From the law of angular motion of the latter its radius vector will run ahead of PQ near A, PQ will overtake and pass it at apocentre, and the two will again coincide at pericentre when the revolution is completed.
The number of partitions of a biweight pq into exactly i biparts is given (after Euler) by the coefficient of a, z xPy Q in the expansion of the generating function 1 - ax.
For write (pq) =sï¿½ and take logarithms of both sides of the fundamental relation; we obtain slox +soot' = + (3ly) S20x 2 +2siixy+s02y 2 = E(aix+(3 ly) 2, &C., and siox+SOly - (S 20 x2 + 2s ii x y+ s ooy 2) +...
From the above D p4 is an operator of order pq, but it is convenient for some purposes to obtain its expression in the form of a number of terms, each of which denotes pq successive linear operations: to accomplish this write d ars and note the general result exp (mlodlo+moldol +...
Recalling the formulae above which connect s P4 and a m, we see that dP4 and Dp q are in co-relation with these quantities respectively, and may be said to be operations which correspond to the partitions (pq), (10 P 01 4) respectively.
Since dp4+(-)P+T1(p +q qi 1)!dd4, the solutions of the partial differential equation d P4 =o are the single bipart forms, omitting s P4, and we have seen that the solutions of p4 = o are those monomial functions in which the part pq is absent.
The i t " power of that appertaining to a x and b x multiplied by the j t " power of that appertaining to a x and c x multiplied by &c. If any two of the linear forms, say p x, qx, be supposed identical, any symbolic expression involving the factor (pq) is zero.
Calling the discriminate D, the solution of the quadratic as =o is given by the formula a: = o (a0+a12_x2 (a0x+aix2 If the form a 2 be written as the product of its linear factors p.a., the discriminant takes the form -2(pq) 2.
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.
We can replace pr or -pr by (+p)r or (- p)r, subject to the conditions that (+p) (+q) = (-p) (- q) = (+pq), (+p) (-q) = (- p) (+ q) = (-pq), and that + (-s) means that s is to be subtracted.
In order to find the difference of optical distances between the courses QAQ', QPQ', we have to express QP-QA, PQ'-AQ'.
B is the point at which the effect is required, distant a+b from 0, so that AB= b, AP=s, PQ ds.
Taking as the standard phase that of the secondary wave from A, we may represent the effect of PQ by cos 27r (_) .ds, where, l = BP - AP is the retardation at B of the wave from P relatively to that from A.
Generally if S denotes any closed surface, fixed in the fluid, M the mass of the fluid inside it at any time t, and 0 the angle which the outward-drawn normal makes with the velocity q at that point, dM/dt = rate of increase of fluid inside the surface, (I) =flux across the surface into the interior _ - f f pq cos OdS, the integral equation of continuity.
Let PQ be the ordinate of the point P FIG.
On the circle, and let M be another point on the circle so related to P that the ordinate PQ moves from A to 0 in the same time as the vector OM describes a quadrant.
Then the locus of the intersection of PQ and OM is the quadratrix of Dinostratus.
Newton m m _ _ there states that (p+ pq) - = p n -}- T naq ?- m 2n n bq -?
5, the volume of the element between these planes, when h is very small, is approximately h X AB X arc PQ = h.
Y where K-=4, X qth moment with regard to plane y =o, Lm yn X pth moment with regard to plane x =o, and R is the volume of a briquette whose ordinate at (x,.,y s) is found by multiplying by pq x r P - 1 ys 4-1 the volume of that portion of the original briquette which lies between the planes x =xo, y =yo, y = ys.
12 represent a horizontal section of the dome through the source P. Let OPA be the radius through P. Let PQ represent a ray of sound making the angle B with the tangent at A.
Since the conditions in the region PQ remain always the same, the momentum perpendicular to AB entering the region at Q is equal to the momentum perpendicular to AB leaving the region at P. But, since the motion at Q is along AB, there is no momentum there perpendicular to AB.
Comparing this equation with ux 2 +vy 2 +w2 2 +22G'y2+2v'zx+2W'xy=0, we obtain as the condition for the general equation of the second degree to represent a circle :- (v+w-2u')Ia 2 = (w +u -2v')/b2 = (u+v-2w')lc2 In tangential q, r) co-ordinates the inscribed circle has for its equation(s - a)qr+ (s - b)rp+ (s - c) pq = o, s being equal to 1(a +b +c); an alternative form is qr cot zA+rp cot ZB +pq cot2C =o; Tangential the centre is ap+bq+cr = o, or sinA +q sin B+rsinC =o.
D'Arzo Cr Pq C Tr ' PgL.
Again if PQ be any segment of the beam which is free from load, Q lying to the right of P, we find FPFQ, MPMQ=--F.PQ; (12)
If PQ be a short segment containing an isolated load W, we have FeFi.=W, MQ=MP; (3) hence F is discontinuous at a p concentrated load, diminishing by an amount equal to the load as we pass the loaded point to the right, whilst M is continuous.
The plane in question is called the null-plane of P. If the null-plane of P pass through Q, the null-plane of will pass through P, since PQ is a null-line.
The weight of PQ and the tensions at P,Q.
A vector OU drawn parallel to PQ, of length proportional to PQ/~I on any convenient scale, will represent the mean velocity in the interval 1t, i.e.
Experience the same resultant displacement PQ in the same time.
For in time t the mutual action between two particles at P and Q produces equal and opposite momenta in the line PQ, and these will have equal and opposite moments about the fixed axis.
The path of a point P in or attached to the rolling cone is a spherical epitrochoid traced on the surface of a sphere of the radius OP. From P draw PQ perpendicular to the instantaneous axis.
Let PQ bubble.
1, curve a); the fixed line LM is termed the base, and the line PQ which divides the curve symmetrically is the axis.
1 AQP be the reflecting circle Caustics having C as centre, P the luminous point, and PQ any incident ray, and we join CQ, it follows, by the law of the b efiection.
Conceive a perpendicular PQ to be dropped from this point on the fundamental plane, meeting the latter in the point Q; PQ will then be parallel to OZ.
Are intersected by the lines RK, RM, RN, we have SA/AX = TP/PQ = SP/PQ, since the angle PST = angle PTS.