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a - tan P. %c

= (a + n)" cos (n.x + pp).

d' (uv) Therefore, expanding by the Theorem of Leibnitz,

d x2

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r (1 - 1)
m (m – 1)."

1),m-2 cos {nx + (r – 2)}
1.2

(a + n°)

+ &c.]

(18) Let u v = 64? X, X being any function of x.

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This result, when generalized, is of great importance in the solution of Differential Equations.

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in the expan

1.2...

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If the function to be differentiated be (a + bx + cx?)", the general differential might be found by resolving the trinomial a + bx + cx into two factors of the first degree, as into (x + a) (20 +ß), and then differentiating the product (x + a)" (x + 3)" by the Theorem of Leibnitz ; but instead of doing so we shall make use of two formulæ given by Lagrange*

Let u = a + bx + cx*, u' = b + 2 cx ; Then substituting x + h for x in u" it becomes

(a + u n + cho);

h" and will be the coefficient of

dx sion of this trinomial.

Developing it as a binomial, of which u + u'h is the first term, we obtain

n(n-1) (u + u'h)" + n (u + uha-lch? +

(a + b)”-coh + &c.

1.2 Again, developing each binomial and taking only the terms which multiply h', we find that the term in

n (n − 1)... (n − p + 1) (u + u'h)" is

"-"u" ;

1.2 in (u + u'h)"-1K2 is (n − 1) ... (n – – + 2)

u"-T+1 u'-; 1.2 (r – 2)

(n (n - p + 3) in (u + u'h)"-?h* is

un--+? U'"-1 ; &c. 1.2 (r - 4)

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2)

Collecting these terms, and multiplying by 1.2 ... r, we obtain for the pth differential coefficient of u"

ď' (u")

CU

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r(r-1) = n(n − 1)... (n-r + 1) u"-" u'r {1 +

1.(n - p + 1) u

u'? r (r - 1) (r - 2) (r – 3) cu

+ &c.} (A) 1.2 (n – p+1) (n − p + 2) u'

+

Mémoires de Berlin, 1772, p. 213.

By developing in a different manner a more convenient formula may be obtained :

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and the p.th differential of un is the coefficient of h in this expansion multiplied by 1.2.... Now expanding each term by the binomial theorem, we have for the coefficient of

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and so on. Collecting these terms and multiplying by 1.2...r,

we find

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+

n(n-1) r(n-1) (r – 2) (r - 3) et

+ &c.}... (B.) 2n (2n - 1) ... (2n - 3) u's

1.2

(19) Let u" = (a® + 2?)".

Here u 2x, e = : 4a', and if we make r = n, we find by formula (B), d" (a® + °)"

no
n-1

a
= 2n(2n – 1)...(n + 1) v"{ 1 +
da

i '2n(2n-1) ** {n(n 1)}? (n − 2) (n 3) af

+ &c.}. 2n...(2n - 3) X

+

1.2

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The pth differential of this function may be found as in the last example, but the following method gives it under a form which is more convenient in practice ;

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1

r(r – 1)...2.1 =(-)"+).

2a(-)" {{x + a(-)}}"+1" {x- a (-)}}' = (-)+"(r – 1)...2.1}{r-a(-)}}++! – {w+ a {

+ a(-))}"+"

}. 2a(-)

(aľ + 20°)?+1

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{« - a(-)}}+1 = (a + v2) 7 {cos (r + 1)2-(-)sin (r + 190}, {v + a(-)}}++'= (a +22)** {cos (r + 1)2 + (-)sin (r+1)0}.

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Hence we have

1

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a

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d

(-)'»(– 1)... 2.1 sin (r + 1) 0 da) a2 + x2

(a? + xo") Liouville, Jour. de l'Ecole Polytechnique, Cab. 21, p. 157.

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3

d

cos (r + 1) 0
= (-)'r(r - 1)...2.1
a + x2

(a + x^)

Liouville, Ib. p. 156. These results are useful in the theory of definite integrals.

In the following examples the functions are reduced to the required forms by differentiation in the same way as in Ex. 11.

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