Oxford, Cambridge and Dublin Messenger of MathematicsMacmillan and Company, 1887 |
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½π a₁ algebra angle B₁ B₂ binomial coefficients C₁ C₂ cd'x cdx sdx ndx centre centroid cn'x cn*x cnx dnx coefficients cos'a cosc cosh cosß cosy cs*x cs2x curve deduce denote dn'x dn³x ds 2x ds'x ds*x dx dx dx log elliptic functions formulæ Fourier's given circles given conic Hence hyperbola integral inverse J. W. L. Glaisher k³k kdex kp snu line at infinity log cdx log snx matrices ns 2x ns*x nsx ds nsx dsx csx obtain orthogonal circle orthoptic locus P₁ parabola q-products q-series quadrics respectively sinh sn*x sn³x snx cnx dnx straight line Sumner line tangents tanh theorem three given triangle Y₁ π π πυ ΣΤ кпа
Populære avsnitt
Side 33 - ... cos a = cos b cos с + sin b sin с cos A ; (2) cos b = cos a cos с + sin a sin с cos в ; ^ A. (3) cos с = cos a cos b + sin a sin b cos C.
Side 21 - ... the element of normal intercepted between this surface and the consecutive one. The principal tangents at the other end of the normal element may be constructed by (1) revolving the principal tangents through an angle...
Side 37 - T ein w = p, •where ш is the angle which the perpendicular makes with the axis of x. Denoting F(x, y) by u, and comparing this form of the equation with that...
Side 180 - If fig. 179 were spun about OA, what figure would be generated (i) by the circle, (ii) by AP, (iii) by PQ? Hence find the locus of the points of contact of tangents from a fixed point to a fixed sphere. Ex.
Side 37 - A point moves on an ellipsoid so that its direction of motion always passes through the perpendicular from the centre of the ellipsoid on the tangent plane at any point ; shew that the curve traced out by the point is given by the intersection of the ellipsoid with the surface xm~" yn~l zl~m = constant, I, m, n being inversely proportional to the squares of the semiaxes of the ellipsoid.
Side 6 - Because the two triangles BFC, AFC have their sides BF, AF equal and in a straight line, and the point C common, the triangles are equal; (Prop. 38, Coroll.) therefore the triangle BFC is half of the triangle ABC. Similarly it can be proved that the...
Side 96 - Д (1, 2, 3) denotes the area of the triangle formed by the centres of the circles (1, 2, 3). 5. If the circle (y} be the inverse of any circle (т) with respect to the circle S, it is easily proved geometrically that 2 11 1fl,t ЧГX,» '"'y,' Formulœfor an Orthogonal System, §§6_10.
Side 102 - ... is equal to the reciprocal of the square of the radius of the sphere which cuts them orthogonally.
Side 21 - GÍJ + a./)0, /St + ßßd, 7, -I- 733^, and at the same time give to the other quantities their values at the other end of the normal element. We have then, effecting these changes, subtracting the corresponding equations in (1) with suffix (1) attached to a...