Page:Laplace - Œuvres complètes, Gauthier-Villars, 1878, tome 9.djvu/333

pourvu que l’on suppose $\alpha =0$ après les différentiations ; on a

(6)

${\frac {\alpha }{e^{\alpha }-1}}={\frac {c{\frac {\alpha }{2}}}{e^{\frac {\alpha }{2}}-1}}-{\frac {\cfrac {\alpha }{2}}{e^{\frac {\alpha }{2}}+1}}\,;$

de plus, il est clair que, si l’on suppose $\alpha =0$ après les différentiations, on a

${\frac {d^{n}{\cfrac {\cfrac {\alpha }{2}}{e^{\frac {\alpha }{2}}\pm 1}}}{\left({\cfrac {d\alpha }{2}}\right)^{n}}}={\frac {d^{n}{\cfrac {\alpha }{e^{\alpha }\pm 1}}}{d\alpha ^{n}}},$

ce qui donne

${\frac {d^{n}{\cfrac {\cfrac {\alpha }{2}}{e^{\frac {\alpha }{2}}\pm 1}}}{d\alpha ^{n}}}={\frac {1}{2^{n}}}{\frac {d^{n}{\cfrac {\alpha }{e^{\alpha }\pm 1}}}{d\alpha ^{n}}}\,;$

l’équation (6) donnera, par conséquent,

${\frac {d^{n}{\cfrac {\alpha }{e^{\alpha }-1}}}{d\alpha ^{n}}}={\frac {1}{2^{n}}}{\frac {d^{n}{\cfrac {\alpha }{e^{\alpha }-1}}}{d\alpha ^{n}}}-{\frac {1}{2^{n}}}{\frac {d^{n}{\cfrac {\alpha }{e^{\alpha }+1}}}{d\alpha ^{n}}}.$

Partant, si l’on suppose toujours $\alpha =0$ après les différentiations, on aura

${\frac {d^{n}{\cfrac {\alpha }{e^{\alpha }-1}}}{d\alpha ^{n}}}=-{\frac {d^{n}{\cfrac {\alpha }{e^{\alpha }+1}}}{\left(2^{n}-1\right)d\alpha ^{n}}}=-{\frac {nd^{n-1}{\cfrac {1}{e^{\alpha }+1}}}{\left(2^{n}-1\right)d\alpha ^{n-1}}},$

parce que

$d^{n}{\frac {\alpha }{e^{\alpha }+1}}=\alpha d^{n}{\frac {1}{e^{\alpha }+1}}+nd\alpha d^{n-1}{\frac {1}{e^{\alpha }+1}}=nd\alpha d^{n-1}{\frac {1}{e^{\alpha }+1}},$

lorsque l’on suppose $\alpha =0\,;$ donc

$q_{n-1}=-{\frac {d^{n-1}{\cfrac {1}{e^{\alpha }+1}}}{1.2.3\ldots (n-1)\left(2^{n-1}-1\right)d\alpha ^{n-1}}}.$