56 Lord Rayleigh on BesseVs Functions as applied 



we find 



(l-J)j'„ +1 +|(^-l-^)J„' + (l-1 2 ^l)j ll " = 0. (4) 



In (4) we suppose that z is a root of JV, so that J ? / = 0. 

 The argument then proceeds as before if we can assume that 

 z 2 — n 2 and z 2 — n(n-\-l) are both positive. Passing over this 

 question for the moment, we notice that J n " and J' w+ i have 

 opposite signs, and that both functions are finite. In fact if 

 J n 11 and J n ' could vanish together, so also by (3) would J«, 

 and again by (2) J n+ i ; and this we have already seen to be 

 impossible. 



At consecutive roots of J»', J n n must have opposite signs, 

 and therefore also J'n+i. Accordingly there must be at 

 least one root of J' M +i between consecutive roots of J,/. It 

 follows as before that the roots of J'»+i separate those of J n '. 



It remains to prove that z 2 necessarily exceeds n(rc + l). 

 That z 2 exceeds n 2 is well known*, but this does not suffice. 

 We can obtain what we require from a formula given in 

 4 Theory of Sound,' 2nd ed. § 339. If the finite roots taken 

 in order be z iy z 2 . ... z s ... , we may write 



log J,/ (z) = const. + - 1 ) log z + X log (1-- z 2 jzs 2 ), 



the summation including all finite values of z s ; or on dif- 

 ferentiation with respect to z 



JJ'(z)_n-l K 2z 



J»'0) " 



This holds for all values of z. If we put z—n } we get 



s r^ =1 > •••••• (5) 



~s n 



since by (3) 



J," (n) -r-J,/(wJ = -n- 1 . 



In (5) all the denominators are positive. We deduce 



f.-' 2 —- n 2 Z x 2 —7l 2 Z-, 2 — n 2 



2n z 2 —n 



and therefore 



z 1 2 >n 2 + 2n>n{n + l). 



Our theorems are therefore proved. 



* Eiemann's Partielle Differential Gleichungen \ 'Theorv of Sound,' 

 §210. 



