401 
of Edinburgh, Session 1876-77. 
</> 3= — A(2 + u) +B(3 + 4w + i> 2 ) 
- A(1 + 3v + v 2 ) + B(1 + 6v5v 2 + v 3 ) 
$4 = — A(3 + 4v + v 2 ) + B(4 + 10v 4- 6u 2 4- v 3 ) , 
from which it is obvious that the coefficient of - A in the expres- 
sion for (f>n , is a transcript of that of B in the preceding expression 
for tf)(n - 1 ). Hence, for the present, we may confine our attention 
to the latter. 
The coefficients of B form the following progression : — 
in cf > 0 0 
in (f> 1 1 
in <£2 2 + v 
in <£3 3 + 4v 4- v 2 
in <£4 4 + 1 Ou + 6u 2 + v 3 
in (f)5 5 + 20w 4- 2 Id 2 + 8n 3 4- v 4 
in <£6 6 + 35v + 56v 2 4- 36n 3 + KM + v 5 
in <j>7 7 + 56v+ 126v 2 + 120v 3 4- 55v 4 + 1 2v 5 4- n 6 
in cfjS 8 4- 84n + 25 2 v 2 x 330v 3 4- 220id 4- 78u 5 + I4v 3 + v 7 
in p 9 4- 120v4- 462u 2 4- 792id 4- 715v 4 4- 364v 5 4- 105v 6 4- 16u 7 4- &c. 
and in general 
in <cf>n 
n t n - 1 n n 4- 1 
I + _ jT ¥ “3“ 
n — 2n - l n n 4- 1 n 4- 2 
“1 2~ 3 ~1 5” 
v 2 4- &c. 
When v is positive the formula belong to the class of catenarian 
functions ; when v is negative, to the circular ones. 
If we put sin pa for <j> 0, sin p 4- 1 a for <£l, and - chord a for v, 
we obtain 
i n _ i 
sin (p 4- n)a = - mi pa j— |— 
n-2 n-1 n , 9 , P 
— _ cho a 2 4- Ac. 
A 2 6 
, • / , 1\ fW »“l?l»+l , 9,0 ) 
4- sin {p+l)at j — - — — cho a 2 4- &c. i 
and in thus putting p = 0 
sin na — n sin a 
1<-*T 
n 2 — 1 cho a 2 n 2 - 1 n* - 4 cho a 3 
3 
4- 
1.2 3.4 
- &c. | 
