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 ∫_0 ^1 p(x)dxand ∫


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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
	$2) we have p(x) =(1+x^2 +ix)^n −(1+x^2 −ix)^n =Σ_(k=0) ^n C_n ^k (ix)^k (1+x^2 )^(n−k) −Σ_(k=0) ^n C_n ^k (−ix)^k (1+x^2 )^(n−k) =Σ_(k=0) ^n C_n ^k (1+x^2 )^(n−k) { i^k −(−i)^k }x^k =Σ_(p=0) ^([((n−1)/2)]) C_n ^(2p+1) (1+x^2 )^(n−2p−1) (2i)(−1)^p x^(2p+1) =2iΣ_(p=0) ^([((n−1)/2)]) (−1)^p C_n ^(2p+1) x^(2p+1) (Σ_(k=0) ^(n−2p−1) C_(n−2p−1) ^k x^(2k) ) =2i Σ_(p=0) ^([((n−1)/2)]) (−1)^p C_n ^(2p+1) (Σ_(k=0) ^(n−2p−1) C_(n−2p−1) ^k x^(2k+2p+1) )$
	$2) we have p (x) = {(1 + x^{2} + ix)}^{n} - {(1 + x^{2} - ix)}^{n}$ $= \sum_{k = 0}^{n} C_{n}^{k} {(ix)}^{k} {(1 + x^{2})}^{n - k} - \sum_{k = 0}^{n} C_{n}^{k} {(- ix)}^{k} {(1 + x^{2})}^{n - k}$ $= \sum_{k = 0}^{n} C_{n}^{k} {(1 + x^{2})}^{n - k} {i^{k} - {(- i)}^{k}} x^{k}$ $= \sum_{p = 0}^{[\frac{n - 1}{2}]} C_{n}^{2 p + 1} {(1 + x^{2})}^{n - 2 p - 1} (2 i) {(- 1)}^{p} x^{2 p + 1}$ $= 2 i \sum_{p = 0}^{[\frac{n - 1}{2}]} {(- 1)}^{p} C_{n}^{2 p + 1} x^{2 p + 1} (\sum_{k = 0}^{n - 2 p - 1} C_{n - 2 p - 1}^{k} x^{2 k})$ $= 2 i \sum_{p = 0}^{[\frac{n - 1}{2}]} {(- 1)}^{p} C_{n}^{2 p + 1} (\sum_{k = 0}^{n - 2 p - 1} C_{n - 2 p - 1}^{k} x^{2 k + 2 p + 1})$
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