1903-4.] Lord Kelvin on a Free Procession of Waves. 
313 
values of any function for equidifferent arguments. Let f(x) be 
any function whatever, periodic or non-periodic ; and let 
(25); 
which makes 
P(z) = P(z + A) (26). 
Let the Fourier harmonic expansion of P(a?) be expressed as 
follows : — 
P(:c) = A 0 + A x cos a + A 2 cos 2a + A 3 cos 3a + 
+ sin a 4- B 2 sin 2a + B 3 sin 3a + • 
where a : 
Denoting b y j any integer, we have by Fourier’s analysis 
i4 = i> p (<V'¥ • • 
which gives 
<UAj - 2 [ dxf(x + iX) cos j/'— 6 = I dxf{x) cos yFf 
itTooJ 0 A J -oo A 
jrAB,-= y [ dxf(x + iX) sin = [ dxf{x) sin j 
o A J -00 
2i rx 
A , 
§16. Take now in (29), as by (19'), (20), 
f(x) = <f>{x,0)-<f>(x + ) ±,0) 
27 TX 
X 
(27). 
(28); 
(29). 
(30). 
This reduces all the B’s to zero ; reduces the A’s to zero for even 
values of j ; and for odd values of j gives, in virtue of (22), 
C +“> O-r/v. 
%XAj=2 dx(f)(x, 0) cosj — .... (31). 
^ J —00 A 
Go back now to §§ 3, 4, (6), (12), above ; and, according to the 
last lines of § 4, take 
0)= -j RS J. - 4(p + z ) 
• (32). 
A,= x. 
J{?.+ !.X) P 
Hence, for the harmonic expansion (24) of P(£, 0), we have 
2irX 
)CO SJ _ 
. . . (33). 
The imaginary form of the last member of this equation facilitates 
the evaluation of the integral. Instead of cos j--- in the last 
A 
f+". 7(p + z) ,2ttx 4 j DO l /■+“ , J2 
I dxyy cos j = — ( RS ■ dx --y — cos j 
J -o o p J p A | ) J -oo sJ{Z+LX) 
