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be considered, so far as may be necessary, in subsequent chapters.

Ex. 1. To develope Uzin in a series consisting of U, and its successive differences (Ex. of Art. 7, resumed). By definition

Eur Urt = Eur, &c.
Therefore
Uzun= E"42 = (1 + A)* Uz

(1),
n (n − 1) n (n − 1) (n − 2)
=1+na+

A%. 2

2.3

Δ3 +

Ux

n(n-1) n(n-1)(n-2)
=Uz + nAuzt
A%ux +

A®u,,+&c.. (2). 2

2.3

Ex. 2. To express A"uz in terms of u, and its successive values.

Since Au, = Uz+1 Uz = Eur Ux, we have

Aur= (E 1)U,, and as, the operations being performed, each side remains a function of .x, A'un = (E- 1)" uz EnEn-1+

n (n − 1)

EN2 1.2

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= Uxan

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Hence, interpreting the successive terms,

n (n - 1) + 1.2

Uz+n_2 ... + (-1)"U.z.....(3). Of particular applications of this theorem those are the most important which result from supposing U, = 2c". We have

n(n-1) A"ccm = (+ n)" – n(+n-1)" + (a+n–2)" - &c...(4)

1.2

Now let the notation A"Om be adopted to express what the first member of the above equation becomes when x=0; then Δ"0" = η* -n(n-1)"

n (n-1) (n − 2)" n(n-1) (1 — 2) (n − 3)" +

+ &c......(5). 1.2

1.2.3 The systems of numbers expressed by A"0" are of frequent occurrence in the theory of series*. From (2) Art. 1, we have

A"O" =1.2... n, and, equating this with the corresponding value given by (6), we have 1.2... n=n" — N (n − 1)" +

n (n 1)

(n − 2)" — &c. ... (6)+. 1.2

Ex. 3. To obtain developed expressions for the nth difference of the product of two functions U, and Vx

Since

AuxV= 4x+1• Vz+1 -Uzun

= Eu . E'v - U_0%) where E applies to U, alone, and E' to v, alone, we have

AuxV: = (EE' 1) UzV22 and generally

A"uzv. = (EE' 1)" UxVx ......... (7).

It now only remains to transform, if needful, and to develope the operative function in the second member according to the nature of the expansion required.

Thus if it be required to express A"uqv, in ascending differ

* A very simple method of calculating their values will be given in Ex. 8 of this chapter.

+ This formula is of use in demonstrating Wilson's Theorem, that 1+[n-1 is divisible by n when n is a prime number,

ences of v,, we must change E' into A' +1, regarding A' as
operating only on v..

We then have
A"uqVx = {E(1+4') – 1}" Uzv.m

(A + EA')"UzV2

={A++

n (n − 1) A--- EA” + &c.} u Vs.

+ NAN-EA' +

1.2

+

AUT

Remembering then that A and E operate only on u, and A' only on vz, and that the accent on the latter symbol may be dropped when that symbol only precedes Vx, we have A"uxVx= A"U;. Vx+nAlU3+2 • Avr

n (n − 1)

A***Uz+2.A%vr+ &c....... (8),

1.2 the expansion required. As a particular illustration, suppose W2 = a*. Then, since

= An-raxtr = a*An-ram

= q*** (a – 1)**, by (14), Art. 2,
we have
A"a*v. = q*{(a – 1)"vr + n (a – 1)*-aAv.

n (n-1)
+ (a – 1)*-*a*A%v, + &c.}......(9).

2 Again, if the expansion is to be ordered according to successive values of , it is necessary to expand the untransformed operative function in the second member of (7) in ascending powers of E' and develope the result. We find A*uxVx=(-1)" {UV: — nu z+j Vz+1+

n (n-1)

- &c.}... (10).

2 Lastly, if the expansion is to involve only the differences of u, and vx, then, changing E into 1+4, and E' into 1+A', we have

A"u.Vr = (A +A'+44')"u. V p. ............. .(11), and the symbolic trinomial in the second member is now to be developed and the result interpreted.

Ux+2Vx+2

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d

dt

Ex. 4. To express Ahu, in terms of the differential coefficients of uz By (10), Art. 9, A=&dx – 1. Hence A"ux = (dx – 1)"Uz...

(12). Now t being a symbol of quantity, we have € " tt -+

(13), 1.21.2.3

= *" + 4,4n+1 + 4 +1+2 +&c., on expansion, A,, A,, being numerical coefficients. Hence

d
+
dx
+

+ &c.,
and therefore
d in
d

a 4,u= Un

Uz+ &c.... (14). do)

dx) The coefficients 4, 4,,... &c. may be determined in various ways, the simplest in principle being perhaps to develope the right-hand member of (13) by the polynomial theorem, and then seek the aggregate coefficients of the successive powers of t. But the expansion may also be effected with complete determination of the constants by a remarkable secondary form of Maclaurin's theorem, which we shall proceed to demonstrate.

Secondary form of Maclaurin's Theorem. PROP. The development of $ (t) in positive and integral powers of t, when such development is possible, may be expressed in the form

d

d ta φ (t) = φ (0) +φ

0.t+

1.2

o

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d
+ф 03.
20

3
d

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2cm becomes when x = 0. dx,

where

om denotes what φ

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(15),

m

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First, we shall shew that if $ (x) and t (w) are any two functions of x admitting of development in the form a + b3 + cx+ &c., d

d then ф

dx provided that æ be made equal to 0, after the implied operations are performed.

For, developing all the functions, each member of the above equation is resolved into a series of terms of the form

d A x", while in corresponding terms of the two members

idx the order of the indices m and n will be reversed.

dim Now ac" is equal to 0 if m is greater than n, to

dac 1.2...n if m is equal to n, and again to 0 if m is less than n and at the same time x equal to 0; for in this case 3*-m is a factor. Hence if x = 0,

d

d
x" =
da) der

) and therefore under the same condition the equation (15) is true, or, adopting the notation above explained,

d

d
Ф
do
do

.... (16). Now by Maclaurin's theorem in its known form

d

d φ (t) = φ (0) + $ (0).t+

+ &C....... (17). do

'1.2 Hence, applying the above theorem of reciprocity,

m

12

2C",

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(0) = $(0) ++ (m) 10.t+$(b) 0-.-.+ &c.... (18),

the secondary form in question. The two forms of Maclaurin's theorem (17), (18) may with propriety be termed conjugate.

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