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Infinite tetration and superroot of infinitesimal
#27
I have been looking at relation for infinitesimal exponentation of base a:

by definition, if we accept I as infinitesimal which is not 0, I^I = (1+J) where J is infinitesimal as well (since I^0=1). Then log base I ( 1+J) = I log base I (I) = I.
Euler made a proposition which he proved that J=kI, where k = 1 for base e so that ln(1+I)=I where I is just infinitesimal.

If we take imaginary Unit I as being the infinitesimal in base I , what is k?

From Eulers derivations, k as function of base a is :
k= 2( (a-1)/(a+1) + (a-1)^3/3*(a+1)^3+ (a-1)^5/5*(a+1)^5 ....

As we work with base I, where I = imaginary Unit, we have to evaluate term

(I-1)/(I+1) = (sqrt(2)* e^I*((3pi/4)+2pik)/ sqrt(2)* e^I* ((pi/4)+2pin) = e^I* (3pi/4-pi/4+2pik-2pin) = e^I*pi/2= i as a single value, not periodic, since all 2pik and 2pin cancel out.

So k= 2 ( I+I^3 /3+ I^5/5+I^7/7 +...) = 2*I ( 1-1/3+1/5-1/7.......) = 2*I* pi where I is JUST e^I*pi/2, single value.

Then J as infinitesimal result J = (I*pi/2 ) * I so J is also infinitesimal but turned by 90 degrees via multiplication of I by I plus it has module pi/2 and of I^I =(1+J) = (1+ (I*pi/2)*I) and

But we know that I^I = e^-pi/2 so that ln(I^I) = -pi/2= (I*pi/2)*I:

But we also know that I^I = (1+I*pi/2*I) so that ln(I^I) = ln(1+I^2*pi/2) = k*I= J= I*pi/2*I= -pi/2 if I is imaginary infinitesimal such that I^2 = -1.

So assumption that I may be imaginary infinitesimal works with logarithms, but the unique note here is that we are not dealing with ALL periodic values of I, but just I, such that is linked to 1 in this scale so that I=(1-I)/(1+I); In a similar vein, unique value of -I = (1+I)/(1-I) can be obtained.

Interestingly, first logarithms of Napier ( 1-10-7) and Burgi ( 1+10-4) were also based on small deviations from 1.

The conclusion is ( or proposition was) : Mathematics works in SCALES which we do not notice for a reason that they are imaginary ( meaning in our minds),so we assume the same scale exists for all operations, which is not true, and in each scale the relation ship between Unit lentgh 1 and imaginary infinitesimal is defined by I=(1-I)/(1+I). This imaginary infinitesimal is Unique for that scale and by rotating it by +2pi we move into next scale, perhaps via multiplying by J= I*pi/2 where 1 is also different, but we do not notice it, as in math, we mix all scales in our heads - which is not possible in physics, and therefore the discrepancy.
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Messages In This Thread
RE: Infinite tetration and superroot of infinitesimal - by Ivars - 12/28/2007, 12:31 PM

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