
120^291 (1.1*10^605) as infinity borderline Chapt13.40085 Maxwell Equations placing demands on mathematics #646 New Physics #766 ATOM TOTALITY 5th ed
Posted:
Jun 18, 2012 5:51 PM


On Jun 18, 4:00 pm, Transfer Principle <david.l.wal...@lausd.net> wrote: > On Jun 18, 11:02 am, Archimedes Plutonium > > <plutonium.archime...@gmail.com> wrote: > > So, does there exist a number that is between 10^603 and 10^604 that > > is a [power] of 120, that is 120^n [...] > > No, no such number exists. The powers of closest to this range are: > > 120^290 = 9.174055302...*10^602 > 120^291 = 1.100886636...*10^605 > > As 291 is a multiple of three, 120^291 is a perfect cube  its > cube root would be 120^97.
Thanks Transfer, I had some inkling that the gap was going to be larger than 10^603 by loosening the reins over floor pi but did not expect it to go out to 10^605. If it is not too difficult, can you compute the perfect cube 120^n immediately below 120^291? Maybe it is 10^600???
If I have to accept 120^291, that leaves me a problem of reconciling pi for the sphere pseudosphere area. They have to be equal in area at infinity. So if infinity starts at 120^291, I would be required to use pi at 10^605
Pi = 3. 1415926535 8979323846 2643383279 5028841971 6939937510 : 50 5820974944 5923078164 0628620899 8628034825 3421170679 : 100 8214808651 3282306647 0938446095 5058223172 5359408128 : 150 4811174502 8410270193 8521105559 6446229489 5493038196 : 200 4428810975 6659334461 2847564823 3786783165 2712019091 : 250 4564856692 3460348610 4543266482 1339360726 0249141273 : 300 7245870066 0631558817 4881520920 9628292540 9171536436 : 350 7892590360 0113305305 4882046652 1384146951 9415116094 : 400 3305727036 5759591953 0921861173 8193261179 3105118548 : 450 0744623799 6274956735 1885752724 8912279381 8301194912 : 500 9833673362 4406566430 8602139494 6395224737 1907021798 : 550 6094370277 0539217176 2931767523 8467481846 7669405132 : 600 0005681271 4526356082 7785771342 7577896091 7363717872 : 650
I would be required to use the 00056
So say we had a sphere of volume 1, would we have a pseudosphere volume of exactly 1/2 and both areas equal by using pi at those 605 digits and where infinity is this 1.1 x 10^605.
Previously I was counting on the three zero digits to allow for the pseudosphere to catch up but now I need to tag on that extra two digits of 56.
The trouble here in mathematics, is that we have no computer or formula to compute the area and volume of pseudosphere in the 10^603 to 10^605 range.
I need to know if the pseudosphere area catches up and matches the attendant sphere in that range.
Archimedes Plutonium http://www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electrondotcloud are galaxies

