输入
Xf =和包含点
Yf =一个包含点方法的y值的数组= 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，其中实际可以使用所选方法（例如，您不能在Xf数组的上限使用前向差分方法，因为之后没有价值）
DF =这些点的衍生物
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，而不使用内置的派生函数，如 diff。我想帮助编写其中一个，然后我想我应该能够弄清楚如何做其余的。
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
答案 0 :(得分：2)
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
...
答案 1 :(得分：0)
#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
如果这段代码让您感到害怕，那么此视频将详细介绍 eli5 正在发生的事情以及每个部分为何起作用： 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。
答案 2 :(得分：-1)
我确信之前一定得到了解答。不过这里有一些来自Matlab文档的东西：
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)