ATimo wrote: Prove that for each natural number $d$, There is a monic and unique polynomial of degree $d$ like $P$ such that $P(1)$≠$0$ and for each sequence like $a_{1}$,$a_{2}$, $...$ of real numbers that the recurrence relation below is true for them, there is a natural number $k$ such that $0=a_{k}=a_{k+1}= ...$ : $P(n)a_{1} + P(n-1)a_{2} + ... + P(1)a_{n}=0$ $n>1$

hint

ATimo wrote: Prove that for each natural number $d$, There is a monic and unique polynomial of degree $d$ like $P$ such that $P(1)$≠$0$ and for each sequence like $a_{1}$,$a_{2}$, $...$ of real numbers that the recurrence relation below is true for them, there is a natural number $k$ such that $0=a_{k}=a_{k+1}= ...$ : $P(n)a_{1} + P(n-1)a_{2} + ... + P(1)a_{n}=0$ $n>1$ I don't understand the last sentence is $n$ a fixed integer?

andria wrote: ATimo wrote: Prove that for each natural number $d$, There is a monic and unique polynomial of degree $d$ like $P$ such that $P(1)$≠$0$ and for each sequence like $a_{1}$,$a_{2}$, $...$ of real numbers that the recurrence relation below is true for them, there is a natural number $k$ such that $0=a_{k}=a_{k+1}= ...$ : $P(n)a_{1} + P(n-1)a_{2} + ... + P(1)a_{n}=0$ $n>1$ I don't understand the last sentence is $n$ a fixed integer? No. $n$ is not a fixed integer. The relation is true for all integers $n>1$.

Assume there is such $k$: $a_k \neq 0$ and $a_i=0 \forall i>k$. We can easily see that $G(x)=a_1P(x)+a_2P(x-1)+...+P(x-k+1)a_k$ have infinitely root $n>k, n \in \mathbb{N}$, so $G(x)=0$. Then $P(k-1)a_1+P(k-2)a_2+...+P(1)a_{k-1}=0=P(k-1)a_1+P(k-2)a_2+...+P(0)a_k$ so $P(0)=0$. We can also prove that $P(k-2)=P(k-1)=...=P(0)=0$ so we have $P(n)$ is in the form $\sum f(n)\alpha _i^n$ ($\alpha _i$ are the roots of $U(x)=a_1x^k+a_2x^{k-1}+...+a_k$). So all the root must equal $1$, $deg P=k-1$ and we can assume $a_1=1$ to find other $a_i$ and there are only such P satisfied the condition

Juviette wrote: Assume there is such $k$: $a_k \neq 0$ and $a_i=0 \forall i>k$. We can easily see that $G(x)=a_1P(x)+a_2P(x-1)+...+P(x-k+1)a_k$ have infinitely root $n>k, n \in \mathbb{N}$, so $G(x)=0$. Then $\color{red} P(k-1)a_1+P(k-2)a_2+...+P(1)a_{k-1}=0=P(k-1)a_1+P(k-2)a_2+...+P(0)a_k$ so $P(0)=0$. We can also prove that $P(k-2)=P(k-1)=...=P(0)=0$ so we have $P(n)$ is in the form $\sum f(n)\alpha _i^n$ ($\alpha _i$ are the roots of $U(x)=a_1x^k+a_2x^{k-1}+...+a_k$). So all the root must equal $1$, $deg P=k-1$ and we can assume $a_1=1$ to find other $a_i$ and there are only such P satisfied the condition How do we get this? I see that the first equality is from the condition given, but i dont see how we add the last term in the second equality

igli.2001 wrote: Juviette wrote: Assume there is such $k$: $a_k \neq 0$ and $a_i=0 \forall i>k$. We can easily see that $G(x)=a_1P(x)+a_2P(x-1)+...+P(x-k+1)a_k$ have infinitely root $n>k, n \in \mathbb{N}$, so $G(x)=0$. Then $\color{red} P(k-1)a_1+P(k-2)a_2+...+P(1)a_{k-1}=0=P(k-1)a_1+P(k-2)a_2+...+P(0)a_k$ so $P(0)=0$. We can also prove that $P(k-2)=P(k-1)=...=P(0)=0$ so we have $P(n)$ is in the form $\sum f(n)\alpha _i^n$ ($\alpha _i$ are the roots of $U(x)=a_1x^k+a_2x^{k-1}+...+a_k$). So all the root must equal $1$, $deg P=k-1$ and we can assume $a_1=1$ to find other $a_i$ and there are only such P satisfied the condition How do we get this? I see that the first equality is from the condition given, but i dont see how we add the last term in the second equality $G(k-1)=0$

Juviette wrote: Assume there is such $k$: $a_k \neq 0$ and $a_i=0 \forall i>k$. We can easily see that $G(x)=a_1P(x)+a_2P(x-1)+...+P(x-k+1)a_k$ have infinitely root $n>k, n \in \mathbb{N}$, so $G(x)=0$. Then $P(k-1)a_1+P(k-2)a_2+...+P(1)a_{k-1}=0=P(k-1)a_1+P(k-2)a_2+...+P(0)a_k$ so $P(0)=0$. We can also prove that $P(k-2)=P(k-1)=...=P(0)=0$ so we have $P(n)$ is in the form $\sum f(n)\alpha _i^n$ ($\alpha _i$ are the roots of $U(x)=a_1x^k+a_2x^{k-1}+...+a_k$). So all the root must equal $1$, $deg P=k-1$ and we can assume $a_1=1$ to find other $a_i$ and there are only such P satisfied the condition why p(1)≠0?could you explain this?

I think this problem is wrong. Bump.

Here is my wording of the original problem. Let $d$ be a positive integer. Prove that there exists a unique monic polynomial of degree $d$ such that: (a) $P(1)\neq 0$. (b) For every sequence of real numbers $a_1, a_2, a_3, \dots$ which satisfies the equation $$P(n)a_1+P(n-1)a_2+\dots +P(1)a_n=0$$for each positive integer $n>1$, the sequence is eventually zero.