Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Cosine = Contraction? (Banach)

  1. Jun 7, 2010 #1
    So in Analysis I we explained the convergence of cos to a fixed value by Banach's contraction theorem. But is the cos a strict contraction? Is that obvious? (What is its contraction factor?)
  2. jcsd
  3. Jun 7, 2010 #2
    It's not obvious to me that cosine is a strict contraction on the whole real line. In fact, I'm not sure it's even true. For instance, around pi/2 seems to pose many problems. Was the contraction restricted to an interval?
  4. Jun 7, 2010 #3
    As Tedjn mentioned you need to restrict cosine to a small enough interval due to the behavior near pi/2 (and kpi+pi/2 for all integers k). In fact it's a well-known trigonometric identity that [itex]\sin(h)/h \to 1[/itex] as [itex]h\to 0[/itex] so this gives us:
    [tex]\left|\frac{\cos(\pi/2-h) -\cos(\pi/2)}{\pi/2-h-\pi/2}\right| = \frac{\cos(\pi/2-h)}{h} = \frac{\sin(h)}{h} \to 1[/tex]
    for [itex]h \to 0[/itex]. This shows in particular:
    [tex]d(\cos(\pi/2),\cos(x)) \to d(\pi/2,x) \qquad \mbox{for }x \to \pi/2[/tex]
    so if you choose a contraction factor k < 1 it would result in a contradiction for x close enough to [itex]\pi/2[/itex].

    However it's possible to show that cos is a contraction mapping defined on any set of the form [itex][-\pi/2+\epsilon,\pi/2-\epsilon][/itex] where [itex]0 < \epsilon < \pi/2[/itex]. In fact we know [itex]|sin h| < |h|[/itex] for all h in [itex][-\pi/2,\pi/2][/itex]. Let [itex]k=\sin(\pi/2-\epsilon) < 1[/itex]. Using the identity:
    [tex]\cos y - \cos x = 2\sin\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right)[/tex]
    we have:
    f(x,y) = \left|\frac{\cos y - \cos x}{y - x}\right| &= \left|2\sin\left(\frac{x+y}{2}\right)\frac{\sin\left(\frac{x-y}{2}\right)}{x-y}\right| \\
    &= \left|\sin\left(\frac{x+y}{2}\right)\frac{\sin\left(\frac{x-y}{2}\right)}{\frac{x-y}{2}}\right|
    Clearly a least upper bound for f(x,y) is equivalent to a contraction factor. We can fix [itex]x=pi/2-\epsilon[/itex] and let y approach x from below. Then the first term of f(x,y) approaches [itex]\sin(\pi/2-\epsilon)[/itex] and the second approaches 1, so f(x,y) will approach [tex]k=sin(\pi/2-\epsilon)[/tex]. This shows that the contraction factor must at least be k. On the other hand we have:
    [tex]\left|\frac{\sin\left(\frac{x-y}{2}\right)}{\frac{x-y}{2}}\right| \leq 1[/tex]
    [tex]|f(x,y)| \leq \left|sin\left(\frac{x+y}{2}\right)\right| \leq \sin(\pi/2-\epsilon) = k[/tex]
    which shows that k is actually the contraction factor for cos defined on [itex][-\pi/2+\epsilon,\pi/2-\epsilon][/itex].

    This should however be sufficient as [itex]|cos(x)| \leq 1[/itex] so clearly any fixed point will lie in [itex][-1,1][/itex] and [itex]\pi/2 > 1[/itex].
    Last edited: Jun 7, 2010
  5. Jun 7, 2010 #4
    A very nice proof that it fails near pi/2. Another way I've seen the second part proved is using the mean value theorem.
  6. Jun 8, 2010 #5
    Ramshop, that's the second time you've made a very informative post, thank you.

    Also thank you to Tedjn for posting :)

    Having seen the proof, it's logical the fixed-point theorem of Banach still applies, as no matter where you start, you'll eventually end up in the strict contraction "zone".
  7. Jun 8, 2010 #6
    One button calculator, cos, cos, cos,...
    Code (Text):
        |    _   _  _  _  _  _  _     _   |
        |   | |   | _||_|| ||_||_   | _|  |
        |   |_|.  | _| _||_||_| _|  | _|  |
        |                             rad |
        |                                 |
        |                                 |
        |           +--------+            |
        |           | cosine |            |
        |           +--------+            |
        |                                 |
        |                                 |
    Last edited: Jun 8, 2010
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook