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Consider recursion $$a_n+c_1a_{n-1}+\cdots+c_ma_{n-m}=0~~~~~~~~~~(1)$$

Let $\lambda^m+c_1\lambda^{m-1}+\cdots+c_m=0$ be the characteristic function and $(\lambda_1,n_1),\cdots,(\lambda_s,n_s)$ be its solutions (the $\lambda_i$ are the roots with multiplictiy $n_i$, so that $n_1+\cdots+n_s=m$).

Now I want to prove that

$\lambda_1^n,~n\lambda_1^n,~\cdots,~n^{n_1-1}\lambda_1^n$
$\lambda_2^n,~n\lambda_2^n,~\cdots,~n^{n_2-1}\lambda_2^n$
$\cdots\cdots$
$\lambda_s^n~,n\lambda_s^n,~\cdots,~n^{n_s-1}\lambda_s^n$

are $m$ independent linear solutions of $(1)$.

Below is my outline of proof.

  • First of all, they are solutions which I have done.

  • Then they must be independent. My thought is to focus on the next question.

Let $P_1(x),\cdots P_s(x)\in\mathbb C[x]$ and $\lambda_1,\cdots\lambda_s\in\mathbb C$. If $$P_1(n)\lambda_1^n+\cdots+P_s(n)\lambda_s^n=0~~~n\in\mathbb N$$ then $P_1(x)=\cdots=P_s(x)=0$.

I get stuck here. Some advice on proof of the question. Thank you.

Greg Martin
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gaoxinge
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    If all the $|\lambda_i|$ are distinct - say listed in decreasing order of size - then you can divide through by $\lambda_1^n$ and take the limit as $n\to\infty$ to show that $P_1(n)=0$, then divide instead by $\lambda_2^n$ to show that $P_2(n)=0$, etc. But you'll have to be more careful if some of the $|\lambda_i|$ are equal. – Greg Martin Apr 07 '14 at 02:56
  • @GregMartin How to show that $P_1(n)=0$? More details, thank you. – gaoxinge Apr 07 '14 at 13:23
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    Have you tried dividing through by $\lambda_1(n)$ and taking the limit as $n\to\infty$? What results? – Greg Martin Apr 07 '14 at 18:02
  • @GregMartin Assume $|\lambda_1|$ is the biggest. We have $$|P_1(n)|=|-\sum P_i(n)(\frac{\lambda_i}{\lambda_1})^n|\leq\sum |P_i(n)||\frac{\lambda_i}{\lambda_1}|^n$$ Is it the way to show $P_1(n)\rightarrow 0$? – gaoxinge Apr 08 '14 at 02:36
  • @GregMartin In addition, does $P_1(n)\rightarrow0$ mean $P_1(n)=0$? – gaoxinge Apr 08 '14 at 02:39
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    That's basically correct. I'll let you verify that if a polynomial $P(x)$ tends to $0$ as $x\to\infty$, then the polynomial is identically $0$. – Greg Martin Apr 08 '14 at 06:30

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