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⟩.](https://www.tinkutara.com/question/Q36912.png)
$${let}\:\:\langle{p},{q}\rangle=\:\int_{−\mathrm{1}} ^{\mathrm{1}} {p}\left({x}\right){q}\left({x}\right){dx}\:\:{with}\:{p}\:{and}\:{q}\:{are} \\ $$$${two}\:{polynoms}\:{fromR}\left[{x}\right] \\ $$$$\left.\mathrm{1}\right){let}\:{p}\left({x}\right)={x}^{{n}} \:\:\:{calculate}\:\langle{p},{p}\rangle \\ $$$$\left.\mathrm{2}\right){let}\:{p}\left({x}\right)=\mathrm{1}+{x}+{x}^{\mathrm{2}} \:+….+{x}^{{n}} \\ $$$${find}\:\langle{p},{p}\rangle. \\ $$
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))}](https://www.tinkutara.com/question/Q37200.png)
$$\left.\mathrm{1}\right)\langle{p},{p}\rangle=\int_{−\mathrm{1}} ^{\mathrm{1}} {p}^{\mathrm{2}} \left({x}\right){dx}=\int_{−\mathrm{1}} ^{\mathrm{1}} {x}^{\mathrm{2}{n}} {dx} \\ $$$$=\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{\mathrm{2}{n}} {dx}\:=\frac{\mathrm{2}}{\mathrm{2}{n}+\mathrm{1}}\left[{x}^{\mathrm{2}{n}+\mathrm{1}} \right]_{\mathrm{0}} ^{\mathrm{1}} \:=\frac{\mathrm{2}}{\mathrm{2}{n}+\mathrm{1}} \\ $$$$\left.\mathrm{2}\right){p}\left({x}\right)=\sum_{{i}=\mathrm{0}} ^{{n}} \:{x}^{{i}} \:\Rightarrow\langle{p},{p}\rangle=\int_{−\mathrm{1}} ^{\mathrm{1}} \left(\sum_{{i}=\mathrm{0}} ^{{n}} {x}_{{i}} \right)^{\mathrm{2}} {dx} \\ $$$$=\int_{−\mathrm{1}} ^{\mathrm{1}} \:\left(\sum_{{i}=\mathrm{0}} ^{{n}} \:\left({x}^{{i}} \right)^{\mathrm{2}} \:+\mathrm{2}\sum_{\mathrm{1}\leqslant{i}<{j}\leqslant{n}} {x}^{{i}} \:{x}^{{j}} \right){dx} \\ $$$$=\sum_{{i}=\mathrm{0}} ^{{n}} \:\int_{−\mathrm{1}} ^{\mathrm{1}} \:{x}^{\mathrm{2}{i}} \:{dx}\:+\mathrm{2}\sum_{\mathrm{1}\leqslant{i}<{j}\leqslant{n}} \int_{−\mathrm{1}} ^{\mathrm{1}} {x}^{{i}+{j}} {dx} \\ $$$$=\mathrm{2}\sum_{{i}=\mathrm{0}} ^{{n}} \:\:\:\frac{\mathrm{1}}{\mathrm{2}{i}+\mathrm{1}}\:+\mathrm{2}\:\sum_{\mathrm{1}\leqslant{i}<{j}\leqslant{n}} \left[\frac{\mathrm{1}}{{i}+{j}+\mathrm{1}}{x}^{{i}+{j}+\mathrm{1}} \right]_{−\mathrm{1}} ^{\mathrm{1}} \\ $$$$=\mathrm{2}\sum_{{i}=\mathrm{0}} ^{{n}} \:\:\frac{\mathrm{1}}{\mathrm{2}{i}+\mathrm{1}}\:+\mathrm{2}\:\sum_{\mathrm{1}\leqslant{i}<{j}\leqslant{n}} \left\{\:\frac{\mathrm{1}}{{i}+{j}+\mathrm{1}}−\frac{\left(−\mathrm{1}\right)^{{i}+{j}+\mathrm{1}} }{{i}+{j}+\mathrm{1}}\right\} \\ $$