Développements limités

Développements limités d'une réciproque

On pose :

f(x)=\sum _{n=0}^{\infty }t_{n}\left(x-x_{0}\right)^{n}=\sum _{n=0}^{\infty }{\frac {I_{1}\left(y_{0}\right)}{n!}}\left(x-x_{0}\right)^{n}\quad {\text{et}}\quad f(x_{0})=y_{0}

{\begin{aligned}I_{n}&=f^{(n)}\circ f^{-1}(y)&=&\;\left(\partial _{x}^{n}f\right)\circ f^{-1}(y)\\J_{n}&=f^{-1(n)}(y)&=&\;\partial _{y}^{n}f^{-1}(y)\end{aligned}}

On a :

I_{n}'={\frac {I_{n+1}}{I_{n}}}=\partial _{y}I_{n}

Donc :

{\begin{aligned}J_{1}&={\frac {1}{I_{1}}}\\J_{2}&={\frac {-I_{2}}{I_{1}^{3}}}\\J_{3}&={\frac {-I_{3}I_{1}+3\cdot I_{2}I_{2}}{I_{1}^{5}}}\\J_{4}&={\frac {-I_{4}I_{1}I_{1}-I_{3}I_{2}I_{1}+5\cdot I_{3}I_{1}I_{2}+3\cdot I_{3}I_{2}I_{1}+3\cdot I_{2}I_{3}I_{1}-5\cdot 3\cdot I_{2}I_{2}I_{2}}{I_{1}^{7}}}\\J_{5}&={\frac {\left[{\begin{aligned}-&I_{5}I_{1}I_{1}I_{1}&-&I_{4}I_{2}I_{1}I_{1}&-&I_{4}I_{1}I_{2}I_{1}&+7\cdot &I_{4}I_{1}I_{1}I_{2}&-&I_{4}I_{2}I_{1}I_{1}&-&I_{3}I_{3}I_{1}I_{1}&-&I_{3}I_{2}I_{2}I_{1}&+7\cdot &I_{3}I_{2}I_{1}I_{2}&\\+5\cdot &I_{4}I_{1}I_{2}I_{1}&+5\cdot &I_{3}I_{2}I_{2}I_{1}&+5\cdot &I_{3}I_{1}I_{3}I_{1}&-7\cdot 5\cdot &I_{3}I_{1}I_{2}I_{2}&+3\cdot &I_{4}I_{2}I_{1}I_{1}&+3\cdot &I_{3}I_{3}I_{1}I_{1}&+3\cdot &I_{3}I_{2}I_{2}I_{1}&-7\cdot 3\cdot &I_{3}I_{2}I_{1}I_{2}&\\+3\cdot &I_{3}I_{3}I_{1}I_{1}&+3\cdot &I_{2}I_{4}I_{1}I_{1}&+3\cdot &I_{2}I_{3}I_{2}I_{1}&-7\cdot 3\cdot &I_{2}I_{3}I_{1}I_{2}&-5\cdot 3\cdot &I_{3}I_{2}I_{2}I_{1}&-5\cdot 3\cdot &I_{2}I_{3}I_{2}I_{1}&-5\cdot 3\cdot &I_{2}I_{2}I_{3}I_{1}&+7\cdot 5\cdot 3\cdot &I_{2}I_{2}I_{2}I_{2}&\end{aligned}}\right]}{I_{1}^{9}}}\end{aligned}}

Ainsi, sous réserve de convergence, on a :

f^{-1}(y)=\sum _{n=0}^{\infty }{\frac {J_{n}\left(y_{0}\right)}{n!}}\left(y-y_{0}\right)^{n}=\sum _{n=0}^{\infty }{\frac {{\underset {i=1}{\overset {n-1}{S}}}\sum \limits _{a_{i}=1}^{i}\prod \limits _{j=1}^{n-1}\left(1-2j\right)^{a_{j}=j}\;I_{1+\left(a_{j}=j\right)}\left(y_{0}\right)}{n!\;I_{1}\left(y_{0}\right)^{2n-1}}}\left(y-y_{0}\right)^{n}=\sum _{n=0}^{\infty }{\frac {{\underset {i=1}{\overset {n-1}{S}}}\sum \limits _{a_{i}=1}^{i}\prod \limits _{j=1}^{n-1}\left(1-2j\right)^{a_{j}=j}\;t_{1+\left(a_{j}=j\right)}\left[1+\left(a_{j}=j\right)\right]!}{n!\;t_{1}^{2n-1}}}\left(y-y_{0}\right)^{n}

Puisqu'on a :

{\frac {1}{\sqrt {1-x}}}=\sum \limits _{n=0}^{\infty }{\frac {C_{2n}^{n}}{2^{2n}}}x^{n}

{\sqrt {1-x}}=\sum \limits _{n=0}^{\infty }{\frac {C_{2n}^{n}}{\left(1-2n\right)2^{2n}}}x^{n}

\int _{\theta =0}^{\varphi }\sin ^{2n}\theta \,\mathrm {d} \theta =\sum \limits _{j=0}^{n-1}{\frac {\left(-1\right)^{n+j+1}C_{2n}^{j}\sin \left[\left(2n-2j\right)\varphi \right]}{2^{2n-1}\left(2n-2j\right)}}+{\frac {C_{2n}^{n}}{2^{2n}}}\varphi

on a :

{\begin{aligned}F(\varphi ,k)&=\int \limits _{\theta =0}^{\varphi }{\frac {1}{\sqrt {1-k^{2}\sin ^{2}\theta }}}\,\mathrm {d} \theta \quad \;=\sum \limits _{n=0}^{\infty }\left[\left[{\frac {C_{2n}^{n}}{2^{2n}}}\right]^{2}\sum \limits _{j=0}^{n-1}{\frac {\left(-1\right)^{n+j+1}n!^{2}\sin \left[\left(2n-2j\right)\varphi \right]}{\left(n-j\right)\left(2n-j\right)!j!}}+\varphi \right]k^{2n}\\E(\varphi ,k)&=\int \limits _{\theta =0}^{\varphi }{\sqrt {1-k^{2}\sin ^{2}\theta }}\,\mathrm {d} \theta \quad \;\;=\sum \limits _{n=0}^{\infty }\left[\left[{\frac {C_{2n}^{n}}{2^{2n}}}\right]^{2}{\frac {1}{1-2n}}\sum \limits _{j=0}^{n-1}{\frac {\left(-1\right)^{n+j+1}n!^{2}\sin \left[\left(2n-2j\right)\varphi \right]}{\left(n-j\right)\left(2n-j\right)!j!}}+\varphi \right]k^{2n}\\\quad K(k)&=\int _{\theta =0}^{\pi /2}{\frac {1}{\sqrt {1-k^{2}\sin ^{2}\theta }}}\,\mathrm {d} \theta ={\frac {\pi }{2}}\sum _{n=0}^{\infty }\left[{\frac {C_{2n}^{n}}{2^{2n}}}\right]^{2}k^{2n}\\\quad E(k)&=\int \limits _{\theta =0}^{\pi /2}{\sqrt {1-k^{2}\sin ^{2}\theta }}\,\mathrm {d} \theta \quad \;={\frac {\pi }{2}}\sum _{n=0}^{\infty }{\frac {1}{1-2n}}\left[{\frac {C_{2n}^{n}}{2^{2n}}}\right]^{2}k^{2n}\end{aligned}}

Si on veut avoir :

F(\varphi ,k)=K(k_{c})

E(\varphi ,k)=E(k_{c})

au moyen de :

{\begin{aligned}k(K)&={\sqrt {\sum _{n=0}^{\infty }{\frac {2^{2n-2}}{n!\;\pi ^{n}}}{\underset {i=1}{\overset {n-1}{S}}}\sum \limits _{a_{i}=1}^{i}\prod \limits _{j=1}^{n-1}\left(1-2j\right)^{a_{j}=j}\left[{\frac {C_{2+2\left(a_{j}=j\right)}^{1+\left(a_{j}=j\right)}}{2^{2+2\left(a_{j}=j\right)}}}\right]^{2}\left[1+\left(a_{j}=j\right)\right]!\left(K-{\frac {\pi }{2}}\right)^{n}}}\\k(E)&={\sqrt {\sum _{n=0}^{\infty }{\frac {-2^{2n-2}}{n!\;\pi ^{n}}}{\underset {i=1}{\overset {n-1}{S}}}\sum \limits _{a_{i}=1}^{i}\prod \limits _{j=1}^{n-1}\left(1-2j\right)^{a_{j}=j}\;{\frac {1}{1-2n}}\left[{\frac {C_{2+2\left(a_{j}=j\right)}^{1+\left(a_{j}=j\right)}}{2^{2+2\left(a_{j}=j\right)}}}\right]^{2}\left[1+\left(a_{j}=j\right)\right]!\left(E-{\frac {\pi }{2}}\right)^{n}}}\end{aligned}}

on aura :

{\begin{aligned}k_{c}(K)&={\sqrt {\sum _{n=0}^{\infty }{\frac {2^{2n-2}}{n!\;\pi ^{n}}}{\underset {i=1}{\overset {n-1}{S}}}\sum \limits _{a_{i}=1}^{i}\prod \limits _{j=1}^{n-1}\left(1-2j\right)^{a_{j}=j}\left[1+\left(a_{j}=j\right)\right]!\left[{\frac {C_{2+2\left(a_{j}=j\right)}^{1+\left(a_{j}=j\right)}}{2^{2+2\left(a_{j}=j\right)}}}\right]^{2}\left(\sum \limits _{n=0}^{\infty }\left[\left[{\frac {C_{2n}^{n}}{2^{2n}}}\right]^{2}\sum \limits _{j=0}^{n-1}{\frac {\left(-1\right)^{n+j+1}n!^{2}\sin \left[\left(2n-2j\right)\varphi \right]}{\left(n-j\right)\left(2n-j\right)!j!}}+\varphi \right]k^{2n}-{\frac {\pi }{2}}\right)^{n}}}\\k_{c}(E)&={\sqrt {\sum _{n=0}^{\infty }{\frac {-2^{2n-2}}{n!\;\pi ^{n}}}{\underset {i=1}{\overset {n-1}{S}}}\sum \limits _{a_{i}=1}^{i}\prod \limits _{j=1}^{n-1}{\frac {\left(1-2j\right)^{a_{j}=j}\;\left[1+\left(a_{j}=j\right)\right]!}{1-2n}}\left[{\frac {C_{2+2\left(a_{j}=j\right)}^{1+\left(a_{j}=j\right)}}{2^{2+2\left(a_{j}=j\right)}}}\right]^{2}\left(\sum \limits _{n=0}^{\infty }\left[\left[{\frac {C_{2n}^{n}}{2^{2n}}}\right]^{2}{\frac {1}{1-2n}}\sum \limits _{j=0}^{n-1}{\frac {\left(-1\right)^{n+j+1}n!^{2}\sin \left[\left(2n-2j\right)\varphi \right]}{\left(n-j\right)\left(2n-j\right)!j!}}+\varphi \right]k^{2n}-{\frac {\pi }{2}}\right)^{n}}}\end{aligned}}

{\underset {i=1}{\overset {n}{S}}}\sum \limits _{a_{i}=1}^{p}\prod \limits _{j=1}^{n}f\left(j,a_{j}\right)=\prod \limits _{j=1}^{n}\sum \limits _{k=1}^{p}f\left(j,k\right)

{\underset {i=1}{\overset {n}{S}}}\sum \limits _{a_{i}=1}^{i}\prod \limits _{j=1}^{n}f\left(j,a_{j}\right)=\prod \limits _{j=1}^{n}\sum \limits _{k=1}^{j}f\left(j,k\right)