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 $lim_{|x|\to\infty }(\sqrt {f(x)}-g(x))=0$

  • $\begingroup$ Maybe if you try using induction? $\endgroup$
    – Ovi
    May 20, 2013 at 1:45
  • 1
    $\begingroup$ 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


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.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.