Number of roots of $(e^x-1)-k\arctan{x}=0$

Question

$\text{Determine the number of real roots of for every case of the constant k, } \ (e^x-1)-k\arctan{x}=0$

How would I go about solving this analytically? What I have so far is that the derivative, $e^x(x^2+1)-k=0$ is greater than zero if and only if $k>0$, then that the second derivative is always positive, hence the stationary point is a minimum, however it hasn't lead me to anything to do with roots. The hint given in the question is to sketch the graph of $y=(e^x-1)-k\arctan{x}$ the cases for important values of k. The cases of k are clearly dependent on the limits and arctan e.g. if k is less than $\frac{2}{\pi}$, then there are no obvious roots. Through desmos, I can see that if k is greater than $\frac{2}{\pi}$, it actually has two roots. This question is intended to be solved analytically, but I cannot figure how I would find this out analytically(without guessing at small values of x).

Answer 1

The number of real roots in

$(e^x-1)-k*\arctan(X)$

I'll suggest you plot a 3D graph of the function, making it

$z = (e^x-1)-y*\arctan(X)$

This means you've taken in account the range of values of $k$ also

Now you can use partial differentiation to differentiate w.r.t both $x$ and $y$ To see their limit, minimum and maximum

Answer 2

Outside the problem of the number of roots, since they suggest to look at what happens for large values of $k$, for that case, consider that you look for the zero of function $$f(x)=\log \left(\frac{e^x-1}{\tan ^{-1}(x)}\right)-\log(k)$$ If you plot the logarithm, as normal, you will notice an almost straight line with a slope very close to $1$.This means that, at least for large values of $k$, $$f(x)\sim x -\log \left(\frac{\pi k}{2}+1\right)\implies x\sim \log \left(1+k\frac{\pi }{2}\right)$$

This could be a good estimate for starting Newton method. Let us try for $k=123456789$ and we shall see the following iterates $$\left( \begin{array}{cc} n & x_n \\ 0 & 19.082984 \\ 1 & 19.049025 \end{array} \right)$$

Answer 3

Let $$f(x)=e^x-1, ~~g(x)=k \tan^{-1} x, ~~ h(x)=e^x-1-k \tan^{-1} x$$ Note that $h(0)$, so $x=0$ is always a root irrespective of that values od $k$. $$h(-\infty)=-1+k \pi/2, ~~ h(\infty)= +\infty$$ So if $k<0 ~or~ k<2/\pi$, then $f(-\infty) f(\infty) <0$ so there will be odd number of real roots. $$h'(x)=e^x-\frac{k}{1+x^2},~~~h''(x)=e^x+\frac{2kx}{1+x^2}.$$ Next if $k<0$ there will be no root of $ h'(x)=0$ so no max/min so we will havw only one root ($x=0$).

Otherwise, when $k>2/\pi$ 0, 2, 4,... (even) number of roots are possible As $h(0)=0$, $x=0$ is always a root, therefore if $k>2/\pi$ one more real root will be possible. So there will be atmost two reals. As $h'(x)=0$ will have at most one real root only because $h''(x)>0$ son only one max/min. This implies at most two real roots of $h(x)=0$ of which $x=0$ is always one root.