I have weighted function

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

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

If target is defined such that $\displaystyle p=\frac{b-a_n}{b-a_0}$
Then iterating I can find e.g that if
$\displaystyle \alpha=\frac{2}{n}$
$\displaystyle \lim_{n\rightarrow\inf}p=0.135$

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