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Question Number 31460 by abdo imad last updated on 08/Mar/18

find in terms of  n the value of  A_n = ∫_0 ^1  Π_(k=1) ^(n−1) (x^2  −2xcos(((kπ)/n)) +1)dx   with n from N^★ .

$${find}\:{in}\:{terms}\:{of}\:\:{n}\:{the}\:{value}\:{of} \\ $$$${A}_{{n}} =\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\prod_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \left({x}^{\mathrm{2}} \:−\mathrm{2}{xcos}\left(\frac{{k}\pi}{{n}}\right)\:+\mathrm{1}\right){dx}\:\:\:{with}\:{n}\:{from}\:{N}^{\bigstar} . \\ $$

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)  .

$${let}\:{decompose}\:{inside}\:{C}\left[{x}\right]\:{p}\left({x}\right)={x}^{\mathrm{2}{n}} \:−\mathrm{1}\:{the}\:{roots}\:{of} \\ $$$$\left.{p}\left({x}\right){are}\:{z}_{{k}} ={e}^{{i}\frac{{k}\pi}{{n}}} \:\:{with}\:{k}\in\left[\mathrm{0},\mathrm{2}{n}−\mathrm{1}\:\right]\right]\:{we}\:{have} \\ $$$${z}_{\mathrm{0}} =\mathrm{1}\:,\:{z}_{\mathrm{1}} ={e}^{{i}\frac{\pi}{{n}}} \:\:,{z}_{\mathrm{2}} ={e}^{{i}\frac{\mathrm{2}\pi}{{n}}} \:,...{z}_{{n}−\mathrm{1}} ={e}^{{i}\frac{\left({n}−\mathrm{1}\right)\pi}{{n}}} \:,{z}_{{n}} =−\mathrm{1}\:, \\ $$$${z}_{{n}+\mathrm{1}} =\:{e}^{{i}\frac{\left({n}+\mathrm{1}\right)\pi}{{n}}} =\:\:\overset{−} {{z}}_{{n}−\mathrm{1}} \:\:\:,\:{z}_{{n}+\mathrm{2}} =\:\:\overset{−} {{z}}_{{n}−\mathrm{2}} \:,\:\:{z}_{\mathrm{2}{n}−\mathrm{1}} =\overset{−} {{z}}_{\mathrm{1}} \:\Rightarrow \\ $$$${p}\left({x}\right)=\prod_{{k}=\mathrm{0}} ^{\mathrm{2}{n}−\mathrm{1}} \:\left({x}−{z}_{{k}} \right)=\left({x}^{\mathrm{2}} −\mathrm{1}\right)\prod_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \left({x}−{z}_{{k}} \right)\left({x}−\overset{−} {{z}}_{{k}} \right) \\ $$$$=\left({x}^{\mathrm{2}} −\mathrm{1}\right)\:\prod_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\left({x}^{\mathrm{2}} \:−\mathrm{2}{Re}\left({z}_{{k}} \right){x}\:\:+\mid{z}_{{k}} \mid^{\mathrm{2}} \right) \\ $$$$=\left({x}^{\mathrm{2}} \:−\mathrm{1}\right)\:\prod_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\left({x}^{\mathrm{2}} \:−\mathrm{2}{cos}\left(\frac{{k}\pi}{{n}}\right){x}\:+\mathrm{1}\right)\:\Rightarrow \\ $$$$\prod_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\left({x}^{\mathrm{2}} \:−\mathrm{2}{cos}\left(\frac{{k}\pi}{{n}}\right){x}\:+\mathrm{1}\right)=\:\frac{{x}^{\mathrm{2}{n}} \:−\mathrm{1}}{{x}^{\mathrm{2}} \:−\mathrm{1}}\:\Rightarrow \\ $$$${A}_{{n}} =\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{x}^{\mathrm{2}{n}} \:−\mathrm{1}}{{x}^{\mathrm{2}} −\mathrm{1}}{dx}=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\left(\mathrm{1}+{x}^{\mathrm{2}} \:+{x}^{\mathrm{4}} \:+...+{x}^{\mathrm{2}{n}−\mathrm{2}} \right){dx} \\ $$$${A}_{{n}} \:=\:\left[{x}\:+\frac{\mathrm{1}}{\mathrm{3}}\:{x}^{\mathrm{3}} \:+\frac{\mathrm{1}}{\mathrm{5}}{x}^{\mathrm{5}} \:\:\:\:\:+....+\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}{x}^{\mathrm{2}{n}−\mathrm{1}} \right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\mathrm{1}\:+\frac{\mathrm{1}}{\mathrm{3}}\:+\frac{\mathrm{1}}{\mathrm{5}}\:+.....+\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}\:{let}\:{find}\:{A}_{{n}} \:{interms}\:{of}\:{H}_{{n}} \\ $$$${A}_{{n}} =\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{3}}\:+\frac{\mathrm{1}}{\mathrm{5}}\:+...+\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}\:+\frac{\mathrm{1}}{\mathrm{2}{n}}\:−\frac{\mathrm{1}}{\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{4}}\:−...−\frac{\mathrm{1}}{\mathrm{2}{n}} \\ $$$${A}_{{n}} ={H}_{\mathrm{2}{n}} \:−\frac{\mathrm{1}}{\mathrm{2}}\:{H}_{{n}\:} \:\:\:\:{with}\:{H}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}}\:\:. \\ $$

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