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  "body": "\n## Counterexample\n\nLet $f(x) = e^{-1/x}$ for $x \u003e 0$ and $f(x) = 0$ for $x \\le 0$.\nThen $f$ is [[smooth|0BE3QDH73A]] on $\\mathbb{R}$, and $f^{(n)}(0) = 0$ for all $n$.\nSo its [[Taylor series|0225QPWRRE]] at $0$ is the zero series. But $f(x) \u003e 0$ for $x \u003e 0$, so $f$ is not [[analytic|0JD5ZNW0F4]] at $0$.\n\n## Remarks\n\nFor $x \u003e 0$, each derivative has the form $p_n(1/x) e^{-1/x}$ for a polynomial $p_n$. It tends to $0$ as $x \\to 0^+$, because $e^{-1/x}$ decays faster than any power of $x$.\nSuch functions give smooth bump functions and smooth partitions of unity.\nMathlib defines this function as `expNegInvGlue`.\n\n## Explanations\n\n### L1 human\n\nThis function is so flat at $0$ that all its derivatives vanish there.\nSo its Taylor series sees only the zero function, but the function is not zero.\n",
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