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Question Number 142893 by mnjuly1970 last updated on 06/Jun/21

$$$$$$.....mathematical.....analysis......$$$$f\in C[0,1]and{\int }_0^1{x}^{n}f\left(x\right)dx=\frac{1}{n+2},n\in \mathbb{N}$$$$provef\left(x\right):=x.....$$

Answered by mindispower last updated on 06/Jun/21

$$\frac{1}{n+2}={\int }_0^1{x}^{n+1}dx$$$$\Rightarrow {\int }_0^1{x}^n(f\left(x\right)-x)dx=0,\mathrm{\forall }n\in \mathbb{N}$$$$sincef\left(x\right)\in C[0,1],\mathrm{\exists }{p}_{m}ofpolynomialsuchthat$$$$lagrangetheorem$$$$\mid f\left(x\right)-p_m\mid \to 0$$$$\Rightarrow \mathrm{\forall }n\in \mathbb{N}{\int }_0^1{x}^n{p}_m\left(x\right)dx=0,p_m\in \mathbb{R}\left[X\right]$$$$p_m\left(x\right)=\underset{k=0}{\sum ^l}{\int }_0^1{a}_k{x}^k.x^{n}dx=0\Rightarrow$$$$\underset{k=0}{\sum ^l}\frac{a_k}{k+n+1}=0wegotinfintielinearequation$$$$\Rightarrow \left(a_k\right)=0,\mathrm{\forall }k\in [0,l]$$$$\Rightarrow p_m=0,\mathrm{\forall }m\in \mathbb{N}$$$$\Rightarrow \mid p_m-x\mid \to 0\Rightarrow p_m\to x$$$$\mid f\left(x\right)-x\mid =\mid f\left(x\right)-p_m+p_m-x\mid ⩽\mid p_m-x{\mid }_0+\mid f\left(x\right)-p_m{\mid }_{=0}$$$$\Rightarrow of\left(x\right)-x=0\Rightarrow f\left(x\right)=x$$

Commented by mnjuly1970 last updated on 06/Jun/21

$$bravo..verynicemrpower...$$

Commented by mindispower last updated on 06/Jun/21

$$pleasurilovemathsbuti/stoppedbefor$$$$mygraduationsosad$$

Commented by Ar Brandon last updated on 17/Jun/21

Oh ! Dommage ! Qu'est-ce qui s'est passé ?

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