Calculus: Analytic Geometry and Calculus, with Vectors

492 Exponential and logarithmic functions

To investigate the factor ah - 1, we employ the method which we used to
prove Theorem 9.15 when we knew practically nothing about exponentials.
Let h be a number, now not necessarily rational, for which 0 < h < 1 and
let n be the greatest integer for which nh <- 1. The equality

(9.23) (ah - 1)a(k-1)h = akh - ack-1>h

still holds when k = 1, 2, 3, , n,and summing over these values of

k gives, as in (9.153),

n++1

(9.231) (ah - 1) C 4 a(k-1)hl = anh- 1

k=

and hence

(9.232)

ah-1h _ nanh- 1

I atk-1)hh

k=1

The numerator of the right side converges to a - 1 as h ---> 0. If to the

denominator of the right side we add the negligible term anh(1 - nh),

the new sum will be a Riemann sum, with partition points

0<h<2h<3h<. <nh<1

in the interval 0 <_ x < 1, which converges to the positive number

f0 1 ax dx as h, the norm of the partition, approaches 0. Because it is

sometimes extremely helpful to be able to recognize Riemann sums when

they appear in somewhat disguised forms, we give careful attention to

the details. If we set f(x) = ax, set xk = kh when 0 < k 5 n, set

xn+1 = 1, set xk* = xk when 1 k 5 n + 1, set Oxk = h when 1 < k <- n,

and set Ax.+1 = 1 - nh, then the new sum becomes the Riemann sum

n+1

(9.233) 1 f(xk) Lxk

k=1

and as h, the norm of the partition, approaches 0 the Riemann sum

approaches fol f(x) dx, that is fol ax dx. Therefore,

(9.24)

limah - = a - 1

= A,

h-o h f axdx

0

where A is the constant defined by the last equality. It is possible to

squeeze information from this formula, but we obtain a formula giving

a simpler relation between a and A before actually showing that there is

one and only one number a for which A = 1.

We have obtained the first of the formulas

(9.25)

TXax = Aax, dx logo x =x