Worksheet08: Dickson 038
Current Work (editable)
Thus if p_{i} y^{n-i} is a term of f(y), also ±p_{i} y^{i} is a term. Hence (18') f(y) $$ y^{n} ± 1 + p_{1}(y^{n-1} ± y) + p_{2}(y^{n-2} ± y^{2}) + .... If n is odd, n = 2t + 1, the final term is p_{t}(y^{t+1} ± y^{t}), and y ± 1 is a factor of f(y). In view of (17), the quotient Q(y) $$ f(y) / y ± 1 has the property that y^{n-1} Q(1/y) $$ Q(y). Comparing this with (17), which implied (18'), we see that Q(y) = 0 is a reciprocal equation of the type (18) y^{2t} + 1 + c_{1} (y^{2t-1} + y) + c_{2} (y^{2t-2} + y^{2}) + ... + c_{t-1} (y^{t+1} + y^{t-1}) + c_{t} y^{t} = 0. If n is even, n = 2t, and if the upper sign holds in (17), then (18') is of the form (18). Next, let the lower sign hold in (17). Since a term p_{t} y^{t} would imply a term -p_{t} y^{t}, we have p_{t} = 0. The final term in (18') is therefore p_{t-1} (y^{t+1} - y^{t-1}). Hence f(y) has the factor y^{2} - 1. The quotient q(y) $$ f(y)/(y^{2} - 1) has the property that y^{n-2} q(1/y) $$ q(y). Comparing this with (17) as before, we see that q(y) = 0 is of the form (18) where now 2t = n - 2. Hence, at least after removing one or both of the factors y ± 1, any reciprocal equation may be given the form (18). The method by which (13) was reduced to a cubic equation may be used to reduce any equation (18) to an equation in x of half the degree. First, we divide the terms of (18) by y^{t} and obtain (y^{t} + 1/y^{t}) + c_{1} (y^{t-1} + 1/y^{t-1}) + ... + c_{t-1} (y + 1/y) + c_{t} = 0. Next, we perform the substitution (14) by either of the following methods: We may make use of the relation y^{k} + 1/y^{k} = x (y^{k-1} + 1/y^{k-1}) - (y^{k-2} + 1/y^{k-2}) to compute the values of y^{k} + 1/y^{k} in terms of x, starting with the special
Answer Key (non-editable)
Thus if $p_{i} y^{n-i}$ is a term of $f(y)$, also $±p_{i} y^{i}$ is a term. Hence \[ \Tag{(18')} f(y) \equiv y^{n} ± 1 + p_{1}(y^{n-1} ± y) + p_{2}(y^{n-2} ± y^{2}) + \dots. \] If $n$ is \emph{odd}, $n = 2t + 1$, the final term is $p_{t}(y^{t+1} ± y^{t})$, and $y ± 1$ is a factor of $f(y)$. In view of \Eq{(17)}, the quotient \[ Q(y) \equiv \frac{f(y)}{y ± 1} \] has the property that \[ y^{n-1} Q\left(\frac{1}{y}\right) \equiv Q(y). \] Comparing this with \Eq{(17)}, which implied \Eq{(18')}, we see that $Q(y) = 0$ is a reciprocal equation of the type \[ \Tag{(18)} y^{2t} + 1 + c_{1} (y^{2t-1} + y) + c_{2} (y^{2t-2} + y^{2}) + \dots + c_{t-1} (y^{t+1} + y^{t-1}) + c_{t} y^{t} = 0. \] If $n$ is \emph{even}, $n = 2t$, and if the upper sign holds in \Eq{(17)}, then \Eq{(18')} is of the form \Eq{(18)}. Next, let the lower sign hold in \Eq{(17)}. Since a term $p_{t} y^{t}$ would imply a term $-p_{t} y^{t}$, we have $p_{t} = 0$. The final term in \Eq{(18')} is therefore $p_{t-1} (y^{t+1} - y^{t-1})$. Hence $f(y)$ has the factor $y^{2} - 1$. The quotient $q(y) \equiv f(y)/(y^{2} - 1)$ has the property that \[ y^{n-2} q\left(\frac{1}{y}\right) \equiv q(y). \] Comparing this with \Eq{(17)} as before, we see that $q(y) = 0$ is of the form \Eq{(18)} where now $2t = n - 2$. Hence, at least after removing one or both of the factors $y ± 1$, \begin{Thm}any reciprocal equation may be given the form \Eq{(18)}\end{Thm}. The method by which \Eq{(13)} was reduced to a cubic equation may be used to reduce any equation \Eq{(18)} to an equation in $x$ of half the degree. First, we divide the terms of \Eq{(18)} by $y^{t}$ and obtain \[ \left(y^{t} + \frac{1}{y^{t}}\right) + c_{1} \left(y^{t-1} + \frac{1}{y^{t-1}}\right) + \dots + c_{t-1} \left(y + \frac{1}{y}\right) + c_{t} = 0. \] Next, we perform the substitution \Eq{(14)} by either of the following methods: We may make use of the relation \[ y^{k} + \frac{1}{y^{k}} = x \left(y^{k-1} + \frac{1}{y^{k-1}}\right) - \left(y^{k-2} + \frac{1}{y^{k-2}}\right) \] to compute the values of $y^{k} + 1/y^{k}$ in terms of $x$, starting with the special
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