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If mi and hence rnk-\ is positive, (9) tells us that all the m&_i members of S: e8k_1+i> ' ' ' » e8k_1+mk_1 belong to C. In particular the last one which is e8fe belongs to C. Moreover mk~n — d(sk) = 0. Hence e8k is the nth member of C. If m\ and hence mk-i is negative, (9) tells us that all the nik-i members of S: etk+1, • • • , e8k_l belong to C. In particular e8h_l is a member of C. But mk-\ = n — d{sk-\) or O(Sk-i) =n — mk-i\ hence e8Je_1 is the (n — mfc_i)st member of C. Thus the proof is complete. If S and C are wholly arbitrary one may show by means of examples that the conclusion of the theorem is the best that can be obtained from the hypothesis. Additions to the hypothesis make the last two statements of the theorem more precise. For example, if we assume that no two consecutive members of C are consecutive members of 5, it is easy to prove that e8k_t is the (w — l)st member of C in case rai<0.

A new class of functions of two variables involving Bessel functions of half an odd integer