Leonhard Euler · 1770 (1828 English ed.) · Internet Archive
Elements of Algebra
All 627 leaves · leaf 547
Transcription
510
ADDITIONS.
CHAP. II.
$A(aq' - p') \times (p'' - bq'') =$
$-Ap'p'' + Aap'q'' + Abp'q'' - Aabq'q'' =$
$-Ap'p'' - cq'q'' - \frac{1}{2}B(p'q'' + q'p'') + \frac{1}{2}\sqrt{E}(p''q' - q''p'),$
$A(p'' - aq'') \times (bq''' - p''') =$
$-Ap''p''' + Aap''q''' + Abp''q''' - Aabq''q''' =$
$-Ap''p''' - cq''q''' - \frac{1}{2}B(p''q''' + q''p''') + \frac{1}{2}\sqrt{E}(p'''q'' - q'''p''),$
and so on. Now, in order to abridge, let us make
$P^0 = A$
$P' = Ap'^2 + Bp'q' + Cq'^2$
$P'' = Ap''^2 + Bp''q'' + Cq''^2$
$P''' = Ap'''^2 + Bp'''q''' + Cq'''^2, \ldots$
$Q^0 = \frac{1}{2}B$
$Q' = A\mu + \frac{1}{2}B$
$Q'' = Ap'p'' + \frac{1}{2}B(p'q'' + q'p'') + Cq'q''$
$Q''' = Ap''p''' + \frac{1}{2}B(p''q''' + q''p''') + Cq''q''', \ldots$
Because
$p''q' - q''p' = 1, p'''q'' - q'''p'' = -1, p^{iv}q''' - q^{iv}p''' = 1, \ldots$
we shall have the following values,
$\mu \angle \frac{-Q^0 + \frac{1}{2}\sqrt{E}}{P^0}$
$\mu' \angle \frac{-Q' - \frac{1}{2}\sqrt{E}}{P'}$
$\mu'' \angle \frac{-Q'' + \frac{1}{2}\sqrt{E}}{P''}$
$\mu''' \angle \frac{-Q''' - \frac{1}{2}\sqrt{E}}{P'''}, \ldots$
Now, if in the expression of $q''$ we put, for $p''$ and $q''$, their values, $\mu'p' + 1$, and $\mu'$, it will become $\mu'P' + Q'$; also, if we substitute in the expression of $q'''$, for $p'''$ and $q'''$, their values $\mu''p'' + p'$, and $\mu''q'' + q'$, it will be changed into $\mu''P'' + Q''$, and so on; so that we shall have
$Q' = \mu P^0 + Q^0$
$Q'' = \mu' P' + Q'$
$Q''' = \mu'' P'' + Q''$
$Q^{iv} = \mu''' P''' + Q''', \ldots$
Likewise, if we substitute the values of $p''$, and $q''$, in the expression of $P''$, it will become $\mu'^2 P' + 2\mu'Q' + A$; and if we substitute the values of $p'''$, and $q'''$, in the expression of $P'''$,
This text was produced by machine and has not been corrected by hand. Where it matters, read the scan. Confidence is the model's own, not an external check.