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How can i prove that there exist some real $a >0$ such that $\tan{a} = a$ ?

I tried compute $$\lim_{x\to\frac{\pi}{2}^{+}}\tan x=\lim_{x\to\frac{\pi}{2}^{+}}\frac{\sin x}{\cos x}$$

We have the situation " $\frac{1}{0}$ " which leads us " $\infty$ "

$$\lim_{x\to\frac{\pi}{2}^{-}}\tan x=\lim_{x\to\frac{\pi}{2}^{-}}\frac{\sin x}{\cos x}$$

We have the situation " - $\frac{1}{0}$" which tells us " $- \infty$"

This means for any real number $y$ there exists $x_1$ and $x_2$ in $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ such that

$f(x_1)<y<f(x_2)$.

Remember the Intermediate Value Theorem. If f is a continuous function and $f(a)<0<f(b)$ then there exist $x \in (a,b)$ such that $f(x) = 0$

So $f(x_1)<y<f(x_2)$ is equivalent to

f(x1) - y < 0 < f(x2) - y (I just subtracted y from each part)

Now we can use the Intermediate Value Theorem (applied to $(f - y)$ to say there exists an $x \in (x_1 , x_2)$ such that $f(x) - y = 0 $

or $f(x) = y$ or $\tan x = y$

We know $x_1 < x < x_2$ and $-\frac{\pi}{2}<x_1<x_2<\frac{\pi}{2}$

So we know that $-\frac{\pi}{2}<x<\frac{\pi}{2}$, then $f(x) - y = 0$.

But how can i prove it for $a >0$ such that $\tan{a} = a$ ?

Dario
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Rosa Maria Gtz.
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  • I think the answer is yes, just by considering the graphs of tanx and x. I just can't figure out the actual x. – user99680 Jun 28 '14 at 08:24
  • On every interval $](2k-1) \frac{\pi}{2}, (2k+1) \frac{\pi}{2}[$, the $\tan$ function attains every value in $\mathbb{R}$, since the limit towards the left it $-\infty$, towards the right is $+\infty$ and $f$ is continuous on such an interval –  Jun 28 '14 at 08:28
  • @Maxerize: Sounds like what I was thinking, but more clear. Why don't you make it an answer (tho you need to show they intersect more than once)? – user99680 Jun 28 '14 at 08:29

2 Answers2

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Use the intermediate value theorem on $\tan x -x$ after plugging in $x={3\pi\over 4}$ and $x={35\pi\over 24}$.

I chose those specific values because (in theory) you can compute them by hand using half-angle formulae and it's more constructive, but one can use the definition of unboundedness near $3\pi/2$ and the boundedness of the identity function on any compact set to make things slicker.

Adam Hughes
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  • Nice and to the point, +1. – user99680 Jun 28 '14 at 08:33
  • the OP mentioned $a \gt 0$, this shows that there is one $a$ between $-\frac {\pi}{4}$ and $+\frac {\pi}{4}$ such that ${\rm tan} \space a=a$. But we already know that ${\rm tan} \space 0 = 0$ and also that ${\rm tan} \space x \gt x$ in $(0,\frac {\pi}{4}]$. Hence, I don't think this is helpful and the DV. – Indrayudh Roy Jun 28 '14 at 08:46
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    Periodicity fixes it. Thanks for noting that was the request. I've edited my answer to reflect a positive solution. – Adam Hughes Jun 28 '14 at 08:56
  • Now observe that ${\rm tan} \frac {3\pi}{4} - \frac {3\pi}{4}$ and ${\rm tan} \frac {5\pi}{4}-\frac {5\pi}{4}$ are both negative. – Indrayudh Roy Jun 28 '14 at 09:07
  • Ah damn, I suppose you're right: you really do have to get closer to the pole. I'll fix it one more time, though it's so much uglier. :-) – Adam Hughes Jun 28 '14 at 09:19
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Hint

You are looking for the intersection of two functions $y_1=\tan(x)$ and $y_2=x$. You also know that $\tan(x)$ has discontinuities at $x=(2k+1) \frac {\pi}{2}$ (to be more precise, as gniourf_gniourf commented, the $\tan$ is not defined at these points). So, you have an infinite number of solutions for $\tan(x)=x$ (in particular because $\tan(k\pi)=0$) and the solutions are closer and closer to the vertical asymptotes.

But, as you know, in the range $(0,\frac{\pi}{2})$ $y_2$ is the tangent to $y_1$ and there is no solution in this interval. The first solution will happen just below $\frac{3\pi}{2}$, the second still closer to $\frac{5\pi}{2}$ and so on. You also can notice that, if $x_n$ is a solution $-x_n$ is another.

Moreover, Taylor series built at $x=(2k+1) \frac {\pi}{2}$ shows then $$\tan(x) \simeq -\frac{1}{x-(2k+1)\frac{\pi}{2}}$$ so you can approximate the solutions solving for $x$ $$x= -\frac{1}{x-(2k+1)\frac{\pi}{2}}$$ which leads to $$x_k \simeq \frac{1}{4}\Big((2k+1)\pi+\sqrt {(2k+1)^2 \pi^2-16}\Big)$$ from which you could show that, for large values of $k$ $$x_k\simeq(2k+1) \frac {\pi}{2}-\frac{2}{\pi (2 k+1)}$$

Claude Leibovici
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  • To be really correct, $\tan$ has no discontinuities at the points of the form $(2k+1)\pi/2$ for $k\in\mathbb{Z}$. It is just not defined at these points! Otherwise, $\tan$ is continuous on its domain ${x\in\mathbb{R};|;\cos(x)\neq0}$. – gniourf_gniourf Jun 28 '14 at 08:58
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    @gniourf_gniourf. You are perfectly right ! I shall edit quoting you.Cheers :) – Claude Leibovici Jun 28 '14 at 09:00