228 Mr Sharpe, Liquid Motion from a Single Source 
expand (say) 6 in a series of sines true from 0 to 7r/2, then in 
another series of sines true from 0 to 7r and subtract one from the 
other. We shall thus get zero, and we can take 
S (a m sin mO) = f3 
22 (- 1)“+' s i^ + 2 (- 1)- s -“^l (14). 
This will be zero from 0 to 7 t /2 excluding 7 t/ 2 itself. Instead 
of expanding 6 in a series of sines &c. we could of course similarly 
expand any function of 6. Again, by Fourier’s Theorem it can 
be shewn that between the limits 6 = 0 and 6 = 7t/ 2 (but excluding 
the actual limit tt/2) 
sin (2 p -{-1)6 — — § sin 2 nd . 
ir i 
sin (2n — 2p — 1) ^ sin (2n + 2p + 1) ^ 
2?i — 2p — 1 
2n + 2p + 1 
= 0 (15), 
where p is an integer and 2 signifies summation with regard to n 
from 1 to oc . By this means, between the given limits, the sine 
of any odd multiple of 6, or any finite expression of the form 
A sin 6 + B sin 3# + C sin 56, 
can be expanded in series of sines of even multiples of 6. We 
thus might use the single series (15) with a particular value of p 
between 6=0 and 6 = a a where a x < 7r/2, or if we wish to use 
expansions that would be true even at the limit 6 = 7t/2, we might 
use expansions of expressions like A sin 6 + B sin 36 + C sin 56 
with the condition A — B + C = 0. Again, to satisfy equation (5) 
w.e might take any linear combination of the expressions (13), (14) 
and (15) multiplied by constants. (5) can therefore be satisfied in 
an infinite number of ways, so the problem proposed admits of an 
infinite number of solutions. 
6. We will now proceed to examine in some detail the sim- 
plest case of equation (5) viz. 
Suppose 
S (a m sin m<9) = bt (- l) n+1 sin (2n - 1) 6 = 0 (16), 
n having all integer values from 1 to oo , and (16) being true 
from 6 = 0 to 6= tt/2 (tt/2 being excluded). Equation (8) for 
finding will then become 
fjbir = c sin otj + pOL x + b 
fsin 2 
sin 4a! sin 6a x 0 
4 1 6 " &C - 
(17). 
