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Infinite tetration and superroot of infinitesimal
#58
Without bothering about pi and signs, in principle we can define lines and surfaces in hyperdimensional space as

Integral (a to b) di is a line, integral (1-infinity ) di is = - i is also a line but infinite.
We also have integral (-infinity to 0?) = i which is also some infinite line. Then Integral (-infnity to plus infinity) = -I + I = 0 Is a linear combination of those 2 lines- the lines itself are most likely undefined spirals of opposing rotation, but perhaps even not so well specifiable in general case.

Probably we can construct "area " of "surface" in hyperdimensional space as
Integral over all surface of dI*dI where dI*dI is a surface element in hyperdimensional space
and so on-volume, 4 volume .......... infinite dimensional volume.

while pi/2 is assumed constant, "area " element dI*dI = -pi/2*-pi/2 = (pi^2)/4


and "volume" element dI*dI*dI = -(pi^3)/8.

If dI is imaginary infinitesimal, it is possible to create (1/dI) = imaginary infinity

1/dI = I/lnI= (-2/pi). So far so good.

In fact, it seems pi is not constant as hyperdimensions shift by dI. So, as long as 1/2 is constant,

dI=d(pi/2) = -1/2d(pi)

With this, element of area is 1/4 d(pi)^2. Interestigly, that relates to black hole area rather nicely, just need to take integral over d(pi) over involved hyperdimensions.

if dpi is a change in pi, than Intergral (-1/2)dpi) = - I = e^(-I*pi/2)

Which seems logical , also the sign- as further we go away from e^(pi/2) by infinite tetration, we get closer to -I, or +I, so increase in I+dI should lead to decrease of 1/2d(pi). In the end, we have just +-I, pi/2 has vanished. Comnig back from both -I, +I ovr infinite number of hyperdimensions, I vanishes, pi/2 is fully achieved.

Interestingly, e seems to fall out from dI based math in open form. But that has to be checked further. e appears only when integrals over certain amount oif dimensions are taken, meaning that e is variable as well, but depends on number dimensions included in integration over dI, so

e =e (function of number of hyperdimensions over which integral is taken).

Since e usually relates to time, one can see, that in individual hyperdimension there is no time. Time only appears when integrals over hyperdimensions are taken, and that is imaginary or perpendicular, if You like, time, if integration is finite.

Also, it seems obvious that lnI has 1 value on each dimension I-dI, I-2dI (may be ) etc , so there is no multivalued functions anymore- each of the values correspond to different hyperdimension.

The question of infinte values of hyperdimensions also arises when taking x root of y. There has to be infinite number of roots if x is real, not rational, even more so if it is irrational or transcendental, or imaginary.

if self root of i leads to e^pi/2, that means that either e^pi/2 has infinite hyperdimensional values, or, it is by itself a hyperdimensional number.
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Messages In This Thread
RE: Infinite tetration and superroot of infinitesimal - by Ivars - 01/24/2008, 03:44 PM

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