Proposition 18.41.6.79BA8CB51C0E There is a constant $C_ r>0$, depending only on $r$, such that for $m>0$, $z\ne 0$ and $T=m|z|^{2r}\geq 1$,
69B0FF116418
18.41.7
\begin{equation} \left|\frac{\pi B_{m,r}^{(q)}(z)}{mr^2|z|^{2r-2}} -\frac1r\sum _{s=0}^{r-1}\mathbb P\{ N_ T\geq L_ s(q)\} \right| \leq C_ r T^{-1/2} \quad (q\geq 0). \end{equation}
In particular, the error is uniform in the vanishing order $q$.
Proof.
Let $\rho =|z|>0$ and put $\alpha _ s=(s+1)/r$ for integers $0\leq s<r$. Grouping the positive series (18.41.2) by the residues $s$ and using $T=m\rho ^{2r}$, we obtain the exact identity
\[ \begin{split} B_{m,r}^{(q)}(z) & =\frac r{\pi \rho ^2}\sum _{s=0}^{r-1} \mathrm{e}^{-T}\sum _{\ell \geq L_ s(q)} \frac{T^{\ell +\alpha _ s}}{\Gamma (\ell +\alpha _ s)}\\ & =\frac{rT}{\pi \rho ^2}\sum _{s=0}^{r-1} \mathbb E\bigl [w_{\alpha _ s,T}(N_ T) \mathbf1_{\{ N_ T\geq L_ s(q)\} }\bigr ]. \end{split} \]
Divide by $mr^2\rho ^{2r-2}/\pi =r^2T/(\pi \rho ^2)$ and apply lemma 18.41.5 to each of the finitely many residues. Since the indicators are bounded by one independently of their thresholds, the resulting error is uniform in $q$.
$\square$
Pome
Comments
Corrections, questions, and mathematical discussion are welcome. Sign in with GitHub to comment. Comments are public.