3 个答案:

答案 0 :(得分：6)

你可以使用这样的实现：

function x = newton(f,dfdx,x0,tolerance)

err = Inf;

x = x0;

while abs(err) > tolerance

xPrev = x;

x = xPrev - f(xPrev)./dfdx(xPrev);

% stop criterion: (f(x) - 0) < tolerance

err = f(x); %

% stop criterion: change of x < tolerance

% err = x - xPrev;

end

并传递函数及其派生的函数句柄。这种衍生物可以通过一些不同的方法获得：手工分化，象征性区分或自动区分。您也可以在数字上执行区分，但这很慢并且需要您使用修改后的实现。所以我假设你已经以任何合适的方式计算了导数。然后你可以调用代码：

f = @(x)((x-4).^2-4);

dfdx = @(x)(2.*(x-4));

x0 = 1;

xRoot = newton(@f,@dfdx,x0,1e-10);

答案 1 :(得分：3)

没有办法代数获取m文件中定义的函数句柄或函数的派生词。您必须通过在多个点评估函数并近似导数来数字。

>> syms x %# Create a symbolic variable x

>> f = (x-4)^2-4; %# Create a function of x to find a root of

>> xRoot = 1; %# Initial guess for the root

>> g = x-f/diff(f); %# Create a Newton-Raphson approximation function

>> xRoot = subs(g,'x',xRoot) %# Evaluate the function at the initial guess

xRoot =

1.8333

>> xRoot = subs(g,'x',xRoot) %# Evaluate the function at the refined guess

xRoot =

1.9936

>> xRoot = subs(g,'x',xRoot) %# Evaluate the function at the refined guess

xRoot =

2.0000

在几次迭代之后，您可以看到xRoot的值接近真根的值(即2)。您还可以将函数求值放在while循环中，其条件可以检查每个新猜测与之前猜测之间的差异有多大，当该差异足够小时停止(即已找到根)：

xRoot = 1; %# Initial guess

xNew = subs(g,'x',xRoot); %# Refined guess

while abs(xNew-xRoot) > 1e-10 %# Loop while they differ by more than 1e-10

xRoot = xNew; %# Update the old guess

xNew = subs(g,'x',xRoot); %# Update the new guess

end

xRoot = xNew; %# Update the final value for the root

答案 2 :(得分：0)

% Friday June 07 by Ehsan Behnam.

% b) Newton's method implemented in MATLAB.

% INPUT:1) "fx" is the equation string of the interest. The user

% may input any string but it should be constructable as a "sym" object.

% 2) x0 is the initial point.

% 3) intrvl is the interval of interest to find the roots.

% returns "rt" a vector containing all of the roots for eq = 0

% on the given interval and also the number of iterations to

% find these roots. This may be useful to find out the convergence rate

% and to compare with other methods (e.g. Bisection method).

%

function [rt iter_arr] = newton_raphson(fx, x, intrvl)

n_seeds = 10; %number of initial guesses!

x0 = linspace(intrvl(1), intrvl(2), n_seeds);

rt = zeros(1, n_seeds);

% An array that keeps the number of required iterations.

iter_arr = zeros(1, n_seeds);

n_rt = 0;

% Since sometimes we may not converge "max_iter" is set.

max_iter = 100;

% A threshold for distinguishing roots coming from different seeds.

thresh = 0.001;

for i = 1:length(x0)

iter = 0;

eq = sym(fx);

max_error = 10^(-12);

df = diff(eq);

err = Inf;

x_this = x0(i);

while (abs(err) > max_error)

iter = iter + 1;

x_prev = x_this;

% Iterative process for solving the equation.

x_this = x_prev - subs(fx, x, x_prev) / subs(df, x, x_prev);

err = subs(fx, x, x_this);

if (iter >= max_iter)

break;

end

end

if (abs(err) < max_error)

% Many guesses will result in the same root.

% So we check if the found root is new

isNew = true;

if (x_this >= intrvl(1) && x_this <= intrvl(2))

for j = 1:n_rt

if (abs(x_this - rt(j)) < thresh)

isNew = false;

break;

end

end

if (isNew)

n_rt = n_rt + 1;

rt(n_rt) = x_this;

iter_arr(n_rt) = iter;

end

end

end

end

rt(n_rt + 1:end) = [];

iter_arr(n_rt + 1:end) = [];