How to solve $\upsilon^\upsilon=\upsilon+1$ 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.
 A: 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.
A: 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
$$\upsilon =  1.77677504009705$$
