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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

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

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.
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First, create the function fun
The storage equations program saves fun. m to the working path as follows:
Function f = fun (X );
% Defines

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

, 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)

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

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

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
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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