Bernstein gives unstable plots

Question

Sahithi Chaganti 2016-1-12

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/263515-bernstein-gives-unstable-plots

评论： Torsten 2016-1-15

Following is my code

syms t
b3 = bernstein(@(t) 1/(1 + 25 * t^2), 3, t);
b10 = bernstein(@(t) 1/(1 + 25 * t^2), 10, t);
b100 = bernstein(@(t) 1/(1 + 25 * t^2), 100, t);
ezplot(1/(1 + 25 * t^2),[0,1])
hold on
ezplot(b3,[0,1])
ezplot(b10,[0,1])
ezplot(b100,[0,1])
legend('sine function','3rd-degree polynomial',...
       '10th-degree polynomial','100th-degree polynomial')
title('Bernstein polynomials')
hold off
figure
ezplot(1/(1 + 25 * t^2),[-1,1])
hold on
ezplot(b3,[-1,1])
ezplot(b10,[-1,1])
ezplot(b100,[-1,1])
legend('sine function','3rd-degree polynomial',...
       '10th-degree polynomial','100th-degree polynomial')
title('Bernstein polynomials')
hold off

Why does Bernstein result in unstable values over the interval [-1,1]? From the documentation of Bernstein function, I understand that Bernstein polynomial approximates the function f over the interval [0,1]. What to do if I need the Bernstein polynomial to approximate a function over the interval [-1,1].

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Torsten 2016-1-15

在 MATLAB Online 中打开

B_[a:b](x)=B_[0:1]((x-a)/(b-a))

Bernstein polynomials on [a:b], evaluated at x, are just the Bernstein polynomials on [0:1], evaluated at t=(x-a)/(b-a).

Best wishes

Torsten.

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Answer 1

Walter Oevel 2016-1-15

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/263515-bernstein-gives-unstable-plots#answer_206280

Hi Saithi,

why do you expect the Bernstein representation to be numerically stable on [-1, 1]? It is designed to be stable on [0, 1] only. However, it is not invariant under the transformation t -> -t, although the original function is and the expanded/simplified version of the Bernstein polynomial is, too (note, however, that the latter is not a numerically stable representation).

The work-around may be the following: subtitute t^2 by y in the original function, do a Bernstein expansion w.r.t. y, then substitute back y by t^2. Of course, this is not the same as the Bernstein expansion of a given degree w.r.t. t, but it might do for your purposes.