bers of this inequality approach a common value, their limit; that is,

x' ť

-

-u, and uo, since

In the important case when u′ =

uo requires that x1, the expression last written be

dicular

Q. E. D. (Cf. Art. 43.)

53. Area of a Hyperbolic Sector. Let the perpenbe dropped from any point P, of the equilateral hyperbola x2-y2= a2, upon its asymptote, meeting the

Assuming these equations as known, lay off upon the asymptote

OS, OS, OS, ..., OS = a | √2, S1, S2, . . ., §,

such that these lengths are in geometrical progression, and let the corresponding perpendiculars upon the asymptote be

AS, P,S,, PS2, ..., PS=Po, P1, P2, ..., p.

I

I

Then if p be the common ratio of the successive terms So, S1, S2, ... s, we have

If now (a, o), (x,, y1),

p1a (2 p s.),

=

p2 a2 (2 p2 s.),

p = a2 | (2 p” s.).

(x21 Y2), . . ., (x, y) be the co-ordinates of A, P1, P2, . . ., P, the area of the triangle OP, P, is (x,y,x,y), and by virtue of the relations

2 3

I

2

Thus the triangles OAP1, O P ̧Ð1⁄2, OP1⁄2P ̧, etc., are equal to one another in area and the area of the whole polygon OAP, P... Pis

When the number of points of division S1, S2, S3, etc., is indefinitely increased the polygon OAP,P,.. P approaches coincidence with the hyperbolic sector OAP, P... P, that is,

but at the same time

area of sector OAP,.. P, when n∞;

54. Agenda. The Addition Theorem for Hyperbolic Ratios. From the foregoing definitions of the hyperbolic ratios deduce the following formulæ :

(r). sinh (u+v)=sinh a cosh u + cosh a sinh v. (2). cosh(u+v)=cosh a cosh a + sinh u sinh

tanh u + tanh v

(3). tanh (u+v)=

I +tanh u tanh

(4). Deduce these formulæ also geometrically from the constructions of Arts. 48, 53, assuming for the definitions of sinh u and cosh u the ratios NP / a and ON a. [Burnside: Messenger of Mathematics, vol. xx, pp. 145-148.]

I

(5). In the figure of Art. 53 show that the trapezoids SAP1S1, SP、P2S2, etc., are equal in area to the corresponding triangles OAP, OP, P2, etc., and consequently to each other.

(6). Show that when the hyperbolic sector OAP (Art. 53) increases uniformly, the corresponding segment OS, laid off on the asymptote, increases proportionately to its own length.

(7). Assuming a=1 in the equilateral hyperbola of Art. 48, and that the area of any sector isu, prove that limit [(sinh u) /u]=1. (Use the method of Art. 55.)

u =

55. An Approximate Value of Natural Base. We may determine between what integers the numerical value of e must lie, by substituting their equivalents in x and y for the terms of the inequality:

S

Triangle OAP > sector OAVP > triangle ORS, as represented in Fig. 28. For our present purpose it will involve no loss of generality and it will simplify the computation to assume OA I, so that the equation of the hyperbola is

RA

N

Fig. 28.

x2 — y2 = 1.

as

The sectorial area OAVP, previously found in Art. 53, is then u, the area of OAP is

obviously, and for that of ORS we may write

OR X (ordinate of S).

To determine OR and this ordinate, write the equations to the tangent RS and the line OP, and find the ordinate of their intersection, and the intercept of the former on the x-axis. The results are:

η

the equation to the tangent RS, έ and ʼn being the current co-ordinates of the line;

the required intercept on the x-axis; and

the ordinate of S, found by eliminating & from the equations to RS and OP. Hence the area ORS is

If y 1, then u <1, x=√2, and

e2=x+y=1+√2=2.4+;

* Had the assumption a = 1 not been made, this inequality would have been

The equations for RS and OP and the expressions for OR and

would have been

correspondingly changed, but the final results would have been the same as those

given above.