Question Number 31460 by abdo imad last updated on 08/Mar/18
$${find}\:{in}\:{terms}\:{of}\:\:{n}\:{the}\:{value}\:{of} \\ $$$${A}_{{n}} =\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\prod_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \left({x}^{\mathrm{2}} \:−\mathrm{2}{xcos}\left(\frac{{k}\pi}{{n}}\right)\:+\mathrm{1}\right){dx}\:\:\:{with}\:{n}\:{from}\:{N}^{\bigstar} . \\ $$
Commented by abdo imad last updated on 10/Mar/18
![let decompose inside C[x] p(x)=x^(2n) −1 the roots of p(x)are z_k =e^(i((kπ)/n)) with k∈[0,2n−1 ]] we have z_0 =1 , z_1 =e^(i(π/n)) ,z_2 =e^(i((2π)/n)) ,...z_(n−1) =e^(i(((n−1)π)/n)) ,z_n =−1 , z_(n+1) = e^(i(((n+1)π)/n)) = z_(n−1) ^− , z_(n+2) = z_(n−2) ^− , z_(2n−1) =z_1 ^− ⇒ p(x)=Π_(k=0) ^(2n−1) (x−z_k )=(x^2 −1)Π_(k=1) ^(n−1) (x−z_k )(x−z_k ^− ) =(x^2 −1) Π_(k=1) ^(n−1) (x^2 −2Re(z_k )x +∣z_k ∣^2 ) =(x^2 −1) Π_(k=1) ^(n−1) (x^2 −2cos(((kπ)/n))x +1) ⇒ Π_(k=1) ^(n−1) (x^2 −2cos(((kπ)/n))x +1)= ((x^(2n) −1)/(x^2 −1)) ⇒ A_n = ∫_0 ^1 ((x^(2n) −1)/(x^2 −1))dx= ∫_0 ^1 (1+x^2 +x^4 +...+x^(2n−2) )dx A_n = [x +(1/3) x^3 +(1/5)x^5 +....+(1/(2n−1))x^(2n−1) ]_0 ^1 =1 +(1/3) +(1/5) +.....+(1/(2n−1)) let find A_n interms of H_n A_n =1+(1/2) +(1/3) +(1/5) +...+(1/(2n−1)) +(1/(2n)) −(1/2) −(1/4) −...−(1/(2n)) A_n =H_(2n) −(1/2) H_(n ) with H_n =Σ_(k=1) ^n (1/k) .](https://www.tinkutara.com/question/Q31592.png)
$${let}\:{decompose}\:{inside}\:{C}\left[{x}\right]\:{p}\left({x}\right)={x}^{\mathrm{2}{n}} \:−\mathrm{1}\:{the}\:{roots}\:{of} \\ $$$$\left.{p}\left({x}\right){are}\:{z}_{{k}} ={e}^{{i}\frac{{k}\pi}{{n}}} \:\:{with}\:{k}\in\left[\mathrm{0},\mathrm{2}{n}−\mathrm{1}\:\right]\right]\:{we}\:{have} \\ $$$${z}_{\mathrm{0}} =\mathrm{1}\:,\:{z}_{\mathrm{1}} ={e}^{{i}\frac{\pi}{{n}}} \:\:,{z}_{\mathrm{2}} ={e}^{{i}\frac{\mathrm{2}\pi}{{n}}} \:,…{z}_{{n}−\mathrm{1}} ={e}^{{i}\frac{\left({n}−\mathrm{1}\right)\pi}{{n}}} \:,{z}_{{n}} =−\mathrm{1}\:, \\ $$$${z}_{{n}+\mathrm{1}} =\:{e}^{{i}\frac{\left({n}+\mathrm{1}\right)\pi}{{n}}} =\:\:\overset{−} {{z}}_{{n}−\mathrm{1}} \:\:\:,\:{z}_{{n}+\mathrm{2}} =\:\:\overset{−} {{z}}_{{n}−\mathrm{2}} \:,\:\:{z}_{\mathrm{2}{n}−\mathrm{1}} =\overset{−} {{z}}_{\mathrm{1}} \:\Rightarrow \\ $$$${p}\left({x}\right)=\prod_{{k}=\mathrm{0}} ^{\mathrm{2}{n}−\mathrm{1}} \:\left({x}−{z}_{{k}} \right)=\left({x}^{\mathrm{2}} −\mathrm{1}\right)\prod_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \left({x}−{z}_{{k}} \right)\left({x}−\overset{−} {{z}}_{{k}} \right) \\ $$$$=\left({x}^{\mathrm{2}} −\mathrm{1}\right)\:\prod_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\left({x}^{\mathrm{2}} \:−\mathrm{2}{Re}\left({z}_{{k}} \right){x}\:\:+\mid{z}_{{k}} \mid^{\mathrm{2}} \right) \\ $$$$=\left({x}^{\mathrm{2}} \:−\mathrm{1}\right)\:\prod_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\left({x}^{\mathrm{2}} \:−\mathrm{2}{cos}\left(\frac{{k}\pi}{{n}}\right){x}\:+\mathrm{1}\right)\:\Rightarrow \\ $$$$\prod_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\left({x}^{\mathrm{2}} \:−\mathrm{2}{cos}\left(\frac{{k}\pi}{{n}}\right){x}\:+\mathrm{1}\right)=\:\frac{{x}^{\mathrm{2}{n}} \:−\mathrm{1}}{{x}^{\mathrm{2}} \:−\mathrm{1}}\:\Rightarrow \\ $$$${A}_{{n}} =\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{x}^{\mathrm{2}{n}} \:−\mathrm{1}}{{x}^{\mathrm{2}} −\mathrm{1}}{dx}=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\left(\mathrm{1}+{x}^{\mathrm{2}} \:+{x}^{\mathrm{4}} \:+…+{x}^{\mathrm{2}{n}−\mathrm{2}} \right){dx} \\ $$$${A}_{{n}} \:=\:\left[{x}\:+\frac{\mathrm{1}}{\mathrm{3}}\:{x}^{\mathrm{3}} \:+\frac{\mathrm{1}}{\mathrm{5}}{x}^{\mathrm{5}} \:\:\:\:\:+….+\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}{x}^{\mathrm{2}{n}−\mathrm{1}} \right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\mathrm{1}\:+\frac{\mathrm{1}}{\mathrm{3}}\:+\frac{\mathrm{1}}{\mathrm{5}}\:+…..+\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}\:{let}\:{find}\:{A}_{{n}} \:{interms}\:{of}\:{H}_{{n}} \\ $$$${A}_{{n}} =\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{3}}\:+\frac{\mathrm{1}}{\mathrm{5}}\:+…+\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}\:+\frac{\mathrm{1}}{\mathrm{2}{n}}\:−\frac{\mathrm{1}}{\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{4}}\:−…−\frac{\mathrm{1}}{\mathrm{2}{n}} \\ $$$${A}_{{n}} ={H}_{\mathrm{2}{n}} \:−\frac{\mathrm{1}}{\mathrm{2}}\:{H}_{{n}\:} \:\:\:\:{with}\:{H}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}}\:\:. \\ $$