If-Cis-2pi-7-and-f-x-A-0-n-1-14-A-n-x-n-Then-prove-that-0-6-f-n-x-7-A-0-A-7-x-7-A-14-x-14-where-Cis-Cos-iSin-

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

If α = Cis (2 π / 7) and f (x) = A_{0} + \sum_{n = 1}^{14} A_{n} x^{n}

Then prove that \sum_{α = 0}^{6} f (α^{n} x) = 7 (A_{0} + A_{7} x^{7} + A_{14} x^{14})

where Cis θ = Cos θ + iSin θ

Commented by arcana last updated on 10/Jun/19

\underset{n = 0}{\sum^{6}} f (α^{n} x) = f (x) + (α x) + \dots + f (α^{6} x)

= (A_{0} + \sum_{n = 1}^{14} A_{n} x^{n}) + (A_{0} + \sum_{n = 1}^{14} A_{n} {(α x)}^{n}) + \dots + (A_{0} + \sum_{n = 1}^{14} A_{n} {(α^{6} x)}^{n})

= {7 A}_{0} + A_{1} \underset{n = 0}{\sum^{6}} α^{n} x + A_{2} \underset{n = 0}{\sum^{6}} α^{2 n} x^{2} + \dots + A_{14} \underset{n = 0}{\sum^{6}} α^{14 n} x^{14}

= {7 A}_{0} + A_{1} x \underset{n = 0}{\sum^{6}} α^{n} + A_{2} x^{2} \underset{n = 0}{\sum^{6}} α^{2 n} + \dots + A_{14} x^{14} \underset{n = 0}{\sum^{6}} α^{14 n}

but prop . raices complex

1 + α + α^{2} + α^{3} + α^{4} + α^{5} + α^{6} = 0

\Rightarrow

= {7 A}_{0} + A_{7} x^{7} \underset{0}{\sum^{6}} α^{7 n} + A_{14} x^{14} \underset{0}{\sum^{6}} α^{14 n} (u s e s p o w e r s m o d 7)

α^{7} = {Cis}^{7} (2 π / 7) = Cis (2 π) = 1

\Rightarrow α^{7 n} = 1 \Rightarrow α^{14 n} = 1

\Rightarrow

= {7 A}_{0} + {7 A}_{7} x^{6} + {7 A}_{14} x^{14}

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