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Question Number 198988 by Safojon last updated on 26/Oct/23

Answered by witcher3 last updated on 26/Oct/23

$$\mathrm{f}\left(\mathrm{x}\right)-\mathrm{g}\left(\mathrm{x}\right)=\mathrm{x}\underset{\mathrm{k}=0}{\sum ^{1010}}\left({\mathrm{x}}^{2\mathrm{k}}\right)>0$$$${\mathrm{u}}_{\mathrm{n}}={\mathrm{f}}^{\left(\mathrm{n}\right)}\left(\frac{1}{2023}\right),{\mathrm{v}}_{\mathrm{n}}={\mathrm{g}}^{\left(\mathrm{n}\right)}\left(\frac{1}{2023}\right)$$$${\mathrm{u}}_{\mathrm{n}+1}=\mathrm{f}\left({\mathrm{u}}_{\mathrm{n}}\right);{\mathrm{v}}_{\mathrm{n}+1}=\mathrm{g}\left({\mathrm{v}}_{\mathrm{n}}\right)$$$${\mathrm{u}}_{\mathrm{n}}>{\mathrm{v}}_{\mathrm{n}}..?\mathrm{by}\mathrm{induction}$$$$...{\mathrm{u}}_1>{\mathrm{v}}_1,\mathrm{cause}(\mathrm{f}-\mathrm{g})\left(\frac{1}{2023}\right)>0$$$${\mathrm{f}}^{\prime }=\frac{1+{\mathrm{x}}^2}{{(1-{\mathrm{x}}^2)}^2}>0\mathrm{supose}\mathrm{\forall }\mathrm{n}>0{\mathrm{u}}_{\mathrm{n}}>{\mathrm{v}}_{\mathrm{n}}$$$${\mathrm{u}}_{\mathrm{n}+1}=\mathrm{f}\left({\mathrm{u}}_{\mathrm{n}}\right)>\mathrm{f}\left({\mathrm{v}}_{\mathrm{n}}\right)>\mathrm{g}\left({\mathrm{v}}_{\mathrm{n}}\right)={\mathrm{v}}_{\mathrm{n}+1}$$$$\mathrm{f}\left({\mathrm{u}}_{\mathrm{n}}\right)>\mathrm{f}\left({\mathrm{v}}_{\mathrm{n})}\right)\mathrm{f}\mathrm{increse}\mathrm{function}$$$$\mathrm{g}\left({\mathrm{v}}_{\mathrm{n}}\right)<\mathrm{f}\left({\mathrm{v}}_{\mathrm{n}}\right),\mathrm{f}-\mathrm{g}>0,\mathrm{\forall }\mathrm{x}⩾0$$$$\Rightarrow {\mathrm{u}}_{\mathrm{n}+1}>{\mathrm{v}}_{\mathrm{n}+1}$$$$\Rightarrow {\mathrm{u}}_{2022}>{\mathrm{v}}_{2022}\iff {\mathrm{f}}^{2022}\left(\frac{1}{2023}\right)>{\mathrm{g}}^{2022}\left(\frac{1}{2023}\right)$$

Commented by Safojon last updated on 26/Oct/23

$$thenks.$$

Commented by witcher3 last updated on 26/Oct/23

$$\mathrm{y}\mathrm{re}\mathrm{welcom}$$

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