Derivative of a function in a particular point

1 次查看（过去 30 天）
george veropoulos
george veropoulos 2021-7-16
评论： george veropoulos 2021-7-19
Hi
I have an external function y= function fa(x)
y=sin(x./pi)
end
i want in the main program to find the derivative of fa in numerical point
thank you
George

请先登录，再回答此问题。

采纳的回答

george veropoulos
george veropoulos 2021-7-16
format long
x = 2 ;
h = sqrt(eps) ;
f1=(sin(x + h)./(x+h) - sin(x)./x) / h % -0.400000
% -0.435397773981094
f2=cos(x)./x-sin(x)./x^2
% -0.435397774979992
  2 个评论
Walter Roberson
Walter Roberson 2021-7-16
Ran in:
format long
x = 2 ;
h = sqrt(eps(x)) ;
f1=(sin(x + h)./(x+h) - sin(x)./x) / h
f1 =
-0.435397778831590
f2=cos(x)./x-sin(x)./x^2
f2 =
-0.435397774979992

请先登录，再进行评论。

更多回答（1 个）

Jan
Jan 2021-7-16
Use one of the quotients of differences to get a numerical approximation of the derivative:
x = 1.2345;
h = sqrt(eps);
dy_right = (fa(x + h) - fa(x)) / (h)
dy_left = (fa(x) - fa(x - h)) / (h)
dy_both = (fa(x + h) - fa(x - h)) / (2 * h)
function y = fa(x)
y = sin(x./pi)
end
  9 个评论
显示 7更早的评论
Jan
Jan 2021-7-18
@george veropoulos: The numerical analysis for the optimal choice of h is still worth to write a PhD thesis. As said before, the 2nd derivative of the function matters. To estimate this, you need a further small variation. Because this is expensive, if the function to be evaluated is huge, some heuristics are useful. This can be an important part of the processing time if you optimize an expensive function. A related field is the optimal choice of temproal and spatial steps sizes in finite element problems, e.g. the simulation of the earth clima.
The rule of thumb: If the function and the argument are about 1 and do not explode nearby, sqrt(eps) is a fair choice between the cancellation and discretization error. The "fairness" can be checkd by testing 10*sqrt(eps) and 0.1*sqrt(eps): If they reply the same derivative, you can assume to be on the right side. For a professional simulation, the explanation must sound more seriously. ;-)

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 MATLAB 的更多信息

标签

产品


版本

R2020b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by