# Math Help - Help to calculate a limit

1. ## Help to calculate a limit

First, a warm "Hello to all!"...and excuse my English, pls

I need very much to calculate the following limit (it's a Lyapunov exponent):

After some numerical estimations, it seems to me that L is a periodic function for r<10 and a linear function for r>10. In both casese I obtained L < r*ln(2).

But I need very much something more exact...

Thank you all!

2. ## Re: Help to calculate a limit

The $y_{i}, i=1,2,...,k$ are discrete variables, i.e. for each is $y_{i}= \frac{1}{2}$ or $y_{i}= - \frac{1}{2}$...

... aren't they?...

Kind regards

$\chi$ $\sigma$

3. ## Re: Help to calculate a limit

Originally Posted by chisigma
The $y_{i}, i=1,2,...,k$ are discrete variables, i.e. for each is $y_{i}= \frac{1}{2}$ or $y_{i}= - \frac{1}{2}$...

... aren't they?...

Kind regards

$\chi$ $\sigma$
Thank you very much for your interrest chisigma!
The yi variables aren't discrete. They are from the interval [-1/2, 1/2] as coordinates of some points from an orbit of a chaotic dynamical system.

Best regards,

4. ## Re: Help to calculate a limit

Thank you very much for your interrest chisigma!
The yi variables aren't discrete. They are from the interval [-1/2, 1/2] as coordinates of some points from an orbit of a chaotic dynamical system.

Best regards,
All right!... so the $y_{i}$ are continous variables and $\forall {i}\ -\frac{1}{2} ... one more question: are the $y_{i}$ uniformely distributed in $[-\frac{1}{2}\ ,\ \frac{1}{2} ]$ or their statistics is different?...

Kind regards

$\chi$ $\sigma$

5. ## Re: Help to calculate a limit

For the ease of reading, I rewrite the original limit in Latex:
$L=\lim_{k \to \infty} \ln\left|2^{rk} \cdot \cos\left(2^{r+1}y_1\right) \cdot \ldots \cdot \cos\left(2^{r+1}y_k\right)\right|^\frac{1}{k}, r>0, y_1,\ldots,y_k \in\left[-\frac{1}{2},\frac{1}{2}\right]$

Thank you very much again chisigma for your help!

As they are coordinates of points from an attractor of a chaotic dynamical system which seems to me to cover very dense a rectangle (and I made NIST tests for the randomness and the results were very good), I think that you can assume that these $y_i$ are uniformly distributed.

Best regards,

6. ## Re: Help to calculate a limit

The proposed problem is very interesting... in order to avoid misundestanding however I have again a question [I do hope the last question...] : what is r?... an integer o a real number?... and if it is a real number, what is its range?...

Kind regards

$\chi$ $\sigma$

7. ## Re: Help to calculate a limit

You're very kind chisigma...thank you a lot!
Parameter r is a real one...positive...but if you need, you can assume that it is from an interval [0,x] with x>=20.

Best regards a million thanks again!

8. ## Re: Help to calculate a limit

Very well!... we can start writing the identity...

$\lambda_{k}= \ln |2^{r k}\ \cdot\ \cos (2^{r+1}\ y_{1}\ \cdot ...\cdot \cos(2^{r+1}\ y_{k})|^{\frac{1}{k}}= r\ \ln 2 + \frac{1}{k}\ \sum_{i=1}^{k} \ln |\cos (2^{r+1}\ y_{i})|$ (1)

The first term in (1) doesn't depends from k and the second term [containing the 'summation'...] for the central limit theorem has as limit the expected value of $f(y)= \ln |\cos (2^{r+1}\ y)|$ where y is uniformely distributed in $[-\frac{1}{2}, \frac{1}{2}]$, so that is...

$L= \lim_{k \rightarrow \infty} \lambda_{k} = r\ \ln 2 + \int_{-\frac{1}{2}}^{\frac{1}{2}} \ln |\cos(2^{r+1}\ y)|\ dy$ (2)

The next step is of course the computation of the integral in (2) and it seems to me that that requires some little effort ...

Kind regards

$\chi$ $\sigma$

9. ## Re: Help to calculate a limit

Wooow...you're a fabulous man chisigma...thank you a billion times!!!
I hope I will be able to estimate the integrale...as I remeber, it cann't be calculated in an explicit manner...but I will do more tries...even an approximation it's OK!

Best regards and infinite thanks!

10. ## Re: Help to calculate a limit

I made some numerical estimations and it seems to me that the integral is between -1 and 0 for values of r between 0 and 10...which it's in accord with my visual estimates...so, the work continues!

Thank you again chisigma!

11. ## Re: Help to calculate a limit

I made more approximations using the trapezoidal method and it seems that the integral is around -0.0007

Looking at the graphic of the function f(x)=ln(abs(cos(2^r*x))), it seems to me to be obvlious that for a big r the area becomes smaller...

12. ## Re: Help to calculate a limit

I wrote this program in Matlab to approximate the integral (the results are in the file "integrala.txt"):

fid = fopen('integrala.txt', 'w');

for r=0:0.001:1000
x = -0.5:0.001:0.5;
y = log(abs(cos(2^r*x)));
z = 0.001*trapz(x,y);
fprintf(fid,'%.12f\n',z);
end

fclose(fid);

It's OK?

13. ## Re: Help to calculate a limit

Rearranging the terms in more tractable form we arrive to write the function as...

$L(r)= r\ \ln 2 + I(r) = r\ \ln 2 + \int_{0}^{1} \ln |\cos (2^{r}\ t)|\ d t$ (1)

The integral in (1) can be 'attacked' numerically and using the Simpson rule with 10000 points we arrive to the following results...

$I(0)= -.187538169021$

$I(1)= -.942815934642$

$I(2)= -.573452255802$

$I(3)= -.733681277224$

$I(4)= -.680552144052$

$I(5)= -.681461932829$

$I(6)= -.685264312695$

$I(7)= -.696704871583$

$I(8)= -.692532754951$

$I(9)= -.693289534457$

$I(10)= -.693024042941$

For low values of r [ $r\le 3$...] the $I(r)$ oscillates but for higher values it tends to -.693 so that for $r>3$ is with good approximation...

$L(r) \sim r\ \ln 2 -.693$ (2)

... as illustrated in the figure where the (2) is the 'red line' and the (1) the 'black line'...

Kind regards

$\chi$ $\sigma$

14. ## Re: Help to calculate a limit

You're a wonderful genius chisigma!!!

I think that I made a stupid mistake, because I obtained almost the same result as you, but divided by 1.000...

Best regards and many thanks again!