# A little Math problem...

I have been studying the Mandelbrot Set for a number of years, and during that time I've learned many wonderful things.  The Set features a filligree of interwoven spiral patterns in seemingly endless variations.  But it turns out that the branch points and centers of spirals display a mathematical regularity that makes them Misiurewicz points - places where the generating function does not immediately repeat, but eventually falls into a repeating pattern.  Such points are called pre-periodic, and they are classified by how often they repeat and how long they take to begin repeating.  So how do we find them?

When we look at the generating equation for the Mandelbrot Set and Julia Sets, it reads z->z^2+c and this becomes z->z^2+z, or more specifically z_n=z_(n-1)^2+z_0 in the case of cc(M).  So we expect that the Misiurewicz points will have terms of this same form, repeated a specific number of times on either side of the equation.  One such point of interest satisfies the equation ((c^2+c)^2+c)^2+c=(c^2+c)^2+c where c in bbb(C) - the complex numbers*.  We see that this is the Mandelbrot formula iterated three times on the left hand side and twice on the right hand side of the equation.

We could expand this out to get c^8+4c^7+6c^6+6c^5+5c^4+2c^3+c^2+c=c^4+2c^3+c^2+c by squaring the terms inside the parentheses and then combining like factors, but we are looking for roots of the equation.

When we solve for c we find a trivial solution (which is actually a periodic or repeating point).

c=0

and we find another solution at the terminus of the main "antenna."

c=-2

We get two non-trivial results with imaginary components.

c=(sqrt(11)/3^(3/2)-17/27)^(1/3)*((sqrt(3)i)/2-1/2)-(2*(-(sqrt(3)i)/2-1/2))/(9*(sqrt(11)/3^(3/2)-17/27)^(1/3))-2/3

and

c=(sqrt(11)/3^(3/2)-17/27)^(1/3)*(-(sqrt(3)i)/2-1/2)-(2*((sqrt(3)i)/2-1/2))/(9*(sqrt(11)/3^(3/2)-17/27)^(1/3))-2/3

But the solution we are interested is a non-trivial result where c in bbb(R)  i.e. - a Real numbered value.

c=(sqrt(11)/3^(3/2)-17/27)^(1/3)-2/(9*(sqrt(11)/3^(3/2)-17/27)^(1/3))-2/3

We note that the expression sqrt(11)/3^(3/2)-17/27 can be simplified.  Since 3^3=27, the first segment becomes sqrt(11)/sqrt(27) and using the fractions of square roots rule, we can write this as sqrt(11/27).  Then our solution becomes

c=(sqrt(11/27)-17/27)^(1/3)-2/(9*(sqrt(11/27)-17/27)^(1/3))-2/3

This has a numerical value around

-1.5436890126920763615708559718

But can we get closer to an exact value?

This extended precision result may suffice:

-1.543689012692076361570855971801747986525203297650983935240 804037831168673927973866485157914576059125462120829226367060 189278756463322141011522909218905292722992781697157582950422 170089518563410700385200128000282366477261779986908996884104 614976976927086847969611978920181446128991348585592145309524 153482664595734117098522176434389326263543393937637861596321 991246778909541806274987275657314799920191066093455727443186 409799911554184677317541006847088177067596744933391559037024 490072743842093947624299476858021814236680891765387460029177 563608220435803992788777820556050367509668431452581000660479 923719409341669347

* - see Peitgen and Richter "The Beauty of Fractals" pg. 59 sec. 4.22 for more details.