How to find the Interval $[a,b]$ for a contraction mapping for $f(x)=x-\cos x$ where $a,b$ exists in $[0,1]$?

Question

So, using Newton's method, I know the fixed point for the contraction mapping is approximately $0.739$. However, I'm not sure how to go about finding the interval for the contraction mapping. Every interval between $0$ to $1$ that I have tried does not work. I assume that the interval must be really close to the approximation, but that's the only thing I've got. Thank you for your help.

Answer 1

The value you have mentioned in your title seems to be the Dottie number.

But this is not the fixed point of $x-\mathrm{cos}(x)$. Rather, it is the fixed point of $\mathrm{cos}(x)$.

So, with $f(x)=\mathrm{cos}(x)$, any closed and bounded interval in $\mathbb R$ mapped into itself by $f$ should work.

For instance, $\left[0,\frac{\pi}{2}\right]$ should be a suitable interval.

EDIT: The 'closed and bounded interval' comment above only works if you're okay dealing with shrinking mappings instead of contraction mappings. These are mappings for which, instead of $|f(x)-f(y)|\leq \alpha |x-y|$ (for some $0\leq\alpha<1$) you have only $|f(x)-f(y)|< |x-y|$, in general. So all contraction mappings are shrinking mappings, but not the other way around. As it turns out, shrinking mappings on any closed bounded set to itself also have fixed points. However, if you wish to specifically find an interval over which $\mathrm{cos}$ is contraction, you should use the suggestion in Robert Israel's answer instead.

Answer 2

I assume you mean you want an interval $[a,b]$ such that $\cos$ is a contraction mapping from $[a,b]$ into $[a,b]$. Note that $\dfrac{d}{dx} \cos(x) = -\sin(x)$ and $|\sin(x)| < 1$ on $[0,1]$, so $|\cos(x) - \cos(y)| < |x-y|$ for $x, y \in [0,1]$. Also $0 \le \cos(x) \le 1$ for $0 \le x \le 1$. Thus you can take the interval to be $[0,1]$.

Answer 3

I think that you are out for a fixed point of the function $g(x):=\cos x$.

I claim that the interval $J:=[0.65, 0.8]$ serves your purpose.

Proof. As $0.8<{\pi\over2}$ the function $\cos$ is monotonically decreasing in $J$. From $\cos0.65=0.796\ldots<0.8$ and $\cos0.8=0.6967>0.65$ we may therefore conclude that $$\cos(J)\subset J\ .\tag{1}$$ Within $J$ we have $$|g'(x)|=\sin x\leq\sin0.8=0.717<1\ .\tag{2}$$ The facts $(1)\wedge(2)$ guarantee that the functions $\cos$ has a fixed point $\xi\in J$.