2.1 DIFFERENTIATION
The pattern emerging is clear and strongly suggests that the results generalise to
f(n)=∑nr=0n!
r!(n−r)!u(r)v(n−r)=∑nr=0nC
ru(r)v(n−r), (2.14)where the fractionn!/[r!(n−r)!] is identified with the binomial coefficientnCr
(see chapter 1). Toprovethat this is so, we use the method of induction as follows.
Assume that (2.14) is valid fornequal to some integerN.Thenf(N+1)=∑Nr=0NCrd
dx(
u(r)v(N−r))=∑Nr=0NC
r[u(r)v(N−r+1)+u(r+1)v(N−r)]=∑Ns=0NCsu(s)v(N+1−s)+N∑+1s=1NCs− 1 u(s)v(N+1−s),where we have substituted summation indexsforrin the first summation, and
forr+ 1 in the second. Now, from our earlier discussion of binomial coefficients,
equation (1.51), we have
NC
s+NC
s− 1 =N+1C
sand so, after separating out the first term of the first summation and the last
term of the second, obtain
f(N+1)=NC 0 u(0)v(N+1)+∑Ns=1N+1C
su(s)v(N+1−s)+NC
Nu(N+1)v(0).ButNC 0 =1=N+1C 0 andNCN=1=N+1CN+1, and so we may write
f(N+1)=N+1C 0 u(0)v(N+1)+∑Ns=1N+1C
su(s)v(N+1−s)+N+1C
N+1u(N+1)v(0)=N∑+1s=0N+1C
su(s)v(N+1−s).This is just (2.14) withnset equal toN+ 1. Thus, assuming the validity of (2.14)
forn=Nimplies its validity forn=N+ 1. However, whenn=1equation
(2.14) is simply the product rule, and this we have already proved directly. These
results taken together establish the validity of (2.14) for allnand prove Leibnitz’
theorem.