Now, when m is very large as compared with i, this be since C+ C+...=1, as may be seen by putting 0 = 0. Hence ["P, cos me sin e de tends to the limit is indefinitely increased. 2 m2, as m The value of the factor involving m has been shewn above to be {m — (i − 2)} {m — (i − 4)} ... (m − 2) m2 (m + 2) ... (m + i − 2) {m − (i + 1)} {m — (i − 1)} if m be even, and (m + i + 1) {m — (i − 2)} {m — (i — 4)} ... (m − 1) (m + 1) ... (m + i− 2) {m − (i + 1)} {m — (i − 1)} if m be odd. Each of these factors contains in its numerator two factors less than in its denominator. It approaches, therefore, when m is indefinitely increased, to the value {m — (i—2)} {m— (i—4)} ... (m −2) m2 (m+2)... {m+ (i −2)} ́{m−(i+1)} {m—(i −1)}... (m − 1) (m + 1) ... {m + (i + 1)} if m and i be even, and In each of these expressions i may be any integer such that mi is even, i being not greater than m. Hence they will always be negative, except when i is equal to m. 20. We may apply these expressions to develop cos me in a series of zonal harmonics.. Assume cos me = BmPm+ Bm-2P, + ... m-2 +B,P+... Multiply by P, sin 0, and integrate between the limits 0 and π, and we get - 2 {m — (i − 2)} {m — (i − 4)} ... {m + (i − 2)} ' {m − (i + 1)} {m − (i − 1)} Hence B1 = − (2i + 1) · ... {m + (i + 1)} ̄ ̄ 2i+1 {m — (i − 2)} {m — (i − 4)} ... {m + (i −2)} - Hence, putting m successively = 0, 1, 2, ... 10, 21. The present will be a convenient opportunity for investigating the development of sin in a series of zonal harmonics. Since sin = (1-2), it will be seen that the series must be infinite, and that no zonal harmonic of an odd order can enter. Assume then Multiplying by P, and integrating with respect to μ between the limits - 1 and + 1, we get supposing P expressed in terms of the cosines of 0 and its multiples For values of i exceeding 2, we observe, that if we write for P; the expression investigated in Art. 18, the only part of the expression [P. (1 - cos 20) de which does not vanish i will arise either from the terms in P; which involve cos 20, or from those which are independent of 0. We have therefore = 2+1 1.3... (¿ − 1) 1. 3 ... (¿– 3) 4 2.4... π — г 2.4... (2) -1, +1. 2 + 2 cos 20) (1-cos 20) de i+2 П i+2/ 4 2.4... 2 2.4...(-2) 2 (2i + 1) π 1 . 3 ... (¿ −1) 1.3 ... (i − 3) i being any even integer. |