(boing!) Cnoocy Mosque O'Witz (
cnoocy
) wrote
2004-01-22 11:51 am (UTC)
no subject
You're starting it when T = 3/2ln(2). The formula for the time taken by the process is:
T
x
= T
0
/2^(x2/3)
So:
T
x
= T
0
/2^((3/2ln(2))2/3)
T
x
= T
0
/2^(1/ln(2))
So the speed has increased by 2^(1/ln(2)), which is e.
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no subject
Tx = T0/2^(x2/3)
So:
Tx = T0/2^((3/2ln(2))2/3)
Tx = T0/2^(1/ln(2))
So the speed has increased by 2^(1/ln(2)), which is e.