Algorithm question 2 (a solid root of the high equation 2x ^ 4-4x ^ 3 + 6x ^ 2-8x-8 = 0)

Obtain a solid root of the high equation 2x ^ 4-4x ^ 3 + 6x ^ 2-8x-8 = 0 (accuracy requirement: 10 ^-3)

Algorithm analysis is as follows: There are many real-root algorithms for High-Level equations. Here we introduce a kind of bipartite method.

If the higher-order equation f (x) is set to 0, a real root algorithm is obtained using the bipartite method as follows:

1. input the initial values X1 and X2 and calculate their corresponding function values Y1 = f (X1), y2 = f (X2 ). if the numbers Y1 and Y2 are the same, there is no real root to re-input between X1 and X2; otherwise there is a real root between X1 and X2, which is converted to step 2.

2. Calculate the intermediate position X = (X1 + x2)/2

3. Calculate the X-point function value Y1 = f (X1), Y = f (x ). If the function value is y = 0, the algorithm ends. X is a solid root of the higher order worker. Otherwise, it is converted into four steps.

4. If the function values Y and Y1 are the same, it indicates that the real root is between X1 and X2, then X1 = x is converted to 2nd. Otherwise, it indicates that the real root is between X1 and X, x2 = x, convert to 2nd

# Include <stdio. h> <br/> # include <math. h>/* The program uses the function FABS () */<br/> int main () <br/> {<br/> float x1, x2, X, y1, Y2, Y; <br/> DO <br/>{< br/> printf ("Enter do"); <br/> scanf ("% F, % F ", & X1, & x2 ); <br/> Y1 = 2 * X1 * X1 * X1 * x1-4 * X1 * X1 * X1 + 6 * X1 * x1-8 * x1-8; <br/> Y2 = 2 * X2 * X2 * X2 * x2-4 * X2 * X2 * X2 + 6 * X2 * x2-8 * x2-8; <br/> printf ("Y1 * y2 = % d Y1 * Y2> 0: % d", Y1 * Y2, (Y1 * Y2)> 0 ); <br/>}while (Y1 * Y2)> 0); <br/> while (1) /* This is an infinite loop under a constant condition */<br/> {<br/> X = (X1 + x2)/2; <br/> Y = 2 * x * X-4 * x + 6 * x * x-8 * x-8; <br/> Y1 = 2 * X1 * X1 * X1 * x1-4 * X1 * X1 * X1 + 6 * X1 * x1-8 * x1-8; <br/> If (FABS (y) <1e-5) break;/* find the root and force exit */<br/> else if (y * Y> 0) x1 = x; <br/> else X2 = x; <br/>}< br/> printf ("x = % F/N", X ); /* This equation has a solid root of 2.0 between 0 and 10 */<br/>}< br/>