Define \(f_0(x)=\sqrt{1-x^2}\).

Now, for \(n>0\) recursively define \(f_n(x)=\frac{1}{2}\left(f_{n-1}(x)+f_{n-1}\left(x-\frac{1}{2^{n-1}}\right)\right)\).

For \(n>0\), the domain of \(f_n(x)\) is \(-1+\displaystyle \sum_{k=0}^{n-1} \frac{1}{2^k}\leq x \leq 1\)

If \(\displaystyle \lim_{n\to \infty} f_n(1)=\frac{\pi^a}{b}\) for positive integers \(a,b\), find \(a+b\).

Helpful hints:

- \(f_2(x)=\dfrac{\dfrac{\sqrt{1-x^2}+\sqrt{1-(x-1)^2}}{2}+\dfrac{\sqrt{1-(x-.5)^2}+\sqrt{1-(x-1.5)^2}}{2}}{2}\)

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