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Math Help - Help to calculate a limit

  1. #1
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    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):

    Help to calculate a limit-limit.gif

    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!
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  2. #2
    MHF Contributor chisigma's Avatar
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    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
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  3. #3
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    Re: Help to calculate a limit

    Quote Originally Posted by chisigma View Post
    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,
    Radu
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  4. #4
    MHF Contributor chisigma's Avatar
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    Re: Help to calculate a limit

    Quote Originally Posted by radubor View Post
    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,
    Radu
    All right!... so the y_{i} are continous variables and  \forall {i}\ -\frac{1}{2}<y_{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
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  5. #5
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    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,
    Radu
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  6. #6
    MHF Contributor chisigma's Avatar
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    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
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  7. #7
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    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!
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  8. #8
    MHF Contributor chisigma's Avatar
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    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
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  9. #9
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    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!
    Radu
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  10. #10
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    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!
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  11. #11
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    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...
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  12. #12
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    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?
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  13. #13
    MHF Contributor chisigma's Avatar
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    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
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  14. #14
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    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!
    Radu
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