NEWTONIAN FUNCTION TIME
by Henryk Szubinski
Newton would have had to define the shared Integrals with the shared differencials as some time between them.
The common denominators are x7 and n7 so the fact that 3n are shared between 3x would total into 6n or 6x meaning that the
n7 as shared between x7 may be computed.
The 3 divisions by 7 define the 3 differencials or 3x as the product needed for the jump from the 6th to 7th n.
so that INV 3 = Integer 6/7+0,709
INV 3 =1,566 Integers
or in basic terms that the
nr of
nr INV (differencials )x= half the x nr INtegers.+- 0,066 error
Meaning that Newton would have defined the differencial side as the very high number to infinity as having the integral side of it as half of a very small number to zero.
That these two sides would equal their multiple s and their divisions as the highest and the lowest would also imply that the error would be the amount of calculus needed to work out their extra error by using up the mathematical volume while they computed. So the number error might be
INV 660 DIFF = 1/2 INTEGER 330
while the multiples used were to change from 0,066 to 660 as x 1000 multiple
So basically the
DIFF INV of 2 = INTEGER 1
or as the 1 DIFFERENCIAL expands the 2 INGERATIONS contract.
So the 1 DIFFERENCIAL would have to be the 1000 expansions as the differencial of the differencial or the largest differencial possible to the edge of infinity or the limits of the universe as 1.
another way to define this that has no problems of the non shared sides of the 1 or 2 as shown below.
so that INV 3 = Integer 6/7+0,709
INV 3 =1,566 Integers
or in basic terms that the
nr of
nr INV (differencials )x= half the x nr INtegers.+- 0,066 error
Meaning that Newton would have defined the differencial side as the very high number to infinity as having the integral side of it as half of a very small number to zero.
That these two sides would equal their multiple s and their divisions as the highest and the lowest would also imply that the error would be the amount of calculus needed to work out their extra error by using up the mathematical volume while they computed. So the number error might be
INV 660 DIFF = 1/2 INTEGER 330
while the multiples used were to change from 0,066 to 660 as x 1000 multiple
So basically the
DIFF INV of 2 = INTEGER 1
or as the 1 DIFFERENCIAL expands the 2 INGERATIONS contract.
So the 1 DIFFERENCIAL would have to be the 1000 expansions as the differencial of the differencial or the largest differencial possible to the edge of infinity or the limits of the universe as 1.
another way to define this that has no problems of the non shared sides of the 1 or 2 as shown below.
Depending on the directions of the verge up or down ,the gravity could power this device to generate free energy by a FIRST START of some motion push or it could be turned in the other direction to lift