?? 構造正交多項式.cpp
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// 構造正交多項式.cpp : Defines the entry point for the console application.
//
#include "stdafx.h"
#include "iostream.h"
#include "math.h"
double f(double x) //求函數(shù)值
{
return x;
}
double sum(double a, double h, int n) //求所有新節(jié)點的函數(shù)和
{
int i;
double s=0;
for (i=0;i<n;i++)
s=s+f(a+i*h+h/2);
return s;
}
double D_abs(double i, double j) //求絕對值
{
if (i-j<0)
return j-i;
else
return i-j;
}
int main(int argc, char* argv[])
{
cout << " *******************************************" << endl;
cout << " ** **" << endl;
cout << " ** 構造正交多項式 **" << endl;
cout << " ** **" << endl;
cout << " *******************************************" << endl << endl;
double a=0.25,b=1.0;
double h=b-a;
double e=0.000001; //精度控制
double T1=h*(1+f(b))/2; //T1初始值
double T2,a0;
int n=1;
while (1) //迭代求解
{
T2=T1/2+h*sum(a, h, n)/2;
if (D_abs(T2, T1)<e)
{
break;
}
T1=T2;
n=n*2;
h=h/2;
}
a0=T2/0.75;
cout << "所求一階正交多項式為:" << endl;
cout << "Q1(x)=x-" << a0 << endl;
return 0;
}
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