Español
Português
中文 (中国)
العربية
हिन्दी
Русский
Français
Deutsch
Ελληνικά
Italiano
日本語
한국어
LatineSummary
This lesson opens the unit devoted to the exponential, logarithmic and hyperbolic functions. It presents its six lessons and what each one proves; it traces the history of these ideas, from the correspondence between progressions that Archimedes glimpsed and Stifel set in order to the logarithms of Napier and Bürgi, the number of Jakob Bernoulli and Euler, and the hyperbolic functions of Riccati and Lambert; it states the problem that gave rise to them, which is to turn products into sums, and translates it into two functional equations; and it describes, finally, the place these functions hold in the continuation of this course, in other courses on the site and in science.
Learning objectives
- Describe the content of each lesson of the unit and the logical order in which they build on one another, from the power with a real exponent to the hyperbolic functions.
- Place in their historical context the invention of logarithms, the discovery of the number and the introduction of the hyperbolic functions, identifying the practical problem that motivated each of them.
- Express the problem of turning products into sums by means of the equations and , and explain with Stifel’s table why its solution requires real exponents.
- Recognize applications of these functions in the Algebra and Geometry II course, in other courses and in physics, chemistry, biology, finance and computer science.
Entry assessment
Before working through the unit, it is advisable to measure the starting point, without any grading. The first ten questions cover the six content lessons of the unit: two for each of the three lessons on which the rest relies (the power with a real exponent, the number and the equations), one for the lesson on the logarithm, another for the one on the hyperbolic functions and two for the one on the models, the second of which asks you to set up a model from data; the same skills are measured again at the end, in the synthesis lesson. The last three check what the unit takes as known from the units on number systems and on analytic geometry of the plane. Each wrong answer indicates the lesson on which it is advisable to dwell.
Before starting the unit, measure your starting point. It is a short self-assessment: it is not graded and its result is stored only in your browser.
Question 1What is the range of the function ?
The range of has been shifted without taking into account the minus sign, which reflects the graph in the -axis and turns the positive values into negative ones.This is what one gets by taking as the bound the value at , which is ; but the function is decreasing, and for negative it takes values greater than , ever closer to .The exponential takes neither negative values nor zero, so never reaches the value , let alone exceeds it.The power takes all positive values, and only those; changing the sign gives all the negative values, and adding gives all the reals less than . The line is a horizontal asymptote that the graph does not reach.Review: The exponential function and powers with real exponents
Question 2Compute .
Correct answer:
By the power of a power theorem, which holds for real exponents, the result is . Whoever gets has added the exponents instead of multiplying them, because ; whoever gets has also made a second error, that of adding the radicands, as if were .Review: The exponential function and powers with real exponents
Question 3A capital of is invested at an annual rate of percent, that is, , compounded times a year. What is the supremum of the capitals after one year?
This is the supremum for . When the rate is doubled, the supremum does not stay the same: it goes from to .The rate does not multiply the supremum; it is its exponent: , and not . For example, with the capital is already , more than , and exceeds , which is less than .The sequence is increasing, but it is bounded: by the inequality , each term is less than . An increasing bounded sequence has a supremum by completeness.By the continuous compounding theorem, the supremum of the terms is , and none of them reaches it; with , it is .Question 4With , what upper bound on the error does the series error theorem give?
Correct answer:
The theorem gives , and with the bound is . Whoever gets has taken the first omitted term, , which is smaller than the error; whoever gets has forgotten the factor .Question 5Compute .
Correct answer:
Since and , by the change of base ; indeed, . Whoever gets has inverted the quotient, and whoever gets has divided by .Question 6What are the real solutions of the equation ?
The root of the trinomial does not give : it would give , which has no solution. The value does not make the left-hand side zero.These are the roots of the trinomial in , not the values of . The change of variable still has to be undone and the negative root discarded.The trinomial has a positive root, , which corresponds to ; indeed, .With , the equation is . The root is discarded, because a power with a positive base is not negative, and gives .Review: Exponential and logarithmic equations and inequalities
Question 7Solve the equation .
Correct answer:
In the domain, , the product law gives , that is, . The candidate is an extraneous solution, outside the domain; the only solution is , and indeed .Review: Exponential and logarithmic equations and inequalities
Question 8A radioactive isotope has a half-life of days. What fraction of the initial mass remains after days?
Correct answer:
By the half-life theorem, the mass is ; after days three half-lives have elapsed, and remains. Whoever gets has divided by the number of half-lives instead of dividing by that many times.Question 9The counts of a bacterial culture, recorded on a semi-log plot, lie on a line, and the line passes through the points corresponding to bacteria at hours and bacteria at hours. How many bacteria were there, according to the model, when the culture began?
Correct answer:
Data that lie on a line in the semi-log plot follow a model , by the linearization theorem. In hours the count is multiplied by , so it doubles every hours; going back hours from hour divides the count by : . Whoever gets a negative number has extended the line through the data on the ordinary scale, as if the growth were linear; whoever gets has gone back hours instead of .Question 10If and , what is ?
Correct answer:
By the fundamental identity, , and since the hyperbolic sine is positive: . Whoever gets has subtracted without squaring; whoever gets has forgotten the sign of the hyperbolic sine for .Review: The hyperbolic functions
Question 11From earlier unitsCompute .
Correct answer:
By the definition of the power with a rational exponent, , and the negative exponent gives the reciprocal: . Whoever gets has multiplied the base by the exponent, and whoever gets has forgotten the sign of the exponent.Question 12From earlier unitsWhat is the supremum of the set of the numbers , with a natural number?
This confuses the supremum with the maximum. The supremum is the least of the upper bounds, and it may not belong to the set; completeness guarantees that it exists for every nonempty set that is bounded above.This is the least element, obtained with : the minimum of the set, not its supremum.This is the element corresponding to , but it is not an upper bound: with one gets , which exceeds it.Every element is less than , so is an upper bound; and no number is one, because as soon as . The least upper bound is , even though it does not belong to the set.Review: Number systems: from the natural to the complex numbers
Question 13From earlier unitsWhat are the asymptotes of the hyperbola ?
The quotient has been inverted: the slope is , with the semi-axis associated with . These are the asymptotes of .The squares of the semi-axes have been taken instead of the semi-axes: from it follows that , by taking roots.These are the vertical lines through the vertices; the hyperbola touches them there, and does not approach them indefinitely.With and , the asymptotes of are the lines , which result from setting the left-hand side equal to zero.
Loading the content…
Completed
Sign in to save your progress