§2. Series expansions
375
(vii) The number 1r. Throughout this section of the text, the least easy,
we work in R The psychological difficulty is to look for the number 7r while
appearing not to have found it already; there is a telling phrase of Pascal's
on this theme. The situation is less ridiculous than it might appear: there are
many "special functions" much more complicated than the circular functions
in order to locate whose roots one is forced to use arguments similar to those
which follow.
Consider the series
1 - cos x = x 2 /2 - x 4 /2.3.4 + x 6 /2.3.4.5.6 - ...
It is alternating, with decreasing terms for 0 ~ x ~ 3: indeed one passes from
one term to the next by multiplying by x 2 /3.4, x 2 /5.6, etc., i.e. by a factor
< 1 since x 2 < 3.4. Its sum therefore lies between x 2 /2 and x 2 /2 - x 4 /2.3.4,
whence
(14.9)
1 - x 2 /2 ~ cos x ~ 1 - x 2 /2 + x 4 /24
for 0 ~ x ~ 3.
In particular we deduce that cos 2 < -1/3 < 0 and that
-1 < cosx < 1 for 0 < x < 2,
as we see by examining the graphs of 1-x 2 /2 and of 1-x 2 /2+x 4 /24 between
o and 2.
Since the function cos x is continuous and equal to 1 for x = 0, the intermediate value theorem of Chapter III shows that the function cos x vanishes
somewhere between 0 and 2. Every limit of solutions of the equation f( x) = 0
being again a solution for every continuous function f, the set of roots 2:: 0
of cos x = 0 contains its greatest lower bound a; this is > 0 since cos 0 = 1.
By definition
(14.10)
7r /2 = least number a > 0 such that cos a = O.
One must have sina = +1 or -1; but the series defining sinx is also alternating with decreasing terms for 0 < x < 2, whence
(14.11)
x> sin x > x - x 3 /6 > 0 for 0 < x < 2,
and consequently
(14.12)
cos a = 0,
sin a = 1.
The addition formulae now show immediately that
(14.13')
(14.13")
cos2a
sin2a
and more generally that
-1, cos3a = 0, cos4a = 1,
0, sin3a = -1, sin4a = 0
375
(vii) The number 1r. Throughout this section of the text, the least easy,
we work in R The psychological difficulty is to look for the number 7r while
appearing not to have found it already; there is a telling phrase of Pascal's
on this theme. The situation is less ridiculous than it might appear: there are
many "special functions" much more complicated than the circular functions
in order to locate whose roots one is forced to use arguments similar to those
which follow.
Consider the series
1 - cos x = x 2 /2 - x 4 /2.3.4 + x 6 /2.3.4.5.6 - ...
It is alternating, with decreasing terms for 0 ~ x ~ 3: indeed one passes from
one term to the next by multiplying by x 2 /3.4, x 2 /5.6, etc., i.e. by a factor
< 1 since x 2 < 3.4. Its sum therefore lies between x 2 /2 and x 2 /2 - x 4 /2.3.4,
whence
(14.9)
1 - x 2 /2 ~ cos x ~ 1 - x 2 /2 + x 4 /24
for 0 ~ x ~ 3.
In particular we deduce that cos 2 < -1/3 < 0 and that
-1 < cosx < 1 for 0 < x < 2,
as we see by examining the graphs of 1-x 2 /2 and of 1-x 2 /2+x 4 /24 between
o and 2.
Since the function cos x is continuous and equal to 1 for x = 0, the intermediate value theorem of Chapter III shows that the function cos x vanishes
somewhere between 0 and 2. Every limit of solutions of the equation f( x) = 0
being again a solution for every continuous function f, the set of roots 2:: 0
of cos x = 0 contains its greatest lower bound a; this is > 0 since cos 0 = 1.
By definition
(14.10)
7r /2 = least number a > 0 such that cos a = O.
One must have sina = +1 or -1; but the series defining sinx is also alternating with decreasing terms for 0 < x < 2, whence
(14.11)
x> sin x > x - x 3 /6 > 0 for 0 < x < 2,
and consequently
(14.12)
cos a = 0,
sin a = 1.
The addition formulae now show immediately that
(14.13')
(14.13")
cos2a
sin2a
and more generally that
-1, cos3a = 0, cos4a = 1,
0, sin3a = -1, sin4a = 0