Page:Elementary algebra (1896).djvu/446

428 Rh To determine this invariable form of the product we may give to m and n any values that are most convenient; for this purpose suppose that m and n are positive integers. In this case f(m) is the expanded form of (1 +x)^n, and f(n) is the expanded form of (1 +x)^n; and therefore

f(m) \times f(n) = (1 +x)^m \times (1 +x)^n = (1 +x)^{m+n};

but when m and » are positive integers, the expansion of (1+x)^n is

This then is the form of the product of f(m) \times f(n) in all cases, whatever the values of m and n may be; and in agreement with our previous notation, it may be denoted by f(m +n); therefore for all values of m and n

f(m) \times f(n) = f(m + n).

Also f(m) \times f(n) \times f(p) = f(m + n) \times f(p)

= f(m +n +p), similarly.

Proceeding in this way we may show that

f(m) \times f(n) \times f(p) \ldots to k factors = f(m +n +p+  \ldots to k terms).

Let each of these quantities, m, n, p, \ldots, be equal to \frac{h}}{k}, where h and k are positive integers ;

{f(\frac{h}}{k})}^k = f(h)

but since h is a positive integer, f(h)=(1+x)^n;

but f(\frac{h}}{k}) stands for the series