Page:Joseph Louis de Lagrange - Œuvres, Tome 9.djvu/19

4. Soit, par exemple, $f(x)={\frac {1}{x}}\,;$ on aura

$f(x+i)={\frac {1}{x+i}}\,;$

donc

${\begin{alignedat}{3}i\mathrm {P} =&\qquad {\frac {1}{x+i}}\quad -\ {\frac {1}{x}}\ =-{\frac {i}{x(x+i)}},&\mathrm {P} =&-{\frac {1}{x(x+i)}},&p=&-{\frac {1}{x^{2}}}\,;\\i\mathrm {Q} =&-\ {\frac {1}{x(x+i)}}\ +{\frac {1}{x^{2}}}=\quad {\frac {i}{x^{2}(x+i)}},&\mathrm {Q} =&\quad \ {\frac {1}{x^{2}(x+i)}},&q=&\quad \ {\frac {1}{x^{3}}}\,;\\i\mathrm {R} =&\quad \ {\frac {1}{x^{2}(x+i)}}-{\frac {1}{x^{3}}}=-{\frac {i}{x^{3}(x+i)}},\quad &\mathrm {R} =&-{\frac {1}{x^{3}(x+i)}},\quad &r=&-{\frac {1}{x^{4}}}\,;\\\end{alignedat}}$

$\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \,;$

ainsi l’on aura

${\frac {1}{x+i}}={\frac {1}{x}}-{\frac {i}{x(x+i)}}={\frac {1}{x}}-{\frac {i}{x^{2}}}+{\frac {i^{2}}{x^{2}(x+i)}}$

$={\frac {1}{x}}-{\frac {i}{x^{2}}}+{\frac {i^{2}}{x^{3}}}-{\frac {i}{x^{3}(x+i)}}=\ldots \,;$

comme il résulte de la division actuelle.

Prenons encore pour exemple la fonction irrationnelle ${\sqrt {x}}.$ On aura donc

$f(x)={\sqrt {x}},f(x+i)={\sqrt {x+i}}={\sqrt {x}}+i\mathrm {P} \,;$

donc

$i\mathrm {P} ={\sqrt {x+i}}-{\sqrt {x}}={\frac {i}{{\sqrt {x+i}}+{\sqrt {x}}}},\quad \mathrm {P} ={\frac {1}{{\sqrt {x+i}}+{\sqrt {x}}}},\quad p={\frac {1}{2{\sqrt {x}}}},$

${\begin{aligned}i\mathrm {Q} =\mathrm {P} -p&={\frac {1}{{\sqrt {x+i}}+{\sqrt {x}}}}-{\frac {1}{2{\sqrt {x}}}}\\&={\frac {{\sqrt {x}}-{\sqrt {x+i}}}{2{\sqrt {x}}\left({\sqrt {x+i}}+{\sqrt {x}}\right)}}=-{\frac {i}{2{\sqrt {x}}\left({\sqrt {x+i}}+{\sqrt {x}}\right)^{2}}},\end{aligned}}$

$\ \,\mathrm {Q} =-{\frac {1}{2{\sqrt {x}}\left({\sqrt {x+i}}+{\sqrt {x}}\right)}}=-{\frac {1}{8x{\sqrt {x}}}}$

${\begin{aligned}i\mathrm {R} =\mathrm {Q} -q&={\frac {1}{2{\sqrt {x}}}}\left[{\frac {1}{4x}}-{\frac {1}{\left({\sqrt {x+i}}+{\sqrt {x}}\right)^{2}}}\right]\\&={\frac {1}{2{\sqrt {x}}}}{\frac {i-2x+2{\sqrt {x}}{\sqrt {x+i}}}{4x\left({\sqrt {x+i}}+{\sqrt {x}}\right)^{2}}}={\frac {1}{2{\sqrt {x}}}}{\frac {{\sqrt {x+i}}+3{\sqrt {x}}}{4x\left({\sqrt {x+i}}+{\sqrt {x}}\right)^{3}}}\end{aligned}}$

$\ \,\mathrm {R} ={\frac {{\sqrt {x+i}}+3{\sqrt {x}}}{8x{\sqrt {x}}\left({\sqrt {x+i}}+{\sqrt {x}}\right)^{3}}},\quad r={\frac {1}{16x^{2}{\sqrt {x}}}},$

$\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .$