prove that for every $0 \le \alpha(n), \beta(n) \le1 \; s.t.$ there exists $c \in \Bbb{N} \;s.t \;\alpha(n)+\beta(n) \le 1- \frac{1}{n^c}$ then $BPP(\alpha(n), \beta(n)) = BPP$.I tried to show that by running k times we can get the error to be exponentially small using the chernoff inequality, for $0\le\delta\le1$$$p(\sum_{i=1}^{n} X_i \gt (1+\delta)pn) \le e^{-\delta^2pn} $$ If we run the algorithm k times, and accept only if we accepted at least $\frac{a(n)+b(n)}{2} k$ times.then we should set $\delta = \frac{\alpha(n) + \beta(n)}{2\alpha(n)} -1$But then I get stuck because it doesnt necessarily hold that $0 \le\delta \le 1$
prove that $BPP(\alpha(n), \beta(n)) = BPP$
complexity theory;complexity classes;randomized algorithms