Question-198988

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

Answered by witcher3 last updated on 26/Oct/23

f(x)−g(x)=xΣ_(k=0) ^(1010) (x^(2k) )>0 u_n =f^((n)) ((1/(2023))),v_n =g^((n)) ((1/(2023))) u_(n+1) =f(u_n );v_(n+1) =g(v_n ) u_n >v_n ..? by induction ...u_1 >v_1 ,cause (f−g)((1/(2023)))>0 f′=((1+x^2 )/((1−x^2 )^2 ))>0 supose ∀n>0 u_n >v_n u_(n+1) =f(u_n )>f(v_n )>g(v_n )=v_(n+1) f(u_n )>f(v_(n)) ) f increse function g(v_n )<f(v_n ),f−g>0,∀x≥0 ⇒u_(n+1) >v_(n+1) ⇒u_(2022) >v_(2022) ⇔f^(2022) ((1/(2023)))>g^(2022) ((1/(2023)))

$$\mathrm{f}\left(\mathrm{x}\right)−\mathrm{g}\left(\mathrm{x}\right)=\mathrm{x}\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{1010}} {\sum}}\left(\mathrm{x}^{\mathrm{2k}} \right)>\mathrm{0} \\ $$$$\mathrm{u}_{\mathrm{n}} =\mathrm{f}^{\left(\mathrm{n}\right)} \left(\frac{\mathrm{1}}{\mathrm{2023}}\right),\mathrm{v}_{\mathrm{n}} =\mathrm{g}^{\left(\mathrm{n}\right)} \left(\frac{\mathrm{1}}{\mathrm{2023}}\right) \\ $$$$\mathrm{u}_{\mathrm{n}+\mathrm{1}} =\mathrm{f}\left(\mathrm{u}_{\mathrm{n}} \right);\mathrm{v}_{\mathrm{n}+\mathrm{1}} =\mathrm{g}\left(\mathrm{v}_{\mathrm{n}} \right) \\ $$$$\mathrm{u}_{\mathrm{n}} >\mathrm{v}_{\mathrm{n}} ..?\:\mathrm{by}\:\mathrm{induction} \\ $$$$…\mathrm{u}_{\mathrm{1}} >\mathrm{v}_{\mathrm{1}} ,\mathrm{cause}\:\left(\mathrm{f}−\mathrm{g}\right)\left(\frac{\mathrm{1}}{\mathrm{2023}}\right)>\mathrm{0} \\ $$$$\mathrm{f}'=\frac{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }{\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} }>\mathrm{0}\:\mathrm{supose}\:\forall\mathrm{n}>\mathrm{0}\:\mathrm{u}_{\mathrm{n}} >\mathrm{v}_{\mathrm{n}} \\ $$$$\mathrm{u}_{\mathrm{n}+\mathrm{1}} =\mathrm{f}\left(\mathrm{u}_{\mathrm{n}} \right)>\mathrm{f}\left(\mathrm{v}_{\mathrm{n}} \right)>\mathrm{g}\left(\mathrm{v}_{\mathrm{n}} \right)=\mathrm{v}_{\mathrm{n}+\mathrm{1}} \\ $$$$\mathrm{f}\left(\mathrm{u}_{\mathrm{n}} \right)>\mathrm{f}\left(\mathrm{v}_{\left.\mathrm{n}\right)} \right)\:\:\mathrm{f}\:\mathrm{increse}\:\mathrm{function} \\ $$$$\mathrm{g}\left(\mathrm{v}_{\mathrm{n}} \right)<\mathrm{f}\left(\mathrm{v}_{\mathrm{n}} \right),\mathrm{f}−\mathrm{g}>\mathrm{0},\forall\mathrm{x}\geqslant\mathrm{0} \\ $$$$\Rightarrow\mathrm{u}_{\mathrm{n}+\mathrm{1}} >\mathrm{v}_{\mathrm{n}+\mathrm{1}} \\ $$$$\Rightarrow\mathrm{u}_{\mathrm{2022}} >\mathrm{v}_{\mathrm{2022}} \Leftrightarrow\mathrm{f}^{\mathrm{2022}} \left(\frac{\mathrm{1}}{\mathrm{2023}}\right)>\mathrm{g}^{\mathrm{2022}} \left(\frac{\mathrm{1}}{\mathrm{2023}}\right) \\ $$

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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