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Jack
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Let $f:[0,2]\rightarrow [0,\infty]$$f:[0,2]\rightarrow [0,\infty)$ be a strictly increasing smooth function. The Weierstrass approximation theorem says that we can uniformly approximate $f$ by polynomials. But my concern is

Question: Can we uniformly approximate $f$ by polynomials with non-negative coefficients?

All the proofs of the Weierstrass approximation theorem I have seen is either existential or the continuous function $f:[0,1]\rightarrow \mathbb{R}$ has been approximated by the sequence of functions $$B_n(x)=\sum_{k=0}^n{n \choose k}x^k(1-x)^{n-k}f(\frac{k}{n}).$$ But there is no assurance that all the $B_n(x)$s are polynomials with non-negative coefficients.

P.s.- For my particular case, I don't care about the constant term of the polynomials, I only need the coefficients of $x^k$ for $k>0$ to be non-negative.

Let $f:[0,2]\rightarrow [0,\infty]$ be a strictly increasing smooth function. The Weierstrass approximation theorem says that we can uniformly approximate $f$ by polynomials. But my concern is

Question: Can we uniformly approximate $f$ by polynomials with non-negative coefficients?

All the proofs of the Weierstrass approximation theorem I have seen is either existential or the continuous function $f:[0,1]\rightarrow \mathbb{R}$ has been approximated by the sequence of functions $$B_n(x)=\sum_{k=0}^n{n \choose k}x^k(1-x)^{n-k}f(\frac{k}{n}).$$ But there is no assurance that all the $B_n(x)$s are polynomials with non-negative coefficients.

P.s.- For my particular case, I don't care about the constant term of the polynomials, I only need the coefficients of $x^k$ for $k>0$ to be non-negative.

Let $f:[0,2]\rightarrow [0,\infty)$ be a strictly increasing smooth function. The Weierstrass approximation theorem says that we can uniformly approximate $f$ by polynomials. But my concern is

Question: Can we uniformly approximate $f$ by polynomials with non-negative coefficients?

All the proofs of the Weierstrass approximation theorem I have seen is either existential or the continuous function $f:[0,1]\rightarrow \mathbb{R}$ has been approximated by the sequence of functions $$B_n(x)=\sum_{k=0}^n{n \choose k}x^k(1-x)^{n-k}f(\frac{k}{n}).$$ But there is no assurance that all the $B_n(x)$s are polynomials with non-negative coefficients.

P.s.- For my particular case, I don't care about the constant term of the polynomials, I only need the coefficients of $x^k$ for $k>0$ to be non-negative.

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Jack
Jack
  • 11
  • 2

Let $f:[0,2]\rightarrow [0,\infty]$ be a strictly increasing smooth function. The Weierstrass approximation theorem says that we can uniformly approximate $f$ by polynomials. But my concern is

Question: Can we uniformly approximate $f$ by polynomials with non-negative coefficients?

All the proofs of the Weierstrass approximation theorem I have seen is either existential or the continuous function $f:[0,1]\rightarrow \mathbb{R}$ has been approximated by the sequence of functions $$B_n(x)=\sum_{k=0}^n{n \choose k}x^k(1-x)^{n-k}f(\frac{k}{n}).$$ But there is no assurance that all the $B_n(x)$s are polynomials with non-negative coefficients.

P.s.- For my particular case, I don't care about the constant term of the polynomials, I only need the coefficients of $x^k$ for $k>0$ to be positivenon-negative.

Let $f:[0,2]\rightarrow [0,\infty]$ be a strictly increasing smooth function. The Weierstrass approximation theorem says that we can uniformly approximate $f$ by polynomials. But my concern is

Question: Can we uniformly approximate $f$ by polynomials with non-negative coefficients?

All the proofs of the Weierstrass approximation theorem I have seen is either existential or the continuous function $f:[0,1]\rightarrow \mathbb{R}$ has been approximated by the sequence of functions $$B_n(x)=\sum_{k=0}^n{n \choose k}x^k(1-x)^{n-k}f(\frac{k}{n}).$$ But there is no assurance that all the $B_n(x)$s are polynomials with non-negative coefficients.

P.s.- For my particular case, I don't care about the constant term of the polynomials, I only need the coefficients of $x^k$ for $k>0$ to be positive.

Let $f:[0,2]\rightarrow [0,\infty]$ be a strictly increasing smooth function. The Weierstrass approximation theorem says that we can uniformly approximate $f$ by polynomials. But my concern is

Question: Can we uniformly approximate $f$ by polynomials with non-negative coefficients?

All the proofs of the Weierstrass approximation theorem I have seen is either existential or the continuous function $f:[0,1]\rightarrow \mathbb{R}$ has been approximated by the sequence of functions $$B_n(x)=\sum_{k=0}^n{n \choose k}x^k(1-x)^{n-k}f(\frac{k}{n}).$$ But there is no assurance that all the $B_n(x)$s are polynomials with non-negative coefficients.

P.s.- For my particular case, I don't care about the constant term of the polynomials, I only need the coefficients of $x^k$ for $k>0$ to be non-negative.

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Jack
Jack
  • 11
  • 2

Approximating a strictly increasing non-negative function on a non-negative domain by polynomials with non-negative coefficients

Let $f:[0,2]\rightarrow [0,\infty]$ be a strictly increasing smooth function. The Weierstrass approximation theorem says that we can uniformly approximate $f$ by polynomials. But my concern is

Question: Can we uniformly approximate $f$ by polynomials with non-negative coefficients?

All the proofs of the Weierstrass approximation theorem I have seen is either existential or the continuous function $f:[0,1]\rightarrow \mathbb{R}$ has been approximated by the sequence of functions $$B_n(x)=\sum_{k=0}^n{n \choose k}x^k(1-x)^{n-k}f(\frac{k}{n}).$$ But there is no assurance that all the $B_n(x)$s are polynomials with non-negative coefficients.

P.s.- For my particular case, I don't care about the constant term of the polynomials, I only need the coefficients of $x^k$ for $k>0$ to be positive.