Page images

.(1 - س)


6. The first process shews, by the aid of Lagrange's Theorem, that



3 i du
Let y denote a quantity, such that


h h? h being less than 1.


[merged small][merged small][merged small][merged small][merged small][subsumed][merged small][subsumed][merged small][subsumed][merged small][ocr errors][subsumed][ocr errors][subsumed][subsumed]
[ocr errors]
[ocr errors]
[ocr errors]
[ocr errors]


[ocr errors]
[ocr errors]
[ocr errors]
[ocr errors]

yja – 1 y =M +

2 Hence, by Lagrange's Theorem, Ma - 1

hồ * y=uth 2 du


1.2 i du-1

2 therefore, differentiating with respect to u and observing that

dy = (1 – 2uh+29+,
d /M2

ha ď /M2 1 )h

du 2

1.2 ?

2 k ď (M2 +

1.2... i due ' 2

[ocr errors]


1 d Hence

i djes

(u? — 1). 7. From this form of P, it may be readily shewn that the values of p, which satisfy the equation P,= 0, are all real, and all lie between - 1 and 1.

[ocr errors]

For the equation

(u? — 1)' = 0 has i roots = 1, and i roots =-1,

[ocr errors]
[ocr errors]


Que? – 1)=0 has ¿ – 1 roots = 1, (i-1) roots =-1, and one root = 0,

[ocr errors]

[ocr errors]
[ocr errors]

(u -1)=0 has (i – 2) roots = 1, one root between 1 and 0, one between 0 and =-1, and (i – 2) roots =-1, and so on.

Hence it follows that d' ¿

i Cu - 1)'= 0 has du

roots be2 roots between 1 and 0, and ;

Z tween 0 and -1, if i be even, 2-1

i-1 roots between 1 and 0, roots between 0 and 2 2

2 1, and one root = 0, if i be odd.

It is hardly necessary to observe that the positive roots of each of these equations are severally equal in absolute magnitude to the negative roots.

and on

8. We may take this opportunity of introducing an important theorem, due to Rodrigues, properly belonging to the Differential Calculus, but which is of great use in this subject.

The theorem in question is as follows:
If m be any integer less than i,
1.2... (i - m)

(ac* — 1)

(2* — 1) 1.2 (i + m)


(022 - 1)m ducitm

[ocr errors]


It may be proved in the following manner.

If (oc2 – 1)' be differentiated i - m times, then, since the equation

(oc* - 1)'= 0 bas i roots each equal to 1, and i roots each equal = -1, it follows that the equation


(0 - 1)' = 0




has i – (i m) roots (i.e. m) roots each =1, and m roots each 1, in other words that (aco – 1)" is a factor of


(- 1)'.

daiWe proceed to calculate the other factor.

For this purpose consider the expression
(x + a) (x+2.) ... (x + a) (x +B.) (x +B.) ... (x+B.).
a c ) B

) Conceive this differentiated (I) i-m times, (II) i + m times. The two expressions thus obtained will consist of an equal number of terms, and to any term in (I) will correspond one term in (II), such that their product will be (c+Q) (x + ax) ... (x + a) (x+B) (x+B.) ... (x+ B.), i.e. the a

a term in (II) is the product of all the factors omitted from the corresponding term in (I) and of those factors only. Two such terms may be said to be complementary to each other.

[ocr errors]

Now, conceive a term in (II) the product of p factors of

P the form x+a, say x +a', a + 3 +0), and of factors

9 of the form x + B, say x+ B, C+ B,,... IC + Big We must have p+q=i- m. The complementary term in (I) will involve

p factors x+B, &+B" ... x + plo),
factors x + a, x+Q,, ... +

a taga

[ocr errors]

Now, every term in (I) is of i + m dimensions. We have accounted for p + q (or i – m) factors in the particular term

p we are considering. There remain therefore 2m factors to be accounted for. None of the letters

a', a" ...CO), B, B,, ... Biebe
B', ß"... Blo)

, QQ,, ... Olgy can appear there. Hence the remaining factor must involve ma's and m ß's,-say,

al, ar
B, B... B.

There will be another term in (II) containing
(x+B) (x + ß') ... (oc + Blo) (x+2,) (x + ay) (v + air)).
The corresponding term in (I) will be, as shewn above,
(x+ a') (x+a") ... (a + c) (x +B.) (x+B.) ... (x +BW)

v ) cc) B
(x+, a) (x + 2) ... (x+ m2) (oc + ) (x+B)... (a + mB).

Hence, the sum of these two terms of (I) divided by the sum of the complementary two terms of (II) is (x + 2) (x+,a) ... (x+ m) (x +,B)(x+) ... (x + mb).

Now, let each of the a's be equal to 1, and each of the B's equal to - 1, then this becomes (2* — 1)". The same factor enters into every such pair of the terms of (I). Hence


(ac* — 1)".

di-m (az – 1)'

ditm (2-1)
(aca - 1)

to a numerical dace-m

daitm factor près. The factor may easily be calculated, by considering that

di-m (- 1) the coefficient of it in

is 2i (21-1)... (i+m+1),

ditm (ac* — 1) and that the coefficient of min

[ocr errors]


[ocr errors]


1" is


2i (2i – 1) ... (i + m + 1)(i + m)... (i m +1).


Hence the factor is

1.2... (i m)
(i+m) (i + m - 1) (i - m +1)' 1.2... (i + m)

d' 9. This theorem affords a direct proof that C

a ca C being any constant, is a value of f () which satisfies the equation

df (u)?

+i(i+1) f(u) =0. du


[ocr errors]




[merged small][merged small][ocr errors][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][subsumed][subsumed][subsumed][ocr errors][subsumed][ocr errors][subsumed][subsumed][subsumed][subsumed][subsumed][subsumed][subsumed]

Hence, the given differential equation is satisfied by putting f (u) = C

[ocr errors]
[ocr errors]
[ocr errors]

Introducing the condition that P, is that value of f (w). which is equal to 1, when u=1, we get

1 d'


i du 10. We shall now establish two very important properties of the function Pı; and apply them to obtain the development of P. in a series.

2.1.2...i dü (m? –1).


« PreviousContinue »