Perhaps you may do this,Originally Posted byDenMac21

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,

thus,

thus,

thus,

thus,

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.

Thus,

becomes,

Now this is a variation of the identity before show, with

Thus,