Several numeric calculation questions Objective C language implementation

Note: For personal habits, the c ++ statement is included.
1. Use the Newton method to find the solid root nearby and take four valid digits.

# Include <iostream>
# Include <cstdio>
# Include <cstdlib>
# Include <cmath>
# Include <algorithm>
 
Using namespace std;
 
Const double eps = 1e-4;
Const double x0 = 2;
 
Double FunOfSet (double x)
{
Return pow (x, 3.0)-3 * X-1;
}
 
Void ProGetRes ()
{
Double Begin = x0-1, End = x0 + 1;
Double Mid;
While (fabs (FunOfSet (Begin)-FunOfSet (End)> eps)
{
Mid = (Begin + End)/2;
If (FunOfSet (Begin) * FunOfSet (Mid) <0)
End = Mid;
Else Begin = Mid;
}
Printf ("The root of fuction beside 2 is: %. 4f \ n", Mid );
}
Int main ()
{
ProGetRes ();
Return 0;
}

Ii. Use Gaussian elimination to solve the following equations

# Include <iostream>
# Include <cstdio>
# Include <cstdlib>
# Include <cmath>
 
Using namespace std;
 
Const int LenOfMatrix = 3;
 
Double a [LenOfMatrix + 1] [LenOfMatrix + 2] = {0, 0}, {,-1,-}, {, 4, -}, {,-, 11 }};
Double Res [LenOfMatrix + 1];
Void GussPro ()
{
Int I, j, k, MaxPos, n = 3;
Double mmax;
For (k = 1; k <n-1; k ++)
{
Mmax =-1;
For (I = k; I <= n; I ++)
If (fabs (a [I] [k])> mmax)
{
Mmax = a [I] [k];
MaxPos = I;
}
Double temp;
For (j = 1; j <= n + 1; j ++)
{
Temp = a [k] [j];
A [k] [j] = a [MaxPos] [j];
A [MaxPos] [j] = temp;
}
For (I = k + 1; I <= n; I ++)
For (j = k + 1; j <= n + 1; j ++)
A [I] [j] = a [I] [j]-(a [I] [k]) * (a [k] [j]) /(a [k] [k]);
}
Res [n] = (a [n] [n + 1])/(a [n] [n]);
Double sum;
For (k = n-1; k> = 1; k --)
{
Sum = 0;
For (j = k + 1; j <= n; j ++)
Sum + = (a [k] [j]) * (Res [j]);
Res [k] = (a [k] [n + 1]-sum)/(a [k] [k]);
}
Cout <"For the follow matrix:" <endl;
Cout <"2x1-1x2-1x3 = 4" <endl;
Cout <"3x1 + 4x2-2x3 = 11" <endl;
Cout <"3x1-2x2 + 4x3 = 11" <endl;
Cout <"The answer is :";
For (I = 1; I <= n; I ++)
Printf ("x [% d] = %. 4f", I, Res [I]);
Cout <endl;
}
Int main ()
{
GussPro ();
Return 0;
}
3. known function table

 

X 1.1275	1.1503	1.1735	1.1972
F (x) 0.1191	0.13954	0.15932	0.17903

 

Calculate the approximate value of f (1.1300) using the Laplace interpolation formula.

# Include <iostream>
# Include <cstdio>
# Include <cstdlib>
# Include <cmath>
# Include <algorithm>
 
Using namespace std;
 
Const int LenOfNum = 4;
Double x [] = {1.1275, 1.1503, 1.1735, 1.1972 };
Double y [] = {0.1191, 0.13954, 0.15932, 0.17903 };
Double Li [LenOfNum];
Int I, j;
Void GetLi (double x0)
{
For (I = 0; I <LenOfNum; I ++)
{
Li [I] = 1;
For (j = 0; j <LenOfNum; j ++)
If (I! = J) Li [I] * = (x0-x [j])/(x [I]-x [j]);
}
}
 
Void ProLag ()
{
Double X 0 = 1.1300;
Double res = 0;
GetLi (x0 );
For (I = 0; I <LenOfNum; I ++)
Res + = (y [I] * Li [I]);
Printf ("f (1.1300) = % f \ n", res );
}
Int main ()
{
ProLag ();
}
# Include <iostream>
# Include <cstdio>
# Include <cstdlib>
# Include <cmath>
# Include <algorithm>

Using namespace std;

Const int LenOfNum = 4;
Double x [] = {1.1275, 1.1503, 1.1735, 1.1972 };
Double y [] = {0.1191, 0.13954, 0.15932, 0.17903 };
Double Li [LenOfNum];
Int I, j;
Void GetLi (double x0)
{
For (I = 0; I <LenOfNum; I ++)
{
Li [I] = 1;
For (j = 0; j <LenOfNum; j ++)
If (I! = J) Li [I] * = (x0-x [j])/(x [I]-x [j]);
}
}

Void ProLag ()
{
Double X 0 = 1.1300;
Double res = 0;
GetLi (x0 );
For (I = 0; I <LenOfNum; I ++)
Res + = (y [I] * Li [I]);
Printf ("f (1.1300) = % f \ n", res );
}
Int main ()
{
ProLag ();
}

From int64Ago's column