I have weighted function

a_1 = a_0(1-\alpha) + b\alpha

"Running-Gag" solved the my previous question:

You can show that
Therefore to get for some given you must have

I have many source a and destination b values i.e. signal but I have to use only one \alpha.

If target is defined such that p=\frac{b-a_n}{b-a_0}
Then iterating I can find e.g that if

So what is function for \alpha if p and n is given ?: