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  "body": "\n## Definition\n\nLet $U \\subseteq \\C$ be a [[domain|097KNJFJ7V]] and let $f$ be [[holomorphic|0DG0Z72CQD]] on $U$.\nA point $\\zeta$ of the [[boundary|0JJ9J1XZ8G]] $\\partial U$ is a *regular point* of $f$ if the following holds.\nThere are an open disc $B$ centered at $\\zeta$ and a holomorphic function $g$ on $B$ such that $g = f$ on some connected component of $B \\cap U$.\nThat component must have $\\zeta$ in its closure.\nThe boundary $\\partial U$ is a *natural boundary* of $f$ if no point of $\\partial U$ is a regular point of $f$.\n\n## Remarks\n\nThe most common case is a power series with radius of convergence $R$, with $U$ its open disc of convergence.\nOn the circle of convergence there is always at least one point that is not regular.\nExample: $f(z) = \\sum_{n \\ge 0} z^{2^n}$ on the unit disc has the unit circle as natural boundary.\nIf $\\zeta^{2^k} = 1$, then $f(r\\zeta) = \\sum_{n \u003c k} (r\\zeta)^{2^n} + \\sum_{n \\ge k} r^{2^n} \\to \\infty$ as $r \\to 1^-$.\nSuch $\\zeta$ are dense in the circle, while $f$ would be bounded near a regular point.\nNon-example: $\\frac{1}{1 - z} = \\sum_{n \\ge 0} z^n$ does not have the unit circle as natural boundary.\nEach point $\\zeta \\neq 1$ with $\\abs{\\zeta} = 1$ is regular, since $\\frac{1}{1 - z}$ is holomorphic on a disc around $\\zeta$.\n",
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