Trigonomania: I

Calculus Level 3

Evaluate

$\lim _{ n\rightarrow \infty }\underbrace { { \left( \sin { \left( \cos { \left( \tan { \left( \sin { \left( \cos { \left( \tan { \left( ... \right) } \right) } \right) } \right) } \right) } \right) } \right) } }_{n}$

rounded to three decimal places.

Other value 0.657 0.656 Does Not Exist

2 solutions

Calvin Lin Staff
Jun 1, 2015

Suppose that the limit existed and is equal to $a$ .
Then we must have $\tan a = a , \cos a = a, \sin a = a$ .
This is not possible, and there are many ways of showing it. For example:
$a = \tan a = \frac{ \sin a } { \cos a } = \frac {a}{a} = 1$ . But clearly $\sin 1 \neq 1$ .

Thus the limit does not exist.

instead, what we get is a cyclic permutation of $\tan a = b, \cos b = c, \sin c = a$ . This will give us $\sin \cos \tan a = a$ , which occurs at approximately 0.657.

So, the graphs do "straighten out", but because of the scale, you are not aware of what levels they straighten out at. Namely, $a \approx 0.657, b \approx 0.771, c \approx 0.717$ .

Why must $\tan{(a)}=a$ etc.? I don't follow. Also I did not intend this limit to be necessarily forced to come in triplets. But I guess since we're going to infinity it really makes no difference.

Is there a simpler way to explain this? I'm not really all proof guru. I'm just up to Multivariable Calculus and I understand Algebra pretty well - but not Brilliant-well, if you will. Cool stuff, either way.

John M. - 6 years ago

It is precisely because the function is not forced into triplets, which is why the limit does not exist. Had the function been forced into triplets, then the limit would exist (except at odd multiple of pi/2).

Put another way, consider the functions $f(x) = x + 1$ and $g(x) = x-1$ . For a fixed $x$ , does the limit

$\underbrace{f( \ g( \ f (\ g (\ldots f (x))\ldots )}_{n \text{ number of times}}$

exist? Clearly, no, because it bounces back and forth between $x+1$ and $x$ .

But, for $h(x) = f(g(x)) = x$ , the limit $\underbrace{h( \ h( \ h (\ h (\ldots h (x))\ldots )}_{n \text{ number of times}}$ exists, and is equal to $x$ .

Calvin Lin Staff - 6 years ago

Okay, I'm back. It seems to me that if we consider any string of embedded trigonometric functions in the order of Sin, Cos, Tan, but starting and ending at any arbitrarily point, e.g. Cos(Tan(Sin(Cos(Tan(Sin(x)))))), i.e., it starts at Sin(x) and ends with Cos(...), for a given number of terms (6 in this example), the final value for all x still exists between two finite bounds, and the window between those bounds shrinks with increasing number of terms. I think that adequately defines a limit, i.e., as n grows, the difference between the bounds shrinks monotonically, and it doesn't depend on which is the "starting" function and which is the "ending" function.

Michael Mendrin - 6 years ago

@Michael Mendrin – Yes, if we are using the entire "trio of functions", and can express the limit as $f^{(n) } (x)$ , then we could study the behavior of $f(x)$ to draw conclusions.

In this case, because $| f'(x) | < 1$ , hence we can apply the contraction mapping theorem to conclude that the (pointwise) limit exists.
Uniform convergence of the function would be a separate consideration (though I do believe that the convergence is uniform)

Calvin Lin Staff - 6 years ago

John M.
May 30, 2015

This is quite tricky. Plotting it on most calculators will reveal (with many, many iterations of $\sin{(\cos{(\tan{(...)})})}$ a convergence to the value of approx 0.65707. However, this is not so. Taking a look at some Desmos graphs (because Wolfie is too slow and keeps complaining to subscribe to Pro version):

Define $n$ as the number of trig compositions in the sin cos tan order. So, $n=2$ will be $\sin{(\cos{(x)})}$ .

(P.S. you can click at any of the $n=$ to get a zoomed view in a new tab)

Now you probably think that those distortions will simply go away if we let $n$ truly go to infinity. In fact, let's try

$n=90$

But this is not so. Let's zoom in to $n=90$ at around $x=1.6$ :

A few distortions can be seen. And this will NOT go away no matter how many iterations we take. Let's make a one last ultra-zoom at $n=90$

(Again, you can click on $n=90$ just above to open a zoomed view in a new tab).

BANG! As you can see, at around $x=1.508$ , the function goes bananas. Think this is argumentum ad nauseam? Think again: Guess what $1.5708$ is close to? That's right: $\pi/2$ . And guess what happens to $\tan{(\frac{\pi}{2})}$ ? Yup.

(Note: I chose to analyze $n=90$ on purpose because it is divisible by $3$ - implying that it is the third iteration in the cycle of sin cos and tan - so the one that encloses $x$ is tan.)

It is the ultimate battle of infinities, but the one infinity that has the zero on its side wins.

Here's a table for approx. $n=800$ . $x$ is in radians.

The agreement breaks down at the 21st digit. That looks very promising.

But looks can be deceiving.

Moderator note:

Indeed, looks can be deceiving. This solution presents you with bunch of pretty graphs, and then claims without any proof that the limit does not exist. Just because there are distortions, doesn't mean that the pointwise limit (IE for each x) doesn't exist.

Ignoring the issue of $x = { (2n+1) \frac{ \pi}{2} }$ and $\tan x$ is undefined, it can be shown that if $f(x) = \sin ( \cos ( \tan (x ) ) )$ and $f^{(n) } (x) = f( f^{ (n-1) } (x) )$ , then $\lim f^{(n) } (x) = 0.657 \ldots$ .

What is the true reason that the limit does not exist?

Your solution, while pretty, misses the point altogether and doesn't prove the lack of convergence.

For example, we still have $\lim_{n \rightarrow \infty } \frac {1}{n} \sin nx = 0$ , even though it bounces back and forth numerous times.

Just because $\frac{ 1}{ 800 } \sin 800 \times 2 \neq \frac{ 1}{800 } \sin 800 \times 3$ doesn't imply that the function doesn't eventually converge.

So what if the distortions exist? What is more important is that the amplitude of the distortions go to 0. And your pictures do suggestion that the amplitude goes to 0, which means that the limit exists.

Calvin Lin Staff - 6 years ago

You know, I thought about that. However, here it's not quite the same because tan will always remain tan and its effect is not modified by $n$ - whereas in $\frac{1}{n} \sin{(nx)}$ $\frac{1}{n}$ is the modifier. (For example, at $n=8000$ , the distortions were still present upon zoom. And the thing is, I really do not see them ever going away for the function at $\pi/2$ is never going to be defined. "At"... Hm... But we're not looking at at, are we? Are you trying to tell me that the distortions squeeze themselves around $\pi/2$ to a line at infinity?)

But this is the best I came up with. I really don't know. But I'd love to see an analytical proof, if you will.

John M. - 6 years ago

So, ignore all of your writeup. That has (almost) nothing to do with the actual solution.

First, as stated above, if the question had been $\lim f^{(n)} (x)$ , then the answer would be 0.657.

Second, the answer to your question is indeed "limit does not exist" (even if we exclude undefined tan x points). Think about what is the difference between your question, and asking for $\lim f^{(n)} (x)$ . That should guide you towards the actual solution.

Calvin Lin Staff - 6 years ago

@Calvin Lin – Well alright. But I really don't understand what is meant by $f^{(n)}(x)$ . nth derivative? nth power? What is $f$ ? And what would $n$ be, then?

Then I may figure it out.

John M. - 6 years ago

@John M. – $f(x) = \sin(\cos(\tan(x)))$ , $f^{(n)} (x) = \underbrace{f( \ f( \ f (\ f (\ldots f (x))\ldots )}_{n \text{ number of times}}$ .

Pi Han Goh - 6 years ago

@John M. – As above (CM note), i defined $f(x) = \sin ( \cos ( \tan (x) ) )$ , and the iterative function $f^{(n) } (x) = f ( f^{(n-1) } (x) )$ . So, this basically restricts you to your $n = 3k$ case, namely we applied equal number of sin, cos, tan.

Calvin Lin Staff - 6 years ago

@Calvin Lin – So it has nothing to do with tan's distortions going away and lining in to $\frac {\pi}{2}$ ? Well if that did happen then we WOULD have a limit now, would we? So my claim that distortions don't go away must stand. Or maybe not. I'm pretty confident it's something like that, but if you're asking for an analytical solution, count me out.

I'm looking forward to the proof.

John M. - 6 years ago

Why can't you just show that it does not exist if $n$ is in the form of $m\pi + \pi /2$ for integer $m$ ?

Pi Han Goh - 6 years ago

I don't quite follow.

$n$ is naturally a set of natural numbers (no pun intended). $x$ is unrestricted - though I've only analyzed the real realm (which suffices).

Also, I saw you were the first "solver" - I think you even liked & reshared. Sorry I had to break it to you :P

John M. - 6 years ago

Ohhhhh there's a $n$ down there, I didn't see that. Nevermind.

Pi Han Goh - 6 years ago

+1 for the hard work you did.

Abhishek Sharma - 6 years ago

Thanks, took a while :)

John M. - 6 years ago

Trigonomania: I

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