Page:Scientific Papers of Josiah Willard Gibbs - Volume 2.djvu/88

72 Hence, since $$[\Phi^2. \beta] \times [\Phi. \beta]$$ and $$[\Phi. \beta] \times \beta$$ are parallel, Since $$a^{-1}\left\vert \Phi \right\vert$$ is positive, we may set  If we also set

the vectors $$\beta, \beta_{1}, \beta_{2},$$ etc., $$\beta_{-1}, \beta_{-2},$$ etc., will all lie in the plane perpendicular to $$\alpha ',$$ and we shall have We may therefore set  Multiplying by $$p^{-1}\Phi,$$ and by $$p\Phi^{-1},$$

Now, if $$n > 1,$$ and we lay off from a common origin the vectors the broken line joining the termini of these vectors will be convex toward the origin. All these vectors must therefore lie between two limiting lines, which may be drawn from the origin, and which may be described as having the directions of $$\beta_{\infty}$$ and $$\beta_{-\infty}.$$ A vector having either of these directions is unaffected in direction by multiplication by $$\Phi.$$ In this case, therefore, $$\Phi$$ is a tonic. If $$n < -1$$ we may obtain the same result by considering the vectors except that a vector in the limiting directions will be reversed in direction by multiplication by $$\Phi,$$ which implies that the two corresponding coefficients of the tonic are negative.

If $$1 > n > -1,$$ we may set Then  Let us now determine $$\gamma$$ by the equation  This gives  Now $$\alpha '$$ is one of the reciprocals of $$\alpha, \beta,$$ and $$\gamma.$$ Let $$\beta '$$ and $$\gamma '$$ be the others. If we set we have