QUESTIONS RÉSOLUES.
Solution des problèmes d’analise transcendante, proposés
à la page 247 du XIII.e volume du présent recueil ;
Par
M. W. H. Talbot, membre de la société philosophique
de Cambridge.
≈≈≈≈≈≈≈≈≈
I. Pour sommer la série
![{\displaystyle S={\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/164b791ccf965fb046edff3c638ce31f04795c99)
posons
avec la relation
nous aurons, comme l’on sait,
![{\displaystyle \operatorname {Cos} .nx={\frac {1}{2}}\left(u^{n}+v^{n}\right)\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9b8991654288065a459e84456120874290a52b12)
de sorte qu’en posant
![{\displaystyle {\begin{aligned}&U={\frac {au}{1}}-{\frac {a^{3}u^{3}}{3}}+{\frac {a^{5}u^{5}}{5}}-{\frac {a^{7}u^{7}}{7}}+\ldots ,\\\\&V={\frac {av}{1}}-{\frac {a^{3}v^{3}}{3}}+{\frac {a^{5}v^{5}}{5}}-{\frac {a^{7}v^{7}}{7}}+\ldots ,\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2d7bbdd9ab2f623a5d8fdc9a8ac96f901d095575)
nous aurons
![{\displaystyle S={\frac {1}{2}}(U+V)\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/18237edb69fe034599fba3ffc7fcc2ff0e5340d5)
cela donne
![{\displaystyle U=\operatorname {Arc} .(\operatorname {Tang} .=au),\qquad V=\operatorname {Arc} .(\operatorname {Tang} .=av)\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8e2b76be6d4fccb508c03dcf43f4731c64ec74d5)
mais, en renversant le théorème connu,
![{\displaystyle \operatorname {Tang} .(p+q)={\frac {\operatorname {Tang} .p+\operatorname {Tang} .q}{1-\operatorname {Tang} .p\operatorname {Tang} .q}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0cbe59dcaa85178df414eb0d036f234301ccfb21)
on a
![{\displaystyle \operatorname {Arc} .(\operatorname {Tang} .=p')+\operatorname {Arc} .(\operatorname {Tang} .=q')=\operatorname {Arc} .\left(\operatorname {Tang} .={\frac {p'+q'}{1-p'q'}}\right),}](https://wikimedia.org/api/rest_v1/media/math/render/svg/14d1df00a5dc0070db55ade8cddbb7db776d0ed6)
et par suite,
![{\displaystyle \operatorname {Arc} .(\operatorname {Tang} .=au)+\operatorname {Arc} .(\operatorname {Tang} .=av)=\operatorname {Arc} .\left(\operatorname {Tang} .={\frac {a(u+v)}{1-a^{2}uv}}\right)\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/baad5e0905b319abb3cfa8e78bccf9d9b49addd0)
remettant donc
pour
et
pour
il viendra finalement
![{\displaystyle S={\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/a85434e835e8bade812a84acf36ff175873c9459)
II. Pour sommer la seconde série
![{\displaystyle S={\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/6ae5c962687b778a1906d19c3d5a5be6b91928e3)
nous poserons encore
avec la condition
alors, en posant
![{\displaystyle {\begin{aligned}&U=u+{\frac {1}{2}}.{\frac {u^{3}}{3}}+{\frac {1.3}{2.4}}.{\frac {u^{5}}{5}}+{\frac {1.3.5}{2.4.6}}.{\frac {u^{7}}{7}}+\ldots =\operatorname {Arc} .(\operatorname {Sin} .=u),\\\\&V=v+{\frac {1}{2}}.{\frac {v^{3}}{3}}+{\frac {1.3}{2.4}}.{\frac {v^{5}}{5}}+{\frac {1.3.5}{2.4.6}}.{\frac {v^{7}}{7}}+\ldots =\operatorname {Arc} .(\operatorname {Sin} .=v),\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/472b83d590ac47da9d189a8acf4d151166b3d268)
nous aurons
![{\displaystyle S={\frac {1}{2}}(U+V)={\frac {1}{2}}\left\{\operatorname {Arc} .(\operatorname {Sin} .=u)+\operatorname {Arc} .(\operatorname {Sin} .=v)\right\}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/434c5ab9e472ce840a3ae4e8a953f23ac7a5f606)
donc
![{\displaystyle \left.{\begin{aligned}&u=\operatorname {Sin} .U\\&v=\operatorname {Sin} .V\end{aligned}}\right\}\quad \mathrm {d'o{\grave {u}}} \quad \left\{{\begin{aligned}&{\sqrt {1-u^{2}}}=\operatorname {Cos} .U,\\&{\sqrt {1-v^{2}}}=\operatorname {Cos} .V,\end{aligned}}\right.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3dc23f2f96c6b4ba714867c0b2298f509de4e061)
donc aussi
![{\displaystyle \operatorname {Sin} .2S=\operatorname {Sin} .(U+V)=u{\sqrt {1-v^{2}}}+v{\sqrt {1-u^{2}}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d267a6bf7862f45b3ce189b5dcd0f0fd463ee4c7)
ou, en quarrant
![{\displaystyle \operatorname {Sin} .^{2}2S=u^{2}\left(1-v^{2}\right)+v^{2}\left(1-u^{2}\right)+2uv{\sqrt {(1-u^{2})(1-v^{2})}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/011ed6a487a2a25f9130179aa369dc19760742f4)
en développant et observant que
et que conséquemment
![{\displaystyle u^{2}+v^{2}=(u+v)^{2}-2uv=(u+v)^{2}-2=4\operatorname {Cos} .^{2}x-2,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7a0e1ac5d893558691694f5cecff7b2c7775366e)
on trouvera
![{\displaystyle \operatorname {Sin} .^{2}2S=4\operatorname {Cos} .^{2}x-4+2{\sqrt {4-4\operatorname {Cos} .^{2}x}}=4\operatorname {Sin} .x-4\operatorname {Sin} .^{2}x,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a4c95e67908aa66bd26260de59adba8caabc16dd)
donc
![{\displaystyle \operatorname {Sin} .2S=2{\sqrt {\operatorname {Sin} .x-\operatorname {Sin} .^{2}x}}\quad }](https://wikimedia.org/api/rest_v1/media/math/render/svg/3e1a979a5f8fe747c63370f43346abf9cd714300)
et
![{\displaystyle \quad 2S=\operatorname {Arc} .\left(\operatorname {Sin} .=2{\sqrt {\operatorname {Sin} .x-\operatorname {Sin} .^{2}x}}\right),}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0a37272d3f67f3b57126954beac7f1666f64e327)
c’est-à-dire,
[1]
III. Pour sommer la troisième série
![{\displaystyle S={\frac {\operatorname {Cos} .x\operatorname {Cos} .y}{1}}-{\frac {\operatorname {Cos} .2x\operatorname {Cos} .2y}{2}}+{\frac {\operatorname {Cos} .3x\operatorname {Cos} .3y}{3}}-\ldots ,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7457e307cf9cffbe248c0a43be2fa9833433651d)
faisant d’abord, comme ci-dessus,
![{\displaystyle \operatorname {Cos} .x={\frac {1}{2}}(u+v),\qquad uv=1\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a8888146746ce6f6a3e721490cc58cba718982aa)
en posant
![{\displaystyle {\begin{aligned}&U={\frac {u\operatorname {Cos} .y}{1}}-{\frac {u^{2}\operatorname {Cos} .2y}{2}}+{\frac {u^{3}\operatorname {Cos} .3y}{3}}-{\frac {u^{4}\operatorname {Cos} .4y}{4}}+\ldots ,\\\\&V={\frac {v\operatorname {Cos} .y}{1}}-{\frac {v^{2}\operatorname {Cos} .2y}{2}}+{\frac {v^{3}\operatorname {Cos} .3y}{3}}-{\frac {v^{4}\operatorname {Cos} .4y}{4}}+\ldots \,;\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9aeb3752cb54f842f217b8f8c56f4cd2e3385105)
nous aurons
![{\displaystyle S={\frac {1}{2}}(U+V)\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/18237edb69fe034599fba3ffc7fcc2ff0e5340d5)
de sorte que tout se réduira à sommer les séries
et ![{\displaystyle V.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1b2661a49b86bd1a5548e527bbfb068aa9f59978)
Pour y parvenir, nous poserons
![{\displaystyle \operatorname {Cos} .y={\frac {1}{2}}(p+q),\qquad pq=1\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2f4e63cae75f11cb87e6a8517a45746e122f2e06)
alors, en faisant
![{\displaystyle {\begin{aligned}&P={\frac {up}{1}}-{\frac {u^{2}p^{2}}{2}}+{\frac {u^{3}p^{3}}{3}}-{\frac {u^{4}p^{4}}{4}}+\ldots =\operatorname {Log} .(1+up),\\\\&Q={\frac {uq}{1}}-{\frac {u^{2}q^{2}}{2}}+{\frac {u^{3}q^{3}}{3}}-{\frac {u^{4}q^{4}}{4}}+\ldots =\operatorname {Log} .(1+uq),\\\\&P'={\frac {vp}{1}}-{\frac {v^{2}p^{2}}{2}}+{\frac {v^{3}p^{3}}{3}}-{\frac {v^{4}p^{4}}{4}}+\ldots =\operatorname {Log} .(1+vp),\\\\&Q'={\frac {vq}{1}}-{\frac {v^{2}q^{2}}{2}}+{\frac {v^{3}q^{3}}{3}}-{\frac {v^{4}q^{4}}{4}}+\ldots =\operatorname {Log} .(1+vq).\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7f3a30d85ed74e81807ca2a0509de1b0ab6b79f5)
nous aurons
![{\displaystyle {\begin{aligned}&U={\frac {1}{2}}(P+Q)={\frac {1}{2}}\left\{\operatorname {Log} .(1+up)+\operatorname {Log} .(1+uq)\right\},\\&V={\frac {1}{2}}(P'+Q')={\frac {1}{2}}\left\{\operatorname {Log} .(1+vp)+\operatorname {Log} .(1+vq)\right\}\,;\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b6d3d4daa0adb139e5484c93f8a72764117313df)
c’est-à-dire,
![{\displaystyle {\begin{aligned}&U={\frac {1}{2}}\operatorname {Log} .(1+up)(1+uq)={\frac {1}{2}}\operatorname {Log} .\left[1+(p+q)u+u^{2}\right],\\\\&V={\frac {1}{2}}\operatorname {Log} .(1+vp)(1+vq)={\frac {1}{2}}\operatorname {Log} .\left[1+(p+q)v+v^{2}\right],\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c85e41813a6051c6b7de72c2f62b1f74adeec852)
ou encore
![{\displaystyle 2U=\operatorname {Log} .\left(1+2u\operatorname {Cos} .y+u^{2}\right),\qquad 2V=\operatorname {Log} .\left(1+2v\operatorname {Cos} .y+v^{2}\right)\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0845fe95b44743944e7b82cea73842845c91b461)
donc
![{\displaystyle 4S=2U+2V=\operatorname {Log} .\left(1+2u\operatorname {Cos} .y+u^{2}\right)+\operatorname {Log} .\left(1+2v\operatorname {Cos} .y+v^{2}\right)\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/88d9c1bdcaa89fc20cfc430636949b6e781ebedb)
c’est-à-dire,
![{\displaystyle 4S=\operatorname {Log} .\left(1+2u\operatorname {Cos} .y+u^{2}\right)\left(1+2v\operatorname {Cos} .y+v^{2}\right)\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/76888b8512510083be376da4132b784805b028e6)
en développant et se rappelant que
cela donnera
![{\displaystyle 4S=\operatorname {Log} .\left[2+u^{2}+v^{2}+4(u+v)\operatorname {Cos} .y+4\operatorname {Cos} .^{2}y\right]\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ca93cf8094d76ad7db85a7b000deaddac481b159)
mais
![{\displaystyle u+v=2\operatorname {Cos} .x,\qquad u^{2}+v^{2}=(u+v)^{2}-2=4\operatorname {Cos} .^{2}x-2\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/07b7b11954be48ae990de16df46f77e5b7d3899a)
donc
![{\displaystyle 4S=\operatorname {Log} .4\left(\operatorname {Cos} .^{2}x+2\operatorname {Cos} .x\operatorname {Cos} .y+\operatorname {Cos} ^{2}y\right)=\operatorname {Log} .4(\operatorname {Cos} .x+\operatorname {Cos} .y)^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/930f5319381045880b1f9c0c9caa24dd6a67c021)
ou
![{\displaystyle 4S=2\operatorname {Log} .(2\operatorname {Cos} .x+2\operatorname {Cos} .y),}](https://wikimedia.org/api/rest_v1/media/math/render/svg/557fa9ef9640fe4a7bdc56750f1179ad53844318)
d’où finalement
![{\displaystyle S={\frac {1}{2}}\operatorname {Log} .2(\operatorname {Cos} .x+\operatorname {Cos} .y)={\frac {1}{2}}\operatorname {Log} .4\operatorname {Cos} .{\frac {1}{2}}(x+y)\operatorname {Cos} .{\frac {1}{2}}(x-y).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/53c2e76f08befe45d37037ede3e04e512db6ebd6)
Nous ayons donc, en résumé,
1.o ![{\displaystyle {\frac {a\operatorname {Cos} .x}{1}}-{\frac {a^{3}\operatorname {Cos} .3x}{3}}+{\frac {a^{5}\operatorname {Cos} .5x}{5}}-\ldots ={\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/eb22f37b1d8b0b820d2d9da9ce8d75174ffac2c4)
2.o ![{\displaystyle {\frac {\operatorname {Cos} .x}{1}}+{\frac {1}{2}}{\frac {\operatorname {Cos} .3x}{3}}+{\frac {1.3}{2.4}}{\frac {\operatorname {Cos} .5x}{5}}+\ldots ={\frac {1}{2}}\operatorname {Arc} .(\operatorname {Cos} .=2\operatorname {Sin} .x-1),}](https://wikimedia.org/api/rest_v1/media/math/render/svg/eb804c24dc4e0883f21e830da08a85a7f903ee80)
3.o ![{\displaystyle {\frac {\operatorname {Cos} .x\operatorname {Cos} .y}{1}}-{\frac {\operatorname {Cos} .2x\operatorname {Cos} .2y}{2}}+{\frac {\operatorname {Cos} .3x\operatorname {Cos} .3y}{3}}-\ldots }](https://wikimedia.org/api/rest_v1/media/math/render/svg/cfb9de4f94011bc8d54ccb8a66cb49052719d0c0)
![{\displaystyle ={\frac {1}{2}}\operatorname {Log} .4\operatorname {Cos} .{\frac {1}{2}}(x+y)\operatorname {Cos} .{\frac {1}{2}}(x-y).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fdd22f1ba408882d719281e95c12b1f546981088)
Il est aisé de parvenir, en suivant la même marche, à des sommes de séries beaucoup plus compliquées. Nous nous bornerons à en rapporter deux exemples, en nous dispensant de développer les calculs qui sont en tout semblables à ceux qu’on a vu ci-dessus.
On trouve, en premier lieu,
![{\displaystyle {\frac {a\operatorname {Cos} .x\operatorname {Cos} .y}{1}}-{\frac {a^{3}\operatorname {Cos} .3x\operatorname {Cos} .3y}{3}}+{\frac {a^{5}\operatorname {Cos} .5x\operatorname {Cos} .5y}{5}}-\ldots }](https://wikimedia.org/api/rest_v1/media/math/render/svg/440041b55761b1bf28b8906d709e934166e59f9c)
![{\displaystyle ={\frac {1}{4}}\operatorname {Arc} .\left\{\operatorname {Tang} .={\frac {4a\left(1-a^{2}\right)\operatorname {Cos} .x\operatorname {Cos} .y}{\left(1+a^{2}\right)-4a^{2}\left(\operatorname {Cos} .^{2}x\operatorname {Cos} .^{2}y\right)}}\right\}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/01bb8ff14de464a515f57e47f33fbd6f9039bab9)
Dans le cas de
ou
cette somme devient celle de la première de nos trois séries, quoique sous une forme un peu différente.
On trouve, en second lieu,
![{\displaystyle {\frac {a\operatorname {Cos} .x}{1}}+{\frac {1}{2}}{\frac {a^{3}\operatorname {Cos} .3x}{3}}+{\frac {1.3}{2.4}}{\frac {a^{5}\operatorname {Cos} .5x}{5}}+\ldots }](https://wikimedia.org/api/rest_v1/media/math/render/svg/b4aa5a10cca02a78a8f213bd6c84964697c8ec52)
![{\displaystyle ={\frac {1}{2}}\operatorname {Arc} .\left(\operatorname {Cos} .={\sqrt {\left(1+a^{2}\right)^{2}-4a^{2}\operatorname {Cos} .^{2}x}}-a^{2}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3a19d44eb41894287493169875156b31f90f9940)
qui se change immédiatement dans la seconde série, lorsqu’on, fait ![{\displaystyle a=1.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c87aaf9fe1b82ddb0ba04b257c04f6d0eb4697a5)
Si, dans ces diverses séries, on attribue à
des valeurs particulières, on en fera dériver une multitude d’autres plus ou moins remarquables. La troisième, par exemple, en supposant
devient
![{\displaystyle {\frac {\operatorname {Cos} ^{2}.x}{1}}-{\frac {\operatorname {Cos} .^{2}2x}{2}}+{\frac {\operatorname {Cos} .^{2}3x}{3}}-\ldots ={\frac {1}{2}}\operatorname {Log} .4\operatorname {Cos} .x.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ccf2bf84249688e54d39c823e18f622996856059)
On trouve, plus généralement,
![{\displaystyle {\frac {a\operatorname {Cos} .^{2}x}{1}}-{\frac {a^{2}\operatorname {Cos} .^{2}2x}{2}}+{\frac {a^{3}\operatorname {Cos} .^{2}3x}{3}}-\ldots }](https://wikimedia.org/api/rest_v1/media/math/render/svg/240fc0e072557404a3a74f8e5c3076d20f7a40d0)
![{\displaystyle ={\frac {1}{4}}\operatorname {Log} .\left\{\left(1-a^{2}\right)^{2}+4a\left(1+a\right)^{2}\operatorname {Cos} .^{2}x\right\}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/475b9e53bc2fcaab0c7552d5ffa82d95a52b78e1)
Rome (mai 1823).