q being equal to e; so that if we can sum 4g", we can express 4q" as a definite integral, the only case requiring care being when the sum of Aq" is ambiguous as, e.g., when

in which the arc intermediate to 0 and must be taken. The integral may also be written by making an evident substitution in the form

so that we can deduce Σ A, from Aq"; and writing

(33) in the form

√n

we see that can be expressed as a definite integral

√n

if Σ Aq" can be summed; thus, for example,

The integral (35) leads us back to the consideration of the series which Cauchy obtained from (1), viz.

This theorem Cauchy speaks of as having appeared to Laplace of so much interest, that the latter verified it numerically when a was small. Cauchy also remarks that Poisson had obtained it also by considering the decomposition of a definite integral into singular integrals. Another investigation of it is to be found in Schlömilch's Analytische Studien, part II., p. 30.

Cauchy proved a more general theorem, viz. '+e(-a)2 + e ̄(z+a)2 + € ̄(5-2)2 +ē ̄(2+2a;?

+...

This result is also obtained by Sir W. Thomson in a far more natural manner, viz. from Fourier's theorem,* and another proof depending on elliptic functions, by Professor Cayley, is added.

The integral

xe cotaxdx, shown in this paper to be

equal to a π ' ne ,has, I have since found, been evaluated by Prof. de Haan (Nouvelles Tables (16), Table 362), who finds that it is equal to a √ Σ, neTM" - π a result identical in form, but of different sign from that given above. From the reference it is clear that Prof. de Haan obtained his evaluation by differentiating the integral

1

[* eTM1 log (sinʼax) dx = √/■ {− log 2 + 2 = e."}

with respect to a.

This last integral was obtained by Schlömilch (Analytische Studien, part I., p. 159), in the form

-4a9

-9a

'log (4 sin3ax) dic = e ̄a2 + } e ̄*«2 + }e ̄3«o +....

The error of sign is caused by Schlömilch having taken the sum of the series

r cosz + r2 cos2z+... to be log (1 − 2r cosz +r3),

instead of

log (1-2r cosz+). The integral in the formula marked (6) in Schlömilch's work has also a wrong sign for the same reason. In the Tables, the two errors in (7) and (8), Table 467, are corrected in the errata; but the integrals (15) and (16), Table 362, which were deduced from them, have not been corrected. The values of these last two integrals should have their signs changed. The value of the integral (10), Table 467, has also a wrong sign.

In the first edition of the Tables (Amsterdam Transactions, vol. IV.), the corresponding integrals are (12), (13), (14), Table 439, and (20), (21), Table 388, all of which are wrong in sign; in the last two the reference also is inaccurate.

July, 1871.

• Quarterly Journal, vol. I., p. 316.

ON A THEOREM RELATING TO EIGHT POINTS ON A CONIC.

By Professor CAYLEY.

THE following is a known theorem:

"In any octagon inscribed in a conic, the two sets of alternate sides intersect in the 8 points of the octagon and in 8 other points lying in a conic."

In fact the two sets of sides are each of them a quartic curve, hence any quartic curve through 13 of the 8+ 8 points passes through the remaining 3 points: but the original conic together with a conic through 5 of the 8 new points form together such a quartic curve; and hence the remaining 3 of the new points (inasmuch as obviously they are not situate on the original conic) must be situate on the conic through the 5 new points, that is the 8 new points must lie on a conic.

We may without loss of generality take (a,,a,, 1), (a ̧2, α,, 1), ..... (a, a, 1), as the coordinates (x, y, z) of the 8 points of the octagon; and obtain hereby an à posteriori verification of the theorem, by finding the equation of the conic through the 8 new points: the result contains cylical expressions of an interesting form.

Calling the points of the octagon 1, 2, 3, 4, 5, 6, 7, 8 the 8 new points are

12.45, 23.56, 34.67, 45.78, 56.81, 67.12, 78.23, 81.34, viz. 12.45 is the intersection of the lines 12 and 45; and so on. The 8 points lie on a conic, the equation of which is to be found.

The equation of the line 12 is

x − (α, + α) y + a‚a ̧2 = 0,

or as it is convenient to write it

viz. 1, 2, &c., are for shortness written in place of

respectively.

The coordinates of the point 12.45 are consequent proportional to the terms of

12 (4+5) — 45 (1 + 2): 12 - 45: 1+ 2 − (4 + 5).

The equation of the line (12.45) (23.56) which joins the points 12.45 and 23.56 thus is

12 (4 + 5) − 45 (1 + 2), 12 + 45, (1+2) − (4+5)

23 (5+6) − 56 (2+3), 23 − 56,

(2+3) − (5 +6)

where the determinant vanishes identically if 2-5=0(a−a ̧=0); it in fact thereby becomes

=

22(14), 2(1-4), (1-4)

22(3-6), 3 (3-6), (3−6)

which is 0; the determinant divides therefore by 2-5; the coefficient of x is easily found to be

=(2-5) (12-23 + 34-45+ 56-61),

and so for the other terms; and omitting the factor 2-5 the equation is

x122334-45+ 56-61}

− y {12(4+ 5) − 23(5+6)+34(6+1)−45(1+2)+56(2+3)−61(3+4)} +z (1234-2345+ 3456 4561 +5612-6123} = 0.

There is now not much difficulty in forming the equation of the required conic; viz. this is

[12-2334-45 +56 61]

−y[12(4+5)−23(5+6)+34(6+1)−45(1+2)+56(2+3)−61(3+4)]

+z[1234-2345+ 3456-4561 +5612 - 6123]

+ (6 − 8) {x − (1 + 2) y + 12z}

x[2334+45-56+67-72]

-y[23(5+6)−34(6+7)+45(7+2)−56(2+3)+67(3+4)−72(4+

+≈[2345 −3456 + 4567 - 5672 +6723-7234]

= 0.

In fact this equation written with an indeterminate coefficient A, say for shortness, thus

67 [(12.45) (23.56)] = λ12. [(23.56) (34.67)] = 0,

is the general equation of the conic through the 4 points 12.67, 34.67, 12.45, and 23.56; and by making the conic

pass through 1 of the remaining 4 of the 8 points, I succeeded

6-8

in finding the value λ=; so that the conic in question

2 8

passes through 5 of the 8 points, and is therefore by the theorem the conic through the 8 points. But as thus written down the equation contains the extraneous factor 2-6, as appears at once by the observation that the left-hand side on writing therein 6=2(a = a,) becomes identically = 0; the value in fact is

− (2−8) [x− (2+7)y+27z] (23 −34+45 −52) [x−(1+2) y+12z] + (2 −8) [x − (1 + 2) y+12z] (23 −34+ 45 −52) [x − (2+7)y+27z] which is = 0; there is consequently the factor 2-6 to be rejected, and throwing this out the equation assumes symmetrical form in regard to the 8 symbols 1, 2, 3, 4, 5, 6, 7, 8. The coefficient of x is very easily found to be

=

= (2 − 6) (12 − 23 + 34 − 45 + 56 − 67 +78 − 81),

and similarly that of z2 to be

= (2 −6) {123456 – 234567 + 345678 - 456781 +567812

a

− 678123 +781234 – 812345},

those of the other terms are somewhat more difficult to calculate; but the final result, throwing out the factor (2-6), and introducing an abbreviated notation

=

12 (12-23+ 34-45+ 56-67+78-81),

and the like in other cases, is found to be

where it is to be observed that 1256 consists of 4 distinct terms each twice repeated: 21256 consists therefore of these 4 terms; and in the coefficient of y' they destroy 4 of the 32 terms of £12 (4+5) (6 +7) so that the coefficient of y contains 324,= 28 terms. In the coefficient of ze there is no destruction, and this contains therefore 12+ 4, 16 terms.

=