How would I go about finding the roots of $(e^x-1)-k\arctan(x) = 0$?

Question

I have the equation $$(e^x-1)-k\arctan(x) = 0$$ where $0<k \leq \frac 2\pi$ and I was wondering how I would go about starting to determine the amount of real roots of this equation. So far I have just manipulated the equation to get different equations of $x$, however I'm unsure what to do with them.

Current equations for $x$ are $x = \ln(k\arctan(x)-1)$ and $x = \frac{\tan(e^x-1)}{k}$

Answer 1

If $k>0$

Let $$f(x)=e^x-1-k \tan^{-1} x \implies f'(x)=e^x-\frac{k}{1+x^2}, f''(x)=e^{x}+\frac{k}{(1+x^2)^2}>0$$ $\implies$ $f'(x)$ is an increasing function $\implies$ $f'(x)>f'(-\infty)>0 $ $\implies$ $f(x)$ is an increasing function. Hence $f(x)=0$, will have at most one real root. As $f(-\infty)=-1+k\pi/2$ and $f(\infty)>0$, for one real root by IVT $f(-\infty) <0$. Finally this gives $0<k<\frac{2}{\pi}.$ The one root is $x=0$.

For $k>2/\pi$, both $f(-\infty)>$ and $f(\infty)>0$ so $f(x)=0$ will have $0,2,4,...$ number of real roots. Since $x=0$ is essentially a root. So there will two real roots if $k>2/\pi$. Further, since $f''(x)>0$, for the function $f(x)$ can have atmost one min, this rules out more than two real roots.

Answer 2

Of course $x=0$ is always a solution. Otherwise, write the equation as $$k = \frac{e^x-1}{\arctan(x)}$$ Call the right side $f(x)$. The singularity at $x=0$ is removable, with $\lim_{x \to 0} f(x) = 1$. We also have $\lim_{x \to -\infty} f(x) = 2/\pi$ and $\lim_{x \to +\infty} f(x) = +\infty$.
It appears that $f(x)$ is increasing. If so, for $2/\pi < k < 1$ and $1 < k < \infty$ there are two real roots ($x=0$ and the root of $f(x)=k$), otherwise there is only $x=0$.

Answer 3

Let $f(x)=e^x-1-k\arctan(x)$. We have $f(0)=0.$ If $k=0$, obviously $x=0$ is the only solution.

If $k\neq0$, then $f'(x)=e^x-\frac k{1+x^2},$ so that $$f'(x)=0\iff (1+x^2)e^x=k.$$ Since $(1+x^2)e^x$ is increasing, goes to $0$ at $-\infty$ and goes to $\infty$ at $\infty$ we see that $f'$ has exactly one real zero, say $f'(x_0) = 0$. We see that $f$ decreases on $(-\infty, x_0)$ and increases on $(x_0, \infty)$, so that $f$ attains its minimum at $x_0$.

Since we know that $f(0)=0$, it must be that $x_0<0$ and $x=0$ is the only solution.