another math curiousity

[Don Shepherd]

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

[Egan Ford]

Quote:

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

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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]

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]

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

Namir

[Thomas Klemm]

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]

This post is beautiful in a sublime mathematical way.

[Bob Patton]

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]

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]

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

- Pauli

[Egan Ford]

Quote:

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For the in-progress 20b scientific firmware I'm fiddling with this is easy (thanks to Egan's formula for X given Y):

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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]

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What's the best procedure for this forum?

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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]

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)]

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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

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Maybe the Lambert function.

Regards,
George Bailey

[Egan Ford]

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Maybe the Lambert function.

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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.