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Question Number 200722 by cortano12 last updated on 22/Nov/23

$$\mathrm{Find}\mathrm{all}\mathrm{polynomials}\mathrm{P}\left(\mathrm{x}\right)\mathrm{with}$$$$\mathrm{real}\mathrm{coefficients}\mathrm{such}\mathrm{that}\mathrm{for}$$$$\mathrm{all}\mathrm{nonzero}\mathrm{real}\mathrm{numbers}\mathrm{x},$$$$\mathrm{P}\left(\mathrm{x}\right)+\mathrm{P}\left(\frac{1}{\mathrm{x}}\right)=\frac{\mathrm{P}(\mathrm{x}+\frac{1}{\mathrm{x}})+\mathrm{P}(\mathrm{x}-\frac{1}{\mathrm{x}})}{2}$$

Commented by Frix last updated on 22/Nov/23

$$P\left(x\right)={ax}^2[\vee P\left(x\right)=0]$$

Answered by witcher3 last updated on 22/Nov/23

$$\mathrm{x}\stackrel{\mathrm{f}}{\to }\mathrm{x}-\frac{1}{\mathrm{x}};$$$$]0,+\mathrm{\infty }[\to ]-\mathrm{\infty },+\mathrm{\infty }[\mathrm{bijective}$$$$\mathrm{p}\left(\frac{1}{\mathrm{x}}\right)+\mathrm{p}\left(\mathrm{x}\right)=\frac{1}{2}(\mathrm{p}(\mathrm{x}+\frac{1}{\mathrm{x}})+\mathrm{p}(-\mathrm{x}+\frac{1}{\mathrm{x}})=\frac{1}{2}(\mathrm{p}(\mathrm{x}+\frac{1}{\mathrm{x}})+\mathrm{p}(\mathrm{x}-\frac{1}{\mathrm{x}}))$$$$\Rightarrow \mathrm{p}(\mathrm{x}-\frac{1}{\mathrm{x}})=\mathrm{p}(-(\frac{1}{\mathrm{x}}-\mathrm{x}))$$$$\iff \mathrm{\forall }\mathrm{a}\in \mathbb{R}\mathrm{since}\mathrm{bijectiln}\mathrm{of}\mathrm{f}.\mathrm{p}\left(\mathrm{a}\right)=\mathrm{p}(-\mathrm{a})\Rightarrow$$$$\mathrm{p}\left(\mathrm{x}\right)=\underset{\mathrm{k}=0}{\sum ^{\mathrm{n}}}{\mathrm{a}}_{\mathrm{k}}{\mathrm{x}}^{2\mathrm{k}};\mathrm{p}\in {\mathbb{R}}_{2\mathrm{n}}\left[\mathrm{X}\right]$$$$\mathrm{x}={\mathrm{e}}^{\mathrm{t}},$$$$\mathrm{p}\left({\mathrm{e}}^{\mathrm{t}}\right)+\mathrm{p}\left({\mathrm{e}}^{-\mathrm{t}}\right)=\frac{1}{2}\mathrm{p}\left(2\mathrm{ch}\left(\mathrm{t}\right)\right)+\mathrm{p}\left(2\mathrm{sh}\left(\mathrm{t}\right)\right)$$$$\iff 2\mathrm{p}\left({\mathrm{e}}^{\mathrm{t}}\right)+2\mathrm{p}\left({\mathrm{e}}^{-\mathrm{t}}\right)=\mathrm{p}\left(2\mathrm{ch}\left(\mathrm{t}\right)\right)+\mathrm{p}\left(2\mathrm{sh}\left(\mathrm{t}\right)\right)$$$$\iff 4\underset{\mathrm{k}=0}{\sum ^{2\mathrm{n}}}{\mathrm{a}}_{\mathrm{k}}\mathrm{ch}\left(2\mathrm{kt}\right)=\underset{\mathrm{k}=0}{\sum ^{2\mathrm{n}}}{\mathrm{a}}_{\mathrm{k}}{2}^{2\mathrm{k}}({\mathrm{ch}}^{2\mathrm{k}}\left(\mathrm{t}\right)+{\mathrm{sh}}^{2\mathrm{k}}\left(\mathrm{t}\right))$$$$\iff \mathrm{\forall }\mathrm{k}\in [0,2]$$$$4\mathrm{ch}\left(2\mathrm{kt}\right)=2^{2\mathrm{k}}({\mathrm{ch}}^{2\mathrm{k}}\left(\mathrm{t}\right)+{\mathrm{sh}}^{2\mathrm{k}}\left(\mathrm{t}\right)$$$$\mathrm{k}=1\mathrm{true},\mathrm{\forall }\mathrm{k}⩾2$$$$4\mathrm{ch}\left(2\mathrm{kt}\right)=2^{2\mathrm{k}}({\mathrm{ch}}^{2\mathrm{k}}\left(\mathrm{t}\right)+{\mathrm{sh}}^{2\mathrm{k}}\left(\mathrm{t}\right);\mathrm{t}=0$$$$4=2^{2\mathrm{k}}\mathrm{false}\mathrm{k}⩾2$$$$\Rightarrow \mathrm{p}\left(\mathrm{x}\right)={\mathrm{a}}_0+{\mathrm{a}}_1{\mathrm{x}}^2;\mathrm{p}\left(1\right)+\mathrm{p}\left(\frac{1}{1}\right)=\frac{1}{2}(\mathrm{p}\left(2\right)+\mathrm{p}\left(0\right))$$$$\iff 2({\mathrm{a}}_0+{\mathrm{a}}_1)={\mathrm{a}}_0+{2\mathrm{a}}_1$$$${\mathrm{a}}_0=0$$$$\mathrm{p}\left(\mathrm{x}\right)={\mathrm{ax}}^2\mathrm{worck}\mathrm{unique}\mathrm{by}\mathrm{construcrion}$$$$;\mathrm{a}\in \mathbb{R}$$

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