Mr Sharpe, On the Reflection of Sound at a Paraboloid. 117 



of A shall vanish in the result. It will be found that we shall 

 get 



w = ± 2(Avf-ilogv.+ i^ + ^^±&c. ...(51), 



the upper signs going together and the lower together. Each will 

 be found to give solutions of (12). This may be shewn thus. 

 Confining ourselves only for the moment to the two first or 



leading terms in (51), if we put for Fin (12) e c{Av) /v*, where c is 

 a constant, and retain only the most important terms, we shall 

 find that the leading term in the result is multiplied by (c 2 — 4), 

 and is therefore caused to vanish by c= + 2. By comparing (51) 

 with (46) we see that the upper sign in the leading term of (51) 

 gives the leading term in V x . It follows therefore that the lower 

 signs in (51) must give us V 2 . The value of G in (49) for the 1st 

 solution of (12) is evidently ^7r - ^ A~~* as we see from (46). The 

 result (51) suggests another method of getting (perhaps in the 

 most convenient form for calculation) solutions of (12) adapted to 

 the case of v moderate and A large, we might assume in (12) 



F=(7e ± 2 u,)Mios yx |i + |l + J + ^ + &c l (52), 



and we could then determine w 1 , w 2 , &c. as functions of v. 



The result (39) can also be obtained by the method employed 

 in this Article. 



23. We will now proceed to find what V 1 in (12) becomes, 

 when v is large and A moderate, and it will be found that the 

 method is applicable to the case of both large, but A 2 /v small. 



In (12) put 



V 1 = € iv w 1 + e- iv w 2 (53), 



where w 1 is the same function of v and i as w 2 is of v and — i. In 

 the result equate to the coefficients of e iv and e~~ iv . Then 



' ^^(^ + l) + {-A^i)w^0 (54), 



w 2 satisfies a similar equation with the sign of i changed. 

 Solving (54) in descending powers of v we shall get 



e iv w 1 = a e iv x v~i {1+Ai ) 



' ( 1 + Aif i (1 + Ai f (3 + Ai) 



K \fv) 2*(3!) \2v) + ® 



2 2 2v 2 4 (2!) 



(55). 



