Klaroline

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4 years, 91 days

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These are replies submitted by Klaroline

@vv Lyapunov exponent

 
\lambda (x_0) = \lim_{n \to \infty}  \frac{1}{n} \sum_{i=0}^{n-1}  \ln | f'(x_i)|

 I'm not sure, it's okey ?

@Kitonum 
Thanks.
First I had a problem with finding a interval where parameter 'a' is changin. So am I right, that 'a' can be any real values ? 

@vv

for example:

g:=x->mu*x*(1-x);
dg:=diff(g(x),x);
solve(dg=0,x);
g(1/2);
         1/4*mu
solve(1/4*mu<1,mu);
                 

@Carl Love 
I know, here I didn't add it;) 
But my question is, what I need to do next - if I want range of "a"

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