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Here @ is a small angle, whose probability is under consideration, but E is in general large and may then be reckoned as if the distribution were uniform. Thus
!T>    _      fin
Q 7/3 __C")    I  \ cn'i   1/v                                  ^99\
a   o
and             = 5 ^Ur- {sin ft (ft - f _x) + . . . + sin s/5 (ft - _,,)} ....... (23)
By the rules of the Theory of Errors, we have
*^?1 =       f       (sin2  + sin2 2/3 + . . . + sin3 s/3) ....... (24)
Mod2|    2nasmsial
In (24) Mod2  = 2w/3/a, as before, and the series of (sin)2 may be replaced by
Thus                           Mod'Qsa   ^BiJia .........................(25)
If a is small, this reduces to a2/6?^, as in (4). If a = TT, that is if the distribution be over a semicircle, we get 7r2/4n. If we make a=2?r in (2o), the result is indeterminate, since, although sin ^ a = 0, n is infinite. There is a like indeterminateness when a is any multiple of 2?r, and this was to be expected. When the arc of distribution consists of entire revolutions, the phase of the resultant is arbitrary. But if the arc differs, even a little, from an integral number of revolutions, there is a definite phase favoured for the resultant, and Mod2  diminishes as n increases.
The case where the arc consists of entire revolutions is exceptional also as regards the amplitude, or intensity, of the resultant. As we know, in that case no definite value is approached, however great n may be, and the expectation of intensity is n. But if there be a fractional part of a revolution outstanding, the intensity does tend to a definite value, that namely which corresponds to a uniform distribution over the arc, and this value is proportional to the square of n.
We may go further and calculate what exactly is the expectation of intensity. We have to evaluate
...------------... [(cos 6 + cos & + cos #"+... )2 + (sin O+smff+sinO" +...)*]
+ 2cos(0-0//)+... + 2cos(0'-0") + ...]> -..(26)
the integration being in each case from + a.    Taking first the integration with respect to 6, we have
- \*dB \n + 2 cos (6 - 6'} + 2 cos (6 - 0") + ... + 2 cos (& - &'} + ...] = 4ft-1 sin ^a {cos ff + cos 6" + . ..} + n + 2 cos (& - 6"} + ....lines followed in equations (1) and (2). Thus with a replacing a and {3 replacing b, we have for the resultant whose amplitude is R arid phase (reckoned from the middle) ,