How can i take ds/dx and ds/dy of a function numerically

Question

esat gulhan 2020-10-4

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https://ww2.mathworks.cn/matlabcentral/answers/604669-how-can-i-take-ds-dx-and-ds-dy-of-a-function-numerically

评论： Ameer Hamza 2020-10-4

采纳的回答： Ameer Hamza

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I tried to take derivatives of s=-x^3+3*x*y^2 numericaly

ds/dx=3*y^2-3*x^2

ds/dy=6*x*y

I tried to gradient of s to find ds/dx and ds/dy but results are far from analtical results. My code is below

clc;clear;
[x y]=meshgrid(0:5,0:5)
s=-x.^3+3.*x.*y.^2
[dsx,dsy]=gradient(s)

dsy should refers ds/dy

0 3 6 9 12 15

0 6 12 18 24 30

0 12 24 36 48 60

0 18 36 54 72 90

0 24 48 72 96 120

0 27 54 81 108 135

ds/dy =6xy analtically

0 0 0 0 0 0

0 6 12 18 24 30

0 12 24 36 48 60

0 18 36 54 72 90

0 24 48 72 96 120

0 30 60 90 120 150

How can i take derivatives numerically in gradient way. where is my fault.

NOTE: It is the simplified example to find the derivatieves numerically. My real project is very complicated and all values in matrices form. So I do not intrested in analtical solutions. It should be solved in numerical. Why gradient results are far away from analtical results how can i solve this numerically.

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esat gulhan 2020-10-4

Oh yes, the error is step size. thanks

Ameer Hamza 2020-10-4

I am glad that this solved the issue for you. I have added it as an answer too.

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Answer 1

Ameer Hamza 2020-10-4

0
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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/604669-how-can-i-take-ds-dx-and-ds-dy-of-a-function-numerically#answer_504937

在 MATLAB Online 中打开

The numerical gradient can not be exactly equal to the analytical gradient, except in very simple cases. For any nonlinear function, there will be a huge difference between both. There is no way around except using a very small step size, which will decrease the difference beween analytical and numerical gradient. For example:

If step size is 1

[x, y] = meshgrid(0:5,0:5);
s = -x.^3+3.*x.*y.^2;
[~, dsy] = gradient(s, 1, 1);
err = mean((dsy - 6.*x.*y).^2, 'all') % mean squared error

Result

>> err
err =
   27.5000

If step size is 0.1

[x, y] = meshgrid(0:0.1:5,0:0.1:5);
s = -x.^3+3.*x.*y.^2;
[~, dsy] = gradient(s, 0.1, 0.1);
err = mean((dsy - 6.*x.*y).^2, 'all') % mean squared error

Result