{
  "aliases": null,
  "backlinks": null,
  "body": "\n## Proof\n\nWe integrate a finite geometric sum with an explicit remainder, and show that the remainder tends to $0$. This handles $|x| \u003c 1$ and $x = \\pm 1$ at once.\n\n1. Fix an integer $n \\ge 0$ and a real $t$. Put $r = -t^2$. Then $r \\le 0$, so $r \\ne 1$ and $1 - r = 1 + t^2$.\n2. By the [[sum of a finite geometric series|05APWSX53S]], $\\sum_{k=0}^{n} r^k = \\frac{1 - r^{n+1}}{1 + t^2}$. Rearranging gives\n$$\\frac{1}{1 + t^2} = \\sum_{k=0}^{n} (-1)^k t^{2k} + (-1)^{n+1} \\frac{t^{2n+2}}{1 + t^2}.$$\n3. Let $0 \u003c x \\le 1$. The functions $t \\mapsto \\frac{1}{1 + t^2}$, $t \\mapsto t^{2k}$ and $t \\mapsto \\frac{t^{2n+2}}{1 + t^2}$ are continuous on $[0, x]$. This holds since [[polynomials are continuous|0C5TP6RYSV]] and [[quotients of continuous functions are continuous|0GR2APWTAC]] where the denominator is nonzero.\n4. So these functions are Riemann integrable on $[0, x]$, since [[continuous functions are Riemann integrable|06SESV7GQY]].\n5. By the [[derivative of arctangent|05P4B4Q22V]], $\\arctan$ is differentiable on $\\R$ with derivative $\\frac{1}{1 + t^2}$. So it is continuous on $[0, x]$, since [[differentiable functions are continuous|07ZMX4ANJS]].\n6. Since $\\tan 0 = 0$ and $0 \\in (-\\pi/2, \\pi/2)$, the definition of the [[arctangent|065XW2V6C3]] gives $\\arctan 0 = 0$.\n7. By the [[fundamental theorem of calculus|0NHGY0N0JB]] and steps 5 and 6, $\\int_0^x \\frac{dt}{1 + t^2} = \\arctan x - \\arctan 0 = \\arctan x$.\n8. By the [[integral of a power|0Q3HYFT9RQ]], $\\int_0^x t^{2k} \\, dt = \\frac{x^{2k+1}}{2k + 1}$ for each $k \\ge 0$.\n9. Let $S_n(x) = \\sum_{k=0}^{n} \\frac{(-1)^k x^{2k+1}}{2k + 1}$ and $R_n(x) = \\int_0^x \\frac{t^{2n+2}}{1 + t^2} \\, dt$.\n10. Integrate the identity of step 2 over $[0, x]$. By [[linearity of the integral|080RWVVSB6]] and steps 7 and 8, $\\arctan x = S_n(x) + (-1)^{n+1} R_n(x)$.\n11. For $t \\in [0, x]$, we have $0 \\le \\frac{t^{2n+2}}{1 + t^2} \\le t^{2n+2}$, since $1 + t^2 \\ge 1$.\n12. By [[monotonicity of the integral|0MBWQKM8QR]], step 11 and step 8, $0 \\le R_n(x) \\le \\frac{x^{2n+3}}{2n + 3} \\le \\frac{1}{2n + 3}$. The last bound uses $0 \u003c x \\le 1$.\n13. By steps 10 and 12, $|\\arctan x - S_n(x)| \\le \\frac{1}{2n + 3} \u003c \\frac{1}{n}$ for each $n \\ge 1$.\n14. Let $\\varepsilon \u003e 0$. There is $N \\ge 1$ with $\\frac{1}{N} \u003c \\varepsilon$, by the [[Archimedean property|0WH916TJNY]]. For $n \\ge N$, step 13 gives $|\\arctan x - S_n(x)| \u003c \\frac{1}{n} \\le \\frac{1}{N} \u003c \\varepsilon$.\n15. So the partial sums $S_n(x)$ tend to $\\arctan x$. By the definition of the [[sum of a series|0K0H5E3Z5N]], the series converges and its sum is $\\arctan x$, for $0 \u003c x \\le 1$.\n16. For $x = 0$, every term of the series is $0$, and $\\arctan 0 = 0$ by step 6. So the formula holds.\n17. Let $-1 \\le x \u003c 0$, and put $w = -x \\in (0, 1]$ and $y = \\arctan w \\in (-\\pi/2, \\pi/2)$.\n18. The series defining [[sine and cosine|0WB57D9488]] contain only odd and only even powers. So $\\sin(-y) = -\\sin y$ and $\\cos(-y) = \\cos y$.\n19. By step 18 and the definition of the [[tangent|06T0HS14PP]], $\\tan(-y) = -\\tan y = -w = x$. Also $-y \\in (-\\pi/2, \\pi/2)$.\n20. By the uniqueness in the definition of the arctangent, $\\arctan x = -y = -\\arctan w$.\n21. Each term of $S_n$ is an odd power of its argument times a constant. So $S_n(x) = -S_n(w)$ for each $n$.\n22. By step 15, $S_n(w) \\to \\arctan w$. So $S_n(x) = -S_n(w) \\to -\\arctan w = \\arctan x$ by step 20.\n23. Steps 15, 16 and 22 give $\\arctan x = \\sum_{n=0}^{\\infty} \\frac{(-1)^n x^{2n+1}}{2n + 1}$ for every real $x$ with $|x| \\le 1$.\n\n## Idea\n\nThe derivative $\\frac{1}{1 + t^2}$ is a geometric sum in $-t^2$ plus a small remainder. Integrating term by term over $[0, x]$ gives the partial sums of the series, plus an error at most $\\frac{|x|^{2n+3}}{2n+3}$. This error tends to $0$ even at $|x| = 1$, so no separate argument with Abel's theorem is needed.\n",
  "commit": "4a3290d299c8b2cd6493fce2296920ba7f82d942",
  "front": {
    "kind": "proof",
    "msc": [
      "30B10",
      "33B10"
    ],
    "name": "Proof of the power series of arctangent by integrating a finite geometric sum",
    "proves": "0T32ME0A6H",
    "sources": [
      {
        "locator": "Proofs",
        "url": "https://en.wikipedia.org/wiki/Leibniz_formula_for_%CF%80"
      }
    ],
    "tag": "Q5F4QDP9FT",
    "uses": [
      "065XW2V6C3",
      "06T0HS14PP",
      "0WB57D9488",
      "05P4B4Q22V",
      "05APWSX53S",
      "0C5TP6RYSV",
      "0GR2APWTAC",
      "06SESV7GQY",
      "0NHGY0N0JB",
      "07ZMX4ANJS",
      "080RWVVSB6",
      "0Q3HYFT9RQ",
      "0MBWQKM8QR",
      "0WH916TJNY",
      "0K0H5E3Z5N"
    ]
  },
  "kind": "proof",
  "link": "https://math2.tamnd.com/node/Q5F4QDP9FT/",
  "links": [
    {
      "field": "proves",
      "target": "0T32ME0A6H"
    },
    {
      "field": "uses",
      "target": "065XW2V6C3"
    },
    {
      "field": "uses",
      "target": "06T0HS14PP"
    },
    {
      "field": "uses",
      "target": "0WB57D9488"
    },
    {
      "field": "uses",
      "target": "05P4B4Q22V"
    },
    {
      "field": "uses",
      "target": "05APWSX53S"
    },
    {
      "field": "uses",
      "target": "0C5TP6RYSV"
    },
    {
      "field": "uses",
      "target": "0GR2APWTAC"
    },
    {
      "field": "uses",
      "target": "06SESV7GQY"
    },
    {
      "field": "uses",
      "target": "0NHGY0N0JB"
    },
    {
      "field": "uses",
      "target": "07ZMX4ANJS"
    },
    {
      "field": "uses",
      "target": "080RWVVSB6"
    },
    {
      "field": "uses",
      "target": "0Q3HYFT9RQ"
    },
    {
      "field": "uses",
      "target": "0MBWQKM8QR"
    },
    {
      "field": "uses",
      "target": "0WH916TJNY"
    },
    {
      "field": "uses",
      "target": "0K0H5E3Z5N"
    }
  ],
  "markdown": "https://math2.tamnd.com/node/Q5F4QDP9FT.md",
  "msc": [
    "30B10",
    "33B10"
  ],
  "name": "Proof of the power series of arctangent by integrating a finite geometric sum",
  "prov": null,
  "role": "",
  "source": "https://github.com/tamnd/math2/blob/main/nodes/Q5F/Q5F4QDP9FT.md",
  "status": null,
  "tag": "Q5F4QDP9FT",
  "uri": "m2://Q5F4QDP9FT",
  "url": "https://math2.tamnd.com/wiki/proof-of-the-power-series-of-arctangent-by-integrating-a-finite-geometric-sum/",
  "writing": {
    "tag": "Q5F4QDP9FT",
    "kind": "proof",
    "level": "W0",
    "next": [
      {
        "what": "an inventory row",
        "note": "the row of the inventory that the node accounts for",
        "have": 0,
        "want": 1
      }
    ]
  }
}