Trying to find roots of $\arctan(2(x − 1)) − \ln |x| = 0$ analytically

Question

I have found the roots graphically and also numerically.

Apparently there is also a calculus root to finding them analytically.

I was thinking to use the derivative but it doesn't seem to work.

Many thanks for any help or advice.

Answer 1

This equation is highly transcendental and analytical solutions cannot be obtained (remember that this is already the case for the simple $x=\cos(x)$).

However, you can get very close approximations using Taylor expansions built around the approximate roots. Graphing (and excluding the trivial solution $x=1$), you noticed a root close to $x=-0.3$, another one close to $x=0.4$ and another one close to $x=4$.

So, around each of these values, using Taylor expansions, we should successively get $$\left(\log \left(\frac{10}{3}\right)-\tan ^{-1}\left(\frac{13}{5}\right)\right)+\frac{1045}{291} \left(x+\frac{3}{10}\right)+\frac{485075 \left(x+\frac{3}{10}\right)^2}{84681}+O\left(\left(x+\frac{3}{10}\right)^3\right )$$ $$\left(\log \left(\frac{5}{2}\right)-\tan ^{-1}\left(\frac{6}{5}\right)\right)-\frac{205}{122} \left(x-\frac{2}{5}\right)+\frac{117025 \left(x-\frac{2}{5}\right)^2}{29768}+O\left(\left(x-\frac{2}{5}\right)^3\right)$$ $$\left(\tan ^{-1}(6)-\log (4)\right)-\frac{29 (x-4)}{148}+\frac{601 (x-4)^2}{43808}+O\left((x-4)^3\right)$$

Solve the quadratic equations to get analytical expressions and keep the proper solutions. Converted to their decimal representations, they would be $-0.300098$, $0.425460$ and $4.09946$ while the "exact" solutions (you already found them probably using Newton method) are $-0.300098$, $0.425412$ and $4.09946$.

Edit

We could have "simpler" (no radical) explicit approximation formulae using the simplest $[1,1]$ Padé approximant built around $x=a$ instead of the corresponding Taylor series.

$$x =a +\frac{2 f(a) f'(a)}{f(a) f''(a)-2 f'(a)^2}$$

This would give $$\frac{3 \left(-43681-21143 \log \left(\frac{10}{3}\right)+21143 \tan ^{-1}\left(\frac{13}{5}\right)\right)}{10 \left(43681-19403 \log \left(\frac{10}{3}\right)+19403 \tan ^{-1}\left(\frac{13}{5}\right)\right)}\approx -0.300098$$

$$\frac{2 \left(3362+321 \log \left(\frac{5}{2}\right)-321 \tan ^{-1}\left(\frac{6}{5}\right)\right)}{5 \left(3362-4681 \log \left(\frac{5}{2}\right)+4681 \tan ^{-1}\left(\frac{6}{5}\right)\right)} \approx 0.425364$$

$$\frac{4 \left(-1682+1545 \log (4)-1545 \tan ^{-1}(6)\right)}{-1682-601 \log (4)+601 \tan ^{-1}(6)}\approx 4.09946$$