450 TRANSACTIONS OF SECTION A. 
and a simple calculation by means of an ordinary table of logarithms suggests that 
the limit exists and does not differ much from 2-718. The next step is the proof 
of the existence of this limit. By help of the binomial theorem for positive integral 
exponents it is easy to prove that, if 2 is a positive integer, (i.) (1+ 1/m)” increases 
as 7 increases, (ii.) (1 + 1/2)" lies between 2 and 3, and therefore (1 +1/m)" has a 
limit when increases indefinitely through positive integral values. Since any 
positive value of x which is not an integer lies between two consecutive integers, 
it is easy to deduce that (1+1/n)" has a limit when x increases indefinitely by 
continuous variation. The statement that the variation is continuous, or, in other 
words, that the index 2 runs through all values, irrational as well as rational, 
implies that powers with irrational exponents have been defined. In the elementary 
theory of indices they are not defined. It would seem to be appropriate to state 
at this stage that such powers are defined by the following property: If A is any 
real positive number, and a is any real number, rational or irrational, which 
lies between two rational numbers @ and 0, then A“ lies between A% and A®, 
The somewhat abstract discussion necessary to prove that this property defines 
A* uniquely when a is irrational may be postponed to a later stage. 
The conclusion which has been reached is that there is a definite number, 
between 2 and 8, which is the limit lim,_.(1+1/n)". We call this number e 
and we set before ourselves the problem of computing e. By the process already 
employed we find 
d 1 de* 
e, — log, ==, and thence — =e. 
Ee BN Stee dx 
Hence e* has a differential coefficient which is continuous. We now apply the 
theorem of mean value to the expression @(), where 
_ (ay, (=a) 
(n-1)! n! 
d 1 
da log,v= . log, 
R 
ba) see (L-COM Shows 
and 
adh BPE at sicte alk | 
DY Se = a 
This expression ¢(x) vanishes when 2 =1 and also when 2=0. We find 
R=n![e-1-1- 
Now #’(x) must vanish for some value of « between 0 and 1. Let a be this 
value; then R=ce%, and we have 
SoS he alee Jt eae where l>a>0. 
Dil ees (n—1)! mn! 
aL Io 
Since e lies between 2 and 3, the term = is less than uF and therefore we 
NM. ns 
may compute e with arbitrarily close approximation by the formula 
ALL 1 
Itl+tait othe is + 
In like manner by writing 
ie aie jee (0-2) 6= 4)" 4 .6=0)% 
(2) =e — ee —(b—2)e*— aa =i es we iy! e = R, 
and 
! b Gr 
R= @[e-1-0-5 -...- a 
bn 2! (n—1)!1’ 
we find that R=e* for some value of a between 0 and 4, and therefore 
6? §r-1 bn 
ee pon eet ae! _— et 
é Labie 54S elit ae ee 
