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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 $\tan 1 \approx \pi/2$.

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  • $\begingroup$ 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
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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

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  • $\begingroup$ 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
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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 $\pi/2$.

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    $\begingroup$ 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
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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 $-0.0000939875$.

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    $\begingroup$ 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
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    $\begingroup$ @user122171. I totally agree. $\endgroup$ – Claude Leibovici Jan 18 '14 at 8:05
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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 $x=0.9999103$.

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