计算方法插值法实验报告(计算方法实验 一)

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拉格朗日插值数学原理

给定平面上 n + 1 n + 1 n+1个不同的数据点 ( x k , f ( x k ) ) (x_{k},f(x_{k})) (xk​,f(xk​))， k = 0 , 1 , ⋯   , n k = 0,1,cdots,n k=0,1,⋯,n， x i ≠ x j x_{i} neq x_{j} xi​​=xj​， i ≠ j i neq j i​=j；则满足条件

P n ( x k ) = f ( x k ) ,    k = 0 , 1 , ⋯   , n P_{n}left( x_{k} right) = fleft( x_{k} right), k = 0,1,cdots,n Pn​(xk​)=f(xk​),k=0,1,⋯,n

的 n n n次拉格朗日插值多项式

P n ( x ) = ∑ k = 0 n f ( x k ) l k ( x ) P_{n}(x) = sum_{k = 0}^{n}{f(x_{k})}l_{k}(x) Pn​(x)=k=0∑n​f(xk​)lk​(x)

是存在唯一的。若 x k ∈ [ a , b ] , k = 0 , 1 , ⋯   , n x_{k} in lbrack a,brbrack,k = 0,1,cdots,n xk​∈[a,b],k=0,1,⋯,n，且函数 f ( x ) f(x) f(x)充分光滑，则当 x ∈ [ a , b ] x in lbrack a,brbrack x∈[a,b]时，有误差估计式

f ( x ) − P n ( x ) = f ( n + 1 ) ( ξ ) ( n + 1 ) ! ( x − x 0 ) ( x − x 1 ) ⋯ ( x − x n ) ,    ξ ∈ [ a , b ] fleft( x right) - P_{n}left( x right) = frac{f^{left( n + 1 right)}left( xi right)}{left( n + 1 right)!}left( x - x_{0} right)left( x - x_{1} right)cdotsleft( x - x_{n} right), xi in lbrack a,brbrack f(x)−Pn​(x)=(n+1)!f(n+1)(ξ)​(x−x0​)(x−x1​)⋯(x−xn​),ξ∈[a,b]

程序设计

核心代码

double x, y =0.0;scanf("%lf",&x);double a[N +1], b[N +1];int n =0;while(scanf("%lf%lf",&a[n],&b[n])>=2) n++;    n--;for(int k =0; k <= n; k++){double l =1.0;for(int j =0; j <= n; j++){if(j != k) l *=(x - a[j])/(a[k]- a[j]);}        y += l * b[k];}printf("x = %.3lfny = %.3lf", x, y);

多个n多个x测试时怎么办呢？

作为真正的程序员！一定要会设计

#include<cmath>#include<cstdio>#define N1 3   // n amount#define N2 4   // x amount#define N3 20  // n maxint Ns[N1]={5,10,20};double x[N2]={-0.95,-0.05,0.05,0.95};double l =-1.0;double r =1.0;doubleX(int k,int n){double h =(r - l)/ n;return l + k * h;}doubleY(double x){return1/(1+ x * x);}intmain(){for(int i =0; i < N2; i++)printf("tx=%.2lf", x[i]);printf("n");for(int i =0; i < N1; i++){double a[N3 +1], b[N3 +1];int n = Ns[i];for(int k =0; k <= n; k++){            a[k]=X(k, n);// x            b[k]=Y(a[k]);// y}printf("n=%d", n);for(int p =0; p < N2; p++){double y =0.0;for(int k =0; k <= n; k++){double l =1.0;for(int j =0; j <= n; j++){if(j != k) l *=(x[p]- a[j])/(a[k]- a[j]);}                y += l * b[k];}printf("t%.6lf", y);}printf("n");}printf("Actual");for(int p =0; p < N2; p++)printf("t%.6lf",Y(x[p]));return0;}