Math Is Fun Forum

Here is an important link that explains things in a similar way: http://www.philosophyforum.com/forum/sh … .php?t=225
let's say that a part of the brain is the self, D, and that this brainpart has the neurons C.

C --> E = C-a, D(C) = D(E), you loose one neuron, but you do not loose your self.
E --> R = E+b, D(C) = D(E), you gain one neuron, but you don't loose yourself.

Let's say Q = C-a+b-c+d....+z

*D(X) is the only function that can simulate self, no matter richards paradox. Note: I only give the neuron counters C,E and D to Illustrate that they are only equal in amount possibly, but they do not belong to eachother, hence you are not a unit of your distant past as a direct function of your neuron identity unless the neurons all have the same identity which would be hard to understand, but possibly they can and then ofcourse D(X) would not simulate the self, because ns=xs (they would have the same identity). An n:dimensional unity. My thesis now is that a p-dimensional world has a p+q dimensional unity, for instance, 2 dots (or more possibly along the 1D surface) has a 1D line as a unity, 2 lines (or more along the same 2D surface) has a 2D surface as a unity. 2 surfaces has  a 3D space as a unity, 2 rooms has a 4D time space as a unity. But if there were more then 2 dots, then possibly a 2D or even nD surface would be the unity of the dots, it is a nD dot. Perhaps the universe started with an n:dimensional dot, and an n:dimensional dot has a unity in the n:th dimension. But perhaps an n:dimensional dot requires that in an n-q dimensional world there are exactly (or likely, since momentum&vektor&energy is preserved) >2^q (n-q):dimensional objects given that there could only be 2 (n-1):dimensional objects, that is 2^(n-X) X-dimensional objects. And since every dimension has atleast a beginning and an end, the n-dimensional particle being the only exception. But then what is a dot? Know that a dot in a fourdimensional world, a particle, is a 4D dot, a 4D object, just like an n:dimensional world had this n:dimensional dot, this unity. A 4D world, as our own, does have p^(n-3) particles, but only p^(n-4) 4D dots, 4D-particles. A 4D dot seen along a one dimensional line is not one dot, it is 2 1D dots, and not all dots can be seen, since they are not all along the line, But it is also true that in a 3D room you can only see the 3D expression of a 4D-particle, hence what we see is a 3D particle, and 4D particles are fewer then the 3D particles, and we cannot see their entirety from where we are. There are 2 3D expression particles along every 4D particle lifetime line, one in the beginning and one in the end. This all given, "the number of the planck particles in the universe" is 2^(D-3) = e^(ln(2)(D-3)), you can get D by logarithming both sides then divide with ln(2) and then add 3. But I just thought this up, it don't have to be right. Comments on this would be appreciated*

And any neuron(s) ns of C (C(ns)) does not belong to Q or is not Q(xs) where xs stands for neuron(s) ie. one or more neurons in a neural network:

C(ns) ≠Q(xs)

But D(C) = D(Q).

So D ≠ f(ns or xs) unless f(ns) = f(xs)
*Sorry, I'm a bit ringrusty, f(ns) etc. are possibly matrixes/part matrixes, ie. M(ns)*
But:

f(ns) has only incommon with f(xs) the force that keeps the self together, the force that binds it, any other function f(ns) would not be f(xs).

If the self was temporary and f(ns) varies from self to not self, then:

D(ns,t1) --> R(ns,t2), D ≠ R. R(ns,t1) ≠ D(ns,t1)
What this would mean can be described as R(t) = w(t)D(t)^v(t) + j(t), not to dismiss anything.

Then the self is still not bound to mass. Since the self do not leave the body anyday, and evolution would not keep it since what's the point, the self would just be a temporeral change, new change, new self. But self in this case would be as temporaral as change, and not bound to a certain matter, it could emerge anywere at any time, so the self is really not bound to temporal change. So given that f is a certain funktion p that remains and is self:

D(f(ns)) = D(f(xs)), D = p(ns) = p(xs) = f(ns) that belongs to f(xs).

That was what I wanted to say. Cause when you analyse these equations you will see that the self will always remain, since the self is force, and not change in force which implies that any force will do (note: if it aint moving at all, and it still is the self, then anything can be the self)

And even if the self was a change, that change would still need to be dy/dx that would in all cases equal the fluctuation in the force. The fluctuation would always have a value, v. This value would be the source of existence. The impact P of the fluctuations would be the size of the self. The self is rather dependent on a v-value then a certain value v. v defines the impact P. Twice the impact =/= half the self, P only defines size, it does not define whether or not the self exist for any value > 0. The self is merely an impact P on an area A.

Everything has the property of self.

Any comments on whether this is true or not are deaply appreciated. I might w8 a week or 2 before answering any questions or making any comment. Until then, feel free to point out flaws etc.

**posted similar stuff before on another existing forum**