§2. Series expansions
359
divine the general term from it; Newton, in fact, did so, starting from the
law of formation of the first coefficients and extrapolating what he knew for
the case of an exponent sEN.
Example 3. Taking s = -1/2, we find the square root of 1/(1 + z), denoted
(1 + Z)-1/2 even for z E C. On identifying the coefficients we get
(11.20)
(1 + z)-1/2 = 1 - z/2 + 3z 2 /8 - 5z 3 /16 +
+ 35z 4 /128 -... (Izl < 1),
or, in different notation,
(11.21)
1
1
1.3 2 1.3.5 3
(I I )
-:----:-::--;;:- = 1 + -z + - z + - - z +... z < 1 .
(1 - z )1/2
2
2.4
2.4.6
12 - The power series for the logarithm
We showed in Chap. III, nO 16, example 1, and again in n° 10, that the
formula
log' x = l/x
leads to
log(1 + x) = x - x 2 /2 + x 3 /3 - ...
for -1 < x < 1. We also know that the formula remains valid for x = 1
because of the continuity of the left hand side and of the fact that the series on
the right hand side converges uniformly on [0,1] (Chap. III, nO 8, example 4).
One can give a proof of this series expansion which is more complicated
and, besides its historical interest, is a beautiful exercise in passing to the
limit in a sequence of series.
Formula (11.7) shows that N 1/ p (x) = (1 + X)l/p for -1 < x < 1 and p a
nonzero integer. This relation can be written, by the definition of N s , as
(12.1) (1 + X)l/p =
= 1 + x/p + (l/p)(l/p - 1)x 2 /2! + (l/p)(l/p - l)(l/p - 2)x 3 /3! + ..
= 1 + x/p - (1 - 1/p)x 2 /p.2 + (1 - 1/p)(2 - 1/p)x 3 /p.2.3 -
- (1 - 1/p)(2 - 1/p)(3 - 1/p)x 4 /p.2.3.4 + ... ,
assuming -1 < x < 1. [For example
(1.5)1/8 = 1 + 1/16 - 7/512 + 105/24576 - 2415/1572864 + ... ,
an example of an alternating series with decreasing terms which even converges absolutely, since 10.51 < 1.] (1) shows that
(12.2)
p [(1 + X)l/p - 1] =
= x - (1 - 1/p)x 2 /2 + (1 - l/p)(l - 1/2p)x 3 /3 -
- (1 - l/p)(l - 1/2p)(1 - 1/3p)x 4 /4 + ....