0
$\begingroup$

I am representing a function by nth order Bernstein polynomial coefficients. I have bounded the coefficients between some $f_{min}$ and $f_{max}$. From what I can see experimentally, it appears that bounding the coefficients has also bounded the polynomial output, but I don't have a proof of this.

Does bounding coefficients of a polynomial basis like Bernstein polynomials also bound the output, while also being able to represent arbitrary bounded polynomials of the nth degree (on the [0,1] interval where Bernstein polynomials are defined)?

If this is true, proofs or citations would be appreciated.

Context: I am trying to find a set of polynomial coefficients to generate a bounded function that minimizes some cost function based on experimental data. The function is generated using Bernstein polynomial coefficients which are used as the parameters of the minimization problem (Chebyshev or similar would also be fine). I have solved the optimization while also bounding the parameters. This seems to work and generates functions bounded the way I would like. However, I have not found proofs that this is an appropriate approach: that bounding the coefficients bounds the output while simultaneously being able to represent any bounded polynomial of the nth order on the interval where the function is defined. Citations or methods to be able to prove this would be very helpful. And if it isn't true, suggestions of alternate methods to generate bounded polynomials would be appreciated.

$\endgroup$
2
  • $\begingroup$ The Bernstein polynomials have the property that the sum of all Bernstein polynomials of level $n$ is $1$. $\endgroup$ Commented Oct 14, 2019 at 16:28
  • 1
    $\begingroup$ To expand on @Bullet51’s comment: combining this with the observation that the Bernstein polynomials are non-negative gives what you want. $\endgroup$ Commented Oct 14, 2019 at 21:08

0

You must log in to answer this question.

Browse other questions tagged .