How can I find the possible values of $x$ for:
$\tan(x)=x$
mathematically?
How can I find the possible values of $x$ for:
$\tan(x)=x$
mathematically?
There is no closed form for the solutions of $ \tan(x) = x $, but let me state a few interesting facts. Let $ (\lambda_{n})_{n \in \mathbb{N}} $ be the sequence that lists the positive solutions of $ \tan(x) = x $ in increasing order. Then
$ \displaystyle \sum_{n=1}^{\infty} \frac{1}{\lambda_{n}} = \infty $.
$ \displaystyle \sum_{n=1}^{\infty} \frac{1}{\lambda_{n}^{2}} = \frac{1}{10} $.
I gave a talk on this equation in April 2006 and (a .pdf version of) the LaTeX slides I used may be of interest to you.
See also the math StackExchange question Derivation of asymptotic solution of $\tan (x)=x$. Finally, see the posts in this January 2006 sci.math thread: Regarding tan(x) = x.
$$\frac{\sin(x)}{\cos(x)}=x$$ $$\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}...$$ $$\cos(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}...$$ Your question is equivalent to solving the equation
$$x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots=x-\frac{x^3}{2!}+\frac{x^5}{4!}-\cdots$$ $$x^3\left(\frac{1}{3!}-\frac{1}{2!}\right)-x^5\left(\frac{1}{5!}-\frac{1}{4!}\right)+x^7\left(\frac{1}{7!}-\frac{1}{6!}\right)+\cdots=0$$ Evidently giving $$x=0$$
The other solutions are given by the equation $$\left(\frac{1}{3!}-\frac{1}{2!}\right)-x^2\left(\frac{1}{5!}-\frac{1}{4!}\right)+x^4\left(\frac{1}{7!}-\frac{1}{6!}\right)-\cdots=0$$
But I don't know if there is a way to get from that to a closed form.