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Let $f(x)=\sum\limits_{i=0}^n a_i x^i$ be a polynomial that given in implicit way, so I dont know $a_i$. I need to know if all its coefficient are positive.

A brute force way to do it is check out its derivative - if $$ \frac{d^i f(x)}{dx^n}>0, $$
then and $a_i>0.$

Question. Is there an elegant way to determine if the polynomial has only positive coefficients?

Leox
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  • You could verify that $\lim_{n\rightarrow\infty}P(x)/x^n>0$ – Alex R. Jul 10 '19 at 06:45
  • if all roots are negative and real, $P(x)=(x+a_1)(x+a_2)...(x+a_n)$ leads to all coefficients are positive. – NoChance Jul 10 '19 at 07:02
  • @ Alex R Well...$$\lim\limits_{n \to \infty} \frac{1-x}{x^n} =\lim\limits_{n \to \infty} \frac{1+x}{x^n}=0.$$ How it can help ? – Leox Jul 10 '19 at 10:12
  • Taking the derivative seems “elegant” enough to me. Why is that so undesirable? – David K Jul 11 '19 at 21:09
  • @David K Of cource the derivative is “elegant” but I cant use it if a polynomial is given in not direct way, for example put $$f_n(x)=e^x \frac{d^n}{dx^n}(e^{-x} x^n)$$ – Leox Jul 12 '19 at 05:06
  • For that example, expansion of the polynomial for a few small values of $n$ suggests you will never have all positive coefficients for $n >0.$ They seem to alternate signs. – David K Jul 12 '19 at 12:12
  • @David K Is there any statement about the "a few small values"? – Leox Jul 12 '19 at 14:25
  • I just made such a statement. I also said it suggests the same will be true for all positive $n$. Of course you cannot really know until you have proved it formally. But “small value” cases can sometimes give clues toward a proof. – David K Jul 12 '19 at 17:04
  • I think $e^x \frac{d^n}{dx^n}(e^{-x} x^n) = \binom n0 \frac{d^n}{dx^n}x^n-\binom n1 \frac{d^{n-1}}{dx^{n-1}}x^n+\binom n2 \frac{d^{n-2}}{dx^{n-2}}x^n-\cdots\pm\binom nn x^n,$ demonstrable by induction, and since $\frac{d^k}{dx^k}x^n=n(n-1)(n-2)\cdots(n-k+1)x^{n-k},$ this is a polynomial with coefficients of alternating signs. – David K Nov 28 '20 at 17:22

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