The conventions adopted in these illustrations are the simplest that have occurred to me. I may in a future article illustrate division from the same diagram, and addition and subtraction from other diagrams. In the mean time, let those who can, give a geometrical illustration of an imaginary exponent. EQUATIONS OF DIFFERENCES. BY WALTER SIVERLY, OIL CITY, PA. I AM not aware that this subject has been discussed in any American work, and, so far as I know, in but few foreign ones. The only works that I have seen that treat upon this subject are "Hymer's Finite Differences" and "De Morgan's Differential and Integral Calculus.” The limits of this article will not allow me to introduce the discussions involved in the integration of the equations; but I will give, 1st, the formula resulting from the integration of an equation of the first degree, in the first order of differences, with an example of its application; and 2nd, the formula resulting from the integration of an equation of the nth degree, with an example also of its application. These formulæ can be used without further knowledge of the subject. 1. Let u, u1, 2, . . . Ux−1, Ux, ux+1, &c., be any series of which u2, a function of x, is the general term; any relation between the terms in functions of x expressed by means of an equation, is called an equation of differences. Let the relation be u1+1 - Au, = B, A and B being constants. The integration of this equation gives C being an arbitrary constant determined as in ordinary integration. EXAMPLE.-One of two casks contains a gallons of wine and the other b gallons of water; e gallons are taked from the first and poured into the second cask, and then c gallons are taken from the second and poured into the first. Required the quantity of wine in the second cask after n such operations. Let uthe wine in the second cask after the nth operation, the first cask will then contain a -u, gallons of wine; at the next operation there will be taken out of the first cask a (a — u2) gallons of wine, which being poured into the second cask it will then contain gallons of wine; but the second cask now contains b + c gallons in all. Hence, taking out c gallons of the mixture in the second cask, there will remain in the second cask b+q{+(1—c)u. } gallons of wine. b Ип b+c -b (a–c) u = Un+1· bc Un+1 Un = a\b+c b+c .... ... 2. In the equation ux+n+Aux+n-1+Bux+n−2+ +Pu=0, putting u1 = 2′′ and substituting, the equation becomes z*+* + Azx+n−1 + B2*+n-8 + + P=0. Dividing by z* we have 2" + Az"-1 + B2-2 +...+P = 0, an equation of n dimensions. Let the roots of this equation be r1, 72, 73,.... r, then the complete integral of the original equation will be uz = C1(r1)* + C2(r2)2 + C3(rz)* +... + Cn(rn)*, C1, C2, C3, &c., being n arbitrary constants. (2) EXAMPLE.-A heifer calf at the age of two years has a heifer calf, and one every year afterward. All the progeny increase in the same manner. How many will there be at the end of the xth year? Let us the number at the end of the xth year. The increase for any year is from all those of two years preceding and from no others, hence But the first three terms of the series u, u1, u2, &c., are u。 1, and u2 = 2. Hence, putting x = 0, we have Au1 =1, and Pu。=1, When x=0 u, C1+C2 = 1. And when x = 1 = PROBLEM.-TO FIND THE RELATION BETWEEN THE MEAN ANOMALY AND THE TRUE ANOMALY. b = BY PROF. JOSEPH FICKLIN, COLUMBIA, MO. LET EDF be the orbit of the planet, having the sun in the focus at S. Put a = CE, CG, and from S as a center and a radius SA= V.(ab), describe the circumference of a circle; then the circle and the ellipse will contain equal areas. At the same time that the planet departs from E, the perihelion, let. a body begin to move with a uniform motion. from A through the circumference ABH, and perform a whole revolution in the same time that the planet describes the ellipse. Suppose the body describing the circle to be at B when the planet is at D; the angle ASB is the mean anomaly, and ESD the true anomaly. Put the angle ESD, and 0 = angle ASB; then the area of ASB is AS2. 0 = ab. 0. S2 2 Putr radius vector of the ellipse, then the area of ESD will be The polar equation of the ellipse, the focus being at the pole, is 1 = (1+e cos x)2 B(1+e cos x) + ; (1 + e cos x)2 1+e cos x Taking the differential coefficients we have A cos x (L+e cos x)+A esin’r 1 = A cos x (1+e cos x)+A e sin3x +B (1+e cos x) (2) Equating like powers of cos x, we have A+Be = 0, and A e+B=1; whence p❜e sin x p2 -- 2(1-e)(1+ecos x) (1—e2) 2(1-e2)(1+ecos x) (1—e2) When x = 0, area ESD tang-1 √(1±e) cot fx + C. 0, area ESD = 0; therefore, from this equation, p2 tang-1√(+6 (1+e 2 cos2) +C and, therefore, the general expression for the area ESD, is Substituting this value of Sr'da in (1) we obtain |