如何在 Octave 中找到函数的导数？

Question

输入：

Xf = 和保存点 x 值的数组

Yf = 保存点的 y 值的数组方法 = 2 点前向差、2 点后向差、3 点中心差、5 点中心差

输出：

X = 包含有效 x 值的数组，其中选择的方法可以实际使用（例如，您不能在Xf数组的上限使用前向差分方法，因为它之后没有值）

DF = 这些点的导数

我需要为脚本提供一组点，然后使用 4 种不同的方法计算这些点的导数，而不使用像diff这样的内置导数函数。 我需要一些帮助来编写其中一个代码，然后我想我应该能够弄清楚如何完成其​​余的工作。

2分前差

我的尝试：

[a, minidx] = min(Xf);
[b, maxidx] = max(Xf);
n = 10;
h = (b-a)/n;
f = (x .^3) .* e.^(-x) .* cos(x);

If method = "forward" #Input by user

    X = [min(Xf), Xf(maxidx-1)];
    for k = min(Xf):n # not sure if this is the right iteration range...

        f(1) = f(x-2*h) + 8*f(x +h);
        f(2) = 8*f(x-h) + f(x+2*h);
        DF = (f1-f2)/(12*h);

    endfor
endif

Answer 1

https://wiki.octave.org/Symbolic_package

% this is just a formula to start with,  
% have fun and change it if you want to.  
f = @(x) x.^2 + 3*x - 1 + 5*x.*sin(x);  
% these next lines take the Anonymous function into a symbolic formula  
pkg load symbolic  
syms x;  
ff = f(x);  
% now calculate the derivative of the function  
ffd = diff(ff, x)  
% answer is ffd = (sym) 5*x*cos(x) + 2*x + 5*sin(x) + 3  
...  

Answer 2

下面是一些关于 Matlab 如何计算导数的文档：

diff Difference and approximate derivative.
     diff(X), for a vector X, is [X(2)-X(1)  X(3)-X(2) ... X(n)-X(n-1)].
     diff(X), for a matrix X, is the matrix of row differences,
           [X(2:n,:) - X(1:n-1,:)].
     diff(X), for an N-D array X, is the difference along the first
          non-singleton dimension of X.
     diff(X,N) is the N-th order difference along the first non-singleton 
          dimension (denote it by DIM). If N >= size(X,DIM), diff takes 
          successive differences along the next non-singleton dimension.
     diff(X,N,DIM) is the Nth difference function along dimension DIM. 
         If N >= size(X,DIM), diff returns an empty array.

Examples:
   h = .001; x = 0:h:pi;
   diff(sin(x.^2))/h is an approximation to 2*cos(x.^2).*x
   diff((1:10).^2) is 3:2:19

   If X = [3 7 5
           0 9 2]
   then diff(X,1,1) is [-3 2 -3], diff(X,1,2) is [4 -2
                                                  9 -7],
   diff(X,2,2) is the 2nd order difference along the dimension 2, and
   diff(X,3,2) is the empty matrix.

这是另一个例子：

xp= diff(xf);
yp= diff(yf);

% derivative:
dydx=yp./xp;
% also try:
dydx1=gradient(yf)./gradient(xf)

Answer 3

演示八度函数来计算导数：

#This octave column vector is the result y axis/results of your given function  
#to which you want a derivative of.  Replace this vector with the results of  
#your function. 
observations = [2;8;3;4;5;9;10;5] 
 
#dy (aka the change in y) is the vertical distance (amount of change) between  
#each point and its prior.  The minus sign serves our purposes to "show me the  
#vertical different per unit x."  The NaN in square brackets does a 1 position 
#shift, since I want my tangent lines to be delimited by 1 step each. 
dy = observations - [NaN; observations(1:end-1,:)] 
 
#dx (aka the change in x) is the amount of horizontal distance (amount of  
#change) between each point and its prior.  for simplicity we make these all 1,  
#however your data points might not be constant width apart, that's okay  
#to populate the x vector here manually using the parameter-inputs to the  
#function that you want the derivative of. 
dx = ones(length(observations), 1) 
 
# -- alternative: if your x vector is subjective, then do something like: -- 
#function_parameters = [1;3;5;9;13;90;100;505] 
#dx = function_parameters 
 
#The derivative of your original function up top is "the slope of the  
#tangent line to the point on your curve.  The slope is calculated with the  
#equation: (rise / run) such that 'Rise' is y, and 'run' is x.  The tangent  
#line is calculated above with the dy variable.  'The point' is each point  
#in the observations, and 'the curve' is simply your function up top that  
#yielded those y results and x input parameters. 
dy_dx_dv1_macd = dy ./ dx 

如果这段代码让你感到害怕，那么这个视频会详细介绍正在发生的事情以及为什么每件作品都起作用： https ://youtu.be/gtejJ3RCddE?t=5393

曲线上一点的切线的斜率就是导数的定义。 您可以使用八度（或任何计算机语言）为给定的函数计算它。 作为练习，绘制这些图，您应该看到 x 平方的导数是 x，而 sin(x) 的导数是 cos(x)。 当您看到它时，它很漂亮，因为这是大多数机器学习算法的核心。 导数会根据您所在的位置告诉您在哪里可以找到奖励。