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(without typos) from the publisher. Not indexed. Not illustrated.
1917 Excerpt: ...+ od--6c = 0. If this equation has two distinct
roots sl, s2, there exist two independent integrals 0 (x), 02(x)
such that we have (102) 0z + 2w) = slx), 02(x + 2 w) = s202(x), and
the relations (101) can be replaced by the two relations of the
same form (103) 0 + 2 /) = By means of the relations (102) and
(103), we can now obtain two different expressions for 0l(x + 2 a +
2 w') and 02(x + 2 a + 2 w'). We have, on the one hand, 0 (x + 2 w
+ 2 /) = Simz + 2 of) = s0 (x) + sx). On the other hand, proceeding
in the inverse order, we may also write 0 (z + 2 w + 2 w') = t0i(x
+ 2 w) + j02(x + 2 -) = fcsx) + fei02(x). Since these two
expressions must be identical, we have I = 0, for sl--s2 is not
zero. Similarly, by considering the two expressions for 02(x + 2w +
2w'), we find m = 0. The integrals 0l(x), 02(x) are therefore
analytic functions except for poles, which reproduce themselves
multiplied by a constant factor when the variable z increases by a
period; these are called d0ubly periodic functions of the second
kind. Every function 0 (x) analytic except for poles which
possesses this property can be expressed in terms of the
transcendental functions p, f, a, since the logarithmic derivative
0'(x)/0 (x) is an elliptic function, and we have seen that the
integration does not introduce any new transcendental (II, Part I,
75). Moreover, we can prove this without any integration. Let 0 (x)
be an analytic function except for poles such that Consider the
auxiliary function (x) = efxa(x--a)/r(x), where a and p are any two
constants. From the properties of the function a (see Vol. II, Part
I, 72) we have where C' is not zero. In the first case the
integrals 0i(x), 02(x) are again doubly periodic functions of the
second kind. In the second case the...
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