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Question Number 61835 by alphaprime last updated on 09/Jun/19
If α = Cis(2π/7) and f(x) = A_0  + Σ_(n=1) ^(14) A_n x^n    Then prove that Σ_(α=0) ^6 f(α^n x)= 7(A_0 +A_7 x^7 +A_(14) x^(14) )  where Cisθ = Cosθ + iSinθ
$$\mathrm{If}\:\alpha\:=\:\mathrm{Cis}\left(\mathrm{2}\pi/\mathrm{7}\right)\:\mathrm{and}\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\mathrm{A}_{\mathrm{0}} \:+\:\sum_{\mathrm{n}=\mathrm{1}} ^{\mathrm{14}} \mathrm{A}_{\mathrm{n}} \mathrm{x}^{\mathrm{n}} \: \\ $$$$\mathrm{Then}\:\mathrm{prove}\:\mathrm{that}\:\sum_{\alpha=\mathrm{0}} ^{\mathrm{6}} \mathrm{f}\left(\alpha^{\mathrm{n}} \mathrm{x}\right)=\:\mathrm{7}\left(\mathrm{A}_{\mathrm{0}} +\mathrm{A}_{\mathrm{7}} \mathrm{x}^{\mathrm{7}} +\mathrm{A}_{\mathrm{14}} \mathrm{x}^{\mathrm{14}} \right) \\ $$$$\mathrm{where}\:\mathrm{Cis}\theta\:=\:\mathrm{Cos}\theta\:+\:\mathrm{iSin}\theta \\ $$
Commented by arcana last updated on 10/Jun/19
Σ_(n=0) ^6 f(α^n x)=f(x)+(αx)+...+f(α^6 x)  =(A_0  + Σ_(n=1) ^(14) A_n x^n )+(A_0  + Σ_(n=1) ^(14) A_n (αx)^n )+...+(A_0  + Σ_(n=1) ^(14) A_n (α^6 x)^n )  =7A_0 +A_1 Σ_(n=0) ^6 α^n x +A_2 Σ_(n=0) ^6 α^(2n) x^2 +...+A_(14) Σ_(n=0) ^6 α^(14n) x^(14)   =7A_0 +A_1 xΣ_(n=0) ^6 α^n  +A_2 x^2 Σ_(n=0) ^6 α^(2n) +...+A_(14) x^(14) Σ_(n=0) ^6 α^(14n)     but prop. raices complex  1+α+α^2 +α^3 +α^4 +α^5 +α^6 =0  ⇒  =7A_0 +A_7 x^7 Σ_0 ^6 α^(7n) +A_(14) x^(14) Σ_0 ^6 α^(14n)  (uses powers mod 7)  α^7 =Cis^7 (2π/7)=Cis(2π)=1  ⇒α^(7n) =1⇒α^(14n) =1  ⇒  =7A_0 +7A_7 x^6 +7A_(14) x^(14)
$$\underset{{n}=\mathrm{0}} {\overset{\mathrm{6}} {\sum}}{f}\left(\alpha^{{n}} {x}\right)={f}\left({x}\right)+\left(\alpha{x}\right)+…+{f}\left(\alpha^{\mathrm{6}} {x}\right) \\ $$$$=\left(\mathrm{A}_{\mathrm{0}} \:+\:\sum_{\mathrm{n}=\mathrm{1}} ^{\mathrm{14}} \mathrm{A}_{\mathrm{n}} \mathrm{x}^{\mathrm{n}} \right)+\left(\mathrm{A}_{\mathrm{0}} \:+\:\sum_{\mathrm{n}=\mathrm{1}} ^{\mathrm{14}} \mathrm{A}_{\mathrm{n}} \left(\alpha\mathrm{x}\right)^{{n}} \right)+…+\left(\mathrm{A}_{\mathrm{0}} \:+\:\sum_{\mathrm{n}=\mathrm{1}} ^{\mathrm{14}} \mathrm{A}_{\mathrm{n}} \left(\alpha^{\mathrm{6}} \mathrm{x}\right)^{{n}} \right) \\ $$$$=\mathrm{7A}_{\mathrm{0}} +\mathrm{A}_{\mathrm{1}} \underset{{n}=\mathrm{0}} {\overset{\mathrm{6}} {\sum}}\alpha^{{n}} {x}\:+\mathrm{A}_{\mathrm{2}} \underset{{n}=\mathrm{0}} {\overset{\mathrm{6}} {\sum}}\alpha^{\mathrm{2}{n}} {x}^{\mathrm{2}} +…+\mathrm{A}_{\mathrm{14}} \underset{{n}=\mathrm{0}} {\overset{\mathrm{6}} {\sum}}\alpha^{\mathrm{14}{n}} {x}^{\mathrm{14}} \\ $$$$=\mathrm{7A}_{\mathrm{0}} +\mathrm{A}_{\mathrm{1}} {x}\underset{{n}=\mathrm{0}} {\overset{\mathrm{6}} {\sum}}\alpha^{{n}} \:+\mathrm{A}_{\mathrm{2}} {x}^{\mathrm{2}} \underset{{n}=\mathrm{0}} {\overset{\mathrm{6}} {\sum}}\alpha^{\mathrm{2}{n}} +…+\mathrm{A}_{\mathrm{14}} {x}^{\mathrm{14}} \underset{{n}=\mathrm{0}} {\overset{\mathrm{6}} {\sum}}\alpha^{\mathrm{14}{n}} \\ $$$$ \\ $$$$\mathrm{but}\:\mathrm{prop}.\:\mathrm{raices}\:\mathrm{complex} \\ $$$$\mathrm{1}+\alpha+\alpha^{\mathrm{2}} +\alpha^{\mathrm{3}} +\alpha^{\mathrm{4}} +\alpha^{\mathrm{5}} +\alpha^{\mathrm{6}} =\mathrm{0} \\ $$$$\Rightarrow \\ $$$$=\mathrm{7A}_{\mathrm{0}} +\mathrm{A}_{\mathrm{7}} {x}^{\mathrm{7}} \underset{\mathrm{0}} {\overset{\mathrm{6}} {\sum}}\alpha^{\mathrm{7}{n}} +\mathrm{A}_{\mathrm{14}} {x}^{\mathrm{14}} \underset{\mathrm{0}} {\overset{\mathrm{6}} {\sum}}\alpha^{\mathrm{14}{n}} \:\left({uses}\:{powers}\:{mod}\:\mathrm{7}\right) \\ $$$$\alpha^{\mathrm{7}} =\mathrm{Cis}^{\mathrm{7}} \left(\mathrm{2}\pi/\mathrm{7}\right)=\mathrm{Cis}\left(\mathrm{2}\pi\right)=\mathrm{1} \\ $$$$\Rightarrow\alpha^{\mathrm{7}{n}} =\mathrm{1}\Rightarrow\alpha^{\mathrm{14}{n}} =\mathrm{1} \\ $$$$\Rightarrow \\ $$$$=\mathrm{7A}_{\mathrm{0}} +\mathrm{7A}_{\mathrm{7}} {x}^{\mathrm{6}} +\mathrm{7A}_{\mathrm{14}} {x}^{\mathrm{14}} \\ $$

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