which contains an arbitrary multiplier; but it is plain that the expansion of u, being unknown, Zu, is also unknown, and the successive differential coefficients would similarly be expressed in forms of unknown functions; and therefore the expansion by the immediate application of Maclaurin's Theorem would be impracticable.

now to find "u, in this case, we must have recourse to the theorem

the coefficients thus found being obviously more complicated than the function itself.

Try again to find u, by a series arranged according to the powers of x, and containing indeterminate coefficients, that is, put

and since u2 = log. U+1; ... U。 = log. u1 = log. e = 1;

Then the equation u+= " becomes

1; .. A = 1.

1 + B (x + 1) + C (x + 1)2 + D (x + 1)3, &c. = €. €1x. €Сx2. €1x3. &c.

and equating like powers of a we get,

1 + B + C + D + &c. = €,

B+ 2C + 3D + 4E + &c. = B. e,

C+ 3D + 6E + 10F + &c.

which clearly show that the coefficients cannot be found but by the resolution of equations of infinitely high degrees.

Now similar difficulties opposing the expansion of other kinds of successive functions, these can only be removed by attending carefully to the equations in finite differences, by which the law of the formation of the functions is expressed. I therefore here propose a means for resolving such equations, of whatever order or degree, in an algebraical form.

FIRST CLASS OF EQUATIONS.

All successive functions such as those above mentioned, are represented by the equation u,+1=4(u), for this manifestly expresses the law of the successive functions,

u12 = a, u1 = $(a), u2 = (a), uz = φφφ (α),

u2 = pppp......(x times)...... {a};

hence the expansion of u, in every such case depends on the solution of this equation.

Put u = f(y) the form of the function f, and the quantity y remaining at present unknown, and also let y2 = 2.

Then, since u, =ƒ(≈), and u,+1 = f(y), we have ƒ(y≈) = $(fx).

Suppose now f(x) = A + B + Cx2 + D3 + &c. the preceding equation becomes

A+By+ Cy2x2 + Dy3≈3, &c,

=

$ {A + B≈ + C≈2 + D≈3, &c.}

and it remains to expand the latter function, in order to compare like powers of, and thus obtain the assumed coefficients and the quantity y.

Now by Taylor's Theorem the development will be of the following nature, viz.

p(A) + Z,p' (A) + Z2q” (A) + Z3p" (A) + &c, where Z1, Z2, Z3, &c.

are functions of ≈ completely independent of the form of the function 4, and p(A), "(4), &c. represent the successive differential coefficients of (A).

ΦΟ

tity.

To determine Z1, Z2, &c. put (A) = e«4, a being an arbitrary quan

Now the value of the left-hand member of this equation is also

And if we equate the coefficients of " in each we get

and comparing this result with the first expansion, viz.

A+ By≈ + Cy2x2 + Dy3ï3 +...... + My ~ + &c.

we have M = Σ

Bb Cc Dd......

1.2...b x 1.2...cx 1.2...d x &c.

where it must be observed, that b, c, d, &c. having previously satisfied the equation b + 2c + 3d, &c. = m, the quantity n is then found by summing b, c, d, &c.

Put Ċ = c,B3‚_ D'= c,B3‚ É = C3 B', &c.

and making m = 1, 2, 3, &c. successively, we get the following identities, by which A, Y, C1, C2, C3, &C. are completely known.

The general law of which equations is thus expressed :

b, c, d, &c. being regulated by the two conditions before mentioned.

From hence we obtain the complete integral of the proposed equation u,+1 (u); for since

=

ƒ(≈) = A + B≈ + Cx2 + D3 + &c.

:. u2 = A + By" + c1(By2)2 + c2(By2)3 + c3(By1)1, &c.

2

The arbitrary constant B is determined as usual by assigning a particular value to x, as

u。 = A + B + c1B2 + c2 B3 + cзB', &c.

by the reversion of which series B is found in a series arranged according to the powers of u。- A,

Before proceeding to any particular applications of this general solution, a few observations will be useful.

I. When p(u,) is of the form u + const., then A becomes generally infinite, and y becomes 1, the solution therefore fails in this case, but more generally it may be remarked, that it also fails when the value of A deduced from the equation A = (4) satisfies the equation p’(A) = 1; for this, by making y=1, renders infinite the coefficients c1, C1, &c. Such cases of failure will shortly be separately considered.

(u) is of the nth degree, the equa

II. When the equation u2+1 tion for finding A is of the same degree, and therefore A has n values; then 11 γ g'(A) has also n corresponding values, which, being represented by 71, 72, ... Y, and putting for abridgment F(By') for the series above found for u,, we have

and the complete solution is found by taking the product of the members on the left side and equating to zero: the result will only contain one arbitrary constant, since B1, B2, &c. are all found in terms of u, as before shewn.

III. u is a known function of B, u, is the same function of By*.

Let - be the function which is inverse to p, that is, such that φ' φ (α) = a, then it follows that

1

The same formula therefore which represents the ath successive direct function of u。, will also give the ath successive inverse function by merely writing - in place of x.

which is a known numerical quantity and may be represented by &;

Thus the number of times it is necessary to take the successive functions of a to arrive at u, as a result, determines $

0

since γ is known, B may be thus also determined.

log. B

and

log. Y

,