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I'm trying to understand whether the following is true:

I have the polynomial $P(x)$ which is non-negative when $-1<x<1$

Is it correct that $P(x)$ could be represented by a sum with NON-negative coefficients of the non-negative set of polynomials (non necessary linearly independent) ? For example: $T_0(x), \{1+T_i(x), 1-T_i(x)\ \forall i \in \mathbb{N}\}$ where $T_i(n)$ are Chebyshev polynomials.

The reason for the question is that I'm trying to approximate an non-negative function by a sum of non-negative basis functions. (Obviously the representation won't be unique, but I don't care).

(Added) The original question got answered by leshik, but the problem still remains, I'm trying to approximate a non-negative function in $-1<x<1$ (which is in fact a density distribution of some objects) by a positive positive linear combinations of some positive functions. I was naively thinking that something like polynomials would work, but apparently not. I know I could do that with say Gaussians ( with small enough width and large enough number of them I can approximate everything), but I would want some family where I can reproduce uniform or almost-uniform distribution with a small number of components.

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  • $\begingroup$ Well, if you are ok to pay the price in some positivity, which of course depends on your applications, you may consider Bernstein polynomial associate with your function, namely $\sum_{k=0}^n {n\choose k}f(k/n)x^k(1-x)^{n-k}.$ These polynomials approximate your function and for $x\ge 0$ provide what you requested. For $x<0$ you have representation with alternating coefficients instead. $\endgroup$ – leshik Jun 27 '13 at 11:59
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As stated, the question does not have a meaningful answer. For each "basic" set of functions, we can find a non-negative polynomial which cannot be represented as a positive linear combination of the basic functions. The idea is that one may always consider something of the form $P=P_{0}(x)-\sum_{i}\varepsilon_iP_i(x)$ and take $P_{0}(x)$ to have zero of the smallest order at $x=-1$ and/or $x=1.$

As to the useful representation, each nonnegative on $[-1,1]$ polynomial is either of the form $P^2(x)+(1-x^2)Q^2(x)$ or $(1-x)P^2(x)+(1+x)Q^2(x)$ depending on the parity of the degree.

Some interesting classes of nonnegative polynomials that can be represented as a positive linear combination of the $(1-x)^k(1+x)^m$ functions include so-called Lorentz classes. You may find Borwein and Erdelyi's book on polynomial inequalities to be a useful reference.

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  • $\begingroup$ Thank you for a detailed answer, But I still fail to see why "for each basic set of functions we can find a non-negative polynomial which cannot be represented...". Maybe that's obvious, but I cannot see it, e.g. for say deg<=2 case and chebyshev-based "basis" ${1, x+1, 1-x, x^2, 1-x^2}$. $\endgroup$ – sega_sai Jun 27 '13 at 0:49
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    $\begingroup$ take $(1-x)-\varepsilon(1-x^2)$ and try to represent it as a linear combination with positive coefficients. If it was possible all basic function had to vanish at $x=1$ and thus you could use only $1-x$ and $1-x^2...$ $\endgroup$ – leshik Jun 27 '13 at 1:02
  • $\begingroup$ You are welcome. Let us know if you need something more specific about representation. In each particular case, one may be able to come up with it. $\endgroup$ – leshik Jun 27 '13 at 1:18
  • $\begingroup$ I updated the question to describe what kind of representation I looking for in the end $\endgroup$ – sega_sai Jun 27 '13 at 1:44
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It's certainly not possible for an arbitrary basis of nonnegative polynomials. For example, change the domain to $0<x<1$ for convenience. Then $\{1,x,x^2,x^3,\dots\}$ is a basis of nonnegative polynomials, but the nonnegative polynomial $1-x$ can't be well-approximated by a nonnegative linear combination of $1,x,x^2,x^3,\dots$. (If it could, what would the coefficient of $1$ be, and how would it approximate $1-x$ both near $x=0$ and $x=1$?)

I don't know a general theorem to this effect, but one might be able to examine specific bases to spot flaws of the same type.

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  • $\begingroup$ Yes, I completely agree with what you've said, but I don't need a linearly independent set of polynomials, e.g I'm fine with $T_0$(x), $T_i$(x)+1 and 1-$T_i$(x) set (for e.g. Chebyshev basis), which has twice the number of functions, normally needed (and my guess that's a smallest set for a given degree) $\endgroup$ – sega_sai Jun 26 '13 at 22:38

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