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Let $f,g$ be polynomials. There are two questions :

(1) If $f(n)\geq 0$ for all $n\gg 0$ and $n$ integer, then the leading coefficient of $f$ is a positive.

(2) If $f(n)=g(n)$ for all $n\gg 0$ and $n$ integer, then $f=g$.

I know that it is true. but I can't prove in detail.

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    how much have you tried? For the first one, what would happen if the leading coefficient wasn't positive?2012-09-10

2 Answers 2

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I'll replace $n$ with $x$ since proving the statements for $\mathbb{R}$ implies they are true for $\mathbb{N}$:

(1) If $f(x)\geq 0$ for all $x$ then the degree of $f$ must be even (including degree 0).

Assume the leading coeff. is negative:

$\bullet$ If the degree of $f$ is greater than 1, $\displaystyle\lim_{\infty}f=-\infty$ therefore there exists $x_0$ such that for all $x>x_0$ we have $f(x)<0$. Contradiction. Therefore the leading coefficient must be positive

$\bullet$ If $f$ is constant the leading coefficient must be either positive or 0 to satisfy $f(x)\geq 0$.

(2) First note that if $u(x)=v(x)$ for $x\geq 0$ and $u,v$ are even, then $u(-x)=v(-x)$. Similarly if $u(x)=v(x)$ for $x\geq 0$ and $u,v$ are odd, then $-u(-x)=-v(-x)\Leftrightarrow u(-x)=v(-x)$

Therefore if two polynomials are either even or odd and equal for $x\geq 0$ they are equal for all $x\in\mathbb{R}$

Now let:

$\bullet$ $f(x)=p_n(x)+p_{n-1}(x)\cdots p_1(x)+p_0(x)$ where $p_i(x)=a_ix^i$

$\bullet$ $g(x)=q_n(x)+q_{n-1}(x)\cdots q_1(x)+q_0(x)$ where $q_i(x)=b_ix^i$

For each $i,\;\;p_i$ and $q_i$ are either both even or both odd. Since $p_i(x)=q_i(x)$ for all $x\geq 0$, necessarily $p_i=q_i$ for all $x\in\mathbb{R}$. Therefore $f=g$

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(1) Let $f(x) = \sum_{k=0}^n a_k x^k$, with $a_n\neq 0$. Write $f(x) = a_nx^n \left(1 + \sum_{k=0}^{n-1}\frac{a_k}{a_n}x^{k-n}\right)$. Note that $x^{k-n} = o(1)$ as $x\to+\infty$ (for $k < n$). Therefore, $f(x) = (1 + o(1)) a_nx^n$. For $x$ large enough the $(1+o(1))$ term is greater than $1/2$. Thus $f(x)$ has the same sign as $a_n x^n$. Since $f(x)$ is positive and $x^n$ is positive, we have that $a_n > 0$.

(2) Consider $h(x) = f(x) - g(x)$. We have that $h(x) = 0$ for all sufficiently large $x$. Assume to the contrary that $h \neq 0$. Then $h(x) = \sum_{k=0}^n b_k x^k$ for some $n\geq 0$, with $b_n\neq 0$. Similarly to (1), we have that $h(x) = (1+o(1)) b_n x^n$, and thus for sufficiently large $x$, $h(x) \neq 0$. We get a contradiction.