Addition à la solution donnée à la page 353
du présent volume ;
Par
M. Querret, chef d’institution à St-Malo.
≈≈≈≈≈≈≈≈≈
Quelques géomètres pouvant objecter, contre la manière dont nous avons sommé les deux premières suites de la page 353, qu’il n’est peut-être pas permis de traiter les lignes trigonométriques des arcs imaginaires comme celles des arcs réels, nous nous empressons de remplacer le procédé dont nous avons fait usage en cet endroit par un autre qui nous paraît à l’abri de toute objection.
1.o Pour sommer la suite
![{\displaystyle {\frac {a\operatorname {Cos} .x}{1}}-{\frac {a^{3}\operatorname {Cos} .3x}{3}}+{\frac {a^{5}\operatorname {Cos} .5x}{5}}-{\frac {a^{7}\operatorname {Cos} .7x}{7}}+\ldots }](https://wikimedia.org/api/rest_v1/media/math/render/svg/54d768ca7c660aa3b11cfda41c482996a278f257)
nous considérerons que
![{\displaystyle \int {\frac {\operatorname {d} a}{1+a^{2}}}={\frac {a}{1}}-{\frac {a^{3}}{3}}+{\frac {a^{5}}{5}}-{\frac {a^{7}}{7}}+\ldots }](https://wikimedia.org/api/rest_v1/media/math/render/svg/c83c57df762c057bd015a2493a5f44f5ca217387)
d’où il suit que
![{\displaystyle \operatorname {f} (a)=\int {\frac {\operatorname {d} a}{1+a^{2}}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e408a96800639332dad15698f376c8ab8d64e09a)
changeant donc, tour à tour,
en
et
on aura, au moyen de notre théorème général, pour la somme de la suite proposée,
![{\displaystyle {\frac {1}{2}}\left\{\int {\frac {a\operatorname {d} .\left(\operatorname {Cos} .x+{\sqrt {-1}}\operatorname {Sin} .x\right)}{1+a^{2}\left(\operatorname {Cos} .2x+{\sqrt {-1}}\operatorname {Sin} .2x\right)}}+\int {\frac {a\operatorname {d} .\left(\operatorname {Cos} .x-{\sqrt {-1}}\operatorname {Sin} .x\right)}{1+a^{2}\left(\operatorname {Cos} .2x-{\sqrt {-1}}\operatorname {Sin} .2x\right)}}\right\},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d92f11c1e94db1829e4809eb37c6c75369555b0d)
ou, en exécutant les différentiations,
![{\displaystyle {\frac {a}{2}}\int \operatorname {d} x\left\{{\frac {-\operatorname {Sin} .x+{\sqrt {-1}}\operatorname {Cos} .x}{\left(1+a^{2}\operatorname {Cos} .2x\right)+{\sqrt {-1}}a^{2}\operatorname {Sin} .2x}}-{\frac {\operatorname {Sin} .x+{\sqrt {-1}}\operatorname {Cos} .x}{\left(1+a^{2}\operatorname {Cos} .2x\right)-{\sqrt {-1}}a^{2}\operatorname {Sin} .2x}}\right\},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d0c160c691e07bc01e8d2f5be117d2ed0466d28c)
ou bien
![{\displaystyle {\frac {1}{2}}\int -{\frac {2\left(1-a^{2}\right)a\operatorname {d} x\operatorname {Sin} .x}{\left(1-a^{2}\right)+4a^{2}\operatorname {Cos} .^{2}x}}={\frac {1}{2}}\int -{\frac {\frac {2a\operatorname {d} x\operatorname {Sin} .x}{1-a^{2}}}{1+\left({\frac {2a\operatorname {Cos} .x}{1-a^{2}}}\right)^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9768fdaed2e35fba7ff03f349fdcd8cc2ad1aafe)
![{\displaystyle ={\frac {1}{2}}\operatorname {Arc} .\left(\operatorname {Tang} .={\frac {2a\operatorname {Cos} .x}{1-a^{2}}}\right),}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9fe247bb10f55835e9786d719a87b03aa2e6178a)
comme nous l’avions déjà trouvé.
2. Pour sommer la suite
![{\displaystyle {\frac {\operatorname {Cos} .x}{1}}+{\frac {1}{2}}.{\frac {\operatorname {Cos} .3x}{3}}+{\frac {1.3}{2.4}}.{\frac {\operatorname {Cos} .5x}{5}}+{\frac {1.3.5}{2.4.6}}.{\frac {\operatorname {Cos} .7x}{7}}+\ldots ,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/69f2bbabeabaa6514a8971659e96089fc6bf3ee9)
nous considérerons que
![{\displaystyle \int {\frac {\operatorname {d} a}{\sqrt {1-a^{2}}}}={\frac {a}{1}}+{\frac {1}{2}}.{\frac {a^{3}}{3}}+{\frac {1.3}{2.4}}.{\frac {a^{5}}{5}}+{\frac {1.3.5}{2.4.6}}.{\frac {a^{7}}{7}}+\ldots }](https://wikimedia.org/api/rest_v1/media/math/render/svg/8d3bdfce2467321f3982823e592d4cd7c47b5996)
d’où il suit que
![{\displaystyle \operatorname {f} (a)=\int {\frac {\operatorname {d} a}{\sqrt {1-a^{2}}}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/39a0da02648c48f5297cfd5173a969c01a8d306e)
changeant donc, tour à tour,
en
et
on aura, au moyen de notre théorème général, pour la somme de la suite proposée,
![{\displaystyle {\frac {1}{2}}\left\{\int {\frac {\operatorname {d} x\left(-\operatorname {Sin} .x+{\sqrt {-1}}\operatorname {Cos} .x\right)}{\sqrt {1-\operatorname {Cos} .2x-{\sqrt {-1}}\operatorname {Sin} .2x}}}-\int {\frac {\operatorname {d} x\left(\operatorname {Sin} .x+{\sqrt {-1}}\operatorname {Cos} .x\right)}{\sqrt {1-\operatorname {Cos} .2x+{\sqrt {-1}}\operatorname {Sin} .2x}}}\right\},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a6c893dad46a8693309f8dc3ddc720ea76a3746d)
ou encore
![{\displaystyle {\frac {1}{2}}\int \left\{-{\frac {\operatorname {d} x}{\sqrt {2\operatorname {Sin} .x}}}\left({\sqrt {\operatorname {Sin} .x-{\sqrt {-1}}\operatorname {Cos} .x}}+{\sqrt {\operatorname {Sin} .x+{\sqrt {-1}}\operatorname {Cos} .x}}\right)\right\},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a4cd9d935cf047c94c13052cf4c56c73aa9b45a1)
ou bien
![{\displaystyle \int -{\frac {{\frac {1}{2}}\operatorname {d} x}{\sqrt {\operatorname {Sin} .x}}}.{\sqrt {1+\operatorname {Sin} .x}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/95fc462587e03235492b84b5e9885210cde98895)
Or, soit
, on aura
![{\displaystyle {\frac {{\frac {1}{2}}\operatorname {d} x\operatorname {Cos} .x}{\sqrt {\operatorname {Sin} .x}}}=\operatorname {d} t,\qquad {\frac {{\frac {1}{2}}\operatorname {d} x}{\sqrt {\operatorname {Sin} .x}}}={\frac {\operatorname {d} t}{\sqrt {1-t^{4}}}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8c144adb4ca4370c08f0159f59e3480bd14e42a3)
donc
![{\displaystyle \int -{\frac {{\frac {1}{2}}\operatorname {d} x}{\sqrt {\operatorname {Sin} .x}}}.{\sqrt {1+\operatorname {Sin} .x}}=\int -{\frac {\operatorname {d} t}{\sqrt {1-t^{4}}}}{\sqrt {1+t^{2}}}=\int -{\frac {\operatorname {d} t}{\sqrt {1-t^{2}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a4d76f74d6a465c09ba90b6e5e3bd339b9add858)
![{\displaystyle =\operatorname {Arc} .(\operatorname {Cos} .=t)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ba196720d16787d29f9258ee10d2bb9ef979bf12)
donc enfin la somme de la série proposée sera
![{\displaystyle \operatorname {Arc} .\left(\operatorname {Cos} .={\sqrt {\operatorname {Sin} .x}}\right)={\frac {1}{2}}\operatorname {Arc} .(\operatorname {Sin} .=2\operatorname {Sin} .x\operatorname {Cos} .x-1)\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b0bdc07c5587b0089067d9526a18481ae926ef12)
comme nous l’avions également trouvé.