let ⟨p,q⟩= ∫_(−1) ^1 p(x)q(x)dx with p and q are two polynoms fromR[x] 1)let p(x)=x^n calculate ⟨p,p⟩ 2)let p(x)=1+x+x^2 +....+x^n find ⟨p,p⟩.


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Question Number 36912 by prof Abdo imad last updated on 07/Jun/18

$l e t ⟨ p, q ⟩ = \int_{- 1}^{1} p (x) q (x) d x w i t h p a n d q a r e$ $t w o p o l y n o m s f r o m R [x]$ $1) l e t p (x) = x^{n} c a l c u l a t e ⟨ p, p ⟩$ $2) l e t p (x) = 1 + x + x^{2} + . . . . + x^{n}$ $f i n d ⟨ p, p ⟩ .$
Commented by abdo.msup.com last updated on 10/Jun/18
$1)⟨p,p⟩=∫_(−1) ^1 p^2 (x)dx=∫_(−1) ^1 x^(2n) dx =2 ∫_0 ^1 x^(2n) dx =(2/(2n+1))[x^(2n+1) ]_0 ^1 =(2/(2n+1)) 2)p(x)=Σ_(i=0) ^n x^i ⇒⟨p,p⟩=∫_(−1) ^1 (Σ_(i=0) ^n x_i )^2 dx =∫_(−1) ^1 (Σ_(i=0) ^n (x^i )^2 +2Σ_(1≤i<j≤n) x^i x^j )dx =Σ_(i=0) ^n ∫_(−1) ^1 x^(2i) dx +2Σ_(1≤i<j≤n) ∫_(−1) ^1 x^(i+j) dx =2Σ_(i=0) ^n (1/(2i+1)) +2 Σ_(1≤i<j≤n) [(1/(i+j+1))x^(i+j+1) ]_(−1) ^1 =2Σ_(i=0) ^n (1/(2i+1)) +2 Σ_(1≤i<j≤n) { (1/(i+j+1))−(((−1)^(i+j+1) )/(i+j+1))}$
$1) ⟨ p, p ⟩ = \int_{- 1}^{1} p^{2} (x) d x = \int_{- 1}^{1} x^{2 n} d x$ $= 2 \int_{0}^{1} x^{2 n} d x = \frac{2}{2 n + 1} {[x^{2 n + 1}]}_{0}^{1} = \frac{2}{2 n + 1}$ $2) p (x) = \sum_{i = 0}^{n} x^{i} \Rightarrow ⟨ p, p ⟩ = \int_{- 1}^{1} {(\sum_{i = 0}^{n} x_{i})}^{2} d x$ $= \int_{- 1}^{1} (\sum_{i = 0}^{n} {(x^{i})}^{2} + 2 \sum_{1 ⩽ i < j ⩽ n} x^{i} x^{j}) d x$ $= \sum_{i = 0}^{n} \int_{- 1}^{1} x^{2 i} d x + 2 \sum_{1 ⩽ i < j ⩽ n} \int_{- 1}^{1} x^{i + j} d x$ $= 2 \sum_{i = 0}^{n} \frac{1}{2 i + 1} + 2 \sum_{1 ⩽ i < j ⩽ n} {[\frac{1}{i + j + 1} x^{i + j + 1}]}_{- 1}^{1}$ $= 2 \sum_{i = 0}^{n} \frac{1}{2 i + 1} + 2 \sum_{1 ⩽ i < j ⩽ n} {\frac{1}{i + j + 1} - \frac{{(- 1)}^{i + j + 1}}{i + j + 1}}$
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