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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?

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up vote 3 down vote accepted

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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Nice :-) A little simplification. $\frac{x_{n+k}}{n}=\pi+o\left(\frac{1}{n^2}\right)$. Substituting $n+k=m$, ($n=m-k$) and using $\frac{1}{m-k}=\frac{1}{m}+o\left(\frac{1}{m^2}\right)$ we obtain $\frac{x_m}{m}=\pi+o\left(\frac{1}{m^2}\right)$. – vesszabo Nov 28 '12 at 21:57
@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. – user17762 Nov 28 '12 at 22:05

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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1) Hm, if $a<0$ then $\frac{ax}{x^2+b}$ is increasing and concave from some $x_0$ and $\tan(x)$ on $(\pi n - \frac{\pi}{2},\pi n)$ is increasing and concave too. In this case we have to prove analyticaly that $\tan(x)$ and $\frac{a x}{x^2 + b}$ meet each other in one point on such intervals. Then I'm not sure that assuming $a>0$ is "without loss of generality". 2) How did you obtain $x_n - n \pi = o(1)$? – Nimza Nov 28 '12 at 21:39
@Nimza 1) As Marvis says, adapt the idea (or method). 2) Which steps is problematic? (We know that $\tan(x_n)=\frac{a x_n}{x_n^2+b}$). – vesszabo Nov 28 '12 at 21:49

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