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Jiawei Gong 2019 年 11 月 6 日
コメント済み: darova 2019 年 11 月 6 日
MATLAB suggests not to call gradient multiple times, but I did it for sin function. As shown in the figure, the first order derivative is fine, the second has one boundary point off, and the third two points off. I really appreciate if someone can quantatively explain it from the mathematical point of view.
h = .1; % time step
t = 0:h:10; % time array
y = sin(t); % sin function
plot(t,[y;dy;ddy;dddy]);

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### 採用された回答

darova 2019 年 11 月 6 日
It's because of calculation first and last point. gradient does it this way:
%% ...
BLack curve and black points is your original data. Red points are points where gradient/derivative is calculated. Pay attention that first and last points are not at the beginning and end
Second time you use gradient you use the same step (again). But for the first and last points step/2 should be used.
Interesting fact
If you have not regular step (h is not equal) gradient can't handle it. Better use diff
% generate some data
x = sort(10*rand(100,1));
y = sin(x);
dy2 = diff(y)./diff(x);
x2 = x(2:end)-diff(x)/2; % actual x for diff/derivative
plot(x,y) % original data
hold on
plot(x,dy1,'g') % derivative obtained with gradient
plot(x2,dy2,'.m') % derivative obtained with diff
plot(x,cos(x),'k') % exact derivative
hold off

#### 4 件のコメント

darova 2019 年 11 月 6 日
Second time using gradient when you want to calculate derivative at point M
M = 1/2*p1 + 1/2*p2;
But it's not correct because of step
Don't know if i explained it clearly
Jiawei Gong 2019 年 11 月 6 日
Thank you, I got your idea. It was caused by inconsisently using forward, central, and backward scheme. I did some derivation; it shows the issue was brought from an introduced term of truncation error.
Substituing Eqs.(1) and (2) into (3) yields
darova 2019 年 11 月 6 日
Don't know what you are talking about =)

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