单位根反演

公式

\[[n\mid k]=\frac 1n\sum_{i=0}^{n-1} \omega_n^{ik} \]

证明

\(n\mid k\) 时，\(\forall i\in \mathbb N\)，\(\omega_n^{ik}=1\)，因此 \(\frac 1n\sum_{i=0}^{n-1} \omega_n^{ik}=\frac 1n\cdot n=1\)

\(n\nmid k\) 时，\(\omega_n^{ik}\) 取遍 \(\omega_n^{i\gcd(n,k)}(0\le i<\frac n{\gcd(n,k)})\) 各 \(\gcd(n,k)\) 次，总和为 \(0\)

应用

\[\begin{aligned} &\sum_{i\ge 0}[x^{ni}]f(x)\\ =&\sum_{k\ge 0}[x^k]f(x)\cdot[n\mid k]\\ =&\sum_{k\ge 0}[x^k]f(x)\frac 1n\sum_{i=0}^{n-1} \omega_n^{ik}\\ =&\frac 1n\sum_{i=0}^{n-1}\sum_{k\ge 0}[x^k]f(x)(\omega_n^i)^k\\ =&\frac 1n\sum_{i=0}^{n-1}f(\omega_n^i)\\ \end{aligned} \]

\[\begin{aligned} &\sum_{i\ge 0}[x^{ni+d}]f(x)\\ =&\sum_{k\ge 0}[x^k]f(x)\cdot[n\mid (k-d)]\\ =&\sum_{k\ge 0}[x^k]f(x)\frac 1n\sum_{i=0}^{n-1} \omega_n^{i(k-d)}\\ =&\frac 1n\sum_{i=0}^{n-1}\sum_{k\ge 0}[x^k]f(x)(\omega_n^i)^{(k-d)}\\ =&\frac 1n\sum_{i=0}^{n-1}\sum_{k\ge 0}[x^k]f(x)(\omega_n^i)^k\omega_n^{-id}\\ =&\frac 1n\sum_{i=0}^{n-1}\omega_n^{-id}f(\omega_n^i)\\ \end{aligned} \]