Worksheet22: Dickson 141
Current Work (editable)
In order to compute \Sigma \alpha_{1}^{4} \alpha_{2}^{3} \alpha_{3}^{2} in terms of the s_{k}, we form the product \Sigma \alpha_{1}^{4} · \Sigma \alpha_{1}^{3} \alpha_{2}^{2} = \Sigma \alpha_{1}^{7} \alpha_{2}^{2} + \Sigma \alpha_{1}^{6} \alpha_{2}^{3} + \Sigma \alpha_{1}^{4} \alpha_{2}^{3} \alpha_{3}^{2}. Making three applications of (22), we get s_{4}(s_{3} s_{2} - s_{5}) = (s_{7} s_{2} - s_{9}) + (s_{6} s_{3} - s_{9}) + \Sigma \alpha_{1}^{4} \alpha_{2}^{3} \alpha_{3}^{2}. Hence \Sigma \alpha_{1}^{4} \alpha_{2}^{3} \alpha_{3}^{2} = s_{2} s_{3} s_{4} - s_{2} s_{7} - s_{3} s_{6} - s_{4} s_{5} + 2s_{9}. EXERCISES For a quartic equation, express in terms of the s_{k} and ultimately in terms of the coefficients c_{1}, ..., c_{4}: 1. \Sigma \alpha_{1}^{2} \alpha_{2}^{2}. 2. \Sigma \alpha_{1}^{3} \alpha_{2}. 3. \Sigma \alpha_{1}^{2} \alpha_{2} \alpha_{3}. 4. \Sigma \alpha_{1}^{2} \alpha_{2}^{2} \alpha_{3}^{2}. 5. If a $$ b > c > 0, prove that \Sigma \alpha_{1}^{a} \alpha_{2}^{b} \alpha_{3}^{c} = 1/m (s_{a} s_{b} s_{c} - s_{a} s_{b + c} - s_{b} s_{a + c} - s_{c} s_{a + b} + 2s_{a + b + c}), where m = 1 if a > b, m = 2 if a = b. 6. \Sigma \alpha_{1}^{a} \alpha_{2}^{b} \alpha_{3}^b = 1/2 (s_{a} s_{b}^{2} - s_{a} s_{2b} - 2s_{b} s_{a + b} + 2s_{a + 2b}), a > b > 0. 7. \Sigma \alpha_{1}^{a} \alpha_{2}^{a} \alpha_{3}^{a} = 1/6 (s_{a}^{3} - 3s_{a} s_{2a} + 2s_{3a}), a > 0. 109. Computation of Symmetric Functions. The method last explained is practicable when a term of the \Sigma-function involves only a few distinct roots, the largeness of the exponents not introducing a difficulty in the initial work of expressing the \Sigma-function in terms of the s_{k}. But when a term of the \Sigma-function involves a large number of roots with small exponents, we resort to a method suggested by §104, which tells us which auxiliary simpler symmetric fuctions[**functions] should be multiplied together to produce our \Sigma-function along with simpler ones. For example, to find \Sigma x_{1}^{2} x_{2} x_{3} x_{4}, when n > 4, we employ E_{1} E_{4} $$ \Sigma x_{1} · \Sigma x_{1} x_{2} x_{3} x_{4} = \Sigma x_{1}^{2} x_{2} x_{3} x_{4} + 5\Sigma x_{1} x_{2} x_{3} x_{4} x_{5}, \Sigma x_{1}^{2} x_{2} x_{3} x_{4} = E_{1} E_{4} - 5E_{5}. To find \Sigma x_{1}^{2} x_{2}^{2} x_{3}^{2} x_{4}, employ E_{3} E_{4} = \Sigma x_{1} x_{2} x_{3} · \Sigma x_{1} x_{2} x_{3} x_{4}. When many such products of \Sigma-functions are to be computed, it will save time in the long run to learn and apply the "method of leaders" explained in the author's Elementary Theory of Equations, pp. 64-65.
Answer Key (non-editable)
In order to compute $\Sigma \alpha_{1}^{4} \alpha_{2}^{3} \alpha_{3}^{2}$ in terms of the $s_{k}$, we form the product \[ \Sigma \alpha_{1}^{4} · \Sigma \alpha_{1}^{3} \alpha_{2}^{2} = \Sigma \alpha_{1}^{7} \alpha_{2}^{2} + \Sigma \alpha_{1}^{6} \alpha_{2}^{3} + \Sigma \alpha_{1}^{4} \alpha_{2}^{3} \alpha_{3}^{2}. \] Making three applications of \Eq{(22)}, we get \[ s_{4}(s_{3} s_{2} - s_{5}) = (s_{7} s_{2} - s_{9}) + (s_{6} s_{3} - s_{9}) + \Sigma \alpha_{1}^{4} \alpha_{2}^{3} \alpha_{3}^{2}. \] Hence \[ \Sigma \alpha_{1}^{4} \alpha_{2}^{3} \alpha_{3}^{2} = s_{2} s_{3} s_{4} - s_{2} s_{7} - s_{3} s_{6} - s_{4} s_{5} + 2s_{9}. \] \begin{Exercises} For a quartic equation, express in terms of the $s_{k}$ and ultimately in terms of the coefficients $c_{1}$, \dots, $c_{4}$: \begin{itemize} \item[1.] $\Sigma \alpha_{1}^{2} \alpha_{2}^{2}$. \item[2.] $\Sigma \alpha_{1}^{3} \alpha_{2}$. \item[3.] $\Sigma \alpha_{1}^{2} \alpha_{2} \alpha_{3}$. \item[4.] $\Sigma \alpha_{1}^{2} \alpha_{2}^{2} \alpha_{3}^{2}$. \item[5.] If $a \geqq b > c > 0$, prove that \[ \Sigma \alpha_{1}^{a} \alpha_{2}^{b} \alpha_{3}^{c} = \frac{1}{m} (s_{a} s_{b} s_{c} - s_{a} s_{b + c} - s_{b} s_{a + c} - s_{c} s_{a + b} + 2s_{a + b + c}), \] where $m = 1$ if $a > b$, $m = 2$ if $a = b$. \item[6.] $\Sigma \alpha_{1}^{a} \alpha_{2}^{b} \alpha_{3}^b = \frac{1}{2} (s_{a} s_{b}^{2} - s_{a} s_{2b} - 2s_{b} s_{a + b} + 2s_{a + 2b})$,\quad $a > b > 0$. \item[7.] $\Sigma \alpha_{1}^{a} \alpha_{2}^{a} \alpha_{3}^{a} = \frac{1}{6} (s_{a}^{3} - 3s_{a} s_{2a} + 2s_{3a})$,\quad $a > 0$. \end{itemize} \end{Exercises} \Par{109. Computation of Symmetric Functions.} The method last explained is practicable when a term of the $\Sigma$-function involves only a few distinct roots, the largeness of the exponents not introducing a difficulty in the initial work of expressing the $\Sigma$-function in terms of the $s_{k}$. But when a term of the $\Sigma$-function involves a large number of roots with small exponents, we resort to a method suggested by §104, which tells us which auxiliary simpler symmetric \DPtypo{fuctions}{functions} should be multiplied together to produce our $\Sigma$-function along with simpler ones. \begin{Remark} For example, to find $\Sigma x_{1}^{2} x_{2} x_{3} x_{4}$, when $n > 4$, we employ \begin{gather*} E_{1} E_{4} \equiv \Sigma x_{1} · \Sigma x_{1} x_{2} x_{3} x_{4} = \Sigma x_{1}^{2} x_{2} x_{3} x_{4} + 5 \Sigma x_{1} x_{2} x_{3} x_{4} x_{5}, \\ \Sigma x_{1}^{2} x_{2} x_{3} x_{4} = E_{1} E_{4} - 5E_{5}. \end{gather*} To find $\Sigma x_{1}^{2} x_{2}^{2} x_{3}^{2} x_{4}$, employ $E_{3} E_{4} = \Sigma x_{1} x_{2} x_{3} · \Sigma x_{1} x_{2} x_{3} x_{4}$. When many such products of $\Sigma$-functions are to be computed, it will save time in the long run to learn and apply the ``method of leaders'' explained in the author's \textit{Elementary Theory of Equations}, pp.~64--65. \end{Remark}
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