§2. Series expansions
379
(xi) Complex roots of the trigonometric functions. The relations (4) show
that sin z = 0 is equivalent to exp( iz) = exp( -iz), Le. to exp(2iz) = 1, Le.
to 2iz = 2ki7f, so that it has no other complex roots than the obvious real
roots. Same result for cos z.
(xii) Expansions of 7f in series. There are as many as one wants. Let us
start for example from the function sin x on I =]-7f/2,7f/2[. Since 7f/2 is the
least root> 0 of cos x = cos( -x), the function cos x, which is > 0 for x = 0,
is> 0 on I (intermediate value theorem). So we have sin' x> 0 on I, so that
sinx is strictly increasing (Chap. III, nO 16), thus maps I onto J =]- 1,1[.
The inverse map
arcsin: J ~ I
exists and is differentiable, with
arcsin' y = 1/ cos x = (1 - y2)-1/2
as we have already remarked in Chap. III, formula (15.7). Since Iyl < 1,
Newton's binomial series gives us
(1 - y2)-1/2 = 1 + y2/2 + 1.3y4/2.4 + 1.3.5y6/2.4.6 + ... ;
see (11.12). The primitive series of the latter is
F(y) = y + y3/3.2 + 1.3y5/5.2.4 + 1.3.5y7/7.2.4.6 + ...
and represents a differentiable function on J such that
F' (y) = arcsin' y
by the general theorem on term-by-term differentiation of power series.
Chap. III, n° 16 shows then that arcsiny = F(y) up to an additive constant; it is zero since the two functions are zero at y = O. In consequence,
(14.24) arcsin y = y + y3/3.2 + 1.3y5/5.2.4 + 1.3.5y 7 /7.2.4.6 + ...
for Iyl < 1; since arcsin(I/2) = 7f/6 we obtain a series for 7f which converges
quite quickly.
Formula (24) is due to Newton, who had a proof very like ours: he integrated the binomial series term-by-term, which we shall not be able to do
until Chap. V. Newton obtained the series for sin z by inverting (24), an easy
exercise if one wants only the first terms.
Another possibility: use the function
tanx = sin x/ cos x,
379
(xi) Complex roots of the trigonometric functions. The relations (4) show
that sin z = 0 is equivalent to exp( iz) = exp( -iz), Le. to exp(2iz) = 1, Le.
to 2iz = 2ki7f, so that it has no other complex roots than the obvious real
roots. Same result for cos z.
(xii) Expansions of 7f in series. There are as many as one wants. Let us
start for example from the function sin x on I =]-7f/2,7f/2[. Since 7f/2 is the
least root> 0 of cos x = cos( -x), the function cos x, which is > 0 for x = 0,
is> 0 on I (intermediate value theorem). So we have sin' x> 0 on I, so that
sinx is strictly increasing (Chap. III, nO 16), thus maps I onto J =]- 1,1[.
The inverse map
arcsin: J ~ I
exists and is differentiable, with
arcsin' y = 1/ cos x = (1 - y2)-1/2
as we have already remarked in Chap. III, formula (15.7). Since Iyl < 1,
Newton's binomial series gives us
(1 - y2)-1/2 = 1 + y2/2 + 1.3y4/2.4 + 1.3.5y6/2.4.6 + ... ;
see (11.12). The primitive series of the latter is
F(y) = y + y3/3.2 + 1.3y5/5.2.4 + 1.3.5y7/7.2.4.6 + ...
and represents a differentiable function on J such that
F' (y) = arcsin' y
by the general theorem on term-by-term differentiation of power series.
Chap. III, n° 16 shows then that arcsiny = F(y) up to an additive constant; it is zero since the two functions are zero at y = O. In consequence,
(14.24) arcsin y = y + y3/3.2 + 1.3y5/5.2.4 + 1.3.5y 7 /7.2.4.6 + ...
for Iyl < 1; since arcsin(I/2) = 7f/6 we obtain a series for 7f which converges
quite quickly.
Formula (24) is due to Newton, who had a proof very like ours: he integrated the binomial series term-by-term, which we shall not be able to do
until Chap. V. Newton obtained the series for sin z by inverting (24), an easy
exercise if one wants only the first terms.
Another possibility: use the function
tanx = sin x/ cos x,