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  "body": "\n## Definition\n\nLet $f \\colon \\R \\to \\R$ be $f(w) = w e^w$, where $e^w$ is the [[exponential function|0J0FCWV32D]].\n1. The *principal branch* $W_0 \\colon [-1/e, \\infty) \\to [-1, \\infty)$ is the [[inverse|0J2WAR83WP]] of the restriction of $f$ to $[-1, \\infty)$.\n2. The *lower branch* $W_{-1} \\colon [-1/e, 0) \\to (-\\infty, -1]$ is the inverse of the restriction of $f$ to $(-\\infty, -1]$.\n\nThus $W(x)\\, e^{W(x)} = x$ for both branches $W$ on their domains.\n\n## Remarks\n\nThe two restrictions are bijections because $f'(w) = (1 + w) e^w$ is positive for $w \u003e -1$ and negative for $w \u003c -1$, $f(-1) = -1/e$, $f(w) \\to 0^-$ as $w \\to -\\infty$, and $f(w) \\to \\infty$ as $w \\to \\infty$.\n\nExample: $W_0(e) = 1$, because $1 \\cdot e^1 = e$ and $1 \\ge -1$.\nAlso $W_0(0) = 0$, $W_0(-1/e) = W_{-1}(-1/e) = -1$, and $W_0(1) = \\Omega \\approx 0.5671$, the omega constant.\n\nNon-example: $x = -1$ is in the domain of neither branch, because $w e^w \\ge -1/e \u003e -1$ for every real $w$.\n\nOver $\\C$, the equation $w e^w = z$ has infinitely many solutions for $z \\ne 0$, and they are grouped into branches $W_k$, $k \\in \\Z$, which extend $W_0$ and $W_{-1}$.\nImplicit differentiation gives $W'(x) = \\frac{W(x)}{x\\,(1 + W(x))}$ for $x \\ne 0, -1/e$.\n",
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