stratsticks57jl

2022-07-20

Show that there is a number $x\in \left[\pi /2,\pi \right]$ such that $\mathrm{tan}\left(x\right)=-x$
Can I solve this by the intermediate value theorem?

Wayne Everett

Expert

Yes, you can. Consider $f\left(x\right)=\mathrm{tan}x+x$. Notice that $f\left(\pi /2+ϵ\right)\approx -\mathrm{\infty }$ for small $ϵ$ (more formally, the right limit as $x\to \pi /2$ of $f\left(x\right)$ is $-\mathrm{\infty }$) while $f\left(\pi \right)=\pi$. Now the intermediate value theorem gives you want you want.

Donna Flynn

Expert

Your problem is equivalent to proving that
$\begin{array}{}\text{(1)}& f\left(z\right)=z\mathrm{cot}z={z}^{2}+\frac{\pi z}{2}=g\left(z\right)\end{array}$
for some $z\in I=\left(0,\frac{\pi }{2}\right)$. That is trivial since $f\left(z\right)$ decreases from 1 to 0 on $I$ while $g\left(z\right)$ increases from 0 to $\frac{{\pi }^{2}}{2}$ on $I$, and both $f\left(z\right)$ and $g\left(z\right)$ are continuous over $I$. By approximating $z\mathrm{cot}z$ with $1-\frac{{z}^{2}}{3}$ we also get:
$z\approx \frac{-3\pi +\sqrt{192+9{\pi }^{2}}}{16}$
hence the solution to the original problem is about:
$\begin{array}{}\text{(2)}& x\approx \frac{5\pi +\sqrt{192+9{\pi }^{2}}}{16}.\end{array}$

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