User blog:Nayuta Ito/PEGG detailed log

I can only update this chart on Wednesdays, Fridays, Satuadays, and Sundays unpredictably randomly. If you want to see the value for today, please check PsiCubed2's Current Value Of PEGG. The page is gone now. Somehow the page is back.

The age of E
This age lasted for 32 days.

This is the first "age" of PEGG. In this age, X is added by 0.01*(The hundredth place of the previous day's DJIA closing price) and PEGG is 10^X.

Therefore, PEGG grows approximately exponentially.

And something interesting happens. Since X gor over 10, the definition's rule (e) is applied.


 * (e-1) Increment the letter in Y by one.
 * Y was "E", so Y becomes "F". In this log, it means proceeding to the next age.
 * (e-2) Convert the new value of PEGG to the form m
 * So we have to convert E10.20 into F. I will use "Lc" for the appreviation of common logarithm.
 * E10.20=EELc10.20(~EE1.0086)=EEELcLc10.20(~EEE0.00372)=F(2+LcLc10.20)
 * (e-3) int(mK)/K→X
 * "m" is what comes after F, which is 2+LcLc10.20 in this case. m starts with 2.003719, so int(mK)=2003. and divide that by K, so X becomes 2.003. (Recall that K=1000)
 * (e-4) 0.7xZ→Z
 * Z was 0.01, so Z becomes 0.007.
 * (e-5) 10K→K
 * Now K becomes 10000.

The age of F
This age lasted for 30 days.

It gets slower because Z gets smaller. Each age will take about 1.4 times longer than te previous one.

Starting Values:


 * X=2.003
 * Y="F"
 * Z=0.007
 * K=10000

And now, by the definition of Fx, PEGG will grow tetrationally.

And something interesting happens. Since X got over 10, the definition's rule (e) is applied.


 * (e-1) Increment the letter in Y by one.
 * Y was "F", so Y becomes "G". In this log, it means proceeding to the next age.
 * (e-2) Convert the new value of PEGG to the form m
 * So we have to convert F10.081 into G. I will use "Lc" for the appreviation of common logarithm.
 * F10.081=FELc10.081=FE(10^LcLc10.081)=FF(1+LcLc10.081)=FF(10^Lc(1+LcLc10.081))=G(2+(Lc(1+LcLc10.081)))
 * (e-3) int(mK)/K→X
 * "m" is what comes after G, which is 2+Lc(1+LcLc10.081) in this case. m starts with 2.000659, so int(mK)=20006. and divide that by K, so X becomes 2.0006. (Recall that K=10000)
 * (e-4) 0.7xZ→Z
 * Z was 0.007, so Z becomes 0.0049.
 * (e-5) 10K→K
 * Now K becomes 100000.

The age of G
Now it will be twice slower than the age of E.

Starting Values:


 * X=2.0006
 * Y="G"
 * Z=0.0049
 * K=100000

And now, by the definition of Gx, PEGG will grow pentationally.

Gx is supposed to be continuous 10^^^x, so let's check the definition for non-integeers:

10^^^x=10^^(10^^^(x-1)) for x>1 and 10^^^x=10^x for x<=1.

And something interesting happens. Since X got over 10, the definition's rule (e) is applied.


 * (e-1) Increment the letter in Y by one.
 * Y was "G", so Y becomes "H". In this log, it means proceeding to the next age.
 * (e-2) Convert the new value of PEGG to the form m
 * So we have to convert G10.1395 into H. I will use "Lc" for the appreviation of common logarithm.
 * G10.1395=GELc10.1395=GE(10^LcLc10.1395)=GF(1+LcLc10.1395)=GF(10^Lc(1+LcLc10.1395))=GG(1+Lc(1+LcLc10.1395))=H(2+Lc(1+Lc(1+LcLc10.1395)))
 * (e-3) int(mK)/K→X
 * "m" is what comes after G, which is 2+Lc(1+Lc(1+LcLc10.1395)) in this case. m starts with 2.00049044, so int(mK)=200049. and divide that by K, so X becomes 2.00049. (Recall that K=100000)
 * (e-4) 0.7xZ→Z
 * Z was 0.0049, so Z becomes 0.00343.
 * (e-5) 10K→K
 * Now K becomes 1000000.