soit le coefficient de
dans le développement de cette fonction. En repassant des coefficients aux fonctions génératrices, l’équation précédente donnera
![{\displaystyle u=u\left({\frac {pt'}{t}}+qtt'\right),}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3e45c39f2affa3db500360a652b277db4ecf17e0)
d’où l’on tire
![{\displaystyle 1={\frac {pt'}{t}}+qtt'\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9f072f13d6140b5cc3df8440edc0016a6109a5f3)
par conséquent,
![{\displaystyle {\frac {1}{t}}={\frac {1}{2pt'}}={\frac {\sqrt {{\frac {1}{t'^{2}}}-4pq}}{2p}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bcf0a78e61f0e3f32bd5054dad91d4aeb514108f)
ce qui donne
![{\displaystyle {\frac {1}{t^{x}}}={\frac {1}{(2p)^{x}}}\left({\frac {1}{t'}}\pm {\sqrt {{\frac {1}{t'^{2}}}-4pq}}\right)^{x}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5a402b8ea875659dd71868cbb28695934a19f246)
donc
![{\displaystyle {\frac {u}{t^{x}t'^{x'}}}={\frac {u}{(2p)^{x}t'^{x'}}}\left({\frac {1}{t'}}\pm {\sqrt {{\frac {1}{t'^{2}}}-4pq}}\right)^{x}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/864fb18d7154d94271df1f387e71a84183e5db70)
Cette équation peut être mise sous la forme suivante :
![{\displaystyle {\frac {u}{t^{x}t'^{x'}}}={\frac {u}{2(2p)^{x}t'^{x'}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4f1bada52f4c74c79b816d0c4dc7649662e7287b)
![{\displaystyle \times \left\{{\begin{aligned}&\left({\frac {1}{t'}}+{\sqrt {{\frac {1}{t'^{2}}}-4pq}}\right)^{x}+\left({\frac {1}{t'}}-{\sqrt {{\frac {1}{t'^{2}}}-4pq}}\right)^{x}\\&\pm {\sqrt {{\frac {1}{t'^{2}}}-4pq}}{\frac {\left({\cfrac {1}{t'}}+{\sqrt {{\cfrac {1}{t'^{2}}}-4pq}}\right)^{x}-\left({\cfrac {1}{t'}}-{\sqrt {{\cfrac {1}{t'^{2}}}-4pq}}\right)^{x}}{\sqrt {{\cfrac {1}{t'^{2}}}-4pq}}}\end{aligned}}\right.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/97cbb02f5eeae395f67468e26d57733263e01fca)
L’expression précédente de
donne
![{\displaystyle \pm {\sqrt {{\frac {1}{t'^{2}}}-4pq}}={\frac {2p}{t}}-{\frac {1}{t'}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/878164286ef966ac8f839372d0afb39c3c911edb)
on a donc
![{\displaystyle {\begin{aligned}&{\frac {u}{t^{x}t'^{x'}}}={\frac {u}{2(2p)^{x}t'^{x'}}}\left[\left({\frac {1}{t'}}+{\sqrt {{\frac {1}{t'^{2}}}-4pq}}\right)^{x}+\left({\frac {1}{t'}}-{\sqrt {{\frac {1}{t'^{2}}}-4pq}}\right)^{x}\right]\\&\quad +{\frac {u\left({\cfrac {1}{t}}-{\cfrac {1}{2pt'}}\right)}{2(2p)^{x-1}t'^{x'}}}{\frac {\left({\cfrac {1}{t'}}+{\sqrt {{\cfrac {1}{t'^{2}}}-4pq}}\right)^{x}-\left({\cfrac {1}{t'}}-{\sqrt {{\cfrac {1}{t'^{2}}}-4pq}}\right)^{x}}{\sqrt {{\cfrac {1}{t'^{2}}}-4pq}}}\,;\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9998820b5c2768f63ae540508372150673a1ccac)
sous cette forme, l’ambiguïté du signe
disparaît.