а which B=0 and D=0, which reduces the equation to the form aʻy=x+2hx+k, when the locus is infinite. It has as curvilinear asymptotes the Proximate twin Parabolas, aʻy=(22+h)? The Species are:-1, Twin Goblets ; 2 (when their vertices unite), Pointed Goblets or Knotted Parabolic Hour-glass; 3, Parabolic Hour-glass ; 4, Perforated Hour-glass (with disk in centre); 6, Hollow-bottomed Goblet. When A=0, we may have asymptotes parallel to the axis of x, and when B=0, to the axis of y. Such curves must be treated apart. When a Quartan Hyperbola is confined between parallel asymptotes, I call it an Arch, Round-headed or Hollowheaded, as the case may be ; they are found, of course, in pairs. A Quartan Hyperbola which is contined within diverging asymptotes like the Conic Hyperbola, I call a Basin; when it crosses or otherwise envelopes its diverging asymptotes, I call it a Cup. Cups and basins may be Round-bottomed or Hollow-bottomed. Again, a Quartan Hyperbola may lie between an oblique and a vertical asymptote; I then call the Hyperbola itself Oblique, equally when it lies between two asymptotes of different systems. Such an Hyperbola may cut one, and only one, asymptote; then I call it Paratomous : if it cut both, it is an Oblique Cup. Cups may be Pointed at bottom and unite; they may be also in Contact at bottom, or they may intersect. Vertical and horizontal asymptotes develope other and simpler forms. Conchoids, grouped in pairs, generate one class, and Arches another. Arches may intersect, Basins also may intersect sideways; I call this Paratomy. Such are the elements (adding only Studs or Conjugate Points) of which all the loci are composed. Four Hyperbolas, of whatever class, are the utmost that can arise as locus of a Quartan equation ; whether in square, each in one quadrant, or as Cross Arches, or as Oblique, or Oblique and Paratomous, or as around and crossing the axes, or between unsymmetrical asymptotes, or it may be Cups instead of Basins. In the midst of these infinite curves, some one of the Monad or Twin Ovals are often found as Satellites. It must be added that when AB=D2 and the locus is infinite, we find Oblique Parallel Asymptotes, and even, related to them, Oblique Paratomous Arches. Such is the general description of the forms. The investigation is simple. We know that a straight line can cut a Quartan at most in four points. This often shows what forms are impossible. A D E Put V=DBF then by Conics we know that if V=0, our general equa E F C tion will degenerate into the product of two quadratic factors. Besides, if A, B, C are positive and E, F negative, and E-=AC, F=BC, the equation degenerates into two ellipses. If F?=BC, and B, F have opposite signs, the curve crosses itself (in a Knot) where x=0 and By2 +F=0; but if B, F have the same sign, the Knots become Studs. Thus if E-=AC, and F=BC, but E, F have opposite signs, there are two Knots on one axis and two Studs on the other. We find where the curve crosses its axis, by putting Axt+2Ex?+C=0 when y=0, and By +2Fy+C=0 when x2=0. Then if AC is positive, E must be negative, if there be any vertex in OX. If AC is negative, there are two vertices. Put T=Ax:+2D.xayo+Bys; ..T+(2EP.22+2F4y+C)=0 is the equation to the curve. When T is essentially positive, the curve is finite. This happens when A, D, B are all of one sign, or w when AB-D2>0. When A=0 and B=0, the curve is finite only when D, E, F are of one sign. If B=0, Art+2Er+C -y= 2022 +F and the curve is finite when A, D, F are of one sign. If D and F are of opposite signs, there are asymptotes parallel to y, viz. 2Dx2 +F=0. Indeed now T=(Ax+2Dya)."; thus, if A and D have opposite signs, A.x2 +2Dy2=0 are oblique asymptotes. When A, D, B are finite, solve the equation for y’, regarding B as positive. Put g=D-AB, h=DF-EB, k=F---CB; .: By+Dx' +F="V (gxt +2hx? +k). For D2- AB>0, we may assume g=1; then for the upper sign we get, as Proximate Conic Hyperbola, if I be <1, By,’ +Dx+F=x2 +h. By, +Dx®+F=-r-h. If D-AB=0, y=0, and the curve is infinite only when h is positive: then if D is negative, By,'+DxP+F=+v(2h)x is two Proximate Conic Hyperbolas, and the asymptotes are oblique and parallel in pairs; they do not pass through the centre, but are equidistant from it. Evidently if C=0 and E, F have opposite signs, the curve crosses itself in the centre; but if C=0 and E, F have the same sign, the centre is a mere Stud. An undulation of the curve implies a double tangent. Such double tangents are always parallel to one axis. _[I desire a general proof. There can be only two pairs parallel to one axis. To ascertain whether there is undulation across OY, put x=0, and try whether y2 is there a maximum or a minimum. Making zo infinitesimal and k positive (which is implied), hx? ✓(k+2hx?+gr)=vk+ 2k ya thus for upper sign, h gk-ha 24 2k h whence ya is a minimum at w=0 if h -D>0, a maximum if -D<0. Yet h Nk when =D, ya is a minimum at x= v k =0 if gk-h2>0, or a maximum if gk-h? Co. Now gk-ho=BV, and we cannot have V=0 without degeneracy. Hence this test is final. Also h=D Nk is equivalent to BE? – 2DEF+CD=0. If the branch we are investigating is infinite and yis a minimum, there is no undulation; but if ya is a maximum, it begins to decrease, yet must afterwards increase; hence there is undulation. On the contrary, if the upper branch be finite and y? be a maximum, there is no undulation; but there is undulation if yž be a minimum. dy In general, for tangents parallel to the axis of x, putting dx=0, we have the obvious solution x=0, when there is a vertex on the axis of y. Besides this, we may have a double tangent where DV (gx' +2hx*+k)=gx' +h, which yields gr?+h /h-gk = V(gx* +2hx?+k)=By +Dx®+F. Hence at the points of contact x 9kh2 x VR if h'=DE-FA. When T=(1972-) (p*r?+o'y), the curve has oblique asymptotes, 1 'x'=u'y! To try whether it ever cut its asymptotes, put y=y.; then at the common point 1°x:=poy?, and 2ExS+2Fy+C=0. If the x, y hence determined is within the limits of the curve, it does thus cut; if not, it does not. If T=(1922 – u?y?) (p2x2 -— oʻy?), there are two pair of oblique asymptotes, A’x?=moy?, pʻx=oʻy?; and by combining either of them with the second equation, 2Ex? +2Fya+C=0, we decide on Paratomy. When the general equation is given to us in this form and we desire to find the Proximate Conics, the most direct method is to assume pa=1, oʻ=1, and (y - Xor? - M)(y-p?r? -N)=T+2(Ex? +Foy')+MN; whence X?N+p'M=2E, N,} 2F1 or -N= 2(E+Fp) pa-12 Then the Proximate Conics are y,' --r?=M, y?-pére=N. These are closer indications of the infinite branches than the asymptotes. Put MN=C', A DE EFC But in general it is expedient to put X=gx'+2hx*+k, and study the variations of X in the equation By? +Dx?+F=+VX. In many cases the lower sign is inadmissible; in most it is more restricted than the upper. When we have only the upper, evidently there is no undulation across the axis of x; for ya has then but one positive value for any given value of x. The forms of X are as follows: X=no(m? —x'), X=n'x+ m), XE = *(co-mo ), X], =(x -mo)+n4 6 10 = 11 13 Remarks on Napier's original Method of Logarithms. By Professor PURSER. On Linear Differential Equations. By W. H. L. RUSSELL, F.R.S. The object of this paper was to explain the progress the author is making in his theory for the solution of Linear Differential Equations, especially when the complete integral involves logarithmic functions. On MacCullagh's Theorem. By W. H. L. RUSSELL, F.R.S. This paper was intended to simplify the process given by Dr. Salmon to prove MacCullagh’s theorem relative to the focal properties of surfaces of the second order. Note on the Theory of a Point in Partitions. By J. J. SYLVESTER, F.R.S. In writing down all the solutions in positive integers of the indefinite Equation of Weight, x+2y+32+ ...=n, or, in other words, in exhibiting all the partitions of n any integer greater than zero, it may sometimes be useful to be provided with an easy test to secure ourselves against the omission of any of them. Sucha a test is furnished by the following theorem : =(1-x+y-Ta....)=0. (1) y=2, (5) x=4, the values of the omitted variables in each solution being zero. The five corresponding values of 1—æ+xy.... are 1, 1, 0, 1, -3, whose sum is zero. The theorem may be proved immediately by expressing the denumerant (which is zero) of the simultaneous equations x+2y+3z+...=n, x+y+z+.....=0, in terms of simple denumerants according to the author's general method, or by virtue of the known theorem, (1-1)(1—ť)(1-t).... ts t6 + + t (1-t)'1 t This gives at once the equation 1 t t ...=1. (1-t)(1—+2)(1 – t)... (1—t)?(1 —t)(1—?)... (1—t)?(1—?)?(1–49)... Hence the coefficient of tr in the above written series for all values of n other than zero is zero. But it will easily be seen that the coefficient of tn in the first term is el, in the second term Ex, in the third Exy, &c.; so that (1-2+xy....)=0, as was to be shown. Thus we have obtained for the problem of indefinite partition a new algebraical unsymmetrical test supplementing the well-known pair of transcendental symmetrical tests expressible by the equations z T7(x+y+2...)=0 II.x [ly [lz... :50;} = =1-1-0) (1-1)(1–16) (1–t)(1-7)(1-6) + (1-1)(1–4) (1–P)(1–4) 1–2*–e... + 1-A3 (1–)... 11(x+y+%...)=2n-1, E(-)+y+ ... + 1/(x+y+. * Subject of course to the conditions that n is greater than 1. If x, y, 2, .. represents any solution in positive integers of the equation x+2y+3z... trw=r, it is easy to see that .tw) =1, -1, or 0, IIx Ily. Ilw according as n, in regard to the modulus r+1, is congruent to 0, 1, or neither to O nor 1, for the left-hand side of the equation is obviously the coefficient of cm in the development of 1 1-3 i.e. 1+x+2...+2P 1-Xti On making r=00, this theorem becomes the one in the text. It obviously affords & remarkable pair of independent arithmetical quantitative criteria for determining whether or not one number is divisible by another. + t...e The identity employed in the text is only a particular case of Euler's identity, tz t2 (1+tz)(1+t*2)(1+tz)....=1+ (1-t)'(1-1)(1--€) which is tantamount to affirming that the number of partitions of n into r distinct integers is the same as the number of partitions of n into any integers none greater than r, in which all the integers from 1 to r appear once at least. It has not, I believe, been noticed that these two systems of partitions are conjugate to each other, each partition of the one system having a correspondent to it in the other. The mode of passing from any partition to its correspondent is by converting each of its integers into a horizontal line of units, laying these horizontal lines vertically under each other, and then summing the columns. Thus, ex. gr., 3, 4, 5 will be first expanded horizontally into 1 1 1, 1 1 1 1 1, and then summed vertically into 3 3 3 2 1. This is the method employed by Mr. Ferrers to show that the number of partitions of n into r, or a less number of parts, is the same as the number of partitions of n into parts none greater than r, and is, in fact, only a generalization of the method of intuitive proof of the fact that mXn=nXm, the difference merely being that we here deal with a parallelogram separated into two conterminous parts by an irregularly stepped boundary-one filled with units, the other left blank, instead of dealing with one entirely filled up with units. On the General Canonical Form of a Spherical Harmonic of the nth Order. By Sir W. Thomson, LL.D., D.C.L., F.R.SS. L. & E. Let H; (x, y, z), H'; (x, y, z)...., or for brevity H, H', &c., denote 2i+1 index , , 1 pendent spherical harmonics of degree i, that is to say, homogeneous functions each fulfilling the Laplace's equation dx2 (1) dya dza The formula AH+A' II', &c., (2) where A, A' are constants, is a general expression for the harmonic of degree i; but it is not a “canonical” expression. Borrowing this designation from Mr. Clifford's previous paper, we may define as canonical constituents for the general spherical harmonic of degree i, any set of particular distinct harmonics fulfilling the following conditions : SSHH'doro, [SH' H"do=0, &c. (3) where sf do denotes integration over any spherical surface having the origin of coordinates for centre. Supposing now that H, H' ... actually fulfil these conditions, let it be required to find, if possible, another canonical form (H, H', ...). Try V=AH+A'H'+&c., H'=BH+B' H'+&c. AB+A' B' +&c.=0. A, A', A", . |