I have weighted function

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

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"Running-Gag" solved the my previous question:

You can show that
Therefore to get for some given you must have
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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
\alpha=\frac{2}{n}
\lim_{n\rightarrow\inf}p=0.135

So what is function for \alpha if p and n is given ?:
\alpha=f(p,n)