# solving nonlinear equations with newton s method

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### Calculation method-c/c++ Newton iterative method for solving the approximate root of nonlinearequation

Spread f (x) around X0 expansions series f (x) = f (x0) +f ' (x0) (x-x0) +f "(x0)/2!* (x-x0) ^2+ ... Then take its linear part, as the nonlinear equation f (x) = 0Approximate equation, that is, the first two of Taylor's expansion, there aref (x) = F ' (x0) x-x0*f ' (x0) + f (x0) = 0F ' (x0) x = X0*f ' (x0)-F (x0)x = x0-f (x0)/F ' (x0)Get an iterative sequence of

### "Qing Orange A1094" "Newton Iterative Method" Newton iterative method to find the root of the equation

Problem description given three times function f (x) =ax3+bx2+cx+d 4 coefficients a,b,c,d, and a number z, use Newton iterative method to find the root of function f (x) =0 near Z, and give the number of iterations required.The principle of Newton's iterative method is as follows (reference):The XK is a guessing solution near the exact solution of the

### Implementing Newton Iteration Method for Solving Nonlinear Equations Using MATLAB

Http://hi.baidu.com/aillieo/blog/item/0800e2a10ac9a59647106493.html The known nonlinear equations are as follows:3 * x1-cos (X2 * X3)-1/2 = 0 X1 ^ 2-81*(X2 + 0.1) ^ 2 + sin (X3) + 1.06 = 0 Exp (-X1 * x2) + 20 * X3 + (10 * pi-3)/3 = 0 The accuracy of the solution must reach 0.00001. -------------------------------- First, create the function fun The storage equations program saves fun. m to the working path as follows: Function f = fun (X ); % Defines

### Real roots of nonlinearequation by dichotomy method

#include #include #include #include float f (float x){float A;a=x*x*x-2*x-5;return A;}/*Www.quzhuanpan.comReprint please be sure to inform*/int main (void){Double low,bottom,prec,mid;//left endpoint, right endpoint, precision, middle valueCLRSCR ();printf ("\nplease input data low=");scanf ("%lf", low);printf ("\nplease input data bottom=");scanf ("%lf", bottom);printf ("\nplease input data precision=");scanf ("%lf", prec);Mid= (Low+bottom)/2;if (f (Low) *f (bottom) {while (f (mid)!=0){Mid= (Low

### Basic algorithm of C language 25-Newton iterative method for approximate root of equation

Newton Iterative Method!/*============================================================Title: Newton Iterative method is used to solve the approximate solution of 3*x*x*x-2*x*x-16=0.============================================================*/#include #include #define E 1e-8Double HS (Double x){return (3*X*X*X-2*X*X-16

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### C Language Implementation Newton iterative method solution equation

, and the other is that the number of iterations required cannot be determined. In the former case, a fixed number of loops can be constructed to control the iterative process, and in the latter case, further analysis is needed to conclude the conditions that can be used to end the iterative process.Next, I present a typical case of an iterative algorithm----Newton-Raphson (Kristinn) methodNewton-Raphson (Kristinn)

### Basic algorithm of C language 25-Newton iterative method for approximate root of equation

Newton Iterative Method!/*============================================================Title: Newton Iterative method is used to solve the approximate solution of 3*x*x*x-2*x*x-16=0.============================================================*/#include #include #define E 1e-8Double HS (Double x){return (3*X*X*X-2*X*X-16

### NOIP2001 three-time equation solution [derivative + Newton iterative method]

formula) to get the maximum value of f (x).There must be an F (x) 0 point for the three intervals consisting of the maximum points, and the 0 points can be obtained by Newton iterative method.Newton's iterative method is the constant use of a point tangent to fit the curve, the derivative of that point is the tangent slopeAnd so on, we can get an iterative method

### C: Using Newton's Iterative method to find the root of the equation near 1.5:2x^3-4x^2+3x-6=0.

Newton's Iterative method was used to find the root of the equation near 1.5:2x^3-4x^2+3x-6=0.Solution: Newton's Iterative method is also called Newton tangent method. Setf =2x^3-4x^2+3x-6,F1 is the derivative of the equation, the

### Using Newton's Iterative method to find the root of the following equation near 1.5:2x^3-4x^2+3x-6=0

Using Newton's Iterative method to find the root of the following equation near 1.5:2x^3-4x^2+3x-6=0 As for the Newton iterative method, in the course of computational methods, the basic formula is: xn+1=xn-f (Xn)/F *(Xn) xn+1 is the n+1 iteration result,Xn is the nth iteration result,f * ( Xn) is the Guide function

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