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Analytic matrices and the base units
#1
Question 
Hi, there!
I have been looking for the number of the dimensions of the hyper- and interdimensional spaces of the Multiverse for long years. This is why my last thread about the Taylor series of i[x] is. I guess the next step is to create analytic matrices in where between two value there is another one. Because we can check the existence of the hyper- and interdimensional spaces by matrix-multiplication, cannot we?
Here is an example:
Let 
F = [ f(x,y) ]
and
G = [ g(x,y) ]
be  so-called functional or analytic matrices. My question is that what are

F*G, F^n, F^-1, |F| (determinant) ?
I suppose that F+G = [ f(x,y)+g(x,y) ]
I think this way leads us to the solutuion of the dimension-question.
Any idea?
Xorter Unizo
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#2
Ah, yes, I forgot something to write: F_{i j} = f(i,j)
And another question: Are these are related to Carleman matrices?
Xorter Unizo
Reply
#3
I have found something: https://math.stackexchange.com/questions...us-indices
Would it be the solution for multiplication functional / analytic matrices?

So F*G = [ f(x,y) ]*[ g(x,y) ] = [ int from alpha to beta f(x,u)*g(u,y) du ]

But I have a question: Why does not it work with constants?!

The next question: what is the analytic representation of the imaginary base units?

We can be sure that: [ 1 ] = [ x, 0; 0, x ]^oinfinity o 1, but what is i?
Could it be [ i ] = [ x, 0; 0, x ]^oinfinity o [ 0, 1; -1, 0 ] ?

And what is about the interdimensional base units, like i[1.5]?
I feel we are closer than ever before.
Xorter Unizo
Reply


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