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How can I find the possible values of $x$ for:



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$x=0{}{}{}{}{}$. – Daryl Jan 3 '13 at 20:29
You can get very nice estimates for the roots. Some non-trivial mathematics is involved. There is as far as I know no "closed form" expression for the family of solutions. – André Nicolas Jan 3 '13 at 20:29
Note that for $-\pi/2 < x < \pi/2$, only $x=0$ is a solution. But there are infinitely many solutions with $x$ outside this region, and, as Andre says, it is unlikely they have a closed form. – Thomas Andrews Jan 3 '13 at 20:31
Let $k$ be any integer. Then there is exactly one solution $x$ in the interval $(k\pi -\pi/2,k\pi+\pi/2)$ – Thomas Andrews Jan 3 '13 at 20:39
Degrees, radians, it doesn't matter much. The analysis for $\tan x=ax$ is quite similar to the one for $a=1$. – André Nicolas Jan 3 '13 at 20:45
up vote 8 down vote accepted

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.

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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} $.

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Guess Im not the only one who knows this :p +1 from me. – mick Jan 3 '13 at 20:56
@mick: Oh, you mean the second identity? – Haskell Curry Jan 3 '13 at 20:59
Yes ... What else ? – mick Jan 3 '13 at 21:00
@HaskellCurry: Can u provide me some links to the explanations of this properties. Right now I am having difficulty understanding how these properties came. – Inquisitive Jan 3 '13 at 21:00
You can refer to this other MSE thread:…. – Haskell Curry Jan 3 '13 at 21:10

$$\frac{\sin(x)}{\cos(x)}=x$$ $$\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}...$$ $$\cos(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}...$$ 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.

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