By (9) or (10) a, d, = a, d,; therefore, when a = 0; d=10, and by (8) or (11) at the same time b c. This circumstance has been noticed by Mr. Horner. If also a =∞, d, = ∞; for = some finite positive or negative quantity. Supposing to, equas. (8) and (11) give a, = 0, d. = o, and bc1, which brings out the obvious case mare di vit

at

4° x = x.
br+t

Because when a1++ = 0, b1++ = + we have (9) and (10) b, b1 = c, c. That is the product of any two, and, therefore, of any other number of the component values of b which give a resultant value for a = o is equal to the product of two, or of the same number of corresponding component values of c; conse= 1.220 7901202

quently if b∞∞, c, = ∞; for

=

bt

Ct

bo

=

Co

· and t = (v-1) r, we obtain by (8)

A, A., b, + C(0-1)+

A.

=

A = A ̧_, b1+d, a (。 − 2) + + C, C (。 − 2) ; — A‚—‚ b ̧ + á‚ d ̧ Ã ̧_‚ +C, C (o− 2) by taking t = (v2) r in (10). Again substituting v — 1 for in the preceding value of A, and putting the value of c thus derived for its equal in the second value of A., we shall have

supposing pb, + c, and q = a, d, b, c,. In like manner "it is found that

(14), (15), is evidently an equa. of differences of the second order with respect to v, the coefficients p, q, being constant in relation to this v. The solution of either of them (13), for instance, by the usual methods, is

in which Q, Q, are the arbitrary constants. Now if we assume

2

() and identify (16) with the well-known theorem

(cos z + √ - 1 sin z) = cos v % ± √

1 sin v z

(2). {Q (cos v2 + √ − 1 sin v 2) + Q, (cós v

cos

z

Determining now Q, Q, from the conditions of (17) when v = 0 and v = 1, we shall find after due reductions

I

(br+c) tanz

And if R, R, be the corresponding arbitrary constants in the solution of (14), we easily perceive that

and C, is the same expression with a negative instead of a plus sign before the second number under the vinculum { }. In the same way if P, P, be the arbitrary constants of (12), appears that P = P, and

it

cr

(b+ √ 1 sin 2)

P =

2 cos z

value 1+brer, and substituting for q its value c,

Having now obtained general values for the coefficients in rx, let us consider a little the limitations of the indeterminate quantity z. In the first place, it is plain that ≈ must vary independently of v; and in the next, it appears from q = (2 cos z)2

&c. because then (cos z) = o, which, unless b, — c1, gives Ja, d1 = — (∞∞), that is, both a, and d, infinite, the one negative, and the other positive. Thirdly, z must never be a 0, 1, 2, 3, &c.

d,

Κλ n

=

2v-1

multiple of λ, for that would give a„, =
va,, a value
which could never become = 0 by any finite value to v, unless
at the same time b, c,; and, therefore, the function could
never be periodic. Fourthly, z must be such that when v = n,
the order of the periodic function, sin v z must be = o. Assume
therefore ≈ = and it is evident the first condition is satisfied
by k being any independent variable; the second by its never
becoming a, &c. multiple of n, and the third by its never
being a 0, 1, 2, 3, &c. multiple of the same index unless simulta-
neously b,c,. Finally, the fourth case is satisfied by k
having integral values only; for otherwise k varying indepen-
dently of v could not generally give sin and, therefore, a.,

=

3 5

=o when v = n.

r+t

υκλ

n

Moreover it may be further observed, since by (13), (14),
a, da, d, and consequently b,+ e − bị b1 = cr + 1 = c, c, that
b, t
k must be the same in a,, as in d.,, and the same in b1, as in ci
but not necessarily the same in a,, or d, as in b, or c... This,
therefore, gives the number of functional roots no for the same
form of, and is another instance of deviation from the algebraic
analogy. It arises from a similar cause to the preceding devia-
tion, an anticipation in part, as it will presently appear, of the
arbitrary function in the constants a, b, c,.

For the sake of brevity, we shall adopt one common indeter-
minate integer which will enable us to give our final result some-
thing simpler.

n

(b,+ e,) sin

br2-2b,c, cos

.sin

n

n

n

n

x

* This formula is in many instances better adapted for practice and printing under the following form

which is produced from the other by merely multiplying the terms of the numerator and denominator 1

n

2kλ

This is another general expression for determining 4" x from xx whatever be the value of v, r, or n, rational, irrational, or imaginary. If we put r = v = 1, we easily deduce Mr. Horner's expression for x, namely,

(22)

C

but

which was investigated for positive integral values only of n, which our general views show to be true for every value real or even imaginary.

Let us now apply these theorems to a few examples. Suppose first that n = 2, and k = 1; then cos =-1, and the value

for d becomes

must have b—

(b + c)2

2 a (cós λ + 1)

2 kλ

n

which since the denominator vanishes

c. Differentiating the numerator and denominator twice with respect to c and a respectively, we obtain d=

2a812

denoting differentiation. Therefore, because & c and d x

are mutually independent, this value of d may be any thing, and hence

is the solution of 2 x = x. It is rather curious that this solution is obtained on the hypothesis of k=which Mr. Horner thinks cannot be, and obtained also from his own theorem. Again, let n = 5 and k = 1, then by Gauss's division of the

which is the first and only solution I have seen of $x = x. I shall not for brevity's sake stop to compute other cases of integral functions, but shall just give an instance, the first, I believe, that has been given, of the solution of a fractional func

which coincides with Mr. Babbage's solution of 3 x = x. And

3

because = 1 - it is evident that ought to be the in

4

verse of (24). Put therefore in (21) r = 1, v =

course making k as above 1. From these data, we have

=

which it may be easily shown is the inverse of (24), or equal to

This method of solution may, for the sake of distinction, be called the algebraic.

(To be continued.)

Oct. 2.

ARTICLE III.

Astronomical Observations, 1824.
By Col. Beaufoy, FRS.

Bushey Heath, near Stanmore.

Latitude 51° 37' 44.3" North. Longitude West in time 1′ 20.93".

Immersion of Jupiter's second (13h 50′ 50′′

satellite.....

satellite....

13.52 11

Mean Time at Bushey.
Mean Time at Greenwich.
Mean Time at Bushey.

02 29

Mean Time at Greenwich.

Oct. 13. Immersion of Jupiter's first S17 01 08 {17