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I am attempting to find a closed form solution or a nice series for $x^{x+1}=(x+1)^{x}$. First of all I looked at $a^b=b^a$. Fixing a, this means finding out when $a^{1/a} = x^{1/x}$. $f(x)=x^{1/x}$ is an interesting function. It has a maximum at e. (Proof due to Jakub Steiner.) $x^{1/x}$ goes to 1 as x goes to infinity. Based on that, I can conclude that $a^{1/a} = x^{1/x}$ has non-trivial solutions when x > 1 and that when a > e the non-trivial solution is between 1 and $e^{1/e}$.

I thought this was pretty neat, but it isn't helping me to solve the original problem. So, I would like to know is there a name, given a, for the value of b (not equal to a) such that $a^b = b^a$? For example, we could call b the "switch-ponent" of a but there must be a better name than that. Anyone know what these are called or maybe a name for the original problem?

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Why would it have a name? It's honestly not that interesting a function. –  Qiaochu Yuan Apr 22 '10 at 18:21
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It's probably possible to write a solution in terms of the Lambert W function (en.wikipedia.org/wiki/Lambert_W_function), which is the inverse of the function $f(z) = ze^z$, but I can't quite get it to work out. –  Michael Lugo Apr 22 '10 at 18:52
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Mathematica gives $b=-\frac{a W(-log(a)/a)}{log(a)}$ –  Jean-Philippe Burelle Apr 22 '10 at 19:00
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There's a lot of interesting math in the answers, but is this a question of research interest? Voting to close. –  Gerry Myerson Feb 22 '12 at 11:55
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7 Answers

As is said, this is too long for a comment, but no more substantial.

The related question in Jonas Meyer's comment seems to have settled almost anything involved, except that, as is common, the original person posting never supplied the claimed continued fraction.

I once read a book intended for high-school students that pointed out that $$ a^b = b^a $$ is equivalent to $$ \frac{\log a}{a} = \frac{\log b}{b} $$ and so any pair can be found by drawing a line through the origin that intersects the curve $ y = \log x$ and taking $a,b$ as the $x$-coordinates of the two intersection points. This was given to motivate the idea that the only integral pair is $2^4 = 4^2.$ Of course there is also the point $(e,1)$ at which the line $y = x / e$ through the origin is tangent to the curve.

Are there any other pairs $a,b$ with both rational?

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The answer to your question is "yes". Rational solutions are characterized in my answer. –  S. Carnahan Apr 23 '10 at 4:29
    
I've used this problem with the geometric approach you describe as an interview question for prospective undergrads (of course, they get lots of help along the way). It's a cute problem. –  Jeffrey Giansiracusa Apr 23 '10 at 13:31
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This doesn't answer your question, but I thought I'd point out that if we assume $1 < a < b$, the solutions to $a^b = b^a$ can be written in the form $a = \left( \frac{s+1}{s} \right)^s$ and $b = \left( \frac{s+1}{s} \right)^{s+1}$ for some unique positive $s$. This is particularly useful when looking for rational solutions, since $s$ then has to be an integer (a homework exercise from BSM). Unfortunately, you're looking for a transcendental value (see D. Savitt's answer).

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Saw your comment. This is excellent, the limit as $s$ increases is $e$ for both. And $b−a$ has a maximum of 2 at $s=1$ then down to $9/8$ for $s=2$ immediately below 1 with $64/81$ for $s=3.$ So this explains the special position of 2 as an integral value of $b-a.$ –  Will Jagy Apr 23 '10 at 5:02
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This is not an answer, but the solution to $x^{x+1}=(x+1)^{x}$ is known as the first Foias Constant.

For more information see: http://mathworld.wolfram.com/FoiasConstant.html

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That article says it cannot be expressed as a root of a function. How did you came to this result? –  Anixx Nov 7 '10 at 8:08
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For the problem of $a^b=b^a$, you can indeed express the complete solution in 'closed' form in terms of the LambertW function. All solutions can be expressed as $$-W \left( j,-\frac{\ln \left( a \right)}{a} {e^{\frac { 2 \pi i n}{a}}} \right) \frac{a}{\ln \left( a \right)}$$

where n is an arbitrary integer and j is an arbitrary positive integer.

If it's a series you want, then $x^{(x+1)} - (x+1)^x$ can be expanded (around 0) as $$-1+x+ \left( -\ln \left( \frac{1}{x} \right) -1 \right) {x}^{2}+ \left( 1/2\ln \left( \frac{1}{x} \right)^{2}+{ \frac {1}{2}} \right) {x}^{3}+ \left( -\frac{1}{6}\ln \left( \frac{1}{x} \right)^{3}-{\frac {5}{6}} \right) {x}^{4}+ \left( \frac{1}{24} \ln \left( \frac{1}{x} \right)^{4}+{\frac {3}{4}} \right) {x}^{5}+O \left( \ln \left( \frac{1}{x}\right) ^{5} {x}^{6}\right) $$

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Not a closed form, but you could write it as $x=\left(1+\frac{1}{x}\right)^x$ and express the solution as $\lim_{n\rightarrow\infty}x_n$, where $x_0=1$ and $x_{n+1}=\left(1+\frac{1}{x_n}\right)^{x_n}$.

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On a curios note, Jacob Bernoulli discovered the number e by examining $\lim_{x\rightarrow\infty}{\left(1+\frac{1}{x}\right)^x}$ –  Halfdan Faber Nov 15 '10 at 2:00
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Given your original function, $f(x) = x^{x+1} - (x+1)^x = 0$, you can define the Taylor Series expansion around some value v to give an expansion such that f( x) = a for some a, which in itself isn't helpful. However, there exists an inversion, using the Lagrange Inversion Theorem, which allows you to specify the a which your f( x) will return and gives you the corresponding x.
The first few terms of the series are:
$a+\frac{x-f[a]}{f'[a]}-\frac{f''[a] (x-f[a])^2}{2 f'[a]^3}+\frac{\left(3 f''[a]^2-f'[a] f^{(3)}[a]\right) (x-f[a])^3}{6 f'[a]^5}+\frac{\left(-15 f''[a]^3+10 f'[a] f''[a] f^{(3)}[a]-f'[a]^2 f^{(4)}[a]\right) (x-f[a])^4}{24 f'[a]^7}$
Remember, though: this is HIGHLY volatile when you're not near the actual value. Using 3 makes the value explode upwards, and using 2 makes it explode in the negative direction. However, a value of 2.2 will converge to the value you're looking for. You can use the Inversion Theorem to calculate more terms in the inverse series and use it to calculate your constant to arbitrary precision, given enough computational time.

Of course, this isn't what you're looking for (i.e. closed form or 'nice' series), but it's the best that I've found for this particular problem. Unfortunately, it converges so slowly with respect to computational time that it probably becomes more convenient for you to use the secant method to get any sort of good approximation.

--Gabriel Benamy

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To find rational solutions to $a^b=b^a$

let $t=b/a \longrightarrow a^{at}=(at)^a \longrightarrow a^t=at \longrightarrow a^{t-1}=t \longrightarrow a=t^{1/(t-1)}$

We want $n=1/(t-1) \in\mathbb{Z}$ to guarantee $a\in\mathbb{Q}$

This generates infinite rational solutions $a = (1+1/n)^{n}, b= (1+1/n)^{n+1}\forall n\in\mathbb{Z}$

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