Question Number 63893 by mathmax by abdo last updated on 10/Jul/19
$$\left.\mathrm{1}\right)\:{simplify}\:{W}_{{n}} \left({z}\right)=\left(\mathrm{1}+{z}\right)\left(\mathrm{1}+{z}^{\mathrm{2}} \right)….\left(\mathrm{1}+{z}^{\mathrm{2}^{{n}} } \right)\:\left({z}\:{from}\:{C}\right) \\ $$$$\left.\mathrm{2}\right)\:{simplify}\:{P}_{{n}} \left(\theta\right)\:=\left(\mathrm{1}+{e}^{{i}\theta} \right)\left(\mathrm{1}+{e}^{\mathrm{2}{i}\theta} \right)…..\left(\mathrm{1}+{e}^{{i}\mathrm{2}^{{n}} \theta} \right)\:{and}\:{sove} \\ $$$${P}_{{n}} \left(\theta\right)=\mathrm{0} \\ $$
Commented by mathmax by abdo last updated on 05/Nov/19
![1) we have W_n (z)=(1+z)(1+z^2 )....(1+z^2^n ) let prove by recurrence that W_n (z)=((1−z^2^(n+1) )/(1−z)) n =0 →W_n (z)=1+z =((1−z^2 )/(1−z))(true) let suppose the relation true W_(n+1) =(1+z)(1+z^2 )....(1+z^2^(n+1) )=(1+z)(1+z^2 )...(1+z^2^n )(1+z^2^(n+1) ) =((1−z^2^(n+1) )/(1−z))×(1+z^2^(n+1) )=((1+z^2^(n+1) −z^2^(n+1) −z^2^(n+2) )/(1−z)) =((1−z^2^(n+2) )/(1−z)) c/c W_n (z)=((1−z^2^(n+1) )/(1−z)) with z≠1 2) P_n (θ)=(1+e^(iθ) )(1+e^(2iθ) ).....(1+e^(i 2^n θ) ) =(1+e^(iθ) )(1+(e^(iθ) )^2 ).....(1+(e^(iθ) )^2^n )=W_n (e^(iθ) ) =((1−(e^(iθ) )^2^(n+1) )/(1−e^(iθ) )) =((1−e^(i2^(n+1) θ) )/(1−e^(iθ) )) (if θ ≠2kπ) P_n (θ)=0 ⇔1−e^(i2^(n+1) θ) =0 ⇒e^(i2^(n+1) θ) =e^(i2kπ) ⇒ θ_k =((2kπ)/2^(n+1) ) =((kπ)/2^n ) with k∈[[1,2^(n+1) −1]]](https://www.tinkutara.com/question/Q72991.png)
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{W}_{{n}} \left({z}\right)=\left(\mathrm{1}+{z}\right)\left(\mathrm{1}+{z}^{\mathrm{2}} \right)….\left(\mathrm{1}+{z}^{\mathrm{2}^{{n}} } \right) \\ $$$${let}\:{prove}\:{by}\:{recurrence}\:{that}\:{W}_{{n}} \left({z}\right)=\frac{\mathrm{1}−{z}^{\mathrm{2}^{{n}+\mathrm{1}} } }{\mathrm{1}−{z}} \\ $$$${n}\:=\mathrm{0}\:\rightarrow{W}_{{n}} \left({z}\right)=\mathrm{1}+{z}\:=\frac{\mathrm{1}−{z}^{\mathrm{2}} }{\mathrm{1}−{z}}\left({true}\right)\:{let}\:{suppose}\:{the}\:{relation}\:{true} \\ $$$${W}_{{n}+\mathrm{1}} =\left(\mathrm{1}+{z}\right)\left(\mathrm{1}+{z}^{\mathrm{2}} \right)….\left(\mathrm{1}+{z}^{\mathrm{2}^{{n}+\mathrm{1}} } \right)=\left(\mathrm{1}+{z}\right)\left(\mathrm{1}+{z}^{\mathrm{2}} \right)…\left(\mathrm{1}+{z}^{\mathrm{2}^{{n}} } \right)\left(\mathrm{1}+{z}^{\mathrm{2}^{{n}+\mathrm{1}} } \right) \\ $$$$=\frac{\mathrm{1}−{z}^{\mathrm{2}^{{n}+\mathrm{1}} } }{\mathrm{1}−{z}}×\left(\mathrm{1}+{z}^{\mathrm{2}^{{n}+\mathrm{1}} } \right)=\frac{\mathrm{1}+{z}^{\mathrm{2}^{{n}+\mathrm{1}} } −{z}^{\mathrm{2}^{{n}+\mathrm{1}} } \:−{z}^{\mathrm{2}^{{n}+\mathrm{2}} } }{\mathrm{1}−{z}}\:=\frac{\mathrm{1}−{z}^{\mathrm{2}^{{n}+\mathrm{2}} } }{\mathrm{1}−{z}} \\ $$$${c}/{c}\:\:\:\:\:{W}_{{n}} \left({z}\right)=\frac{\mathrm{1}−{z}^{\mathrm{2}^{{n}+\mathrm{1}} } }{\mathrm{1}−{z}}\:\:\:\:{with}\:{z}\neq\mathrm{1} \\ $$$$\left.\mathrm{2}\right)\:{P}_{{n}} \left(\theta\right)=\left(\mathrm{1}+{e}^{{i}\theta} \right)\left(\mathrm{1}+{e}^{\mathrm{2}{i}\theta} \right)…..\left(\mathrm{1}+{e}^{{i}\:\mathrm{2}^{{n}} \theta} \right) \\ $$$$=\left(\mathrm{1}+{e}^{{i}\theta} \right)\left(\mathrm{1}+\left({e}^{{i}\theta} \right)^{\mathrm{2}} \right)…..\left(\mathrm{1}+\left({e}^{{i}\theta} \right)^{\mathrm{2}^{{n}} } \right)={W}_{{n}} \left({e}^{{i}\theta} \right) \\ $$$$=\frac{\mathrm{1}−\left({e}^{{i}\theta} \right)^{\mathrm{2}^{{n}+\mathrm{1}} } }{\mathrm{1}−{e}^{{i}\theta} }\:=\frac{\mathrm{1}−{e}^{{i}\mathrm{2}^{{n}+\mathrm{1}} \theta} }{\mathrm{1}−{e}^{{i}\theta} }\:\:\:\left({if}\:\theta\:\neq\mathrm{2}{k}\pi\right) \\ $$$${P}_{{n}} \left(\theta\right)=\mathrm{0}\:\Leftrightarrow\mathrm{1}−{e}^{{i}\mathrm{2}^{{n}+\mathrm{1}} \theta} =\mathrm{0}\:\Rightarrow{e}^{{i}\mathrm{2}^{{n}+\mathrm{1}} \theta} ={e}^{{i}\mathrm{2}{k}\pi} \:\Rightarrow \\ $$$$\theta_{{k}} \:=\frac{\mathrm{2}{k}\pi}{\mathrm{2}^{{n}+\mathrm{1}} }\:=\frac{{k}\pi}{\mathrm{2}^{{n}} }\:\:\:{with}\:\:{k}\in\left[\left[\mathrm{1},\mathrm{2}^{{n}+\mathrm{1}} −\mathrm{1}\right]\right] \\ $$