a 2 = Vi — — 2 c x 4 q (A x 2 + c A 2 ) A 2 + A 3 (— 2 c 2 c ^ 4 A 4 p — 2 c c x 4 A x A 3 p) + 
+ (2 c x 4 p 2 — c 2 c 2 q 2 + 2 c c 4 A, q — ± c 2 c 4 A 2 — 8 c x 4 A 2 q) + 
+ A 5 ( 2 c 4 c í 4 A 4 — 6c 3 c l 2 A 1 A 3 —2c 2 c 1 4 A 4 —2c 1 4 A 1 cA 3 )+A 6 (c 2 c 1 2 A 1 2 — 
— 2 c 3 c 2 A 2 + 2cc 4 A 2 — 2 c 4 q — 4c 4 A 2 ) + A 8 ( 2 c 2 c 2 —2 c 4 ), 
a 3 = s 1 — 4 c c j 3 A 4 A 3 q 2 A~ 4 c 2 c x 3 A 3 A 4 q + 6 c c x 3 A 4 A 5 q + (2 c c x 3 — 
— 2 c 3 c x A x ) A\ 
a 4 — d x = č x 2 (c x 2 — c 2 ) q 2 A 4 — 2 c x 2 (c 2 — c 2 ) q A q Ar (c 2 — c x 2 ) 2 A 8 , 
b 0 = p 2 + 2 c 1 q /2 X = c x q A 4 (6 c 2 c x 3 A 4 2 A 3 —-2 c 2 c x 3 A 3 q — 2 c 3 c x 3 A x A^j + 
+ c x q A 5 (2 c 2 c x 3 A x A 2 — 2 c c x 3 A x q + 2 c c x 3 ^4 X 3 ) + 2 c 2 c x 4 A 6 ^4 X 2 A 3 + 
+ 2 c 2 c 4 A 1 A X A 2% 
b\ = V 2 2 c x $ y 1 = A 3 (2 c 2 c x 5 A 4 q + 6 c c x 5 A x 3 A 3 q — 6 c c ± 5 A x A 3 q 2 —- 
— 2 c 2 c x 5 A 2 A 4 q) + A 4 (2 c 2 c x 5 A x A 5 q + 4 cf A 4 q — 3 c 2 c 3 q 3 — 
— 2 c 4 cf A 2 + 10 c 3 c x á A x A 3 A 4 — 12 c 2 c* A 2 A 2 — c 3 c x 3 ,4 2 p 2 + 
+ c 2 c 3 ^4 X 2 q 2 + 2 c 2 c x 5 ^4 2 2 q + 4 c x 5 q 3 + 2 c c x 5 A 2 q 2 — 8 c x 5 ^4 X 2 q 2 ) + 
A- A 5 (8 c c x 5 A x A 3 q + 2 c 3 c x 5 A 3 A 5 + 2 c 3 c x 5 ^4 2 A 4 + 2 c 5 c x 3 ^4 2 — 
— 6 c 4 č x 3 A x A 2 A 3 — 2 c 4 c x 3 ,4 X 2 i 4 + 2 c 2 p* A x 2 A 4 + 6 c 3 c x 3 A 3 A 3 - 
— 6 c c x 5 A 3 A 3 + 5 c 4 c x 3 A 4 q — 15 c 3 c x 3 A x A 3 q — 4 c 2 c x 5 A 4 q) + 
+ A 6 (2 c 2 c x 5 A x A 5 —4 c 2 c x 5 A 3 2 — 8 c c x 5 A 2 A 2 + 10 c x b A 4 —c 2 c 3 A 2 q + 
+ 2 c 3 c x 3 ^4 2 q — 4 c 2 c 3 A 2 qA~^ c x 5 A x 2 q — 5 c 3 c x 3 A 2 qA~ 2 č c x 5 A 2 q — 
— 2 c x 5 q 2 — 12 c^A-^q) + A 7 (2 c c-fAj A 3 — 6 c 3 c 1 3 ^4 1 ^4 3 — 2 c 2 c x 5 ^4 4 ) + 
+ ^4 8 (2 c 3 c x 3 ^ 2 — 8 c 2 c x 3 A x 2 + 4 c x 5 ^4 X 2 + 5 c 2 c x 3 q — 4 c x 5 q), 
b 2 = s 2 A- 2 cq 6 x = A 4 (3 c 3 c x A x A 4 q 13 c 2 c/^^^ 3 ^ + 7 c 2 c 4 A 3 q 2 + 
+ 4 c 3 c x A 2 A 3 q A~ 2 c 3 c x 4 A x 3 A 4 — 2 c 4 c x 4 A x A 2 A 4 — 6 c 2 c x 4 A x 4 A 3 + 
+ 6 c 3 c 4 A 2 A 2 A 3 ) A- A 5 (c 4 c x 4 A 3 A 4 — 13c c 4 ^ x 3 q — 3 c 3 c 4 A x A 3 2 + 
+ 10 c c x 4 A x q 2 + 2 c 2 c x 4 A x A 2 q — c 3 c x 4 A 5 q — 4 c 2 c x 4 A x 3 A 2 + 
+ 2 c 3 c 4 A x A 2 2 A~ 2 c c 4 A x 5 ) A- A 6 (3 c 3 c x 4 A t A 4 — 13 c 2 c x 4 ^ x 2 A a — 
— c 3 c x 4 A 2 A 3 ) +AT(c 3 c 2 A x <jA- 3cc 4 A x Q — hcc 4 A 3 — 4c 2 c 4 A x A 2 + 
+ c 3 c 1 2 ^4 1 3 — c 4 c 1 2 A 1 A 2 ) + ^4 8 (4c 4 c 1 2 ^4 3 — c 2 Cj 4 ^4 3 ) + A 9 (4c 3 c 1 2 ^4 1 — 
— cc x 4 ^i), 
^3 = ^2 + 2 c x q á x = A 5 (6 c 3 c x 3 A x A 3 q — 2 c 4 c x 3 A 4 q A~ 2 c 2 cf A 4 q) + 
+ A 6 (8 c 2 c 1 3 ^4 1 2 q + 6 c 2 c 1 5 ^4 3 2 — 2 c x 5 A x 4 A~ 2 c c x 5 A x 2 A 2 A~ 2 c 3 c x 3 A 2 ^ — 
— 2 cc x 5 A 2 q — 2 c 2 c 3 (> 2 + 2 c 4 5 q 2 ) A~ A 7 (15 c 3 c x 3 A X A 3 — 2 c 4 c x 3 A 4 A- 
+ 2 c 2 c x 5 A 4 ) + A 8 (11 c 2 c 3 A 2 + 2 c 3 c x 3 A 2 — 2 c c x 5 A 2 — 2 c x 6 ^ x 2 + 
+ c 4 c 1 ()- 6 c 2 c 1 3 (» + 2 c x 5 q), 
b 4 = fj 2 = —3 c c x 2 (c 2 — c x 2 ) A x A 1 q —-3 c 2 c x 2 (c 2 — c x 2 ) ^4 ? ,^4 8 — 3 c c x 2 (c 2 — 
— c 2 ) A X A 9 . 
Koěfficienty au jsou stupně 12 -ho, bk stupně 15-ho vzhledem na 
všechny veličiny v nich přicházející. 
Eliminací sin co z rovnic (7) obdržíme vztah mezi Ak, poněvadž pak 
dle vztahů (4) platL 
XXIX. 
