| 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 |
Which is larger, e^pi or pi^e ? Demonstrate WITHOUT NUMERICAL
APPROXIMATION.
/Eric
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 515.1 | CLT::GILBERT | Juggler of Noterdom | Mon Jun 16 1986 21:55 | 23 | |
Well, we know that e^(1/e) > pi^(1/pi) > 1 so raising each of these to the power e*pi > 0 gives: e^pi > pi^e > 1. You may well ask how we 'know' that first equation. Well, the function x^(1/x) is continuous over the range epsilon to infinity, and has a local maxima at x = e, since the derivative of x^(1/x) is: d x^(1/x) /dx = d e^(ln(x)/x) /dx = e^(ln(x)/x) [ ln(x)(-x^-2) + x^-2 ] = x^(1/x) (1-ln(x)) / x�, and this is zero only at x = e (in the range epsilon to infinity), and so x = e is either a local minima or maxima. Furthermore, because 1^(1/1) < 2^(1/2) = 4^(1/4), we conclude that x^(1/x) is concave downward, so x = e is a maxima. Thus, e^(1/e) >= x^(1/x) for *any* positive x, including pi. | |||||
| 515.2 | LATOUR::JMUNZER | Tue Jun 17 1986 14:24 | 8 | ||
I believe that 508.6 and 512.2 both supply the answer to .0.
If .1 is hidden from all but its author and the moderator, is it
a lot like a WOM[*]?
John
[*] write only memory
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| 515.3 | An alternative proof | AUSSIE::GARSON | Wed Jun 17 1992 00:23 | 22 | |
e^x = 1 + x + x^2/2! + x^3/3! + ... (1)
so for x>0, e^x > 1 + x (2)
pi > e (an assumption)
=> pi/e > 1
=> pi/e - 1 > 0
so let x = pi/e - 1 in (2)
e^(pi/e - 1) > 1 + pi/e - 1
=> e^(pi/e) / e > pi/e
=> e^(pi/e) > pi
=> e^pi > pi^e
(This proof is not mine but it seemed suitably elementary by comparison
with the previous proof to be worth posting. Naturally from (1) it is
easy to see that d/dx e^x = e^x and from there the result using
calculus follows.)
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