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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}) \dots (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, \dots) \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,

Missing or unrecognized delimiter for \left

2 + 4 + 8 + \dots + 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}) \dots . (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}) \dots . (1 + x^{2^{n + 1}})

= (1 + x^{2}) (1 + x^{4}) \dots . (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, \dots . m - 1} \Rightarrow

z_{k} = e^{i (\frac{k π}{2^{n}})} and k \in {1, \dots . 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)