Irrerbthist6n

Answered

2022-01-14

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

Answer & Explanation

reinosodairyshm

Expert

2022-01-16Added 36 answers

If k>0

Let

$f\left(x\right)={e}^{x}-1-k{\mathrm{tan}}^{-1}x\Rightarrow f\prime \left(x\right)={e}^{x}-\frac{k}{1+{x}^{2}},f\prime \prime \left(x\right)={e}^{x}+\frac{k}{{(1+{x}^{2})}^{2}}>0$

$\Rightarrow$ f′(x) is an increasing function $\Rightarrow f\prime \left(x\right)>f\prime (-\mathrm{\infty})>0\Rightarrow f\left(x\right)$ is an increasing function. Hence f(x)=0, will have at most one real root. As $f(-\mathrm{\infty})=-1+\frac{k\pi}{2}$ and $f\left(\mathrm{\infty}\right)>0$ , for one real root by IVT $f(-\mathrm{\infty})<0$ . Finally this gives $0<k<2\pi$ . The one root is x=0.

For$k>\frac{2}{\pi}$ , both $f(-\mathrm{\infty})>$ 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.

Let

For

RizerMix

Expert

2022-01-20Added 437 answers

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

alenahelenash

Expert

2022-01-24Added 366 answers

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

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