Page:Annales de mathématiques pures et appliquées, 1818-1819, Tome 9.djvu/269

263

FRACTIONS CONTINUES.

$1+a{\frac {z}{1}}+a(a+k){\frac {z^{2}}{1.2}}+a(a+k)(a+2k){\frac {z^{3}}{1.2.3}}+\ldots$

qu’il soit question de développer en fraction continue ; pour procéder à ce développement, nous emploîrons la méthode indiquée par Euler ; c’est-à-dire que nous poserons successivement

$1+a{\frac {z}{1}}+a(a+k){\frac {z^{2}}{1.2}}+a(a+k)(a+2k){\frac {z^{3}}{1.2.3}}$

$+a(a+k)(a+2k)(a+3k){\frac {z^{4}}{1.2.3.4}}+\ldots ={\frac {1}{1-A}},$

$A=az.$

${\frac {1+{\frac {1}{2}}(a+k){\frac {z}{1}}+{\frac {1}{3}}(a+k)(a+2k){\frac {z^{2}}{1.2}}+{\frac {1}{4}}(a+k)(a+2k)(a+3k){\frac {z^{3}}{1.2.3}}+\ldots }{1+a{\frac {z}{1}}+a(a+k){\frac {z^{2}}{1.2}}+a(a+k)(a+2k){\frac {z^{3}}{1.2.3}}+a(a+k)(a+2k)(a+3k){\frac {z^{4}}{1.2.3.4}}+\ldots }}$

$={\frac {az}{1+B}},$

$B=(a-k)z.$

${\frac {1+{\frac {2}{3}}(a+k){\frac {z}{1}}+{\frac {2}{4}}(a+k)(a+2k){\frac {z^{2}}{1.2}}+{\frac {2}{5}}(a+k)(a+2k)(a+3k){\frac {z^{3}}{1.2.3}}+\ldots }{2+{\frac {1.2}{2}}(a+k){\frac {z}{1}}+{\frac {1.2}{3}}(a+k)(a+2k){\frac {z^{2}}{1.2}}+{\frac {1.2}{4}}(a+k)(a+2k)(a+3k){\frac {z^{3}}{1.2.3}}+\ldots }}$

$={\frac {(a-k)z}{2-C}},$

$C=(a+k)z.$

${\frac {1+{\frac {2.3}{3.4}}(a+k){\frac {z}{1}}+{\frac {2.3}{4.5}}(a+2k)(a+3k){\frac {z^{2}}{1.2}}+{\frac {2.3}{5.6}}(a+2k)(a+3k)(a+4k){\frac {z^{3}}{1.2.3}}+\ldots }{3+{\frac {2.3}{3}}(a+k){\frac {z}{1}}+{\frac {2.3}{4}}(a+k)(a+2k){\frac {z^{2}}{1.2}}+{\frac {2.3}{5}}(a+k)(a+2k)(a+3k){\frac {z^{3}}{1.2.3}}+\ldots }}$

$={\frac {(a+k)z}{3+D}},$

$D=2(a-2k)z.$

${\frac {1+{\frac {3.4}{4.5}}(a+k){\frac {z}{1}}+{\frac {3.4}{5.6}}(a+2k)(a+3k){\frac {z^{2}}{1.2}}+{\frac {3.4}{6.7}}(a+2k)(a+3k)(a+4k){\frac {z^{3}}{1.2.3}}+\ldots }{4+{\frac {2.3.4}{3.4}}(a+2k){\frac {z}{1}}+{\frac {2.3.4}{4.5}}(a+2k)(a+3k){\frac {z^{2}}{1.2}}+{\frac {2.3.4}{5.6}}(a+2k)(a+3k)(a+4k){\frac {z^{3}}{1.2.3}}+\ldots }}$

$={\frac {2(a-2k)z}{4-E}},$

$E=2(a+2k)z.{\frac {1+{\frac {3.4.5}{4.5.6}}(a+3k){\frac {z}{1}}+{\frac {3.4.5}{5.6.7}}(a+3k)(a+4k){\frac {z^{2}}{1.2}}+\ldots }{5+{\frac {3.4.5}{4.5}}(a+2k){\frac {z}{1}}+{\frac {3.4.5}{5.6}}(a+2k)(a+3k){\frac {z^{2}}{1.2}}+\ldots }}={\frac {2(a+2k)z}{5+F}},$

$F=3(a-3k)z.{\frac {1+{\frac {4.5.6}{5.6.7}}(a+3k){\frac {z}{1}}+{\frac {4.5.6}{6.7.8}}(a+3k)(a+4k){\frac {z^{2}}{1.2}}+\ldots }{6+{\frac {3.4.5.6}{4.5.6}}(a+3k){\frac {z}{1}}+{\frac {3.4.5.6}{5.6.7}}(a+3k)(a+4k){\frac {z^{2}}{1.2}}+\ldots }}={\frac {3(a-3k)z}{6-G}},$

$G=3(a+3k)z.{\frac {1+{\frac {4.5.6.7}{5.6.7.8}}(a+4k){\frac {z}{1}}+{\frac {4.5.6.7}{6.7.8.9}}(a+4k)(a+5k){\frac {z^{2}}{1.2}}+\ldots }{7+{\frac {4.5.6.7}{5.6.7}}(a+3k){\frac {z}{1}}+{\frac {4.5.6.7}{6.7.8}}(a+3k)(a+4k){\frac {z^{2}}{1.2}}+\ldots }}={\frac {3(a+3k)z}{7+H}},$