Question

Newtons-Raphsons方法很容易在Mathematica中实现，但在Matlab中似乎有点困难。如果我可以将函数传递给函数以及如何将导数作为函数使用，我不会得到。

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]]]

似乎你既不能导出函数句柄也不能导出文件中定义的函数，但我可能错了。

Answer 1

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);

Answer 2

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。

您可能希望做的是differentiation of symbolic equations，您需要Symbolic Math Toolbox。以下是使用Newton-Raphson method找到根的示例：

>> 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

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

Answer 3

% 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) = [];

Matlab中的Newton Raphsons方法？

3 个答案: