Given a root of $x\tan x-1 = 0$, how to approximate the next one?

Question

Let $f(x) = x\tan x-1$. Let's consider when $f(x)=0$.

In the neighbourhood of every $k\pi$, where k is an integer, there should be a solution for $f(x)=0$.

Assuming $x_0$ approximately satisfies $f(x)$, is it correct that the next solution can be approximated to be at $x_0 + \pi - g(x)$ ?

If so, how can I find $g(x)$ to approximate the position of the next root to an arbitrary precision?

Answer 1

You are looking for the zero's of function $$f(x) = x\tan( x)-1$$ which are close to multiples of $\pi$.

I think that it is better to consider instead the problem of the zero's of function $$g(x) = x\sin( x)-\cos(x)$$

What can easily be done is to expand $g(x)$ as a Taylor series around $x=k \pi$ and to proceed later to a series resversion.

This would give $$\color{blue}{x_{(k)}=q+\frac{1}{q}-\frac{4}{3 q^3}+\frac{53}{15 q^5}-\frac{1226}{105 q^7}+\frac{13597}{315 q^9}-\frac{1531127}{8960 q^{11}}+\frac{84665251}{120960 q^{13}}+O\left(\frac{1}{q^{15}}\right)}$$ where $\color{blue}{q=k \pi}$.

Using the above truncated series (which could easily be extended), below are reproduced some results $$\left( \begin{array}{ccc} k & \text{approximation} & \text{exact solution} \\ 1 & 3.4256889465952269449 & 3.4256184594817281465 \\ 2 & 6.4372981816942647058 & 6.4372981791719471204 \\ 3 & 9.5293344053667719230 & 9.5293344053619636030 \\ 4 & 12.645287223856678278 & 12.645287223856643104 \\ 5 & 15.771284874815881554 & 15.771284874815882047 \\ 6 & 18.902409956860023945 & 18.902409956860024151 \\ 7 & 22.036496727938565037 & 22.036496727938565083 \\ 8 & 25.172446326646664703 & 25.172446326646664714 \\ 9 & 28.309642854452012361 & 28.309642854452012364 \\ 10 & 31.447714637546233552 & 31.447714637546233553 \\ 11 & 34.586424215288923664 & 34.586424215288923664 \end{array} \right)$$

Answer 2

A preliminary remark: The equation $f(x)=0$ is equivalent to $\tan(x)=\frac1x$. Now, $\tan$ is $\pi$-periodic and continuous, so for large $x$, the solutions will be arbitrarily close to the solutions of $\tan(x)=0$ which is just $x=\pi n$ for $n\in\mathbb Z$. This is why (almost) all your solutions are close to multiples of $\pi$.

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Using Newton's method we can give the following approximations for the next root:

If $x_0$ is a solution to $f(x_0)=0$, then the next solution $x_1$ is approximately $$x_1=h_n(x_0+\pi),$$

where $h_n$ denotes $n$ applications of Newton's method. For example (sorry for the ugly $h_3$ expression), $$h_1(x)=x-\frac{f(x)}{f'(x)}=x-\frac{x \tan (x)-1}{\tan (x)+x \sec ^2(x)}$$ and

$$h_2(x)=x-\frac{x \tan (x)-1}{\tan (x)+x \sec ^2(x)}-\frac{\left(x-\frac{x \tan (x)-1}{\tan (x)+x \sec ^2(x)}\right) \tan \left(x-\frac{x \tan (x)-1}{\tan (x)+x \sec ^2(x)}\right)-1}{\left(x-\frac{x \tan (x)-1}{\tan (x)+x \sec ^2(x)}\right) \sec ^2\left(x-\frac{x \tan (x)-1}{\tan (x)+x \sec ^2(x)}\right)+\tan \left(x-\frac{x \tan (x)-1}{\tan (x)+x \sec ^2(x)}\right)}$$ and $$h_3(x)=x+\frac{1-x \tan (x)}{x \sec ^2(x)+\tan (x)}-\frac{\frac{\left(x^2 \sec ^2(x)+1\right) \tan \left(\frac{2 x^2+\cos (2 x)+1}{2 x+\sin (2 x)}\right)}{x \sec ^2(x)+\tan (x)}-1}{\frac{\left(x^2 \sec ^2(x)+1\right) \sec ^2\left(\frac{2 x^2+\cos (2 x)+1}{2 x+\sin (2 x)}\right)}{x \sec ^2(x)+\tan (x)}+\tan \left(\frac{2 x^2+\cos (2 x)+1}{2 x+\sin (2 x)}\right)}-\frac{\left(x+\frac{1-x \tan (x)}{x \sec ^2(x)+\tan (x)}-\frac{\frac{\left(x^2 \sec ^2(x)+1\right) \tan \left(\frac{2 x^2+\cos (2 x)+1}{2 x+\sin (2 x)}\right)}{x \sec ^2(x)+\tan (x)}-1}{\frac{\left(x^2 \sec ^2(x)+1\right) \sec ^2\left(\frac{2 x^2+\cos (2 x)+1}{2 x+\sin (2 x)}\right)}{x \sec ^2(x)+\tan (x)}+\tan \left(\frac{2 x^2+\cos (2 x)+1}{2 x+\sin (2 x)}\right)}\right) \tan \left(x+\frac{1-x \tan (x)}{x \sec ^2(x)+\tan (x)}-\frac{\frac{\left(x^2 \sec ^2(x)+1\right) \tan \left(\frac{2 x^2+\cos (2 x)+1}{2 x+\sin (2 x)}\right)}{x \sec ^2(x)+\tan (x)}-1}{\frac{\left(x^2 \sec ^2(x)+1\right) \sec ^2\left(\frac{2 x^2+\cos (2 x)+1}{2 x+\sin (2 x)}\right)}{x \sec ^2(x)+\tan (x)}+\tan \left(\frac{2 x^2+\cos (2 x)+1}{2 x+\sin (2 x)}\right)}\right)-1}{\left(x+\frac{1-x \tan (x)}{x \sec ^2(x)+\tan (x)}-\frac{\frac{\left(x^2 \sec ^2(x)+1\right) \tan \left(\frac{2 x^2+\cos (2 x)+1}{2 x+\sin (2 x)}\right)}{x \sec ^2(x)+\tan (x)}-1}{\frac{\left(x^2 \sec ^2(x)+1\right) \sec ^2\left(\frac{2 x^2+\cos (2 x)+1}{2 x+\sin (2 x)}\right)}{x \sec ^2(x)+\tan (x)}+\tan \left(\frac{2 x^2+\cos (2 x)+1}{2 x+\sin (2 x)}\right)}\right) \sec ^2\left(x+\frac{1-x \tan (x)}{x \sec ^2(x)+\tan (x)}-\frac{\frac{\left(x^2 \sec ^2(x)+1\right) \tan \left(\frac{2 x^2+\cos (2 x)+1}{2 x+\sin (2 x)}\right)}{x \sec ^2(x)+\tan (x)}-1}{\frac{\left(x^2 \sec ^2(x)+1\right) \sec ^2\left(\frac{2 x^2+\cos (2 x)+1}{2 x+\sin (2 x)}\right)}{x \sec ^2(x)+\tan (x)}+\tan \left(\frac{2 x^2+\cos (2 x)+1}{2 x+\sin (2 x)}\right)}\right)+\tan \left(x+\frac{1-x \tan (x)}{x \sec ^2(x)+\tan (x)}-\frac{\frac{\left(x^2 \sec ^2(x)+1\right) \tan \left(\frac{2 x^2+\cos (2 x)+1}{2 x+\sin (2 x)}\right)}{x \sec ^2(x)+\tan (x)}-1}{\frac{\left(x^2 \sec ^2(x)+1\right) \sec ^2\left(\frac{2 x^2+\cos (2 x)+1}{2 x+\sin (2 x)}\right)}{x \sec ^2(x)+\tan (x)}+\tan \left(\frac{2 x^2+\cos (2 x)+1}{2 x+\sin (2 x)}\right)}\right)}$$

Answer 3

Although there is an accepted answer, Lagrange reversion gives all but the 2 smallest roots:

$$\cot(x)=x=\pi k+\sum_{n=1}^\infty\left.\frac{d^{n-1}}{dw^{n-1}}\frac{\cot^{-1}(w)^n}{n!}\right|_{\pi k},k\in\Bbb Z$$

Apply Stirling S1:

$$\left.\frac{d^{n-1}}{dw^{n-1}}\frac{\cot^{-1}(w)^n}{n!}\right|_{\pi k}=\left(\frac i2\right)^n\sum_{m=n}^\infty\frac{S_m^{(n)}(-2i)^m}{m!}\left.\frac{d^{n-1}}{dw^{n-1}}(w+i)^{-m}\right|_{\pi k}$$

Therefore:

$$\boxed{\cot(x)=x=\pi k+\sum_{n=1}^\infty\sum_{m=n}^\infty\frac{S_m^{(n)}(-m)^{(n-1)}(-2i)^{m-n}}{m!(\pi k +i)^{m+n-1}}}$$

shown here and both sums are interchangeable