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For any monic polynomial $f \in \mathbb {Q[x]}$ of even degree,how to prove, there exists polynomial $g \in \mathbb {Q[x]}$ such that $lim_{|x|\to\infty }(\sqrt {f(x)}-g(x))=0$

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  • $\begingroup$ Maybe if you try using induction? $\endgroup$
    – Ovi
    May 20, 2013 at 1:45
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    $\begingroup$ Is it true?What about $$f(x)=2x^2 ?$$ $\endgroup$
    – Chung. J
    May 20, 2013 at 2:35
  • $\begingroup$ @JaeyoungChung: The statement specifies that $f$ must be a monic polynomial, which saves it from such counterexamples. $\endgroup$ May 20, 2013 at 2:44

1 Answer 1

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Let $f$ have degree $2k$.

Solution 1: Write $f(x) = x^{2k}(1 + c_1x^{-1} + c_2x^{-2} + \cdots + c_{2k}x^{-2k}$. Then $\sqrt{f(x)} = x^k(1 + d_1x^{-1} + d_2x^{-2} + \cdots)$, where $1 + d_1x^{-1} + d_2x^{-2} + \cdots$ is the Maclaurin series for $\sqrt{1 + c_1x^{-1} + c_2x^{-2} + \cdots + c_{2k}x^{-2k}}$. Show that $g(x) = x^k (1 + d_1x^{-1} + d_2x^{-2} + \cdots + d_kx^{-k})$ works.

Solution 2: It suffices to find a monic polynomial $g(x)$ of degree $k$ such that $f(x)-g(x)^2$ has degree at most $k-1$. (To see this, bound $\sqrt{f(x)}-g(x) = (f(x)-g(x)^2)/(\sqrt{f(x)}+g(x))$ when $x$ is large.) Writing $g(x) = x^k + b_{k-1}x^{k-1} + \cdots + b_1x + b_0$, one can recursively solve for $b_{k-1},\dots,b_1,b_0$ in terms of the previous $b_j$ and the coefficients of $f$.

(The two solutions give the same polynomial $g(x)$; indeed, it's easy to show from first principles that such a $g(x)$ must be unique.)

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