She relinquished the ax for the glass of tea.
He swung the ax once more, now only half a body length away.
Again she swung the ax, and this time it went half way through the log.
"That's called an ice ax, or piolet," Ryland answered.
Dean leaped on Shipton, clawing away at the soft snow, pummeling him like an eighth grade schoolyard brawler while Shipton, still clutching his ice ax in one hand, swung at Dean, catching him on the cheek and face with the side of the solid handle.
Shipton flailed out at him with his ax, missing his head by inches as Dean leaned sideways and frantically fumbled with his line to drop again.
He swung the ax again.
I suppose you've got a good reason why you tried to beat the brains out of a guy holding an ice ax, in the middle of the street with a bunch of people watching.
He swung his ice ax into the wall in front of him, dug in the toes of his crampons and began to ascend toward Dean.
Shipton swung his ice ax again, inching up closer to Dean.
Shipton's ax bit the ice scarcely a foot below Dean as the man glared up at him, a snarl on his face.
The leg wound from Shipton's flailing ice ax had been an eight-stitcher of no permanent consequence.
His muscular back glistened with perspiration as he swung the ax, expertly splitting a chunk of wood.
This time the ax sank about four inches into the wood - in another spot.
After a full minute of tugging and grunting she managed to dislodge the ax from the wood.
Ax., Green axis creeping on the surface of damp soil; rh., colorless rhizoids penetrating the soil; asc. ax., ascending axes of green cells.
Such curves are given by the equation x 2 - y 2 = ax 4 -1bx 2 y 2 +cy 4 .
- 2 ay2' ax, Ux2 ï¿½ï¿½ï¿½ a y n ay.
Taking two of the equations ax + +cz) x"' 1 +...
Other forms are n-1 n-2 2 ax +nbx x +n(n-i)cx x +..., 1121 2 the binomial coefficients C) being replaced by s!(e), and n 1, n-1 1 n-2 2 ax 1 +l i ox l 'x 2 + L ?cx 1 'x2+..., the special convenience of which will appear later.
As between the original and transformed quantic we have the umbral relations A1 = A1a1 d-A2a2, A2 = /21a1+/22a2, and for a second form B1 =A 1 b 1+ A 2 b 2, B 2 =/21bl +ï¿½2b2ï¿½ The original forms are ax, bi, and we may regard them either as different forms or as equivalent representations of the same form.
L ax 2 2 ax i l This is called the kth transvectant of f over 4); it may be conveniently denoted by (f, (15)k.
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.
But his cry came an instant too late as Shipton plummeted past him, his ice ax swinging in a rip across Dean's calf as he plummeted backward into space, and down to the rocks and churning river below.
Taking a few steps back she gripped the ax half way down on the handle and slammed it down against the block of wood with a dull whack.
He lifted the ax, taking aim at a new block of wood.
In the theory of surfaces we transform from one set of three rectangular axes to another by the substitutions 'X=' by+ cz, Y = a'x + b'y + c'z, Z =a"x+b"y-l-c"z, where X 2+Y2+Z2 = x2+ y2+z2.
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+ ?
From the three equations ax = alxl+ a2x2, b.= blxl+b2x2, cx = clxi+c2x2, we find by eliminating x, and x 2 the relation a x (bc)+b x (ca) +c x (ab) =0.
32 ax l ay 2 ax2ay1' which, operating upon any polar, causes it to vanish.
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.
Since, If F = An, 4) = By, 1 = I (Df A4) Of A?) Ab A"'^1Bz 1=, (F, Mn Ax I Ax 2 Axe Ax1) J The First Transvectant Differs But By A Numerical Factor From The Jacobian Or Functional Determinant, Of The Two Forms. We Can Find An Expression For The First Transvectant Of (F, ï¿½) 1 Over Another Form Cp. For (M N)(F,4)), =Nf.4Y Mfy.4), And F,4, F 5.4)= (Axby A Y B X) A X B X 1= (Xy)(F,4))1; (F,Ct)1=F5.D' 7,(Xy)(F4)1.
It is (f = (ab) 2 a n-2 r7 2 =Hx - =H; unsymbolically bolically it is a numerical multiple of the determinant a2 f a2f (32 f) 2ï¿½ It is also the first transvectant of the differxi ax axa x 2 ential coefficients of the form with regard to the variables, viz.
In general for a form in n variables the Hessian is 3 2 f 3 2 f a2f ax i ax n ax 2 ax " ï¿½ï¿½ ' axn and there is a remarkable theorem which states that if H =o and n=2, 3, or 4 the original form can be exhibited as a form in I, 2, 3 variables respectively.
If the forms be ax, b2, cy,...
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.
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.
The Binary Quadratic.-The complete system consists of the form itself, ax, and the discriminant, which is the second transvectant of the form upon itself, viz.: (f, f') 2 = (ab) 2; or, in real coefficients, 2(a 0 a 2 a 2 1).
=y la xl -i-y2a x2 must also vanish for the root a, and thence ax, and a must also vanish for the same root; which proves that a is a double root of f, and f therefore a perfect square.
If a, a be the linear forms, above defined, he raises the identity ax(0) =ax(aJ3) - (3x(aa) to the fifth power (and in general to the power n) obtaining (aa) 5 f = (a13) 5 az - 5 (a0) 4 (aa) ax?3 -F...
When C vanishes j has the form j = pxg x, and (f,j) 3 = (ap) 2 (aq)ax = o.
Hence, from the identity ax (pq) = px (aq) -qx (ap), we obtain (pet' = (aq) 5px - 5 (ap) (aq) 4 pxg x - (ap) 5 gi, the required canonical form.
When a z and the invariants B and C all vanish, either A or j must vanish; in the former case j is a perfect cube, its Hessian vanishing, and further f contains j as a factor; in the latter case, if p x, ax be the linear factors of i, f can be expressed as (pa) 5 f =cip2+c2ay; if both A and j vanish i also vanishes identically, and so also does f.
Thus what have been called seminvariants are not all of them invariants for the general substitution, but are invariants for the particular substitution xl = X11 + J-s12, X 2 = 112 Again, in plane geometry, the most general equations of substitution which change from old axes inclined at w to new axes inclined at w' =13 - a, and inclined at angles a, l3 to the old axis of x, without change of origin, are x-sin(wa)X+sin(w -/3)Y sin w sin ' _sin ax y sin w a transformation of modulus sin w' sin w' The theory of invariants originated in the discussion, by George Boole, of this system so important in geometry.
Then of course (AB) = (ab) the fundamental fact which appertains to the theory of the general linear substitution; now here we have additional and equally fundamental facts; for since A i = Xa i +,ia2, A2= - ï¿½ay + X a2, AA =A?-}-A2= (X2 +M 2)(a i+ a z) =aa; A B =AjBi+A2B2= (X2 +, U2)(albi+a2b2) =ab; (XA) = X i A2 - X2 Ai = (Ax i + /-Lx2) (- /-jai + Xa2) - (- / J.x i '+' Axe) (X a i +%Ga^2) = (X2 +, u 2) (x a - = showing that, in the present theory, a a, a b, and (xa) possess the invariant property.
(ab), aa, ab, (xa), ax, xx.
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.
There is no linear covariant, since it is impossible to form a symbolic product which will contain x once and at the same time appertain to a quadratic. (v.) is the Jacobian; geometrically it denotes the bisectors of the angles between the lines ax, or, as we may say, the common harmonic conjugates of the lines and the lines x x .
So they filled a small boat with the things that he would need the most--an ax, a hoe, a kettle, and some other things.
An ax will be useful, a hunting spear not bad, but a three-pronged fork will be best of all: a Frenchman is no heavier than a sheaf of rye.
He was armed with a musketoon (which he carried rather as a joke), a pike and an ax, which latter he used as a wolf uses its teeth, with equal ease picking fleas out of its fur or crunching thick bones.
Tikhon with equal accuracy would split logs with blows at arm's length, or holding the head of the ax would cut thin little pegs or carve spoons.
He had a musketoon over his shoulder and an ax stuck in his girdle.
So I went for them with my ax, this way: 'What are you up to?' says I.
She gripped the ax handle.
The ax blade went about an inch into the wood.
He hefted the ax to his shoulder and grinned down at her.
The ï¿½th polar of ax with regard to y is n-ï¿½ a aye i.e.
The complete covariant and contravariant system includes no fewer than 34 forms; from its complexity it is desirable to consider the cubic in a simple canonical form; that chosen by Cayley was ax 3 +by 3 + cz 3 + 6dxyz (Amer.
Hit him with an ax, eh!...