#### Sentence Examples

• Y1 = x 15+f2n; fï¿½ y2 =x2-f?n, f .a b = ax+ (a f) n, l; n u 2 " 2 22 2 +` n) u3 n-3n3+...+U 2jnï¿½ 3 n Now a covariant of ax =f is obtained from the similar covariant of ab by writing therein x i, x 2, for yl, y2, and, since y?, Y2 have been linearly transformed to and n, it is merely necessary to form the covariants in respect of the form (u1E+u2n) n, and then division, by the proper power of f, gives the covariant in question as a function of f, u0 = I, u2, u3,...un.
• In order to obtain the seminvari ants we would write down the (w; 0, n) terms each associated with a literal coefficient; if we now operate with 52 we obtain a linear function of (w - I; 8, n) products, for the vanishing of which the literal coefficients must satisfy (w-I; 0, n) linear equations; hence (w; 8, n)-(w-I; 0, n) of these coefficients may be assumed arbitrarily, and the number of linearly independent solutions of 52=o, of the given degree and weight, is precisely (w; 8, n) - (w - I; 0, n).
• The Number Of Linearly Independent Seminvariants Of The Given Type Will Then Be Denoted By (W; 0, P; 0', Q) (W; 0, P; 0', Q); And Will Be Given By The Coefficient Of A E B E 'Z W In L Z 1 A.
• If we take a fixed point (x',y',z') and a curve u = o of order m, and suppose the axes of reference altered, so that x', y', z' are linearly transformed in the same way as the current x, y, z, the curves (x' - 'x' + y z' 2 u = o, (r = I, 2, ...