Relations among different boolean approximations
The new parameters are polynomially related to the old parameters. In fact, for every $0&lt;\delta,\epsilon&lt;1$ it is the case that $d_{0,f,\epsilon} = \Theta_{\delta,\epsilon}(d_{0,f,\delta})$. This is proved in the same way that the usual relation regarding $d_{f,\epsilon}$ is proved, as detailed below.

Suppose we are given $0&lt;\delta,\epsilon&lt;1$. Let $g$ be some continuous function satisfying $g(0) = 0$ and $g(x) = 1$ for $1-\delta \leq x \leq 1+\delta$. The Weierstrass approximation theorem shows that there is some polynomial $G$ such that $|G(0)| \leq \epsilon/2$ and $|G(x)-1| \leq \epsilon/2$ for all $1-\delta \leq x \leq 1+\delta$. Let $G_0(x) = G(x) - G(0)$, so that $G_0(0) = 0$ and $|G_0(x)-1| \leq \epsilon$ for all $1-\delta \leq x \leq 1+\delta$. The polynomial $P = G_0(p_{0,\delta})$ has degree $(\deg G_0)d_{0,f,\delta}$ and satisfies $P(x) = 0$ whenever $f(x) = 0$, and $|P(x)-1| \leq \epsilon$ whenever $f(x) = 1$. This shows that $d_{0,f,\epsilon} \leq (\deg G_0) d_{0,f,\delta}$. Since $G_0$ only depends on $\delta,\epsilon$, this proves that $d_{0,f,\epsilon} = O_{\delta,\epsilon}(d_{0,f,\delta})$.