Steps / Historical AI

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 $P'''$,
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