拉格朗日插值

这篇文章存在极其严重的伪证现象，请就情况往下翻。

在平面直角坐标系中，给出$n+1$个函数在不同的坐标的点，求其解析式

即设$n+1$个点坐标分别为$:(x_0,y_0),(x_1,y_1),......,(x_n,y_n)$

有$:\sum _{i=0}^n{y}_i\frac{\prod _{j=0}^n(x-x_j)(i\ne j)}{\prod _{j=0}^n(x_i-x_j)(i\ne j)}$

引理$:$

设$n+1$个点得出的解析式可表达为$L_n(x)$

$L_n(x)=a_0+a_1(x-x_0)\dots a_n(x-x_0)(x-x_1)\dots (x-x_{n-1})$

有$:a$有唯一解且一定有解,并且$L_n(x_j)=y_j$

可得以下方程$:$

$$\{\begin{array}{rlr}{y}_0& =& a_0\\ y_1& =& a_0& +a_1(x_1-x_0)\\ ⋮\\ y_n& =& a_0& +a_1(x_1-x_0)+\dots +a_n(x_n-x_0)(x_n-x_1)\dots (x_n-x_{n-1})\end{array}$$

我们首先证明一下一个定理$:$克拉默法则

即$:$

当

$$\{\begin{array}{rl}{a}_{11}{x}_1+a_{12}{x}_2+\dots +a_{1n}{x}_n& =b_1\\ a_{21}{x}_1+a_{22}{x}_2+\dots +a_{2n}{x}_n& =b_2\\ ⋮\\ a_{n1}{x}_1+a_{n2}{x}_2+\dots +a_{nn}{x}_n& =b_n\end{array}$$

令$:$

$$D=\left|\begin{array}{cccc}{a}_{11}& a_{12}& \dots & a_{1n}\\ a_{21}& a_{22}& \dots & a_{2n}\\ ⋮& ⋮& \ddots & ⋮\\ a_{n1}& a_{n2}& \dots & a_{nn}\end{array}\right|$$

称$D$为方程组的系数行列式

有$:x_1=\frac{D_1}{D},x_2=\frac{D_2}{D},\dots ,x_n=\frac{D_n}{D}$

其中$D_j(j=1,2,\dots ,n)$是把西施行列式$D$中的第$j$列元素用方程组右端的常数项代替所得到的$n$阶行列式，即

$$D_j=\left|\begin{array}{cccccccc}{a}_{11}& a_{12}& \dots & a_{1,j-1}& b_1& a_{1,j+1}& \dots & a_{1n}\\ a_{21}& a_{22}& \dots & a_{2,j-1}& b_2& a_{2,j+1}& \dots & a_{2n}\\ ⋮& ⋮& & ⋮& ⋮& ⋮& & ⋮\\ a_{n1}& a_{n2}& \dots & a_{n,j-1}& b_n& a_{n,j+1}& \dots & a_{nn}\end{array}\right|$$

$Proof:$ 我们分2步证明

$(1):$把方程写成

$$\sum _{j=1}^n{a}_{ij}{x}_j=b_i(i=1,2,\dots ,n)$$

把$x_1=\frac{D_1}{D},x_2=\frac{D_2}{D},\dots ,x_n=\frac{D_n}{D}$带入式子$:$

$$\sum _{j=1}^n{a}_{ij}\frac{D_j}{D}=\frac{1}{D}\sum _{j=1}^n{a}_{ij}{D}_j$$

因为$:D_j=b_1{A}_{1j}+b_2{A}_{2j}+\dots +b_n{A}_{nj}=\sum _{s=1}^n{b}_s{A}_{sj}($其中$A_{sj}$为元素$a_{sj}$的代数余子式$)$

且

$$\sum _{j=1}^n{a}_{ij}{A}_{sj}=\{\begin{array}{r}D,s=i\\ 0,s\ne i\end{array}$$

所以

$$\frac{1}{D}\sum _{j=1}^n{a}_{ij}{D}_j=\frac{1}{D}\sum _{j=1}^n{a}_{ij}\sum _{s=1}^n{b}_s{A}_{sj}=\frac{1}{D}\sum _{j=1}^n\sum _{s=1}^n{a}_{ij}{A}_{sj}{b}_s$$

$$=\frac{1}{D}\sum _{s=1}^n\sum _{j=1}^n{a}_{ij}{A}_{sj}{b}_s=\frac{1}{D}\sum _{s=1}^n(\sum _{j=1}^n{a}_{ij}{A}_{sj})b_s$$

$$=\frac{1}{D}D{b}_i=b_i$$

这相当于此式确为方程组的解

$(2):$用$D$中第$j$列的代数余子式$A_{1j},A_{2j},\dots ,A_{nj}$依次乘方程组的$n$个方程，再把它们相加，得$:$

$$(\sum _{k=1}^n{a}_{k1}{A}_{kj})x_1+\dots +(\sum _{k=1}^n{a}_{kj}{A}_{kj})x_j+(\sum _{k=1}^n{a}_{kn}{A}_{kj})x_n=\sum _{k=1}^n{b}_k{A}_{kj}$$

于是有

$$D{x}_j=D_j(j=1,2,3,\dots ,n)$$

当$D\ne 0$时，得解一定满足式子，综上所述，方程组有唯一解.

我们回到之前的式子$:$

其系数行列式

$$D=\left|\begin{array}{cccc}1& 0& \dots & 0\\ 1& (x_1-x_0)& \dots & 0\\ ⋮& ⋮& \ddots & ⋮\\ 1& (x_n-x_0)& \dots & (x_n-x_0)(x_n-x_1)\dots (x_n-x_{n-1})\end{array}\right|$$

$\because x_i\ne x_j(i\ne j)$

$\therefore D\ne 0$

$\therefore a$一定有解，且有唯一解。

引理得证

接下来我们可以用数学归纳法证明等式成立

当$n=2$时

即$:L_2(x)=a_0+a_1(x-x_0)$

有

$$\{\begin{array}{r}{y}_0=L_n(x_0)=a_0\\ y_1=L_n(x_1)=a_0& +a_1(x_1-x_0)\end{array}$$

$\therefore a_0=y_0,a_1=\frac{y_1-y_0}{x_1-x_0}$

即$:$

$$L_n(x)=y_0+\frac{y_1-y_0}{x_1-x_0}(x-x_0)$$
$$=y_0+y_1\frac{x-x_0}{x_1-x_0}-y_0\frac{x-x_0}{x_1-x_0}$$
$$=y_0(1-\frac{x-x_0}{x_1-x_0})+y_1\frac{x-x_0}{x_1-x_0}$$
$$=y_0\frac{x-x_1}{x_0-x_1}+y_1\frac{x-x_0}{x_1-x_0}$$

证毕

即，当$n=2$等式成立

设$n=p-1$等式成立

证$:n=p$，等式成立

$\because L_p(x)=a_0+a_1(x-x_0)\dots a_p(x-x_0)(x-x_1)\dots (x-x_{p-1})$

易得$:$

$$a_p=\frac{L_p(x_p)-L_p(x_p)}{(x_p-x_0)(x_p-x_1)\dots (x_p-x_{p-1})}$$

将本式代入原式

$$L_p(x)=L_{p-1}(x)+\frac{y_p-L_{p-1}(x_p)}{(x_p-x_0)(x_p-x_1)\dots (x_p-x_{p-1})}(x-x_0)(x-x_1)\dots (x-x_{p-1})$$

$$=\sum _{k=0}^{p-1}{y}_k\frac{(x-x_0)\dots (x-x_{k-1})(x-x_{k+1})\dots (x-x_{p-1})}{(x_k-x_0)\dots (x_k-x_{k-1})(x_k-x_{k+1})\dots (x_k-x_{p-1})}+[y_p-\sum _{k=0}^{p-1}{y}_k\frac{(x_p-x_0)\dots (x_p-x_{k-1})(x_p-x_{k+1})\dots (x_p-x_{p-1})}{(x_k-x_0)\dots (x_k-x_{k-1})(x_k-x_{k+1})\dots (x_k-x_{p-1})}]\frac{(x-x_0)\dots (x-x_{p-1})}{(x_p-x_0)\dots (x_p-x_{p-1})}$$

$$=\sum _{k=0}^{p-1}{y}_k\frac{(x-x_0)\dots (x-x_{k-1})(x-x_{k+1})\dots (x-x_{p-1})}{(x_k-x_0)\dots (x_k-x_{k-1})(x_k-x_{k+1})\dots (x_k-x_{p-1})}-\sum _{k=0}^{p-1}{y}_k\frac{(x-x_0)\dots (x-x_{k-1})(x-x_{k+1})\dots (x-x_{p-1})}{(x_k-x_0)\dots (x_k-x_{k-1})(x_k-x_{k+1})\dots (x_k-x_{p-1})}\cdot \frac{x-x_k}{x_p-x_k}+y_p\frac{(x-x_0)\dots (x-x_{p-1})}{(x_p-x_0)\dots (x_p-x_{p-1})}$$

$$=\sum _{k=0}^{p-1}{y}_k\frac{(x-x_0)\dots (x-x_{k-1})(x-x_{k+1})\dots (x-x_{p-1})}{(x_k-x_0)\dots (x_k-x_{k-1})(x_k-x_{k+1})\dots (x_k-x_{p-1})}\cdot (1-\frac{x-x_k}{x_p-x_k})+y_p\frac{(x-x_0)\dots (x-x_{p-1})}{(x_p-x_0)\dots (x_p-x_{p-1})}$$

$\because 1-\frac{x-x_k}{x_p-x_k}=\frac{x_p-x_k-x+x_k}{x_p-x_k}=\frac{x-x_p}{x_k-x_p}$

$$\therefore L_p(x)=\sum _{k=0}^{p-1}{y}_k\frac{(x-x_0)\dots (x-x_{k-1})(x-x_{k+1})\dots (x-x_p)}{(x_k-x_0)\dots (x_k-x_{k-1})(x_k-x_{k+1})\dots (x_k-x_p)}+y_p\frac{(x-x_0)\dots (x-x_{p-1})}{(x_p-x_0)\dots (x_p-x_{p-1})}$$

$$\therefore L_p(x)=\sum _{i=0}^n{y}_i\frac{\prod _{j=0}^n(x-x_j)(i\ne j)}{\prod _{j=0}^n(x_i-x_j)(i\ne j)}$$