The degrees of polynomials $P$ and $Q$ with real coefficients do not exceed $n$. These polynomials satisfy the identity \[ P(x) x^{n + 1} + Q(x) (x+1)^{n + 1} = 1. \]Determine all possible values of $Q \left( - \frac{1}{2} \right)$.
Problem
Source: Tuymaada 2020 Senior, P8
Tags: algebra, polynomial
IndoMathXdZ
06.10.2020 05:10
Here's my solution involving bashing by induction and binomial manipulation.
The answers are all values of $2^n$, where $n \in \mathbb{N}_0$.
We'll now prove that these values are achievable, and these are the only ones.
Before that, we define $\binom{0}{0} = 1$, and
\[ A_n = \binom{n}{n} + \frac{1}{2} \binom{n + 1}{n} + \dots + \frac{1}{2^n} \binom{2n}{n} \]Claim 01. $A_n = 2^n$ for all $n$.
We'll prove this via induction. Before that, we'll prove a few supporting lemmas.
Lemma. $\binom{a}{n} - \binom{a - 1}{n} = \binom{a - 1}{n - 1}$ for all $a \ge n + 1$.
Proof. To prove this, just notice that $\binom{a}{n} - \binom{a - 1}{n} = \frac{(a - 1)!}{(a - n - 1)! n!} \cdot \frac{n}{a - n} = \binom{a -1}{n - 1}$.
Now, we have
\begin{align*}
A_n - \frac{1}{2} A_n &= \binom{n}{n} + \frac{1}{2} \left( \binom{n + 1}{n} - \binom{n}{n} \right) + \frac{1}{2^2} \left( \binom{n + 2}{n} - \binom{n + 1}{n} \right) + \dots + \frac{1}{2^n} \left( \binom{2n}{n} - \binom{2n - 1}{n} \right) - \frac{1}{2^{n + 1}} \binom{2n}{n} \\
&=1 + \frac{1}{2} \binom{n}{n - 1} + \frac{1}{2^2} \binom{n + 1}{n - 1} + \dots + \frac{1}{2^{n - 1}}\binom{2n - 2}{n - 1} + \frac{1}{2^n} \binom{2n - 1}{n - 1} - \frac{1}{2^{n + 1}} \binom{2n}{n} \\
&= A_{n - 1} + \frac{1}{2^{n + 1}} \left( 2 \binom{2n - 1}{n - 1} - \binom{2n}{n} \right) \\
&= A_{n - 1}
\end{align*}Thus, we established $A_n = 2A_{n - 1}$ for all $n \in \mathbb{N}$. It is also easy to check that $A_0 = \binom{0}{0} = 1 = 2^0$. Therefore, by induction we are done.
Claim 02 (Main Claim). $Q(x) = \sum_{i = 0}^n \binom{n + i}{n} (-x)^i $
Proof. To prove this, we'll consider the equation for all terms with degree $\le n$.
Thus, we will attain the equality
\[ Q(x) \left( \sum_{i = 0}^n \binom{n + 1}{i} x^i \right) = 1 \]by comparing coefficient of both sides.
Now, we let $Q(x) = a_n x^n + a_{n - 1} x^{n - 1} + \dots + a_0$.
\[ \left( \sum_{j = 0}^n a_j x^j \right) \left( \sum_{i = 0}^n \binom{n + 1}{i} x^i \right) = 1 \]This gives us $a_0 = 1$ and
\[ \sum_{i + j = k} \binom{n + 1}{i} a_j = 0 \]for all $k \ge 1$ by observing the coefficient of $x^k$.
We'll now prove by induction that $a_i = (-1)^i \binom{n + i}{n}$, which is what we wanted for $0 \le i \le n$.
Base case is true as for $i = 0$, $a_0 = \binom{n}{n} = 1$, which is true.
Now, notice that $a_i$ is uniquely determined from all $a_j$, $0 \le i \le j - 1$. Thus, it suffices to just verify that
\[ \sum_{j = 0}^{k + 1} (-1)^j \binom{n + j}{n} \binom{n + 1}{k + 1 - j} = 0 \]However,
\[ \sum_{j = 0}^{k + 1} (-1)^j \binom{n + j}{n} \binom{n + 1}{k + 1 - j} = \frac{n + 1}{k + 1} \sum_{j = 0}^{k + 1} (-1)^j \binom{k + 1}{j} \binom{n + j}{k} = 0 \]The last part is true by observing that the polynomial $P(n + j) = \binom{n + j}{k} = \frac{1}{k!} (n + j) (n + j - 1) \dots (n + j -k + 1)$ has degree $k$, and therefore by finite difference we have that
\[ \Delta^{k + 1}(P(n + j))= \sum_{j = 0}^{k + 1} (-1)^j \binom{k + 1}{j} P(n + j) = 0\]
Claim 03. $Q \left( - \frac{1}{2} \right) = A_n$ if $Q$ is of degree $n$.
Proof. By Claim 02, we notice that
\[ Q \left( - \frac{1}{2} \right) = \sum_{i = 0}^n \binom{n + i}{n} \left( \frac{1}{2} \right)^i = A_n \]by definition.
Claim 04. For every such $n$, $P(x)$ exists.
Proof. To prove this, notice that we now have
\[ P(x) x^{n + 1} + \sum_{i + j \ge n + 1} \binom{n + i}{n} (-1)^i \binom{n + 1}{j} x^{i + j} = 0 \]Let $P(x) = p_n x^n + \dots + p_1 x + p_0$. We can just take
\[ p_k = -\sum_{i + j = n + k + 1} \binom{n + i}{n} (-1)^i \binom{n + 1}{j} \]and this satisfies.
ACGNmath
06.10.2020 13:06
Claim: The polynomials $P$ and $Q$ are uniquely determined. Proof: Suppose otherwise, i.e. suppose $P_1, P_2, Q_1, Q_2$ are polynomials such that $$P_1(x)x^{n+1}+Q_1(x)(x+1)^{n+1}=P_2(x)x^{n+1}+Q_2(x)(x+1)^{n+1}=1$$Then $$(P_1(x)-P_2(x))x^{n+1}+(Q_1(x)-Q_2(x))(x+1)^{n+1}=0$$Since $(x+1)^{n+1}$ divides $(Q_1(x)-Q_2(x))(x+1)^{n+1}$ and the right hand side, it follows that $$(x+1)^{n+1}|(P_1(x)-P_2(x))x^{n+1}$$Since $x+1$ and $x$ are coprime, we have $$(x+1)^{n+1}|P_1(x)-P_2(x)$$But the degree of $P$ is less than $n$. This means that $P_1(x)-P_2(x)=0$, or $P_1(x)=P_2(x)$. Similarly, we can get $Q_1(x)=Q_2(x)$. Back to the question. We want to "swap" the terms on the left hand side. We can do this by letting $x=-y-1$. $$Q(-y-1)(-1)^{n+1}y^{n+1}+P(-y-1)(-y-1)^n(-1)^{n+1}=1$$By the lemma, $$P(x)=(-1)^{n+1}Q(-x-1)$$$$Q(x)=(-1)^{n+1}P(-x-1)$$Setting $x=-\frac{1}{2}$, we have $$P\left(-\frac{1}{2}\right)=(-1)^{n+1}Q\left(-\frac{1}{2}\right)$$$$P\left(-\frac{1}{2}\right)\cdot\left(-\frac{1}{2}\right)^{n+1}+Q\left(-\frac{1}{2}\right)\cdot\left(\frac{1}{2}\right)^{n+1}=1$$Hence, $$Q\left(-\frac{1}{2}\right)=2^n$$
BOBTHEGR8
31.05.2021 21:11
IndoMathXdZ wrote: The degrees of polynomials $P$ and $Q$ with real coefficients do not exceed $n$. These polynomials satisfy the identity \[ P(x) x^{n + 1} + Q(x) (x+1)^{n + 1} = 1. \]Determine all possible values of $Q \left( - \frac{1}{2} \right)$. Solved together with L567, PUjnk, Rg230403 Differentiate to get $$(n+1)P(x)x^n + P'(x)x^{n+1}+(n+1)Q(x)(x+1)^n+Q'(x)(x+1)^{n+1} = 0$$$$\implies x^n | ((n+1)Q(x)+Q'(x)(x+1))(x+1)^n \implies x^n|(n+1)Q(x)+Q'(x)(x+1)$$As degree of $Q$ is at most $n$ we have $(n+1)Q(x)+Q'(x)(x+1) = cx^n$ $$\implies (Q(x)(x+1)^{n+1})' = ((n+1)Q(x)+Q'(x)(x+1))(x+1)^n = cx^n(x+1)^n$$Integrating from $-1$ to $x$ we have that $$Q(x)(x+1)^{n+1} =c \int_{-1}^{x} x^n(x+1)^n= c(F(x)-F(-1))$$Where $F(x) = \int_{0}^{x} x^n(x+1)^n$, now by similar logic we have $(P(x)x^{n+1})' = dx^n(x+1)^n$ $$\implies P(x)x^{n+1} = d F(x)$$Plugging in original equation we get $(d+c)F(x) - cF(-1) = 1 \implies d+c = 0 , c = \frac{-1}{F(-1)}$ Hence $Q(\dfrac{-1}{2})(\frac{1}{2})^{n+1} = \frac{(F(-1)-F(\frac{-1}{2})}{F(-1)} = 1-\frac{F(\frac{-1}{2})}{F(-1)}$ But due to symmetry of $F$ around $\frac{-1}{2}$ we have $F(-1) = 2 F(\frac{-1}{2})$ $$\implies Q\left(\frac{-1}{2}\right ) = 2^n$$$$QED$$
ILOVEMYFAMILY
26.07.2022 17:20
ACGNmath wrote: Claim: The polynomials $P$ and $Q$ are uniquely determined. Proof: Suppose otherwise, i.e. suppose $P_1, P_2, Q_1, Q_2$ are polynomials such that $$P_1(x)x^{n+1}+Q_1(x)(x+1)^{n+1}=P_2(x)x^{n+1}+Q_2(x)(x+1)^{n+1}=1$$Then $$(P_1(x)-P_2(x))x^{n+1}+(Q_1(x)-Q_2(x))(x+1)^{n+1}=0$$Since $(x+1)^{n+1}$ divides $(Q_1(x)-Q_2(x))(x+1)^{n+1}$ and the right hand side, it follows that $$(x+1)^{n+1}|(P_1(x)-P_2(x))x^{n+1}$$Since $x+1$ and $x$ are coprime, we have $$(x+1)^{n+1}|P_1(x)-P_2(x)$$But the degree of $P$ is less than $n$. This means that $P_1(x)-P_2(x)=0$, or $P_1(x)=P_2(x)$. Similarly, we can get $Q_1(x)=Q_2(x)$. Back to the question. We want to "swap" the terms on the left hand side. We can do this by letting $x=-y-1$. $$Q(-y-1)(-1)^{n+1}y^{n+1}+P(-y-1)(-y-1)^n(-1)^{n+1}=1$$By the lemma, $$P(x)=(-1)^{n+1}Q(-x-1)$$$$Q(x)=(-1)^{n+1}P(-x-1)$$Setting $x=-\frac{1}{2}$, we have $$P\left(-\frac{1}{2}\right)=(-1)^{n+1}Q\left(-\frac{1}{2}\right)$$$$P\left(-\frac{1}{2}\right)\cdot\left(-\frac{1}{2}\right)^{n+1}+Q\left(-\frac{1}{2}\right)\cdot\left(\frac{1}{2}\right)^{n+1}=1$$Hence, $$Q\left(-\frac{1}{2}\right)=2^n$$ something 's wrong. (x+1)^(n+1) divides (Q1(x)-Q2(x))(x+1)^(n+1) if Q1(x)-Q2(x) is integer
M11100111001Y1R
11.10.2022 22:10
Here is another solution by Arvin Sahami; Plugging $x-1$ instead of $x.$ We obtain $$P(x-1)(x-1)^{n+1}+Q(x-1)x^{n+1}=1.$$It follows that $(x-1)^{n+1}\mid T(x)=x^{n+1}Q(x-1)-1$ and $deg\ T(x)\le2n+1$ Now, we have following lemma: Lemma: If $deg\ P(x)\le d$ and for all $k\ge d,$ if $(x-1)^a\mid P(x)$ then $(x-1)^a\mid x^kP(\frac{1}{x}).$ Proof: Write $P(x)=(x-1)^aR(x).$ Then, $$x^kP(\frac{1}{x})=x^k(\frac{1}{x}-1)^aR(\frac{1}{x})=x^{k-a}(1-x)^a(x^{d-a}R(\frac{1}{x})).$$This completes our proof. Now, setting $Q(x-1)=T(x)$ it follows that $(x-1)^{n+1} \mid x^{n+1}T(x)-1,$ putting $k=2n+1$ in the lemma, it follows that $(x-1)^{n+1}\mid x^nT(\frac{1}{x})-x^{2n+1}.$ $S(x)=x^nT(\frac{1}{x})$ then $deg\ S(x)\le n.$ Moreover, $Q(-\frac{1}{2})=T(\frac{1}{2})=2^{-n}S(2).$ Whence, $(x-1)^{n+1}\mid S(x)-x^{2n+1}.$ That is, $x^{n+1} \mid S(x+1)-(x+1)^{2n+1}.$ Letting $A(x)=S(x+1),$ it follows that $x^{n+1} \mid A(x)-(x+1)^{2n+1}.$ Since $deg\ A(x)\le n,$ it follows that $A(x)$ is the remainder of the division of $(x+1)^{2n+1}$ by $x^{n+1}.$ Thus: $$A(x)=\sum_{i=0}^n \binom {2n+1} i x^i.$$Thus, $A(1)=S(2)= \sum_{i=0}^n \binom {2n+1} i=2^{2n}.$ Whence $$Q(-\frac{1}{2})=T(\frac{1}{2})=2^{-n}S(2)=2^n.$$