This is the particular solution of equation (1), and the complete solution is found by adding the complementary function to the expression on the right side of the above equation. d In the equation (sin(.)}~ dx y=$(x), the solution takes a simpler form than in the first equation. The complementary function is the primitive of {sin(x. d)) Y=0; y=em.x will be a particular solution of this if m be such as to make sin(\.m)=0. The required values of m are found by giving to r all integral values (including zero) from r.π. to ∞ in λ Hence Y=Σ Υπα the expression on the right side of this equation is the complementary function in the solution of the second equation. To find the particular integral we write equation (2) in the The complete solution is found by adding the complementary function to the expression on the right of the above equation. The solution of the equation (2) is therefore Eighth Meeting, June 8th, 1894. Dr C. G. KNOTT, President, in the Chair. On the Highest Wave of Permanent Type. By JOHN M'Cowan, D.Sc. On a Problem in Tangency. FIGURE 17. Let AOB, DOE be any two intersecting chords of a circle. Required to inscribe a circle in one of the compartments, as BOE. Draw OF bisecting the angle BOE and let it cut the circle in F. Draw FM perpendicular to AB. Join C' (the centre of the circle) to O and produce to cut the circle in G. [Draw GH in any direction equal to FM, and join C'H. Draw OK parallel to GH cutting C'H in K.] Cut off OL=LQ=OK. Join LF and with Q as centre and radius FM or GH describe a circle cutting LF in S. Draw QSR and draw a parallel to it through C' cutting OF in F' and the circle in P'. Then F' shall be the centre of the required circle. PROOF.-Draw F'M' perpendicular to OB, draw through F a parallel CP to C'P', and join OP', producing it to cut CF in P. Then ... LQ QS OL: FM :: OK : GH :: C'O : C'G :: C'O: C'P' the A's LQS and OC'P' have an angle common and the sides about that angle proportional. and C is the centre of circle AGB a circle with centre F' and radius F'M' will touch both OE and OB and will touch the circle AGB at P'. The Algebraic Solution of the Cubic and Quartic in x by means of the Substitution λα + μ 1 + x By CHARLES TWEEDIE, M.A., B.Sc. E. Carpenter's Proof of Taylor's Theorem. By R. F. MUIRHEAD. The idea of the following proof was communicated to me some years ago by Mr Edward Carpenter of Millthorpe, Derbyshire, formerly Fellow of Trinity Hall, Cambridge; who remarked that it seemed to afford a demonstration of Taylor's Theorem which came very naturally and directly from the definition of a differential coefficient. The chief difficulty seemed to arise in dealing with the negligible small quantities which are produced in great numbers. However, I found it not difficult to complete the proof for the case when all the successive differential coefficients of f(x) are finite and continuous. It occurred to me lately that this proof might interest the Society and it is here given with the addition of a modified proof leading to an expansion in m terms with a remainder. : 1. If f(x) possesses a differential coefficient f'(x), then where a is a quantity which vanishes with h. Similarly ƒ' (x + h) =ƒ' (x)+hf" (x)+ha2 ƒ"(x+h)=ƒ"(x)+hƒ'''(x)+ha ̧ and so on, a ag etc. being quantities which vanish with h. Writing x+h for x in (1) we have f(x+2h) = f(x+h) + hf'(x+h) + ha' where a, also vanishes with h; and hence, by (2) f(x+2h) = f(x) + 2hf'(x) + h2ƒ"(x) +h(a1 + a1') + h2a |