586 Prof, F. D. Murnaghan on the Deflexion of a 



the perpendicular distance from the origin being A/C). 



Hence j is the largest value of u ; so that it is a small 



quantity of the order ^-.10~ 5 . We shall denote this 

 small quantity by a, and for the next approximation shall 

 keep in any expression only the lowest powers of a occurring 

 in it. 



The discriminant of the literal cubic equation 



a x z + a\ x 2 -\-a 2 x + a 3 = 



is a 2 a 2 + 18a a 1 a 2 a r — 4oa a 2 3 — 4a 1 3 <2 3 — 21a 2 a 2 . 



For the cubic in question 



2mu 3 — u 2 +u 2 = 0; 



this is 4a 2 (1— 27m 2 a 2 ), which is positive since a is small. 

 Hence the three roots of the cubic are real ; when a = 



they are 0, 0, — , and writing as first approximations when 



a is small, A^a, k 2 ot, — -f- k s a, we find from the symmetric 

 functions of the roots 



h + h + & 3 = ; (Jc x + k 2 ) -— = ; tifac? = —a 2 , 



so that i7i 



& 3 = 0, A?! = — 1, & 2 = l, 



and we may write the three roots as 



1 * 

 u 1 = —a ; u 2 — + cl ; w 3 = — - . 



The next approximation is found in the same way or by 

 direct substitution in the cubic to be 



Mi — — ot + ma 2 ; 2* 2 = a + ?na 2 ; u 3 = — — 2ma 2 . 



Now the cubic (-tt I = 2mu 3 — u 2 4- a 2 cannot be negative 



in our physical problem, whilst also u starts out very small 

 and is always positive. As the ray of light is traced out, 

 u starts at zero (let us say) and increases to v 2 , at which 



value I -7-7 J is zero and u has a maximum value corre- 

 sponding to the perihelion of the light-path. The variable u 

 must then retrace its values (it cannot lie between u 2 and m 3 , 



* It is just as convenient to substitute the trial solutions - — \-~ka, etc.. 



directly in the cubic, and thus determine the k. 



2m 



