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Consider the polynomial $p(x)$ such that:

$$ p(x) = 1 + 2x + 3x^2 + 4x^3 + \cdots + (2n+2)x^{2n+1} $$ where $ n $ is a natural number. Because $p(x)$ has odd degree, then from the Complex conjugate root theorem it follows that it has at least one real root. My question is then, is this root unique ? This question is actually related to this other question. I plotted the graph for $ n = 0 $ to $ n = 7 $, and in all cases the polynomial has a single real root.

HJE
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1 Answers1

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Yes, the root is unique. If we define $q(x) = p(x)(x-1)^2$, we can verify that $$ q(x) = (2n+2)x^{2n+3} - (2n+3)x^{2n+2} + 1. $$ By Decartes' rule of signs, $q(x)$ has at most two positive roots; since we know that $q(x)$ has a double root at $x=1$, there can't be any others. (We knew this anyway since $p(x)>0$ for $x>0$.) On the other hand, $$ q(-x) = -(2n+2)x^{2n+3} - (2n+3)x^{2n+2} + 1 $$ has exactly one sign change in its coefficients, so $q(x)$ has exactly one negative root, which is therefore the unique root of $p(x)$ (since clearly $x=0$ is not a root).

Greg Martin
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  • Really elegant solution! I was not aware of Descartes' rule of signs, definitely will check it out. – HJE Jun 27 '24 at 23:13
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    Alternatively: $q'(x) = (2n+3)(2n+2)x^{2n+1}(x-1) > 0$ for $x < 0$, so that $q$ cannot have more than one negative root. – Martin R Jun 28 '24 at 09:01