Is the simpler proof for Descartes' Rule of Signs valid?

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The discussion centers on the validity of a simpler proof for Descartes' Rule of Signs, which contrasts with the traditional six-lemma proof that is more rigorous and algebra-based. The simpler proof, which is only two pages long, relies on properties of continuous functions that the author claims are "obvious," but this lack of rigor raises concerns. Participants note that while the simpler proof may appear plausible through visual representations, it does not adequately demonstrate that certain continuous functions cannot be polynomials. The conversation highlights the importance of rigorous proof in mathematical discussions, especially when dealing with fundamental concepts like polynomials and their properties. Ultimately, the validity of the simpler proof remains in question due to its reliance on non-rigorous assumptions.
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People who are actually supposed to answer this question are those who know about the Descartes' Rule of Signs, so I will not go about explaining it. The well-known proof for the Rule includes somewhat 6 lemmas and covers 7 pages or so, presented http://homepage.smc.edu/kennedy_john/POLYTHEOREMS.PDF , starting from (17). However, I came across a simpler proof that is presented http://www.math.tamu.edu/~rojas/wangdescartes.pdf , which covers somewhat 2 pages. My question is that is this last proof valid? I did not spot any mistakes so far, I am curious if you will.
 
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The big difference is that the first proof is rigorous and uses only algebra, but the second one uses properties of continuous functioms which the author claims are "obvious".

They are "obvious" in the sense that you can draw some pictures to show they are plausble, but that isn't a rigorous proof. For example it is easy to invent a continuous function p(x) with p(0) > 0 and ##p(x) \rightarrow \infty## as ##x \rightarrow \infty##, which crosses the positive x-axis an infinite number of times. He doesn't attempt to prove that such a function can not be a polynomial.
 
Maybe, but a function that has finite maxima/minima is bound to have finite x-intercepts. Polynomials have finite maxima/minima because their derivatives are also polynomials, and a polynomial of a finite degree has finite solutions.
 
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