1034 Proceedings of Royal Society of Edinburgh. [sess. 
throughout space, we find that the term involved contributes to 
the force at the pole the amount 
2- 2 ^ 5 Ma2 [4 + 105 cos 4 e - 75 cos 2 0](2p)f* , 
where cos0 has the value (aX + /3p + yv)/(2p) 112 as before. 
Terms involving the first power of a/p vanish because they contain 
odd powers of X, etc., so that they cancel in pairs in the summation 
for each value of p, positive and negative values for X, etc., 
occurring symmetrically. It is necessary also to note that 
P = oo 
2(3 cos 2 6 — 1) = 0. The proof is given in next section. 
p = i 
To the quantity just obtained we have to add, from the expres- 
sion for F cos (6 + <£), the terms involving « 2 /r 2 , for these have 
practically the same value at the pole as at the centre. Thus we 
get finally, for the component of the force, at the pole, taken 
parallel to the magnet, the expression 
[245 cos 4 <9-165 cos 2 6 + ll](2p)-» 2 . 
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8. Proof that 2 (3 cos 2 6 - 1) = 0. — The sum vanishes for every 
p=i 
value of p independently. Consider 2(aX + /3g + yv) 2 = 2 (a 2 A 2 + /3 2 /x 2 
+ y V 2 ) + 21(a /3Xp + fiygv + yavX). The second sum on the right 
hand side of the equation vanishes since, with a given value of p, 
positive and negative values of A, /x, v, occur symmetrically, so that 
each positive product is cancelled by an equal negative product. 
Again, by the symmetry of the expression 2 p = X 2 + p? + v 2 , we see 
that the number of points having a given X , /x, or v respective^ 
are equal. Thus 2-A 2 = 2-^, 2 = 2-v 2 = J2(A 2 + /x 2 + 1 / 2 ). Therefore 
2(aA + /3p + yv) 2 — J2(A 2 + f 2 + v 2 ). This proves the theorem. 
It is interesting to find that the truth of this theorem in topology, 
regarding points in any homogeneous cubic arrangement, necessi- 
tates also the condition that these points are loci of zero force if 
magnetic or electric, similarly directed, double-points are centred 
at them. 
9. Modified Expression for the Parallel Components. — It is 
