76 : REPORT—1846. . 
Then z and y are both given as functions of « and v. We may therefore 
put 
r=Aa(uvr), y=A (uv); 
and similarly, 
zl =A(u! v'), ae (u! v'); : 
and as gat ¢@b=u+u! 
?,a+¢% b=v+I, ; 
we shall have a=A(ut+ul,v+v') . 
b=A (u+ul, v+v'). 
Hence the functions A (w+ u', v + v') anda, (u + u!, v + v!) are expres- 
sible as algebraical functions of A (uv), A, (wv), A (ul v'), A, (ul v!). 
These then are functions to which the integrals are in a certain sense in- 
verse, and which have the same fundamental property as circular and elliptic 
functions. 
In the general case of Abel’s theorem, we introduce (when the degree of 
the polynomial is 2m or 2 m—1), m—1 functions analogous to A, each 
being a function of m — 1 variables. These functions will, it may easily be 
shown, have the fundamental property just pointed out for the case in which 
m is equal to three. 
Again, the differential equations of which Abel’s theorem gives us alge- 
braical integrals, are, if the degree of the polynomial X be five or six, the 
following : 
dx + dy | dz _ 0. 
VX VME YZ 
udx ydy  zdz 
Fo Ws Gu 
and generally, if the degree of the polynomial be 2m or 2m — 1, there are 
m — 1 such equations, the numerators of the last containing the (m — 2)th 
power of the variables. ; 
M. Jacobi concludes by suggesting as a problem the direct integration of 
these differential equations, so as to obtain a proof of Abel’s theorem cor- 
responding to that which Lagrange gave of Euler’s (vide ante, p. 36). 
30. Another important paper by M. Jacobi is that which is entitled ‘De 
Functionibus duarum Variabilium quadrupliciter periodicis, etc.’ (Crelle, xiii. 
p-55). It is here shown that a periodic function of one variable cannot 
have two distinct real periods. In the case of a circular function, though we 
have for all values of x 
sina = sin(x + 2m) 
= sin(a +2nzn), 
m and m being any integers, yet 2m and 2m do not constitute two 
distinct periods, since each is merely a multiple of 27, which is the funda- 
mental period of the function. But if we had for all the values of x 
S@)=Sf {x + a} =f {x + B}, 
we should also have 
f(z) =f(e@+ma+nfp), 
where m and may be any integers, positive or negative. Hencema-+nB 
