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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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2  
Newtons method? –  picakhu Dec 21 '12 at 20:15
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I think only numerical solution. –  Patrick Li Dec 21 '12 at 20:23
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No, LambertW says nothing about this. I doubt very much that there is a closed-form solution. –  Robert Israel Dec 21 '12 at 20:39
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$x^x = x + 2$ is easier. –  Will Jagy Dec 21 '12 at 20:54
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See 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. –  Yury Jan 11 '13 at 2:24

2 Answers 2

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. –  AmirHosein SadeghiManesh Jan 7 '13 at 8:21
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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. –  hombre Jan 7 '13 at 22:41

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 $$\upsilon = 1.77677504009705$$

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