if $m/N^{1/\eta}$ is bounded away from zero and infinity as $N{{\to}\infty}$. When $m$ has the latter property, (\[eq2.28\]) is satisfied, for all sufficiently large $N$, if $$\label{eq2.32} |\rho_{12}-\rho_0|=O \bigl(N^{-\{1+(1/\eta)+\delta_2\}}
\bigr)$$ for some $\delta_2>\delta_1$.

Note that it is at (\[eq2.30\]) that the type, $\eta$, of the irrational number $\rho_0$ enters into consideration. In the case $\eta=1$ the exponent $\delta$ in (\[eq2.30\]) could not be removed or reduced, perhaps by replacing the implicit factor $m^\delta$ in (\[eq2.30\]) by $(\log m)^C$ for some $C>0$, without an analogous strengthening of Roth’s Theorem. Formula (\[eq2.30\]) also marks the step at which it becomes apparent that a poorer bound will be obtained for an irrational number of type 1, relative to one of type $\eta>1$.

Applying the bound (\[eq2.31\]), for several versions of the polynomial $q$, in the case $N=2 \lfloor n^{\alpha}\rfloor+1$, we deduce that $$\label{eq2.33} n^{-r/2} \sup_{-\infty<z<\infty} \biggl|\sum
_{\nu\in{{\mathcal N}}_\ell} \nu_\ell^r \psi \biggl(z-
\frac{e_2 n_1}{ e_1 n_2} \nu \biggr) \biggr| =O \bigl(n^{r \{{\alpha}-(1/2)\}+{\alpha}\{1-(1/\eta)+\delta_1\}} \bigr) ,$$ provided that (\[eq2.32\]) holds, i.e., $|\rho_{12}-\rho_0|=O(n^{-{\alpha}\{
1+(1/\eta)+\delta_2\}})$. Now, the only constraint on ${\alpha}$ is $0<{\alpha}<{\frac{1}{2}}$, and so we can choose ${\alpha}$ as close to ${\frac{1}{2}}$, but less than ${\frac{1}{2}}$, as we desire. In particular, if $\delta_3>0$ is given, and we 