2

Find the asymptotic behaviour as $u \to \infty$ of the solutions of $x^2-\log x = u$. Is there a standard method to solve this kind of problems? May the fact that we obviously know the derivative of $f^{-1}(x)$ help? A similar problem was: "given $u=x+\tanh x$ show that, if $u \to \infty$ then $x=u-1+e^{2-2u}+O(e^{-4u})$

d. zeffiro
  • 415
  • There is indeed a standard method to solve these kinds of problems. You can find an example of it in this answer and the in the answers linked therein. I'll be happy to give a complete answer here if you need additional help. – Antonio Vargas Dec 11 '13 at 18:16

1 Answers1

1

I would do this by Newton's method. Obviously, for $x^2 - \log x = u,$ the first guess would be $x = \sqrt{u},$ (since $\log x \ll x,$ for $x$ large). The first Newton step ought to give you an excellent approximation. As for your "similar problem", what you wrote makes no sense (since there is a mix of $x$s and $u$s), so you might want to edit. That said, the solution should be very similar, since for $x\gg 1,$ we know that $\tanh x$ is very close to $1,$ so obviously $x = u-1$ is the first guess, and you can get the second by Newton.

Igor Rivin
  • 25,994
  • 1
  • 19
  • 40
  • @d.zeffiro, this approach can be modified to obtain the behavior of the other root which tends to $0$ as $u \to \infty$. – Antonio Vargas Dec 11 '13 at 18:33
  • ok, thank you Antonio and Igor. I would also add that the hypothesis to apply Newton method hold, since the function is clearly convex and we start with a guess in which the function is positive. – d. zeffiro Dec 11 '13 at 19:42