Perhaps you may do this,Originally Posted by DenMac21
Non of the could be more than 1, thus,
Do this in terms of a function of one, where is a changing variable with . Thus,
Now solve these two for it does not matter because we can consider them to be the same here. Thus, by substitution,
by Quadradic Formula, thus,
Thus, as a function of for
You can check that,
Know all is left is to find what,
as see whether it is invaraint of .
Here it gets messy, notice that
This is confirmed by simple expansion.
Now this is a variation of the identity before show, with