Limit of Cos(Cos( ... Cos(x) ... )))
Date: 02/24/2001 at 13:46:28 From: Cyndi Dellinger Subject: Limit of repeating trigonometric function I'm trying to find the limit of a repeating trigonometric function as the number of times of repetition goes to infinity. For example, lim cos(cos( ... cos(x) ... )) x is obviously less than cos(x) when x is between 0 and pi/2, so I can see that the limit is going to 0. Can this be rewritten as: lim cos(x)^n n->oo to take the limit?
Date: 02/25/2001 at 13:54:59 From: Doctor Fenton Subject: Re: Limit of repeating trigonometric function Hi Cyndi, Thanks for writing to Dr. Math. Be careful with your assertions: >x is obviously less than cos(x) when x is between 0 and pi/2, so I >can see that the limit is going to 0. If you graph the cosine function on [0,pi/2], it starts at height 1, since cos(0) = 1, and decreases to the x-axis, since cos(pi/2) = 0. The graph of y = cos(x) crosses the line y = x in this interval, at a point I'll call x0. For x > x0, cos(x) is below the line y = x, so your statement "x is obviously less than cos x when x is between 0 and pi/2" is false for x0 < x < pi/2. The point x0 is called a "fixed point" of the mapping y = cos(x), because: x0 = cos(x0) Thus, all iterates cos(x0) = x0 cos(cos(x0)) = x0 cos(cos(cos(x0))) = x0 : These iterates clearly don't approach 0. There's a good way to visualize such function iteration, i.e. the process of generating a sequence x(n) by repeated applying cos to the result: x(n+1) = cos(x(n)) Start with an initial value x(0) (this is not x0, the fixed point: the index is in parentheses, and x(0) can be any point in (0,pi/2)). To evaluate x(1) = cos(x(0)), go from x(0) on the x-axis vertically until you reach the graph of y = cos(x), at (x(0),cos(x(0)) which is (x(0),x(1)). To apply cos to x(1) to find x(2), you need to start from x(1) on the x-axis. This seems tricky, but since you are at the right height at (x(0),x(1)), go horizontally to the line y = x, which will be the point (x(1),x(1)). Then go vertically until you reach the graph of y = cos(x), which will be at (x(1),x(2)). Keep repeating this, going vertically to the graph y = cos(x), horizontally to the line y = x, then vertically to y = cos(x), horizontally to y = x, and so on. This generates a "cobweb" plot, and will illustrate what happens under repeated function iteration. If you have a TI-83+, there is a "cobweb plot" option for sequences; other graphing calculators may have something similar. If you try this, I think you'll see that the sequence does not approach 0. You can also get some analytic results. Apply the Mean Value Theorem to f(x) = cos(x) on [a,b], where 0 < a < b < pi/2. It says that there is a c between a and b such that: f(b)-f(a) --------- = f'(c) b - a If we take b = x0, the fixed point, then for f(x) = cos(x), we have: cos(x0) - cos(a) ---------------- = sin(c) x0 - a where c is in (0,pi/2). So: |sin(c)| < 1 Then: | cos(x0) - cos(a)| < |x0 - a| If I take a to be some point in my iterated sequence x(n), then this says that: | cos(x0) - cos(x(n))| < |x0 - x(n)| or |x0 - x(n+1)| < |x0 - x(n)| since x0 = cos(x0) and x(n+1) = cos(x(n)). Note that this says that each term in the sequence x(n) is closer to x0 than the previous term x(n) was. So, what do you think the limit of the x(n)'s will be? If you have a scientific calculator, you can put in any number in the display between 0 and pi/2, make sure the calculator is in radian mode, and start pressing cos repeatedly, watching the display. If you're using a Windows computer, the calculator on the Start Menu (Under accessories) does this nicely. You might want to check a book such as Alligood, Sauer, and Yorke's _Chaos_, or Robert Devaney's _An Introduction to Chaotic Dynamical Systems_. Both have extensive discussions of iterated functions. If you have further questions, please write again. - Doctor Fenton, The Math Forum http://mathforum.org/dr.math/
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