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From an elementary question in differential entropy for decision sequences...

Numerical solutions is: x = 2.293166287408052...

The equality is only well defined (with respect to its origination) in the real domain for x>0. There are obviously an infinite number of solutions in the complex domain.

I have found a continued fraction solution, but have not been able to come up with a closed form variant.

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  • $\begingroup$ I think you need to add a lot of background for this to be a good question for MO. $\endgroup$
    – GMRA
    Oct 30, 2009 at 3:35
  • $\begingroup$ (1) The body of your question is indecipherable and (2) there is no reason I can think of that this equation should have a closed form solution. Your best bet is probably to solve this numerically. $\endgroup$
    – David E Speyer
    Oct 30, 2009 at 3:36
  • $\begingroup$ Why do you want an "algebraic" solution? What range of values is x confined to? What have you tried thus far? $\endgroup$
    – Yemon Choi
    Oct 30, 2009 at 3:38
  • $\begingroup$ Related: mathoverflow.net/questions/22230/ab-ba-when-a-is-not-equal-to-b $\endgroup$
    – Jonas Meyer
    Apr 23, 2010 at 5:13

3 Answers 3

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If I'm not mistaken think it's not so hard to see that the solution is transcendental. One is asking if there's algebraic x such that log(x+1)/(x+1) - log(x)/x = 0. Suppose this is so. One of Alan Baker's famous theorems is that if a1,...,an are algebraic and log(a1),...,log(an) are linearly dependent over the algebraic numbers then log(a1),...,log(an) are linearly dependent over the rationals. So we also have p log(x+1) + q log(x)=0 for rational p,q. It follows that x/(x+1) is rational, so x is rational. Put x=r/s in lowest terms; the equation becomes r^(r+s) = (r+s)^r * s^s, which has no solutions.

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  • $\begingroup$ How does it follow that x/(x+1) is rational? $\endgroup$
    – David E Speyer
    Oct 30, 2009 at 12:17
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    $\begingroup$ x/(x+1) = log(x)/log(x+1) = -p/q $\endgroup$
    – D. Savitt
    Oct 30, 2009 at 15:28
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Really? If you've found a continued fraction expression for this solution, I'd love to see it. The sequence I get

2, 3, 2, 2, 3, 4, 2, 3, 2, 130, 1, 1, 1, 1, 1, 6, 3, 2, 1, 15, 1, 1, 1, 8, 10, 3, 1, 5, 6, 4, 39, 2, 1, 1, 2, 2, 1, 1, 1, 4, 8, 4, 2, 5, 1, 2, 1, 5, 12, 1, 1, 5, 1, 5, 1, 1, 4, 2, 3, 3, 2, 3, 4, 4, 1, 1, 1, 4, 2, 1, 7, 11, 2, 4, 2, 39, 1, 2, 1, 29, 3, 2, 6, 8, 3, 7, 1, 7, 1, 2, 2, 1, 3, 1, 3, 7, 1, 1, 79, 2, 1, 11, 2, 4, 1, 1, 1, 1, 1, 9, 3, 6, 9, 1, 2, 1, 2, 2, 1, 1, 3, 1, 3, 2, 6, 26, 6, 4, 3, 1, 1, 4, 1, 3, 9, 296, 6, 1, 1, 96, 1, 2, 1, 3, 7, 4, 86, 4, 168, 19, 34, 21, 3, 2, 6, 1, 1, 1, 18, 1, 9, 2, 1, 1, 6, 2, 3, 2, 1, 1, 3, 1, 2, 1, 6, 7, 9, 5, 1, 2, 4, 1, 1, 5, 117, 3, 5, 1, 40, 1, 1, 3, 2, 26, 8, 22, 2, 1, 1, 1, 12, 1, 5, 6, 2, 1, 5, 1, 2, 2, 5, 1, 1, 1, 39, 2, 2, 2, 1, 6, 2, 13, 1, 71, 1, 4, 3, 1, 11, 1, 7, 2, 4, 5, 4, 1, 1, 5, 4, 2, 12, 2, 91, 1, 1, 2, 25, 1, 1, 24, 1, 18, 1, 2, 1, 4, 1, 2, 2, 1, 2, 1, 1, 1, 2, 92, 2, 1, 1, 35, 1, 1, 1, 9, 2, 2, 1, 1, 2, 2, 1, 9, 1, 1, 4, 1, 1, 1, 1, 38, 2, 41, 2, 1, 3, 1, 1, 27, 18, 4, 20, 9, 1, 2, 9, 37, 1, 1, 1, 1, 10, 1, 11, 1, 2, 1, 1, 2, 1, 1, 2, 6, 14, 1, 17, 1, 1, 5, 2, 2, 1, 2, 5, 2, 1, 131, 1, 8, 12, 1, 2, 3, 1, 3, 1, 5, 5, 1, 1, 1, 7, 1, 2, 1, 5, 1, 1, 8, 2, 2, 4, 1, 1, 17, 3, 14, 2, 11, 2, 3, 2, 3, 2, 1, 1, 1, 4, 6, 2, 37, 2, 1, 1, 201, 6, 11, 1, 113, 1, 8, 7, 18, 1, 2, 2, 17, 4, 3, 5, 2, 1, 1, 1, 2, 2, 2, 1, 3, 1, 9, 4, 2, 2, 2, 3, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 8, 3, 1, 17, 1, 1, 1, 2, 1, 4, 1, 3, 4, 11, 2, 7, 15, 1, 1, 2, 1, 1, 18, 1, 1, 1, 3, 7, 2, 1, 1, 2, 12, 6, 1, 2, 1, 2, 2, 1, 2, 4, 2, 1, 1, 3, 1, 108, 1, 1, 2, 6, 4, 1, 4, 4, 7, 1, 6, 2, 19, 1, 4, 4, 1, 1, 10, 1, 12, 1, 2, 2, 5, 8, 2, 1, 1, 6, 1, 1, 1, 21, 2, 1, 13, 1, 3, 19, 1, 1, 27, 2, 8, 183, 3, 2, 1, 1, 2, 1, 11, 7, 2, 1, 1, 4, 1, 4, 2, 26, 2, ...

looks pretty darn structureless!

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  • $\begingroup$ Scott, Are you able to find structure in the sequence for Pi? If so, I would love to hear about it... $\endgroup$
    – Halfdan
    Oct 30, 2009 at 4:51
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    $\begingroup$ @Halfdan: I don't understand your comment. Do you have a continued fraction solution or don't you? If you do, perhaps you could edit your question to include some information about the continued fraction solution. What Scott is saying here is that whatever your continued fraction solution is, it's non-obvious. $\endgroup$
    – Anton Geraschenko
    Oct 30, 2009 at 5:07
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    $\begingroup$ Halfdan, what exactly did you mean by "I have found a continued fraction solution" then? $\endgroup$
    – Scott Morrison
    Oct 30, 2009 at 5:35
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    $\begingroup$ Scott, out of curiousity, how the heck did you get that many convergents? You have about 680 terms, which should mean that you computed the root of this equation with accuracy at least (GoldenRatio)^{-680} (and probably closer to 2^{-680}.) That's hundreds of digits, for an equation whose root is probably transcendental. $\endgroup$
    – David E Speyer
    Oct 30, 2009 at 12:16
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    $\begingroup$ In mathematica, for example, define: X[n_] := ContinuedFraction[x /. FindRoot[x^(1 + x) - (1 + x)^x, {x, 2}, WorkingPrecision -> 2n, PrecisionGoal -> n]] and then calculate whatever you want: X[10000]. Of course the parameter here is a number of decimal places, not the number of correct convergents, but as long as you ignore enough of the tail you're safe. Newton's method is nice and fast :-) $\endgroup$
    – Scott Morrison
    Oct 30, 2009 at 16:37
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If you set ab=ba, you get that (log a)/a=(log b)/b. This function is increasing on (0,e) and decreasing on (e,∞), so it seems like there should only be one x-value for which this could work anyways. I don't know exactly what it is, but I'd guess that it's probably not algebraic...

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