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solve-x-x-2-x-3-




Question Number 205680 by mnjuly1970 last updated on 27/Mar/24
           solve            ⌊x ⌋ + ⌊ x^2 ⌋ = ⌊ x^3  ⌋
$$ \\ $$$$\:\:\:\:\:\:\:\:\:{solve}\: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\lfloor{x}\:\rfloor\:+\:\lfloor\:{x}^{\mathrm{2}} \rfloor\:=\:\lfloor\:{x}^{\mathrm{3}} \:\rfloor \\ $$$$ \\ $$
Answered by Frix last updated on 27/Mar/24
Obviously x=0  −1<x<0: ⌊x^3 ⌋=⌊x⌋=−1∧⌊x^2 ⌋=0  0<x<1: ⌊x^3 ⌋=⌊x^2 ⌋=⌊x⌋=0  ⇒  −1<x<1 ★    x=2^(1/3) : ⌊x^3 ⌋=2∧⌊x^2 ⌋=⌊x⌋=1  The change is at x=2^(1/2) : ⌊x^2 ⌋=2  ⇒  2^(1/3) ≤x<2^(1/2)  ★    x=3^(1/3) : ⌊x^3 ⌋=3∧⌊x^2 ⌋=2∧⌊x⌋=1  The change is at x=4^(1/3) : ⌊x^3 ⌋=4  ⇒  3^(1/3) ≤x<4^(1/3)  ★
$$\mathrm{Obviously}\:{x}=\mathrm{0} \\ $$$$−\mathrm{1}<{x}<\mathrm{0}:\:\lfloor{x}^{\mathrm{3}} \rfloor=\lfloor{x}\rfloor=−\mathrm{1}\wedge\lfloor{x}^{\mathrm{2}} \rfloor=\mathrm{0} \\ $$$$\mathrm{0}<{x}<\mathrm{1}:\:\lfloor{x}^{\mathrm{3}} \rfloor=\lfloor{x}^{\mathrm{2}} \rfloor=\lfloor{x}\rfloor=\mathrm{0} \\ $$$$\Rightarrow \\ $$$$−\mathrm{1}<{x}<\mathrm{1}\:\bigstar \\ $$$$ \\ $$$${x}=\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{3}}} :\:\lfloor{x}^{\mathrm{3}} \rfloor=\mathrm{2}\wedge\lfloor{x}^{\mathrm{2}} \rfloor=\lfloor{x}\rfloor=\mathrm{1} \\ $$$$\mathrm{The}\:\mathrm{change}\:\mathrm{is}\:\mathrm{at}\:{x}=\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} :\:\lfloor{x}^{\mathrm{2}} \rfloor=\mathrm{2} \\ $$$$\Rightarrow \\ $$$$\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{3}}} \leqslant{x}<\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} \:\bigstar \\ $$$$ \\ $$$${x}=\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{3}}} :\:\lfloor{x}^{\mathrm{3}} \rfloor=\mathrm{3}\wedge\lfloor{x}^{\mathrm{2}} \rfloor=\mathrm{2}\wedge\lfloor{x}\rfloor=\mathrm{1} \\ $$$$\mathrm{The}\:\mathrm{change}\:\mathrm{is}\:\mathrm{at}\:{x}=\mathrm{4}^{\frac{\mathrm{1}}{\mathrm{3}}} :\:\lfloor{x}^{\mathrm{3}} \rfloor=\mathrm{4} \\ $$$$\Rightarrow \\ $$$$\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{3}}} \leqslant{x}<\mathrm{4}^{\frac{\mathrm{1}}{\mathrm{3}}} \:\bigstar \\ $$

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