Worksheet14: Dickson 073
Current Work (editable)
Lemma. The number of variations of sign of F(x) is equal to that of f(x) increased by some positive odd integer. For, the sequence A_{0}, A_{1}, ..., A_{k_{1}} has an odd number of variations of sign since its first and last terms are of opposite sign; and similarly for the v sequences A_{k_{1}}, A_{k_{1}+1}, ..., A_{k_{2}}; ............ A_{k_{v}}, A_{k_{v}+1}, ..., A_{l+1}. The total number of variations of sign of the entire sequence A_{0}, A_{1}, ..., A_{l+1} is evidently the sum of the numbers of variations of sign for the v + 1 partial sequences indicated above, and is thus the sum of v + 1 positive odd integers. Since each such odd integer may be expressed as 1 plus 0 or a positive even integer, the sum mentioned is equal to v + 1 plus 0 or a positive even integer, i.e., to v plus a positive odd integer. To prove Descartes' rule of signs, consider first the case in which f(x) = 0 has no positive real roots, i.e., no real root between 0 and +$$. Then f(0) and f($$) are of the same sign (§63), and hence the first and last coefficients of f(x) are of the same sign.[1] Thus f(x) has either no variations of sign or an even number of them, as Descartes' rule requires. Next, let f(x) = 0 have the positive real roots r_{1}, ..., r_{k} and no others. A root of multiplicity m occurs here m times, so that the r's need not be distinct. Then f(x) $$ (x - r_{1}) ... (x - r_{k}) \phi(x), where \phi(x) is a polynomial with real coefficients such that \phi(x) = 0 has no positive real roots. We saw in the preceding paragraph that \phi(x) has either no variations of sign or an even number of them. By the Lemma, the product (x - r_{k})\phi(x) has as the number of its variations of sign the number for \phi(x) increased by a positive odd integer. Similarly when we introduce each new factor x - r_{i}. Hence the number of variations of sign of the final product f(x) is equal to that of \phi(x) increased by k positive odd integers, i.e., by k plus 0 or a positive even integer. Since \phi(x) has either no variations of sign or an even number of them, the number of variations of sign of f(x) is k plus 0 or a positive even integer, a result equivalent to our statement of Descartes' rule. 1 In case f(x) has a factor x^{n-l}, we use the polynomial f(x)/x^{n-l} instead of f(x) in this argument.
Answer Key (non-editable)
\begin{Lemma} The number of variations of sign of $F(x)$ is equal to that of $f(x)$ increased by some positive odd integer. \end{Lemma} For, the sequence $A_{0}$, $A_{1}$, \dots, $A_{k_{1}}$ has an odd number of variations of sign since its first and last terms are of opposite sign; and similarly for the $v$ sequences \[ \begin{array}{*{4}{c}} A_{k_{1}}, & A_{k_{1}+1}, & \dots, & A_{k_{2}}; \\ \hdotsfor{4} \\ A_{k_{v}}, & A_{k_{v}+1}, & \dots, & A_{l+1}. \end{array} \] The total number of variations of sign of the entire sequence $A_{0}$, $A_{1}$, \dots, $A_{l+1}$ is evidently the sum of the numbers of variations of sign for the $v + 1$ partial sequences indicated above, and is thus the sum of $v + 1$ positive odd integers. Since each such odd integer may be expressed as $1$ plus $0$ or a positive even integer, the sum mentioned is equal to $v + 1$ plus $0$ or a positive even integer, i.e., to $v$ plus a positive odd integer. To prove Descartes' rule of signs, consider first the case in which $f(x) = 0$ has no positive real roots, i.e., no real root between $0$ and $+\infty$. Then $f(0)$ and $f(\infty$) are of the same sign (§63), and hence the first and last coefficients of $f(x)$ are of the same sign.\footnote {In case $f(x)$ has a factor $x^{n-l}$, we use the polynomial $f(x)/x^{n-l}$ instead of $f(x)$ in this argument.} Thus $f(x)$ has either no variations of sign or an even number of them, as Descartes' rule requires. Next, let $f(x) = 0$ have the positive real roots $r_{1}$, \dots, $r_{k}$ and no others. A root of multiplicity $m$ occurs here $m$ times, so that the $r$'s need not be distinct. Then \[ f(x) \equiv (x - r_{1}) \dots (x - r_{k}) \phi(x), \] where $\phi(x)$ is a polynomial with real coefficients such that $\phi(x) = 0$ has no positive real roots. We saw in the preceding paragraph that $\phi(x)$ has either no variations of sign or an even number of them. By the Lemma, the product $(x - r_{k})\phi(x)$ has as the number of its variations of sign the number for $\phi(x)$ increased by a positive odd integer. Similarly when we introduce each new factor $x - r_{i}$. Hence the number of variations of sign of the final product $f(x)$ is equal to that of $\phi(x)$ increased by $k$ positive odd integers, i.e., by $k$ plus $0$ or a positive even integer. Since $\phi(x)$ has either no variations of sign or an even number of them, the number of variations of sign of $f(x)$ is $k$ plus $0$ or a positive even integer, a result equivalent to our statement of Descartes' rule.
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