问题描述
我有以下函数(Viviani 曲线):
I have the following function (Viviani's curve):
Phi = @(t)[ cos(t)^2, cos(t)*sin(t), sin(t) ]
只是检查它是否有效:
s = linspace(0,T,1000);
plot3(cos(s).^2, cos(s).*sin(s), sin(s));
如何推导函数Phi
(可能多次),它代表了点t
处的Viviani曲线,其中t
来自0
到 2*pi
?我是否定义了适用于这种衍生的 Phi
?我已经尝试过 diff
,但它没有保留我需要的 Phi
.
How to derivate the function Phi
(maybe multiple times), which represents Viviani's curve in a point t
where t
goes from 0
to 2*pi
? Did I defined Phi
suitable for such a derivative? I've tried diff
, but it did not keep the Phi
as I would need it.
如果二阶导数是 Phi_d2
,我需要获取它的值(例如在 t = 0
中).
If the second derivative would be Phi_d2
, I need to get it's value (for example in t = 0
).
我怎样才能做到这一点?
How can I achieve this?
推荐答案
您可以通过以下三种方式完成此任务.第一个使用 subs
,第二个使用 symfun
,第三个使用复阶微分:
Here are three ways you can accomplish this. The first uses subs
, the second uses a symfun
, and the third uses complex step differentiation:
% Using subs
syms t
Phi = [cos(t) cos(t).*sin(t) sin(t)];
Phi_d2 = diff(Phi,t)
double(subs(Phi_d2,t,0))
% Using symfun
syms t
Phi(t) = [cos(t) cos(t).*sin(t) sin(t)];
Phi_d2 = diff(Phi,t)
double(Phi_d2(0))
% Using complex step differentiation
Phi = @(t)[cos(t) cos(t).*sin(t) sin(t)];
h = 2^-28;
cdiff = @(f,x)imag(f(x(:)+1i*h))/h;
Phi_d2 = cdiff(Phi,0)
你可以在我的 GitHub 上找到一个用于执行一阶和二阶复阶微分的函数:cdiff
.请注意,复杂的阶跃微分不适用于高阶导数.当只有不可微函数或需要快速数值一阶导数时最好.
You can find a function for performing first- and second-order complex step differentiation on my GitHub: cdiff
. Note that complex step differentiation won't work well for higher order derivatives. It's best when one only has a non-differentiable function or needs fast numerical first derivatives.
这篇关于计算向量的导数的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!