Hyperbolic Functions: with Configuration Theorems and Equivalent and Equidecomposable Figures

Hyperbolic Functions

V. G. Shervatov

Configuration Theorems

B. I. Argunov and L. A. Skornyakov

Equivalent and

Equidecomposable Figures

V. G. Boltyanskii

THREE VOLUMES BOUND AS ONE

Dover Publications, Inc.

Mineola, New York

Bibliographical Note

This Dover edition, first published in 2007, is an unabridged republication in one volume of three works originally published in English in 1963 by D. C. Heath and Company, Boston. Hyperbolic Functions was translated and adapted from the second Russian edition (1958) by A. Gordon Foster and Coley Mills, Jr. Configuration Theorems was translated and adapted from the first Russian edition (1957) by Edgar E. Enochs and Robert B. Brown. Equivalent and Equidecomposable Figures was translated and adapted from the first Russian edition (1956) by Alfred K. Henn and Charles E. Watts.

International Standard Book Number eISBN 13: 978-0-486-17005-3

Manufactured in the United States of America

Dover Publications, Inc., 31 East 2nd Street, Mineola, N.Y. 11501

Hyperbolic Functions

V. G. Shervatov

PREFACE TO THE AMERICAN EDITION

IN THIS BOOKLET the hyperbolic sine, hyperbolic cosine, and hyperbolic tangent are introduced in connection with the geometric properties of the hyperbola, and the fundamental properties of these functions are derived. Natural logarithms are then discussed.

With the aid of complex numbers, basic formulas are derived which interrelate hyperbolic functions, trigonometric functions, and the number e. Finally, infinite series expansions of sinh t, cosh t, sin t, cos t, and ea are given.

Two years of high school algebra and an acquaintance with trigonometry are sufficient to enable the reader to understand Chapters 1 and 2. To understand fully the last chapter it is necessary to have some acquaintance with the theory of limits. However, those who do not possess this background will be able to follow much of the discussion.

CONTENTS

CHAPTER 1.  The Hyperbolic Rotation

1.  Homothetic transformations

2.  Strains

3.  The hyperbola

4.  The hyperbolic rotation

5.  Properties of the hyperbola

CHAPTER 2.  Hyperbolic Functions

6.  The unit hyperbola and its equation

7.  The hyperbolic functions

8.  Identities between hyperbolic functions

CHAPTER 3.  Natural Logarithms

9.  Logarithms and the hyperbola

10.  The base e of natural logarithms

11.  Analytic expressions for the hyperbolic functions

12.  Infinite series expansions for ea, sinh t, cosh t

13.  Euler’s formulas

1. The Hyperbolic Rotation

1.  HOMOTHETIC TRANSFORMATIONS

In solving problems in geometry we often use the following type of transformation. Pick a point O in the plane and a positive number k. If A is any other point, let us transform the point A into the point A′ such that A′ lies on the half-line OA (Fig. 1a, b). We leave the point O fixed by the transformation. Such a transformation is called a homothetic transformation, O is called the homothetic center (or center of similitude), and k is called the homothetic ratio (or ratio of similitude). Obviously, if k < 1, then OA′ < OA (Fig. 1a), while if k > 1, then OA′ > OA (Fig. 1b).

Fig. 1

Under a homothetic transformation every figure F is transformed into a figure F′, similar to F (Fig. 2). If k < 1, F′ is smaller than F, while if k > 1, F′ is larger than F.

Fig. 2

A homothetic transformation transforms a straight line into a straight line (Fig. 3a), parallel lines into parallel lines (Fig. 3b), and circles into circles (Fig. 3c).

Fig. 3

A homothetic transformation changes the lengths of all line segments in the plane by a constant ratio, the homothetic ratio k. It also changes the areas of all figures by the constant ratio k², the square of the homothetic ratio.

To see this, let F be a figure in the plane. Consider also a grid of small squares in the plane (Fig. 4). The area of F is approximately equal to the number of squares which are contained in F multiplied by the area of one square; the smaller the squares of the grid, the smaller will be the error. (In fact, if we choose any small positive number ε whatsoever, we can make the error smaller than ε by making the squares sufficiently small.) Under a homothetic transformation the grid of squares is transformed into a new grid of squares and the figure F into the figure F′, which will contain exactly as many squares of the new grid (the squares will be smaller if k < 1 and larger if k > 1) as there were squares of the first grid inside the figure F. The area of F′ is approximately equal to the number of squares contained in it multiplied by the area of one square.¹

Fig. 4

But the area of each new square is equal to the area of one of the original squares multiplied by k², since the lengths of the sides of the original squares are multiplied by k. Therefore, the area of F′ is equal to the area of F multiplied by k².

As an example of the application of homothetic transformations, let us solve the following problem: Inscribe in a given right triangle ABC a rectangle BDEF so that the lengths of its sides are in a given ratio (Fig. 5).

Fig. 5

First let us construct an arbitrary rectangle BD′E′F′, with the lengths of its sides in the given ratio and such that the vertices D′ and F′ lie respectively on the sides BC and AB of the given triangle. Let E be the point of intersection of the line BE′ and the side AC of the triangle. It is easy to show that the homothetic transformation with B as the homothetic center and

as the homothetic ratio will transform the rectangle BD′E′F′ into the required rectangle BDEF. From this we can easily construct, the desired rectangle. When the given triangle ABC is not a right triangle, the problem can be solved in an analogous way. We shall not dwell on this here.

2.  STRAINS

We shall need to use another transformation called a strain. Let l be a straight line in the plane. Let us transform the point A in the plane into the point A′ such that A′ lies on the perpendicular PA dropped from A to l , some positive constant (Fig. 6a, b). Every point of the line l is left fixed by the transformation. The straight line l is called the axis of the strain, and k is called the coefficient of the strain. If k is greater than 1, then PA′ > PA (Fig. 6b), and the strain is sometimes called an elongation. If k is less than 1, then PA′ < PA (Fig. 6a), and the strain is sometimes called a compression. A figure F is transformed by a strain into a new figure F′, which is usually not similar to F; that is,

Fig. 6

Fig. 7

Several properties of strains are analogous to the properties of homothetic transformations.

(a) A straight line is transformed by a strain into a straight line.

Suppose that a strain has axis l and coefficient k; let m be any straight line. If m is parallel to l and at a distance d from l, then m is transformed into a straight line m′ parallel to l and at a distance kd from l (Fig. 8a). On the other hand, suppose that m and l intersect, say at the point O (Fig. 8b). Under the strain the point O remains fixed. Let A be any point different from O on the line m, and let A′ be the point into which A is transformed by the strain. Then PA′ = k · PA. Let B be another point on the straight line m. Let B′ be the point where the perpendicular BQ from B to l intersects the straight line OA′. The triangles OQB and OPA are similar, and the triangles OQB′ and OPA′ are similar; hence,

Fig. 8

that is, QB′ = k · QB. From this last equality we see that the strain transforms the point B into the point B′. Since B was an arbitrary point of the line m, this line is transformed by the strain into the line OA′, which we shall call m′.

(b) Parallel lines are transformed by a strain into parallel lines.

Let the lines m and n be parallel; then they have no point of intersection. In that case the lines m′ and n′ into which m and n are transformed also have no point of intersection (since a point common to m′ and n′ could only result from a point common to the lines m and n); this shows that the lines m′ and n′ are also parallel (Fig. 9). Note that if φ and φ′ are the angles formed by the original line m and the image line m′ with the axis of the strain l, then from Fig. 8b it is easy to see that

Fig. 9

Hence, it also follows from this that parallel lines (intersecting l with the same angle φ) are transformed into parallel lines (intersecting l with the same angle φ′).

(c) The ratios of lengths of line segments lying on the same straight line are unchanged by a strain.

(Fig. 10).

Fig. 10

(d) A strain changes the areas of all figures by the constant ratio k, the coeficient of strain.

To see this, let F be a figure in the plane. Consider also a grid of small squares whose sides are parallel and

perpendicular to the axis of the strain. The area of F is approximately equal to the number of squares which are contained in F multiplied by the area of one square (Fig. 11). The squares are transformed by the strain into a grid of rectangles, the area of each of which is the area of one of the original squares multiplied by k (one of the dimensions of the square is unchanged, while the other is multiplied by k). The figure F′ contains exactly the same number of rectangles as the figure F did of squares, and the area of F′ is approximately equal to the number of rectangles contained in it multiplied by the area of one rectangle. Hence, the area of F′ is equal to the area of F multiplied by k.

Fig. 11

In order to illustrate the use of strains, let us solve the following problem (compare this with the problem on p. 3): In a given right triangle ABC inscribe a rectangle BDEF of given area T (Fig. 12). For the solution, apply a strain with axis BC and coefficient

Fig. 12

to the triangle ABC. This strain transforms the triangle ABC into an isosceles right triangle A′BC. Since

the area of the triangle A′BC is kS, where S is the area of the triangle ABC. By property (d), any rectangle BDEF of area T inscribed in ABC would be transformed into a rectangle BDE′F′ with area kT. Therefore, in the isosceles right triangle A′BC let us inscribe a rectangle BDE′F of area kT.

It is easy to do this, for

therefore, SΔE′DC + SΔA′F′E′ = SΔA′BC – SBDE′F′ = kS – kT.

On the other hand,

(Here we make use of the fact that the triangle A′BC, and consequently the similar triangles E′DC and A′F′E′, are isosceles.) Thus, we obtain

Now that we know the length of the segment BE′, we can at once find the point E′. After that we can construct the rectangle BDE′F′ inscribed in the triangle A′BC, and finally the rectangle BDEF inscribed in the triangle ABC.

Depending on the value of T, the problem may have two,