How can I find the possible values of $x$ for:
$\tan(x)=x$
mathematically?
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2$x=0{}{}{}{}{}$. – Daryl Jan 03 '13 at 20:29
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3You 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 03 '13 at 20:29
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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 03 '13 at 20:31
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@ThomasAndrews: Is infinte many solutions possible? When I tried to do this graphically , I got only three solutions. – Inquisitive Jan 03 '13 at 20:36
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3Let $k$ be any integer. Then there is exactly one solution $x$ in the interval $(k\pi -\pi/2,k\pi+\pi/2)$ – Thomas Andrews Jan 03 '13 at 20:39
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When you did it graphically, you may not have set your calculator window to be large enough. – Shivam Sarodia Jan 03 '13 at 20:39
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I was talking about degrees not radian. – Inquisitive Jan 03 '13 at 20:41
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1Degrees, radians, it doesn't matter much. The analysis for $\tan x=ax$ is quite similar to the one for $a=1$. – André Nicolas Jan 03 '13 at 20:45
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I guess there are no rational solutions except 0. – Amr Jan 03 '13 at 20:52
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In degrees, there are three solutions between the two vertical asymptotes closest to the origin. In radians, there is only the one trivial solution between those two asymptotes and all others are farther from the origin. – Michael Hardy Jan 04 '13 at 00:19
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There is a Bessel J zero function solution and a series solution with Stirling numbers. If you want the this solution posted, then please say so as there are already 3 answers here – Тyma Gaidash Jun 12 '23 at 20:43
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Does this answer your question? Derivation of asymptotic solution of $\tan(x) = x$. – user Mar 26 '24 at 14:32
3 Answers
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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} $.
Haskell Curry
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@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 03 '13 at 21:00
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You can refer to this other MSE thread: http://math.stackexchange.com/questions/75206/sum-of-the-squares-of-the-reciprocals-of-the-fixed-points-of-the-tangent-functio. – Haskell Curry Jan 03 '13 at 21:10
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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.
hbghlyj
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Dave L. Renfro
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$$\frac{\sin(x)}{\cos(x)}=x$$ $$\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}...$$ $$\cos(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}...$$ 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.
Meow
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