Solving $\ln{x}=\tan{x}$ with infinitely many solutions

Question

Lets take $f(x)=\ln{x}$ and $g(x)=\tan{x}$

When $f(x)=g(x)$ that is $\ln{x}=\tan{x}$, we see that the graph is like:

Hence we see that there are infinitely many solutions to $x$ but the two graphs do not coincide (like while solving "$x=x$"!)

The solutions as given by WolframAlpha were like :

---

So i decided to use Newton's method to solve this but due to having infinitely many solutions, all solutions given by that method were close to the approximations given by WolframAlpha but were not as accurate.

$x_{n+1} = x_{n} - \dfrac{\ln{x_{n}}-\tan{x_{n}}}{{\dfrac{1}{x_{n}}}-\sec^2{x_{n}}}$

giving us x = 4.02 , 7.31 , ... (and hence not correct)

So is there any other way to solve these system of equations? Maybe using Lambert-W or maybe it does have some simple solution which I'm missing?

Please help, Thanks!

Answer 1

Hint

The problem is better conditioned if, instead, you look for the roots of $$h(x)=\log(x)\cos(x)-\sin(x)=0$$ and the convergence of Newton method is much faster.

For example, using $x_0=4$, the iterates are : $4.09701$, $4.09546$ which is the solution for six significant digits. Using $x_0=7$, the iterates are : $7.42088$, $7.39041$, $7.39037$.

Using this transform, you will find easily the $k^{th}$ root starting at $x_0=(2k+1)\frac \pi 2$. The first iterate will just be $$x_1=\pi \left(k+\frac{1}{2}\right)-\frac{1}{\log \left(\pi \left(k+\frac{1}{2}\right)\right)}$$

Edit

What is interesting is to look at the value of the solution $x_k$ as a function of $k$. A totally empirical model leads to the following approximation $$x_k=(2k+1)\frac \pi 2-\Big(0.0962696 +\frac{0.489665}{k^{0.409352}}\Big)$$ For example, the estimate of the $10^{th}$ root is $32.6997$ while the solution is $32.7075$; the estimate of the $50^{th}$ root is $158.455$ while the solution is $158.456$.

Just for comparison wtih the results from Wolfram Alpha, the five first roots are estimated as $4.12645,7.38901,10.5870,13.7633,16.9291$.

Edit

There is another way to generate nice approximate values of the $k^{th}$ solution. The idea is to build for function $h(x)$ its simplest Pade approximant around $\theta_k=(2k+1)\frac \pi 2$ and to compute the value of $x$ which cancels the numerator. This leads to the simple approximation $$x_k\approx\theta_k \left(1-\frac{2 \log (\theta_k)}{\theta_k \left(2 \log ^2(\theta_ k)+1\right)-2}\right)$$ Just for comparison wtih the results from Wolfram Alpha, the five first roots are estimated as $4.13630,7.40792,10.6062,13.7814,16.9459$. The $10^{th}$ root is estimated as $32.7113$ and the $50^{th}$ root is estimated as $158.457$.

We could still improve using a Pade[1,2] approximant which would lead to $$x_k\approx \theta_k\Big(1 -\frac{3 \left(\theta_k +2 \theta_k \log ^2(\theta_k )-2\right)}{\theta_k \log (\theta_k ) \left(5 \theta_k +6 \theta_k \log ^2(\theta_k )-12\right)-3}\Big)$$ Just for comparison wtih the results from Wolfram Alpha, the five first roots are estimated as $4.09189,7.38912,10.5942,13.7725,16.9387$. The $10^{th}$ root is estimated as $32.7074$ and the $50^{th}$ root is estimated as $158.455$.

Answer 2

In each interval $\left(\pi n - \frac{\pi}{2}, \pi n + \frac{\pi}{2}\right)$ there is exactly one solution $x_n$ (i.e. $\tan x_n = \ln x_n$), and, when $n$ is large, it appears that $x_n$ is approximately $\pi n + \frac{\pi}{2}$. Let's show this.

Since $\tan$ is $\pi$-periodic we have

$$\tan\left(\pi n + \frac{\pi}{2} - x_n\right) = \tan\left(\frac{\pi}{2} - x_n\right)$$ $$\hspace{2.4 cm} = \frac{1}{\tan x_n}$$ $$\hspace{2.6 cm} = \frac{1}{\ln x_n} \to 0$$

as $n \to \infty$, where the second-to-last equality follows from the identites $$\sin\left(\frac{\pi}{2} - \theta\right) = \cos \theta,$$ $$\cos\left(\frac{\pi}{2} - \theta\right) = \sin \theta.$$

Since $-\frac{\pi}{2} < \pi n + \frac{\pi}{2} - x_n < \frac{\pi}{2}$ and since $\tan$ is continuous in this interval we have $\pi n + \frac{\pi}{2} - x_n \to 0$ as $n \to \infty$.

So, we know that

$$ x_n = \pi n + \frac{\pi}{2} + o(1). $$

Let's get an estimate for the error term. If we set $w_n = \left(\pi n + \frac{\pi}{2}\right)^{-1}$ and $z_n = w_n^{-1} - x_n$ then

$$ \tan x_n = \frac{1}{\tan z_n} $$

by the above calculation and

$$ \ln x_n = \ln w_n^{-1} + \ln(1+w_n z_n), $$

so the equation $\tan x_n = \ln x_n$ becomes

$$ \frac{1}{\tan z_n} = \ln w_n^{-1} + \ln(1+w_n z_n). \tag{$*$} $$

Now $w_n,z_n \to 0$ as $n \to \infty$, so

$$ \frac{1}{\tan z_n} \sim \frac{1}{z_n} $$

and

$$ \ln w_n^{-1} + \ln(1+w_n z_n) \sim \ln w_n^{-1} $$

as $n \to \infty$. Thus, from $(*)$,

$$ \frac{1}{z_n} \sim \ln w_n^{-1}, $$

or

$$ z_n \sim \frac{1}{\ln w_n^{-1}} = \frac{1}{\ln(\pi n + \pi/2)}. $$

By definition of $z_n$ we therefore get the asymptotic

$$ x_n = \pi n + \frac{\pi}{2} - \frac{1}{\ln(\pi n + \pi/2)} + o\left(\frac{1}{\ln n}\right). $$

Answer 3

If you do not mind a contour integral expression for your roots, one can use the Delves-Lyness scheme to represent your roots:

$$x_{\ast}=\frac1{2\pi i}\oint_\gamma \frac{z f^\prime(z)}{f(z)}\,\mathrm dz$$

where $f(x_{\ast})=0$ and $\gamma$ is any anticlockwise contour encircling the desired root and has no poles of $f(x)$ within.

In particular, borrowing Claude's strategy of reformulating the equation as $\log x\cos x-\sin x=0$ and localizing the roots around $x=\pi\left(k+\frac12\right)$, we can use the contour $\gamma=\pi\left(k+\frac12\right)+\frac{\pi}{4}\exp(it)$ to evaluate the Delves-Lyness contour integral. A realization in Mathematica for finding the first $10$ roots goes like this:

Table[Re[NIntegrate[Exp[I t] (Cos[#] - # Log[#] Sin[#] - # Cos[#])/
                    (Log[#] Cos[#] - Sin[#]) &[π (k + 1/2) + π Exp[I t]/4],
                    {t, 0, 2 π},
                    Method -> {"Trapezoidal", "SymbolicProcessing" -> 0}]/8],
      {k, 10}]

   {4.095461605910074, 7.390369571112949, 10.59483666075098, 13.772896612324644,
    16.93905539111342, 20.098667030734298, 23.254225459730314, 26.40706896072724,
    29.55798827765332, 32.70748457864333}

and these can be polished further with Newton-Raphson or some other iterative method if need be.

Answer 4

A simple recursion for $x^{(n)}$ (the $n-$th root, around $n \pi$, with $n\ge 1$) is

$$ x^{(n)}_{i+1}= \tan^{-1} \left( \log x^{(n)}_{i} \right) + n \pi$$

starting with $x^{(n)}_{0}= n \pi$.