$$\begin{bmatrix} a_0 \\ a_1 \\ a_2 \\ a_3 \\ a_4 \\ a_5 \\ a_6 \\ a_7 \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & \omega_8^1 & \omega_8^2 & \omega_8^3 & \omega_8^4 & \omega_8^8 & \omega_8^6 & \omega_8^7 \\ 1 & \omega_8^2 & \omega_8^4 & \omega_8^6 & \omega_8^8 & \omega_8^{10} & \omega_8^{12} & \omega_8^{14} \\ 1 & \omega_8^3 & \omega_8^6 & \omega_8^9 & \omega_8^{12} & \omega_8^{15} & \omega_8^{18} & \omega_8^{21} \\ 1 & \omega_8^4 & \omega_8^8 & \omega_8^12 & \omega_8^{16} & \omega_8^{20} & \omega_8^{24} & \omega_8^{28} \\ 1 & \omega_8^5 & \omega_8^{10} & \omega_8^{15} & \omega_8^{20} & \omega_8^{25} & \omega_8^{30} & \omega_8^{35} \\ 1 & \omega_8^6 & \omega_8^{12} & \omega_8^{18} & \omega_8^{24} & \omega_8^{30} & \omega_8^{36} & \omega_8^{42} \\ 1 & \omega_8^7 & \omega_8^{14} & \omega_8^{21} & \omega_8^{28} & \omega_8^{35} & \omega_8^{42} & \omega_8^{49} \end{bmatrix}^{-1} \begin{bmatrix} f(\omega _8^0) \\ f(\omega _8^1) \\ f(\omega _8^2) \\ f(\omega _8^3) \\ f(\omega _8^4) \\ f(\omega _8^5) \\ f(\omega _8^6) \\ f(\omega _8^7) \end{bmatrix} $$
看起来是不是像范德蒙德行列式。 我们只要能够得到范德蒙德矩阵的逆，一切就解决了。 事实是，普通的范德蒙德矩阵求逆很麻烦。 但，在复数领域一切都变得简单了呢。 矩阵求逆：$$\begin{bmatrix} 1 & 1 & \cdots & 1 \\ 1 & \omega _n^1 & \cdots &\omega _n^{n-1} \\ \vdots & \vdots & & \vdots \\ 1 & \omega _n^{n-1} & \cdots & \omega _n^{(n-1)(n-1)} \end{bmatrix}^{-1} = \frac 1n \begin{bmatrix} 1 & 1 & \cdots & 1 \\ 1 & \omega _n^{-1} & \cdots &\omega _n^{-(n-1)} \\ \vdots & \vdots & & \vdots \\ 1 & \omega _n^{-(n-1)} & \cdots & \omega _n^{-(n-1)(n-1)} \end{bmatrix} $$
所以最后得到公式：
$$a(x)=f(\omega _8^0)+f(\omega _8^1)x^{1}+f(\omega _8^2)x^{2}+f(\omega _8^3)x^{3}+f(\omega _8^4)x^{4}+f(\omega _8^5)x^{5}+f(\omega _8^6)x^{6}+f(\omega _8^7)x^{7}$$
怎么感觉和最上面的$f(x)=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5+a_6x^6+a_7x^7$有几分相似。 只是将点，相应的变成原来的共轭复数$\begin{align*} a(x)&=f(\omega _8^0)+f(\omega _8^1)x^{1}+f(\omega _8^2)x^{2}+f(\omega _8^3)x^{3}+f(\omega _8^4)x^{4}+f(\omega _8^5)x^{5}+f(\omega _8^6)x^{6}+f(\omega _8^7)x^{7} \\ &=f(\omega _8^0)+f(\omega _8^2)x^2+f(\omega _8^4)x^4+f(\omega _8^6)x^6+x(f(\omega _8^1)+f(\omega _8^3)x^4+f(\omega _8^5)x^4+f(\omega _8^7)x^6) \end{align*}$
$令A^{[0]}(x)=f(\omega _8^0)+f(\omega _8^2)x+f(\omega _8^4)x^2+f(\omega _8^6)x^3,A^{[1]}(x)=f(\omega _8^1)+f(\omega _8^3)x+f(\omega _8^5)x^2+f(\omega _8^7)x^3$
$则a(x)=A^{[0]}(x^2)+xA^{[1]}(x^2)$
$\begin{align*} & a(\omega _8^0) =A^{[0]}(\omega _8^0) +\omega _8^0 A^{[1]}(\omega _8^0) =A^{[0]}(\omega _4^0)+\omega _8^0A^{[1]}(\omega _4^0) \\ & a(\omega _8^{-1})=A^{[0]}(\omega _8^{-2})+\omega _8^{-1}A^{[1]}(\omega _8^{-2})=A^{[0]}(\omega _4^{-1})+\omega _8^{-1}A^{[1]}(\omega _4^{-1}) \\ & a(\omega _8^{-2})=A^{[0]}(\omega _8^{-4})+\omega _8^{-2}A^{[1]}(\omega _8^{-4})=A^{[0]}(\omega _4^{-2})+\omega _8^{-2}A^{[1]}(\omega _4^{-2}) \\ & a(\omega _8^{-3})=A^{[0]}(\omega _8^{-6})+\omega _8^{-3}A^{[1]}(\omega _8^{-6})=A^{[0]}(\omega _4^{-3})+\omega _8^{-3}A^{[1]}(\omega _4^{-3}) \\ & a(\omega _8^{-4})=A^{[0]}(\omega _8^{-8})+\omega _8^{-4}A^{[1]}(\omega _8^{-8})=A^{[0]}(\omega _4^0)-\omega _8^0A^{[1]}(\omega _4^0) \\ & a(\omega _8^{-5})=A^{[0]}(\omega _8^{-10})+\omega _8^{-5}A^{[1]}(\omega _8^{-10})=A^{[0]}(\omega _4^{-1})-\omega _8^{-1}A^{[1]}(\omega _4^{-1}) \\ & a(\omega _8^{-6})=A^{[0]}(\omega _8^{-12})+\omega _8^{-6}A^{[1]}(\omega _8^{-12})=A^{[0]}(\omega _4^{-2})-\omega _8^{-2}A^{[1]}(\omega _4^{-2}) \\ & a(\omega _8^{-7})=A^{[0]}(\omega _8^{-14})+\omega _8^{-7}A^{[1]}(\omega _8^{-14})=A^{[0]}(\omega _4^{-3})-\omega _8^{-3}A^{[1]}(\omega _4^{-3}) \end{align*}$
$\begin{align*} A^{[0]}(x)&=f(\omega _8^0)+f(\omega _8^2)x+f(\omega _8^4)x^2+f(\omega _8^6)x^3 \\ &=f(\omega _8^0)+f(\omega _8^4)x^2+x(f(\omega _8^2)+f(\omega _8^6)x^2) \end{align*}$
令$B^{[0]}(x)=f(\omega _8^0)+f(\omega _8^4)x,B^{[1]}(x)=f(\omega _8^2)+f(\omega _8^6)x$
则$A^{[0]}(x)=B^{[0]}(x^2)+xB^{[1]}(x^2)$