Fixed points of $\tan\sqrt{x}$

Question

This question came in my class test in an MCQ format.

$\DeclareMathOperator{\N}{\mathbb N}$ Let $X=\{x\in\mathbb R^+: \tan(\sqrt{x})= x\}$. Consider the sequence $(b_n)_{n\in\N}$ of real numbers defined by $b_1=\inf(X)$, $b_{n+1}=\inf (X\setminus\{b_1,b_2,\ldots, b_{n}\})$. Does the series $\sum_{n=1}^\infty \frac{1}{b_n}$ converge?

I looked at the graph of $y=\tan\sqrt{x}$, it cuts the line $y=x$ exactly once in every interval of the form $\big(\left(\frac{\pi}{2}+n\pi\right)^2,\left(\frac{3\pi}{2}+n\pi\right)^2\big)$. Thus, each $b_n>\left((2n+1)\frac\pi2\right)^2$ so, by comparison test with $\sum \frac{1}{n^2}$, it follows that $\sum \frac{1}{b_n}$ converges.

My question is: How do I prove that there are countably many fixed points of $\tan\sqrt{x}$ and exactly one each in intervals of the form $I_n=\big(\left(\frac{\pi}{2}+n\pi\right)^2,\left(\frac{3\pi}{2}+n\pi\right)^2\big)$?

I made various attempts. I observed that $\tan\sqrt{\cdot}:I_n\to\mathbb R$ is a continuous surjection and that it's a composition of two strictly increasing functions. However, this doesn't seem to get anywhere.

Another thought I have is that $\tan\sqrt{x}=x$ may be equivalently written as $f(x)=0$ for some function $f$ s.t. $f'(x)=0\implies x\in \{\big(\frac{\pi}2+n\pi\big)^2:n\in\mathbb Z\}$. By application Rolle's theorem, we would conclude that there is exactly one root of $f$ between every pair of consecutive roots of $f'$.

Answer 1

Let $f(x) = \tan\sqrt{x}$, then $f'(x) = \frac{\sec^2 \sqrt{x}}{2\sqrt{x}}$ and $f''(x) = \frac{\sec^2\sqrt{x}\left(2\tan\sqrt{x} - \frac{1}{\sqrt{x}}\right)}{4x^{2}}$. Also define $g(x) = f(x) - x$. Now notice the following facts:

1. $\forall n$, $f'(x)>0$ for all $x\in I_n$, so $f$ is strictly increasing in each of these intervals.

2. $\forall n$, $f$ has exactly one inflection point in each interval, called $x_n$. To see this, notice that $f''$ changes sign in $I_n$ only when $2\tan\sqrt{x} - \frac{1}{\sqrt{x}}=0$. However, this a strictly increasing function, which can have at most 1 root. This function ranges from $-\infty$ to $\infty$ on each subinterval, so it does obtain this one root. Notice also that this fact is true for $g$ since $f''=g''$.

3. $g$ has a largest and smallest root in each subinterval. To see this, write each subinterval as $I_n = (a_n,b_n)$. Now notice that $\lim_{x\to b_n^-}f(x)=\lim_{x\to b_n^-}f'(x)=\infty$. Therefore there exists some root (at least one must exist by the intermediate value theorem) $x_{\text{max}}$ such that $\tan\sqrt{x}>x$ $\forall x\ > x_{\text{max}}$. We can use the same argument to show the existence of a minimum root, $x_{\text{min}}$ such that all roots of $g$ must lie in $[x_{\text{min}}, x_{\text{max}}]$.

Now pick some $I_n$ with corresponding inflection point, maximum root of $g$, and minimum root of $g$: $x_n$, $x_{\text{max}}$, and $x_{\text{min}}$, respectively. If $x_{\text{min}} = x_{\text{max}}$, there is a unique root in this subinterval and we add it to the sum. Now suppose that $x_{\text{min}}<x_{\text{max}}$. Notice that $g(x)$ is strictly convex on $[x_n+\varepsilon,x_{\text{max}}]$ for all $\varepsilon>0$ since $g''>0$ here. Therefore, $g$ can have at most two roots in this closed interval, which comes from Rolle's theorem. We can apply the same result to the interval $[x_{\text{min}},x_n-\varepsilon]$ since $-g$ is strictly convex here. This gives us that $g$ has at most 4 roots in $[x_{\text{min}},x_n-\varepsilon]\cup[x_n+\varepsilon,x_{\text{max}}]$ for all $\varepsilon>0$. Additionally, we could have $g(x_n)=0$, which leads to an upper bound of $5$ roots in each $I_n$, which proves the claim by the comparison test.

Answer 2

Too long for a comment.

If you want to approximate the non trivial roots of $$f(x)=\tan(\sqrt{x})- x$$ it could be better to let $x=y^2$ and to consider the zero's of $$g(y)=y^2 \cos (y)-\sin (y)$$ which does not present any discontinuity.

Because of the leading $y^2$ term, the roots will be closer and closer to $q=(2n+1)\frac \pi 2$ where $n$ is a positive integer.

Using series around this point we have $$0=-1-q^2 (y-q)+\frac{1-4 q}{2} (y-q)^2+\frac{q^2-6}{6} (y-q)^3+\frac{8 q-1}{24} (y-q)^4+\frac{20-q^2}{120} (y-q)^5+\frac{1-12 q}{720} (y-q)^6+O\left((y-q)^{7}\right)$$

Now, using power series reversion $$y=q-\frac 1 {q^2}-\frac{2}{q^5}+\frac{1}{3 q^6}-\frac{7}{q^8}+\frac{8}{3 q^9}-\frac{1}{5 q^{10}}-\frac{30}{q^{11}}+O\left(\frac{1}{q^{12}}\right)$$ Back to $x=y^2$

$$x=q^2-\frac{2}{q}-\frac{3}{q^4}+\frac{2}{3 q^5}-\frac{10}{q^7}+\frac{14}{3 q^8}-\frac{2}{5q^9}-\frac{42}{q^{10}}+ \frac{30}{q^{11}}+O\left(\frac{1}{q^{12}}\right)$$ This is very accurate $$\left( \begin{array}{ccc} n & \text{estimate} & \text{solution} \\ 1 & 21.7762185187785 & 21.7762181756291 \\ 2 & 61.4296083297709 & 61.4296083292618 \\ 3 & 120.720560969037 & 120.720560969030 \\ 4 & 199.717944050799 & 199.717944050799 \\ \end{array} \right)$$