4*58 
Mr. Ivory on the astronomical refractions. 
And it follows, from what has been said, that the subsidiary 
part of this expression is no other than the expansion of R 
deprived of its first term. In like manner if we separate the 
parts of A^°\ A^\ which involve c~ m , we shall get 
f—~ = * {*.+( 1 — £b 3 + ( 1 — = + «’+a < 3 V+&c. | 
+ 
2 C 
-m 
V 
zim 
{R_«+(l+i)*‘-(l+i+^)* , }5 
And here the subsidiary part is the expansion of R wanting 
the three first terms. On account of the factor c~ m , the sub- 
sidiary parts decrease without limit as m increases ; and 
thus the value of the integral can always be found to any 
required degree of exactness, in a series coinciding with the 
rigorous expression in its first terms, at the same time that it 
converges in its remaining terms. 
Now, let m = 8 : then 
a ( 3 ) _ *3 235 -8 
A — c 
64 
0.204357 
A ( 4 ) 21 
A 256' 
*9 
( 5 ) 
1024 
(6) 121 
2141 -8 
=0.079225 
. 23029 -8 
+ =0.026099 
287575 -8 
— ‘-^-c =0.0074 53 
A 
A _ 
4096 4096 
A <7, =-^ + ««-'=o.001876 
.(8) __ 2223 7 
A —65536 
16384 
66205285 -8 
5 c =0.000422 
65536 
And hence, neglecting the parts of A^°\ A (1 \ A^ z \ that in 
volve c~ m , we get 
4\A 
Cos. © 
1 — e 1 
