ON LAGRANGE’S BALLISTIC PROBLEM. 
207 
On putting R 2 for r', we have the value of x 0 along the junction of the second middle 
wave and the second reflected wave from the left expanded in a series of powers of S', 
or (s' — S 2 )/lS 2 , in the form 
= - 5 /*. [( b\ - 1 ) +{ b ' 2 - (3 b\ - 2) ( b'i - 1 )} r 
+ {B / 3 -3B , 2 (2B / 1 -l) + (7B?-6B' a + 2)(B / 1 -l)}r 
+ {B / 4 -2B , 3 C3 B'i — 1) -3B' 2 2 -±B' 2 (42B?-33B / 1 + 5) 
-|(28B' 1 3 -2lB , 1 2 +9B' 1 -2) (B\-l)} T 
+ {B' 5 -iB / 4 (30B / 1 -7)-6B / 2 B / 3 + T 1 oB / 3 (210B / 1 2 -108B / 1 + 7) 
+ }B' 2 2 (10 5 B'i - 2 7) - ^B' 2 (5 6 OB? - 4 62 B? + 81 B' x - 7) 
+ Hl26B / 1 4 -56B / 1 3 +2lB / 1 2 -6B' 1 + l)(B , 1 -l)}r+ ...]. 
Now at any point (R 2 , s') on the same junction the value of x 0 can be obtained by 
forming — II 3Z/3cr, where 
Z — K x -r B 4 u + ( — _ 
\cr dcr 
1 3 \ 4 f Fi (cr + u) 
cr 
, V io 
~r ^2 
— —) j— {n^ h —>n£‘ + m£ s — 
cr dcr/ [cr 
and 
II = a (cr/cr 0 ) 10 , £ = (s-S 2 )/2Ei, 
and putting therein 
cr — R 2 + 6 y 3 u — R 2 — s', s — s', s'— S 2 = t 2 S', 
and the result can be expressed in terms of S' in the form 
Xo 
a 
94 5 F i (2R 2 ) _ 945 ( 1+ ^IY^2B*) +42Q ( 1 + ^ F 1 (2 Y2R 2 ) 
2 ,2 
a 2 
-h> 5 ( , +sr ?J^A +15 (1+ , y rq2E a ) _ (1+y) . F.y 
■ 
+ m ( w ) { 945 ^ 6 - 945 x 3 ( 1 +^)^ 5 + 4 20 xV -( i +< 5 / ) 2 ^ 4 
-105 x 15 (1 + S'fS' 3 + 15 x *£- (1+S'Y S'*-^(l+S'f S'} 
- m ( r) 1 {945<i /7 —945 x^ (l +8') S' 6 + 420x^- (l + S') 2 S'» 
\2Ri/ 
_ 105 X (1 + S') 3 S ri + 15 X (1 + S'Y r-4^ (1 + S') 5 S' 2 } 
+ m l^j {945<T 8 - 945 X 4 (1 + S') S n + 420 x 14 (1 + S') 2 S' 6 
-105 x 42(1+^) 3 <5 /5 +15 x 105(1+ ^) 4 ^ /4 -210 (l + S') 5 S' 3 } 
_>79 (^r) y l 945 ^ 9 — 94 5 x f (l +#)#*+ 420x 18 {l + S') 2 S n 
- 105 X 63(1 + <r) 3 <i /6 + 15 X 189(1 + S'YS' b -^(l +S') 5 S' i } . 
2 g 
VOL. CCXXII.-A. 
