function in his peculiar way, and who has brought out many curious properties.

We can find an equation between three successive values of the function as follows: Let

k sin ε),

Since u is zero at the limits o and л we have by integrating this value of du,

k Ji-1 -2 i Jik Ji+1 = 0.

This equation gives the value of the function from the values of the two lower orders, but it is not well adapted to numerical calculation, since it gives the value of a small quantity from the difference of two greater ones. It can however be easily transformed into a continued fraction well suited to such calculation.

From equation (3) we have

ᅲ

1

J; = 1S"cos i e cos (k sin e). de + sin i e sin (k sin e). d e.

π

dε

k sin ε
+

1.2.3.4.5.6

de

1.2.3

sin e sin (k sin €) . d ɛ =

× { ✯ sin3 e

1.2.3.4.5

Integrating the terms of the series between the limits o and π we have

These values and equation (4) enable us to compute all values of the function. An elegant derivation of the general series for J will be found in Schlömilch's Compendium of the Higher Analysis, Vol. 2, p. 157.

This derivation depends on the following formula given by Jacobi,

BY IRVING P. CHURCH, B. C. E., NEWBURGH, N. Y.

Construction by Points.-Let OC c be the base of any logarithmic system. On O U and C V, perpendicular to O C, lay off O D=u, and CR v, such that u= log v in the system whose base

is c. The intersection of O R and D P, (parallel to O C) fixes P, a point of the curve.

From the construction we can immediately derive the Kpolar equation of the curve:

y

Substituting from y = 3 sin, and tan = 2, we have

for rectangular co-ordinates,

By reference to the equation (2) we see that if c>unity the curve has

the form indicated by our first diagram, which will give

a maximum for x; also if c = unity, (2) becomes x = y, the equation of a straight line bisecting the first angle, while if c<unity, the curve passes to the other side of the line just mentioned, and is tangent to P at y = , thus, and gives a minimum for x.

It is now our object to find, when c> 1, its particular value, such that'

or M = (base). (3), solved by successive approximations I find to give, to six decimals

c = 1.444667.... = ∞ max.,

for which value

1
=
y=
Το

2. 71828....=e=

Nap. base.

Now by reference to the first figure and construction of this article, we easily perceive that, if CV cut the curve, then for any such point of intersection we must have u vor v = log v in the system whose basis is c. Moreover, when c = x max., C V is tangent to the curve, giving

Also noticing that if c> 1.444667.... CV cannot intersect the curve; that for

C Vintersects it but once, we are prepared to make the following deductions:

I. In logarithmic systems with bases greater than e

1.444667.... there can be no logarithm equal to its antilogarithm or natural number. II. In the system whose base is e

1.444667....

e

there is but one such logarithm and its value is 2.718....e Nap. base, and it is at the same time the modulus (M) of the system,

III. In systems whose bases are less than unity, there is but one such logarithm and it is less than unity.

y

아

for point of inflection. That is, the tangent to the curve at the point of inflection is parallel to a diagonal of the rectangle formed by the co-ordinates of the point of inflection and the co-ordinate axes.

SOLUTIONS OF PROBLEMS IN NO. 3.

Solutions have been received as follows: S.J. Child solved 12; Theo. L. DeLand solved 11; Prof. A. B. Evans solved 11, 12, 13 and I4; Prof. J. M. Greenwood solved 13 and' 14; Henry Gunder solved 11 and 12; William Hoover solved 12; Artemas Martin solved 11, 12, 13 and 14; L, Regan solved 12; Walter Siverly solved 11, 12, 13 and 14; S. W. Salmon solved 11 and 12; Prof. J. Scheffer solved 11 and 12; and E. B. Seitz solved 15.

11. "Borrowed a sum of money at 8 per cent. simple interest and loaned it out again at 5 per cent. compound interest; in what time will I gain the amount borrowed."

SOLUTION BY HENRY GUNDER, GREENVILLE, OHIO.

Let the sum borrowed be $1, and put t for the time to gain this sum. Then by the conditions we have (1.05)' =2.08 t. By trial we find t = 30 years+x, a fraction of a year. To find this fraction we have the following equation: (1.05) + (1.05)30 x .05 x 4.40 + .08 x, from which we find x = .5735 years; hence the time is 30y. 6m. 26d.

12. "Given the base A C of a triangle and the ratio of A B to B C to find the locus of the point B by Geometry."

SOLUTION BY ARTEMAS MARTIN, ERIE, PA.

Put A C=a and the ratio of A B to B C as m to n; then (A B)2 : (B C)2:: m3 : n2.

Take the origin at A and let A D = ∞, BD=y; then A B= √ x2 + y3 and BC= √(a √ (a — x)2 + y3;

· · · x2 + y2 : (a — x)2 + y3 :: m3 : n3;

13. "A body is projected at a given distance from the center of force with a given velocity, and in a direction perpendicular to that distance: When the force is repulsive and varies inversely as the cube of the distance, find the path of the body."

SOLUTION BY ASHER B. EVANS, LOCKPORT, N. Y.

Let r and be the angular co-ordinates of the body at the time t from the commencement of motion, the origin being the center of force. Let v be the velocity of projection, and let the prime radius be so taken that r = a, and 0 = 0 at the instant of projection.

From the usual equations of motion we have, μ being the intensity of the repulsive force at a unit's distance,

Eliminating dt between these equations, we find for the differential equation of the curve