159 
This always gives a -f result for n small, as may be readily 
seen from the formula in its original shape. Its complete 
solution in x for every value of n -f i would require us to 
solve an algebraical equation of the form 
A Jiz^=( Bx — c y— D 
which would be difficult, if possible. But we may see that 
the second term can never become negative, unless F is 
positive, since E is essentially positive. No negative result 
can be obtained, therefore, for tan «, unless 
87696 
315 2 
n 7 
212-1 
315 
Whenever n is greater than this value, E — F may become 
negative, and if it become so for some value of x (=sin 2 a) 
less than 1, we may always adjust i so as to give * — ~ ^^ na 
an equal value, where a is the negative angle in the first 
quadrant the square of whose sine is defined as above. 
Apparently, therefore, there ought to be at all events one 
solution implying deviation opposite to that for a small mag 1 
l 2 212*1 
net, for some value of i. whenever — 7 — 
r 2 315 
The results in several cases in which different values 
have been assigned to are given below. 
212*1 
315 
(limit value) 
then E = 0, for x ~ f f \ \ = sin 2 a. 
boo*3 
but in order that the original equation should hold, we 
must have i indefinitely great. 
