(Oral X-Cachan Psi) On pose {x \in[0,1[} et {x_{1} = \lfloor 2x\rfloor}. Pour tout {n\in\mathbb{N}^{*}}, on pose :{x_{n+1}= \lfloor 2^{n+1}(x - S_{n})\rfloor\;\text{où}\;S_{n} = \displaystyle\sum_{k=1}^{n}\dfrac{x_{k}}{2^{k}}}
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(Oral X-Cachan Psi) On pose {x \in[0,1[} et {x_{1} = \lfloor 2x\rfloor}. Pour tout {n\in\mathbb{N}^{*}}, on pose :{x_{n+1}= \lfloor 2^{n+1}(x - S_{n})\rfloor\;\text{où}\;S_{n} = \displaystyle\sum_{k=1}^{n}\dfrac{x_{k}}{2^{k}}}
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