The only positive solution of the equation $\sin (\tan x) = x$ is at a number $a = 0.999906...$. Is it a coincidence that the number $a$ is so close to $1$, or is there a conceptual explanation?

It was obvious $a$ was going to have to be less than $1$, but not this close. The answer seems to be related to $\tan 1 \approx \pi/2$, but that merely shifts the problem to explaining why that is the case.

(I became interested in this question after reading this other question: Prove: $\sin (\tan x) \geq {x}$ )

Edit: I have thought about the question some more, and added some ideas in the form of an answer partially addressing why we have $\tan 1 \approx \pi/2$.

  • $\begingroup$ I'm using the word "positive" to mean $x > 0$. $\endgroup$ – user122171 Jan 18 '14 at 7:41
  • $\begingroup$ That is, it should probably say $x\in(0,\pi/4]$. $\endgroup$ – dfeuer Jan 18 '14 at 7:42
  • $\begingroup$ That was the formulation of the previous problem, which led me to consider this one. Do you think I should put that context at the end? $\endgroup$ – user122171 Jan 18 '14 at 7:43
  • $\begingroup$ I guess it doesn't really matter much; either you can chop off the endpoint or consider the continuous extension. $\endgroup$ – dfeuer Jan 18 '14 at 7:45
  • $\begingroup$ Look at the post math.stackexchange.com/questions/640325/fun-logarithm-question/…. The solution is 1.444667364812 while e^(1/e) is 1.444667861010. Much closer than for your case. $\endgroup$ – Claude Leibovici Jan 18 '14 at 8:04

J Swanson and Christian Blatter's answers address the question of why, given that $\tan 1 \approx \pi/2$ (within $1 \%$), we have $\sin \tan 1 \approx 1$ within $0.01\%$. Taylor's formula for $\sin x$ about $x = \pi/2$ shows why the accuracy would be approximately squared.

That leaves open the question of why $\tan 1 \approx \pi/2$.

Gauss's continued fraction for $\tan z$ (see http://en.wikipedia.org/wiki/Gauss_continued_fraction ) is: $$\tan z = \cfrac{z}{1 - \cfrac{z^2}{3 - \cfrac{z^2}{5 - \cfrac{z^2}{7 - {}\ddots}}}}.$$ In particular, for $z = 1$, this yields $$ \tan 1 \approx \cfrac{1}{1 - \cfrac{1}{3 - \cfrac{1}{5}}} = \frac{14}{9}.$$ This is an excellent approximation, valid to within just over $0.1\%$. So we need only consider the question of why $\pi \approx 28/9$.

I don't have a very good answer for this part, as I haven't been able to locate any standard approximating sequences of $\pi$ one of whose first terms is $28/9$. However, you can obtain $28/9$ as an approximation of $\pi$ in the following way, as explained in A History of Pi by Beckmann. (If the circle has area $\pi$, then the octagon has area $28/9$.)

enter image description here

  • $\begingroup$ That's cute. Thanks especially for the octagon picture. I wish the continued fraction could be given geometric meaning instead. This would be a great example for certain calculus classes if it didn't need to rely on a "magic formula" whose proof would probably be too much of a digression. $\endgroup$ – J Swanson Jan 19 '14 at 8:25

It happens that $\pi/2 - \tan 1 \approx 0.0133886$. This may just be a numerical coincidence or there may be a good reason. But in any case, $\sin$ is very flat close to $\pi/2$, so even values that are just pretty close to $\pi/2$ get translated to values very close to $1$, hence $\sin \tan 1 \approx 0.9999103$. By continuity/smoothness the fixed point must be very close to $1$, as you say.

I'll make this more quantitative in an effort to make it more "explanatory". Suppose $f(x)$ is smooth near $x=1$ and $f(1) - \pi/2 = \delta$. Then $\sin(f(1)) = \cos(\delta) = 1 - \delta^2/2 + \cdots$. Hence $\sin f(x) - x$ at $x=1$ is $-\delta^2/2 + \cdots$. If $|\delta| \approx 0$, the difference is approximately $0$, so if a fixed point exists, it should be near here, which can be made rigorous by comparing the derivative of $\sin f(x) - x$ to $-\delta^2/2 + \cdots$ at $x=1$.

In the present case with $f = \tan$, we find $\delta \approx -0.0133886$, we have $-\delta^2/2 \approx -0.0000896273$, and this is tiny compared to $(\sin \tan x - x)'(1) \approx -0.954138$, so we'll definitely get a fixed point very close to here. Indeed, the actual fixed point differs from $1$ by $0.0000939875$, which is very close to my estimate, and moreover $0.0000896273/0.954138 \approx 0.0000939354$. That is, the closeness of the fixed point to $1$ is just an amplified version of the closeness of $\tan 1$ to $\pi/2$ thanks to the sine function.

Some thoughts on $\pi/2 - \tan 1 \approx 0.0133886$: we can change the problem in at least two ways--replace $\sin$, $\pi/2$, $1$; replace $\tan$--and sometimes the above reasoning will still go through. There are a lot of potential tweaks, so I'd be satisfied by the coincidence explanation. On the other hand, it would be more interesting if was something really was going on with $\tan$ and $\pi/2$.

  • 2
    $\begingroup$ Note the function $\sin \tan x \approx x$ is rather good throughout the entire interval $[-1, 1]$, so a satisfying explanation would need to explain why the fixed points should be near $\pm 1$ and $0$ rather than any of the other points. $\endgroup$ – Hurkyl Jan 18 '14 at 11:00
  • $\begingroup$ Sorry but I fail to see the explanatory content of this answer. $\endgroup$ – Did Jan 18 '14 at 11:13
  • $\begingroup$ @Hurkyl I don't really agree that the approximation is good throughout the interval. Already at $x=0.9$, $\sin \tan x - x \approx 0.05$ is orders of magnitude larger than the difference at $x=1$, hence the fixed point will be very close to $x=1$ and will not be that close to $0.9$. (It doesn't seem worthwhile to make "closeness" quantitative here, but it's of course easy with some calculus.) $\endgroup$ – J Swanson Jan 18 '14 at 11:37
  • $\begingroup$ @Did I've updated the answer to be quantitative. I'm able to predict the first 7 digits after the decimal point of the fixed point and explain why they're so close to $1$ for problems that are similar to this one, which seems explanatory to me. The only thing I don't explain is why $\tan 1$ is moderately close to $\pi/2$. $\endgroup$ – J Swanson Jan 18 '14 at 12:04
  • $\begingroup$ The update is definitely more explanatory. Thanks. $\endgroup$ – Did Jan 18 '14 at 13:32

Consider that you try to solve your equation using Newton method starting at $x=1$. The first iterate is given by $$x = 1 + \frac {-1 + \sin(\tan(1))} {1 - \cos(\tan(1)) \sec(1)^2};$$ the change is $-0.000093934$.

If you use Halley method, the formula becomes much more complex and the change is $-0.0000939875$.

  • 9
    $\begingroup$ The reason the change is small is because $-1 + \sin \tan 1$ is close to zero. This fact is nearly equivalent to the fact I'm asking for an explanation of. $\endgroup$ – user122171 Jan 18 '14 at 7:58
  • 2
    $\begingroup$ @user122171. I totally agree. $\endgroup$ – Claude Leibovici Jan 18 '14 at 8:05

From $${4\over3}=1+{(1)^3\over3}<\tan 1<\tan{\pi\over3}=\sqrt{3}$$ it follows that $$\tan 1={\pi\over2}+\tau$$ for a $\tau$ of small absolute value. That in fact $\tau\doteq-0.0134$ and therefore $|\tau|\ll1$ is pure luck.

For a $\delta$ with $0<\delta\ll 1$ we therefore have $$\tan(1-\delta)\doteq{\pi\over 2}+\tau-\delta\left(1+{\pi^2\over4}\right)\doteq {\pi\over2}+\tau-{7\over2}\delta\ ,$$ where we have made use of $\pi^2\doteq10$. It follows that $$\sin\bigl(\tan(1-\delta)\bigr)\doteq 1-{1\over2}\left(\tau-{7\over2}\delta\right)^2\ ,$$ and this should be equal to $1-\delta$ for some positive $\delta\ll1$. The equation $$\left(\tau-{7\over2}\delta\right)^2=2\delta$$ has two solutions $\delta_1\doteq{\tau^2\over2}$ and $\delta_2\doteq{8\over49}$, the first of which is the one we are interested in. Given the value of $\tau$ from above we obtain $\delta_1\doteq0.0000896$, which leads to $x=0.9999103$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.