Exact value of the limit of the Mandelbrot iteration for c=1/4? I know and understand the proofs, that c $\to c^2+c \to(c^2+c)^2+c \to ...$ converges, if $0<=c<=\frac{1}{4}$. But I wonder: is the exact value (a closed expression) for the limit known for the case $c=\frac{1}{4}$?
Edit, clarification :
  I thought of a squeeze-like argument. Or of the usage of techniques as used to show the divergence of the harmonic series. So, that one (in the special case of $c=\frac{1}{4}$) sees, that the limit is both at least and at most $\frac{1}{2}$ (without refering to fixpoints).
 A: You want to solve $z^2+1/4=z$. This is a quadratic equation with double root $z=1/2$.
A: If $c, c^2+c, (c^2+c)^2+c, \ldots$ converges to $z_0$, then $z_0$ is a fixed point of $f(z) = z^2+c =  z^2+\frac{1}{4}$. Thus
$z_0^2+\frac{1}{4} = z_0 \Rightarrow z_0^2-z_0+\frac{1}{4} = 0 \Rightarrow \left(z_0-\frac{1}{2}\right)^2 = 0 \Rightarrow z_0 = \frac{1}{2}$
A: I find the value c = .25 instructive. z converges to .5 but only comparatively slowly. After ten iterations of $ f(z) = z^2 + c $ from z = 0 we reach z = .4305. For some c I find it impossible to determine whether z converges to a point or not because of  slowness. "Moving slowly" defined by 0 < modulus (z - f(z)) < 0.01 is easy to calculate. I find the interesting observation that no z with c on the boundary of the Mandelbrot set are moving slowly after an arbitrary number of iterations, apart from c=.25  and c = -2 where z stops at +2.   
If all of the z moving slowly are converging to a point, this implies that if z is cycling it never moves slowly, which does indeed seem to be the case. Moving slowly is an efficient way of testing whether z is converging to a point.            
