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another math curiousity
Message #1 Posted by Don Shepherd on 12 Jan 2009, 4:28 p.m.

OK, consider: xy = yx. Is this equation true for any values other than 2 and 4?

      
Re: another math curiousity
Message #2 Posted by Egan Ford on 12 Jan 2009, 4:31 p.m.,
in response to message #1 by Don Shepherd

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.

            
Re: another math curiousity
Message #3 Posted by Bob Patton on 13 Jan 2009, 8:07 p.m.,
in response to message #2 by Egan Ford

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.

      
Re: another math curiousity
Message #4 Posted by Namir on 12 Jan 2009, 6:02 p.m.,
in response to message #1 by Don Shepherd

I agree with Egan. You have one equation and two variables. So you have a family of solutions.

Namir

      
Re: another math curiousity
Message #5 Posted by Thomas Klemm on 12 Jan 2009, 8:26 p.m.,
in response to message #1 by Don Shepherd

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.

            
Re: another math curiousity
Message #6 Posted by bill platt on 13 Jan 2009, 1:27 a.m.,
in response to message #5 by Thomas Klemm

This post is beautiful in a sublime mathematical way.

      
Re: another math curiousity
Message #7 Posted by Bob Patton on 12 Jan 2009, 8:27 p.m.,
in response to message #1 by Don Shepherd

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?

            
Re: another math curiousity
Message #8 Posted by Paul Dale on 12 Jan 2009, 9:51 p.m.,
in response to message #7 by Bob Patton

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

                  
Re: another math curiousity
Message #9 Posted by Paul Dale on 12 Jan 2009, 9:52 p.m.,
in response to message #8 by Paul Dale

Missed a common factor there which would have saved a step :-(

- Pauli

                  
Re: another math curiousity
Message #10 Posted by Egan Ford on 13 Jan 2009, 12:00 a.m.,
in response to message #8 by Paul Dale

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.
            
Re: another math curiousity
Message #11 Posted by Egan Ford on 12 Jan 2009, 11:40 p.m.,
in response to message #7 by Bob Patton

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.
      
Math Curiousity...an old fashioned approach
Message #12 Posted by Hal Bitton in Boise on 13 Jan 2009, 1:53 a.m.,
in response to message #1 by Don Shepherd

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

            
Re: Math Curiousity...an old fashioned approach
Message #13 Posted by George Bailey (Bedford Falls) on 13 Jan 2009, 2:59 a.m.,
in response to message #12 by Hal Bitton in Boise

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

                  
Re: Math Curiousity...an old fashioned approach
Message #14 Posted by Egan Ford on 13 Jan 2009, 9:17 a.m.,
in response to message #13 by George Bailey (Bedford Falls)

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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