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I conjecture, that the function $f(x)=x^x-\tan x$ has exactly one root in any of the intervals $\left[\dfrac{2n+1}{2}\pi,\dfrac{2n+3}{2}\pi\right]$ , where $n$ is a nonnegative integer. Does anyone know a proof?

I tried the trick using the function $g(x)=\log\left(\dfrac{x^x}{\tan x}\right)$ , which has the same roots, but it did not help either.

kimtahe6
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Peter
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  • Is your function $x^x - tan(x)$ or $x^{x-tan(x)}$? – Hassan Hayat Oct 18 '13 at 11:02
  • It is (x^x)-tan(x) – Peter Oct 18 '13 at 11:03
  • ok. oh, and by the way, just as a tip, if you want your math functions to have that math display, just surround your math with dollar signs. (i.e.: one dollar sign before and one dollar sign after) – Hassan Hayat Oct 18 '13 at 11:05
  • I tried this for integrals, and it did not work. How could I display integral(x=0,1,exp(x)) in math display, for example ? – Peter Oct 18 '13 at 11:06
  • Sorry, there is no edit-button. Seems that only the user itself can edit the comment. – Peter Oct 18 '13 at 11:13
  • IMO this is a direct consequence of the intermediate value theorem and the fact that $x^x$ is monotone increasing for $x \ge 3\pi/2$, see Antonio Vargas' answer in the related topic at http://math.stackexchange.com/questions/110256/derivation-of-asymptotic-solution-of-tanx-x – gammatester Oct 18 '13 at 11:14
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    oops sorry. Surround this with dollar signs: \int\limits_{x = 0}^{1}e^{x}dx – Hassan Hayat Oct 18 '13 at 11:15
  • For some basic information about writing math at this site see e.g. here, here, here and here. – Daniel R Oct 18 '13 at 11:15
  • The indermediate value theorem only guarantees the existence of a root, it does not proof that there is exactly one. – Peter Oct 18 '13 at 11:21
  • @Peter: This comes from mononoty. – gammatester Oct 18 '13 at 11:25
  • BTW: Are the intervals correct after edit? IMO: $\left[\frac{2n+1}{2\pi},\frac{2n+3}{2\pi}\right]$ should be replaced by $\left[\frac{2n+1}{2}\pi,\frac{2n+3}{2}\pi\right]$. – gammatester Oct 18 '13 at 11:25
  • Yes, but to clarify, I did not edit it :) – Peter Oct 18 '13 at 11:27
  • If I search the roots of x^x-tan(x), what does the monotony of x^x help ? If tan(x) would decrease, so that x^x-tan(x) increases, that would be right. But tan(x) increases as does x^x. – Peter Oct 18 '13 at 11:30

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Hint: the equation is equivalent to $$ \cot x=x^{-x} $$

egreg
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  • Short and simple ! :-) – Lucian Oct 18 '13 at 15:43
  • A very good idea! The function $x^{-x}$ decreases very fast indeed, so my conjecture should be provable using this idea. – Peter Oct 19 '13 at 08:10
  • @Peter I guess so: the slopes of $x^{-x}$ and $\cot x$ near the intersection points are very far apart from each other. – egreg Oct 19 '13 at 17:07
  • Why has my comment been deleted ? Did it belong to a deleted answer ? – Peter Oct 22 '13 at 10:37
  • Just to repeat my comment : The fixpoint therem from banach can be used for the function g(x):=arctan($x^x$)+n$\pi$, where the nonnegative integer n depends on the wanted solution. It is not difficult to show, that the derivation of g is between 0 and 1 for x>=$e^{(-1)}$. For the interval [0,$e^{(-1)}$] , it is virtually trivial to show, that there is no solution for the equation tan(x)=$x^x$. – Peter Oct 22 '13 at 10:44