U .+k .-(L .+k )
so n +1
 A.

 U.-L. k.u-k .
 S2 + 3 +1
 A. A.
 1 2
 k .-k .
 =n f+ 3 3


but since k.u, k and A. are all constants the difference between

number of functions and the number of variables is always the same and

hence there is no acceleration of variables.

 There can be no acceleration of variables for a variable index

defined by a list since the difference between the number of functions

(or function index values) and variables (or variable index values) is

always the number of list elements minus one.