{
  "aliases": null,
  "backlinks": null,
  "body": "\n## Proof\n\n1. Fix a real $x$. Let $u_n = \\frac{(-1)^n x^{2n+1}}{(2n+1)!}$ for $n \\ge 0$, where $k!$ is the [[factorial|0WFCGVMRRC]].\n2. If $x = 0$, every $u_n$ is $0$, so $\\sum \\abs{u_n}$ converges.\n3. If $x \\ne 0$, then $u_n \\ne 0$ and $\\frac{\\abs{u_{n+1}}}{\\abs{u_n}} = \\frac{x^2}{(2n+2)(2n+3)} \\to 0 \u003c 1$.\n4. By the [[ratio test|0QNY6ZBXPN]], $\\sum u_n$ converges [[absolutely|0W3K3QPEWT]] in this case too.\n5. By the [[definition of sine|0WB57D9488]], $\\sin x = \\sum_{n=0}^{\\infty} u_n = x - \\frac{x^3}{3!} + \\frac{x^5}{5!} - \\cdots$.\n6. It remains to identify this series as the Maclaurin series. By the [[derivative of sine|00R4MWW532]] and the [[derivative of cosine|0V2DV29EGT]], $\\sin' = \\cos$ and $\\cos' = -\\sin$ on $\\mathbb{R}$.\n7. By induction on $k$, the [[higher derivatives|084NGSTWM7]] exist for all orders: $\\sin^{(4k)} = \\sin$, $\\sin^{(4k+1)} = \\cos$, $\\sin^{(4k+2)} = -\\sin$ and $\\sin^{(4k+3)} = -\\cos$. So $\\sin$ is smooth.\n8. At $0$, every term of the sine series vanishes, and only the first term of the cosine series is nonzero. So $\\sin 0 = 0$ and $\\cos 0 = 1$.\n9. By steps 7 and 8, $\\sin^{(m)}(0) = 0$ for even $m$, $\\sin^{(m)}(0) = 1$ for $m = 4k + 1$, and $\\sin^{(m)}(0) = -1$ for $m = 4k + 3$.\n10. So for $m = 2n + 1$ we get $\\sin^{(m)}(0) = (-1)^n$, since $n$ is even exactly when $m = 4k + 1$.\n11. The [[Maclaurin series|043SJ5RF59]] $\\sum_m \\frac{\\sin^{(m)}(0)}{m!} x^m$ thus has zero terms for even $m$, and the term $\\frac{(-1)^n x^{2n+1}}{(2n+1)!}$ for $m = 2n + 1$.\n12. Removing the zero terms, the Maclaurin series of $\\sin$ is $\\sum_{n=0}^{\\infty} \\frac{(-1)^n x^{2n+1}}{(2n+1)!}$, the series of the statement.\n\n## Idea\n\nThe terms shrink like $\\frac{1}{(2n+1)!}$, which beats any power of $x$, so the ratio test gives absolute convergence. With sine defined by this series, the equality is the definition. The derivatives of $\\sin$ at $0$ cycle through $0, 1, 0, -1$, and these values give back the same coefficients.\n",
  "commit": "4a3290d299c8b2cd6493fce2296920ba7f82d942",
  "front": {
    "kind": "proof",
    "msc": [
      "33B10",
      "30B10"
    ],
    "name": "Proof of the power series of sine",
    "proves": "02KTRHQMHN",
    "sources": [
      {
        "locator": "Series and polynomials",
        "url": "https://en.wikipedia.org/wiki/Sine_and_cosine"
      }
    ],
    "tag": "6M016JBDF0",
    "uses": [
      "0WB57D9488",
      "0QNY6ZBXPN",
      "0W3K3QPEWT",
      "00R4MWW532",
      "0V2DV29EGT",
      "084NGSTWM7",
      "043SJ5RF59",
      "0WFCGVMRRC"
    ]
  },
  "kind": "proof",
  "link": "https://math2.tamnd.com/node/6M016JBDF0/",
  "links": [
    {
      "field": "proves",
      "target": "02KTRHQMHN"
    },
    {
      "field": "uses",
      "target": "0WB57D9488"
    },
    {
      "field": "uses",
      "target": "0QNY6ZBXPN"
    },
    {
      "field": "uses",
      "target": "0W3K3QPEWT"
    },
    {
      "field": "uses",
      "target": "00R4MWW532"
    },
    {
      "field": "uses",
      "target": "0V2DV29EGT"
    },
    {
      "field": "uses",
      "target": "084NGSTWM7"
    },
    {
      "field": "uses",
      "target": "043SJ5RF59"
    },
    {
      "field": "uses",
      "target": "0WFCGVMRRC"
    }
  ],
  "markdown": "https://math2.tamnd.com/node/6M016JBDF0.md",
  "msc": [
    "33B10",
    "30B10"
  ],
  "name": "Proof of the power series of sine",
  "prov": null,
  "role": "",
  "source": "https://github.com/tamnd/math2/blob/main/nodes/6M0/6M016JBDF0.md",
  "status": null,
  "tag": "6M016JBDF0",
  "uri": "m2://6M016JBDF0",
  "url": "https://math2.tamnd.com/wiki/proof-of-the-power-series-of-sine/",
  "writing": {
    "tag": "6M016JBDF0",
    "kind": "proof",
    "level": "W0",
    "next": [
      {
        "what": "an inventory row",
        "note": "the row of the inventory that the node accounts for",
        "have": 0,
        "want": 1
      }
    ]
  }
}