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查看完整版本 : [MATLAB数学相关] Matlab数值分析里插值方法的问题

Zz小猪
2009-04-22, 16:06
Matlab数值分析里插值方法的问题
比较插值法的运算耗时为什么用TIC,TOC算出来都是0.00的?
这是拉格朗日,牛顿和hermit插值算法编在一起的函数
function jisuan(n)
x=-1:0.25:1;
y=1./(1+9*x.^2);
x0=linspace(min(x),max(x),n);
tic
p=zeros(size(x0));
for i=1:length(x0)

p(i)=lagrint(x,y,x0(i));
end
toc
tic
p2=newtint(x,y,x0);
toc
tic
if nargin<1,n=4;
end
yp=(1-x).*exp(-x);
p3=hermint(x,y,yp,x0);
toc
为什么运行出来的时间都是0.000呢?
为什么运行出来的时间都是0.000呢?
>> jisuan(100)
Elapsed time is 0.000000 seconds.
Elapsed time is 0.015000 seconds.
Elapsed time is 0.000000 seconds.
>> jisuan(200)
Elapsed time is 0.000000 seconds.
Elapsed time is 0.000000 seconds.
Elapsed time is 0.000000 seconds.
>> jisuan(300)
Elapsed time is 0.000000 seconds.
Elapsed time is 0.000000 seconds.
Elapsed time is 0.000000 seconds.
>> jisuan(10)
Elapsed time is 0.000000 seconds.
Elapsed time is 0.000000 seconds.
Elapsed time is 0.000000 seconds.
>> jisuan(20)
Elapsed time is 0.000000 seconds.
Elapsed time is 0.000000 seconds.
Elapsed time is 0.015000 seconds.
>> jisuan(30)
Elapsed time is 0.000000 seconds.
Elapsed time is 0.000000 seconds.
Elapsed time is 0.016000 seconds.
>> jisuan(40)
Elapsed time is 0.000000 seconds.
Elapsed time is 0.000000 seconds.
Elapsed time is 0.000000 seconds.
>> jisuan(50)
Elapsed time is 0.000000 seconds.
Elapsed time is 0.000000 seconds.
Elapsed time is 0.000000 seconds.
>> jisuan(60)
Elapsed time is 0.000000 seconds.
Elapsed time is 0.000000 seconds.
Elapsed time is 0.000000 seconds.
>> jisuan(70)
Elapsed time is 0.000000 seconds.
Elapsed time is 0.000000 seconds.
Elapsed time is 0.000000 seconds.
>> jisuan(80)
Elapsed time is 0.000000 seconds.
Elapsed time is 0.000000 seconds.
Elapsed time is 0.000000 seconds.
>> jisuan(90)
Elapsed time is 0.000000 seconds.
Elapsed time is 0.000000 seconds.
Elapsed time is 0.000000 seconds.
>> jisuan(100)
Elapsed time is 0.000000 seconds.
Elapsed time is 0.000000 seconds.
Elapsed time is 0.015000 seconds.
同样是100运行出来还是不一样的 时间?!这是为什么啊
关于lagrange,Newton,Hermit的代码应该没问题的
这是lagrange的function yi=lagrint(x,y,xi)
%输入:X和Y是插值节点的横坐标和纵坐标构成的向量
%输出:y是拉格朗日插值多项式的系数构成的向量
dxi=xi-x;%xi-x(k)的向量值
n=length(x);%求出X的元素个数
L=zeros(size(y));%产生一个和向量y同阶的零矩阵
L(1)=prod(dxi(2:n))/prod(x(1)-x(2:n));%j=1 ,prod计算向量元素的乘积
L(n)=prod(dxi(1:n-1))/prod(x(n)-x(1:n-1)); %j=n
for j=2:n-1
num=prod(dxi(1:j-1))*prod(dxi(j+1:n));
den=prod(x(j)-x(1:j-1))*prod(x(j)-x(j+1:n));
L(j)=num/den;
end
yi=sum(y.*L); %估计插值:y(j)*L(j)的和 ,j=1,...,n
NEWTON
function[yhat,dy,cout]=newtint(x,y,xhat)%x,y定义要插值的表格数据向量,参数xhat是在该处要计算差值的标量或向量值
n=length(y);
if length(x)~=n,error('x and y are not compatible');
end
c=y(:);
for j=2:n
for i=n:-1:j
c(i)=(c(i)-c(i-1))/(x(i)-x(i-j+1));%计算系数c(i)即均差f[x1,x2,...xi]
end
end
yhat=c(n);
for i=n-1:-1:1
yhat=yhat.*(xhat-x(i))+c(i);%计算牛顿插值多项式的值,插值结果返回给yhat
end
if nargout>1
yn2=c(n-1);
for i=n-2:-1:1;
yn2=yn2.*(xhat-x(i))+c(i);
end
dy=yhat-yn2;%dy是yhat中n-1阶和n-2阶多项式插值的差
if nargout>2,cout =c;%c是牛顿多项式的系数
end
end


谁能帮帮我哦...
Zz小猪
2009-04-22, 16:08
分段三阶hermit
function yhat=hermint(x,f,fp,xhat)%分段三阶hermite插值
n=length(x);
if length(f)~=n,error('x and f are not compatible');
elseif length(fp)~=n,error('x and fp are not compatible');end
x=x(: );xhat=xhat(: );f=f(: );fp=fp(: );
dx=diff(x);
divdif=diff(f)./dx;
a=f(1:n-1);
b=fp(1:n-1);
c=(3*divdif-2*fp(1:n-1)-fp(2:n))./dx;
d=(fp(1:n-1)-2*divdif+fp(2:n))./dx.^2;
i=zeros(size(xhat));
for m=1:length(xhat)
i(m)=binSearch(x,xhat(m));
end
xx=xhat-x(i);
yhat=a(i)+xx.*(b(i)+xx.*(c(i)+xx.*d(i)));
Zz小猪
2009-04-22, 16:23
已经解决了,不用回答了...