Find all polynomials with real coefficients, for which the equality \[ P(2P(x)) = 2P(P(x)) + 2(P(x))^{2}\] holds for any real number $ x$.
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Tags: algebra, polynomial, algebra proposed
04.03.2009 22:18
Let $ P(x) = x^n + a_1x^{n - 1} + \dots + a_n$ $ \Rightarrow P(2P(x)) = 2P(P(x)) + 2(P(x))^{2} \Leftrightarrow 2^nx^{n^2} + \cdots = 4x^{2n} + \cdots$ $ \Rightarrow n = 2\Leftrightarrow P(x) = x^2 + ax + b$ $ 4(x^2 + ax + b)^2 + 4a(x^2 + ax + b) + 2b =$ $ 2(x^2 + ax + b)^2 + 4a(x^2 + ax + b) + 2b + 2(x^2 + ax + b)^2$
05.03.2009 11:58
How about the cases for $ P(x)=a_0 x^n + a_1 x^{n-1}+ \cdots+a_n$ with $ a_0 \neq 1$?
05.03.2009 17:21
BG Yoda wrote: $ 2^nx^{2n} + \cdots = 4x^{2n} + \cdots$ $ deg(P(2P(x)))=n^2$ and not $ 2n$
14.03.2009 16:23
let$ P(x) = a_n x^n + a_{n - 1} x^{n - 1} + \cdots + a_0$, let $ P(2P(x)) - 2P(P(x)) = g(x)$, when $ deg(P(x))\ge3$,i.e. $ a_n\neq0$ $ \deg(P(2P(x))),\deg(P(P(x)) = n^2 > deg(P(x))^2 = 2n$ but $ P(2P(x)) - 2P(P(x)) = 2(P(x))^{2}$ $ \therefore deg(g(x)) = 2n$ the coefficient of $ x^{n^2}$ in $ g(x)$ is $ (2^n - 2)a_{n}^{n + 1}$, $ \therefore (2^n - 2)a_{n}^{n + 1} = 0, a_{n} = 0$which is contradictory to $ a_n\neq0$ when $ \deg(P(x)) = 2$, then$ 2(a_2 - 1)[P(x)]^2 = a_0,\forall x\in R$, $ \therefore a_2 = 1, a_0 = 0$, $ P(x) = x^2 + a_1x$ where a_1 is any real number when $ \deg(P(x)) = 1$, then$ 2(a_{1}x + a_0)^2 = - a_0,\forall x\in R$ which is impossible. therefore there is only one solution $ P(x) = x^2 + ax,\forall a \in R$
19.03.2009 12:41
If $ | \{ P(x) : x \in \mathbb{R} \} | < \infty$ then $ P(x) = c \forall x$. Substitute into the equation, one has \[ c = 2c + 2c^2 \iff c = 0 \vee c = -\frac 12\] Now, in the other case, since $ P(x)$ consumes infinitely many values, it must be the case that \[ P(2y) = 2P(y) + 2y^2 \forall y \in R\] Let $ Q(x) = P(x) - x^2$. It turns out to be $ Q(2x) = 2Q(x) \forall x$. The only possible case are $ Q(x) \equiv 0$ or $ Q(x) \equiv \alpha x$ where $ \alpha \not= 0$. These two cases could be merged to $ Q(x) \equiv \alpha x$. Therefore, \[ P(x) = \begin{cases} 0 \\ - \frac 1 2 \\ x^2 + \alpha x \end{cases}\]
02.03.2014 20:46
I know this is a major revive, but I was doing this problem today, and I appear to be getting something different. Can someone check my solution please?
22.03.2016 09:25
AlcumusGuy wrote: I know this is a major revive, but I was doing this problem today, and I appear to be getting something different. Can someone check my solution please?
there is error at these lines: \[(2(x^2 + bx + c))^2 + b[2(x^2 + bx + c)] + c = 2[(x^2 + bx +c)^2 + b(x^2 + bx + c) + c] + 2(x^2 + bx + c)^2\]\[4(x^2 + bx + c)^2 + 2b(x^2 + bx + c) + c = 2(x^2 + bx + c)^2 + 2b(x^2 + bx + c) + >2c< + 2(x^2 + bx + c)^2\]so c=0 and answer will $\boxed{P(x) = x^2 + bx}$