Worksheet16: Dickson 080
Current Work (editable)
8. Prove that if one of Sturm's functions has p imaginary roots, the initial equation has at least p imaginary roots. 9. State Sturm's theorem so as to include the possibility of a, or b, or both a and b being roots of f(x) = 0. 71. Sturm's Functions for a Quartic Equation. For the reduced quartic equation f(z) = 0, (5) { f = z^{4} + qz^{2} + rz + s, { f_{1} = 4z^{3} + 2qz + r, { f_{2} = -2qz^{2} - 3rz - 4s. Let q $$ 0 and divide q^{2} f_{1} by f_{2}. The negative of the remainder is (6) f_{3} = Lz - 12rs - rq^{2}, L = 8qs - 2q^{3} - 9r^{2}. Let L $$ 0. Then f_{4} is a constant which is zero if and only if f = 0 has multiple roots, i.e., if its discriminant \Delta is zero. We therefore desire f_{4} expressed as a multiple of \Delta. By §50, (7) \Delta = -4P^{3} - 27Q^{2}, P = -4s - q^{2}/3, Q = 8/3qs - r^{2} - 2/27 q^{3}. We may employ P and Q to eliminate (8) 4s = -P - q^{2}/3, r^{2} = -Q - 2/3 qP - 8/27 q^{3}. We divide L^{2} f_{2} by (9) f_{3} = Lz + 3rP, L = 9Q + 4qP. The negative of the remainder[1] is (10) 18r^{2} qP^{2} - 9r^{2} LP + 4sL^{2} = q^{2} \Delta. The left member is easily reduced to q^{2} \Delta. Inserting the values (8) and replacing L^{2} by L(9Q + 4qP), we get -18qQP^{2} - 12q^{2} P^{3} - 16/3 q^{4} P^{2} + 2qP^{2} L + 4/3 q^{3} PL - 3q^{2} QL. Replacing L by its value (9), we get q^{2} \Delta. Hence we may take (11) f_{4} = \Delta. Hence if qL\Delta $$ 0, we may take (5), (9), (11) as Sturm's functions. 1 Found directly by the Remainder Theorem (§14) by inserting the root z = -3rP/L of f_{3} = 0 into L^{2} f_{2}.
Answer Key (non-editable)
\begin{Exercises} \begin{itemize}%[** Continued] \item[8.] Prove that if one of Sturm's functions has $p$ imaginary roots, the initial equation has at least $p$ imaginary roots. \item[9.] State Sturm's theorem so as to include the possibility of $a$, or $b$, or both $a$ and $b$ being roots of $f(x) = 0$. \end{itemize} \end{Exercises} \Par{71. Sturm's Functions for a Quartic Equation.} For the reduced quartic equation $f(z) = 0$, \[ \Tag{(5)} \left\{ \begin{aligned} f &= z^{4} + qz^{2} + rz + s, \\ f_{1} &= 4z^{3} + 2qz + r, \\ f_{2} &= -2qz^{2} - 3rz - 4s. \end{aligned} \right. \] Let $q \neq 0$ and divide $q^{2} f_{1}$ by $f_{2}$. The negative of the remainder is \[ \Tag{(6)} f_{3} = Lz - 12rs - rq^{2},\qquad L = 8qs - 2q^{3} - 9r^{2}. \] Let $L \neq 0$. Then $f_{4}$ is a constant which is zero if and only if $f = 0$ has multiple roots, i.e.,\DPnote{** Not ital. elsewhere} if its discriminant $\Delta$ is zero. We therefore desire $f_{4}$ expressed as a multiple of $\Delta$. By §50, \[ \Tag{(7)} \Delta = -4P^{3} - 27Q^{2},\qquad P = -4s - \frac{q^{2}}{3},\qquad Q = \tfrac{8}{3} qs - r^{2} - \tfrac{2}{27} q^{3}. \] We may employ $P$ and $Q$ to eliminate \[ \Tag{(8)} 4s = -P - \frac{q^{2}}{3},\qquad r^{2} = -Q - \tfrac{2}{3} qP - \tfrac{8}{27} q^{3}. \] We divide $L^{2} f_{2}$ by \[ \Tag{(9)} f_{3} = Lz + 3rP,\qquad L = 9Q + 4qP. \] The negative of the remainder\footnote {Found directly by the Remainder Theorem (§14) by inserting the root $z = -3rP/L$ of $f_{3} = 0$ into $L^{2} f_{2}$.} is \[ \Tag{(10)} 18r^{2} qP^{2} - 9r^{2} LP + 4sL^{2} = q^{2} \Delta. \] The left member is easily reduced to $q^{2}\Delta$. Inserting the values \Eq{(8)} and replacing $L^{2}$ by $L(9Q + 4qP)$, we get \[ -18qQP^{2} - 12q^{2} P^{3} - \tfrac{16}{3} q^{4} P^{2} + 2qP^{2} L + \tfrac{4}{3} q^{3} PL - 3q^{2} QL. \] Replacing $L$ by its value \Eq{(9)}, we get $q^{2}\Delta$. Hence we may take \[ \Tag{(11)} f_{4} = \Delta. \] Hence if $qL\Delta \neq 0$, we may take \Eq{(5)}, \Eq{(9)}, \Eq{(11)} as Sturm's functions.
Comparison of Current Work and Answer Key
Text to be deleted.
Text to be inserted.
¶
symbols signify newlines.