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What is the real positive $\upsilon$ that satisfies $\upsilon^\upsilon=\upsilon+1$? I think the Lambert-W function might be relevant here, but I have no idea how to use it.

$\upsilon\approx 1.775678$

I just really like the letter upsilon. It doesn't get enough love.

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    Fyi, Mathematica 9 has no built-in methods to solve it symbolically.2012-12-21
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    Newtons method?2012-12-21
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    I think only numerical solution.2012-12-21
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    No, LambertW says nothing about this. I doubt very much that there is a closed-form solution.2012-12-21
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    $x^x = x + 2$ is easier.2012-12-21
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    @Will Jagy: $x^x=x+1$ is also not difficult. Try $x=0$...2012-12-21
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    yeah, but I meant the positive one.2012-12-21
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    That's only if you define $0^0$ to be $1$. Sometimes it is defined to be indeterminate.2012-12-21
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    In this case, since $\lim_{x \to 0} x^x = 1$, I think we can safely define $0^0 = 1$ for this one thing.2013-01-05
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    It is possible to write this in terms of Lambert W, but it doesn't help in solving the problem at all. $ \ln \upsilon = \mathrm{W}(\ln(\upsilon + 1)) $2013-01-06
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    @hombre How do you mean "closed form" solution?2013-01-06
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    Heh, good question, vesszabo. This is a bit vague, but I was hoping for a solution in terms of functions that we already know. For example, the error function, Lambert-W, all of that good stuff. I think it might have to do with Galois Groups or something, whether it can be written in terms of functions we already know, or it has to be done in terms of something else. I'm not sure.2013-01-06
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    The inverse symbolic calculator gives nothing useful.2013-01-07
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    See http://oeis.org/A124930 . According to this web page, this number is transcendental. Most likely, there is no closed form expression for it in terms of elementary functions.2013-01-11

2 Answers 2

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I used a very simple method, "iterative fixed point method", look that $x^{x}=x+1$ is equivalent to $x^{x}-x-1=0$ and $-2x=x^{x}-3x-1$ and $x=\frac{x^{x}-3x-1}{-2}$. Define $g(x)=\frac{x^{x}-3x-1}{-2}$ so the answer of equation $x^{x}=x+1$ is the fixed point of the function $g(x)$. For finding it we note that because $g'(x)=\frac{(1+ln(x))x^{x}-3}{-2}$ is decreasing in interval $[1.65,1.9]$ and $g'(1.65)=-0.2124....$ and $g'(1.9)=-0.667....$ so $g'(x)$ is negative in interval $[1.65,1.9]$ and $g(x)$ is decreasing in this interval and as $g(1.65)=1.83....$ and $g(1.9)=1.65....$ we have $$\forall x\in [1.65,1.9] \; :\;g(x)\in [1.65,1.9]$$ and also we have seen that $$\forall x\in [1.65,1.9] \; : \; |g'(x)|\leq 0.67<1$$ so by a theorem in Numerical Analysis for iterative fixed point method the sequence $\{g(x_{n})\}_{n=1}^{\infty}$ is convergence to requested point, if we choose $x_{1}\in [1.65,1.9]$. But if you want something else like only using "Lambert W function" and such things, please say me.

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    @hombre, Perhaps You don't want numerical solutions and looking to this number as a limit of a sequence, from above discussions it is brightly that you want to earn this number as some form of some well-known functions.2013-01-07
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    Yeah, you're right. Thank you; I appreciate your answer, but I was hoping for a solution that wasn't a recursive sequence or an infinite sum or anything like that.2013-01-07
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If you plan to use newton's method, it helps to simplify first: Take $\log$ on both sides to get $$ \upsilon \log(\upsilon) = \log(\upsilon+1)$$

Let $$f(\upsilon) = \upsilon \log(\upsilon) - \log(\upsilon+1)$$ Then $$ f'(\upsilon) = \log(\upsilon) - \frac{\upsilon}{\upsilon+1}$$ The Newton's method gives

$$\upsilon_{n+1} = \upsilon_n - \frac{f(\upsilon)}{f'(\upsilon)}$$

Starting with $$\upsilon_0 =1$$ one gets the solution to machine precision in 5 steps as $$\upsilon = 1.77677504009705$$