OK, consider: x^{y} = y^{x}. Is this equation true for any values other than 2 and 4?
another math curiousity


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01122009, 04:28 PM
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01122009, 04:31 PM
Quote:x = y. Perhaps you meant to state x != y.
How about: y*W(ln(y)/y)Set y = 4 and then x = 2. So set anything you like for y and see what you get for x.
Edited: 12 Jan 2009, 5:11 p.m. ▼
01132009, 08:07 PM
There is a parametric solution to x^y = y^x namely:
01122009, 06:02 PM
I agree with Egan. You have one equation and two variables. So you have a family of solutions. Namir
01122009, 08:26 PM
Taking the logarithm on both sides yields:
log(x)/x = log(y)/y(Assuming both x and y are > 0)
plot(log(x)/x, (1,20))
It's is clear, that for each value x  1 < x < e you can find another value y  e < y which gives the same function value as x. The only exception is x = e since that gives the maximum. Maybe you had solutions in integers in mind. But as there is only one integer between 1 and e (namely 2), the solution you mentioned is the only one.
01122009, 08:27 PM
x^y = y^x has a continuous set of values for x>1 and y>1. ▼
01122009, 09:51 PM
For the inprogress 20b scientific firmware I'm fiddling with this is easy (thanks to Egan's formula for X given Y):
001 LBL F1 I could have optimised things a bit and avoided the register use but I'm feeling lazy.
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01122009, 09:52 PM
Missed a common factor there which would have saved a step :(  Pauli
01132009, 12:00 AM
Quote:I should have stated that with that equation if y <= e, then x will always equal y, so keep y > e as Thomas has already pointed out.
01122009, 11:40 PM
Quote:I've always been a fan of plain text. Just make sure your put [pre] and [/pre] around your code for readability.
01132009, 01:53 AM
I put x^y=y^x into the numerical solver, and plugged various values of x in. For integer values of x>1, the solver seemed to be able to arrive at a viable solution that approached 1 (from the right) as x increased. Futhermore, for even integer values of x, there seemed to also be a viable negative solution for y, which approached 1 (from the right) as x increased. Mathematically it seems to make sense, but I admit I would be at a loss to prove it algebraically. ▼
01132009, 02:59 AM
Quote: Maybe the Lambert function.
Regards, ▼
01132009, 09:17 AM
Quote:Correct. Various calculator (12C, 33s/35s, 42S, 50g) W implementations can be had here: http://www.hpmuseum.org/cgisys/cgiwrap/hpmuseum/archv018.cgi?read=132639. Edited: 13 Jan 2009, 9:26 a.m. 
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