{
  "aliases": null,
  "backlinks": null,
  "body": "\n## Proof\n\nWe show that $\\cos x \u003e 0$ for $0 \\le x \\le 3/2$. Then the zero $\\pi/2$ of $\\cos$ must lie beyond $3/2$.\n\n1. Let $0 \u003c x \\le 3/2$, and let $u = x^2$. Then $0 \u003c u \\le 9/4$.\n2. By the [[definition of cosine|0WB57D9488]], $\\cos x = \\sum_{n=0}^{\\infty} v_n$ with $v_n = \\frac{(-1)^n x^{2n}}{(2n)!}$. The series converges by the [[power series of cosine|08W5RN375X]].\n3. The partial sums of the tail $\\sum_{n \\ge 2} v_n$ are those of $\\sum v_n$ minus $v_0 + v_1$. By the [[limit of a sum|0RAM4E07NR]], the tail converges and $\\cos x = 1 - \\frac{u}{2} + T$, where $T = \\sum_{n \\ge 2} v_n$.\n4. Write $T = \\sum_{k=0}^{\\infty} (-1)^k c_k$ with $c_k = \\frac{x^{2k+4}}{(2k+4)!} \u003e 0$. Here $c_k = \\abs{v_{k+2}}$ and $(-1)^{k+2} = (-1)^k$.\n5. We have $\\frac{c_{k+1}}{c_k} = \\frac{u}{(2k+5)(2k+6)} \\le \\frac{9/4}{30} \u003c 1$. So $(c_k)$ is decreasing.\n6. Since $\\sum v_n$ converges, the [[term test|0VVY4H92G2]] gives $v_n \\to 0$. So $c_k \\to 0$.\n7. By the [[alternating series test|0PK8QH1W1V]] with $N = 0$, $\\abs{T - c_0} \\le c_1$. So $T \\ge c_0 - c_1 = \\frac{u^2}{24} - \\frac{u^3}{720}$.\n8. By steps 3 and 7, $\\cos x \\ge p(u)$, where $p(u) = 1 - \\frac{u}{2} + \\frac{u^2}{24} - \\frac{u^3}{720}$.\n9. By the [[derivative of a polynomial|0MH1WK89PP]], $p'(u) = -\\frac{1}{2} + \\frac{u}{12} - \\frac{u^2}{240}$. For $0 \\le u \\le 9/4$ this is at most $-\\frac{1}{2} + \\frac{3}{16} \u003c 0$.\n10. So $-p$ has positive derivative on $[0, 9/4]$. By [[this result|07J5X2T1ZX]], $-p$ is strictly increasing there, so $p$ is strictly decreasing.\n11. Hence $p(u) \\ge p(9/4)$. Over the common denominator $46080$, $p(9/4) = \\frac{46080 - 51840 + 9720 - 729}{46080} = \\frac{3231}{46080} \u003e 0$.\n12. So $\\cos x \u003e 0$ for $0 \u003c x \\le 3/2$. Also $\\cos 0 = 1 \u003e 0$, since only the first term of the series is nonzero at $0$.\n13. By the [[definition of $\\pi$|0C1VYVNV2N]], $\\cos(\\pi/2) = 0$ and $\\pi/2 \\in [0, 2]$. By step 12, $\\pi/2 \\notin [0, 3/2]$.\n14. So $\\pi/2 \u003e 3/2$, that is, $\\pi \u003e 3$.\n\n## Idea\n\nThe cosine series alternates, and from the term $x^4/4!$ on its terms decrease when $x \\le 3/2$. So cutting the series after $x^6/6!$ gives a lower bound for $\\cos x$. This lower bound stays positive up to $x = 3/2$, so the first zero $\\pi/2$ of cosine comes later.\n",
  "commit": "4a3290d299c8b2cd6493fce2296920ba7f82d942",
  "front": {
    "kind": "proof",
    "msc": [
      "11Y60"
    ],
    "name": "Proof that pi is greater than 3 by the alternating series estimate",
    "proves": "0R7SVNYTJS",
    "sources": [
      {
        "locator": "Alternating series estimation theorem",
        "url": "https://en.wikipedia.org/wiki/Alternating_series_test"
      }
    ],
    "tag": "YPPK15N2PC",
    "uses": [
      "0C1VYVNV2N",
      "0WB57D9488",
      "08W5RN375X",
      "0RAM4E07NR",
      "0VVY4H92G2",
      "0PK8QH1W1V",
      "0MH1WK89PP",
      "07J5X2T1ZX"
    ]
  },
  "kind": "proof",
  "link": "https://math2.tamnd.com/node/YPPK15N2PC/",
  "links": [
    {
      "field": "proves",
      "target": "0R7SVNYTJS"
    },
    {
      "field": "uses",
      "target": "0C1VYVNV2N"
    },
    {
      "field": "uses",
      "target": "0WB57D9488"
    },
    {
      "field": "uses",
      "target": "08W5RN375X"
    },
    {
      "field": "uses",
      "target": "0RAM4E07NR"
    },
    {
      "field": "uses",
      "target": "0VVY4H92G2"
    },
    {
      "field": "uses",
      "target": "0PK8QH1W1V"
    },
    {
      "field": "uses",
      "target": "0MH1WK89PP"
    },
    {
      "field": "uses",
      "target": "07J5X2T1ZX"
    }
  ],
  "markdown": "https://math2.tamnd.com/node/YPPK15N2PC.md",
  "msc": [
    "11Y60"
  ],
  "name": "Proof that pi is greater than 3 by the alternating series estimate",
  "prov": null,
  "role": "",
  "source": "https://github.com/tamnd/math2/blob/main/nodes/YPP/YPPK15N2PC.md",
  "status": null,
  "tag": "YPPK15N2PC",
  "uri": "m2://YPPK15N2PC",
  "url": "https://math2.tamnd.com/wiki/proof-that-pi-is-greater-than-3-by-the-alternating-series-estimate/",
  "writing": {
    "tag": "YPPK15N2PC",
    "kind": "proof",
    "level": "W0",
    "next": [
      {
        "what": "an inventory row",
        "note": "the row of the inventory that the node accounts for",
        "have": 0,
        "want": 1
      }
    ]
  }
}