About equation tg x=x

suadnovic · New Member Offline Joined: 29 Oct 2009 Posts: 14

Let {x(n)} be increasing seqences positive solutions of eqation tg x = x.
Find
lim {x(n)-x(n-1)} , n tends to infinity.

P.S. I hope that's clear. I will use TeX or LaTeX in future. And about problem: I think this must tends to pi.

**Posted:** Thu Oct 29, 2009 4:58 am

zhoraster · Riemann Hypothesis Offline Joined: 03 Aug 2008 Posts: 257

Yes, your guess is correct.

Try to prove that $x_n - \pi (n - 1/2)\to 0$ , $n\to\infty$ .
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**Posted:** Thu Oct 29, 2009 6:00 am

Moubinool

Use this $x(n)=nPi +Pi/2-1/(nPi)+1/(2Pi*n^2)+o(1/n^2)$

**Posted:** Sat Oct 31, 2009 12:31 pm

suadnovic · New Member Offline Joined: 29 Oct 2009 Posts: 14

Here is, I think, elementary solutiom of my posted problem. It is obviousus that
$n\pi < x_{n} < (n + 1/2)\pi\Rightarrow |x_{n} - x_{n - 1} - \pi| < \frac {\pi}{2},\quad n = 2,3,\dots$
By inequality
$|x| < |\tan(x)|,\ \ \text {for}\quad |x| < \frac {\pi}{2}$
periodicity of function $\tan(x)$ , fact that $x_{n}$ are solutions of equation $\tan(x) = x$ and adition formula we have
$|x_{n} - x_{n - 1} - \pi| < |\tan(x_{n} - x_{n - 1} - \pi})| = |\tan(x_{n} - x_{n - 1})| = \big |\frac {\tan(x_{n}) - \tan...$
follow requred
$\lim\limits_{n\to\infty}(x_{n} - x_{n - 1}) = \pi$
Remark

In fact, since

$\tan(x_{n} - x_{n - 1} - \pi) = \tan(x_{n} - x_{n - 1}) = \frac {\tan(x_{n}) - \tan(x_{n - 1})}{1 + \tan(x_{n})\tan(x_{n - 1}...$

we see that must be $x_{n} - x_{n - 1} - \pi > 0$ , and hence
$\pi < x_{n} - x_{n - 1} < \frac {3\pi}{2}$
And one more question!. While $\tan(x)$ is increasing function, together with previousus, is this sufficient to conclude that sequence

$\{x_{n} - x_{n - 1}\}$

is decreasing?

**Posted:** Mon Nov 23, 2009 2:19 pm