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Question Number 36912 by prof Abdo imad last updated on 07/Jun/18
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⟩.
let⟨p,q⟩=∫−11p(x)q(x)dxwithpandqaretwopolynomsfromR[x]1)letp(x)=xncalculate⟨p,p⟩2)letp(x)=1+x+x2+….+xnfind⟨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⟩=∫−11p2(x)dx=∫−11x2ndx=2∫01x2ndx=22n+1[x2n+1]01=22n+12)p(x)=∑i=0nxi⇒⟨p,p⟩=∫−11(∑i=0nxi)2dx=∫−11(∑i=0n(xi)2+2∑1⩽i<j⩽nxixj)dx=∑i=0n∫−11x2idx+2∑1⩽i<j⩽n∫−11xi+jdx=2∑i=0n12i+1+2∑1⩽i<j⩽n[1i+j+1xi+j+1]−11=2∑i=0n12i+1+2∑1⩽i<j⩽n{1i+j+1−(−1)i+j+1i+j+1}

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