Page:Joseph Louis de Lagrange - Œuvres, Tome 5.djvu/638

Donc : 1^o En différentiant et faisant ensuite $y=\alpha ,$ on aura

$\mathrm {{\frac {dP}{d\alpha }}=N} (\alpha -\beta )(\alpha -\gamma )(\alpha -\delta )\ldots \,;$

2^o Si $\alpha =\beta ,$ on a

$\mathrm {P} =\mathrm {N} (y-\alpha )^{2}(y-\gamma )(y-\delta )\ldots .$

Soit

$\mathrm {P} '=\mathrm {N} (y-\gamma )(y-\delta )\ldots ,$

on aura

$\mathrm {P} =(y-\alpha )^{2}\mathrm {P} '\,;$

donc, différentiant et faisant ensuite $y=\alpha ,$ on aura

$\mathrm {{\frac {dP}{d\alpha }}=0,\quad {\frac {d^{2}P}{d\alpha ^{2}}}=2P',\quad {\frac {d^{3}P}{d\alpha ^{3}}}=2.3{\frac {dP'}{d\alpha }},\quad {\frac {d^{4}P}{d\alpha ^{4}}}=3.4{\frac {d^{2}P'}{d\alpha ^{2}}}} \,;$

et ainsi de suite. D’où l’on tire

$\mathrm {P'={\frac {1}{2}}{\frac {d^{2}P}{d\alpha ^{2}}},\quad {\frac {dP'}{d\alpha }}={\frac {1}{2.3}}{\frac {d^{3}P}{d\alpha ^{3}}},\quad {\frac {d^{2}P'}{d\alpha ^{2}}}={\frac {1}{3.4}}{\frac {d^{4}P}{d\alpha ^{4}}}} ,\ldots \,;$

3^o Si $\alpha =\beta =\gamma ,$ on a

$\mathrm {P=N} (y-\alpha )^{3}(y-\delta )(y-\varepsilon )\ldots .$

Soit

$\mathrm {P''=N} (y-\delta )(y-\varepsilon )\ldots ,$

on aura

$\mathrm {P} =(y-\alpha )^{3}\mathrm {P} ''.$

Différentiant et faisant ensuite $y=\alpha ,$ on aura

$\mathrm {{\frac {dP}{d\alpha }}=0,\ \ {\frac {d^{2}P}{d\alpha ^{2}}}=0,\ \ {\frac {d^{3}P}{d\alpha ^{3}}}=2.3P'',\ \ {\frac {d^{4}P}{d\alpha ^{4}}}=2.3.4{\frac {dP''}{d\alpha }}} ,$

$\mathrm {{\frac {d^{5}P}{d\alpha ^{5}}}=3.4.5{\frac {d^{2}P''}{d\alpha ^{2}}},\ldots ,} \,;$

d’où l’on tire

$\mathrm {P''={\frac {1}{2.3}}{\frac {d^{3}P}{d\alpha ^{3}}},\quad {\frac {dP''}{d\alpha }}={\frac {1}{2.3.4}}{\frac {d^{4}P}{d\alpha ^{4}}},\quad {\frac {d^{2}P''}{d\alpha ^{2}}}={\frac {1}{3.4.5}}{\frac {d^{5}P}{d\alpha ^{5}}}} ,$

$\mathrm {{\frac {d^{3}P''}{d\alpha ^{3}}}={\frac {1}{4.5.6}}{\frac {d^{6}P}{d\alpha ^{6}}}} ,\ldots \,;$

et ainsi de suite.