Page:Joseph Louis de Lagrange - Œuvres, Tome 4.djvu/550

c’est-à-dire, en réduisant,

${\begin{aligned}{\frac {f''}{f'}}&={\frac {1}{\alpha '}}{\frac {\varpi '-x(1+\alpha ')}{\varpi ''-(\varpi '-x)(1+\alpha '')}},\\{\frac {f'''}{f''}}&={\frac {1}{\alpha ''}}{\frac {\varpi ''-(\varpi '-x)(1+\alpha '')}{\varpi ''-(\varpi ''-\varpi '+x)(1+\alpha ''')}},\\{\frac {f^{\scriptscriptstyle {\text{IV}}}}{f'''}}&={\frac {1}{\alpha '''}}{\frac {\varpi '''-(\varpi '''-\varpi ''+x)(1+\alpha ''')}{\varpi ^{\scriptscriptstyle {\text{IV}}}-(\varpi '''-\varpi ''+\varpi '-x)\left(1+\alpha ^{\scriptscriptstyle {\text{IV}}}\right)}},\\\ldots &\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots ,\end{aligned}}$

Donc, à cause de $f'={\frac {h}{1+\alpha '}},$ on aura, en substituant successivement les valeurs de $f',f'',\ldots ,$

${\begin{aligned}f'\ \ &={\frac {h}{1}}+{\alpha '}{\frac {\varpi '-x(1+\alpha ')}{\varpi '-x(1+\alpha ')}},\\f''\ &={\frac {h}{\alpha '(1+\alpha ')}}{\frac {\varpi '-x(1+\alpha ')}{\varpi ''-(\varpi '-x)(1+\alpha '')}},\\f'''&={\frac {h}{\alpha '\alpha ''(1+\alpha ')}}{\frac {\varpi '-x(1+\alpha ')}{\varpi '''-(\varpi ''-\varpi '+x)\left(1+\alpha '''\right)}},\\f^{\scriptscriptstyle {\text{IV}}}&={\frac {h}{\alpha '\alpha ''\alpha '''(1+\alpha ')}}{\frac {\varpi '-x(1+\alpha ')}{\varpi ^{\scriptscriptstyle {\text{IV}}}-(\varpi '''-\varpi ''+\varpi '-x)\left(1+\alpha ^{\scriptscriptstyle {\text{IV}}}\right)}},\\\ldots &\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .\end{aligned}}$

Substituant enfin ces valeurs dans les expressions de $h',h'',\ldots$ et réduisant, on aura

${\begin{aligned}h'\ \ &={\frac {h}{\alpha '(1+\alpha ')}}{\frac {\left[\varpi '-x(1+\alpha ')\right]\left[(1+\alpha ')\varpi ''-\alpha '(1+\alpha '')\varpi '\right]}{\left[\varpi '-x(1+\alpha ')\right]\left[\varpi ''-(\varpi '-x)(1+\alpha '')\right]}},\\h''\ &={\frac {h}{\alpha '\alpha ''(1+\alpha ')}}\\&\qquad \times {\frac {\left[\varpi '-x(1+\alpha ')\right]\left[(1+\alpha '')\varpi '''-\alpha ''(1+\alpha ''')\varpi ''\right]}{\left[\varpi ''-(\varpi '-x)(1+\alpha '')\right]\left[\varpi '''-(\varpi ''-\varpi '+x)(1+\alpha ''')\right]}},\\h'''&={\frac {h}{\alpha '\alpha ''\alpha '''(1+\alpha ')}}\\&\qquad \times {\frac {\left[\varpi '-x(1+\alpha ')\right]\left[(1+\alpha ''')\varpi ^{\scriptscriptstyle {\text{IV}}}-\alpha '''(1+\alpha ^{\scriptscriptstyle {\text{IV}}})\varpi '''\right]}{\left({\begin{aligned}&\left[\varpi '''-(\varpi ''-\varpi '+x)(1+\alpha ''')\right]\\&\qquad \qquad \times \left[\varpi ^{\scriptscriptstyle {\text{IV}}}-(\varpi '''-(\varpi '''-\varpi ''+\varpi '-x)(1+\alpha ^{\scriptscriptstyle {\text{IV}}})\right]\end{aligned}}\right)}},\\\ldots &\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .\end{aligned}}$

Ainsi les distances focales $f',f'',\ldots$ et les intervalles entre les len-