Page:A History of Mathematics (1893).djvu/213

 is infinite. But in the case when m is greater than unity and negative, Wallis was unable to interpret correctly his results. For example, if $\scriptstyle{m=-3}$, then the ratio becomes $\scriptstyle{1:-2}$, or as unity to a negative number. What is the meaning of this? Wallis reasoned thus: If the denominator is only zero, then the area is already infinite; but if it is less than zero, then the area must be more than infinite. It was pointed out later by Varignon, that this space, supposed to exceed infinity, is really finite, but taken negatively; that is, measured in a contrary direction.[31] The method of Wallis was easily extended to cases such as $$\scriptstyle{y=ax^\frac{m}{n}+bx^\frac{y}{z}}$$ by performing the quadrature for each term separately, and then adding the results.

The manner in which Wallis studied the quadrature of the circle and arrived at his expression for the value of $$\scriptstyle\pi$$ is extraordinary. He found that the areas comprised between the axes, the ordinate corresponding to x, and the curves represented by the equations $$\scriptstyle{y=(1-x^2)^0,~y=(1-x^2)^1,y=(1-x^2)^2,~y=(1-x^2)^3,}$$ etc., are expressed in functions of the circumscribed rectangles having x and y for their sides, by the quantities forming the series

When $\scriptstyle{x=1}$, these values become respectively $$\scriptstyle{1,~\frac{2}{3},~\frac{8}{15},~\frac{48}{105},}$$ etc. Now since the ordinate of the circle is $$\scriptstyle{y=(1-x^2)^\frac{1}{2},}$$ the exponent of which is $$\scriptstyle{\frac{1}{2}}$$ or the mean value between 0 and 1, the question of this quadrature reduced itself to this: If 0, 1, 2, 3, etc., operated upon by a certain law, give $$\scriptstyle{1,~\frac{2}{3},~\frac{8}{15},~\frac{48}{105},}$$ what will $$\scriptstyle{\frac{1}{2}}$$ give, when operated upon by the same law? He attempted to solve this by interpolation, a method first brought into prominence by him, and arrived by a highly complicated