Solve equation $\lfloor\arcsin x\rfloor+\lfloor\arccos x\rfloor+\lfloor\arctan x\rfloor=\ln x$

Question

Solve over reals: $$\lfloor\arcsin x\rfloor+\lfloor\arccos x\rfloor+\lfloor\arctan x\rfloor =\ln x$$

I believe it has no solution, but I don't know how to prove it.

Answer 1

Hints:

As @bjoyn93 points out, we need only solve over $(0,1]$. All three functions, $\arcsin x, \arccos x, \arctan x$ are continuous and increasing or decreasing, so we can use this to divide the interval into sub-intervals where the LHS takes particular values. The easiest case is $\arctan x$, which has a range of $(-\pi/2,\pi/2)$ and thus only passes near the three integers $-1,0,1$. In fact, over $(0,1]$, its floor is identically $0$.

When do $\arcsin x, \arccos x$ pass by integers? What constant value would the LHS take between each of these crossing-points?

Answer 2

Here's a picture over the interval $(0,1]$. I've marked the critical points of the floored inverse trigonometrical functions to better see how this functions behave.

Simply put, there are three cases:

• $x \in (0, \cos 1]:$ Then $0+1+0=\ln x \Rightarrow x = e \notin (0, \cos 1]$.
• $x \in (\cos 1, \sin 1):$ Then $0+0+0=\ln x\Rightarrow x = 1 \notin (\cos 1, \sin 1)$
• $x \in [\sin 1, 1]:$ Then $1+0+0=\ln x\Rightarrow x = e \notin [\sin 1, 1]$.

In conclusion, no solution.