« PreviousContinue »
.(1 - س)
6. The first process shews, by the aid of Lagrange's Theorem, that
3 i du
h h? h being less than 1.
yja – 1 y =M +
2 Hence, by Lagrange's Theorem, Ma - 1
hồ * y=uth 2 du
2 therefore, differentiating with respect to u and observing that
dy = (1 – 2uh+29+,
ha ď /M2 1 )h
2 k ď (M2 +
1.2... i due ' 2
1 d Hence
(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.
For the equation
(u? — 1)' = 0 has i roots = 1, and i roots =-1,
Que? – 1)=0 has ¿ – 1 roots = 1, (i-1) roots =-1, and one root = 0,
(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.
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:
(2* — 1) 1.2 (i + m)
(022 - 1)m ducitm
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
daiWe proceed to calculate the other factor.
For this purpose consider the expression
) 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.
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),
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
, QQ,, ... Olgy can appear there. Hence the remaining factor must involve ma's and m ß's,-say,
v ) cc) B
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)".
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
2i (2i – 1) ... (i + m + 1)(i + m)... (i – m +1).
Hence the factor is
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
+i(i+1) f(u) =0. du
Hence, the given differential equation is satisfied by putting f (u) = C
Introducing the condition that P, is that value of f (w). which is equal to 1, when u=1, we get
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).