Take the 2-minute tour ×
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.

I can show that $\cos(\sin(x))$ is a contraction on $\mathbb{R}$ and hence by the Contraction Mapping Theorem it will have a unique fixed point. But what is the process for finding this fixed point? This is in the context of metric spaces, I know in numerical analysis it can be done trivially with fixed point iteration. Is there a method of finding it analytically?

share|improve this question
3  
Probably not. ${{{{{}}}}}$ –  Michael Hardy Nov 28 '12 at 17:36
1  
The number $0.76816915673679597746208623955865641813208731218273718569186715\ldots$ does not ring a bell, at least, and it seems that we need a new Plouffe's inverter. –  Hagen von Eitzen Nov 28 '12 at 17:51
1  
The Inverse Symbolic Calculator at isc.carma.newcastle.edu.au/advancedCalc returns a blue square with a white question mark inside it. I don't know what that means. –  Gerry Myerson Nov 29 '12 at 5:00
1  
@Gerry, I believe it means you are pregnant. Or possibly someone you know. –  Will Jagy Dec 21 '12 at 2:12

3 Answers 3

The Jacobi-Anger expansion gives an expression for your formula as:

$\cos(\sin(x)) = J_0(1)+2 \sum_{n=1}^{\infty} J_{2n}(1) \cos(2nx)$.

Since the "harmonics" in the sum rapidly damp to zero, to second order the equation for the fixed point can be represented as:

$x= J_0(1) + 2[J_2(1)(\cos(2x)) + J_4(1)(\cos(4x))]$.

Using Wolfram Alpha to solve this I get $x\approx 0.76868..$

share|improve this answer

First off, it is clear that the fixed point $x$ will lie in $[0,1]$
$\cos{(\sin{(x)})}=x$ $\Rightarrow$
$\cos^{-1}(x) = \sin{(x)}$
Let $F(m) = \sin{(m)} -\cos^{-1}{(m)}$ $\Rightarrow$
$F'(m) = \cos{(m)} + \dfrac{1}{\sqrt{1-m^{2}}}$
Then you can also find the fixed point using Newton's method.
With $x_0$ = 0
$x_{n+1} = x_n - \dfrac{F(m)}{F'(m)}$
Using MATLAB we find with $x_0:=0$
$x_{4}$ = $0.768169156736796$
which is correct to 15 decimal places.
As far as solving analytically, I do not believe there is a way.
Newton's method proves to be superior to functional iteration in this case, because we observe faster convergence.
With $x_0 = 0$,
$x_{n+1}$ = $f(x_n)$
$x_{40} \approx 0.768169156736780$ which is not nearly as accurate of an approximation.

share|improve this answer

Since it's a contraction, just iterate. Pick an arbitrary starting point and keep applying the operator until you achieve the desired accuracy.

share|improve this answer

Your Answer

 
discard

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.