DIURNAL VARIATION OF TERRESTRIAL MAGNETISM. 
193 
Adopting the definition (23) for Qy, where <x is negative, we may return to the 
original form by replacing the /x coefficients with the help of 
K 
— a 
n 
(-1 y 
(n + o-)! 
(n — <r)\ 
/A, 
(31) 
If each term of (30) be subjected separately to the operation (28) and the results 
collected, so that all terms depending on any one value Qy are brought together, we 
may express the result of the operation by a series of the form 
— S E/Q/ sin 6 cos (cr\ — a) 
cr = — oo 
which must be equal to (26). 
If we put 
we find 
Ey = Ay+Byy+Cyy 
(32), 
(33), 
Ay=n(w+i)/cy. 
D <r_ (n-l)(n+l)(n-<r) n (n + 2) (n + cr+l) 
-V-- K n -1 i V-m- K n+1 
2 n— 1 
2 n + 3 
(34) , 
(35) , 
n <r ( n ~ 1 ) ( n + 1) (T-t n .71 + 2 ^ o—i t n . (n + 2) (n + cr+l ) (n + g- + 2) ^ a+1 
" 2.2n-l 2.2n + 3 n+1 2.2n + S ” +1 
_ (n— 1) (n+ 1) (n—a) (n —cr— 1) 
2.2n—1 
K 
„_y +l . . . (36). 
The values of k which determine S may now be found by equating the factors of 
sin (9Qy in (33) to the corresponding factor of sin Oxh/ in (26), remembering, however, 
that in the latter equation cr and n have the definite value belonging to the assumed 
velocity potential. If, for the sake of clearness, the type and degree of this velocity 
potential be now denoted by r and m, we find 
Ey=r 2(2m+ 1)E,„_ 1 T_1 = — (m + r)(m + T— l)(m—2r+ l)y 
(2m+ l)E T m _ 1 = 2T(m + r)y’ 2(2m+ l)E m+1 T_1 = —(m+2r)(w—r+ l)(w—r + 2)y 
(2w+l)E T m+1 = 2r(rn— t+ l)y’ 2(2m+ l)E m _ 1 T+1 = — (m+2r+ l)y 
2(2m+ l)E„ l+1 T+1 = —(m—2r)y 
If in any of these values of E the index is less than the suffix, that value may be 
put equal to zero, as the tessera! function to which that coefficient applies is zero. 
All other values of Ey are zero. If we take the case t = 1, m = 1, as an example, 
we obtain from (33), (34), (35), and (36) a number of equations in which all values 
of Ey may be put equal to zero excepting Ex 1 , E 2 X , E 2 °, E 2 2 ; the values of these are 
VOL. CCV1II.-A. 2 C 
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