150 MR JOHN DOUGALL ON AN ANALYTICAL THEORY OF 
7. Position of the zeroes of the functions sinh CC 
The nature of the infinite series occurring in the above solution will be made more 
intelligible by a short discussion of the position of the roots of the functions. 
sinh 2xh+2«h. These are obviously found from the corresponding roots of sinh (+¢ 
by dividing by 2h. | 
(i) smh (-C=0. 
¢=0 is a triple root, and the remaining roots are all complex, falling into sets of — 
4 of the form +p+iq, where p, q are real. If, then, (=&+7, & and » being real, we 
need only consider the case of & and » both positive. We have then 
sinh €cosy=€ and cosh ésiny=7. 
Cos y and sin y are therefore both positive, and 7 must lie between 2n7 and 2n7 +4 7/2. 
It is easy to prove that there is no root between 0 and z/2. For €>tanh&, or 
E/ sinh > 1/ cosh & so that 
cosy>siny/y or y>tanyn or >7/2. 
For every positive integral value of n, however, beginning with n=1, there is one 
root, and one root only, with » between 2nz7 and 2n7+7/2. This will be readily seen 
on roughly tracing the graphs sinh € cos = and cosh € sin 7=y, or it may be proved 
by an elementary application of the Theory of Functions. ‘Thus, if we make the variable 
¢ describe the contour of the rectangle formed by the four lines 
g=2nr, €=N, y=2nur+7/2, €=0, 
where N is a large positive number, it will be found that the function y= sinh (-—¢ 
describes once a contour enclosing the point y=0 in they plane. There is therefore 
just one point within the rectangle at which v becomes zero. : 
For the large roots cos 7 must be small, or 
n= 2nm + 77/2 —«€, where ¢ is small. 
Hence 
cosh €=n=2nr+7/2, 
or €=log (4n + lr) approximately. 
Then 
e=€/sinh €=2 log (4n4+ 1 7)/(4n+1 =). 
By successive approximation we may now find the roots as nearly as we wish, but 
exact values are not at all necessary, the first approximation being quite sufficient for 
our purpose, 
6, =log (4n +1 3) +(2n+$)xt ; : ‘ . Ce 
(ii) smh €+(=0. 
In this case ¢=0 is a simple root, and the rest of the roots are complex. If 
C=&+1, we have 
sinh €cosy7+&=0 and coshésinn+7n=0. 
Cos 7, sin are both negative when & and » are positive ; hence » lies between (2n—1)r 
