let-p-x-1-ix-x-2-n-1-ix-x-2-n-1-determine-roots-of-p-x-2-find-p-x-at-form-a-i-x-i-3-ddtermne-p-x-at-form-arctan-4-factorize-p-x-inside-C-x-5-calculate-0-1-p-x-dxand-1-

Question Number 95837 by mathmax by abdo last updated on 28/May/20

let p (x) = {(1 + ix + x^{2})}^{n} - {(1 - ix + x^{2})}^{n}

1) determine roots of p (x)

2) find p (x) at form Σ a_{i} x^{i}

3) ddtermne p (x) at form \arctan

4) factorize p (x) inside C [x]

5) calculate \int_{0}^{1} p (x) dxand \int_{1}^{\infty} \frac{dx}{p (x)}

Answered by mathmax by abdo last updated on 02/Jun/20

1) p (x) = 0 \Leftrightarrow {(1 + ix + x^{2})}^{n} = {(1 - ix + x^{2})}^{n} \Rightarrow {(\frac{1 + ix + x^{2}}{1 - ix + x^{2}})}^{n} = 1 let

z = \frac{1 + ix + x^{2}}{1 - ix + x^{2}} \Rightarrow 1 + ix + x^{2} = z - izx + {zx}^{2} \Rightarrow x^{2} + ix + 1 - {zx}^{2} + izx - z = 0 \Rightarrow

(1 - z) x^{2} + i (z + 1) x + 1 - z = 0

Δ = - {(z + 1)}^{2} - 4 {(1 - z)}^{2} \Rightarrow x_{1} = \frac{- i (z + 1) + i \sqrt{{(z + 1)}^{2} + 4 {(z - 1)}^{2}}}{2 (1 - z)}

x_{2} = \frac{- i (z + 1) - i \sqrt{{(z + 1)}^{2} + 4 {(z - 1)}^{2}}}{2 (1 - z)}

z^{n} = 1 \Rightarrow z^{n} = e^{i 2 k π} \Rightarrow z_{k} = e^{\frac{i 2 k π}{n}} with k \in [[0, n - 1]] \Rightarrow the roots are

x_{1 k} = \frac{- i (z_{k} + 1) + i \sqrt{{(z_{k} + 1)}^{2} + 4 {(z_{k} - 1)}^{2}}}{2 (1 - z_{k})} or

x_{2 k} = \frac{- i (z_{k} + 1) - i \sqrt{{(z_{k} + 1)}^{2} + 4 {(z_{k} - 1)}^{2}}}{2 (1 - z_{k})}

Answered by mathmax by abdo last updated on 02/Jun/20

2) we have p (x) = {(1 + x^{2} + ix)}^{n} - {(1 + x^{2} - ix)}^{n}

= \sum_{k = 0}^{n} C_{n}^{k} {(ix)}^{k} {(1 + x^{2})}^{n - k} - \sum_{k = 0}^{n} C_{n}^{k} {(- ix)}^{k} {(1 + x^{2})}^{n - k}

= \sum_{k = 0}^{n} C_{n}^{k} {(1 + x^{2})}^{n - k} {i^{k} - {(- i)}^{k}} x^{k}

= \sum_{p = 0}^{[\frac{n - 1}{2}]} C_{n}^{2 p + 1} {(1 + x^{2})}^{n - 2 p - 1} (2 i) {(- 1)}^{p} x^{2 p + 1}

= 2 i \sum_{p = 0}^{[\frac{n - 1}{2}]} {(- 1)}^{p} C_{n}^{2 p + 1} x^{2 p + 1} (\sum_{k = 0}^{n - 2 p - 1} C_{n - 2 p - 1}^{k} x^{2 k})

= 2 i \sum_{p = 0}^{[\frac{n - 1}{2}]} {(- 1)}^{p} C_{n}^{2 p + 1} (\sum_{k = 0}^{n - 2 p - 1} C_{n - 2 p - 1}^{k} x^{2 k + 2 p + 1})