380
IV - Powers, Exponentials, Logarithms, Trigonometric Functions
still on the interval I. We have tan' x = 1/ cos 2 X = 1/(1 + tan 2 x). The
function is again strictly increasing and this time maps I onto lR in view of
its limit values at the end points of I. Whence a function
arctan y : lR ~ I =] - 7r /2, 7r /2[,
with
the argument used for the function arcsin shows here that
(14.25)
arctan y = y - y3/3 + y5/5 _ ....
For y = 1, we obtain
7r /4 = 1 - 1/3 + 1/5 - ... ,
Leibniz' series; for numerical calculations one should choose something else,
as Newton observed at the time. Moreover he himself had a method based
on his binomial series 26 . To do this he considered a circle, with equation
x 2 + y2 - X = 0, centre (1/2,0), and radius 1/2, whence
(14.26)
X 1 / 2 _ x 3 / 2 /2 _x 5 / 2 /8_
- x 7 / 2 /16 - 5x 9 / 2 /128 - ...
by (11.19). Now he knew that the area of the curve y = xm contained between
the verticals 0 and x is xm+1/(m + 1) and "thus" that the analogous area of
the curve (26) is
(14.27) z = X 1 / 2 (2x/3 - x 2 /5 - x 3 /28 - x 4 /72 - 5x 5 /704 - ... ).
He took x = 1/4 and found, without any difficulty, z = 0.0767731061630473,
the area of the curved triangle AdB in the figure 2. Since the angle ACd is
equal to 7r /3, and since the area of the triangle BdC is equal to V3/32, the
area of the circular sector ACd (namely 7r /24 since the circle considered has
radius 1/2) is equal to 0.07677 ... + V3/32, whence Newton deduced
26 See Vol. III of Mathematical Papers edited by D. T. Whiteside for the "tract"
De methodis serierum et fiuxionum of the winter of 1670--1671 and its English
translation, pp. 223-227. And since Newton clearly did not care to squander
his energy, he calculated simultaneously the areas of the equilateral hyperbola
x 2 - y2 + X = 0, the expansions of y in series being identical to that for the circle
up to changes of signs. The essential point is that if a curve is given by a power
series in x (with maybe non-integral rational exponents), the area bounded by
the curve and the verticals a and b is F(b) - F(a) where F is the primitive
series of y. In fact, Newton showed (in terms of fluents and of fluxions) that
the derivative of the area with respect to x is y and so considered the result in
question as obvious.
IV - Powers, Exponentials, Logarithms, Trigonometric Functions
still on the interval I. We have tan' x = 1/ cos 2 X = 1/(1 + tan 2 x). The
function is again strictly increasing and this time maps I onto lR in view of
its limit values at the end points of I. Whence a function
arctan y : lR ~ I =] - 7r /2, 7r /2[,
with
the argument used for the function arcsin shows here that
(14.25)
arctan y = y - y3/3 + y5/5 _ ....
For y = 1, we obtain
7r /4 = 1 - 1/3 + 1/5 - ... ,
Leibniz' series; for numerical calculations one should choose something else,
as Newton observed at the time. Moreover he himself had a method based
on his binomial series 26 . To do this he considered a circle, with equation
x 2 + y2 - X = 0, centre (1/2,0), and radius 1/2, whence
(14.26)
X 1 / 2 _ x 3 / 2 /2 _x 5 / 2 /8_
- x 7 / 2 /16 - 5x 9 / 2 /128 - ...
by (11.19). Now he knew that the area of the curve y = xm contained between
the verticals 0 and x is xm+1/(m + 1) and "thus" that the analogous area of
the curve (26) is
(14.27) z = X 1 / 2 (2x/3 - x 2 /5 - x 3 /28 - x 4 /72 - 5x 5 /704 - ... ).
He took x = 1/4 and found, without any difficulty, z = 0.0767731061630473,
the area of the curved triangle AdB in the figure 2. Since the angle ACd is
equal to 7r /3, and since the area of the triangle BdC is equal to V3/32, the
area of the circular sector ACd (namely 7r /24 since the circle considered has
radius 1/2) is equal to 0.07677 ... + V3/32, whence Newton deduced
26 See Vol. III of Mathematical Papers edited by D. T. Whiteside for the "tract"
De methodis serierum et fiuxionum of the winter of 1670--1671 and its English
translation, pp. 223-227. And since Newton clearly did not care to squander
his energy, he calculated simultaneously the areas of the equilateral hyperbola
x 2 - y2 + X = 0, the expansions of y in series being identical to that for the circle
up to changes of signs. The essential point is that if a curve is given by a power
series in x (with maybe non-integral rational exponents), the area bounded by
the curve and the verticals a and b is F(b) - F(a) where F is the primitive
series of y. In fact, Newton showed (in terms of fluents and of fluxions) that
the derivative of the area with respect to x is y and so considered the result in
question as obvious.