Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 63893 by mathmax by abdo last updated on 10/Jul/19

1) simplify W_n (z)=(1+z)(1+z^2 )....(1+z^2^n  ) (z from C)  2) simplify P_n (θ) =(1+e^(iθ) )(1+e^(2iθ) ).....(1+e^(i2^n θ) ) and sove  P_n (θ)=0

$$\left.\mathrm{1}\right)\:{simplify}\:{W}_{{n}} \left({z}\right)=\left(\mathrm{1}+{z}\right)\left(\mathrm{1}+{z}^{\mathrm{2}} \right)....\left(\mathrm{1}+{z}^{\mathrm{2}^{{n}} } \right)\:\left({z}\:{from}\:{C}\right) \\ $$$$\left.\mathrm{2}\right)\:{simplify}\:{P}_{{n}} \left(\theta\right)\:=\left(\mathrm{1}+{e}^{{i}\theta} \right)\left(\mathrm{1}+{e}^{\mathrm{2}{i}\theta} \right).....\left(\mathrm{1}+{e}^{{i}\mathrm{2}^{{n}} \theta} \right)\:{and}\:{sove} \\ $$$${P}_{{n}} \left(\theta\right)=\mathrm{0} \\ $$

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]]

$$\left.\mathrm{1}\right)\:{we}\:{have}\:{W}_{{n}} \left({z}\right)=\left(\mathrm{1}+{z}\right)\left(\mathrm{1}+{z}^{\mathrm{2}} \right)....\left(\mathrm{1}+{z}^{\mathrm{2}^{{n}} } \right) \\ $$$${let}\:{prove}\:{by}\:{recurrence}\:{that}\:{W}_{{n}} \left({z}\right)=\frac{\mathrm{1}−{z}^{\mathrm{2}^{{n}+\mathrm{1}} } }{\mathrm{1}−{z}} \\ $$$${n}\:=\mathrm{0}\:\rightarrow{W}_{{n}} \left({z}\right)=\mathrm{1}+{z}\:=\frac{\mathrm{1}−{z}^{\mathrm{2}} }{\mathrm{1}−{z}}\left({true}\right)\:{let}\:{suppose}\:{the}\:{relation}\:{true} \\ $$$${W}_{{n}+\mathrm{1}} =\left(\mathrm{1}+{z}\right)\left(\mathrm{1}+{z}^{\mathrm{2}} \right)....\left(\mathrm{1}+{z}^{\mathrm{2}^{{n}+\mathrm{1}} } \right)=\left(\mathrm{1}+{z}\right)\left(\mathrm{1}+{z}^{\mathrm{2}} \right)...\left(\mathrm{1}+{z}^{\mathrm{2}^{{n}} } \right)\left(\mathrm{1}+{z}^{\mathrm{2}^{{n}+\mathrm{1}} } \right) \\ $$$$=\frac{\mathrm{1}−{z}^{\mathrm{2}^{{n}+\mathrm{1}} } }{\mathrm{1}−{z}}×\left(\mathrm{1}+{z}^{\mathrm{2}^{{n}+\mathrm{1}} } \right)=\frac{\mathrm{1}+{z}^{\mathrm{2}^{{n}+\mathrm{1}} } −{z}^{\mathrm{2}^{{n}+\mathrm{1}} } \:−{z}^{\mathrm{2}^{{n}+\mathrm{2}} } }{\mathrm{1}−{z}}\:=\frac{\mathrm{1}−{z}^{\mathrm{2}^{{n}+\mathrm{2}} } }{\mathrm{1}−{z}} \\ $$$${c}/{c}\:\:\:\:\:{W}_{{n}} \left({z}\right)=\frac{\mathrm{1}−{z}^{\mathrm{2}^{{n}+\mathrm{1}} } }{\mathrm{1}−{z}}\:\:\:\:{with}\:{z}\neq\mathrm{1} \\ $$$$\left.\mathrm{2}\right)\:{P}_{{n}} \left(\theta\right)=\left(\mathrm{1}+{e}^{{i}\theta} \right)\left(\mathrm{1}+{e}^{\mathrm{2}{i}\theta} \right).....\left(\mathrm{1}+{e}^{{i}\:\mathrm{2}^{{n}} \theta} \right) \\ $$$$=\left(\mathrm{1}+{e}^{{i}\theta} \right)\left(\mathrm{1}+\left({e}^{{i}\theta} \right)^{\mathrm{2}} \right).....\left(\mathrm{1}+\left({e}^{{i}\theta} \right)^{\mathrm{2}^{{n}} } \right)={W}_{{n}} \left({e}^{{i}\theta} \right) \\ $$$$=\frac{\mathrm{1}−\left({e}^{{i}\theta} \right)^{\mathrm{2}^{{n}+\mathrm{1}} } }{\mathrm{1}−{e}^{{i}\theta} }\:=\frac{\mathrm{1}−{e}^{{i}\mathrm{2}^{{n}+\mathrm{1}} \theta} }{\mathrm{1}−{e}^{{i}\theta} }\:\:\:\left({if}\:\theta\:\neq\mathrm{2}{k}\pi\right) \\ $$$${P}_{{n}} \left(\theta\right)=\mathrm{0}\:\Leftrightarrow\mathrm{1}−{e}^{{i}\mathrm{2}^{{n}+\mathrm{1}} \theta} =\mathrm{0}\:\Rightarrow{e}^{{i}\mathrm{2}^{{n}+\mathrm{1}} \theta} ={e}^{{i}\mathrm{2}{k}\pi} \:\Rightarrow \\ $$$$\theta_{{k}} \:=\frac{\mathrm{2}{k}\pi}{\mathrm{2}^{{n}+\mathrm{1}} }\:=\frac{{k}\pi}{\mathrm{2}^{{n}} }\:\:\:{with}\:\:{k}\in\left[\left[\mathrm{1},\mathrm{2}^{{n}+\mathrm{1}} −\mathrm{1}\right]\right] \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com