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- Newsgroups: sci.math
- Path: sparky!uunet!caen!zaphod.mps.ohio-state.edu!magnus.acs.ohio-state.edu!regeorge
- From: regeorge@magnus.acs.ohio-state.edu (Robert E George)
- Subject: Re: DERIVATIVE OF X^X
- Message-ID: <1992Nov6.204539.3644@magnus.acs.ohio-state.edu>
- Sender: news@magnus.acs.ohio-state.edu
- Nntp-Posting-Host: bottom.magnus.acs.ohio-state.edu
- Organization: The Ohio State University
- References: <1992Nov4.200746.11729@aio.jsc.nasa.gov> <1992Nov5.004445.21327@inf
- Date: Fri, 6 Nov 1992 20:45:39 GMT
- Lines: 49
-
- In article <13421@ecs.soton.ac.uk> dbc@ecs.soton.ac.uk (Bryan Carpenter) writes
- :
- [deletions]
- >
- >I think ``first principles'' ought to allow you to take
- >
- > Lim (1 + y/n)^n
- > n -> inf
- >
- >as the definition of e^y. So, if y = log x and n is infinitely large
- >(if you know what I mean),
- >
- > x = (1 + y/n)^n
- >
- >so
- >
- > y = n (x^(1/n) - 1)
- >
- >and you can put h = 1/n (so to speak). It gets the right answer, anyway.
- >
- [deletions]
-
- I can see a potential problem with putting h = 1/n in cases where
- the limit did not exist. If you know the derivative (limit) does exist,
- then the limit of
- f(x + a ) - f(x)
- n
- ------------------ (*)
- a
- n
-
- where an --> 0 *will* equal f'(x). If the derivative doesn't exist,
- there can be problems: To give a simple example, let
- f = abs. value, x = 0, and a = 1/n
- n
-
- Then the limit as n --> oo of (*) will equal 1, giving one the false
- notion that f is differentiable at x = 0.
- Now it may well be possible to *show* that a particular function is
- differentiable at a point x. Then one can use (*) to find the derivative.
-
-
- Robert George
- (speaking only for myself)
-
- "The very essence of individual freedom is equal justice under a rule of
- law, a law to which every man shall be subject and which no executive
- can modify."
- Senator Robert Taft , March 11, 1944
-