# Thread: the zero of the third derivative of iterated sines

1. ## the zero of the third derivative of iterated sines

Let $f_n\colon[0,\pi/2]\longrightarrow[0,1]$ for $n=0,1,...$ and

$\begin{cases}f_0=\sin\\f_{n+1}=\sin\circ f_n\end{cases}$

I would like to prove that for $n=1,2,...$ the third derivative $f_n^{(3)}$ has a unique zero in $(0,\pi/2)$. I thought calculating the third derivative and solving an equation, even if doable, doesn't seem very useful. I already "know" it's true after plotting many graphs, so all I would like to know is an elegant proof, not a proof that no sane person would like to read. (I have almost finished calculating the third derivative and if someone thinks it's a good idea, I can post the calculations. I can't guarantee they're correct though.)

The idea is to later take the sequence $g_n=\frac{f_n}{f_n(\pi/2)}$ and show that if $x_n=(g_n^{(2)})^{-1}(\max g_n^{(2)})$, then $\lim(x_n,g_n(x_n))=(0,1)$.

If somebody sees how to prove these in a way that requires a reasonable amount of paper, I would be very glad to see it. I've come up with the problem myself, so I don't know if there is a neat solution.

PS: I thought I would post the graph of a sample $g_n$. Here it is for n=1000000+/-1 (I don't remember how I set the indices in the plotting program): View image: beztytu?u

2. Hold on there trogdorkilla, your first function is the regular sin function and has a domain from 0 to pi/2 and a range of 0 to 1, but the second one USES THE FIRST as an input but it too has a domain from 0 to pi/2. The problem is, the first function doesn't have a range extending to pi/2, it has a range extending to 1. The first function is not onto [0,pi/2] so the second can't be defined over that whole range, it can only be defined from [0,1] and the third can only be defined over [0,sin(1)] and the fourth over [0,sin(sin(1))].

3. I'd just define a new function $\text{Sin}(x):=\sin(\pi x/2)$, and use it instead of sine. You'll get a bunch of scalar factors in your derivatives, but if you only care about zero or nonzero, it shouldn't be too much of a problem.

4. Originally Posted by zortharg
Hold on there trogdorkilla, your first function is the regular sin function and has a domain from 0 to pi/2 and a range of 0 to 1, but the second one USES THE FIRST as an input but it too has a domain from 0 to pi/2. The problem is, the first function doesn't have a range extending to pi/2, it has a range extending to 1. The first function is not onto [0,pi/2] so the second can't be defined over that whole range, it can only be defined from [0,1] and the third can only be defined over [0,sin(1)] and the fourth over [0,sin(sin(1))].
Thank you for your reply, but I don't understand what you are talking about. All of my functions are well defined on $(0,\pi/2)$.

5. Originally Posted by Tinyboss
I'd just define a new function $\text{Sin}(x):=\sin(\pi x/2)$, and use it instead of sine. You'll get a bunch of scalar factors in your derivatives, but if you only care about zero or nonzero, it shouldn't be too much of a problem.
Thank you, Tinyboss, but I must admit I don't understand your advice either. The problem with the third derivative of $f_n$ is that it has an awfully long formula with a lot of sums and pruducts. I don't understand why scaling the functions should help. Could you please explain what you mean?

6. Originally Posted by ymar
Let $f_n\colon[0,\pi/2]\longrightarrow[0,1]$ for $n=0,1,...$ and

$\begin{cases}f_0=\sin\\f_{n+1}=\sin\circ f_n\end{cases}$

I would like to prove that for $n=1,2,...$ the third derivative $f_n^{(3)}$ has a unique zero in $(0,\pi/2)$. I thought calculating the third derivative and solving an equation, even if doable, doesn't seem very useful. I already "know" it's true after plotting many graphs, so all I would like to know is an elegant proof, not a proof that no sane person would like to read. (I have almost finished calculating the third derivative and if someone thinks it's a good idea, I can post the calculations. I can't guarantee they're correct though.)
So we define $f_0(x) = \sin x$ for $x\in[0,\pi/2]$ and then f_n is given inductively by $f_{n+1}(x) = \sin(f_n(x))$. (I can't see what is wrong in that, and I don't understand zortharg's comment.)

I think that it would be easier to make the inductive definition by taking the composition the other way round, $f_{n+1}(x) = f_n(\sin x)$. That gives the same sequence of functions, but in a way that is technically easier to handle. The derivatives are given by

$f_{n+1}'(x) = f_n'(\sin x) \cos x,$
$f_{n+1}''(x) = f_n''(\sin x) \cos^2 x - f_n'(\sin x)\sin x ,$
$f_{n+1}'''(x) = f_n'''(\sin x) \cos^3 x - 3f_n''(\sin x)\cos x\sin x - f_n'(\sin x)\cos x.$

From those equations it easy to show by induction that (in the interval $[0,\pi/2]$) $f_n'(x)\geqslant0$ and $f_n''(x)\leqslant0$. Also, $f_n'''(0) = -2$ and $f_n'''(\pi/2) = 0$, for $n\geqslant1.$

With a bit of effort you can even compute the fourth derivative $f_{n+1}^{(\mathrm{iv})}(x)$ in terms of the derivatives of $f_n(x)$, and check that $f_n^{(\mathrm{iv})}(\pi/2)<0$. That implies that $f_n'''(x)$ is positive when x is a bit less than $\pi/2$. Together with the fact that $f_n'''(0)<0$, that tells you by the intermediate value theorem that $f_n'''(x)$ has at least one zero in the interval.

But that is still a long way from showing that the third derivative has a unique zero in the interval. I have no idea at all how one could reasonably hope to prove that.

7. Thank you very much, Opalg. I noticed those fact about the first and the second derivative too using my definition, but I didn't think of calculating $f_n^{(3)}$ in zero and $\pi/2$.

I'm an undergraduate student so I'm always uncertain when I see a problem whether there aren't some misterious ways in which it can be easily solved. It's good to have an opinion of an experienced mathematician.