### M Theory Lesson 246

The inverses of $C^{n}$ (that is, powers of the circulant $C = (0,1,1)$) also behave very nicely. A little algebra shows that for $C^{n} = (x,y,y)$, $C^{-n} = (a,b,b)$, where

$a = \frac{x+y}{x^2 - 2y^2 + xy}$

$b = \frac{1 - ax}{2y}$

In the special cases of interest, $(x,y) = (n,n+1)$ or $(n+1,n)$, we find (respectively) that

$C^{-n} = (\frac{-(2n+1)}{3n+2}, \frac{n+1}{3n+2}, \frac{n+1}{3n+2})$ or

$C^{-n} = (\frac{2n+1}{3n+1}, \frac{-n}{3n+1}, \frac{-n}{3n+1})$

and $|a| + |b| = 1$ in all cases. For example, the inverse of $(2,3,3)$ is the circulant $(1/8) (-5,3,3)$. These forms for the inverse of a positive circulant hold even when $n$ is not an ordinal. For the more general case of a positive circulant of the form $(n,n+d,n+d)$, the sum $|a| + |b| = 1/d$.

$a = \frac{x+y}{x^2 - 2y^2 + xy}$

$b = \frac{1 - ax}{2y}$

In the special cases of interest, $(x,y) = (n,n+1)$ or $(n+1,n)$, we find (respectively) that

$C^{-n} = (\frac{-(2n+1)}{3n+2}, \frac{n+1}{3n+2}, \frac{n+1}{3n+2})$ or

$C^{-n} = (\frac{2n+1}{3n+1}, \frac{-n}{3n+1}, \frac{-n}{3n+1})$

and $|a| + |b| = 1$ in all cases. For example, the inverse of $(2,3,3)$ is the circulant $(1/8) (-5,3,3)$. These forms for the inverse of a positive circulant hold even when $n$ is not an ordinal. For the more general case of a positive circulant of the form $(n,n+d,n+d)$, the sum $|a| + |b| = 1/d$.

## 0 Comments:

Post a Comment

<< Home