## Bx Sentence Examples

- It may be written in the form n n-1 2 ax 1 +bx1 x2 +cx 1 x 2 + ...; or in the form n n n=1 n n-2 2 +(1)
**bx**x2+ ? - May be a simultaneous invariant of a number of different forms az',
**bx**2, cx 3, ..., where n1, n 2, n3, ... - (III.) Again in (I.) transposing a x (bc) to the other side and squaring, we obtain 2(ac) (bc)axbx = (bc) 2 a'+(ac) 2
**bx**- (ab) 2 c1. - (ab)(ac)
**bxcx**= - (ab)(bc)axcx = 2(ab)c x {(ac)**bx**-(bc)axi = 1(ab)2ci; so that the covariant of the quadratic on the left is half the product of the quadratic itself and its only invariant. - - We have seen that (ab) is a simultaneous invariant of the two different linear forms a x,
**bx**, and we observe that (ab) is equivalent to where f =a x, 4)=b. - The two forms ax,
**bx**, or of, 0, may be identical; we then have the kth transvectant of a form over itself which may, or may not, vanish identically; and, in the latter case, is a covariant of the single form. - The process of transvection is connected with the operations 12; for?k (a m b n) = (ab)kam-kbn-k, (x y x y or S 2 k (a x by) x = 4))k; so also is the polar process, for since f k m-k k k n - k k y = a x by, 4)y =
**bx**by, if we take the k th transvectant of f i x; over 4 k, regarding y,, y 2 as the variables, (f k, 4)y) k (ab) ka x -kb k (f, 15)k; or the k th transvectant of the k th polars, in regard to y, is equal to the kth transvectant of the forms. Moreover, the kth transvectant (ab) k a m-k b: -k is derivable from the kth polar of ax, viz. - Put M 1 For M, N I For N, And Multiply Through By (Ab); Then { (F, C6) } = (Ab) A X 2A Y B X 1 M N I 2 (Xy), ?) 2, = (A B)Ax 1B X 2B Y L I Multiply By Cp 1 And For Y L, Y2 Write C 2, C1; Then The Right Hand Side Becomes (Ab)(Bc)Am Lbn 2Cp 1 M I C P (F?) 2 M { N2 X, Of Which The First Term, Writing C P =, ,T, Is Mn 2 A B (Ab)(Bc)Axcx 1 M 2 N 2 P 2 2222 2 2 _2 A X B X C (Bc) A C
**Bx**M N 2 2 2 M2°N 2 N 2 M 2 2 A X (Bc) B C P C P (Ab) A B B(Ac) Ax Cp 2 = 2 (04) 2 1 (F,0) 2.4 (F,Y') 2 ï¿½?; And, If (F,4)) 1 = Km " 2, (F??) 1 1 M N S X X X Af A _Af A Ax, Ax Ax Ax1 Observing That And This, On Writing C 2, C 1 For Y 11 Y 21 Becomes (Kc) K X 'T 3C X 1= (F,0 1 ', G 1; ï¿½'ï¿½1(F,O) 1 M 1=1 M 2 0`,4)) 2 0, T (Fm 2.4 (0,0 2 .F ' And Thence It Appears That The First Transvectant Of (F, (P) 1 Over 4) Is Always Expressible By Means Of Forms Of Lower Degree In The Coefficients Wherever Each Of The Forms F, 0, 4, Is Of Higher Degree Than The First In X 1, X2. - First observe that with f x =a: = b z = ï¿½ï¿½ï¿½,f1 = a l a z ', f 2 = a 2 az-', f x =f,x i +f 2 x i, we find (ab) - (a f)
**bx**- (b f) ax. - Moreover the second term on the left contains (a f)' c -2b z 2 = 2 (a f) k-2b x 2 - (b) /0-2a 2 ï¿½ if k be uneven, and (af)?'
**bx**(i f) of) '-la if k be even; in either case the factor (af)**bx**- (bf) ax = (ab) f, and therefore (n-k),bk+1 +Mï¿½f = k(n-2)f.(uf)uxn-2k-1; and 4 ' +1 is seen to be of the form f .14+1. - Determinant factors (abc), (abd), (bce), etc...., and other factors az,
**bx**, cx,... - If ai,
**bx**, cx be different forms we obtain, after development of the squared determinant and conversion to the real form (employing single and double dashes to distinguish the real coefficients of**bx**and cz), a(b'c"+b"c'-2 f'f") +b(c'a"+c"a'-2g'g") +c(a' +a"b'-2h'h")+2f(g'h"+g"h'-a' + 2g (h ' f"+h"f'-b'g"-b"g')+2h(f'g"+f"g'-c'h"-c"h'); a simultaneous invariant of the three forms, and now suppressing the dashes we obtain 6 (abc+2fgh -af t - bg 2 -ch2), the expression in brackets being the S well-known invariant of az, the vanishing of which expresses the condition that the form may break up into two linear factors, or, geometrically, that the conic may represent two right lines. - To assist us in handling the symbolic products we have not only the identity (ab) cx + (bc) a x + (ca)
**bx**=0, but also (ab) x x+ (b x) a + (ax) b x = 0, (ab)a+(bc)a s +(ca)a b = 0, and many others which may be derived from these in the manner which will be familiar to students of the works of Aronhold, Clebsch and Gordan. **BX**(2a b).- M _ c Q du dO dt - rrqu 2r qu AB _ Q du L bq (u -a') +V (b -a')1,l (a -u)11/ndu) L 1,1 (a-a/),,1 (u-b), f u Along the wall
**Bx**, cos n0 =I, sin n0 =o, b >u>o ch nSt = ch log () n=, , fb-a?, ? - - 2 To show that the area of a cross-section of a - prismoid is of the form ax e -{-
**bx**-{- c, where x is the distance of the section from one end, we may proceed as in § 27. - Diverging parabolas are cubic curves given by the equation y 2 = 3 -f-
**bx**2 -cx+d. - The load on an element
**bx**of the beam may be represented by wbx, where - w is in general a function of x.