502 Mr. Ivory on the Theory of the Astronomical Refractions. 



The horizontal refraction answers to cos Q = 0, e = 1 ; and 

 the part of it depending on Qo is found by adding all the co- 

 efficients, viz. 



«^l + «^ X 2'8024.736 = 2036"-52. 

 V 5 I 

 If we take the integral between the limits x = 0, a; = oo, the 

 result is not sensibly different, viz. 



a (I +«) fo^dxc-' ^ «(1+«W^ ^ 2036"-52. 



Investigation o/ X x Qi . 

 For this purpose we must find the value of 



A ~*/ 



^(1-^2)2^4. g3,_^ 



m 





2 f? a; c— 2 * 



\/('-^^' 



2a: * 



2»i 



this integral has therefore the same form as Qq , the quantities 

 2 X and 2 w taking the place of x and m. Wherefore, if we 

 assume 



— -— — = a^e + a^e^ -{- a^er + &c., 



the value of a2„+i will be found merely by writing 2 m for m 

 in the expression of Agn+i; but as c-^'" = c-^o jg extremely 

 minute, the part multiplied by it may be neglected. Thus, 

 n+1 , n—\ w-fl.w + 2 o 



The numerical coefficients are as follows : 

 a, = 1 



03 = 0*9 



as = 0-73 



37 = 0*535 



Eg = 0-3555 



a,! = 0*21505 



ai3= 0'118945 



ai5 = 0"0604.215 



a,7 = 0-0283127 



aj9 = 0-0122898 



a.2i = 0-0049621 



a^g = 0-0018695 



035= 0-0006623. 



