经典案例
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幻方游戏
将正整数$1, 2, ⋯, 25$填入$5 × 5$阶幻方，使每行每列以及每条对角线上$5$个数之和相等。注意：循环数列之和必相等。
\[\begin{aligned} \lbrace 1, 2, ⋯, 25 \rbrace &= 5 · \lbrace 0, 1, 2, 3, 4 \rbrace + \lbrace 1, 2, 3, 4, 5 \rbrace \\ \begin{array}{|c|c|c|c|c|} \hline 15 & 20 & 0 & 5 & \fbox{10} \\ \hline 5 & 10 & 15 & \fbox{20} & 0 \\ \hline 20 & 0 & \fbox{5} & 10 & 15 \\ \hline 10 & \fbox{15} & 20 & 0 & 5 \\ \hline \fbox{0} & \fbox{5} & \fbox{10} & \fbox{15} & \fbox{20} \\ \hline \end{array} &= 5 · \begin{array}{|c|c|c|c|c|} \hline 3 & 4 & 0 & 1 & \fbox{2} \\ \hline 1 & 2 & 3 & \fbox{4} & 0 \\ \hline 4 & 0 & \fbox{1} & 2 & 3 \\ \hline 2 & \fbox{3} & 4 & 0 & 1 \\ \hline \fbox{0} & \fbox{1} & \fbox{2} & \fbox{3} & \fbox{4} \\ \hline \end{array} \\ \begin{array}{|c|c|c|c|c|} \hline 17 & 23 & 4 & 5 & \fbox{11} \\ \hline 9 & 10 & 16 & \fbox{22} & 3 \\ \hline 21 & 2 & \fbox{8} & 14 & 15 \\ \hline 13 & \fbox{19} & 20 & 1 & 7 \\ \hline \fbox{0} & \fbox{6} & \fbox{12} & \fbox{18} & \fbox{24} \\ \hline \end{array} &= 5 · \begin{array}{|c|c|c|c|c|} \hline 3 & 4 & 0 & 1 & \fbox{2} \\ \hline 1 & 2 & 3 & \fbox{4} & 0 \\ \hline 4 & 0 & \fbox{1} & 2 & 3 \\ \hline 2 & \fbox{3} & 4 & 0 & 1 \\ \hline \fbox{0} & \fbox{1} & \fbox{2} & \fbox{3} & \fbox{4} \\ \hline \end{array} + \begin{array}{|c|c|c|c|c|} \hline 2 & 3 & 4 & 0 & \fbox{1} \\ \hline 4 & 0 & 1 & \fbox{2} & 3 \\ \hline 1 & 2 & \fbox{3} & 4 & 0 \\ \hline 3 & \fbox{4} & 0 & 1 & 2 \\ \hline \fbox{0} & \fbox{1} & \fbox{2} & \fbox{3} & \fbox{4} \\ \hline \end{array} \\ \begin{array}{|c|c|c|c|c|} \hline 18 & 24 & 5 & 6 & \fbox{12} \\ \hline 10 & 11 & 17 & \fbox{23} & 4 \\ \hline 22 & 3 & \fbox{9} & 15 & 16 \\ \hline 14 & \fbox{20} & 21 & 2 & 8 \\ \hline \fbox{1} & \fbox{7} & \fbox{13} & \fbox{19} & \fbox{25} \\ \hline \end{array} &= 5 · \begin{array}{|c|c|c|c|c|} \hline 3 & 4 & 0 & 1 & \fbox{2} \\ \hline 1 & 2 & 3 & \fbox{4} & 0 \\ \hline 4 & 0 & \fbox{1} & 2 & 3 \\ \hline 2 & \fbox{3} & 4 & 0 & 1 \\ \hline \fbox{0} & \fbox{1} & \fbox{2} & \fbox{3} & \fbox{4} \\ \hline \end{array} + \begin{array}{|c|c|c|c|c|} \hline 2 & 3 & 4 & 0 & \fbox{1} \\ \hline 4 & 0 & 1 & \fbox{2} & 3 \\ \hline 1 & 2 & \fbox{3} & 4 & 0 \\ \hline 3 & \fbox{4} & 0 & 1 & 2 \\ \hline \fbox{0} & \fbox{1} & \fbox{2} & \fbox{3} & \fbox{4} \\ \hline \end{array} + \begin{array}{|c|c|c|c|c|} \hline 1 & 1 & 1 & 1 & \fbox{1} \\ \hline 1 & 1 & 1 & \fbox{1} & 1 \\ \hline 1 & 1 & \fbox{1} & 1 & 1 \\ \hline 1 & \fbox{1} & 1 & 1 & 1 \\ \hline \fbox{1} & \fbox{1} & \fbox{1} & \fbox{1} & \fbox{1} \\ \hline \end{array} \\ \end{aligned}\]剩余定理
取余代数方程$\left\lbrace\begin{aligned}
r_3 &= x \mathop{\%} 7
r_2 &= x \mathop{\%} 5
r_1 &= x \mathop{\%} 3
\end{aligned}\right.$的解集$\lbrace ∀x; x \rbrace$。注意：解集$\lbrace ∀x; x \rbrace$不是线性空间而是模余空间，其同态敛射不是线性空间而是模余空间。
迭代数列
迭代数列$f_{n+2} ≡ f_{n} + f_{n+1}$初始值的解集$\left\lbrace ∀f_0;∀f_1; \langle f_0, f_1 \rangle \right\rbrace$。注意：解集$\left\lbrace ∀f_0;∀f_1; \langle f_0, f_1 \rangle \right\rbrace$是线性空间。
\[\begin{aligned} \mathrm{Iso}\left\langle \langle f_0, f_1 \rangle \right\rangle &≡ \langle f_0, f_1, ⋯, f_n \rangle \\ \mathrm{Iso}\left\langle \langle f_0, f_1 \rangle + \langle g_0, g_1 \rangle \right\rangle &≡ \langle f_0 + g_0, f_1 + g_1, ⋯, f_n + g_n \rangle ≡ \langle f_0, f_1, ⋯, f_n \rangle + \langle g_0, g_1, ⋯, g_n \rangle \\ \mathrm{Iso}\left\langle k · \langle f_0, f_1 \rangle \right\rangle &≡ \langle k · f_0, k · f_1, ⋯, k · f_n \rangle ≡ k · \langle f_0, f_1, ⋯, f_n \rangle \\ 2 = \mathrm{Dime}\left\langle \left\lbrace ∀f_0;∀f_1; \langle f_0, f_1 \rangle \right\rbrace \right\rangle &= \mathrm{Dime}\langle \left\lbrace ∀f_0;∀f_1; \langle f_0, f_1, ⋯, f_n \rangle \right\rbrace \rangle \\ \hline \mathrm{Iso}\langle \bold{e}_1 \rangle = \mathrm{Iso}\left\langle \langle 1, q_1 \rangle \right\rangle &\mathop{=======}\limits_{q_1 = \frac{1 - \sqrt{5}}{2}}^{q^{n+2} = q^{n} + q^{n+1}} \langle 1, q_1, ⋯, q_1^n \rangle \\ \mathrm{Iso}\langle \bold{e}_2 \rangle = \mathrm{Iso}\left\langle \langle 1, q_2 \rangle \right\rangle &\mathop{=======}\limits_{q_2 = \frac{1 + \sqrt{5}}{2}}^{q^{n+2} = q^{n} + q^{n+1}} \langle 1, q_2, ⋯, q_2^n \rangle \\ \mathrm{Iso}\left\langle \bold{e}_1 · x_1 + \bold{e}_2 · x_2 \right\rangle &= \langle x_1 + x_2, q_1 · x_1 + q_2 · x_2, ⋯, q_1^n · x_1 + q_2^n · x_2 \rangle \\ \langle f_0, f_1 \rangle &= \bold{e}_1 · x_1 + \bold{e}_2 · x_2 = \langle 1, q_1 \rangle · x_1 + \langle 1, q_2 \rangle · x_2 = \left\langle 1, \frac{1 - \sqrt{5}}{2} \right\rangle · x_1 + \left\langle 1, \frac{1 + \sqrt{5}}{2} \right\rangle · x_2 \\ \left[\begin{matrix} x_2 \\ x_1 \\ \end{matrix}\right] &= \left[\begin{matrix} \frac{f_1 - f_0 · q_1}{q_2 - q_1} \\ \frac{f_0 · q_2 - f_1}{q_2 - q_1} \\ \end{matrix}\right] \\ \hline f_n &\mathop{===}\limits_{f_0 = 1}^{f_1 = 1} q_1^n · x_1 + q_2^n · x_2 \mathop{=====}\limits_{x_1 = \frac{q_2 - 1}{q_2 - q_1}}^{x_2 = \frac{1 - q_1}{q_2 - q_1}} \dfrac{q_1^n · (q_2 - 1) + q_2^n · (1 - q_1)}{q_2 - q_1} \mathop{=====}\limits_{q_2 - q_1 = \sqrt{5}}^{q_2 + q_1 = 1} \dfrac{q_2^{n+1} - q_1^{n+1}}{q_2 - q_1} = \dfrac{(1 + \sqrt{5})^n + (1 - \sqrt{5})^n}{2^n · \sqrt{5}} \\ \end{aligned}\]拟合方程
拟合$1$元$n$方多项式方程。采用线性代数方程组。当$m = n$时才有唯一解。
\[\begin{aligned} \left[\begin{matrix} y_m \\ ⋮ \\ y_1 \\ y_0 \\ \end{matrix}\right] &= \mathop{\left[\begin{matrix} x_m^0 & x_m^1 & ⋯ & x_m^n \\ ⋮ & ⋮ & ⋰ & ⋮ \\ x_1^0 & x_1^1 & ⋯ & x_1^n \\ x_0^0 & x_0^1 & ⋯ & x_0^n \\ \end{matrix}\right]}\limits_{(1+m)×(1+n)} \rlap{×}{+} \left[\begin{matrix} a_n \\ ⋮ \\ a_1 \\ a_0 \\ \end{matrix}\right] = \left[\begin{matrix} a_0 · x_m^0 + a_1 · x_m^1 + ⋯ + a_n · x_m^n \\ ⋮ \\ a_0 · x_1^0 + a_1 · x_1^1 + ⋯ + a_n · x_1^n \\ a_0 · x_0^0 + a_1 · x_0^1 + ⋯ + a_n · x_0^n \\ \end{matrix}\right] \\ f\langle x \rangle &≡ a_0 · x^0 + a_1 · x^1 + ⋯ + a_n · x^n \\ \end{aligned}\]拟合$1$元$n$方多项式方程。采用同态敛射泛函数。当$m = n$时为同构映射。
\[\begin{aligned} \mathrm{Homo}\langle f \rangle &≡ \left[\begin{matrix} f\langle x_m \rangle \\ ⋮ \\ f\langle x_1 \rangle \\ f\langle x_0 \rangle \\ \end{matrix}\right] \\ \mathrm{Homo}\langle f + g \rangle &= \left[\begin{matrix} f\langle x_m \rangle + g\langle x_m \rangle \\ ⋮ \\ f\langle x_1 \rangle + g\langle x_1 \rangle \\ f\langle x_0 \rangle + g\langle x_0 \rangle \\ \end{matrix}\right] = \left[\begin{matrix} f\langle x_m \rangle \\ ⋮ \\ f\langle x_1 \rangle \\ f\langle x_0 \rangle \\ \end{matrix}\right] + \left[\begin{matrix} g\langle x_m \rangle \\ ⋮ \\ g\langle x_1 \rangle \\ g\langle x_0 \rangle \\ \end{matrix}\right] \\ \mathrm{Homo}\langle k · f \rangle &= \left[\begin{matrix} k · f\langle x_m \rangle \\ ⋮ \\ k · f\langle x_1 \rangle \\ k · f\langle x_0 \rangle \\ \end{matrix}\right] = k · \left[\begin{matrix} f\langle x_m \rangle \\ ⋮ \\ f\langle x_1 \rangle \\ f\langle x_0 \rangle \\ \end{matrix}\right] \\ \hline \mathrm{Homo}\langle e_i \rangle &= \bold{η}_i \\ e_i\langle x \rangle &\mathop{≡≡≡≡≡≡}\limits^{e_i\langle x_j \rangle = \mathop{0}\limits_{i≠j}; \mathop{1}\limits_{i=j}} \prod\limits_{j=0;i≠j}^{n} \dfrac{x - x_j}{x_i - x_j} \\ \mathrm{Homo}\langle e_0 · y_0 + e_1 · y_1 + ⋯ + e_m · y_m \rangle &= \left[\begin{matrix} y_m \\ ⋮ \\ y_1 \\ y_0 \\ \end{matrix}\right] \\ f\langle x \rangle ≡ e_0 · y_0 + e_1 · y_1 + ⋯ + e_m · y_m &= \sum\limits_{i=0}^{m} y_i · \prod\limits_{j=0;i≠j}^{n} \dfrac{x - x_j}{x_i - x_j} \\ \end{aligned}\]