mathematical-analysis-f-C-0-1-and-0-1-x-n-f-x-dx-1-n-2-n-N-prove-f-x-x-

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

\dots . . m a t h e m a t i c a l \dots . . a n a l y s i s \dots \dots

f \in C [0, 1] a n d \int_{0}^{1} x^{n} f (x) d x = \frac{1}{n + 2}, n \in N

p r o v e f (x) := x \dots . .

Answered by mindispower last updated on 06/Jun/21

\frac{1}{n + 2} = \int_{0}^{1} x^{n + 1} d x

\Rightarrow \int_{0}^{1} x^{n} (f (x) - x) d x = 0, \forall n \in N

s i n c e f (x) \in C [0, 1], \exists p_{m} o f p o l y n o m i a l s u c h t h a t

l a g r a n g e t h e o r e m

∣ f (x) - p_{m} ∣\to 0

\Rightarrow \forall n \in N \int_{0}^{1} x^{n} p_{m} (x) d x = 0, p_{m} \in R [X]

p_{m} (x) = \underset{k = 0}{\sum^{l}} \int_{0}^{1} a_{k} x^{k} . x^{n} d x = 0 \Rightarrow

\underset{k = 0}{\sum^{l}} \frac{a_{k}}{k + n + 1} = 0 w e g o t i n f i n t i e l i n e a r e q u a t i o n

\Rightarrow (a_{k}) = 0, \forall k \in [0, l]

\Rightarrow p_{m} = 0, \forall m \in N

\Rightarrow∣ p_{m} - x ∣\to 0 \Rightarrow p_{m} \to x

∣ f (x) - x ∣=∣ f (x) - p_{m} + p_{m} - x ∣⩽∣ p_{m} - x ∣_{0} + ∣ f (x) - p_{m} ∣_{= 0}

\Rightarrow o f (x) - x = 0 \Rightarrow f (x) = x

Commented by mnjuly1970 last updated on 06/Jun/21

b r a v o . . v e r y n i c e m r p o w e r \dots

Commented by mindispower last updated on 06/Jun/21

p l e a s u r i l o v e m a t h s b u t i / s t o p p e d b e f o r

m y g r a d u a t i o n s o s a d

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

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