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Immediately applying the general formulæ above found, they give

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(By")2

+

(By)3

u2 = by + By* + 2by (y − 1) † 2b3y2 (y − 1) (y2 — 1)

+

(By2)1. (y + 5)

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and B is known by reverting the series

&c.;

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To facilitate the determination of B in every case of the general

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A + By + c1 (By*)2 + c2 (By*)3, &c, we may apply

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1 . 2...( − 1) (@, – 4) ( − 1)^- .cc &c.
1.2...b x 1.2...b1 × 1.2...b2 × &c.

where the indeterminate indices are subject to the two conditions

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which I have given in my Memoir on the Resolution of Algebraic Equations.

With respect to the inverse function in this particular example, we have

Ux =

√(a2 + 2bu+1); U-x

=

√ {a2 + 2bu _ (x-1)};

and putting for a successively 1, 2, 3 ... x, we get

u_x = √/ {a2 + 2b √ {a2 + 2b√{a2 + + 2b√(a2 + 2bu)},

...

the number of roots being ; and for the value of this successive function we have

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and B is very readily found by putting a = 1, and thence the required value of u, is obtained in a series extremely convergent.

As another numerical example, let u+1 = 2u2 – 1.

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and the continued surd which is the inverse function may be similarly expressed.

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III. To expand the function a. a, the indices being continued a times.

Denoting this quantity by u,, we have

Ux+1 = a. a1• = $ (U2),

and adapting the general formulæ to this case, we have

A = a. a4, whence A is known,

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and then u,= A + By + c, (By")2 + c2 (By')3 + &c.,

u = A + B + c1 B2 + c2 B3 + &c.

0

1

For the inverse function if we put a = and take logarithms in

β'

the system of which the base is a, we have

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u_, = log. {B.log. {Blog. ẞ..............log. ßu}, æ times;

-3

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the quantities A, B, y, C1, C2, &c. being the same as before.

Before leaving this example, we may observe, that the most rapid way of finding A, is this, let log. B = b, i. e. a' ẞ; and since

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ß

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=

a first approximation,

take its log. and add to b for a second; take the log. of the second approximation, and add to b for a third; and when b>1, we shall get a very converging series of values for A.

If we had applied Lagrange's theorem in this case to the equation A = a4-b, we should have A expressed in a divergent series; it is necessary therefore to limit the announcement, that this general series gives the least root, to the case of real roots, for when there are some imaginary, we see that it may express one of these instead of the least real root.

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CASE OF FAILURE OF THE GENERAL SOLUTION.

When the equations (A) = A, 4′ (4) = 1 are simultaneously true, the terms in the expansion of u, become infinite, as before remarked, we shall therefore give a solution in a different form, for this case.

The equation (4) = A must here have two equal roots, suppose each = c; transform the proposed equation by putting u, v, + c,

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the corresponding equation A = F(A) has then two roots each equal to zero, and consequently F(A) must be of the form

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and accordingly F(v.) or v11 is of the form v ̧ + mv22 + nv33 + &c.

Hence if v vanish, v1, v2, v3, &c. successively vanish, and therefore we put generally

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in which series the coefficients are unknown functions of x, but independent of v。.

2

Hence, v1+1 = A ̧. v1 + B ̧. v122 + C2. v13 + D,. v, &c.......(1),
Vr+1

4

3

1

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C„ v ̧1
F1.vo + F2. v2 + F3v3 + F. v1 &c.
F) (0)
1.2.3...n'

by Maclaurin's Theorem.

viz.

Beside the foregoing form of expressing +1, there are two others,

2

3 t

4

F(v.) = V1+1 = F1. v2+ F. v2 + F. v2 + F.. v,' &c.......(2),

2

Vx+1 = Ar+1.00 + B+1 · vo2 + Cr+1. vo3 + Dr+1 · vo1 &c....... (3),

and since F is manifestly unity, if we compare the expressions 1 and 2, when the latter is arranged according to the powers of , we obtain A1 =1, which is obvious by the law of the successive formation of the quantities v1, v2, &c.; also putting x = 0 in the general value of v., we have v1 = v。 + B。 . v22 + Co. v3 + Dov1 &c.,

3

which shews that B = 0, C = 0, &c.; this being premised, we have by comparing the expressions (1) and (3), the following identity;

2

v. + B+1. v2 + C1+1 v3 + D2+1. vo1 &c. = {v。 + F1⁄2 . v2 + F ̧v3 + F, vo1 &c.}

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