Consider an equation $ \tan (x) = \frac{a x}{x^2+b} $ where $a,b \neq 0$. Plotting $\tan(x)$ and function on the RHS we can see that this equation has infinitely many positive solutions $x_{n}$ and $x_{n} \sim \pi n$ as $n \to \infty$, i.e. $ \lim\limits_{n \to \infty} \frac{x_{n}}{n} = \pi. $ But is it possible to show the latter equality analitically?
Compute the limit
2 Answers
Let us assume $a,b>0$. The proof can be easily adapted if $a,b$ are not positive.
Consider $f(x) = \dfrac{ax}{x^2+b} - \tan(x)$. Note that $f(x)$ is nice except at $x = m \pi + \pi/2$ i.e. $f((m \pi + \pi/2)^-) = - \infty \,\,\,\,\,\, f((m \pi + \pi/2)^+) = \infty$ $f'(x) = \dfrac{a(b-x^2)}{(b+x^2)^2} - \sec^2(x)$ For large enough $x$ i.e. $x > \sqrt{ab}$, we have that $f'(x) < 0$.
Hence, for large enough $n$ within $\left( (n \pi - \pi/2), (n \pi + \pi/2) \right)$, the function is decreasing and changes sign. Hence, there is exactly one root in this interval.
Further, we have that $f(n \pi) = \dfrac{an \pi}{n^2 \pi^2+b} > 0$. \begin{align} f \left(n \pi + \dfrac{a}{n \pi} \right) & = \dfrac{a\left(n \pi + \dfrac{a}{n \pi} \right)}{\left(n \pi + \dfrac{a}{n \pi} \right)^2+b} - \tan\left(n \pi + \dfrac{a}{n \pi} \right)\\ & = \dfrac{a\left(n \pi + \dfrac{a}{n \pi} \right)}{\left(n \pi + \dfrac{a}{n \pi} \right)^2+b} - \tan\left(\dfrac{a}{n \pi} \right)\\ & \leq \dfrac{a\left(n \pi + \dfrac{a}{n \pi} \right)}{\left(n \pi + \dfrac{a}{n \pi} \right)^2} - \tan\left(\dfrac{a}{n \pi} \right)\\ & = \dfrac{a}{\left(n \pi + \dfrac{a}{n \pi} \right)} - \tan\left(\dfrac{a}{n \pi} \right)\\ & \leq \dfrac{a}{n \pi} - \tan\left(\dfrac{a}{n \pi} \right)\\ & \leq 0. \end{align} Hence, the root in the interval $\left( (n \pi - \pi/2), (n \pi + \pi/2) \right) $ in fact lies within $\left( n \pi, n \pi + \dfrac{a}{n \pi} \right)$. Hence, we have that $\left \vert x_{n+k} - n\pi \right \vert \leq \dfrac{a}{n \pi}$ where $k$ is a fixed natural number and takes into account some initial ugliness in the function where there could be possibly more or less roots. Hence, $\left \vert \dfrac{x_{n+k}}{n} - \pi \right \vert \leq \dfrac{a}{n^2 \pi}$ This gives your desired result.
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0@vesszabo Yes. Thanks. Or one could right $\dfrac{x_{n+k}}{n}$ as $\dfrac{x_{n+k}}{x+k} \dfrac{n+k}{n}$ and take limit to obtain the same conclusion. – 2012-11-28
Without loss of generality we may assume that $a>0$. Assume that $x$ is "large enough" depending on $a,b$. Then $\frac{a x}{x^2+b}$ is strictly decreasing on $(x_0,\infty)$ so it meets $\tan(x)$ exactly one on the intervals $(n\pi,n\pi+\pi/2)$. Denote it by $x_n$. Then $\lim_{n\to\infty} x_n=\infty$. Furthermore $\lim_{n\to\infty} \frac{a x_n}{x_n^2+b}=0$. It implies $\lim_{n\to\infty}\tan(x_n)=0$. Hence $x_n-n\pi=o(1)$. From this $\frac{x_n}{n}=\pi+o\left(\frac{1}{n}\right)$.
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0@Nimza 1) As Marvis says, adapt the idea (or method). 2) Which steps is problematic? (We know that $\tan(x_n)=\frac{a x_n}{x_n^2+b}$). – 2012-11-28