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

l e t g i v e ξ \in C a n d ξ^{n} = 1 (ξ i s t h e n^{m e} r o o t o f 1)

s i m p l i f y A = 1 + ξ^{p} + ξ^{2 p} + \dots + ξ^{(n - 1) p}

a n d B = 1 + 2 ξ + 3 ξ^{2} + \dots + n ξ^{n - 1} .

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

A = \frac{ξ^{n p} - 1}{ξ^{p} - 1} = \frac{{(ξ^{n})}^{p} - 1}{ξ^{p} - 1} = 0 if ξ^{p} \neq 1

A = n if ξ^{p} = 1

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

v a l u e o f A

i f ξ^{p} = 1 A = n

i f ξ^{p} \neq 1 A = \sum_{k = 0}^{n - 1} ξ^{k p} = \frac{1 - {(ξ^{p})}^{n}}{1 - ξ^{p}}

= \frac{1 - {(ξ^{n})}^{p}}{1 - ξ^{p}} = 0

v a l u e o f B

i f ξ = 1 B = 1 + 2 + 3 + \dots . + n = \frac{n (n + 1)}{2}

i f ξ \neq 1 B = p (x) w i t h p (x) = 1 + 2 x + 3 x^{2} + \dots . + {n x}^{n - 1}

a n d x \neq 1 w e h a v e

\int p (x) d x = x + x^{2} + x^{3} + \dots . + x^{n} + λ

λ = p (0) = 1 \Rightarrow p (x) = 1 + x + x^{2} + \dots + x^{n}

= \frac{x^{n + 1} - 1}{x - 1} \Rightarrow p (x) = \frac{d}{d x} (\frac{x^{n + 1} - 1}{x - 1})

= \frac{(n + 1) x^{n} (x - 1) - (x^{n + 1} - 1)}{{(x - 1)}^{2}}

= \frac{(n + 1) x^{n + 1} - (n + 1) x^{n} - x^{n + 1} + 1}{{(x - 1)}^{2}}

= \frac{{n x}^{n + 1} - (n + 1) x^{n} + 1}{{(x - 1)}^{2}} a n d w i t h ξ^{n} = 1

B = p (ξ) = \frac{n ξ^{n + 1} - (n + 1) ξ + 1}{{(ξ - 1)}^{2}}

B = \frac{n ξ - n ξ + 1 - ξ}{{(1 - ξ)}^{2}} \Rightarrow B = \frac{1}{1 - ξ} .

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

i f ξ = 1 B = \frac{n (n + 1)}{2} .