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I have the following question:

Given a polynom $p(x)$ I need to prove the the equation $\tan x=p(x)$ has at least one solution.

I defined the difference as a function $f(x)=\tan x-p(x)$ and I know that $f(x)$ is continues in any interval where $\tan x$ is defined, so I am trying to use the Intermediate value theorem, but I don't know how.

J.G.
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Daniel
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  • Well, $p(0)$ and $p(\pi/2)$ have to be finite. So $f(x)$ is positive at one and negative at the other of those points (allowing infinity to be a value.) – B. Goddard Apr 01 '22 at 18:46

2 Answers2

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Consider $$g(x)=\cos x\, p(x)-\sin x$$ We have $g(-\pi/2)=1$ and $g(\pi/2)=-1.$ Hence $g(x_0)=0$ for a point $x_0,$ $-\pi/2<x_0<\pi/2.$ Thus $$p(x_0)={\sin x_0\over \cos x_0}=\tan x_0$$

Ryszard Szwarc
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  • Sure $\tan$ is surjective in $(-\frac{\pi}2,\frac{\pi}2)$ (ie. has range $(-\infty,+\infty)$) and $p(x)$ bounded so there must be an intersection point, but I like how your method avoid discussing infinity at all. – zwim Apr 02 '22 at 01:48
  • @zwim Thanks.I have applied a custom from elementary school math classes: when you have an equation with fractions multiply both sides by their denominators. – Ryszard Szwarc Apr 02 '22 at 03:56
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Here is how I would approach the problem, although there may be a more clever method.

We will restrict our view to the interval $I=(-\frac{\pi}{2}, \frac{\pi}{2})$. Since $p$ is a polynomial, it must be continuous on the closure $\overline{I}=[-\frac{\pi}{2}, \frac{\pi}{2}]$. By the extreme value theorem, $p$ attains both a maximum and a minimum on $\overline{I}$. Therefore, there is some $m\in \mathbb{R}$ such that for all $x\in I, |p(x)|\leq m$.

On the other hand, $tan(x)$ is not bounded on $I$ and has the range $(-\infty, \infty)$.

Let $n=2m$. There exists some $x$ such that $tan(x)=n$. For this fixed $x$, $f(x)=tan(x)-p(x)\geq tan(x)-m=m\geq 0$. Similarly, there exists an $x$ such that $tan(x)=-n$. For this fixed $x$ we have $f(x)=tan(x)-p(x)\leq tan(x)+m=-m\leq 0$.

From the equations above, we see that there is and $x$ such that $f(x)\geq 0$ and an $x$ such that $f(x) \leq 0$. As you suggested, this is where we can apply the intermediate value theorem. $f$ is continuous on $I$ and attains values above and below $0$ so there must be some $x'$ between the two $x$ we found such that $f(x)=0$.

elevensor
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