Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How do I show that $\cos(x)$ is a contraction mapping on $[0,\pi]$? I would normally use the mean value theorem and find $\max|-\sin(x)|$ on $(0,\pi)$ but I dont think this will work here.

So I think I need to look at $|\cos(x)-\cos(y)|$ but I can't see what to do to get this of the form $|\cos(x)-\cos(y)|\leq\alpha|x-y|$?

Thanks for any help

share|cite|improve this question
You won't find a constant $\alpha<1$ which works for all $x,y$. However, if you fix some $0\leq x<y\leq\pi$, the mean value theorem gives you such an $\alpha<1$ specific to $x$ and $y$. – Olivier Bégassat May 5 '12 at 10:50
$\cos$ doesn't define a mapping of $[0, \pi]$ into itself at all. – Chris Eagle May 5 '12 at 10:50
@ChrisEagle Oh ok, that may be causing me a bit of confusion, it is a contraction mapping on $[0,1]$ though right (I can use the mean value theorem for this? should I delete my question now? (the reason I was asking my question was to show that $\cos(x)=x$ has a unique solution in $[0,\pi]$ but I see I have made a mistake- I just need to show that it has a unique solution in $[0,1]$ and then it is obvious that it does not have one in $(1,\pi]$-sorry about that – hmmmm May 5 '12 at 10:54
@Olivier: to be a contraction mapping, by definition there has to be a single constant that works. However, if we restrict to $[0,1]$ then we are far enough away from $\pi/2$ that we can get a single constant to work on $[0,1]$. – Carl Mummert May 5 '12 at 11:46
@hmmmm: you can just edit your question, no need to replace it. – Carl Mummert May 5 '12 at 11:46
up vote 7 down vote accepted

To show that $\cos(x)$ is a contraction mapping on $[0,1]$ you just need to show that it is Lipschitz with a Lipschitz constant less than $1$. Because $\cos(x)$ is continuously differentiable, there is a maximum absolute value of the derivative on each closed interval, and the mean value theorem can be used to show that maximum absolute value works as a Lipschitz constant. Since the derivative of $\cos(x)$ is bounded below $1$ in absolute value on $[0,1]$, that will give the desired result.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.