Find-all-polynomials-P-x-with-real-coefficients-such-that-for-all-nonzero-real-numbers-x-P-x-P-1-x-P-x-1-x-P-x-1-x-2-

Question Number 200722 by cortano12 last updated on 22/Nov/23

Find all polynomials P (x) with

real coefficients such that for

all nonzero real numbers x,

P (x) + P (\frac{1}{x}) = \frac{P (x + \frac{1}{x}) + P (x - \frac{1}{x})}{2}

Commented by Frix last updated on 22/Nov/23

P (x) = {a x}^{2} [\lor P (x) = 0]

Answered by witcher3 last updated on 22/Nov/23

x \overset{f}{\to} x - \frac{1}{x};

] 0, + \infty [\to] - \infty, + \infty [bijective

p (\frac{1}{x}) + p (x) = \frac{1}{2} (p (x + \frac{1}{x}) + p (- x + \frac{1}{x}) = \frac{1}{2} (p (x + \frac{1}{x}) + p (x - \frac{1}{x}))

\Rightarrow p (x - \frac{1}{x}) = p (- (\frac{1}{x} - x))

\Leftrightarrow \forall a \in R since bijectiln of f . p (a) = p (- a) \Rightarrow

p (x) = \underset{k = 0}{\sum^{n}} a_{k} x^{2 k}; p \in R_{2 n} [X]

x = e^{t},

p (e^{t}) + p (e^{- t}) = \frac{1}{2} p (2 ch (t)) + p (2 sh (t))

\Leftrightarrow 2 p (e^{t}) + 2 p (e^{- t}) = p (2 ch (t)) + p (2 sh (t))

\Leftrightarrow 4 \underset{k = 0}{\sum^{2 n}} a_{k} ch (2 kt) = \underset{k = 0}{\sum^{2 n}} a_{k} 2^{2 k} ({ch}^{2 k} (t) + {sh}^{2 k} (t))

\Leftrightarrow \forall k \in [0, 2]

4 ch (2 kt) = 2^{2 k} ({ch}^{2 k} (t) + {sh}^{2 k} (t)

k = 1 true, \forall k ⩾ 2

4 ch (2 kt) = 2^{2 k} ({ch}^{2 k} (t) + {sh}^{2 k} (t); t = 0

4 = 2^{2 k} false k ⩾ 2

\Rightarrow p (x) = a_{0} + a_{1} x^{2}; p (1) + p (\frac{1}{1}) = \frac{1}{2} (p (2) + p (0))

\Leftrightarrow 2 (a_{0} + a_{1}) = a_{0} + {2 a}_{1}

a_{0} = 0

p (x) = {ax}^{2} worck unique by construcrion

; a \in R

Facebook Tweet Pin