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Newtons-Raphsons 方法在 Mathematica 中很容易实现，但在 Matlab 中似乎有点困难。我不知道是否可以将函数传递给函数以及如何将导数用作函数。

newtonRaphson[f_, n_, guess_] := 
 If[n == 0, guess, newtonRaphson[f, n - 1, guess - f[guess]/f'[guess]]]
newtonRaphsonOptimize[f_, n_, guess_] := 
 If[n == 0, guess, 
  newtonRaphsonOptimize[f, n - 1, guess - f'[guess]/f''[guess]]]

似乎您既不能派生函数句柄也不能派生文件中定义的函数，但我可能错了。

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3 回答 3

6

您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

并将函数及其导数的函数句柄传递给它。这种导数可以通过一些不同的方法获得：手动微分、符号微分或自动微分。您也可以通过数值执行微分，但这既慢又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);
于 2011-04-12T19:07:42.440 回答
3

没有办法以代数方式获取 m 文件中定义的函数句柄或函数的导数。您必须通过在多个点上评估函数并逼近导数来以数值方式执行此操作。

你可能想要做的是符号方程的微分，你需要符号数学工具箱。这是使用Newton-Raphson 方法查找根的示例：

>> 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

您可以看到，xRoot经过几次迭代后， 的值就接近于真正根的值（即 2）。您还可以将函数评估放在 while 循环中，条件是检查每个新猜测和前一个猜测之间的差异有多大，当差异足够小时（即已找到根）时停止：

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
于 2011-04-12T18:03:14.840 回答
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) = [];
于 2013-06-08T07:11:09.487 回答