Page:A History of Mathematics (1893).djvu/214

and difficult analysis at the following very remarkable expression:

$\scriptstyle {{\frac {\pi }{2}}={\frac {2.3\cdot 4.4\cdot 6.6\cdot 8.8\cdots }{1\cdot 3.3\cdot 5.5\cdot 7.7\cdot 9\cdots }}}$

He did not succeed in making the interpolation itself, because he did not employ literal or general exponents, and could not conceive a series with more than one term and less than two, which it seemed to him the interpolated series must have. The consideration of this difficulty led Newton to the discovery of the Binomial Theorem. This is the best place to speak of that discovery. Newton virtually assumed that the same conditions which underlie the general expressions for the areas given above must also hold for the expression to be interpolated. In the first place, he observed that in each expression the first term is x, that x increases in odd powers, that the signs alternate + and —, and that the second terms $\scriptstyle {{\frac {0}{3}}x^{3},~{\frac {1}{3}}x^{3},~{\frac {2}{3}}X^{3},{\frac {3}{3}}X^{3},}$ are in arithmetical progression. Hence the first two terms of the interpolated series must be $\scriptstyle {x-{\frac {{\frac {1}{2}}x^{3}}{3}}}$. He next considered that the denominators 1, 3, 5, 7, etc., are in arithmetical progression, and that the coefficients in the numerators in each expression are the digits of some power of the number 11; namely, for the first expression, $\scriptstyle {11^{0}}$ or 1; for the second, $\scriptstyle {11^{1}}$ or 1, 1; for the third, $\scriptstyle {11^{2}}$ or 1, 2, 1; for the fourth, $\scriptstyle {11^{3}}$ or 1, 3, 3, 1; etc. He then discovered that, having given the second digit (call it m), the remaining digits can be found by continual multiplication of the terms of the series $\scriptstyle {{\frac {m-0}{1}}\cdot {\frac {m-1}{2}}\cdot {\frac {m-2}{3}}\cdot {\frac {m-3}{4}}\cdot }$etc. Thus, if $\scriptstyle {m=4}$, then $\scriptstyle {4\cdot {\frac {m-1}{2}}}$ gives 6; $\scriptstyle {6\cdot {\frac {m-2}{3}}}$ gives 4; $\scriptstyle {4\cdot {\frac {m-3}{4}}}$ gives 1. Applying this rule to the required series, since the second term is $\scriptstyle {{\frac {{\frac {1}{2}}x^{2}}{3}},}$ we have $\scriptstyle {m={\frac {1}{2}},}$ and then get for the succeeding co-