let p(x)=(1+x^2 )(1+x^4 )...(1+x^2^n ) with n integr natural 1) find a simple form of p(x) 2) find roots of p(x)and decompose p(x) inside C[x] 3)calculate ∫_0 ^1 p(x)dx 4) decompose the fraction F(x)=(1/(p(x))) .


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Question Number 54808 by turbo msup by abdo last updated on 11/Feb/19
$let p(x)=(1+x^2 )(1+x^4 )...(1+x^2^n ) with n integr natural 1) find a simple form of p(x) 2) find roots of p(x)and decompose p(x) inside C[x] 3)calculate ∫_0 ^1 p(x)dx 4) decompose the fraction F(x)=(1/(p(x))) .$
$l e t p (x) = (1 + x^{2}) (1 + x^{4}) . . . (1 + x^{2^{n}})$ $w i t h n i n t e g r n a t u r a l$ $1) f i n d a s i m p l e f o r m o f p (x)$ $2) f i n d r o o t s o f p (x) a n d d e c o m p o s e$ $p (x) i n s i d e C [x]$ $3) c a l c u l a t e \int_{0}^{1} p (x) d x$ $4) d e c o m p o s e t h e f r a c t i o n$ $F (x) = \frac{1}{p (x)} .$
Commented by maxmathsup by imad last updated on 13/Feb/19

$1) w e c a n u s e r e c u r r e n c e t o p r o v e t h a t f o r x \neq \bar{+} 1$ $P (x) = \frac{1 - x^{2^{n + 1}}}{1 - x^{2}}$ $2) P (z) = 0 \Leftrightarrow 1 - z^{2^{n + 1}} = 0 \Leftrightarrow 1 - z^{q} = 0 w i t h m = 2^{n + 1} t h e r o o t s o f$ $z^{m} = 1 a r e z_{k} = e^{\frac{i 2 k π}{m}} w i t h k \in [[0, m - 1]] \Rightarrow z_{k} = e^{\frac{i 2 k π}{2^{n + 1}}} = e^{\frac{i k π}{2^{n}}}$ $w i t h k \in [[0, 2^{n + 1} - 1]] b u t b u t e l i m i n a t e v a l u e s o f k / z_{k}^{2} = 1 .$
Commented by maxmathsup by imad last updated on 13/Feb/19

$P (x) = \prod_{k = 0_{z_{k} \neq \overset{- 1}{+}}}^{2^{n + 1} - 1} (x - z_{k})$
	Answered by mr W last updated on 12/Feb/19

	$1)$ $p (x) = (1 + x^{2}) (1 + x^{4}) . . . (1 + x^{2^{n}})$ $(1 - x^{2}) p (x) = (1 - x^{2}) (1 + x^{2}) (1 + x^{4}) . . . (1 + x^{2^{n}})$ $(1 - x^{2}) p (x) = (1 - x^{4}) (1 + x^{4}) . . . (1 + x^{2^{n}})$ $(1 - x^{2}) p (x) = (1 - x^{8}) . . . (1 + x^{2^{n}})$ $. . . .$ $(1 - x^{2}) p (x) = (1 - x^{2^{n}}) . . . (1 + x^{2^{n}})$ $(1 - x^{2}) p (x) = (1 - x^{2^{n + 1}})$ $\Rightarrow p (x) = \frac{1 - x^{2^{n + 1}}}{1 - x^{2}}$
	Commented by maxmathsup by imad last updated on 12/Feb/19

	$c o r r e c t s i r b u t x \neq \bar{+} 1$
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