A simple proof of Descartes's rule of sign

Question

I search all over the Internet for a proof of Descartes's rule of sign. Found a pdf file which has page-long proof that a high schooler has to no way to understand.

Can somebody talented here give me a proof of the rule, in geometry preferably, or if impossible to be done in geometry in simple algebra or trigonometry.

Answer 1

This is an old question that I think deserves an answer that is at the level of a high-schooler. There are however a couple of details that, although intuitively obvious, require some basic results from Calculus:

• The intermediate value theorem which in the context of the problem states that if $p$ is a polynomial with real coefficients such that in an interval $[a,b]$, $p(a)p(b)<0$, then there is a point $a<x^*<b$ at which $p(x^*)=0$.
• The notion of limit which the context of the problem implies that if $p(x)=p_n x^n+ p_{n-1}x^{n-1}+\ldots + p_1 x +p_0$, with $p_n\neq0$, and $p_j\in\mathbb{R}$ for $1\leq j\leq n$, then the values of $p$ for large positive values of $x$ have the same sign as that of $p_n$.

Other than these Calculus facts, the proof presented here is at the level of high-school Mathematics and requires a good understanding of algebraic manipulations, and some counting skills.

The proof discussed here follows the one developed by Gauss in Beweis eines algebraischen Lehrsatzes (Proof of certain algebraic Theorems). Crelle’s Journal für die reine und angewandte Mathematik, 1828 v. 3, no. 1, p. 1–4., and a simplified version in the classic book Dickson, L.E., First Course in the Theory of Equations, Wiley, New York 1922.

Before producing a simple proof of the result, I will state some definitions:

• Suppose $p(x)=p_nx^n+p_{n-1}x^{n-1}+\ldots +p_1x+p_0$, with $p_n\neq0$, and $p_j\in\mathbb{R}$ for $0\leq j\leq n$.

1. For each $1\leq j< n$, If (a) $p_jp_{j-1}<0$ or (b) $p_kp_{j-1}<0$ for some $k>j-1$ such that $0=p_{k-1}=\ldots=p_j$, then we say that there is a change of sign between the consecutive nonzero coefficients $p_k$ and $p_{j-1}$ of $p$.
2. If $0<x_1<\ldots<x_s$, are all the distinct positive roots of $p$ (i.e. for each $x>0$, $p(x)=0$ iff $x\in\{x_1,\ldots,x_s\}$) and each $x_j$, $1\leq j\leq s$ has multiplicity $m_j\geq1$ (i.e. for each $x_j$, there is a polynomial $q_j$ such that $p(x)=(x-x_j)^{m_j}q_j(x)$ and $q_j(x_j)\neq0$), then we say that $p$ has a total of $m_1+\ldots+m_s$ positive roots.

The result we are going to prove is the following

Theorem (Descartes) Suppose $p(x)=p_nx^n+\ldots +p_1x+p_0$ is a polynomial with real coefficients and $p_n\neq0$. Let $V$ denote total number of changes of sign amongst the consecutive nonzero coefficients of $p$, and let $r$ denote the total number of positive solutions of $p$. Then $V-r=2m$ for some integer $m\geq0$.

Observations:

1. It suffices to assume that $p$ is a monic polynomial, that is, $$ p(x)=x^n+p_{n-1}x^{n-1}+\ldots +p_1x+a_0$$ Indeed, if $p$ is any polynomial of degree $n$, and $q=\frac{1}{p_n}p$ is a monic polynomial of degree $n$ which has the same solutions as $p$, including multiplicity, as well as the same number of changes among consecutive nonzero coefficients.

2. It suffices to assume that $p_0\neq0$. Indeed, if $p_n\neq0$ and $p_j=0$ for all $0\leq j<k\leq n$, $$p(x)=x^n+ \ldots +p_{k+1}x^{k+1}+p^kx^k=x^k(x^{n-k}+\ldots+p_k)=x^kP(x)$$ The polynomial $P(x)=x^{n-k}+\ldots+a_{k+1}x+a_k$ has the same positive solutions, including multiplicity, as $p$, and the total number of changes of sign between consecutive nonzero coefficients in $P$ and $p$ is the same.

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For the remaining of the proof, we consider monic polynomials $p$ with $p(0)\neq0$.

The proof the Theorem will be based on the following three claims:

Claim I: Suppose that $(p_{k_\ell},p_{j_\ell-1})$, $1\leq \ell\leq V$, where $n\geq k_\ell>j_\ell-1\geq k_{\ell+1}\geq 1$, are all the pair of consecutive nonzero coefficients of $p$ where there is a change in sign, i.e.,
$$n\geq k_1>k_2>\ldots > k_V\geq1,$$ $$p_{k_\ell}p_{j_\ell-1}<0,\qquad 1\leq \ell\leq V$$ and either $$k_\ell=j_\ell,\quad\text{or}\quad p_{k_\ell-1}=\ldots p_{j_\ell}=0.$$ Then the signs of the components ordered sequence $(p_{j_1-1},p_{j_2-1},\ldots,q_{j_V-1})$ alternate.

Indeed, all the coefficients $p_{j_\ell-1}, p_{{j_\ell}-2},\ldots,p_{k_{\ell+1}}$ are either $0$ or have the same sign as $p_{j_\ell-1}$.

Claim II Suppose $[a_n,a_{n-1},\ldots, a_0]$, $a_n\neq0$, is an order $n+1$-tuple of real numbers. If $[a_{m_1},\ldots,a_{m_k}]$, $n\geq m_1>\ldots>m_k\leq0$ is a subsequence which has $U$ changes of sign between its consecutive nonzero elements, then the original sequence has at least $U$ changes of sign between its consecutive nonzero elements.

This can be shown first by adding one single element to the ordered subsequence $[a_{m_1},\ldots,a_{m_k}]$ before $a_{m_1}$, after $a_{m_k}$, or somewhere in between, and then by indiction in the number elements added to the ordered sequence.

Claim III: If $p(x)=x^n+\ldots+p_1x+p_0$, $p_0\neq0$, is a monic polynomial with real coefficients and there are $V_p$ changes of sign between consecutive nonzero coefficients, then for any $\alpha>0$ $$q(x)=(x-\alpha)p(x)$$ is a monic polynomial whose total number of changes in sign amongst its consecutive nonzero coefficients, $V_q$, satisfies $V_q-V_p\geq1$, and $V_p-V_p-1$ is even.

To see this, notice that $$ Q(x)=(x-\alpha)p(x)=x^{n+1} +\Big((p_{n-1}-\alpha p_n)x^n+\ldots+(p_{k-1}-\alpha p_k)x^k_\ldots +(p_0-\alpha p_1)x\Big) -\alpha p_0$$

Notice that if there is a change of sign between the consecutive nonzero coefficients $p_k$ and $p_{j-1}$ $k\geq j$ (either $k=j$ and $p_{j+1}p_j<0$ or $k>j$, $p_{k-1}=\ldots =p_j=0$ and $p_kp_{j-1}<0$), then the coefficient $q_j=(p_{j-1}-\alpha p_j)$ has the same sign as $p_{j-1}$. It follows from Claim I and Claim II that there are at least $V$ changes of sign amongst the consecutive coefficients $[q_n=1,q_{n-1},\ldots,q_1]$. Finally, notice that $q_0=-\alpha p_0$ has opposite sign than $p_0$ and so, than $q_{V}$. Therefore, $Q$ has at least $V+1$ changes of sign amongst its consecutive nonzero coefficients.

To conclude, notice that if $[p_{m_0},p_{m_1},\ldots,p_{m_k}]$, $m_0=n>m_1>\ldots>m_k=0$, contains all the nonzero coefficients of the polynomial $p$. Notice that the total number of changes of sign of consecutive elements of $[p_{m_0},p_{m_1},\ldots,p_{m_k}]$ is the same as $V$. Moreover, if $\sigma(p_{m_j})$ be the sign of $p_{m_j}$, then $$(-1)^V=\prod^{k-1}_{j=0}\frac{\sigma(p_{m_{j+1}})}{\sigma(p_{m_j})}=\frac{\sigma(p_n)}{\sigma(p_0)}=\frac{1}{\sigma(p_0)}$$ Thus, if $V$ is even, then $\sigma(p_0)=1$ whereas if $V$ is odd, then $\sigma(p_0)=-1$.

Consequently, as $q_0=-\alpha p_0$ and $p_0$ have opposite signs, $V_q-V_p-1$ is even.

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Proof of Descartes' theorem: We first prove that $0\leq V-r$. We proceed by induction on the number $r$ of positive roots of a monic polynomial. If $r=0$, then it is clear that $r\leq V$. Furthermore, $V$ must be even. Indeed, as in the proof of Claim II, $(-1)^V=\sigma(p_0)$. We claim that $p_0>0$; otherwise, since $p_1=1>0$, $p(x)>0$ for all large positive values of $x$. This would imply that for some $x^*>0$, $p(x^*)=0$, which is not possible since $r=0$.

Suppose now that the result holds for all monic polynomials with less than $r$ positive roots. Suppose $p$ is a monic with $r$ positive roots. Let $\alpha$ be one such root. Then there is a polynomial $h$ of degree $\operatorname{deg}(h)=\operatorname{deg}(p)-1$ such that $$p(x)=(x-\alpha)h(x)$$ Then $h$ has $r-1$ positive roots and by induction hypothesis, the number of changes of sign amongst consecutive nonzero coefficients of $h$, say $V_h$, satisfies $r-1\leq V_h$, and $V_h-(r-1)=2m'$ for some integer $m'\geq0$. By Claim III, the number of changes of sign amongst consecutive nonzero coefficients of $p$, $V_p=V$, satisfies $V_p-V_h-1=2m$ for some integer $m\geq0$. Consequently, $r\leq V_h+1\leq V_p$, and $V_p-r$ is even.