 Irrerbthist6n

2022-01-14

How would I go about finding the roots of $\left({e}^{x}-1\right)-k\mathrm{arctan}\left(x\right)=0$? reinosodairyshm

Expert

If k>0
Let
$f\left(x\right)={e}^{x}-1-k{\mathrm{tan}}^{-1}x⇒f\prime \left(x\right)={e}^{x}-\frac{k}{1+{x}^{2}},f\prime \prime \left(x\right)={e}^{x}+\frac{k}{{\left(1+{x}^{2}\right)}^{2}}>0$
$⇒$ f′(x) is an increasing function $⇒f\prime \left(x\right)>f\prime \left(-\mathrm{\infty }\right)>0⇒f\left(x\right)$ is an increasing function. Hence f(x)=0, will have at most one real root. As $f\left(-\mathrm{\infty }\right)=-1+\frac{k\pi }{2}$ and $f\left(\mathrm{\infty }\right)>0$, for one real root by IVT $f\left(-\mathrm{\infty }\right)<0$. Finally this gives $0. The one root is x=0.
For $k>\frac{2}{\pi }$, both $f\left(-\mathrm{\infty }\right)>$ and $f\left(\mathrm{\infty }\right)>0$ 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 $k>\frac{2}{\pi }$. Further, since f′′(x)>0, for the function f(x) can have atmost one min, this rules out more than two real roots. RizerMix

Expert

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

Expert

Let $f\left(x\right)={e}^{x}-1-k\mathrm{arctan}\left(x\right)$. We have f(0)=0. If k=0, obviously x=0 is the only solution. If $k\ne 0$, then $f\prime \left(x\right)={e}^{x}-\frac{k}{1+{x}^{2}}$, so that $f\prime \left(x\right)=0⇔\left(1+{x}^{2}\right){e}^{x}=k$. Since $\left(1+{x}^{2}\right){e}^{x}$ is increasing, goes to 0 at $-\mathrm{\infty }$ and goes to we see that f′ has exactly one real zero, say f′(x0)=0. We see that f decreases on $\left(-\mathrm{\infty },{x}_{0}\right)$ and increases on $\left({x}_{0},\mathrm{\infty }\right)$, 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.