Question Number 142893 by mnjuly1970 last updated on 06/Jun/21
![.....mathematical .....analysis...... f ∈ C [0,1] and ∫_0 ^( 1) x^n f(x)dx=(1/(n+2)) , n∈N prove f(x):=x .....](Q142893.png)
$$\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:.....{mathematical}\:.....{analysis}...... \\ $$$$\:\:\:\:\:\:\:{f}\:\in\:{C}\:\left[\mathrm{0},\mathrm{1}\right]\:{and}\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{{n}} {f}\left({x}\right){dx}=\frac{\mathrm{1}}{{n}+\mathrm{2}}\:,\:{n}\in\mathbb{N} \\ $$$$\:\:\:\:\:\:\:\:{prove}\:\:{f}\left({x}\right):={x}\:..... \\ $$
Answered by mindispower last updated on 06/Jun/21
![(1/(n+2))=∫_0 ^1 x^(n+1) dx ⇒∫_0 ^1 x^n (f(x)−x)dx=0,∀n∈N since f(x)∈C[0,1],∃p_m of polynomial such that lagrange theorem ∣f(x)−p_m ∣→0 ⇒∀n∈N ∫_0 ^1 x^n p_m (x)dx=0,p_m ∈R[X] p_m (x)=Σ_(k=0) ^l ∫_0 ^1 a_k x^k .x^n dx=0⇒ Σ_(k=0) ^l (a_k /(k+n+1))=0we got infintie linear equation ⇒(a_k )=0,∀k∈[0,l] ⇒p_m =0,∀m∈N ⇒∣p_m −x∣→0⇒p_m →x ∣f(x)−x∣=∣f(x)−p_m +p_m −x∣≤∣p_m −x∣_0 +∣f(x)−p_m ∣_(=0) ⇒of(x)−x=0⇒f(x)=x](Q142895.png)
$$\frac{\mathrm{1}}{{n}+\mathrm{2}}=\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}+\mathrm{1}} {dx} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}} \left({f}\left({x}\right)−{x}\right){dx}=\mathrm{0},\forall{n}\in\mathbb{N} \\ $$$${since}\:{f}\left({x}\right)\in{C}\left[\mathrm{0},\mathrm{1}\right],\exists{p}_{{m}} \:{of}\:{polynomial}\:{such}\:{that} \\ $$$${lagrange}\:{theorem} \\ $$$$\mid{f}\left({x}\right)−{p}_{{m}} \mid\rightarrow\mathrm{0} \\ $$$$\Rightarrow\forall{n}\in\mathbb{N}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}} {p}_{{m}} \left({x}\right){dx}=\mathrm{0},{p}_{{m}} \in\mathbb{R}\left[{X}\right] \\ $$$${p}_{{m}} \left({x}\right)=\underset{{k}=\mathrm{0}} {\overset{{l}} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} {a}_{{k}} {x}^{{k}} .{x}^{{n}} {dx}=\mathrm{0}\Rightarrow \\ $$$$\underset{{k}=\mathrm{0}} {\overset{{l}} {\sum}}\frac{{a}_{{k}} }{{k}+{n}+\mathrm{1}}=\mathrm{0}{we}\:{got}\:{infintie}\:{linear}\:{equation}\: \\ $$$$\Rightarrow\left({a}_{{k}} \right)=\mathrm{0},\forall{k}\in\left[\mathrm{0},{l}\right] \\ $$$$\Rightarrow{p}_{{m}} =\mathrm{0},\forall{m}\in\mathbb{N} \\ $$$$\Rightarrow\mid{p}_{{m}} −{x}\mid\rightarrow\mathrm{0}\Rightarrow{p}_{{m}} \rightarrow{x} \\ $$$$\mid{f}\left({x}\right)−{x}\mid=\mid{f}\left({x}\right)−{p}_{{m}} +{p}_{{m}} −{x}\mid\leqslant\mid{p}_{{m}} −{x}\mid_{\mathrm{0}} +\mid{f}\left({x}\right)−{p}_{{m}} \mid_{=\mathrm{0}} \\ $$$$\Rightarrow{of}\left({x}\right)−{x}=\mathrm{0}\Rightarrow{f}\left({x}\right)={x} \\ $$
Commented by mnjuly1970 last updated on 06/Jun/21
$$\:\:{bravo}\:..\:{very}\:{nice}\:{mr}\:{power}... \\ $$
Commented by mindispower last updated on 06/Jun/21
$${pleasur}\:{i}\:{love}\:{maths}\:{but}\:{i}/{stopped}\:{befor} \\ $$$${my}\:{graduation}\:{so}\:{sad} \\ $$
Commented by Ar Brandon last updated on 17/Jun/21
Oh ! Dommage ! Qu'est-ce qui s'est passé ?