r+ pL 
nt =q 
Mp 
we find 
qa, + a, + + an =0\ 
a, + ge, + @, = 0 
a, + qa, + a, a (10) 
An—2 + On—1 q + &= | 
a, — + Og) Se ang = 0 
In order to get a solution differing from zero, it is necessary that 
gade Ort 
kg de 0 
OE Aken: 
Anil ea EN RE IND 
00 lg 1 
Aad 01 q| 
Putting again 
: q = 2 cos 0 
we can reduce (11) to determinants of the same kind as (4), and 
to determinants equal to unity. Finally, we find: 
DS AE EPE ys RQ) 
where we have to distinguish between the cases even or n/ odd. 
From (12) we have 
: ka 
for n =even: cosn g == — 1, therefore 6 = — 
n 
ka 
for n’ = odd: cosn'A’'=— 1, Ô=— 
n 
whereas for the two cases we can put 
Beh n +2, kh, =n +4 ey, SS on — 2 
As we have put g=2cos 0, we find for n and also for n’: 
gi 2-cos 6, = - Zeer =D 
2 
Go ACRO — 2 cot = or 
n 
qe — 20080, = — 2 cos— 2 
n 
9 
Gn = 2 cos Oy = — 2 cos — 2% ; 
n 
For the numbers g we find: 
