Question Number 34020 by prof Abdo imad last updated on 29/Apr/18
![let p(x)=cos(2n arccos(x)) with x∈[−1,1] find the roots of p(x) and factorize p(x)](https://www.tinkutara.com/question/Q34020.png)
$${let}\:{p}\left({x}\right)={cos}\left(\mathrm{2}{n}\:{arccos}\left({x}\right)\right)\:\:{with}\:{x}\in\left[−\mathrm{1},\mathrm{1}\right] \\ $$$${find}\:{the}\:{roots}\:{of}\:{p}\left({x}\right)\:{and}\:{factorize}\:\:{p}\left({x}\right) \\ $$
Commented by math khazana by abdo last updated on 02/May/18
![let put arccosx=t ⇔ x=cost p(x)=0 ⇔ cos(2nt)=0 ⇔ 2nt =(π/2) +kπ k∈ Z ⇔ t_k = (π/(4n)) +((kπ)/(2n)) k ∈Z but card{t_k }<+∞ we can prove that k∈[0,2n−1]] so the roots of p(x)are x_k =cos(t_k ) =cos((π/(4n)) +((kπ)/(2n))) and p(x)=λ Π_(k=0) ^(2n−1) (x −cos((π/(4n)) +((kπ)/(2n)))) .](https://www.tinkutara.com/question/Q34212.png)
$${let}\:{put}\:{arccosx}={t}\:\Leftrightarrow\:{x}={cost} \\ $$$${p}\left({x}\right)=\mathrm{0}\:\Leftrightarrow\:{cos}\left(\mathrm{2}{nt}\right)=\mathrm{0}\:\Leftrightarrow\:\mathrm{2}{nt}\:=\frac{\pi}{\mathrm{2}}\:+{k}\pi\:\:\:{k}\in\:{Z} \\ $$$$\Leftrightarrow\:{t}_{{k}} \:=\:\frac{\pi}{\mathrm{4}{n}}\:+\frac{{k}\pi}{\mathrm{2}{n}}\:\:\:{k}\:\in{Z}\:\:\:\:{but}\:{card}\left\{{t}_{{k}} \right\}<+\infty\:{we}\:{can} \\ $$$$\left.{prove}\:{that}\:{k}\in\left[\mathrm{0},\mathrm{2}{n}−\mathrm{1}\right]\right]\:{so}\:{the}\:{roots}\:{of}\:{p}\left({x}\right){are} \\ $$$${x}_{{k}} ={cos}\left({t}_{{k}} \right)\:={cos}\left(\frac{\pi}{\mathrm{4}{n}}\:+\frac{{k}\pi}{\mathrm{2}{n}}\right)\:{and} \\ $$$${p}\left({x}\right)=\lambda\:\prod_{{k}=\mathrm{0}} ^{\mathrm{2}{n}−\mathrm{1}} \:\left({x}\:−{cos}\left(\frac{\pi}{\mathrm{4}{n}}\:+\frac{{k}\pi}{\mathrm{2}{n}}\right)\right)\:. \\ $$