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This series does not converge so rapidly as the ordinary series, as given by Professor CAYLEY, above referred to.

The following series, obtained in a similar way to the above, possesses a greater degree of convergency than either :

8n-3 E

1 + Σ 2k"2

(2n–3) (2n –1)

[ocr errors]

=

A2n k2n

K4 +

}

T

II. Solution by J. HAMMOND, M.A.; J. O. JELLY, M.A.; and others. The general term of the expansion in powers of ka of

12.3 12. 32.5 (1–5+) {1+ k2 +

22

22. 42 is

{coefft. of ki?” – coefft. of km2–2}k24
{
12. 32. 52 ... (2n – 1)2. (2n + 1)
22.42.62 ... (2n)2

kan

22.42.62 (2)
k2n § 12. 32. 52 ... (2n–3)2 (2n — 1) (2n-1) (2n+1)
2 22.42.62 ... (2n— 2)?

2n2
12. 32.52
(2n- 3)2 (2n-1)

-k2n.
2 22. 42.62 ... (2n)
The expansion is therefore

1

12.3 12.32.5 k? - k4

k6 2 22 22.4°

22.42.62 which is a well-known expression for E.

These expansions only give good results when k is small, since the series are divergent when k 1 and only converge very slowly when k is near 1.

12.39:52,0 (21–3)2--2n – 1}}

+1) – 1)

5563. (By Professor SYLVESTER, F.R.S.)-If at two points in a cubic curve, lying in a straight line with a point of inflexion, tangents be drawn to meet the curve again, prove that their intersections with it will also be in a straight line with the same point of inflexion.

I. Solution by W. J. C. SHARP, M.A.; Prof. CASEY, M.A. ; and others.

It is shown in Salmon's Higher Plane Curves (Art. 149), that the equation to any cubic may be thrown into the form ABC = kD2E, where A, B, and C are tangents, D the chord of contact, and E its satellite.

Now, if D pass through an inflexion, and A be the inflexional tangent, E = 0 must also be satisfied when 1=0 and D= are so, or E passes

through the point of inflexion. But E is the line joining the two other points in which B and C, the tangents at the points where D cuts the curve, meet the curve.

II. Solution by J. L. McKenzie, B.A. This theorem, like that of Question 4869, may be proved by the principle of “residuation.” (See Salmon's Higher Plane Curves, 2nd Ed., p. 133.) Take four fixed points on a given cubic, and draw through them any curve of the second order; the line joining the two remaining points in which the two curves intersect will pass through a fixed point on the cubic- the residual of the four fixed points. In the present question take for the four points, the points P and Q, at which the tangents are drawn, each twice. One system of the second order through the four points consists of the tangents PP and QQ, which meet the cubic again in the points X and Y ; then line XY must pass through the point-residual. Another system of the second order through the four points is the line PQ taken twice; and this line meets the cubic again in I—the point of inflexion; hence the line II must pass through the point-residual. But II is the inflexional tangent, and the third point in which it meets the cubic coincides with I, which is therefore the point-residual. Therefore XY passes through I. [Another Solution is given on p. 52 of this volume.]

138. NOTE ON QUESTION 5458. By the EDITOR. (See pp. 55-56 of this

Volume.) On p. 125 of his Aperçu Historique des methodes en Géométrie (Ed. 1875), M. CHABLES states that “LA HIRE a donné le lieu des angles égaux, aigus ou obtus, circonscrits à une conique, lequel est une courbe du quatrième degré ... ;” that LA HIRE has also “traité la même quastion pour la cycloïde, et est parvenu à ce résultat curieux, savoir, que tous les angles égaux, droits, aigus, ou obtus, circonscrits à cette courbe, ont leurs sommets sur une seconde cycloïde raccourcie ou allongée ;” and that he himself has found que les épicycloïdes du cercle jouissent de la même propriété ; c'est à dire que, Si à une epicycloïde, engendrée par un point d'une circonference de cercle qui roule un autre cercle fixé, on circonscrit des angles tous egaux entre eux, leurs sommets seront situés sur une épicycloïde allongée ou raccourcie.”

5540. (By Professor LLOYD TANNER, M.A.)—If M, N are two numbers of n digits each, and the numbers formed by prefixing M to N and N to M are as a to b; find M, N, and indicate the conditions required to ensure (1) at least one solution, and (2) only one solution.

I. Solution by S. TEBAY, B.A.; Prof. Erans, M.A.; and others.

N+10MM The required statement is

i ; and, K being a conditional

M +10"N factor, we can take KM 10"ab, KN = 10"a. If n, a, b are given, these expressions may be prime to one another, and the solution impossible. In general K will be a factor of 102n – 1; and the number of solutions will depend upon the number of factors answering the conditions of the problem. If M, N be the least possible numbers, and Mor N> š x 10”, there is but one solution; if less, there may be two solutions.

Let n=1, a=19, b=25; then 10 x 19-25=5 x 33, 10 x 25 – 19=7 x 33 ; here the only value of K is 33, and therefore M=5, N=7. Let n=2, d=248, b=257 ; therefore

102 x 248–257 = 909 x 27, 102 x 257 — 248 909 x 28. Take K =909, then M = 27, N=28. Take K=303, then M = 81, N=84.

or

II. Solution by the PROPOSER.
Let r be the radix, then Myn+N; a = Non + M :b,

(6M - aN) yn aM — N, but 3M - aN = C .........(1, 2); then, since M, N are each less than rn, we have, by (1), a>c>-b.

.....(3). arn _b

brn — a Solving (1), (2), we have M -c; N

a2 62

cb whence Mrn + N = (ran – 1); Nyn+M (21-1) .........(4). a2 — 62

a2 - 62

[ocr errors]

ac

.(5)

If we assume a>b, and remember that M, N, r, a, b are all positive, it is clear that c is positive also, or (3) may be written a>c>0 Let us also suppose a : 6 to be in its lowest terms, so that both a, b will be prime to a? — 62. Then, since the expressions are integral, a2- 62 must be a divisor of < (3:21 — 1); and c will be found in the following manner. Suppose a to be the denominator of q2n–1: a?- 62 when reduced to its lowest terms, then c must be a multiple of a; and it may be any multiple of a which is less than a. Should a be greater than a, there can of course be no solution.

It is not difficult to determine (when a, b, r. only are given) for what values of n, if any, the problem admits of solution. Consider aa— 62 as composed of two factors; one, a, a product of divisors of r; the other, B, prime

If a> a there is no solution for any value of n. For a is prime to run – 1, since this latter quantity is prime to r; hence a cannot divide c (3-21 – 1), since, by (5), c<a, <a; bence c (-21 – 1) is not divisible by or as— 63, and the expressions on the left of (4) cannot be integral.

If, on the other hand, a<a, we can always find values of n which make the problem soluble. For we may take c, in (4), to be any multiple of a less than a, and of such multiples there is at least one, viz. a itself. Also, B being prime to r, we can always find values for n such that p21 - 1 may be a multiple of B. [See a paper by Mr. GLAISHER, in the Messenger of Mathematics, Vol. V., p. 4.]

to r.

5580. (By S. ROBERTS, M.A.)-If the sides of a variable triangle pass through three fixed points in a straight line, while one vertex moves on another straight line, and a second vertex describes a given curve, prove that the locus of the third vertex is a homographic transformation of the given curve.

I. Solution by Professor TOWNSEND, F.R.S. Denoting by A, B, C the three vertices of the variable triangle, by P, Q, R the three collinear points through which its opposite sides by hypothesis pass, by L the right line on which its vertex A by hypothesis moves, and by E and F the two figures traced out by its two remaining vertices B and C; then, taking any two positions ABC and A'B'C' of the variable triangle, since, by hypothesis, their three pairs of corresponding sides, BC and B'C', CA and C'A', AB and A'B', intersect at three collinear points P, Q, R, therefore, by the fundamental property of triangles in homology in a plane, their three pairs of corresponding vertices, A and A', B and B', C and C', connect by three concurrent lines L, M, N; hence, the two figures E and F, being such that all pairs of their corresponding points B and C connect through a fixed point P, and that all pairs of their corresponding lines M and N intersect on a fixed line L, are in homology in their plane to the centre P and axis L; and therefore, &c.

B

[ocr errors]

P

[ocr errors]

R

II. Solution by the PROPOSER; E. B. ELLIOTT, M.A.; and others. Take the line PQR through the

y fixed points for the axis of x, and the line directrix Oy for the axis of yon which the vertex A moves; the vertex B, whose coordinates are x, y, moves on the curve F (x, y, 1) = 0. Let X, Y be the coordinates of the third vertex C, and write k for OA, and a, b, c for OP, OQ, OR. Then

aY+X-a) k = 0, by + (1-6) k = 0; or

aY (xb) by (X –a) = 0. Hence aYx-6 (X-a)y-abY = 0, and Y:-(x -c) y-Y = 0; giving

x [(a-6) X+a (6-c)] - (a -c) X,

y[(a-) X+a (6-c)] = a (6-c) Y. The required locus is F [6 (a -c) X, a (6-c) Y, (a - b)X + a (6-c)] = 0.

5625. (By Professor CAYLEY, F.R.S.)—The equation

{q? (x + y + 2)2—Y2–2x– xy}? = 4 (2q + 1) xyz (x + y +z) represents a trinodal quartic curve having the lines x=0, y=0, x=0,

3+ y +z = 0 for its four bitangents; it is required to transform to the coordinates X, Y, Z, where X=0, Y=0, Z=0 represent the sides of the triangle formed by the three nodes.

+

29+1, becomes

Orλ =

(441)*(x2+ZX + XY)}

Solution by J. J. WALKER, M.A. Assume x = 1X+Y+Z, y X+1Y+Z, z= - X+Y+XZ, giving

x+y+z = (1 + 2) (X+Y+Z), y2 + 2x + xy = (21 + 1)(X2 + Y2 +Zx) +(12 + 2x + 3)(YZ+ZX + XY),

xyz = 2(X3 + Y3 + Z3) +(12+2 + 1)(X3Y+ ...) +(13 + 32 +2)XYZ; then, by substitution, the given equation becomes [{(1+ 2)2 q? – (2x + 1)} (x2 + Y2 + Zo)

+ 2(x + 2) -(18+ 2x + 3) (YZ+ZX + XY)]
= 4 (+ 2)(29+1) {X(X3+ Y3 + Z3) + (12+2 + 1)(X3Y+ ...)

+(13 + 32 +2) XYZ) (X+Y+Z);

1 which, for the value 1 + 2

9

9 | 2 (29 + 1)

(XR+Y? + ZP) -
9
4 (29 + 1) { 29 +1

+39

(X2Y + ...)
2
9

q?
12q8 + 150° +6q+1 xyz}(X+Y+z).

q3 It is evident that, on multiplying out, the coefficients of X4... are identi

4 (2q + 1)2. Those of X3Y... cally equal on both sides; viz., to

q2 4 (29+1)(a + 1)2

and

4 (29 + 1) (29 +1_378 +39 +1
03

9
9

q2 respectively, which are also equal. Hence the equation reduces to the form u (YoZ* + Z2X2 + XoY2) +v(XRYZ+Y2ZX + ZXY) = 0, and X, Y, Z are the lines joining the nodes. The values of u and v are and -2 (3q + 1)3 (9+1)

respectively.

94 Hence the equation, transformed as required, is (39+1)(Y-Z2 +Z2X’ + X2Y2) – 2 (2+1)(X2YZ+YoZX + Z*XY) = 0.

+

are

)

ע

4

[ocr errors]

5548. (By ELIZABETH BLACKwoon.) — P, Q, R are three random points within a circle, and their respective distances from the centre are 2,9,r; show that the chance that the roots of the equation pxia — qx+r = 0 are real is táa + s loge 2.

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