[proofplan] We apply the Erdős-Turán discrepancy inequality to the finite real sequence $P(1),\dots,P(N)$ with truncation parameter $H$. Each frequency $h$ produces the polynomial phase $hP(n)$, whose leading coefficient is $h\alpha$. The assumed reduced rational approximation $a_h/q_h$ to $h\alpha$ lets us apply Weyl's inequality to the corresponding exponential sum. Substituting these Weyl bounds into the Erdős-Turán inequality gives the claimed estimate after absorbing constants depending only on $k$ and $\varepsilon$. [/proofplan] [step:Reduce discrepancy to polynomial exponential sums] Define the additive character $e:\mathbb R\to\mathbb C$ by $e(t)=\exp(2\pi i t)$. For each integer $n$ with $1\le n\le N$, set $x_n:=P(n)\in\mathbb R$. By [citetheorem:9093] applied to the [real numbers](/page/Real%20Numbers) $x_1,\dots,x_N$ and to the integer truncation parameter $H\ge 1$, we have \begin{align*} D_N(P(1),\dots,P(N))\le \frac{3}{H+1}+\frac{3}{N}\sum_{h=1}^{H}h^{-1}\left|\sum_{n=1}^{N}e(hP(n))\right|. \end{align*} The hypotheses of [citetheorem:9093] are exactly that $x_1,\dots,x_N$ are real numbers and $H\ge 1$ is an integer, which hold because $P:\mathbb Z\to\mathbb R$ and $H\in\mathbb N$ with $H\ge 1$. [guided] We first convert the discrepancy problem into a finite list of exponential sum estimates. Define the additive character $e:\mathbb R\to\mathbb C$ by $e(t)=\exp(2\pi i t)$. The values $P(1),\dots,P(N)$ are real numbers because all coefficients of $P$ are real. Since $H\in\mathbb N$ with $H\ge 1$, the hypotheses of [citetheorem:9093] are satisfied with $x_n=P(n)$ for $1\le n\le N$. Therefore \begin{align*} D_N(P(1),\dots,P(N))\le \frac{3}{H+1}+\frac{3}{N}\sum_{h=1}^{H}h^{-1}\left|\sum_{n=1}^{N}e(hP(n))\right|. \end{align*} This is the key reduction: after this point, proving equidistribution quantitatively is reduced to bounding each exponential sum $\sum_{n=1}^{N}e(hP(n))$. The weight $1/h$ and the term $3/(H+1)$ come directly from the Erdős-Turán inequality, so the remaining work is to insert a Weyl bound at each frequency $h$. [/guided] [/step] [step:Apply Weyl's inequality at each frequency] Fix an integer $h$ with $1\le h\le H$. Define the polynomial map $P_h:\mathbb Z\to\mathbb R$ by $P_h(n)=hP(n)$. Because $h\ne 0$ and $\alpha\ne 0$, the coefficient $h\alpha$ is nonzero. Hence $P_h$ has degree $k$ and leading coefficient $h\alpha$, with \begin{align*} P_h(n)=h\alpha n^k+h\alpha_{k-1}n^{k-1}+\cdots+h\alpha_0. \end{align*} By hypothesis, $\gcd(a_h,q_h)=1$, $q_h\ge 1$, and \begin{align*} \left|h\alpha-\frac{a_h}{q_h}\right|\le q_h^{-2}. \end{align*} Thus [citetheorem:9054] applied to the degree-$k$ polynomial $P_h$ gives, for a constant $W_{k,\varepsilon}>0$ depending only on $k$ and $\varepsilon$, the bound \begin{align*} \left|\sum_{n=1}^{N}e(hP(n))\right|\le W_{k,\varepsilon}N^{1+\varepsilon}\left(q_h^{-1}+N^{-1}+q_hN^{-k}\right)^{2^{1-k}}. \end{align*} The lower-order coefficients $h\alpha_{k-1},\dots,h\alpha_0$ are real, so they satisfy the remaining coefficient hypotheses of [citetheorem:9054] without imposing any additional arithmetic condition. [guided] Fix an integer $h$ with $1\le h\le H$. The frequency-$h$ exponential sum from the Erdős-Turán bound has phase $hP(n)$, so we introduce the polynomial map $P_h:\mathbb Z\to\mathbb R$ by $P_h(n)=hP(n)$. Expanding $P_h$ gives \begin{align*} P_h(n)=h\alpha n^k+h\alpha_{k-1}n^{k-1}+\cdots+h\alpha_0. \end{align*} Because $h\ne 0$ and $\alpha\ne 0$, the coefficient $h\alpha$ is nonzero, so $P_h$ has degree $k$ and leading coefficient $h\alpha$. The theorem [citetheorem:9054] requires a degree-$k$ polynomial with a reduced rational approximation to the leading coefficient with error at most the reciprocal square of the denominator. The hypotheses provide exactly this data for the present frequency: $a_h\in\mathbb Z$, $q_h\in\mathbb N$, $\gcd(a_h,q_h)=1$, and $\left|h\alpha-a_h/q_h\right|\le q_h^{-2}$. The lower-order coefficients $h\alpha_{k-1},\dots,h\alpha_0$ are real numbers, which is the only lower-order coefficient condition in [citetheorem:9054]. Therefore Weyl's inequality applies to $P_h$ and gives a constant $W_{k,\varepsilon}>0$, depending only on $k$ and $\varepsilon$, such that \begin{align*} \left|\sum_{n=1}^{N}e(hP(n))\right|=\left|\sum_{n=1}^{N}e(P_h(n))\right|\le W_{k,\varepsilon}N^{1+\varepsilon}\left(q_h^{-1}+N^{-1}+q_hN^{-k}\right)^{2^{1-k}}. \end{align*} This is the needed estimate for the single frequency $h$. [/guided] [/step] [step:Insert the Weyl bounds and collect constants] Substituting the preceding estimate into the Erdős-Turán bound gives \begin{align*} D_N(P(1),\dots,P(N))\le \frac{3}{H+1}+3W_{k,\varepsilon}N^\varepsilon\sum_{h=1}^{H}h^{-1}\left(q_h^{-1}+N^{-1}+q_hN^{-k}\right)^{2^{1-k}}. \end{align*} Define $C_{k,\varepsilon}:=3W_{k,\varepsilon}$. Then $C_{k,\varepsilon}>0$ depends only on $k$ and $\varepsilon$, and the displayed estimate becomes \begin{align*} D_N(P(1),\dots,P(N))\le \frac{3}{H+1}+C_{k,\varepsilon}N^{\varepsilon}\sum_{h=1}^{H}h^{-1}\left(q_h^{-1}+N^{-1}+q_hN^{-k}\right)^{2^{1-k}}. \end{align*} This is the claimed quantitative discrepancy bound. [guided] We now substitute the frequency-wise Weyl estimates into the discrepancy inequality. From the Erdős-Turán step, \begin{align*} D_N(P(1),\dots,P(N))\le \frac{3}{H+1}+\frac{3}{N}\sum_{h=1}^{H}h^{-1}\left|\sum_{n=1}^{N}e(hP(n))\right|. \end{align*} For each integer $h$ with $1\le h\le H$, the Weyl step gives \begin{align*} \left|\sum_{n=1}^{N}e(hP(n))\right|\le W_{k,\varepsilon}N^{1+\varepsilon}\left(q_h^{-1}+N^{-1}+q_hN^{-k}\right)^{2^{1-k}}. \end{align*} Substituting this bound term by term yields \begin{align*} D_N(P(1),\dots,P(N))\le \frac{3}{H+1}+3W_{k,\varepsilon}N^\varepsilon\sum_{h=1}^{H}h^{-1}\left(q_h^{-1}+N^{-1}+q_hN^{-k}\right)^{2^{1-k}}. \end{align*} The factor $N^{1+\varepsilon}$ from Weyl's inequality becomes $N^\varepsilon$ because the Erdős-Turán inequality contributes the prefactor $3/N$. Define $C_{k,\varepsilon}:=3W_{k,\varepsilon}$. Then $C_{k,\varepsilon}>0$ depends only on $k$ and $\varepsilon$, and the preceding estimate becomes \begin{align*} D_N(P(1),\dots,P(N))\le \frac{3}{H+1}+C_{k,\varepsilon}N^{\varepsilon}\sum_{h=1}^{H}h^{-1}\left(q_h^{-1}+N^{-1}+q_hN^{-k}\right)^{2^{1-k}}. \end{align*} This is exactly the asserted quantitative discrepancy estimate. [/guided] [/step]