# For `a in [0,pi]` we define the sequence `x(n)` ,` n in NN` : `x_1=a, x_(n+1) = sin( x_n )`  We must show that `x(n)` converges and `lim_(n->oo) x(n)` how do we evaluate sin in this context?

txmedteach | High School Teacher | (Level 3) Associate Educator

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Well, let's take this in a direction away from the abstract and more to the concrete. We can model the above easily as a discrete dynamical system.

Let's look at the graph of our function, `sin(x)` juxtaposed with a graph of `y = x`:

Now, `sin(x)` is graphed over the range of `a` given in the problem. We will see that we needn't concern ourselves with other possible values in the domain soon.

In discrete dynamical systems, one of the ways you model the iteration of a function is to take advantage of the `y=x` line. You start with your initial value (x(0) = a), your next x-value will be y(a).

Let me show this to you graphically. Let's zoom in. We're going to let a be 0.95. I'll show the trace of the iteration for a few steps below:

This graphical representation gives us a good idea where points of convergence may lie for a sequence like the one defined in the problem. In fact, in discrete dynamical systems, such points can be solved for using a simple equation:

`x(n) = x(n+1)`

Here, we can easily see the fixed point is at 0, and only 0, but let's solve to demonstrate the method:

`x = sin(x)`

Well, we know graphically that these two already intersect on only one point, and we think it's going to be x = 0. Let's check:

`0 = sin(0)`

Sounds good to me. I know, the face that we figure there is only one fixed point based on graphical evidence is sketchy, however, I'm not quite sure how to prove the tail end of the Taylor series cannot be 0 unless x = 0. I'll leave that to you as an exercise at home!

Therefore, the only fixed point is at (0,0).

The problem here, now, is that we need to confirm the fixed point is a point of convergence, not divergence. In other words, we need to show that the iterative function approaches the point (like a sink) instead of getting further and further from it (like a source). Another way to put this is that we need to show that (0,0) is a "stable" point as opposed to an "unstable" point.

Well, in order to show that a fixed point in a discrete dynamical system is stable, we must show that the derivative of the iterative function is less than 1 at the fixed point. However, in our case `d/dx(sin(x)) = 1` when `x=0`. So, we have to take another route.

Well, we know that the derivative of sin(x) is maximal at x = 0 (it's an inflection point). This means that in the neighborhood of 0, the derivative will be less than one. You, therefore, can make a form of Epsilon-Delta proof (recall those proofs for limits, this is roughly the same) that would demonstrate convergence to x = 0. The proof would roughly look like your using the mean value theorem to find a point whose derivative matches the slope between (0,0) and `(x_n, x_(n+1))` and choosing a new point which is much closer to (0,0). This closer point would give you a derivative less than 1 that you can use to show that `x_(n+1) < x_n`.

In other words, if you're ok with hand-waving, since the derivative is less than 1 all the way until you hit our fixed point, it will be stable. If you don't like handwaving, I don't have the room to show you!

Because (0,0) is the only fixed point for the above function, and it is stable, we know that `x_n` necessarily converges. We also know, based on our finding the fixed point:

`lim_(n->oo) x_n = 0`

And there you have it!

By the way, you don't need to limit `x_0`. You can show using this method that `x_n` converges regardless of initial value.

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