Question Number 54808 by turbo msup by abdo last updated on 11/Feb/19
![let p(x)=(1+x^2 )(1+x^4 )...(1+x^2^n ) with n integr natural 1) find a simple form of p(x) 2) find roots of p(x)and decompose p(x) inside C[x] 3)calculate ∫_0 ^1 p(x)dx 4) decompose the fraction F(x)=(1/(p(x))) .](https://www.tinkutara.com/question/Q54808.png)
$${let}\:{p}\left({x}\right)=\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{x}^{\mathrm{4}} \right)…\left(\mathrm{1}+{x}^{\mathrm{2}^{{n}} } \right) \\ $$$${with}\:{n}\:{integr}\:{natural} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{a}\:{simple}\:{form}\:{of}\:{p}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{find}\:{roots}\:{of}\:{p}\left({x}\right){and}\:{decompose} \\ $$$${p}\left({x}\right)\:{inside}\:{C}\left[{x}\right] \\ $$$$\left.\mathrm{3}\right){calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{p}\left({x}\right){dx} \\ $$$$\left.\mathrm{4}\right)\:{decompose}\:{the}\:{fraction} \\ $$$${F}\left({x}\right)=\frac{\mathrm{1}}{{p}\left({x}\right)}\:. \\ $$
Commented by maxmathsup by imad last updated on 13/Feb/19
![1) we can use recurrence to prove that for x≠+^− 1 P(x) =((1−x^2^(n+1) )/(1−x^2 )) 2) P(z)=0 ⇔ 1−z^2^(n+1) =0 ⇔ 1−z^q =0 with m=2^(n+1) the roots of z^m =1 are z_k =e^((i2kπ)/m) with k ∈ [[0,m−1]] ⇒z_k =e^((i2kπ)/2^(n+1) ) = e^((ikπ)/2^n ) with k ∈[[0,2^(n+1) −1]] but but eliminate values of k /z_k ^2 =1 .](https://www.tinkutara.com/question/Q54873.png)
$$\left.\mathrm{1}\right)\:{we}\:{can}\:{use}\:{recurrence}\:{to}\:{prove}\:{that}\:\:{for}\:{x}\neq\overset{−} {+}\mathrm{1} \\ $$$${P}\left({x}\right)\:=\frac{\mathrm{1}−{x}^{\mathrm{2}^{{n}+\mathrm{1}} } }{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$$\left.\mathrm{2}\right)\:{P}\left({z}\right)=\mathrm{0}\:\Leftrightarrow\:\mathrm{1}−{z}^{\mathrm{2}^{{n}+\mathrm{1}} } =\mathrm{0}\:\:\:\Leftrightarrow\:\mathrm{1}−{z}^{{q}} \:=\mathrm{0}\:\:{with}\:{m}=\mathrm{2}^{{n}+\mathrm{1}} \:\:{the}\:{roots}\:{of}\: \\ $$$${z}^{{m}} =\mathrm{1}\:{are}\:{z}_{{k}} ={e}^{\frac{{i}\mathrm{2}{k}\pi}{{m}}} \:\:{with}\:{k}\:\:\in\:\left[\left[\mathrm{0},{m}−\mathrm{1}\right]\right]\:\Rightarrow{z}_{{k}} ={e}^{\frac{{i}\mathrm{2}{k}\pi}{\mathrm{2}^{{n}+\mathrm{1}} }} \:\:=\:{e}^{\frac{{ik}\pi}{\mathrm{2}^{{n}} }} \\ $$$${with}\:{k}\:\in\left[\left[\mathrm{0},\mathrm{2}^{{n}+\mathrm{1}} −\mathrm{1}\right]\right]\:\:{but}\:\:{but}\:{eliminate}\:{values}\:{of}\:{k}\:/{z}_{{k}} ^{\mathrm{2}} =\mathrm{1}\:. \\ $$
Commented by maxmathsup by imad last updated on 13/Feb/19
$${P}\left({x}\right)\:=\prod_{{k}=\mathrm{0}_{{z}_{{k}} \neq\overset{−\mathrm{1}} {+}} } ^{\mathrm{2}^{{n}+\mathrm{1}} −\mathrm{1}} \left({x}−{z}_{{k}} \right) \\ $$
Answered by mr W last updated on 12/Feb/19
$$\left.\mathrm{1}\right) \\ $$$${p}\left({x}\right)=\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{x}^{\mathrm{4}} \right)…\left(\mathrm{1}+{x}^{\mathrm{2}^{{n}} } \right) \\ $$$$\left(\mathrm{1}−{x}^{\mathrm{2}} \right){p}\left({x}\right)=\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{x}^{\mathrm{4}} \right)…\left(\mathrm{1}+{x}^{\mathrm{2}^{{n}} } \right) \\ $$$$\left(\mathrm{1}−{x}^{\mathrm{2}} \right){p}\left({x}\right)=\left(\mathrm{1}−{x}^{\mathrm{4}} \right)\left(\mathrm{1}+{x}^{\mathrm{4}} \right)…\left(\mathrm{1}+{x}^{\mathrm{2}^{{n}} } \right) \\ $$$$\left(\mathrm{1}−{x}^{\mathrm{2}} \right){p}\left({x}\right)=\left(\mathrm{1}−{x}^{\mathrm{8}} \right)…\left(\mathrm{1}+{x}^{\mathrm{2}^{{n}} } \right) \\ $$$$…. \\ $$$$\left(\mathrm{1}−{x}^{\mathrm{2}} \right){p}\left({x}\right)=\left(\mathrm{1}−{x}^{\mathrm{2}^{{n}} } \right)…\left(\mathrm{1}+{x}^{\mathrm{2}^{{n}} } \right) \\ $$$$\left(\mathrm{1}−{x}^{\mathrm{2}} \right){p}\left({x}\right)=\left(\mathrm{1}−{x}^{\mathrm{2}^{{n}+\mathrm{1}} } \right) \\ $$$$\Rightarrow{p}\left({x}\right)=\frac{\mathrm{1}−{x}^{\mathrm{2}^{{n}+\mathrm{1}} } }{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$
Commented by maxmathsup by imad last updated on 12/Feb/19
$${correct}\:{sir}\:{but}\:{x}\neq\overset{−} {+}\mathrm{1} \\ $$