Question Number 199033 by essaad last updated on 27/Oct/23
Commented by essaad last updated on 27/Oct/23
$${mr}\:\:{W}\:{could}\:{you}\:{please}\:{help}\:{us}\: \\ $$
Commented by mr W last updated on 27/Oct/23
$${what}\:{is}\:{to}\:{do}? \\ $$$${can}\:{you}\:{recognise}\:{what}\:{the}\:{question} \\ $$$${is}?\:{i}\:{can}'{t}! \\ $$
Commented by mr W last updated on 27/Oct/23
$${btw}:\:{this}\:{app}\:{was}\:{specially}\:{developed} \\ $$$${for}\:{you}\:{to}\:{write}\:{nice}\:{mathematical} \\ $$$${formulas}.\:{why}\:{don}'{t}\:{you}\:{use}\:{it}\:{for} \\ $$$${writing}\:{nice}\:{formulas}\:{instead}\:{of} \\ $$$${posting}\:{bad}\:{images}? \\ $$
Answered by witcher3 last updated on 27/Oct/23
$$\mid\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{n}}} \right)−\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{y}^{\frac{\mathrm{1}}{\mathrm{n}}} \right)\mid\leqslant\frac{\mathrm{1}}{\mathrm{2n}}\mid\mathrm{x}−\mathrm{y}\mid \\ $$$$\mathrm{f}\left(\mathrm{t}\right)=\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{n}}} \right),\forall\mathrm{n}\in\mathbb{N}^{\ast} \frac{\mathrm{1}}{\mathrm{n}}−\mathrm{1}\leqslant\mathrm{0} \\ $$$$\mathrm{f}'\left(\mathrm{t}\right)=\frac{\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{n}}−\mathrm{1}} }{\mathrm{n}}.\frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{2n}}} \right)},\mathrm{f}'\left(\mathrm{t}\right)\leqslant\frac{\mathrm{1}}{\mathrm{2n}},\forall\mathrm{t}\geqslant\mathrm{1} \\ $$$$\mid\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{n}}} \right)−\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{y}^{\frac{\mathrm{1}}{\mathrm{n}}} \right)\mid=\mid\int_{\mathrm{y}} ^{\mathrm{x}} \mathrm{f}'\left(\mathrm{t}\right)\mathrm{dt}\mid \\ $$$$\leqslant\mid\frac{\mathrm{1}}{\mathrm{2n}}\int_{\mathrm{y}} ^{\mathrm{x}} \mathrm{1dt}\mid=\frac{\mathrm{1}}{\mathrm{2n}}\mid\mathrm{x}−\mathrm{y}\mid \\ $$