Principles of Mathematics in Operations Research

13.4 Exponential and Logarithmic Functions 181

(^00) yn °° ,.,m °° n yk..,n-k
E(z)E(w)v ; v ' ^ n\ ^ = E^E^ TO! ^ ^ = EE kUn - k) 7^-Tn
n=0 m=0 n=0fc=0 V '
°°in/\ °° / , \rj
n=0 fc=0 V ' n=0
This yields

• E(z)E(-z) = E(z -z) = E(0) = 1, V* € C.


• J5(«) ^ 0, V2 G C. E(x) > 0, Vx G ffi.
  -B(a;) —> +00 asx-4 +00.
  0 < x < y =» £?(ar) < £(y), £(-?/) < £(-x).
  Hence, £?(ir) is strictly increasing on the real axis.
  . lim^ 0 <«+0-*('> = E(z).


• E{z\ + • • • + zn) = E{z) • • • E(zn). Let us take z\ = • • • = zn = 1. Since
  E(l) = e, we obtain E(n) — e", n = 1,2,3,... Furthermore, if p — n\m,
  where n,m e N, then [£(p)]m = £(mp) = £(n) = en so that E(p) =
  ep, p 6 Q+. Since E(—p) = e~p, p e Q+, the above equality holds for all
  rational p.


• Since xy — snpp€QBp<y xp, Vx, j/£l, a; > 1, we define ex — supp6Q9p<;2. ep.
  The continuity and monotonicity properties of E show that

E(x) = ex = exp(x).

Thus, as a summary, we have the following proposition:

Proposition 13.4.1 The following are true:

(a) ex is continuous and differentiable for all x,

(b) (e*)' = e*,

(c) ex is a strictly increasing function of x, and ex > 0,

(d) ex+y = exey,

(e) ex -> +00 as x -) +00, ex —> 0 as x —> —00,

(f) limx^+00 xne~x = 0, Vn.

Proof. We have already proved (a) to (e). Since ex > ?n+1y, for x > 0, then

xne-x < {"±1)1 and (f) follows. D

Since E is strictly increasing and differentiable on R, it has an inverse
function L which is also strictly increasing and differentiable whose domain is
E(R) = K+.

E(L(y)) = y, y > 0 <£> L(E(x)) =I,I£R.

Differentiation yields

£'(£(*)) • E[x) = 1 = L'(y) • y «• L'(y) = -, j, > 0.

y