Using C # To compile a bipartite solution to a non-linear equation

In the calculation method, there is a solution for finding a non-linear equation, called a bipartite method.

A brief introduction is as follows:

The f (x) function is continuous in the range [a, B]. Assume that there is a unique real root in the range, which is counted as x *

The idea of the Bipartite method is: first determine the root interval, divide the second-class interval, and gradually narrow down the root interval by judging the symbol of f (x) until the root interval is small enough, you can find the approximate root that meets the precision requirements.

Then we will give an example and a programming solution for C:

// Returns the root of the equation f (x) = x ^ 4-x-10.27 = 0 on (1, 2. Accurate to 10 ^ (-2)
Using System;

Class Dichotomy
{
Static   Void Main ()
{
Int K =   - 1 ; // Record loop index
Double A =   1 ; // Lower limit
Double B =   2 ; // Upper Limit
Double X =   0 ; // Root

Console. Write ( " K A \ t B \ t x \ t f (x) \ t a-B \ n " );

// First computing
X = ( + B) / 2 ;
K ++ ;

Console. Write ( " {0,-4} {1,-13} {2,-13} {3,-13} {4,-20} {5} \ n " , K, a, B, X, f (x), - B );

While ( True )
{

K ++ ;
Double F = F (x );

If (F <   0 )
{
A=X;
}
Else   If (F >   0 )
{
B=X;
}
X = ( + B) / 2 ;
If (B - A) <   0.01 )
{
Console. Write ("{0,-4} {1,-13} {2,-13} {3,-13} {4,-20} {5} \ n", K, a, B, X, F,-B );
Break;
}

Console. Write ( " {0,-4} {1,-13} {2,-13} {3,-13} {4,-20} {5} \ n " , K, a, B, X, F, - B );

}
Console. Readline ();
}

Private   Static   Double F ( Double X)
{
ReturnMath. Pow (X,4)-X- 10.27;
}  
}

Output result:

K A B x F (x) A-B
0 1 2 1.5-6.7075-1
1 1.5 2 1.75-6.7075-0.5
2 1.75 2 1.875-2.64109375-0.25
3 1.75 1.875 1.8125 0.214619140625-0.125
4 1.8125 1.875 1.84375-1.29024841308594-0.0625
5 1.84375 1.875 1.859375-0.557734031677246-0.03125
6 1.859375 1.875 1.8671875-0.176621873378753-0.015625
7 1.859375 1.8671875 1.86328125 0.0177218760550026-0.0078125

It can be seen that B7-A7 is about 0.0078 <10 ^ (-2), and thus | x *-X7 | <= 0.5 * (B7-A7) <= 0.5*10 ^ (-2), that is, to meet the accuracy requirements of the question, so the root of the requirements is: x * is about 1.863