Cosine = Contraction? (Banach)

[nonequilibrium]

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?)

 

[Tedjn]

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?

 

[rasmhop]

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 \sin(h)/h \to 1 as h\to 0 so this gives us:
\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
for h \to 0. This shows in particular:
d(\cos(\pi/2),\cos(x)) \to d(\pi/2,x) \qquad \mbox{for }x \to \pi/2
so if you choose a contraction factor k < 1 it would result in a contradiction for x close enough to \pi/2.

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

This should however be sufficient as |cos(x)| \leq 1 so clearly any fixed point will lie in [-1,1] and \pi/2 &gt; 1.

 

[Tedjn]

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.

 

[nonequilibrium]

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".

 

[Xitami]

One button calculator, cos, cos, cos,...

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