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Tuesday, December 22, 2020

day 381 of the apocalypse 13, 17 Math for the speech

Still no word, I suppose I'm waiting for the 4th now.
Today was a confused day although I'm getting some direction.
I have a new medical bent to things although I have no way to apply it yet.
Speaking of medical, I wasn't getting any exercise so I rode my bike to work and that meant only a mediocre exercise to go with the walk from yesterday and those silly pullups, silly because they are so pitiful.
And the editing continues, a little vague to me, I'm not excited about how its going, but its going and I have the others to get ready this week now that I have the ideas for what has to be the last one to file.
I went on another walk, pretty hard this time, but still just a walk and not too far.
Here is something for you, if only you loved me like I love you.




photo by Umesh Ghule of the Saturn Jupiter conjunction

Not sure about this math or where it was being used or how; but for some reason I put this together a long time ago and for reasons I don't understand now:


Derivatives:
M(t)=2^t
Easy dM/dt(t)=(2^(t+dt)-2^t)/dt gives teh change over a chosen size dt
The 2^(t+dt) can be broken down to 2^t*2^dt so you get:
[(2^t*2^dt)-2^t]/dt=2^t(2^dt-1)/dt
The derivative of 2^t is what this expression equals as dt approaches zero.
the 2^dt-1/dt approachs a constant (as an infinite series) .6931472... as dt gets very small.
As the "base" (e.g. 2 here) changes (e.g to 3) the constant changes, for 3 it goes to 1.0986...
for base 8 it is 3 times the limit number for 2 which derives from 2^n, here 3.

Where d(a^t)/dt=a^t is euler's number.  d(e^t)/dt(t)=e^t (1).
See the sin equation where we get the .14... number that takes us to a ratio of 256/26 for n=1 with the sin equation. (note that I now have 256/27 which is confusing for n=-1:n=1; what was I thinking here?)
for e, the slope=location on the x axis.
e^3t derives to 3e^t (chain rule):de^ct/dt=3e^ct
2=e^ln(2):  e^?=2 is how ln is derived (natural log)
2^t=e^ln(2)t
So taking the derivative of the equation above
ln(2)2^t=ln(2)*e^ln(2)t


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