Pagina:Werkecarlf03gausrich.djvu/190

tes negativas ipsius ${\textstyle u}$ gignet. Quam ob rem ${\textstyle U^{\prime }}$ nihil aliud erit quam ${\textstyle 2^{n}T^{\prime }}$ per hanc substitutionem transformata, ac perinde ${\textstyle U^{\prime \prime }}$ producta erit ex ${\textstyle 2^{n}T^{\prime \prime }.}$ Nihil itaque intererit, sive in ${\textstyle {\frac {T^{\prime }}{({\frac {\operatorname {d} T}{\operatorname {d} t}})}}}$ substituamus ${\textstyle t=a,}$ sive in ${\textstyle {\frac {U^{\prime }}{({\frac {\operatorname {d} U}{\operatorname {d} u}})}}}$ faciamus ${\textstyle u=b,}$ unde colligimus, ${\textstyle {R},}$ ${\textstyle {R}^{\prime },}$ ${\textstyle {R}^{\prime \prime },}$ ${\textstyle {R}^{\prime \prime \prime }}$ etc. etiam esse valores determinatos functionis ${\textstyle {\frac {U^{\prime }}{({\frac {\operatorname {d} U}{\operatorname {d} u}})}}}$ pro ${\textstyle u=b,}$ ${\textstyle u=b^{\prime },}$ ${\textstyle u=b^{\prime \prime },}$ ${\textstyle u=b^{\prime \prime \prime }}$ etc.

10.

Antequam ulterius progrediamur, haecce praecepta per exemplum illustrabimus. Sit ${\textstyle n=5,}$ statuamusque ${\textstyle a=0,}$ ${\textstyle a^{\prime }={\frac {1}{5}},}$ ${\textstyle a^{\prime \prime }={\frac {2}{5}},}$ ${\textstyle a^{\prime \prime \prime }={\frac {3}{5}},}$ ${\textstyle a^{\prime \prime \prime \prime }={\frac {4}{5}},}$ ${\textstyle a^{\prime \prime \prime \prime \prime }=1.}$ Hinc fit

$$T=t^{6}-3t^{5}+{\frac {17}{5}}t^{4}-{\frac {9}{5}}t^{3}+{\frac {274}{625}}tt-{\frac {24}{625}}t$$

Multiplicando per ${\textstyle t^{-1}+{\frac {1}{2}}t^{-2}+{\frac {1}{3}}t^{-3}+{\frac {1}{4}}t^{-4}+}$ etc. obtinemus

$$T^{\prime }=t^{5}-{\frac {5}{2}}t^{4}+{\frac {67}{30}}t^{3}-{\frac {17}{20}}tt+{\frac {913}{7500}}t-{\frac {19}{7500}}$$

Valores itaque coëfficientium ${\textstyle {R},}$ ${\textstyle {R}^{\prime },}$ ${\textstyle {R}^{\prime \prime },}$ ${\textstyle {R}^{\prime \prime \prime },}$ ${\textstyle {R}^{\prime \prime \prime \prime },}$ ${\textstyle {R}^{\prime \prime \prime \prime \prime }}$ exprimuntur per functionem fractam

$${\frac {t^{5}-{\frac {2}{5}}t^{4}+{\frac {67}{30}}t^{3}-{\frac {17}{20}}tt+{\frac {913}{7500}}t-{\frac {19}{7500}}}{6t^{5}-15t^{4}+{\frac {68}{5}}t^{3}-{\frac {27}{5}}tt+{\frac {548}{625}}t-{\frac {24}{625}}}}$$

in qua pro ${\textstyle t}$ deinceps substituendi sunt valores ${\textstyle 0,}$ ${\textstyle {\frac {1}{5}},}$ ${\textstyle {\frac {2}{5}},}$ ${\textstyle {\frac {3}{5}},}$ ${\textstyle {\frac {4}{5}},}$ ${\textstyle 1.}$ Aliquanto brevior est methodus altera, quae suppeditat ${\textstyle b=-1,}$ ${\textstyle b^{\prime }=-{\frac {3}{5}},}$ ${\textstyle b^{\prime \prime }=-{\frac {1}{5}},}$ ${\textstyle b^{\prime \prime \prime }={\frac {1}{5}},}$ ${\textstyle b^{\prime \prime \prime \prime }={\frac {3}{5}},}$ ${\textstyle b^{\prime \prime \prime \prime \prime }=1}$

$${\begin{aligned}&U=u^{6}-{\frac {7}{5}}u^{4}+{\frac {259}{625}}uu-{\frac {9}{625}}\\&U^{\prime }=u^{5}-{\frac {16}{15}}u^{3}+{\frac {277}{1875}}u\end{aligned}}$$

unde ${\textstyle R,}$ ${\textstyle R^{\prime },}$ ${\textstyle R^{\prime \prime }}$ etc. erunt valores functionis fractae

$${\frac {u^{4}-{\frac {16}{15}}uu+{\frac {277}{1875}}}{6u^{4}-{\frac {28}{5}}uu+{\frac {518}{625}}}}$$

pro ${\textstyle u=-1,}$ ${\textstyle u=-{\frac {3}{5}},}$ ${\textstyle u=-{\frac {1}{5}}}$ etc. Utraque methodus eosdem numeros profert, quos in art. 4. ex Harmonia Mensurarum tradidimus. Ceterum in casu tali, qualem hocce exemplum sistit, ubi ${\textstyle a,}$ ${\textstyle a^{\prime },}$ ${\textstyle a^{\prime \prime }}$ etc. sunt quantitates rationales, valores denominatoris ${\textstyle {\frac {\operatorname {d} T}{\operatorname {d} t}}}$ commodius in forma primitiva computantur, puta ${\textstyle (a-a^{\prime })(a-a^{\prime \prime })(a-a^{\prime \prime \prime })\ldots (a-a^{(n)})}$ pro ${\textstyle t=a}$ ac perinde de reliquis. Idem valet de denominatore ${\textstyle {\frac {\operatorname {d} U}{\operatorname {d} u}},}$ qui pro ${\textstyle u=b}$ fit ${\textstyle =(b-b^{\prime })(b-b^{\prime \prime })(b-b^{\prime \prime \prime })\ldots (b-b^{(n)}).}$