| MadSci Network: Other |
The way you go about this one is to use the "function of a function" rule. For the 2nd natural log, you have
f(x) = ln(ln(x)) = ln(y) where y = ln(x) df/dx = df/dy * dy/dx = 1/y *1/x = 1/(x*ln(x))For the 3rd natural log,
f(x) = ln(ln(ln(x))) = ln(z) where z = ln(y) and y = ln(x) df/dx = df/dz * dz/dy * dy/dx ...Well, you did ask me how to do the calculation, not to actually do it, so I'll leave the details to you. Maybe when you have done that one, and think about how the fourth natural log might work out, you will start to see a pattern emerging.
Moderator Note: You are looking for a recursion relation between the (n-1)th repeated log and the nth repeated log. (I'm not sure I have heard that phrase "repeated log" before, so please don't assume it is standard). As Dr. Christie mentioned, the task involves repeated application of the Chain Rule for derivatives of composite functions. (Those terms are standard; you can learn more by searching on them.) The answer will appear a bit more obvious if we recast the solution to the second log in the form:
d(ln(ln(x))/dx = 1/ln(x) * 1/x
The third log will then be:
d(ln(ln(ln(x)))/dx = 1/ln(ln(x)) * 1/ln(x) * 1/x
I think the pattern is fairly clear. With each iteration there will be
one more term of 1/ln(...(ln(x))...) in the derivative. What is not so
clear is the proper domain of the nth repeated log to the
base e. ln(x) is only defined for x>0, so ln(ln(x)) will only be
defined where ln(x) > 0 which means on that iteration that x must
be > 1. If I am not mistaken, then on the next iteration, it will
only be defined for x > e. On the fourth log, the domain of
valid x will be x > ee. After a bit more of this, the
domain will only begin at a reasonably large number,
e(n-1).
The graph of the nth repeated log is somewhat weird as well. It looks very flat and approaches the x axis after just a few iterations. I suspect it will not have a lot of utility in describing the natural world.
David Winsemius, MD, MPH
Try the links in the MadSci Library for more information on Other.