单位根反演

定理

\([n\mid a] = \dfrac1n\sum_{k=0}^{n-1}\omega_n^{ak}\)

证明

使用等比数列求和

a ≠ 0 (mod n)

公比不为 1

原式 = \(\dfrac1n \times\dfrac{\omega_n^{an}-1}{\omega_n^a-1} = \dfrac1n \times\dfrac{1-1}{\omega_n^a-1} = 0\)

a = 0 (mod n)

公比为 1

原式 = \(\dfrac1n\times n\times \omega_n^0 = 1\)

应用

\([a\equiv b\mod n] = [a-b\equiv 0\mod n] = [n\mid(a-b)] = \dfrac1n\sum_{k=0}^n\omega_n^{(a-b)k}\)

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LJJ学二项式定理

\[\begin{align} &\sum_{i=0}^n\binom ni\cdot s^i\cdot a_{i\% 4}\\ &= \sum_{i=0}^n\binom ni\cdot s^i\sum_{j=0}^3a_j[i=j\mod4]\\ &= \sum_{i=0}^n\binom ni\cdot s^i\sum_{j=0}^3a_j\frac14\sum_{k=0}^3\omega_4^{k(i-j)}\\ &= \frac14\sum_{j=0}^3a_j\sum_{i=0}^n\binom nis^i\sum_{k=0}^3\omega_4^{ki}\omega_4^{-kj}\\ &= \frac14\sum_{j=0}^3a_j\sum_{k=0}^3\omega_4^{-kj}\sum_{i=0}^n\binom nis^i\omega_4^{ki}\\ &= \frac14\sum_{j=0}^3a_j\sum_{k=0}^3\omega_4^{-kj}(1+s\omega_4^k)^n \end{align} \]

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小猪佩奇学数学

\(\left\lfloor\dfrac ik\right\rfloor = \dfrac {i - i\mod k}k\)

原式等于：

\[\frac1k\sum_{i=0}^n\binom ni p^i (i-i\mod k)\\ \frac1k\left[\sum_{i=0}^n\binom ni p^ii-\sum_{i=0}^n\binom ni p^i(i\mod k)\right] \]

分开考虑：

\[ \sum_{i=0}^n\binom nip^ii\\ np\sum_{i=0}^n\binom {n-1}{i-1}p^{i-1}\\ np\sum_{i=0}^{n-1}\binom {n-1}ip^i\\ np(p+1)^{n-1} \]

\[ \sum_{i=0}^n\binom nip^i(i\mod k)\\ \sum_{i=0}^n\binom nip^i\sum_{j=0}^{k-1}j[i\mod k =j]\\ \frac1k\sum_{i=0}^n\binom nip^i\sum_{j=0}^{k-1}j\sum_{t=0}^{k-1}\omega_k^{t(i-j)}\\ \frac1k\sum_{j=0}^{k-1}j\sum_{t=0}^{k-1}\omega_k^{-tj}\sum_{i=0}^n\binom nip^i\omega_k^{ti}\\ \frac1k\sum_{j=0}^{k-1}j\sum_{t=0}^{k-1}\omega_k^{-tj}(1+p\omega_k^t)^n\\ \frac1k\sum_{t=0}^{k-1}(1+p\omega_k^t)^n\sum_{j=0}^{k-1}j(\omega_k^{-t})^j \]

总复杂度是 O(k log n)

附：

\[\begin{align} S(n,k) &= \sum_{i=0}^{n-1}ik^i\\ kS(n,k)-S(n,k) &= \sum_{i=0}^{n-1}ik^{i+1} - \sum_{i=0}^{n-1}ik^i\\ &= \sum_{i=1}^n(i-1)k^i - \sum_{i=1}^{n-1}ik^i\\ &= -\sum_{i=1}^{n-1}k^i + (n-1)k^n\\ &= \frac{1-k^n}{k-1} + 1 + (n-1)k^n \end{align} \]

于是 \(S(n,k) = \dfrac{\dfrac{1-k^n}{k-1} + 1 + (n-1)k^n}{k-1}\)

由于 k 总是 n 次单位根， 于是 \(S(n,k) = \dfrac n{k-1}\)

若 k = 1， \(S(n,1) = \sum_{i=0}^{n-1} i = \dfrac{n(n-1)}2\)

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