## -yl Sentence Examples

- Chem., 1832, 3, p. 2 49) is to be regarded as a most important contribution to the radical theory, for it was shown that a radical containing the elements carbon, hydrogen and oxygen, which they named benzoyl (the termination yl coming from the Gr.
- Suppose n dependent variables yl, y2,ï¿½ï¿½ï¿½yn, each of which is a function of n independent variables x1, x2 i ï¿½ï¿½ï¿½xn, so that y s = f s (x i, x 2, ...x n).
- ï¿½ Oxl d 2x 77n If we have new variables z such that zs=4s(yl, Y2,...yn), we have also z s =1 Y 8(x1, x2,ï¿½ï¿½ï¿½xn), and we may consider the three determinants which i s 7xk, the partial differential coefficient of z i, with regard to k .
- Zl, z2,...zn) yl, y2,.
- X n / I l yl, y2,ï¿½ï¿½ï¿½yn Theorem.-If the functions y 1, y2,ï¿½ï¿½ï¿½ y n be not independent of one another the functional determinant vanishes, and conversely if the determinant vanishes, yl, Y2, ...y.
- We can solve these, assuming them independent, for the - i ratios yl, y2,...yn-iï¿½ Now a21A11 +a22Al2 ï¿½ ï¿½ ï¿½ = 0 a31A11+a32Al2 +ï¿½ ï¿½ï¿½ +a3nAln = 0 an1Al1+an2Al2 +ï¿½ï¿½ï¿½+annAln =0, and therefore, by comparison with the given equations, x i = pA11, where p is an arbitrary factor which remains constant as i varies.
- Yl, y 2,...yn) (zl, z2,...zn z1, z 2, ï¿½ï¿½ï¿½zn xi, 'X' 2,...
- X n/ yl, Y2,...y n j ' x 1, ï¿½ Forming the product of the first two by the product theorem, we obtain for the element in the ith row and kth column aZ, ayl az i ayz azi ayn ayl + e +...+ where or as a21 a22 ï¿½ï¿½ï¿½a2,i -1 a2,i +1 .ï¿½ï¿½a2n a31 ï¿½ï¿½ï¿½a3,i -1 a3,ti+ 1 ï¿½ï¿½ï¿½a3n ï¿½ï¿½ï¿½yi -)tin,and a7,2 ï¿½ï¿½ï¿½a,,,i -1 an,i+1 ...anï¿½' a21 a22 ï¿½ï¿½ï¿½a2, -1 a32 -1 I.
- Y m (xly2 - x2y1) 2 (x0,3 - x 3 yl) 2...
- Moreover, instead of having one pair of variables x i, x2 we may have several pairs yl, y2; z i, z2;...
- 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.
- Ch (2n +1)I 7 ry /a yl-R3ct (2n +I)3.ch(2n+I)17b /a ' 16 cos(2n+I) 2 7 z /a w1=4,i+ 4, ii = iR ?3a2 an elliptic-function Fourier series; with a similar expression for 1,'2 with x and y, a and b interchanged; and thence 4, = '1 +h.
- Now (x1 - x21) (y 1 +y21) = xl l +x2y2 + - (' r 1 2 - x2y1) = FG-x3y3+iV X3, yi+3 7 21_FG-x3y3+2V X3 xl+x21 X12 +X22 (x 1 +x 2 i) = - i{(q' - q)x3+r'y3]+irx3(y1+y21), = FG - x3y3 +ZJ X3 dt2log(x1+x22) - - (q g) x 3- r y3+rx3 F2x32 (12) d dl2 log V x1 ± x2 2 (q'-q)x3-(r'-r) y3FrFF2-x 2 3 ' (13) requiring the elliptic integral of the third kind; thence the expression of x1-f -x21 and yl-}-y21.
- If yl is known this gives 72.