用递归方法求n阶勒让德多项式的值
/*
Date: 07/03/19 15:40
Description: 用递归法求n阶勒让德多项式的值
{ 1 n=0
Pn(x)= { x n=1
{ ((2n-1).x-Pn-1(x)-(n-1).Pn-2(x)/n n>=1
*/
Date: 07/03/19 15:40
Description: 用递归法求n阶勒让德多项式的值
{ 1 n=0
Pn(x)= { x n=1
{ ((2n-1).x-Pn-1(x)-(n-1).Pn-2(x)/n n>=1
*/
#include<stdio.h> float Legendre(int x,int n); int main(void) { int x,n; float value; printf("Enter the order of the polynomials:\n"); scanf("%d %d",&n,&x); printf("n=%d,x=%d\n\n",n,x); value=Legendre(x,n); printf("P%d(%d)=%6.3f\n",n,x,value); return 0; } float Legendre(int x,int n) { float value; if(n==0) value=1; else if(n==1) value=x; else value=((2*n-1)*x-Legendre(x,n-1)-(n-1)*Legendre(x,n-2))/n; return value; }