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I needed to find, using the bisection method, the first positive value that satisfy $x = \tan(x)$. So I went to Scilab, I wrote the bisection method and I got $1.5707903$. But after some reasoning I came to the conclusion that this value is wrong:

  1. $\tan(1.5707903) \approx 1.6x10^5$. Not even close to $1.5707903$.
  2. Forget for a moment the above. $x = \tan(x)$ is actually to find fixed points of $f(x) = \tan(x)$; $(x, f(x))$ must be in the line $y = x$. Here is the plot: plot(tan(x), x)

In $(0, \frac{3}{2}\pi)$ I can only see a fixed point to the right of $x = 4$, therefore $1.5707903$ is wrong.

Here comes the interesting part. If you go to Wolfram Alpha and type $x = \tan(x)$, you will see $1.5708$ in the Plot section: x = tan(x)

However there is no $1.5708$ in the Numerical solutions section. Wolfram Alpha found $0, \pm 4.49340945790906, \ldots$.

But if you type $\tan(x) = x$, you will not see $1.5708$ in the Plot section!: tan(x) = x

To summarize:

  1. Is $4.49340945790906$ the first positive value that satisfy $x = \tan(x)$?
  2. Do you know why Wolfram Alpha is showing $1.5708$ as a solution when you type $x = \tan(x)$ but not when you type $\tan(x) = x$?

Thanks.

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related/possible duplicate –  Parth Kohli Feb 16 '13 at 14:32
    
(1) $\tan x$ maps $(-\pi/2,\pi/2)$ bijectively onto $\mathbb{R}$. (2) Moreover, $\tan$ is periodic with period $\pi$. From (1) and (2) it follows that the $x$ we are looking for is in $(\pi/2,3\pi/2)$. –  AD. Feb 16 '13 at 14:36
    
BTW bisection is not very fast at all, I would suggest you tried with Newton-Raphson with a starting point like $\pi$. –  AD. Feb 16 '13 at 14:37
    
@AD.: I'm interested in your reasoning but I'm not able to follow it. Could you please explain how from (1) and (2) you concluded that the $x$ we are looking for is in $(\pi/2,3\pi/2)$? –  David Robert Jones Feb 16 '13 at 15:08
    
Bisection would definitively work depending on the choose of starting points they need both to belong in $I_n = (-\pi/2,\pi/2)+\pi\cdot n$ and surround the fixed point. –  AD. Feb 16 '13 at 15:23

2 Answers 2

up vote 3 down vote accepted

The reason you are getting this "solution" is because the bisection method assumes the function is continuous in the range, which it's not. Since the function at both sides of $x=\pi/2$ is $\pm \infty$, the bisection method will always converge to this "solution".

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Nor is continuous at $[\pi/2,3\pi/2]$ however the solution is there. –  David Robert Jones Feb 16 '13 at 15:29
    
@DavidRobertJones - it's continuous on the open interval. –  nbubis Feb 16 '13 at 15:33

As you see from the plot of $\tan x$, you're intercepting the asymptote, which is not really the desired behavior. Bisection is not the best method to use.

However, if you're required to use bisection, then instead note that $\tan x = \frac{\sin x}{\cos x}$, so, for relevant values of $x$,

$$x = \tan x \implies x\cos x - \sin x = 0$$

The latter function is continuous, and you should get the desired solution of $x \approx 4.49$.

share|improve this answer
    
I like your idea. Now I'm wondering: What function would you use as $g(x)$ if you're required to use fixed-point iteration method? –  David Robert Jones Feb 16 '13 at 22:28
    
@DavidRobertJones Offhand, I suspect $\sin^{-1}(x\cos x) = x$ should work, as long you start the iteration in $(0,2\pi)$. –  Arkamis Feb 16 '13 at 23:38
    
I'm not sure since $\sin^{-1}(x\cos x)$ has no fixed points in the interval. Also don't forget the conditions for convergence. –  David Robert Jones Feb 17 '13 at 21:15
    
@DavidRobertJones It should. I'm not sure if it meets the convergence criteria, but the fixed point should be in $(0,2\pi)$, if the solution is in the interval. –  Arkamis Feb 18 '13 at 3:34
    
Ah, I know what's wrong... I'm failing to consider the domain of arcsin properly. Maybe substitution exploiting the periodicity of cos will work... I don't know :) –  Arkamis Feb 18 '13 at 3:46

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