代码搜索:lagrange

找到约 623 项符合「lagrange」的源代码

代码结果 623
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lagrange

Lagrange插值2007-07-16 18:36#include #include using namespace std; #define N 5 //以五点为例 void main() { int i,j,k; float l[N],t; float result=0,w1,w2; float x[N],y[N];
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lagrange

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m lagrange.m

function y=lagrange(x0,y0,x) n=length(x0); m=length(x); for i=1:m z=x(i); s=0.0; for k=1:n p=1.0; for j=1:n if j~=k p=p*(z-x0(j))/(x0(k)-x0(j)); end
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m lagrange.m

function y=lagrange(x0,y0,x); n=length(x0);m=length(x); for i=1:m z=x(i); s=0.0; for k=1:n p=1.0; for j=1:n if j~=k p=p*(z-x0(j))/(x0(k)-x0(j)); end end s=p*y0(k)+s; end y(i)=s; end
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m lagrange.m

function y=lagrange(x0,y0,x) % YINSERT=LAGRANGE(XDATA,YDATA,XINSERT) % 对给定的n个插值点x1,x2,...,xn 及对应的函数值y1,y2,…,yn, % 利用n次LAGRANGE插值多项式公式,则对插值区间内任意x的函数值y可通过下式求得: % EXAMPLE : % >>x0=[.4 .5 .8 .9 1.
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c lagrange.c

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c lagrange.c

#include #include #include double LAG(int,double *,double *,double); void main() { int n; double *x,*y,t,lag; t = 0.15; n = 6; x = (double*)calloc(n,
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c lagrange.c

#include #include #include double LAG(int,double *,double *,double); void main() { int n; double *x,*y,t,lag; t = 0.15; n = 6; x = (double*)calloc(n,
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cpp lagrange.cpp

#include #include void main() {
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cpp lagrange.cpp

//Lagrange's interpolation formula #include #include main() { float x[20],y[20],p1,p2,s,xv; int n,i,j; printf("How many sets of values do you want to give:"); scanf("