at one time there were 100,000 inhabitants, we would like to know the population after 100 years.
Introductio in analysin infinitorum (1748) Example II on p. 85 Leonhard Euler(1707-1783)
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Logistic Growth Teacher Notes
Introduction: This teacher-guided activity serves as a superb way to show students what can happen as time goes by and more and more data becomes available to use in making predictions. It helps explain why a mathematical model may need to change with the passing decades. Level: Algebra 2 or Precalculus/Trigonometry (Prerequisites: Using various regression equations on a graphing utility to describe data) Objective: To study various regression equations culminating in the need for the logistic equation to model real-world data Materials: Overhead projector; transparency made from the Population Data; Graphing calculators with regression equation capabilities, including the logistic function Time Frame: One class period How to Use: This activity works very well as a teacher-guided discussion with the everexpanding data set being modeled by successive functions. It helps the students see that as more and more data is available a model may need to change in order to make better predictions. 1. Perhaps begin by asking students which statistics about New York City might be valuable and why. Lead them to realize that population trends might prove useful in city planning in such areas as number and size of schools, sewage treatment and garbage collection, road construction and maintenance. 2. Using the historical information in the lesson discuss past efforts in history to solve similar prediction problems. 3. Ask students to look at the initial data set and imagine that they are city planners for NYC after the 1820 census and that they need to predict the population for the next few decades. Have them enter the data into lists on their calculators or computers, make a scatterplot of the data and fit an appropriate regression function to the data. Then have them predict the population for the next decade. 4. Ask students to pretend that a couple decades have passed and ask if their prediction was good for 1840. Now have them add this data to their previous lists, make another scatterplot incorporating all the data from 1790 through 1880, choose an appropriate regression function and predict the population for the next decade. Some students may try a quadratic regression function. It actually appears to fit the data better than the exponential function. You may want to tell them that fitting the data is not the only criterion for choosing a particular regression equation. For example, the quadratic function would imply that at first the population of NYC would decrease during the first decade under consideration. While that is
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possible, it doesn't fit the real data that is ever increasing. They may point out that the exponential function overestimates the population in the final decade of this data set. That will point the way to needing a different regression function, a suspicion that is confirmed by the next data set. 5. For the third data set, after plotting additional points, students may suggest that it is composed of several parts, such as exponential for the first third, quadratic for the second third, and logarithmic for the final third. You can then suggest that this S-shape is characteristic of the logistic function and students may use their TI-83 graphing calculators to fit a regression function to the data. 6. For the final data set, after plotting the additional points, students may decide it is best described as a cubic function, but you may want to point out that a cubic equation would have predicted a negative population for the first three decades of the 1800's. 7. A discussion of possible reasons why the population growth rate slows down in the final set is a lively finale to this activity. Students may bring up such possibilities as population density, insufficient water/sewage treatment plants, insufficient medical services for a greater population, not enough roads or public transportation, not enough schools to support a greater population. Bibliography: John L. Harper, Population Biology of Plants, Academic Press, 1977 Thomas Malthus, An essay on the principle of population, etc., Penguin, 1985 Raymond Pearl and Lowell Reed, On the rate of growth of the population of the United States since 1790 and its mathematical representation, Proceedings of the National Academy of Sciences, 6 (6): 275-288, 1920. Bonnie Shulman, Using Original Sources to Teach the Logistic Equation, The UMAP Journal, 18 (4): 377-402, 1997 _____________, Math Alive! Using Original Sources to Teach Mathematics in Social Context, PRIMUS, VIII (1): 1-13, 1998 U. S. Bureau of the Census, Population at Selected Ranks from 1st to 100th of the Largest Urban Places: 1790 1990, 1998
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Logistic Growth Student Pages
Overview:
Using census data on the population of New York City from 1790 until 1990, we will consider three different modeling techniques and see how well they fit the given data. We will also consider the entire concept of using mathematics to model a real-world situation. An important use of mathematics is its ability to take current numerical information, or data, and then make predictions of future behavior based on that data. A useful field of investigation is that of population growth, with its obvious impact on city planning, environmental concerns, allocation of resources, and many other areas. Data from previous years can indicate important trends; social scientists, economists, and many other professionals have discovered that one can also use data to predict future growth patterns. We will use data from the U. S. Bureau of the Census on the population of New York City to investigate three different predictive models. One simple possibility for population growth is that the growth is constant; in other words, that the population grows by the same amount each year. The population can then be depicted mathematically with an "arithmetic" sequence, a sequence of numbers where any two successive terms differ by the same amount. An example of an arithmetic sequence is 1, 5, 9, 13, 17... A more complicated model uses the premise that the growth is proportional to the present population; in other words, the larger the population, the more people are added each year. In this situation, we often say that the growth rate, say r%, is constant. The population in this case can be represented by a "geometric" sequence, one where any two successive terms have the same ratio. An example of a geometric sequence is 1, 2, 4, 8, 16... A third model uses the idea that the growth of the population will be proportional to two factors: the current population and also how close the population gets to some ceiling or maximal size. This model implies that initial growth may be faster, but as the constraints of resources come into play, then the growth will slow down. The full mathematical development of these three models requires techniques of calculus, and therefore we will not dwell unnecessarily on the methods used to obtain the equations. However, we can briefly summarize the central mathematical conclusions. In the first model the population y will be a linear function of time t, that is, the population y will be given by the familiar formula y = mt + b. The second model results in an exponential model of growth, in which the population y is expressed by y = y 0 e rt where y 0 is some given initial value, r is the rate, t is the time and e is the familiar 2.718... The third model results in the logistic model of c , where for the additional variables, c represents the maximal population growth, y = 1 + ae ( - bt ) which can be sustained (called the "carrying capacity") and a and b are constants which emerge from a particular set of data.
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Let's consider the use of these three models with the same overall set of data, restricted to different time periods. The U. S. Bureau of the Census contains a myriad of information about populations of various segments of the United States from 1790 until the present. (The Constitution of the United States orders that a count of the population be held every 10 years in order to correctly apportion the seats in the House of Representatives; this raw data can then be used in variety of ways to gain interesting information about patterns of growth in the U. S.) One such set of information is the population of New York City as measured every 10 years in the Census. For the period of 1790 to 1820, the Census gives the following counts: Year Population 1790 33,131 1800 60,515 1810 96,373 1820 123,706
Using your graphing calculator, plot those four points using the Year as the x-coordinate and the Population as the y-coordinate and a suitable viewing rectangle; you should obtain something which resembles Chart 1. It should be pretty clear that these four data points look like they almost fall on a line; you can use the capabilities of your particular calculator to discover the regression equation line which will fit the data which gives you an answer of y = 3075.83x 5,473,441.9. Graph that equation over the set of data to observe how good a fit it is. However, you test a model by not only how well it fits given data but also how well it will predict the future. The count for 1830 was 202,589, and the count for 1840 was 312,710. According to the equation above you should expect the count for 1830 to be 155,327 and the count for 1840 to be 186,085; clearly, the mathematical model is at odds with the real data. The linear model does not account for the complexity of the information and significantly undercounts the real population. Let's consider more of the set of data next; to the data given above for 1790 to 1820, expand the data set to include the following through 1880: Year Pop. 1830 202,589 1840 312,710 1850 515,547 1860 813,669 1870 942,292 1880 1,206,299
Graphing these in a suitable viewing rectangle should yield something resembling Chart 2. This graph should look like the graph of an exponential function, examples of which you have already seen. You can again use the capabilities of your calculator to obtain an exponential function that fits the data, yielding y = (11162 10 -27 )e .040578 x . Graph that function over the data set, and . again note the fit. Using our previous approach, the census count for 1890 was 1,515,301 and the count for 1900 was 3,437,202; our equation predicts that the figure for 1890 should be 2,265,258 and for 1900 it should be 3,398.969. The exponential equation fits the given data well, but again it is less useful when we try to expand it to cover a broader scope. Now we will consider more data; we expand the data set to include the population through 1950:
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1890 1,515,301
1900 3,437,202
1910 4,766,883
1920 5,620,048
1930 6,930,446
1940 7,454,995
1950 7,891,957
Graph the population from 1790 through 1950, and note how the graph initially increases very rapidly, but then levels off and increases much more slowly. Compare your graph to Chart 3. This `S' shape is highly suggestive of the model of logistic growth. If your calculator is advanced enough, you can use the set of data from 1790 to 1950 to obtain a logistic equation that 8,592,944.078 models the information; if not use y = . Again, map this function 1 + (12224 10 50 )e ( -.060438 x ) . over the data set, and observe the fit. To get a better sense of the implication of such a model, extend the domain of your graph to 2050 and the range to 9,000,000 and now consider it. Notice how the curve flattens out. What is the implication of such a graph? Now, let's consider the full range of data through the 1990 census. Include in your data the following: Year Population 1960 7,781,984 1970 7,894,862 1980 7,071,639 1990 7,322,564
The graph of the full set of data is in Chart 4; make sure your graph is similar. Note the strong fit of the curve through the data of 1950, and then how the curve fails to model the data, particularly in 1980 and 1990. Some conclusions: The first matter for us to consider here is the history of some of these models. The simplest concept of growth is one that results in an arithmetic pattern; early scientific consideration of population growth of crops and domesticated animals demonstrated this model was not sufficient to explain the observed growth. Thomas Malthus (1766 1834) was an English clergyman and political economist who wrote An Essay on the Principle of Population in 1798. In there, he wrote that "population, when unchecked, increases in a geometrical ratio." When this idea was investigated mathematically, it resulted in the exponential model expressed above. Pierre Francois Verhulst (1804 1849) was a Belgian mathematician and statistician who is credited with first writing about the logistic model of growth in 1845, when he attempted to take into account factors, such as limited resources, that would also tend to slow down growth after time. (Our logistic equation is written in the form y = f(t); Verhulst wrote it in the form t = h(y). This form requires the use of logarithms, and apparently he coined the word `logistic' to suggest the connection between the curve and the characteristic shape of a logarithm.) Next, we need to consider the whole concept of a model in and of itself. First, one point is to recognize that while a mathematical model may fit a particular set of data, it is a somewhat dangerous perspective to feel that the model will always work. We considered three models linear, exponential, and then logistic which did a reasonable job with one set of data, but then became less accurate as the data encompassed a larger time frame. There is a danger in assuming that we can ever exactly describe a living situation with a mathematical model. There are
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techniques of chaos theory that we could call on to account for the fluctuations in the data we considered in the periods of time following 1950, but what is to say that the future data will fit this model as well? It may be that all we can ever hope for is a rough description of a living entity without ever attaining any sort of permanent exactness. Second, mathematics expands to generate more sophisticated techniques as more data is made available. Malthus had a reasonable perspective, given his time; Verhulst developed a better one, but not necessarily because he was somehow smarter he had access to more information. Complicated models require equally complicated mathematics even to obtain approximate correlation with real-world data. We can even incorporate the new field of chaos theory to further refine our model. Third, we need to be careful about assuming that just because we can describe a set of data mathematically, we have somehow discovered what is creating the data. The Americans Raymond Pearl and Lowell Reed independently rediscovered Verhulst's analysis of population growth in 1920. They cautioned: "No process of empirically graduating raw data with a curve can in and of itself demonstrate the fundamental law which causes the occurring change." In other words, do not mistake an ability to describe a set of data with a mathematical formula with understanding what is creating the data itself. It may be that the mathematics helps us to see some underlying principles, but the ability to extrapolate the future from a set of data is at best an inexact procedure. However, in the article which they wrote to discuss the mathematics of logistic growth and its application to population modeling, Reed and Pearl concluded: "...the hypothesis here advanced as to the law of population growth [that of logistic growth]...so closely describes the known facts regarding the past history of that growth, as to make it potentially profitable to continue the mathematical development and refinement of the hypothesis further." It may still be very useful for us to study the particular model, realizing that we still do not have a full mathematical description of the phenomena being studied. In an article appearing in the March, 1998 edition of the journal Primus on the logistic curve, Bonnie Shulman warns that while the difference between the mathematics of a model and the underlying dynamics may be clear to mathematicians, "important policy decisions are made based on the predictions arising from models." It is perfectly understandable that we would use mathematical models to try to predict the future, but those predictions must always be taken with a sense of caution. Biologists have studied population growth of various species extensively, and as such have repeatedly come across examples of logistic growth. Even at that, in Population Biology of Plants, John L. Harper states: "The logistic curve remains central to population biology, not because it describes how populations behave but rather because it provides a standard base of ideal behavior against which the reality can be judged and measured." (Harper, p. 3.)
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Chart 1:
150000 100000 50000 0 1790 1795 1800
1790-1820
1805 year
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1815
1820 Series1
Chart 2:
1400000 1200000 1000000 800000 600000 400000 200000 0 1780 1790 1800 1810 1820
1790-1880
Series1
1830 year
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Chart 3:
8000000 7000000 6000000 5000000 4000000 3000000 2000000 1000000 0 1750 1800
1790-1950
Series1
1850 year
1900
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1790 1800 1810 1820 1830 1840 1850 1860 1870 1880 1890 1900 1910 1920
33131 60515 96373 123706 202589 312710 515547 813669 942292 1206299 1515301 3437212 4766883 5620084
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Newton's Law of Cooling Teacher Notes
Introduction: In this activity students read a little history behind Newton's Law of Cooling as well as using it. An experiment using the CBL to collect data and then mathematics to analyze it could be incorporated here as well; many books, such as the Texas Instruments book by Chris Brueningsen relating mathematics and science using the CBL collection device, are available. Level: Algebra 2 or Precalculus (Prerequisites: Exponential and logarithmic functions; Solving exponential and logarithmic equations; Best fit curves (optional) Objective: The goal of this activity is to give students an opportunity to learn the history behind Newton's Law of Cooling and then to use this law in several exercises. Materials: Graphing calculator; Internet access (optional) Time Frame: This activity can be completed in one class period with the exercises started in the class, the rest to be assigned as homework. The amount of time needed for this activity may be decreased or increased depending on the mathematical level of your students. How to Use: The teacher may want to discuss the history of Newton's Law of Cooling with the class after the students have had time to read the information. It might be necessary to do several examples of solving exponential and logarithmic equations before assigning this activity. If additional activities are desired, the teacher should visit the cited websites (or others that may be available) to determine the appropriateness for his/her class. Additionally, if the teacher decides to run an experiment using CBL equipment, there would need to be additional preparation on the part of the teacher. Bibliography: 1. Edwards, G. H. and David E. Penney, Elementary Differential Equations with Applications, Third Edition, Prentice Hall, 1994. 2. French, A. P., "Isaac Newton's Thermometry", The Physics Teacher, Vol 31, April 1993, pp208 211. 3. French, A. P., "Newton's Thermometry: The Role of Radiation", The Physics Teacher, Vol 31, May 1993, pp310 312. 4. http://www.southwestern.edu/~richards/cool.html July 1999 5. http://faculty.millikin.edu/~jaskill.nsm.faculty.mu/newtcool.html July 1999 6. Smith, Karl J, Precalculus Mathematics, A Functional Approach, Fourth Edition, Brooks/Cole Publishing Company, 1990. 7. Stewart, James, Calculus, Early Transcendentals, Third Edition, Brooks/Cole Publishing Company, 1995. 8. Zill, Dennis G, A First Course in Differential Equations with Applications, Fourth Edition, PWS-Kent, 1989.
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Newton's Law of Cooling Solutions
1. a. Substituting T0 = 340, T(1) = 100, and Tm = 70 into Newton's Law of Cooling gives: T = Tm + (T0 - Tm )e kt 100 = 70 + (340 - 70)e k (1) 30 = 270e k 30 = ek 270 1 1 = e k ln = k . 9 9 1 k can be left in this form, or it can be evaluated : k = ln -2.1972 9 kt ln(1/9)t ln(1/9) t t b. First notice that e = e = (e ) = (1/9) . Now substitute the values into Newton's Law of Cooling T = Tm + (T0 - Tm )e kt 1 80 = 70 + (340 - 70) 9 1 10 = 270 9 10 1 = 270 9
t t t
3 1 1 1 1 = = 3 = 2t t = hours. 2 27 9 3 3 2. This time the Tm is the unknown. T0 = 200, T(10) = 150 and k = -0.05. Substituting into Newton's Law of Cooling yields: T = Tm + (T0 - Tm )e kt 150 = Tm + (200 - Tm )e -0.5(10) 150 = Tm + (200 - Tm )e -0.5 Multiply both sides by e0.5 . 150e0.5 = Tm e0.5 + 200 - Tm
t
3
2t
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150e0.5 + 200 = Tm (e0.5 - 1) 150e0.5 + 200 Tm = 72.92 e0.5 - 1 3. a. Using the points (0,0) and (34,100), the slope, m, of the line is m= 100 - 0 50 = . 34 - 0 17
The equation of the line relating the linseed oil temperatures to the centigrade temperatures is 50 C = N 17 b. If we substitute into the previous formula, we get C = 50 (1 2 ) 3 5 .2 9 17
This value is lower than the 37 C that we use. c. Again, substituting the given value for N gives us 50 C= (71) 208.82 17 Newton's value is lower than the actual value. d. There are several possible explanations. One is that Newton's values were approximations; another is that the relationship between the linseed oil temperatures and the Celsius temperatures is not linear. A suggested additional activity would be to use the three points (0,0), (12, 37) and (34, 100) in a different model a best fit line or a quadratic equation, for example. And then see how the new model works with the melting point of tin.
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Newton's Law of Cooling Student Pages
History
Besides discovering calculus, Newton experimented in chemistry and alchemy. One of the Newton results discovered is what is now called Newton's Law of Cooling. He wrote anonymously "A Scale of the Degrees of Heat" in the 1701 volume of the Philosophical Transactions of the Royal Society of London. Due to Newton's fame, Roger Cotes easily deduced that Newton was the author of the anonymous paper. Using a linseed oil thermometer, because linseed oil has a high boiling point of 240 C, Newton studied the melting points of various metals and alloys. Newton devised a method for measuring temperatures of up to 1000 C. his law of cooling. Newton's Law of Cooling states that the rate at which the temperature T(t) changes in a cooling body is proportional to the difference between the temperature in the body and constant temperature Tm of the surrounding medium. If one writes this statement as a differential equation and solves, one gets Newton's Law of Cooling in the form T = Tm + (T0 - Tm)ekt where T0 is the initial temperature of the heated body at time t = 0 and where k is a constant which depends on the item being cooled. Notice that this equation is an example of exponential decay decay since k turns out to be negative because the body's temperature is decreasing.
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Exercises: 1. When a cake is removed from the oven, it is 340. One hour later, it has cooled to 100. If the temperature of the room is 70, a. find the constant k in Newton's Law of Cooling; b. find how long will it take for the cake to cool to 80? 2. Suppose the constant k Newton's Law of Cooling -0.05. If the temperature of a cup of coffee is 200 when it is poured and 150 10 minutes later, how warm is the room? 3. According to Newton's paper, "A Scale of the Degrees of Heat", he chose the following temperatures using his linseed oil thermometer: the freezing temperature of water was 0 and the boiling point of water was 34. The Celsius temperature for the freezing of water is 0 and the boiling point of water is 100. a. Find the linear equation that represents the relationship between the linseed oil temperatures and the Celsius temperatures. Let the linseed oil temperatures be the independent variable. b. Newton said that body temperature using his linseed oil thermometer was 12. Using the linear relationship you found in part a, find the Celsius temperature for body temperature. How does this compare with the 37 C that we use? c. Newton found that the melting point of tin was 71. Using the linear relationship you found in part a, find the Celsius temperature for the melting point of tin. According to the Smithsonian Physical Tables, the melting point of tin is 232 C. How does this compare with Newton's value? d. From the two examples you just did, how would you explain the discrepancies between the linseed oil thermometer values that Newton obtained, and the Celsius values that we use?
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Euler and Exponential Functions: An Investigation Teacher Notes
Introduction: This activity highlights a mathematical problem taken from Leonhard Euler's Introductio in analysin infinitorum of 1748. Level: Algebra I, Algebra II, Precalculus (Prerequisites: Exposure to exponential functions and their graphs) Objective: To develop students' understanding of exponential functions by using a problem taken from one of Euler's most famous works, Introductio in analysin infinitorum. Materials: Student Pages; Calculator; Graphing calculator (optional) Time Frame: One 45-minute class period How to Use: Consider using cooperative groups to do this activity. At the very least, consider pairing up students to work on this activity. If students have not had much work with exponential functions, you might need to help them find the function that represents the problem set forth by Euler. This activity can be completed without a graphing calculator. However, more exploration of the exponential function and its curve can occur if students have access to graphing calculators or a graphing utility. Have students consider the following historical problem taken from Leonhard Euler's highly influential pre-calculus book, Introductio in Analysin Infinitorum, written in 1748: If the progression 2, 4, 16, 256, ... is formed by letting each term be the square of the preceding term, find the value of the twenty-fifth term. This example comes from Chapter VI: On Exponentials and Logarithms of Euler's book.. Euler initially solved the problem using logarithms. We simply want to graph the curve that models this progression. It is preferable to have students graph this curve using a graphing calculator. If there is no access to graphing calculators, have students make a table of values and graph the exponential function on graph paper. A suggested window for graphing is x: (0, 10) and y: (0, 300). Students must first determine the function that models this progression. The students should arrive at
y = f ( x) = 2 2 .
x
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A sample table the students may make is as follows:
x 0 1 2 3 4 5 6 7
y = f ( x) = 2 2
0 1
x
2 2 = 21 = 2 22 = 22 = 4 2 2 2 = 2 4 = 16 2 2 = 2 8 = 256 2 2 = 216 = 65,536 2 2 = 2 32 = 4,294,967,296 2 2 = 2 64 2 2 = 2128
7 6 5 4 3
At this time, students should find the twenty-fifth term is
) f (25 = 22 = 216,777,216.
Students will quickly discover this expression cannot be evaluated on a calculator. It is important to stress that Euler solved the problem using logarithms--a topic that will be discussed later and connected with exponential functions. This example is a prelude to logarithmic functions--the inverse of exponential functions. Once the students have discovered the function and curve that modeled Euler's original problem, you can discuss the relationship between this function and the standard exponential function
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y = f ( x) = 2 x.
Then have the students make a comparison between the two functions. Note: Euler's original solution with logarithms to this example is provided (See Logarithmic Functions: Euler and Exponential Functions: An Investigation Revisited). Later in the unit when you introduce logarithms, it would be appropriate to revisit this problem and show students its connection to logarithms and Euler's logarithmic solution. This will make students aware of the inverse relationship between exponential and logarithmic functions. Bibliography: Euler, Leonhard, translated. by John D. Blanton. Introduction to Analysis of the Infinite, Book 1, pp. 75-91 (New York: Springer-Verlag, 1988). Katz, Victor. A History of Mathematics: An Introduction (New York: HarperCollins, 1998).
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Euler and Exponential Functions: An Investigation Solutions
1.
x 0 1 2 3 4 5 6 7
The graph of f(x) looks like this:
y = f ( x) = 2 2
0 1
x
2 2 = 21 = 2 22 = 22 = 4 2 2 2 = 2 4 = 16 2 2 = 2 8 = 256 2 2 = 216 = 65,536 2 2 = 2 32 = 4,294,967,296 2 2 = 2 64 2 2 = 2128
7 6 5 4 3
The graphing window is x: (0,4) and y: (0, 200). 2. The curve starts off relatively flat but increases very rapidly. Students will notice that the curve shoots up quickly around x = 3. 3. The twenty-fifth term is
) f (25 = 22 = 216,777,216.
4. f(25) cannot be evaluated on a standard calculator because it is a very large number. 5. f ( x) = 2 2
x
24
6. Student answers will vary.
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Euler and Exponential Functions: An Investigation Student Pages
Consider the following historical problem taken from Leonhard Euler's highly influential precalculus book, Introductio in Analysin Infinitorum, which was written in 1748. This example comes from Chapter VI: On Exponentials and Logarithms as seen on page 91 of the original manuscript.
Introductio in Analysin Infinitorum Chapter VI On Exponentials and Logarithms Example on page 91 Leonhard Euler (1707-1783)
If the progression 2, 4, 16, 256, ... is formed by letting each term be the square of the preceding term, find the value of the twenty-fifth term.
1. Graph the curve that models the progression of numbers. First, make a table of values. Then, using your graphing calculator or manually plotting values, graph the curve that models the progression of numbers as seen in Euler's pre-calculus book, Introductio in Analysin Infinitorum.
x
y = f ( x) = ?
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Graph of f(x):
2. Describe what the curve looks like?
3. Based on the table of values, identify the twenty-fifth term?
4. What problem do you encounter when evaluating the 25th term on your calculator?
5. Write a general equation for this function. f(x) = ?
6. See if you can find other functions that look like the curve you graphed.
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Introductio in Analysin Infinitorum(1748) Chapter VI On Exponentials and Logarithms Leonhard Euler(1707-1783)
Example on page 91
If the progression 2, 4, 16, 256, ... is formed by letting each term be the square of the preceding term, find the value of the twenty-fifth term.
Transparency
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Calculating ex Teacher Notes
Introduction: We present a slight modification of Euler's method for calculating e x as a power series, a generalized polynomial. Level: Precalculus. A prerequisite is the Binomial Theorem and the calculation of e as in the section "What Will Happen If..." Objective: To show how to calculate the power series for ex Time Frame: One or two class periods, with a homework assignment at the end. How to Use: This is a teacher-directed lesson. The Student Pages should be given to the students, but the teacher will need to present this material and answer questions about it.
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Calculating ex Student Pages
We present here a method of calculating ex by means of a power series, a polynomial of "infinite degree." We first recall the Binomial Theorem: (a + b) n = a n + na n -1b + n(n - 1) n- 2 2 n(n - 1)n - 2) n -3 3 a b + a b + 1 2 1 2 3 + bn .
You should make up several examples of this for practice. In particular, note that the formula is somewhat simpler when a = 1. We then have (1 + b) n = 1 + nb + n(n - 1) 2 n(n - 1)(n - 2) 3 b + b + 1 2 1 2 3 + bn .
Next, recall the result, discussed in an earlier section of this module, that e can be found by considering the expression 1 1 + n
n
and letting n get larger and larger. Today, we generally write this as 1 e = lim 1 + . n n This means simply that as we let n get larger and larger, the given expression gets closer and closer to a particular number, which we have called e. Euler, however, writing before the notion of limit had been explained, would simply write 1 e = 1 + , n where n is an infinitely large number. Today, we do not generally use such concepts, and probably Euler himself knew that such numbers did not really exist. But what he meant by the phrase "infinitely large number" was simply a very very very large number; he used his intuition then to determine what properties such very very very large numbers would have. We will see some of these properties as we work through his argument.
n n
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1 Since e = 1 + , where n is infinitely large, we can also calculate e r . First, we note n that if n is infinitely large and r is a finite quantity, then n/r is also infinitely large. We can therefore rewrite e as 1 e = 1 + n/r It follows that
n/r 1 . e = 1 + n / r r r n/r
n
.
By one of the basic rules for exponents, we can rewrite this in the form 1 e = 1 + . n/r
r n
But we know that
1 r = .. We therefore have, finally, n/r n r er = 1 + , n
n
again where n is infinitely large. r . Of n course, we know that the binomial theorem is only valid for finite values of n, but we will assume that it is also true when n is infinitely large. After all, we are thinking of such values n simply as very very very large positive integers. However, since we do not know how large n actually is, we will only calculate the first few terms of the binomial expansion and use dots to indicate that this expression simply goes on for a long long time. Technically, this "infinite polynomial" is called a power series. We now apply the binomial theorem to this last expression, taking a = 1 and b = We get
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r e = 1 + n
r
n
r n(n - 1) r n(n - 1)(n - 2) r n(n - 1)(n - 2)(n - 3) r = 1+ n + + + + 1 2 n 1 2 3 1 2 3 4 n n n = 1+ r + n(n - 1)r 2 n(n - 1)(n - 2)r 3 n(n - 1)(n - 2)(n - 3)r 4 + + + 1 2n 2 1 2 3n3 1 2 3 4n 4
2
3
4
We want to simplify the various terms of this expression. So we have to use a property of ``infinitely large numbers.'' In particular, we note that if n is infinitely large, subtracting 1 n -1 from n will have no appreciable result. Thus = 1 . Similarly, if we subtract 2 or 3 or indeed n any finite number from n, this will not really change that infinitely large number. We therefore n-k = 1 for any finite number k. What this means, then, is that in the expression above, have n the quantities n, n 1, n 2, n 3, ... in the numerators always "cancel" the factors of n in the denominators. It follows that our expression reduces to er = 1 + r + r2 r3 r4 + + + 1 2 1 2 3 1 2 3 4 ( k - 1) k is usually called
You may know that if k is any integer, the expression 1 2 3
r
k factorial and is written k! Thus, our expression for e , usually called its power series expansion, is written as er = 1 + r + r2 r3 r4 + + + 2! 3! 4!
The value of the power series expansion for Euler was that it enabled him to calculate values for e r . In particular, he could calculate the value for e1 = e . We can do it as well, by substituting r = 1 into the series. We get e = 1+1+ 1 1 1 1 1 1 1 + + + + + + + 2! 3! 4! 5! 6! 7! 8! 1 1 1 1 1 1 1 = 1+1+ + + + + + + + 2 6 24 120 720 5040 40320 = 1 + 1 + 0.5 + 0.166666667 + 0.041666667 + 0.008333333 + 0.001388889 + 0.000198413 + 0.000024802 + 2.718278771 2.71828
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Your calculator can, of course, calculate e as well as e r , for any r in a fraction of a second. You may wonder how it does it. Although different calculators use different procedures for doing this, one of the ways your calculator may work is, in fact, by using the series for e r ,. Your calculator is basically hard-wired to perform additions and multiplications; since calculating with a power series just involves these operations, all that has to happen is that the calculator must perform very many of these operations very quickly. And that is precisely what it can do. Although a normal graphing calculator can probably only calculate e to 10 decimal places, it has been calculated to far more places by a computer. The first 100 decimal places are as follows: e 2.71828 18284 59045 23535 02874 71352 66249 77572 47093 69995 95749 66967 62772 40766 30353 54759 45713 82178 52516 64274 Like another important number , the number e is irrational and even transcendental. The first statement was proved by Euler in 1737. That e is transcendental, that is, that it cannot be a solution to a polynomial equation with integer coefficients, was proved in 1873 by Charles Hermite. Interestingly, these two important numbers are closely related, as we will see in the next activity. Exercises: 1. Use your calculator to determine the decimal form of the next three terms in the series for e, namely, 1/9, 1/10, and 1/11. Add these values to the value already calculated and check that you now have e calculated correctly to 8 decimal places. 2. Use the power series for e r to calculate e0.1. Namely, replace r by 0.1 in that series and evaluate using your calculator. Check that using the terms through the fifth power term gives you accuracy to 8 decimal places. 3. Use the power series for e r to calculate e0.1 and compare with the answer given directly by your calculator. 4. Use the power series for e r to calculate e1 and compare with the answer given directly by your calculator. How many terms do you need to use to get e1 to 8-place accuracy?
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Equations using e Teacher Notes
Introduction: Students are intrigued by e and want to know more about where it is used. Here are two equations that help to satisfy their curiosity. Level: Algebra 2 or Precalculus/Trigonometry (Prerequisites: discussion of e) Objective: to introduce students to two equations using e Materials: Student Pages for each student Time Frame: 20 minutes How to Use: The teacher may want to use this short activity to amplify students' work involving the transcendental number e Bibliography: Kasner, Edward and Newman, James, Mathematics and the Imagination. New York: Simon and Shuster, 1940. Maor, Eli. e: The Story of a Number. Princeton University Press. 1994.
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Equations using e Solutions
1. Answers may vary. Many students will answer parabola. 2. e is the basis of natural logarithms and is used in many real-life applications, such as population growth or radioactive decay. i is the -1 which is used for complex numbers and in many real-life applications, such as electricity. is used, for example, in the area formula for a circle and volume formulas for cylinders, cones and circles. 1 is the identity element for multiplication, just to name one reason it is special. 0 is a place holder, just to name one reason it is special.
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Equations using e Student Pages
1. Hanging link chains and two brothers Imagine a chain made of identical links, hanging from the two endpoints. You've probably seen such a curve as a barrier around a grassy plot or the wire suspended from telephone poles along a highway. Sketch below the shape of the curve formed by the chain. What shape does the curve look like? ________________________________
You are not alone in thinking about this. Over three hundred years ago two brothers of the famous Bernoulli family of mathematicians tried to figure out the equation that described it. In May, 1691, Jakob Bernoulli stated the problem in a scientific journal, Acta Eruditorum. A year later his younger brother Johann as well as two other famous mathematicians, Christian Huygens and Gottfried Wilhelm Leibniz, correctly solved this problem. In fact Jakob Bernoulli was not able to solve it; he and his brother Johann had a strong rivalry as shown in this letter from Johann to a friend. "You say that my brother proposed this problem; that is true, but does it follow that he had a solution of it then? Not at all. When he proposed this problem at my suggestion (for I was the first to think of it), neither the one nor the other of us was able to solve it; we despaired of it as insoluble, until Mr. Leibniz gave notice to the public in the Leipzig journal of 1690, p. 360, that he had solved the problem but did not publish his solution, so as to give time to other analysts, and it was this that encouraged us, my brother and me, to apply ourselves afresh. The efforts of my brother were without success; for my part, I was more fortunate, for I found the skill (I say it without boasting, why should I conceal the truth?) to solve it in full.... The next morning, filled with Joy, I ran to my brother, who was still struggling miserably with this Gordian knot without getting anywhere, always thinking like Galileo that the catenary was a parabola. Stop! Stop! I say to him, don't torture yourself anymore to try to prove the identity of the catenary with the parabola, since it is entirely false." quoted from e the Story of a Number, p. 141 Actually this curve is called a catenary from the Latin word catena meaning chain. The equation e ax + e - ax that describes the catenary is y = where a is a constant that is determined by the 2a density of the chain and the tension as it is held. Calculus is required to prove that this is correct.
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2. The Most Beautiful Equation
e i + 1 = 0
Many consider this to be one of the most beautiful equations ever discovered since it incorporates five of the special numbers of mathematics. It was discovered by Leonhard Euler, a mathematician who studied and contributed to many different branches of mathematics. Why is each of the five numbers special in mathematics?
e _______________________________________________________________ i _______________________________________________________________ _______________________________________________________________
1________________________________________________________________ 0________________________________________________________________ To Benjamin Pierce, one of Harvard's leading mathematicians in the nineteenth century, this formula of Euler came as something of a revelation. Having discovered it one day, he turned to his students and said: "Gentlemen, that is surely true, it is absolutely paradoxical; we cannot understand it, and we don't know what it means. But we have proved it, and therefore we know it must be the truth." (Kasner and Newman, pp. 103-104)
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Introduction to Logarithms Teacher Notes
Level: Algebra 2 or Precalculus/Trigonometry properties of exponents) (Prerequisites: Properties of logarithms,
Objective: To use original sources dealing with logarithms Materials: Transparency of Napier's justification for logarithms. Copy of Paragraphs 102 - 107 from Euler's Introductio in analysin infinitorum for each student; Student Pages for each student Time Frame: 1 to 3 class periods, depending on how used How to Use: The transparency can be used when you first introduce logarithms. Then, consider the excerpt from Euler's Introductio in analysin infinitorum (1748). In these paragraphs, Euler introduces all the information about logarithms that we usually cover in our modern day courses - in two pages! Be sure to find out if the students think that this is a sufficient amount of explanation for understanding. In the student Activity that follows, the students will read and then do examples demonstrating Euler's explanations on logarithms. It would be good after the students finish the activity to point out that there probably is enough information about logarithms here for mastering the basics of logarithmic functions. You may want to use each separate paragraph of the Student Activity on different days, depending on your planned lesson, or the whole Student Activity after you have completed logarithms. Euler uses "log n" to mean "log base a of n." Bibliography: Euler, Leonhard. Introduction to Analysis of the Infinite Book1. Translated by John D. Blanton. Springer Verlag, 1988
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Seeing there is nothing...that is so troublesome to mathematical practice, nor that doth more molest and hinder calculators, than the multiplication, division, square and cubical extractions of great numbers, which besides the tedious expense of time are for the most part subject to many slippery errors, I began therefore to consider in my mind by what certain and ready art I might remove those hindrances. John Napier, A Description of the Admirable Table of Logarithms, 1616
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Introduction to Logarithms Solutions
Part 1 1. If az = y, then z = log y. 2. Yes 3. "a" is the base of the logarithm and it must be a number greater than 1. 4. The domain is positive real numbers, if the logarithm is to be real also. Part 2 5. If y = 1 in az = y, then az = 1 which implies that z = 0. But a0 = 1 corresponds to loga 1 = 0. 6. By definition, a1 = a loga a= 1, a2 = a2 loga a2 = 2, ..., an = an loga an = n 7. Since a is a positive number, 1/a, 1/a2, ... will also be positive. 1/a, 1/a2,...are less than one since the reciprocal of a positive number greater than one is a positive number less than one. (There are other explanations.) loga 1/a = loga a-1 and by the definition of logarithms, loga a-1 = y ay = a-1. Therefore, y = -1 = loga a-1. The same argument holds for the other logarithms. Part 3 8. Power rule: log yn = nz Product rule: log vy = x + y = log v + log y Quotient rule: log (y/v) = z x = log y log v 9. Let log y = z and log v = x, By the definition of logarithms, y = az and v = ax. vy = axaz = ax+z so log vy = x + y. It follows that log vy = x + y = log v + log y. Hence, the logarithm of the product of two numbers is equal to the sum of the logarithms of the factors. 10. Let log y = z and log v = x, By the definition of logarithms, y = az and v = ax. v/y = ax /az = ax-z so log v/y = x - y. It follows that log v/y = x - y = log v - log y. Hence, the logarithm of the quotient of two numbers is equal to the difference of the logarithms of the factors.
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11. If you know the logarithm of 2 and the logarithm of 3, you know the logarithms of 6 = (2)(3), 9 = 32, 27 = 33,... , 4 = 22, 8 = 23, ..., 1.5 = 3/2, 2/3, ... 36 = 2232,...and an infinite number of other examples using just these two values. See the Student Activity Calculation of Logarithms - Using the Method of Napier and Briggs for more examples. Part 4 12. The only way for the value of a logarithm to be rational is if the number is a power of the base. There are many examples: log4 8 = 3/2 (since 8 = 43/2, i.e. 8 is a power of the base 4). 13. A surd is the square root of a number that is not a perfect square such as 2 , 5 , 7 ,...
14. Since the logarithms of numbers which are not the powers of the base are neither rational nor irrational, it is with justice that they are called transcendental quantities. For this reason, logarithms are said to be transcendental. Part 5 15. The rule usually looks like this: log a n = 16. loga n = p ap = n logb n = q bq = n log10 n log10 a ap = bq a = bq/p
Since b, p and q are all constant, a is also a constant and this value has nothing to do with n, the number whose logarithm we are evaluating. 17. Looking for log2 n = q, log10 n = p (we would know this value, since it is a base 10 logarithm) log102 = 0.3010300, also log2 2 = 1 Then p/q= 0.3010300 and q = p/0.3010300 = 3.3219277p. So, if every common logarithm is multiplied by 3.3219277 then we will have produced a table of logarithms for base 2. Note that this is equivalent to the usual method you gave in number 15: log10 n log10 n = = (3.3219277) log10 n log 2 n = log10 2 .3010300
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Introduction to Logarithms Student Pages
Part 1 Read paragraph 102. Then answer the questions. 1. In this paragraph, Euler gives the definition of the LOGARITHM of y. What is it?
2. Is this the same definition that you have been given in your class?
3. In the paragraph, what is "a"? What properties must "a" have?
4. What does the last sentence say about the domain of the logarithmic function?
Part 2 Read paragraph 103. Then answer the questions. 5. Explain why Euler says that, no matter the base, log 1 = 0. (We would write loga 1 = 0.)
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6. Explain why "log a = 1, log a2 = 2, log a3 = 3, log a4 = 4, etc".
7. Explain why 1/a, 1/a2, ... are positive numbers less than 1 and why their logarithm is negative.
Part 3. Read paragraph 104. Then answer the questions. 8. What does Euler say the product rule, the quotient rule and the power rule for logarithms are?
9. Give the explanation Euler uses for the product rule.
10. Give a similar explanation for the quotient rule.
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11. How can these rules be used to "find the logarithms of many numbers from a knowledge of the logarithms of a few"?
Part 4 Read paragraph 105. Then answer the questions. 12. What does the first sentence mean? Give an example of a logarithm of a number that is rational.
13. What is a surd? You may have to use a dictionary.
14. Why, according to Euler, is the logarithmic function a transcendental function?
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Part 5 Read paragraph 107. Then answer the questions. 15. Give the change of basis rule that you have learned in your class.
16. Translate into your own words: "If the base of one system is a and that of the other is b, if also the number n has logarithm p in the first system and logarithm q in the second, then ap = n and bq = n. Therefore ap = bq, so that a = bq/p and the value of p/q is constant, no matter what the value of n may be".
17. Show that Euler is describing how to change base 10 logarithms to base 2 logarithms his example.
in
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Introduction to Logarithms From Introductio in analysin infinitorum
Taken from Introductio in analysin infinitorum (1748) by Leonhard Euler
Chapter VI. On Exponentials and Logarithms
102. Just as, given a number a, for any value of z, we can find the value of y [= az], so, in turn, given a positive value for y, we would like to give a value for z, such that az = y. This value of z, insofar as it is viewed as a function of y, it is called the LOGARITHM of y. The discussion about logarithms supposes that there is some fixed constant to be substituted for a, and this number is the base for the logarithm. Having assumed this base, we say the logarithm of y is the exponent in the power az such that az = y. It has been customary to designate the logarithm of y by the symbol log y. If az = y, then z = log y. From this we understand that the base of the logarithms, although it depends on our choice, still it should be a number greater than 1. Furthermore, it is only of positive numbers that we can represent the logarithm with a real number. 103. Whatever logarithmic base we choose we always have log1 = 0, since in the equation az = y, which corresponds to z = log y, when we let y = 1 we have z = 0. From this it follows that the logarithm of a number greater than 1 will be positive, depending on the base a. Thus log a = 1, log a2 = 2, log a3 = 3, log a4 = 4, etc. and, after the fact, we know what base has been chosen, that is the number whose logarithm is equal to 1 is the logarithmic base. The logarithm of a positive number less than 1 will be negative. Notice that log 1/a = -1, log 1/a2 = -2, log 1/a3 = -3, etc., but the logarithms of negative numbers will not be real, but complex, as we have already noted. 104. In like manner if log y = z, then log y2 = 2z, log y3 = 3z, etc., and in general log yn = nz or log yn = n log y, since z = log y. If follows that the logarithm of any power of y is equal to the product of the exponent and the logarithm of y. For example logy = (z) = (log y), log1/y = log y-1/2 = -(log y), and so forth. It follows that if we know the logarithms of any number, we can find the logarithms of any power of that number. If we already know the logarithms of two numbers, for example log y = z and log v = x, since y = az and v = ax, it follows that log vy = x + y = log v + log y. Hence, the logarithm of the product of two numbers is equal to the sum of the logarithms of the factors. In like manner log (y/v) = z x = log y log v, that is, the logarithm of a quotient is equal to the logarithm of the numerator diminished by the logarithm of the denominator. These rules can be used to find the logarithms of many numbers from a knowledge of the logarithms of a few. 105. From what we have seen, it follows that the logarithm of a number will not be a rational number unless the given number is a power of the base a. That is, unless the number b is a power of the base a, the logarithm of b cannot be expressed as a rational number. In case b is a power of the base a, then the logarithm of b cannot be an irrational number. If, indeed, log b =
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n, then an = b, but this is impossible if both a and b are rational. It is especially desirable to know the logarithms of rational numbers, since from these it is possible to find the logarithms of fractions and also surds. Since the logarithms of numbers which are not the powers of the base are neither rational nor irrational, it is with justice that they are called transcendental quantities. For this reason, logarithms are said to be transcendental. 107. There are as many different systems of logarithms as there are different numbers which can be taken as the base a. It follows that there are an infinite number of systems of logarithms. Given two different systems of logarithms, there is a constant which relates the logarithms of the same number. If the base of one system is a and that of the other is b, if also the number n has logarithm p in the first system and logarithm q in the second, then ap = n and bq = n. Therefore ap = bq, so that a = bq/p and the value of p/q is constant, no matter what the value of n may be. If the logarithms of all numbers have been computed in one system, then it is an easy task, by means of this golden rule for logarithms, to find the logarithms in any other system. For example, we have logarithms for the base 10. From these we can find the logarithms with any other base, for instance the base 2. We look for the logarithm of a number n for base 2, which will be q, while the logarithm with base 10 of the same number n will be p. Since for base 10, log 2 = 0.3010300 and for base 2, log 2 = 1, then p/q = 0.3010300/1 and q = p/0.3010300 = 3.3219277. If every common logarithm is multiplied by 3.3219277 then we will have produced a table of logarithms for base 2.
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Development of Logarithms Using Sequences Teacher Notes
Level: Algebra 2 or Precalculus (Prerequisites: The students should be familiar with the properties of logarithms and exponents and have at least an intuitive understanding of sequences.) Objective: In this activity, the students will explore the ideas and background of the historical invention of the logarithm. Materials/Resources: Student Pages for each student; calculator (optional) Time Frame: This entire activity will probably take the equivalent of three class periods, with part of the activity sheet assigned as homework. Part 4 is the most difficult part and may be omitted if necessary. How to Use: An important thing to keep in mind as you work through this is that John Napier, the inventor of logarithms, was looking for a tool that would simplify complex mathematical calculations which were done laboriously by hand. In this project, you will explore the ideas and background of the historical invention of the logarithm. This is an easy-to-follow activity that provides an excellent development of the meaning of logarithms. The teacher may need to do some of the exercises as guided practice to provide a direction for student answers. Bibliography: Adapted from Lawrence Morales' on line project found at www.sccd.ctc.edu/~lmorales/home/loglab July 1999
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Development of Logarithms Using Sequences Solutions
Part I: Relationships 1. S2: 2 4 8 16 32 S2: 21 22 23 24 25 64 128 26 27 256 28 512 29 1024 210 2048 211 4096 212 8192 213
The sequence S1 gives the power of 2 for the number in S2. 2. (8)(32) = 256 (512)(16) = 2924 = 213 = 8192 (128)( 4) = 2722 = 29 = 512
(There are many other possible examples.)
The product is always another term of S2. This is because each element of S2 is a power of 2, so the product of elements in S2 will also be an element of S2. 3. 8 in S2 corresponds to 3 in S1; 32 in S2 corresponds to 5 in S1; the product 256 corresponds to 8 = 3 + 5 in S1. In the second example, 512 corresponds to 9, 16 corresponds to 4, and the product, 8192, corresponds to 13 = 9 + 4. In the third example, 128 corresponds to 7, 4 corresponds to 2, and the product, 512, corresponds to 7 + 2 = 9. When multiplying the elements of S2, the product will always be an element of S2. This is because 2n2m = 2n+m. In words, multiplication in S2 corresponds to addition in S1. 32/8 = 25/23 = 25-3 = 22 = 8 512/16 = 29/24 = 29-4 = 25= 32 128/4 = 27/22 = 27-2 = 25= 32
4. 5.
(There are many other possible examples.)
The quotient is always another term of S2. This is because each element of S2 is a power of 2, so the quotient of elements in S2 will also be an element of S2. 6. Any two terms in S2 can be divided by considering the exponents, that is, the corresponding numbers in S1, subtracting these numbers, and then finding the corresponding number in S2. 322= (25)2 = 210 = 1026 84 = (23)4 = 212 = 4096 163 = (24)3 = 212 = 4096
7.
(There are many other possible examples.)
The power of a term in S2 is always another term of S2. This is because each element of S2 is a power of 2, so the power of elements in S2 will also be an element of S2.
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8.
Any term in S2 can be raised to a power by finding its exponent, that is, the corresponding number in S1, multiplying this number by the power, and then finding the corresponding number in S2. The easiest way would be to rewrite the number or numbers in S2 as powers of 2 and use the rules of exponents to evaluate the answer. When multiplying the numbers, add the exponents; when dividing the numbers, subtract the exponents; when raising to a power, multiply the exponent by the power.
9.
Part 2 Extending the Results 10. Yes the results from part 1 hold here also. The rules of exponents hold no matter what the base is. Let u = bn and v = bm in each of the following: L(u) + L(v) = L(bn) + L(bm) =n+m = L(bn+m ) = L(bn bm ) = L(uv) L(u) - L(v) = L(bn)/L(bm) = n/m = L(bn/m ) = L(bn /bm ) = L(u/v) L(uk) = L((bn)k) = L(bnk) = nk definition of u and v definition of the function l definition of the function l rules of exponents definition of u and v definition of u and v definition of the function l definition of the function l rules of exponents definition of u and v definition of u rules of exponents definition of the function l
11.
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= k L(bn ) = k L(u)
definition of the function l definition of u
Part 3: Terminology and a Formal Introduction to Napier 12. If we read sequence S3 from right to left, each number is 1 smaller than the one to its right. Thus we can extend it to the left by putting 0 to the left of 1 and then the negative integers to the left of 0. If we read sequence S4 from right to left, we see that each number comes from the one to its right by dividing by b. Since the number in S4 under 1 is b, the number to be put under 0 must be b/b = 1. Similarly, the number to be put under -1 must be 1/b. The number under -2 is then (1/b)/b = 1/b2, and so on. Yes the results from number 10 still hold. The rules of exponents hold for all integers. There are many examples that would be appropriate. We write 1/b as b-1; then 1/b2 = b-2. In general, we have 1/bn = b-n, because 1/bn in S4 is under -n in S3. The 1 in S4 under the 0 in S3 can be written as 1 = b0. So a 0 exponent always means 1, while a negative exponent means 1 over the corresponding positive power of the base b. log 31 = __0_ log381 = __4_ log33 = _1__ log31/3 = _-1__ log39 = __2_ log3 1/243 = __-5_ log327 = _3__ log31/2187 = _-7
13. 14.
15.
Part 4: Filling up the sequences 16. 1.0000001 1.0000002 1.0000003 1.0000004 1.0000005
Raising 1.0000001 to a small power seems just to be 1 with 6 decimal places after the 0 and then the small power, although with more decimal places, we would find more distinctions between the numbers. In any case, these numbers are very close together, differing by only .0000001. To get to 3, however, we would need to find n such that (1.0000001)n = 3. The calculator shows that n needs to be something near 11,000,000. It is therefore not very feasible to continue the sequence in this way. 17. We must have tt = 2 because the 1 (from + ) in S1 corresponds to 2 in S2. The solution to the equation t2 = 2 is t = 2. The calculator shows that this is equal approximately to 1.414. And since we use exponents of 2 to designate numbers in S2, we also have t = 21/2. That is, 21/2 = 2. Then log2 t = log2 2 = . Since u3 = 2, we have u = 3 2 = 1.260 = 21/3. Since v3 = 4, because 2/3 + 2/3 + 2/3 = 2 and 2 in S1 corresponds to 4 in S2, we get v = 3 4 = 1.587 = 22/3. We could also find v by
18.
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noting that v = u2, because 1/3 + 1/3 = 2/3. It follows that v = ( 3 2 )2. So there are two different ways of expressing v in radical form. Also, log2 u = 1/3 and log2 v = 2/3. 19. The number corresponding to is 4 2 = 1.189 = 21/4. The number corresponding to is 4 8 = ( 4 2 )3 = 1.682 = 23/4. The number corresponding to 3/2 is 8 = (2)3 = 2.828 = 23/2. The number corresponding to 4/3 is 3 16 = ( 3 2 )4 = 2.520 = 24/3 and the number corresponding to 5/4 is 4 32 = ( 4 2 )5 = 2.378 = 25/4. The logarithms to the base 2 of these numbers are, in each case, the fraction that we began with. The number in S2 which corresponds to the fraction m/n in S1 is 2m/n. Since the nth power of this number is 2m, corresponding to the fact that the product of m/n by n is m, we can write this number as n 2 . Also, since m/n is m times 1/n, we know that 2m/n is 21/n raised to the power m. Since 21/n is equal to n 2 , we can also write 2m/n as ( n 2 )m. The calculator shows that 2317/200 = 3 (to three decimal places) and also that 2306/109 = 7 (also to three decimal places). The sum of 317/200 and 306/109 is 95753/21800 and 2 raised to that power is equal to 21, again to three decimal places. Thus 95753/21800 in S1 corresponds to 21 in S2. Answers may vary.
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Development of Logarithms Using Sequences Student Pages
Part I: Relationships Consider the two following sequences: S1 S2 1 2 2 4 3 8 4 16 5 32 6 64 7 128 8 256 9 512 10 1024 11 2048 12 4096 13 8192
1. Rewrite the numbers in sequence S2 as powers of a single number. How are the sequences related? 2. Note that when two terms of S2 are multiplied, such as 64 and 128, their product, (64)(128) = 8192, is another term of S2. Pick a few pairs of your own to multiply. Is the product always another term of S2? You may have to extend the terms in each sequence to verify your observations. 3. Note that the number in S1 corresponding to 64 is 6, while the number in S1 corresponding to 128 is 7. The number corresponding to their product, 8192, is 13, the sum of 6 and 7. In other words, multiplication in S2 corresponds to addition in S1. Check that this holds for the other pairs you picked in exercise 2. 4. In a brief paragraph, summarize what you have observed. Most importantly, give a justification for your results using rules of exponents. We will now see if similar laws hold for other mathematical operations. In particular, we will examine the operations of division and taking powers. 5. Take any two terms in S2 and divide the smaller one into the larger. Do at least three cases. In each case, check the corresponding numbers in S1 of your two terms as well as their quotient. 6. After trying several pairs of terms in S2, describe how you could divide any two terms in S2 without using a calculator or doing any computations by hand. 7. Take any term in S2 and raise it to a positive integer power. Do at least three cases. Again, observe the corresponding numbers in S1 of the terms you choose and calculate. 8. After trying several cases, describe how you could raise any term in S2 to a positive integral power without using a calculator or doing any computations by hand.
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9. From the previous problems, we have seen how to multiply, divide and raise to powers by relating terms in S1 and S2 in appropriate ways. Assume that you had no calculator or computer device at your disposal. How would these relationships help you to make complex computations? Explain! Part 2 Extending the Results At this point, we should try to generalize our findings and describe them with some more general mathematical symbols or notation. To do this, let's look at the two following sequences where b > 0. S3: S4: 1 b 2 b2 3 b3 4 b4 5 b5 6 b6 7 b7 8 b8 9 ... b9 ... n bn
We begin by defining a function, L(x), which has as its domain the elements of S4. The "value" of L(x) is its corresponding term in S3. For example L(b6)=6. In general, L(bn)=n for any positive integer n. 10. Review your results from Part 1 of the project. Do you think they hold here? In particular, use the general terms of S3 and S4 to check if the properties hold. Use algebra only to show that they are true. (Do not plug in values for b. That way, you will be showing the general relationships hold.) 11. Verify that your conclusions in number 10 are actually the following in function notation. Let u and v be terms in S4. Use algebra to show the following three properties hold. L(u) + L(v) = L(uv) L(u) - L(v) = L(u/v) L(uk) = k L(u)
Part 3: Terminology and a Formal Introduction to Napier Now extend the Sequence S3 and S4 to the left as follows: S3 ... -5 -4 5 S4 ... 1/b 1/b4 -3 1/b3 -2 1/b2 -1 1/b 0 1 1 b1 2 b2 3 b3 4 b4 5 ... b5 ...
12. Why is it reasonable to extend S4 to the left in the way we did? In particular, why should the number in S4 under 0 in S3 be 1? And why should the numbers in S4 under the negative values in S3 be 1 divided by powers of b? Your explanation may well be by analogy to the development of the sequences initially.
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13. Do the general results from number 11 still hold? Test them out by trying a few terms in S3 and S4. Show your examples. Move on when you're convinced that the rules hold for the extended sequences. 14. How can you write the quotients at the left side of S4 in exponential notation? How would you write the 1 in S4 under the 0 in S3 in exponential notation? Give a brief definition of what a negative exponent means and what a 0 exponent means. At this point, recall our original sequences, S1 and S2: S1 S2 1 2 2 4 3 8 4 16 5 32 6 64 7 128 8 256 9 512 10 1024 11 2048 12 4096 13 8192
John Napier, the Scottish mathematician who invented logarithms, called the terms of the sequence S1 the "logarithm" of the corresponding terms of the sequence S2. The word "logarithm" is a translation of the Greek word logarithmos which comes from the Greek words meaning ratio and number; it therefore means "the number that counts the ratios." For example, he called 7 the logarithm of 128. This example shows how many ratios are required for the term 128 to be found. (Note that Napier did not use base b = 2 as in S2, but the ideas are the same.) The function developed in number 11 above is a logarithmic function for base b = 2. We now replace our L notation relating numbers in S1 to those in S2 by the new notation "log2" (read as logarithm to the base 2). Thus we have the following results: log21 = 0, log22 = 1, log24 = 2 , log28 = 3, etc.
15. Use this "log" notation above to associate the corresponding terms of the following two sequences: S5 S6 0 1 1 3 2 9 3 27 4 81 5 243 6 729 7 2187
Find the following... log 31 = ___ log381 = ___ In general, log33 = ___ log31/3 = ___ log39 = ___ log3 1/243 = ___ log327 = ___ log31/2187 = ___
logba = c if and only if bc = a.
This is the classic modern definition of a logarithm.
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At this point it is very important to remember that Napier did not use this kind of notation! What we are doing here is trying to see the link between our modern logarithms and the theoretical ideas behind Napier's logarithms. Again, Napier started with two related sequences and then went in a very different direction than we do. Part 4 Filling up the sequences Recall that Napier's original motivation for developing logarithms was to replace multiplication and division by addition and subtraction. The use of these sequences enable us to do this on a very limited scale. We can, for example, by the use of sequences S1 and S2 replace the multiplication 16 times 32 by the addition 4 plus 5. And we can replace the division of 256 by 64 by the subtraction of 6 from 8. But what happens if we want to multiply 3 by 7? Neither of these numbers is in S2 and will not be even if we extend the row further to the left or right. What can we do about this? What we need is to have more numbers in S2. One way to accomplish this and this is essentially Napier's idea is to use a base much closer to 1 than our base of 2. Napier, in fact, used 1.0000001. So the first few terms of his sequences, which we will label S7 and S8 looked like the following: S7 S8 1 1.0000001 2 (1.0000001)2 3 (1.0000001)3 4 (1.0000001)4 5 (1.0000001)5
16. Use your calculator to rewrite the numbers in S8 in decimal form. What do you notice about these numbers? Given that we wanted to multiply 3 by 7, about how many numbers would we need to take in the sequence S8 in order for 3 (or even a number very close to 3) to appear? Is this feasible? Napier used some very clever ideas to shorten what looks like an impossible task. Yet he nevertheless worked over twenty years on his table. We will look a somewhat different method for putting more numbers in S2. We will first fill in S1 and then try to figure out the appropriate correspondences. We begin by putting in between the 0 and the 1. We now must determine what the corresponding number in S2 must be. We will call it t. S1 S2 0 1 1 t 2 2 4 3 8 4 16 5 32 6 64
17. To determine t, we need to use the properties of these sequences. In particular, we know that addition in S1 corresponds to multiplication in S2. Therefore, since + = 1 in S1, we must have tt = 2. (Why?) We can rewrite this equation as t2 = 2. What is t? Write the answer in radical form, in decimal form, and in exponential form. In terms of the log2 notation, what is log2 t? 18. Now put 1/3 and 2/3 in their appropriate positions in S1. Let u be the number in S2 corresponding to 1/3 and v be the number corresponding to 2/3. Again using the property
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that addition in S1 corresponds to multiplication in S2, write an equation which u must satisfy. Solve the equation to determine u in radical form, in decimal form, and in exponential form. Then determine v similarly. (Note that instead of using the property that addition in S1 corresponds to multiplication in S2, you may wish to use the property that multiplication in S1 corresponds to taking powers in S2.) What are log2 u and log2 v? 19. Now put and in their appropriate positions in S1. As in question 18, determine the corresponding numbers in S2 in radical form, in decimal form, and in exponential form. Do the same for the numbers 3/2, 4/3, and 5/4 in S1. Determine the logarithm to the base 2 of each of the numbers you have now found in S2. 20. By analogy to your solutions to problems 17, 18, and 19, determine in exponential form and in radical form the number in S2 corresponding to the fraction m/n in S1. Use explicitly one or more of the properties relating S1 to S2. Since we have now filled in S1 with as many fractions as we like, we have also filled in S2 with lots of numbers. If we want to multiply 3 by 7, we now need to determine the numbers in S1 corresponding to 3 and 7 in S2. 21. The numbers in S1 corresponding to 3 and 7 are log2 3 and log2 7, respectively. Show that log2 3 is approximately equal to 317/200 and that log2 7 is approximately 306/109. You will need to use your calculator and the general definition of a logarithm. Find the sum of 317/200 and 306/109. Show that this number in S1 corresponds to 21 in S2. 22. Find two other numbers to multiply by the method of exercise 21. You may have to experiment a bit with your calculator to do this. 23. Consult at least two reliable sources (other than the Internet) in the library and write a short biography of John Napier. Your account should include information not contained in this document. Also, find at least one major application of logarithms (not discussed in class) and briefly describe it. Your write up must be typed and be no more than two pages doublespaced. See me if you need resources to consult.
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Calculation of Logarithms Using the Method of Napier and Briggs Teacher Notes
Level: Algebra 2 or Precalculus (Prerequisites: Properties of exponents; Properties of logarithms; Definition of logarithm) Objective: The purpose of this activity is to give the students an opportunity to investigate, on a introductory level, one way that logarithms can be calculated. Materials: Student Pages; Calculator (optional) Time Frame: One class period with homework assigned to finish the student activity. How to Use: The teacher may want to discuss briefly the included Historical Background with the class. The teacher may also want to review the definitions and properties of logarithms and exponents. It would be a beneficial for the teacher to work the activity before presenting it to the class especially to think about different ways to get the solutions, since some of the methods of solution are not unique. Additionally, the order in which to calculate the logarithms is not necessarily the numerical order, so the teacher will want to warn the students of this fact and perhaps give some hints about what to do when. Also note that these are base 10 logarithms that are being calculated. Bibliography Katz, Victor J., A History of Mathematics: An Introduction, Addison Wesley, 1998.
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Calculation of Logarithms Using the Method of Napier and Briggs Solutions
Part I: log 1 log 1 = y 10y = 1 (definition of logarithm base 10) 10y = 1 y = 0. Therefore log 1 = 0 (this is true for logarithms of any base) At this point it might be best to find log 10 next since it is also easy to calculate: log 10 = y 10y = 10 (definition of logarithm base 10) 10y = 10 y = 1. Therefore log 10 = 1 log 2 Consider the following relationship: 210 = 1024 1000 = 103 210 103 Take the logarithm of both sides of the approximation: log 210 log 103 Now use the power property of logarithms: 10 log 2 3 log 10 And solve for log 2: log 2 (3 log 10)/10 = (3 1) / 10 = 0.3 So log 2 0.30 log 3 Consider the following relationship: 37 2187 2 1093 2 103 37 2 103
Take the logarithm of both sides of the approximation: log 37 log (2 103) Now use the sum property of logarithms: log 37 log 2 + log 103
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Now use the power property of logarithms: 7log 3 log 2 + 3log 10 Solve for log 3: log 3 (log 2 + 3 log 10)/7 (0.30 + 31)/7 = 3.3/7 = 0.47 So log 3 0.47 log 4 Since 4 = 22, So log 4 0.6 log 5 Since 5 = 10/2, log 5 = log (10/2) = log 10 log 2 (by the difference property of logarithms.) 1 0.30 = 0.70 So log 5 0.70 log 6 Since 6 = 23 , log 6 = log (23) = log 2 + log 3 (by the sum property of logarithms.) 0.3 + 0.47 = 0.77 So log 6 0.77 log 7 Consider the following relationship: 72 = 49 50 = 102/2 72 102/2 Take the logarithm of both sides of the approximation: log 72 log 102/2 Now use the difference property of logarithms: log 72 log 102 log 2 Now use the power property of logarithms: 2 log 7 2 log 10 log 2 Solve for log 7: log 7 (2 log 10 log 2)/ 2 (21 0.30 )/2 = 1.70/2 = 0.85 So log 7 0.85 log 4 = log 22 = 2 log 2 2(0.3) = 0.6
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log 8 Since 8 = 23, So log 8 0.90 log 9 Since 9 = 32, So log 9 0.94 log 10 log 10 = 1 (see above) Solutions for Part 2: 1. 2. Since log 1 and log 10 are exact values, an equal sign is used and since the other values are approximations, the approximately equal sign is used. Since the other numbers, such as 4, 6 and 8 are products of these prime numbers, we are able to use the approximate values for the logarithms of 2, 3 and 7. Another reason would be that there is no easy way to use the definition of logarithm to find, say log 2 since by the definition, log 2 = y 10y = 2. (Suggestion for additional activity using a graphing calculator, find the intersection of the graphs of y1 = 10x and y2 = 2, to find a value for log 2.) The values are not the same because of the approximations used in the calculations in the project. The values are not exact since approximations were used for log 2, log 3 and log 7. As an example, log 6 was found using both of the approximations for log 3 and log 2. Using a calculator, log 6 = 0.7781513, while the approximation is 0.77. The difference in values is more than 0.008. Using the calculator, log 3 = 0.4771213 and log 2 = 0.30103 while the approximations were log 3 0.47 and log 2 0.30. Both of these approximations were closer to the calculator values than that of log 6. There are many examples of other relationships such as 72 = 49 = 5 10 and so on. 5. The estimate for log 2 is too large because 210 is larger than 103 log 9 = log 32 = 2 log 3 2(0.47) = 0.94 log 8 = log 23 = 3 log 2 3(0.30) = 0.90
3. 4.
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The estimate for log 5 is too small because the approximation for log 2 is too large and this value is subtracted from an exact value resulting in an approximation that is too small. 6. There are many approaches. One possible solution is that 112 = 121 120 = 456 Taking the log of both sides gives : 2 log 11 = log 456 = log 4 + log 5 + log 6 0.60 + 0.70 + 0.77 = 2.07 So log 11 2.07/2 log 11 1.035 (the calculator value is 1.0413927)
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Calculation of Logarithms Using the Method of Napier and Briggs Student Pages
Historical Background Briggs and Napier each spent many years in doing lengthy computations to determine logarithms. Since Napier did not use base 10 for his logarithms, but Briggs did, this project will demonstrate Briggs' approach. Keep in mind though, that we will be finding approximations for only a few common logarithms and will not have nearly the accuracy that Briggs found when he was developing his logarithm tables. Recall the definitions and properties that you know for logarithms: Definition: logb x = y by = x Sum Property of logarithms: logb(xy) = logb x + logb y Difference Property of logarithms: logb(x/y) = logb x - logb y Power Property of logarithms: logb xn = n logb x Part 1: Find approximations for the common logarithms for the positive integers 1 through 10. Recall that common logarithms are base 10 logarithms. To help get started, there are several hints of approaches to use for finding the logarithms of the prime numbers 2, 3 and 7. log 1 = _________ log 2 __________ log 2 req...