| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
This is a quickie.
Proposed by Ismor Fischer, University of Wisconson, Oshkosh, Wisconsin.
1/(x-1)
(2^x - 1)
Show that 1 < [---------] < 2, for all x != 0, 1.
x
[That's 1 < ((2^x-1)/x)^(1/(x-1)) < 2.]
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 1945.1 | FLOYD::YODER | MFY | Fri Mar 31 1995 17:25 | 2 | |
I have what I think is a proof of this, but it isn't short or pretty. Could you just post the quick proof? | |||||
| 1945.2 | RUSURE::EDP | Always mount a scratch monkey. | Wed Apr 05 1995 15:31 | 8 | |
Here's the solution from Mathematics Magazine:
For fixed x != 0, 1, apply the Mean Value Theorem to f(t) = t^x on
[1,2]. Remark. The lower and upper bounds on the function in the
problem are sharp, and represent the limiting values as x -> -/+
infinity, respectively. Moreover, the function can be extended
continuously through the points 0 and 1 by defining it to take on the
values 1/ln 2 and 4e, respectively.
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