En+2=0, a complete integral of (10) will be given by y= P(S_ES_n)(x), the arbitrary constant C being added after the έ integration, and C, after the 7 integration. Applying this to the primary (8), we find = η C and n= Hence C=n gives n=έ; and if we take the particular integral √(x)=0, we find y= =P (S) ndx 0. For simplicity add unity after each in tegration. Then, whether we evaluate by series, or by treating dy n y as an integral of the linear equation dx √x2 + y = 0, we 2 have y=C+1(x+√x2+1)"; and in like manner the second corresponding particular integral may be found. 7. Boole seems to have considered the regular and the primary forms as of distinct species. This I attribute in part to his not recognizing, in the theory of the former, any change of the independent variable other than that from a to ka". Under appropriate changes there is a certain reciprocity which appears to indicate that all the forms are but varieties of one species. 8. All binomial (op. cit. p. 430) biordinals may be included in d2y 2 a+ ex1 dy 1f+g” + dx2 xh+kx dx Taking the criticoid, we have + day L+M+Ngân x2 h+kxn y=0. y=0; for in this paper I make no explicit change of variable, the form alone of the results being material. The last equation is a trinomial, wherein When ==, the terordinal is soluble as a primordinal. When (e) is insoluble, a transformation of (d) may possibly have a cæsura which is soluble. expressions which give M-L-N the value (k-h) (f-g+a-e) +n(ka—he) + (a—e)2. 9. If h and k are finite, then by changing x into 2 x, dividing, changing x into x", and making appropriate changes in the constants, we may without loss of real generality replace the binomial by (1+x2) x2 d2+2(a+ex2)x the trinomial by d2y dx2 (1+x2)2x2 +(L+Mx2+Nx1)y=0, '. and the accompanying system by L=f+a-a2, N=g+e-e2, M=f+g+3a-e-2ae, 10. Boole's process reduces (11) to {D(D−1)+2aD+f}y +{(D−2)(D−3)+2e(D−2)+g}x2y=0, . (17) and if we put A(F-1)=—L, E(E+1)=-N, then (11) is solved through Boole's reductions : First, if A and E are both integers; Secondly, if Æ-A-E is an odd integer. And (11) is primary : First, if A is an integer and 2 an odd integer; 11. In (12) change the independent variable from ≈ to tan x and take the criticoid of the result. Then (12) is replaced by d2y dx2 +(L cot2 x+M+1+N tan2 x)y=0... (18) Now, if by a factorial substitution we pass from a form in which the last coefficient is variable to one in which such coefficient is constant, the two forms may be called conjugate. A conjugate of (18) is day +2(A cota + E tan x) dy where Q=M+1—A+E+2AE=Æ2 - (A — E)2. Next differentiate (19), replace dx dy by y, and take the criticoid the result. We have dey dx2 + {M + 1 − (A + 1) A cot2 x − (E-1) E tan2 x Q1 =M+1-2A+2E − (A+1)+(E− 1) + 2(A + 1) (E—1) =Q-4A+4E-4. Hence Q1 = Æ2 — (A—E+2)2; (22) and, if neither A nor E is an integer, we shall, after performing this process n times, transform (19) into But Q will vanish if (A-E) is an even integer. Now the two values of A are connected by A1+A2=1; so that if, for instance, Æ+A, -E be even, then E-A-E will be odd; and the condition coincides with one obained from Boole's process. 12. For clearness I have supposed that neither A nor E is entire. But if both or either be so the process is not stopped. The identities c2+c= (c+1)2 — (c+1) and c2-c=(c−1)2+c-1 give us a choice of conjugates. Let E=n, then tan x disappears from (20) and its conjugate; but, since c=-1 satisfies c2+c=0, it may be made to reappear. When A and E are both entire, P may be made to vanish; and Qn is always a constant. Í believe that the results of this process are coextensive with the regular results of that of Boole, and that it applies to such forms which, though of the soluble form of art. 6, is regular. It does not apply to day cot x dy 2 dx2 which is a coresolvent. + 3 dx+y=0, 13. But the reciprocity of forms will appear if in (12) we change x into x√-1, thus obtaining R=L-N-Æ2 + 1 + L cot2 x − (Æ2-1) tan2 x = L cot2 x +m+1+n tan2 x, suppose. The last differential equation will be solved regularly if A be an integer, and if Æ2−}=j(j+]), where j is an integer, i. e. if Æ is half an odd integer. Hence a primary form will have become regular. Again, let a2=m−L−n+1={−N=(E+ })2; then, if E be an integer, 2a will be half an odd integer, and, A being entire, a regular form will have become primary. 14. If in (25) we change a into sec z and take the criticoid, we obtain a result in which the above value of R is replaced by R=N-L-E2+1-(2-1) cot2 + N tan2 x=λ cot2 x +μ+1+N tan2, suppose. Here œ2=μ-λ-N+1={−L=(A—})o, and corresponding inferences may be drawn. 15. The theory of coresolvents shows that if in (10) we take where the modulus of the elliptic integral is cos 15°, then, if a=0, the equation is soluble*. If we rationalize it by changing x into cos am x, it becomes (1 − x2) { 1 —c2(1−x2) } d2 + Pd+Qy=0, dy dx2 and c= cos 15°. This form is not binomial. There is another soluble form when, with the same value of z, we have the sign of integration including an arbitrary constant. "Oakwal" near Brisbane, Queensland, Australia, November 20, 1874. XVIII. Projection of the Fraunhofer Lines of Diffraction and Prismatic Spectra on a Screen. By Prof. JOHN C. DRAPER, College of the City of New York†. H AVING been engaged during the past year in making photographs of absorption-spectra of organic bodies, in which a solar spectrum with Fraunhofer lines was formed by a diffractiongrating, I have resorted to the following method of forming such solar spectra, a description of which may prove of interest to those who are experimenting in the same field, The grating generally used was made by Mr. L. M. Rutherfurd: it is ruled on speculum-metal, 6481 lines to the inch; it gives spectra by reflection. Other gratings on glass, now in my possession, give spectra by reflection and by transmission. The method answers equally well for both. It may be briefly stated as follows:: See the Educational Times,' September 1874, p. 137, and the 'Messenger of Mathematics' there referred to. From the American Journal of Science and Arts, vol. ix. 1875. |