Prove that for any positive integer $ n$, there exists only $ n$ degree polynomial $ f(x),$ satisfying $ f(0) = 1$ and $ (x + 1)[f(x)]^2 - 1$ is an odd function.
Problem
Source: Chinese TST 2007 1st quiz P3
Tags: algebra, polynomial, function, induction, quadratics, limit, integration
08.01.2009 18:12
This method might not be efficient enough. Denote $ f_n(x)$ as the function with degree $ n$, we can have $ f_n(x)=g_n(x)+h_n(x)$, where $ g_n$ is even and $ h_n$ is odd. $ (x+1)f(x)^2-1$ being odd implies $ (1+x)f(x)^2+(1-x)f(-x)^2=2\Rightarrow g(x)^2+h(x)^2+2xg(x)h(x)=1$ (*) Now we define $ g_n,h_n$ as follows: $ g_0(x)=1$,$ h_0(x)=0$. $ g_{2n+1}(x)=g_{2n}(x)$, $ h_{2n+1}(x)=-2xg_{2n}(x)-h_{2n}(x)$; $ g_{2n}(x)=g_{2n-1}(x)+2xh_{2n-1}(x)$, $ h_{2n}(x)=-h_{2n-1}(x)$. It is not hard to verify (*), $ g_n(0)=1$, $ g_n$ being even, and $ h_n$ being odd. We proved existence. Now we prove uniqueness by induction. Uniqueness of $ f_0(x)$ is trivial. Suppose $ f_{n-1}(x)$ is unique and we look at $ f_n(x)=g_n(x)+h_n(x)$. Assume $ n$ is even. $ (g_n,h_n)$ is any pair which meets the requirement. Then we see $ g_n$ has degree $ n$. We also have $ g_n^2+h_n^2+2xg_nh_n=1$. If we look at the coefficient of $ x^{2n}$ of the above equation, we can show $ h_n$ has degree $ n-1$ and $ g_n+2xh_n$ has degree no greater than $ n-2$. Now we find out $ g=g_n+2xh_n$, $ h=-h_n$ also solves (*). But we notice $ g+h$ has degree $ n-1$. Then by induction assumption, we have $ g_n+2xh_n=g_{n-1}$ and $ -h_n=h_{n-1}$. This shows uniqueness of $ f_n$ Similarly we can handle the case when $ n$ is odd. Q.E.D.
15.02.2009 08:52
All I want to say is:a perfect solution! Siince the official solution is based on Pell Equation(defined in polynomial not integer) which is not elementary at all
11.10.2009 08:29
Could you post the official solution of this problem,dear hxy09?
22.09.2011 00:28
We will show that up to sign, there is in fact a unique polynomial in $\mathbb{R}[x]$. Let $f(x)=p(x^2)+xq(x^2)$, so $(x+1)f(x)^2-1=-[(-x+1)f(-x)^2-1]$ rewrites as $x[p(x)+q(x)]^2+(1-x)p(x)^2=1$. (*) Let $r(x)=p(x)+q(x)$. Differentiating both sides, we have \[r(x)[r(x)+2xr'(x)]=p(x)[p(x)+2(x-1)p'(x)].\]Clearly $\deg{p}=\deg{r}$ or else (*) can't hold (the leading coefficients must be the same up to sign), so because $\gcd(p(x),r(x))=1$, $p(x)$ must divide $r(x)+2xr'(x)$ and so there exists a constant $c$ such that $p(x)=c[r(x)+2xr'(x)]$. Plugging this into (*), we get the differential equation \[1=[x+c^2(1-x)]r(x)^2+4c^2x(1-x)r'(x)r(x)+4c^2x^2(1-x)r'(x)^2.\]From the quadratic formula, we find \[r'(x)=-\frac{c(x-1)r(x)\pm\sqrt{(x-1)[xr(x)^2-1]}}{2cx(x-1)},\]or letting $s(x)=r(x)\sqrt{x}$ so that $2xr'(x)\sqrt{x}=2xs'(x)-s(x)$ and simplifying, \[\frac{2cs'(x)}{\sqrt{s(x)^2-1}}=\sqrt{\frac{1}{(2x-1)^2-1}},\]where WLOG $\lim_{x\to\infty}s(x)=\infty$. Anyway, it's easy to find that \[\int\frac{1}{\sqrt{x^2-1}}\;dx=\ln(x+\sqrt{x^2-1})+C,\]so for some constant $C_0$ we have \[c\ln(s(x)+\sqrt{s(x)^2-1})=\ln(\sqrt{x}+\sqrt{x-1})+C_0,\]i.e. \[\frac{1}{s(x)-\sqrt{s(x)^2-1}}=s(x)+\sqrt{s(x)^2-1}=e^{C_0/c}(\sqrt{x}+\sqrt{x-1})^{1/c},\]so there exist constants $C_1,C_2$ such that \[r(x)=\frac{e^{C_1}(\sqrt{x}+\sqrt{x-1})^{C_2}+e^{-C_1}(\sqrt{x}-\sqrt{x-1})^{C_2}}{2\sqrt{x}}.\]Since $r$ is a polynomial, we clearly have $C_2\in\mathbb{Z}$ and thus $C_1=0$. Now we can easily compute $f(x)=(1-x)p(x^2)+xr(x^2)=T_{n+1}(x)-(x-1)U_n(x)$ (up to sign of course). Note that (*) rearranges to the Pell equation $u(x)^2-(x^2-1)v(x)^2=1$ upon the substitutions $u(x)=xr(x^2)$, $v(x)=p(x^2)$, which is a bit cleaner to do with differentiation.
09.03.2024 08:18
Solution from Twitch Solves ISL: Suppose the polynomial $P$ obeys \[ 2 = (x+1)P(x)^2 + (-x+1)P(-x)^2. \]Work in ${\mathbb R}[y, y^{-1}]$ and substitute \[ x = \frac{1}{2}\left( y + \frac 1y \right). \]Then the above equation can be rewritten as \begin{align*} 4 &= \left( y + \frac 1y + 2 \right) P(x)^2 - \left( y + \frac 1y - 2 \right) P(-x)^2 \\ \iff 4y &= \left( (y+1) P(x) \right)^2 - \left( (y-1) P(-x) \right)^2 \\ \iff 4y &= \Big( (y+1) P(x) + (y-1) P(-x) \Big) \Big( (y+1) P(x) - (y-1) P(-x) \Big). \end{align*}Using factorization in ${\mathbb R}[y, y^{-1}]$, we see there must exist an integer $d \ge 0$ and a nonzero real number $\varepsilon$ such that we have identities \begin{align*} 2\varepsilon \cdot y^{d+1} &= (y+1) P(x) \pm (y-1) P(-x) \\ \frac 2\varepsilon \cdot y^{-d} &= (y+1) P(x) \mp (y-1) P(-x) \end{align*}for some opposite choice of signs $\pm$. Replacing $y$ with $1/y$, it follows $\varepsilon = \varepsilon^{-1} \implies \varepsilon = \pm 1$. So the system of equations is equivalent to \begin{align*} y^{d+1} + y^{-d} &= \varepsilon (y+1) P(x) \\ y^{d+1} - y^{-d} &= \pm \varepsilon (y-1) P(-x) \end{align*}Now, the first equation implies that \begin{align*} \varepsilon P(x) &= \frac{y^{d+1}+y^{-d}}{y+1} \\ &= \left( y^d + \frac{1}{y^d} \right) - \left( y^{d-1} + \frac{1}{y^{d-1}} \right) + \dots + (-1)^{d-1} \left( y + \frac 1y \right) + (-1)^d. \end{align*}The theory of Chebyshev polynomials guarantees that $y^k + \frac{1}{y^k}$ is a degree-$k$ polynomial in $x$ for each $k \ge 0$. Hence we see there is a unique choice of polynomial $P$ of degree $n$ (corresponding to $d=n$) which makes the identity $P(x) = \varepsilon \frac{y^{d+1}+y^{-d}}{y+1}$ true, for each $\varepsilon \in \{\pm1\}$. When we add the condition $P(0) = 1$ (set $y=i$ so $x=0$), we find only one $\varepsilon \in \{\pm1\}$ is valid. Conversely, we have the implication \[ \varepsilon (y+1)P(x) = y^{d+1} + y^{-d} \implies \varepsilon (y-1)P(-x) = \pm (y^{d+1} - y^{-d}) \]which follows readily by making the substitution $y \mapsto iy$. So the choice of polynomial $P$ we just mentioned will satisfy the entire system, ergo (reversing the logic above) will satisfy the desired condition. In other words, the unique possibility for $P$ does indeed work. Remark: The first few actual polynomials $P$ are: \begin{align*} P(x) &= 1 \\ P(x) &= 1 - 2x \\ P(x) &= 1 + 2x - 4x^2 \\ P(x) &= 1 - 4x - 4x^2 + 8x^3 \\ P(x) &= 1 + 4x - 12x^2 - 8x^3 + 16x^4. \end{align*}