let P(x) =(1+x^2 )(1+x^4 )...(1+x^2^n ) 1) solve inside C the equation P(x)=0 2) factorize P(x) inside C[x] 3) calvulate P^′ (x) 4) decompose F =(1/(P(x)))


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Question Number 96375 by mathmax by abdo last updated on 01/Jun/20

$let P (x) = (1 + x^{2}) (1 + x^{4}) . . . (1 + x^{2^{n}})$ $1) solve inside C the equation P (x) = 0$ $2) factorize P (x) inside C [x]$ $3) calvulate P^{^{'}} (x)$ $4) decompose F = \frac{1}{P (x)}$
	Answered by 1549442205 last updated on 01/Jun/20

	$1 / P (x) = 0 \Leftrightarrow x^{2^{n}} + 1 = 0 (n = 1, 2, 3, . . .) \Leftrightarrow$ $x^{2^{n}} = - 1 = [\cos (2 k + 1) π + isin (2 k + 1) π]$ $\Rightarrow x = {[\cos (2 k + 1) π + isin (2 k + 1) π]}^{\frac{1}{2^{n}}}$ $= \cos \frac{(2 k + 1) π}{2^{n}} + i . \sin \frac{(2 k + 1) π}{2^{n}} (k = 0, 1, 2,$ $. . ., 2^{n} - 1) . It follows that eq . P (x) = 0 has$ $2 + 4 + 8 + . . . + 2^{n} = 2 (2^{n} - 1) roots$ $2 / P (x) = \underset{n = 1}{\overset{2^{n}}{Π}} \underset{k = 0}{\overset{2^{n} - 1}{Π}} (\cos \frac{(2 k + 1) π}{2^{n}} + i . \sin \frac{(2 k + 1) π}{2^{n}})$ $3 / P (x) = \frac{1 - x^{2 n + 1}}{1 - x^{2}} \Rightarrow P^{'} (x) = \frac{- (2 n + 1) x^{2 n} (1 - x^{2}) + 2 x (1 - x^{2 n + 1})}{{(1 - x^{2})}^{2}}$ $= \frac{(2 n - 1) x^{2 n + 2} - (2 n + 1) x^{2 n} + 2 x}{{(1 - x^{2})}^{2}}$
	Answered by abdomathmax last updated on 01/Jun/20
	$1) let prove by recurrence that (1+x^2 )(1+x^4 )....(1+x^2^n ) =((1−x^2^(n+1) )/(1−x^2 )) (x^2 ≠1) n=1 (1+x^2 )=((1−x^4 )/(1−x^2 )) (true) let suppose (p_n )true ⇒(1+x^2 )(1+x^4 )....(1+x^2^(n+1) ) =(1+x^2 )(1+x^4 )....(1+x^2^n )(1+x^2^(n+1) ) =((1−x^2^(n+1) )/(1−x^2 ))×(1+x^2^(n+1) ) =((1−(x^2^(n+1) )^2 )/(1−x^2 )) =((1−x^2^(n+2) )/(1−x^2 )) so the relation is true at term (n+1) ⇒P(x)=0 ⇔((1−x^2^(n+1) )/(1−x^2 )) =0 and x≠+^− 1 ⇒x^2^(n+1) =1 let m =2^(n+1) (e) ⇒x^m =1 =e^(i(2kπ)) ⇒ the roots are z_k =e^(i(((2kπ)/m))) k∈{0,....m−1} ⇒ z_k =e^(i(((kπ)/2^n ))) and k∈{1,....2^(n+1) −1} and k≠2^n$
	$1) let prove by recurrence that$ $(1 + x^{2}) (1 + x^{4}) . . . . (1 + x^{2^{n}}) = \frac{1 - x^{2^{n + 1}}}{1 - x^{2}} (x^{2} \neq 1)$ $n = 1 (1 + x^{2}) = \frac{1 - x^{4}}{1 - x^{2}} (true) let suppose$ $(p_{n}) true \Rightarrow (1 + x^{2}) (1 + x^{4}) . . . . (1 + x^{2^{n + 1}})$ $= (1 + x^{2}) (1 + x^{4}) . . . . (1 + x^{2^{n}}) (1 + x^{2^{n + 1}})$ $= \frac{1 - x^{2^{n + 1}}}{1 - x^{2}} \times (1 + x^{2^{n + 1}}) = \frac{1 - {(x^{2^{n + 1}})}^{2}}{1 - x^{2}}$ $= \frac{1 - x^{2^{n + 2}}}{1 - x^{2}} so the relation is true at term (n + 1)$ $\Rightarrow P (x) = 0 \Leftrightarrow \frac{1 - x^{2^{n + 1}}}{1 - x^{2}} = 0 and x \neq \bar{+} 1$ $\Rightarrow x^{2^{n + 1}} = 1 let m = 2^{n + 1} (e) \Rightarrow x^{m} = 1 = e^{i (2 k π)} \Rightarrow$ $the roots are z_{k} = e^{i (\frac{2 k π}{m})} k \in {0, . . . . m - 1} \Rightarrow$ $z_{k} = e^{i (\frac{k π}{2^{n}})} and k \in {1, . . . . 2^{n + 1} - 1} and k \neq 2^{n}$
	Commented by abdomathmax last updated on 01/Jun/20

	$2 . P (x) = \prod_{k = 1 and k \neq 2^{n}}^{2^{n + 1} - 1} (x - e^{i (\frac{k π}{2^{n}})})$
	Commented by abdomathmax last updated on 01/Jun/20

	$3 . we have P (x) = \frac{x^{2^{n + 1}} - 1}{x^{2} - 1} let 2^{n + 1} = m \Rightarrow$ $P (x) = \frac{x^{m} - 1}{x^{2} - 1} \Rightarrow P^{^{'}} (x) = \frac{{mx}^{m - 1} (x^{2} - 1) - 2 x (x^{m} - 1)}{{(x^{2} - 1)}^{2}}$ $= \frac{{mx}^{m + 1} - {mx}^{m - 1} - {2 x}^{m + 1} + 2 x}{{(x^{2} - 1)}^{2}}$ $= \frac{(m - 2) x^{m + 1} - {mx}^{m - 1} + 2 x}{{(x^{2} - 1)}^{2}}$ $\Rightarrow P^{^{'}} (x) = \frac{(2^{n + 1} - 2) x^{2^{n + 1} + 1} - 2^{n + 1} x^{2^{m + 1} - 1} + 2 x}{{(x^{2} - 1)}^{2}}$
	Commented by abdomathmax last updated on 01/Jun/20

	$4 . \frac{1}{P (x)} = \frac{1}{\prod_{k = 1 and k \neq 2^{n}}^{2^{n + 1} - 1} (x - e^{i (\frac{k π}{2^{n}})})}$ $= \sum_{k = 1 and k \neq 2^{n}}^{2^{n + 1 - 1}} \frac{a_{k}}{x - e^{i (\frac{k π}{2^{n}})}}$ $a_{k} = \frac{1}{P^{^{'}} (x_{k})}$
	Commented by abdomathmax last updated on 01/Jun/20

	$P^{^{'}} (x_{k}) = \frac{(2^{n + 1} - 2) {(x_{k})}^{2^{n + 1} + 1} - 2^{n + 1} {(x_{k})}^{2^{m + 1} - 1} + {2 x}_{k}}{{(x_{k}^{2} - 1)}^{2}}$ $x_{k} roots of P (x)$
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