70 Solution of a Functional Equation. 
We propose to determine the nature of gz by rigorous analysis. 
From the nature of the function ¢(¢+z2)+9(«~z), which 
constitutes the right hand member of (1), we know that its se- 
cond differential, with reference to z, is the same as its second 
differential, with reference to z; therefore the second differential 
of gxpz, which constitutes the left hand member of (1), with re- 
ference to z, is the same as its second differential, with reference 
to z; hence we have the following condition : 
d? 9a. d? pz 
bho oe nc | = (2); 
d? gr d?.g. 
or, which is ‘the same thing, dr? t= soe we (3). 
‘Now, since the left hand member of (3) is a function of x alti, 
and the right hand member is a function of z alone, it follows 
that each member is equal to a constant eel whioti we = 
d*.¢. 
denote by a?; then we shall have =, * 90 =a (4). 
2 
Multiply (4) by 2¢z. via and we get 2. 5am = 
d.qz 
2a°gx. a (5). 
d.g¢x\? 
Integrating (5) and adding the constant c, we get ea os 
a?(gr)?+c (6), or, by a slight reduction, we obtain dr= 
(7). 
Va*(¢x)?+e an 
; 1 eee 
- Integrating (7) we get r=~ log. c’ (Var ay rete. ga] (8), 
where ¢ is another constant. 
Multiplying (8) by a, and passing from logarithms to exponen- 
tials, we get e =c’ (v7. a* (gx)? +e+a. 7) (9), vite e is 
such, that hyp. log.e=1. 
Dividing (9) by ce’ and transposing, we get /a?(gx)* attes) teas 
bw 
a pcs x _ (10). 
Squaring (10) and reducing, We gPb as! 
| et aire yt es 
Siete da, 
