PHILLIPS AND MOORE. — LINEAR DISTANCE AND ANGLE. 75 



We assume that 



[F p'"] ^ 0. 



Then p^ and \F p^\ r ^ m, are not zero. 



Since [;> •;.»'"] = o, there can be no terms in the expansion of 

 \p-¥ p"^\ which have F outside the parenthesis. Hence 



p {Fp"^) = m {Fp^) p + (^'l) [Fp-'' • r] - 2 (^2^) (F P'") P- 



Consequently 



t>? — 1 



[F p-^-\ /] = 2 (F p"") p. 



Ill 



Similarly 



[F p™"!-;/] = Q \F p^-'-p^-p^-'] - [2 Q - r] {F p"^) p -i 



m 



Solving this for p^~^ and changing r into r + 1, we obtain 



,^ m[Fp"'-'-p'^'] 



^ (Fp"*) (r+ 1) (m-r)' 



Repeated use of this formula gives finally 



p' fmFp"'~^\"'~'' p' 



-.m 



r\ V F p"^ J ml (m^ r)\ 

 If we choose the magnitudes of F and p such that 



iFp^) = m\ .... (A) 



and let 



we have 



(5) 



(32) 



where r = 0, 1, ...?«. 



Letting r = in (32) we have 



ml = 9"'-(/) (^0 



