編碼找到x^2的限制C++

Question

找不到我的程序問題。我試圖計算邊緣內的限制，由於某種原因，我不認爲while循環正在初始化。我很新手。請幫忙。我不知道我做錯了什麼。

#include <iostream> 
#include <iomanip> 
#include <cmath> 

double f(double); 

using namespace std; 

void main() 
{ 
    //main program variable declarations 
    double x_0, h, prev, newapprox, diffapprox; double const tol = 0.000001; 

    cout << setiosflags(ios::showpoint | ios::fixed) << setprecision(7); 

    cout << "\n Enter the value at which the limit will be calculated: "; 
    cin >> x_0; 

    h = 1; 
    prev = f(x_0 + h); 

    cout << "\n x0+h" << setw(25) << "f(x0+h)" << endl; 
    cout << x_0 + h << setw(25) << prev << endl; 

    newapprox = f(x_0 + h/2); 
    cout << x_0 + h/2 << setw(25) << newapprox << endl; 

    diffapprox = fabs(prev - newapprox); 

    while (diffapprox > .00001); 
    { 
     prev = newapprox; 

     cout << "\n x0+h" << setw(25) << "f(x0+h)" << endl; 
     cout << x_0 + h << setw(25) << prev << endl; 

     newapprox = f(x_0 + h); 
     cout << x_0 + h << setw(25) << newapprox << endl; 

     diffapprox = fabs(prev - newapprox); 

     h /= 2; 
    } 

     if (fabs(diffapprox) <= 0.00001) 
     { 
      cout << "\n\n The limit exists and has the value" << diffapprox; 
     } 
     else 
     { 
      cout << "\n\n The limit does not exist at " << x_0; 
     } 


     system("pause"); 

    } 


double f(double x_0) 
{ 
    return pow(x_0, 2.0); 
} 

Answer 1

條件爲您while循環是diffapprox > .00001。從fabs(prev - newapprox)計算出diffapprox。 prev從f(x_0 + h)計算，並且newapprox從f(x_0 + h/2)計算，其中h = 1.0,f(x)是x^2，並且x_0從用戶輸入獲得。

由此，我們可以說prev是(x_0 + 1.0)^2 = x_0^2 + 2(x_0) + 1而newapprox是(x_0 + 1.0/2)^2 = (x_0 + 0.5)^2 = x_0^2 + x_0 + 0.25。這給出了(x_0^2 + 2(x_0) + 1) - (x_0^2 + x_0 + 0.25) = x_0 + 0.75的區別。

因此，對於while循環的初始條件開始，fabs(prev - newapprox) > .00001，失敗時fabs(x_0 + 0.75) <= .00001，或當x_0是內的0.75 +/- 0.00001。