Sequences, Series, and Special Functions 591
- (a) Prove that a convex function e over [xi, X2] is nece~sarily continuous
in (x1, x2).
(b) Prove that e" ( x) > 0 is a sufficient condition for e to be convex on
· [x1,x2].
4. (a) Letting x = 1/ 2 (x 1 + x 2 ) in (8.14-4), we have
e ( X1; X2) < e(x1); e(x2)
Following Jensen [19], this inequality is often taken as the definition of
convexity. However, it is less restrictive than (8.14-4) since it does not
imply continuity.
(b) Letting x = .Ax1 + (1 - .A)x2 in (8.14-4), show that the inequality
becomes
8.15 Series of Negative Integral Powers of z - a
Definition 8.3 Series of the form
00
23 cn(z - a)-n =Co+ c1(z - a)-^1 + · · · + cn(z - a)-n + .' · · (8.15-1)
n=O
where a is a constant and {en} is a sequence of constants, are called series
of negative integral powers of z - a (more precisely, series of powers z - a
with negative integral exponents).
Theorem 8.41 Let r =.limn-+oo V\cJ. Then we have:.
1. If r = O, the series (8.15-1) converges absolutely for every z E C* -{a}.
2. If 0 < r < oo, the series (8.15-1) converges absolutely for every z such
that lz -al > r, the convergence being uniform on lz -al ;:::: r' > r,
and the series is divergent for lz -al < r..
3. If r = oo, the series (8.15-1) is divergent for all finite z.
Proof Letting ( = 1/(z - a), the series (8.15-1) becomes
00
L cnC =Co+ c1( + · · · + cnC + · · · (8.15-2)
n=O
which is an ordinary power series in ( with radius of convergence 1/r. By
Theorem 4.19 we have: