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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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