Why does the parameter of a logistic map influence its behaviour? I am looking at choatic behaviour in discrete dynamical systems. As an example, I have taken the iteration map $x_{n+1} = rx_{n}(1-x_{n})$, where $r$ is the parameter. 
By plotting a bifurcation map, I can see for what value of $r$ the system becomes chaotic (the 'shaded' parts):

My question is, why do these values of $r$ yield chaos: what is special about them? So far I have only found texts which deal with describing the phenomenon, but which do not actually explain the mathematics behind it. 
 A: The key issue that you need to understand is the relationship between slope and dynamical stability. To be clear, let's suppose that $f:\mathbb R \to \mathbb R$ is a function and that $x_0$ is a fixed point of that function, i.e. $f(x_0) = x_0$. Then we say that 


*

*$x_0$ is attractive if $|f'(x_0)| < 1$,

*$x_0$ is repulsive if $|f'(x_0)| > 1$, and

*$x_0$ is neutral if $|f'(x_0)| = 1$.


You might also see the term super-attractive if $f'(x_0)=0$, though that's not particularly important for your specific question. This classification is justified by theorems such as 


*

*If $x_0$ is an attractive fixed point of $f$, then there is a neighborhood of $x_0$ such that the orbit of any seed in that neighborhood will converge to $x_0$ under iteration of $f$.


By contrast, repulsive fixed points are, well, repulsive and there are examples showing that orbits might converge or not for neutral fixed points. 
These types of theorems can be proven with the mean value theorem but we can get a very simple intuitive understanding by examining the function $\ell_r(x) = rx$. Note that the derivative is simply 
$\ell'(x) = r$ and the $n^\text{th}$ iterate of $\ell$ is $\ell^n(x)=r^nx$. Thus, it's easy to see that the the fixed point zero behaves exactly as described by the classification. 
Often, this is illustrated with a cobweb plot:

In this picture, the blue line is the line $y=x$ and the yellow line is the graph of $\ell_r$. If a point $(x_i,x_i)$ on the line $y=x$ and moves vertically to the graph of the function, it arrives at the point 
$$(x_i,f(x_i)) = (x_i,x_{i+1}).$$
If it then moves horizontally back to the line, it arrives at $(x_{i+1},x_{i+1})$. As a result, the process of moving vertically from the line $y=x$ to the graph of $y=f(x)$ and then horizontally back to the line is a geometric representation of one iteration of $f$. Furthermore, you can see from the picture exactly what attraction and repulsion look like.

Let's now apply these ideas to the logistic function:
$$f_r(x) = rx(1-x).$$
As you point out in your comment, 

at 2.9 there is 1 steady value, at 3.4 there are 2 steady values, and at 3.6 there are a lot of steady values

More precisely, I would say that for $r$ a little less than 3, there is a single, attractive fixed point, and for $r$ a little more than 3, there is an attractive orbit of period two. At $r=3$ the fixed point is neutral; the family undergoes a bifurcation as $r$ increases through 3. To see why, take a look at the cobweb plots:

Note that for $r=2.9$, the slope of the graph at the fixed point appears to be close to $-1$ but a little less in absolute value. For $r=3.1$ it appears that the slope is larger than $-1$ in absolute value. In fact, you can show by direct computation that for $r=3$ exactly, $x_0=2/3$ is a fixed point and that $f_3'(2/3) = -1$.
That explains why the fixed point disappeared but, perhaps, not why the new attractive orbit appeared. To understand this next point it helps to add the graph of 
$$f_r^2(x) = r^2 (1-x) x (1-r (1-x) x)$$
to the picture because the points in an attractive orbit of period 2 are fixed points of $f_r^2$.

Note that, after the bifurcation occurs, the graph of $f_r^2$ crosses the line $y=x$ twice with small slope.

As $r$ increases further, the same thing happens to the slope of $f_r^2$. You can prove that when $r=1+\sqrt{6}\approx3.44949$, $f_r^2$ has a neutral fixed point and another bifurcation occurs as $r$ passes through this value. Near that point, the picture looks something like so:

