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Question Number 61835 by alphaprime last updated on 09/Jun/19

$$\mathrm{If}\alpha =\mathrm{Cis}\left(2\pi /7\right)\mathrm{and}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{A}}_0+\sum _{\mathrm{n}=1}^{14}{\mathrm{A}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}$$$$\mathrm{Then}\mathrm{prove}\mathrm{that}\sum _{\alpha =0}^6\mathrm{f}\left({\alpha }^{\mathrm{n}}\mathrm{x}\right)=7({\mathrm{A}}_0+{\mathrm{A}}_7{\mathrm{x}}^7+{\mathrm{A}}_{14}{\mathrm{x}}^{14})$$$$\mathrm{where}\mathrm{Cis}\theta =\mathrm{Cos}\theta +\mathrm{iSin}\theta$$

Commented by arcana last updated on 10/Jun/19

$$\underset{n=0}{\sum ^6}f\left({\alpha }^{n}x\right)=f\left(x\right)+\left(\alpha x\right)+...+f\left({\alpha }^{6}x\right)$$$$=({\mathrm{A}}_0+\sum _{\mathrm{n}=1}^{14}{\mathrm{A}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}})+({\mathrm{A}}_0+\sum _{\mathrm{n}=1}^{14}{\mathrm{A}}_{\mathrm{n}}{\left(\alpha \mathrm{x}\right)}^n)+...+({\mathrm{A}}_0+\sum _{\mathrm{n}=1}^{14}{\mathrm{A}}_{\mathrm{n}}{\left({\alpha }^6\mathrm{x}\right)}^n)$$$$={7\mathrm{A}}_0+{\mathrm{A}}_1\underset{n=0}{\sum ^6}{\alpha }^{n}x+{\mathrm{A}}_2\underset{n=0}{\sum ^6}{\alpha }^{2n}{x}^2+...+{\mathrm{A}}_{14}\underset{n=0}{\sum ^6}{\alpha }^{14n}{x}^{14}$$$$={7\mathrm{A}}_0+{\mathrm{A}}_{1}x\underset{n=0}{\sum ^6}{\alpha }^n+{\mathrm{A}}_2{x}^2\underset{n=0}{\sum ^6}{\alpha }^{2n}+...+{\mathrm{A}}_{14}{x}^{14}\underset{n=0}{\sum ^6}{\alpha }^{14n}$$$$$$$$\mathrm{but}\mathrm{prop}.\mathrm{raices}\mathrm{complex}$$$$1+\alpha +{\alpha }^2+{\alpha }^3+{\alpha }^4+{\alpha }^5+{\alpha }^6=0$$$$\Rightarrow$$$$={7\mathrm{A}}_0+{\mathrm{A}}_7{x}^7\underset{0}{\sum ^6}{\alpha }^{7n}+{\mathrm{A}}_{14}{x}^{14}\underset{0}{\sum ^6}{\alpha }^{14n}\left(usespowersmod7\right)$$$${\alpha }^7={\mathrm{Cis}}^7\left(2\pi /7\right)=\mathrm{Cis}\left(2\pi \right)=1$$$$\Rightarrow {\alpha }^{7n}=1\Rightarrow {\alpha }^{14n}=1$$$$\Rightarrow$$$$={7\mathrm{A}}_0+{7\mathrm{A}}_7{x}^6+{7\mathrm{A}}_{14}{x}^{14}$$

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