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Math Help - the zero of the third derivative of iterated sines

  1. #1
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    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
    Last edited by ymar; June 1st 2011 at 05:51 AM. Reason: a small correction in the problem. fn is now defined for pi/2
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    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))].
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    Senior Member Tinyboss's Avatar
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    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.
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    Quote Originally Posted by zortharg View Post
    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).
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    Quote Originally Posted by Tinyboss View Post
    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?
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  6. #6
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    Quote Originally Posted by ymar View Post
    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.
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    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.
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