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let-U-n-z-C-z-n-1-simplify-p-0-2n-1-w-p-with-w-U-n-and-p-0-2n-1-2w-1-p-




Question Number 148501 by mathmax by abdo last updated on 28/Jul/21
let U_n ={z∈C /z^n  =1}  simplify  Σ_(p=0) ^(2n−1)  w^p         with w∈U_n      and  Σ_(p=0) ^(2n−1) (2w +1)^p
$$\mathrm{let}\:\mathrm{U}_{\mathrm{n}} =\left\{\mathrm{z}\in\mathrm{C}\:/\mathrm{z}^{\mathrm{n}} \:=\mathrm{1}\right\}\:\:\mathrm{simplify} \\ $$$$\sum_{\mathrm{p}=\mathrm{0}} ^{\mathrm{2n}−\mathrm{1}} \:\mathrm{w}^{\mathrm{p}} \:\:\:\:\:\:\:\:\mathrm{with}\:\mathrm{w}\in\mathrm{U}_{\mathrm{n}} \:\:\: \\ $$$$\mathrm{and}\:\:\sum_{\mathrm{p}=\mathrm{0}} ^{\mathrm{2n}−\mathrm{1}} \left(\mathrm{2w}\:+\mathrm{1}\right)^{\mathrm{p}} \\ $$
Answered by Olaf_Thorendsen last updated on 28/Jul/21
• w = 1  S_n (1) = Σ_(p=0) ^(2n−1) 1^p  = 2n  (trivial)    • w^n  = 1 and w ≠ 1  S_n (w) = Σ_(p=0) ^(2n−1) w^p  =((1−w^(2n) )/(1−w))  = ((1−1)/(1−w))  = 0    T_n (w) = Σ_(p=0) ^(2n−1) (2w+1)^p   T_n (w) =((1−(2w+1)^(2n) )/(1−(2w+1)))  = (((2w+1)^(2n) −1)/(2w))
$$\bullet\:{w}\:=\:\mathrm{1} \\ $$$$\mathrm{S}_{{n}} \left(\mathrm{1}\right)\:=\:\underset{{p}=\mathrm{0}} {\overset{\mathrm{2}{n}−\mathrm{1}} {\sum}}\mathrm{1}^{{p}} \:=\:\mathrm{2}{n}\:\:\left(\mathrm{trivial}\right) \\ $$$$ \\ $$$$\bullet\:{w}^{{n}} \:=\:\mathrm{1}\:\mathrm{and}\:{w}\:\neq\:\mathrm{1} \\ $$$$\mathrm{S}_{{n}} \left({w}\right)\:=\:\underset{{p}=\mathrm{0}} {\overset{\mathrm{2}{n}−\mathrm{1}} {\sum}}{w}^{{p}} \:=\frac{\mathrm{1}−{w}^{\mathrm{2}{n}} }{\mathrm{1}−{w}}\:\:=\:\frac{\mathrm{1}−\mathrm{1}}{\mathrm{1}−{w}}\:\:=\:\mathrm{0} \\ $$$$ \\ $$$$\mathrm{T}_{{n}} \left({w}\right)\:=\:\underset{{p}=\mathrm{0}} {\overset{\mathrm{2}{n}−\mathrm{1}} {\sum}}\left(\mathrm{2}{w}+\mathrm{1}\right)^{{p}} \\ $$$$\mathrm{T}_{{n}} \left({w}\right)\:=\frac{\mathrm{1}−\left(\mathrm{2}{w}+\mathrm{1}\right)^{\mathrm{2}{n}} }{\mathrm{1}−\left(\mathrm{2}{w}+\mathrm{1}\right)}\:\:=\:\frac{\left(\mathrm{2}{w}+\mathrm{1}\right)^{\mathrm{2}{n}} −\mathrm{1}}{\mathrm{2}{w}} \\ $$
Answered by mathmax by abdo last updated on 28/Jul/21
if w=1   A_n =Σ_(p=0) ^(2n−1)  w^p  =2n  if w≠1   Σ_(p=0) ^(2n−1)  w^p  =((1−w^(2n) )/(1−w))=((1−(w^n )^2 )/(1−w))=((1−1)/(1−w))=0  alsoB_n = Σ_(p=0) ^(2n−1)  (2w+1)^p   the roots of z^n  =1 are z_k =e^(i((2kπ)/n))   and 0≤k≤n−1 ⇒w≠−(1/2) ⇒B_n =((1−(2w+1)^(2n) )/(1−(2w+1)))  =(1/(2w)){(2w+1)^(2n) −1}  =(1/(2w)){Σ_(k=0) ^(2n) C_(2n) ^k  (2w)^k −1}  =(1/(2w))Σ_(k=0) ^(2n)  C_(2n) ^k  2^k  w^k
$$\mathrm{if}\:\mathrm{w}=\mathrm{1}\:\:\:\mathrm{A}_{\mathrm{n}} =\sum_{\mathrm{p}=\mathrm{0}} ^{\mathrm{2n}−\mathrm{1}} \:\mathrm{w}^{\mathrm{p}} \:=\mathrm{2n} \\ $$$$\mathrm{if}\:\mathrm{w}\neq\mathrm{1}\:\:\:\sum_{\mathrm{p}=\mathrm{0}} ^{\mathrm{2n}−\mathrm{1}} \:\mathrm{w}^{\mathrm{p}} \:=\frac{\mathrm{1}−\mathrm{w}^{\mathrm{2n}} }{\mathrm{1}−\mathrm{w}}=\frac{\mathrm{1}−\left(\mathrm{w}^{\mathrm{n}} \right)^{\mathrm{2}} }{\mathrm{1}−\mathrm{w}}=\frac{\mathrm{1}−\mathrm{1}}{\mathrm{1}−\mathrm{w}}=\mathrm{0} \\ $$$$\mathrm{alsoB}_{\mathrm{n}} =\:\sum_{\mathrm{p}=\mathrm{0}} ^{\mathrm{2n}−\mathrm{1}} \:\left(\mathrm{2w}+\mathrm{1}\right)^{\mathrm{p}} \:\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{z}^{\mathrm{n}} \:=\mathrm{1}\:\mathrm{are}\:\mathrm{z}_{\mathrm{k}} =\mathrm{e}^{\mathrm{i}\frac{\mathrm{2k}\pi}{\mathrm{n}}} \\ $$$$\mathrm{and}\:\mathrm{0}\leqslant\mathrm{k}\leqslant\mathrm{n}−\mathrm{1}\:\Rightarrow\mathrm{w}\neq−\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\mathrm{B}_{\mathrm{n}} =\frac{\mathrm{1}−\left(\mathrm{2w}+\mathrm{1}\right)^{\mathrm{2n}} }{\mathrm{1}−\left(\mathrm{2w}+\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2w}}\left\{\left(\mathrm{2w}+\mathrm{1}\right)^{\mathrm{2n}} −\mathrm{1}\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2w}}\left\{\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{2n}} \mathrm{C}_{\mathrm{2n}} ^{\mathrm{k}} \:\left(\mathrm{2w}\right)^{\mathrm{k}} −\mathrm{1}\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2w}}\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{2n}} \:\mathrm{C}_{\mathrm{2n}} ^{\mathrm{k}} \:\mathrm{2}^{\mathrm{k}} \:\mathrm{w}^{\mathrm{k}} \\ $$$$ \\ $$

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