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Question Number 26997 by abdo imad last updated on 01/Jan/18

let give ξ ∈C and ξ^n =1 (ξ is the n^(me) root of 1) simplify A= 1+ξ^p +ξ^(2p) +... +ξ^((n−1)p) and B= 1+2ξ +3ξ^2 +...+nξ^(n−1) .

$${let}\:{give}\:\xi\:\in\mathbb{C}\:{and}\:\xi^{{n}} =\mathrm{1}\:\left(\xi\:{is}\:{the}\:{n}^{{me}} \:{root}\:{of}\:\mathrm{1}\right) \\ $$$${simplify}\:\:{A}=\:\mathrm{1}+\xi^{{p}} +\xi^{\mathrm{2}{p}} +...\:+\xi^{\left({n}−\mathrm{1}\right){p}} \\ $$$${and}\:{B}=\:\mathrm{1}+\mathrm{2}\xi\:+\mathrm{3}\xi^{\mathrm{2}} +...+{n}\xi^{{n}−\mathrm{1}} . \\ $$

Commented by prakash jain last updated on 01/Jan/18

A=((ξ^(np) −1)/(ξ^p −1))=(((ξ^n )^p −1)/(ξ^p −1))=0 if ξ^p ≠1 A=n if ξ^p =1

$${A}=\frac{\xi^{{np}} −\mathrm{1}}{\xi^{{p}} −\mathrm{1}}=\frac{\left(\xi^{{n}} \right)^{{p}} −\mathrm{1}}{\xi^{{p}} −\mathrm{1}}=\mathrm{0}\:\mathrm{if}\:\xi^{{p}} \neq\mathrm{1} \\ $$$${A}={n}\:\mathrm{if}\:\xi^{{p}} =\mathrm{1} \\ $$

Commented by abdo imad last updated on 02/Jan/18

value of A if ξ^p =1 A= n if ξ^p ≠1 A= Σ_(k=0) ^(n−1) ξ^(kp) = ((1− (ξ^p )^n )/(1−ξ^p )) =((1−(ξ^n )^p )/(1−ξ^p )) =0 value of B if ξ=1 B=1+2+3+....+n =((n(n+1))/2) if ξ≠1 B= p(x) with p(x)= 1+2x +3x^2 +....+nx^(n−1) and x≠1 we have ∫p(x)dx= x +x^2 +x^3 +....+x^n +λ λ=p(0)=1⇒p(x)= 1+x+x^2 +...+x^n =((x^(n+1) −1)/(x−1)) ⇒p(x)=(d/dx)(((x^(n+1) −1)/(x−1))) =(((n+1)x^n (x−1)−(x^(n+1) −1))/((x−1)^2 )) =(((n+1)x^(n+1) −(n+1)x^n −x^(n+1) +1)/((x−1)^2 )) =((nx^(n+1) −(n+1)x^n +1)/((x−1)^(2 ) )) and with ξ^n =1 B=p(ξ)= ((nξ^(n+1) −(n+1)ξ +1)/((ξ−1)^2 )) B= ((nξ −nξ +1−ξ)/((1−ξ)^2 )) ⇒B= (1/(1−ξ)) .

$${value}\:{of}\:{A} \\ $$$${if}\:\xi^{{p}} =\mathrm{1}\:\:{A}=\:{n} \\ $$$${if}\:\xi^{{p}} \neq\mathrm{1}\:\:{A}=\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\xi^{{kp}} \:=\:\frac{\mathrm{1}−\:\left(\xi^{{p}} \right)^{{n}} }{\mathrm{1}−\xi^{{p}} } \\ $$$$=\frac{\mathrm{1}−\left(\xi^{{n}} \right)^{{p}} }{\mathrm{1}−\xi^{{p}} }\:=\mathrm{0} \\ $$$${value}\:{of}\:{B} \\ $$$${if}\:\xi=\mathrm{1}\:\:\:{B}=\mathrm{1}+\mathrm{2}+\mathrm{3}+....+{n}\:=\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$${if}\:\xi\neq\mathrm{1}\:\:{B}=\:{p}\left({x}\right)\:{with}\:\:{p}\left({x}\right)=\:\mathrm{1}+\mathrm{2}{x}\:+\mathrm{3}{x}^{\mathrm{2}} +....+{nx}^{{n}−\mathrm{1}} \\ $$$${and}\:{x}\neq\mathrm{1}\:{we}\:{have} \\ $$$$\int{p}\left({x}\right){dx}=\:{x}\:+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} \:+....+{x}^{{n}} \:+\lambda \\ $$$$\lambda={p}\left(\mathrm{0}\right)=\mathrm{1}\Rightarrow{p}\left({x}\right)=\:\mathrm{1}+{x}+{x}^{\mathrm{2}} +...+{x}^{{n}} \\ $$$$=\frac{{x}^{{n}+\mathrm{1}} −\mathrm{1}}{{x}−\mathrm{1}}\:\:\:\Rightarrow{p}\left({x}\right)=\frac{{d}}{{dx}}\left(\frac{{x}^{{n}+\mathrm{1}} −\mathrm{1}}{{x}−\mathrm{1}}\right) \\ $$$$=\frac{\left({n}+\mathrm{1}\right){x}^{{n}} \left({x}−\mathrm{1}\right)−\left({x}^{{n}+\mathrm{1}} −\mathrm{1}\right)}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\frac{\left({n}+\mathrm{1}\right){x}^{{n}+\mathrm{1}} −\left({n}+\mathrm{1}\right){x}^{{n}} \:−{x}^{{n}+\mathrm{1}} +\mathrm{1}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\frac{{nx}^{{n}+\mathrm{1}} −\left({n}+\mathrm{1}\right){x}^{{n}} +\mathrm{1}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}\:} }\:{and}\:{with}\:\xi^{{n}} =\mathrm{1} \\ $$$${B}={p}\left(\xi\right)=\:\frac{{n}\xi^{{n}+\mathrm{1}} −\left({n}+\mathrm{1}\right)\xi\:+\mathrm{1}}{\left(\xi−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${B}=\:\frac{{n}\xi\:−{n}\xi\:+\mathrm{1}−\xi}{\left(\mathrm{1}−\xi\right)^{\mathrm{2}} }\:\Rightarrow{B}=\:\frac{\mathrm{1}}{\mathrm{1}−\xi}\:. \\ $$$$ \\ $$

Commented by abdo imad last updated on 01/Jan/18

if ξ=1 B=((n(n+1))/2) .

$${if}\:\xi=\mathrm{1}\:\:{B}=\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\:. \\ $$

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