let-P-n-X-n-X-n-1-X-2-X-1-R-X-1-prove-that-P-n-have-one-root-x-n-inside-0-2-study-the-sequence-x-n-

Question Number 73052 by mathmax by abdo last updated on 05/Nov/19

l e t P_{n} = X^{n} + X^{n - 1} + \dots . + X^{2} + X - 1 \in R [X]

1) p r o v e t h a t P_{n} h a v e o n e r o o t x_{n} i n s i d e] 0, + \infty [

2) s t u d y t h e s e q u e n c e x_{n}

Answered by mind is power last updated on 06/Nov/19

P_{n}^{'} = \underset{k = 1}{\sum^{n}} {kx}^{k - 1} > 0

pn (0) = - 1

P_{n} (1) = n - 1 ⩾ 0 if n ⩾ 2

n = 0, p = 0

p_{1} = x - 1 \Rightarrow x_{1} = 1

\forall n ⩾ 2 x_{n} < 1

\Rightarrow P_{n + 1} (x_{n}) = \underset{k = 1}{\sum^{k = n + 1}} x_{n}^{k} - 1 = x_{n}^{n + 1} > 0

p_{n + 1} (x_{n + 1}) = 0 \Rightarrow x_{n + 1} < x_{n}

we have x_{n} decrease sequence and x_{n} \in [0, 1 [

x_{n} cv

p_{n} (x) = \frac{x^{n + 1} - 1}{x - 1} - 2

\Rightarrow p_{n} (x) = 0 \Leftrightarrow x^{n + 1} - 2 x + 1 = 0

\Rightarrow x^{n + 1} - 2 x = - 1

let x_{2}

x^{2} + x - 1 = 0

\Rightarrow x_{2} = \frac{- 1 + \sqrt{5}}{2} < 1

for n ⩾ 2

x_{n}^{n + 1} ⩽ x_{2}^{n + 1} = {(\frac{- 1 + \sqrt{5}}{2})}^{n + 1} \to 0

p_{n} (x_{n}) = 0 \Leftrightarrow x_{n}^{n + 1} - 2 x + 1 = 0

tack n \to + \infty

withe X_{n}^{n + 1} \to 0

\Rightarrow x_{n} \to \frac{1}{2}