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dvanmali

I noticed that Bernstein Basis seems a lot like applying a Taylor Series expansion with an increase in basis. Were these derived from a similar notion? Is so, couldn't we use a Taylor Series this?

theyComeAndGo

Why did we choose Bernstein basis instead of Taylor series, Fourier series etc.?

keenan

For one thing, because Bernstein bases are fairly localized in space and hence have a natural relationship with the control points. For instance, the coefficients of the first and last bases will directly determine the endpoints of the curve. For the cubic basis in particular, the other two coefficients give direct control over tangents at endpoints. In short: because this basis is natural for manipulating curves.

keenan

@dvanmali Also, if you take the 3rd-order Taylor series of a given function, you can always express it in the Bernstein basis. This basis is, after all, a basis for all cubic polynomials, not a special class of them. I would think of the Taylor series as more of a "procedure" for approximating a given function in terms of its kth derivatives; this (truncated) series can then be written down in any basis. The typical way to write it, of course, is just in the monomial basis 1, x, x^2, x^3, ..., but this basis isn't very good from the perspective of locality.