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 ▼ Don Shepherd Posting Freak Posts: 1,392 Threads: 142 Joined: Jun 2007 01-12-2009, 04:28 PM OK, consider: xy = yx. Is this equation true for any values other than 2 and 4? ▼ Egan Ford Posting Freak Posts: 1,619 Threads: 147 Joined: May 2006 01-12-2009, 04:31 PM Quote: OK, consider: xy = yx. Is this equation true for any values other than 2 and 4? x = y. Perhaps you meant to state x != y. How about: ``` y*W(-ln(y)/y) x = - ------------- ln(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. ▼ Bob Patton Junior Member Posts: 33 Threads: 5 Joined: Jan 2009 01-13-2009, 08:07 PM There is a parametric solution to x^y = y^x namely: x = a^((1/(a-1)), y = ax = a^((a/a-1)) which provides real solutions for a>0 and a /= 1. This leads to the parametric form for rational solutions: x = ((k+1)/k)^k, y = ((k+1)/k)^(k+1) for k = 1,2,3... This gives the x=2,y=4 and x=2.25,y=3.375 solutions. Namir Posting Freak Posts: 2,247 Threads: 200 Joined: Jun 2005 01-12-2009, 06:02 PM I agree with Egan. You have one equation and two variables. So you have a family of solutions. Namir Thomas Klemm Senior Member Posts: 735 Threads: 34 Joined: May 2007 01-12-2009, 08:26 PM Taking the logarithm on both sides yields: ```log(x)/x = log(y)/y ``` (Assuming both x and y are > 0) Now have a look at: ```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. ▼ bill platt Posting Freak Posts: 2,448 Threads: 90 Joined: Jul 2005 01-13-2009, 01:27 AM This post is beautiful in a sublime mathematical way. Bob Patton Junior Member Posts: 33 Threads: 5 Joined: Jan 2009 01-12-2009, 08:27 PM x^y = y^x has a continuous set of values for x>1 and y>1. Another exact pair is x=2.25, y=3.375 and vice versa. Curiously the hyperbola (x+2)/(x-1) fits the above curve very closely. I have calculator programs to compute all such real values. I can post if there is any interest. What's the best procedure for this forum? ▼ Paul Dale Posting Freak Posts: 3,229 Threads: 42 Joined: Jul 2006 01-12-2009, 09:51 PM For the in-progress 20b scientific firmware I'm fiddling with this is easy (thanks to Egan's formula for X given Y): ``` 001 LBL F1 002 STO 00 003 LN 004 +/- 005 STO 01 006 RCL/ 00 007 W 008 RCL* 00 009 RCL/ 01 010 RTN ``` I could have optimised things a bit and avoided the register use but I'm feeling lazy. - Pauli ▼ Paul Dale Posting Freak Posts: 3,229 Threads: 42 Joined: Jul 2006 01-12-2009, 09:52 PM Missed a common factor there which would have saved a step :-( - Pauli Egan Ford Posting Freak Posts: 1,619 Threads: 147 Joined: May 2006 01-13-2009, 12:00 AM Quote: For the in-progress 20b scientific firmware I'm fiddling with this is easy (thanks to Egan's formula for X given Y): 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. Egan Ford Posting Freak Posts: 1,619 Threads: 147 Joined: May 2006 01-12-2009, 11:40 PM Quote: What's the best procedure for this forum? I've always been a fan of plain text. Just make sure your put [pre] and [/pre] around your code for readability. Hal Bitton in Boise Senior Member Posts: 291 Threads: 43 Joined: Jun 2007 01-13-2009, 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. I wonder if someone could explain to me what the "W" variable was in Egan's solution of x^y=y^x, as that might help me see the light. Best regards, Hal ▼ George Bailey (Bedford Falls) Senior Member Posts: 335 Threads: 12 Joined: Dec 2007 01-13-2009, 02:59 AM Quote: I wonder if someone could explain to me what the "W" variable was in Egan's solution of x^y=y^x, as that might help me see the light. Best regards, Hal Maybe the Lambert function. Regards, George Bailey ▼ Egan Ford Posting Freak Posts: 1,619 Threads: 147 Joined: May 2006 01-13-2009, 09:17 AM Quote: Maybe the Lambert function. Correct. Various calculator (12C, 33s/35s, 42S, 50g) W implementations can be had here: http://www.hpmuseum.org/cgi-sys/cgiwrap/hpmuseum/archv018.cgi?read=132639. Edited: 13 Jan 2009, 9:26 a.m.

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