Here are a few questions on the fractional part and the greatest integer function.
- Find out $[\sqrt[3]{2022^2}-12\sqrt[3]{2022}]$
- If $\{x\}=x-[x],$ find out $[255\cdot x\{x\}]$ for $x=\sqrt[3]{15015}.$
For question 1, using a calculator, I know that the answer is $8.$ But I really do not know how to proceed in a mathematical way.
For question 2, $$255\cdot x\{x\}=255\cdot x( x-[x])=255\cdot \sqrt[3]{15015}(\sqrt[3]{15015}-24)$$
As $25^3>15015>24^3.$ So we have to find
$$[255\cdot \sqrt[3]{15015}(\sqrt[3]{15015}-24)].$$
We can bound it from below as $[xy]\ge [x]\cdot [y].$
So $$[255\cdot \sqrt[3]{15015}(\sqrt[3]{15015}-24)]\ge 255[\sqrt[3]{15015^2}-\sqrt[3]{15015}\cdot 24].$$ This again looks like question 1 but I do not know how to solve it.
We can upper bound it too using $[x-y]\le [x]-[y].$
Now, $$[255\cdot \sqrt[3]{15015}(\sqrt[3]{15015}-24)]=[255\cdot\sqrt[3]{15015^2}-255\cdot\sqrt[3]{15015}\cdot 24]=[255\cdot {15015^2}]-[255\cdot\sqrt[3]{15015}\cdot 24].$$
Any solutions?
Best Answer
Solution for part 1. Let $a:=\sqrt[3]{2002}$. Then $0<a-12<1$ and $$0<(a-12)^3 = a^3-36(a^2-12a)-12^3<1.$$ Rearranging the inequality we get $$ 8<\frac{2022-12^3-1}{36}<a^2-12a<\frac{2022-12^3}{36}<9 $$