{
  "aliases": null,
  "backlinks": null,
  "body": "\n## Proof\n\nWe bound the error of the difference quotient by $\\abs{h}$ times a convergent series.\n\n1. Let $q \\in [0, 1)$. The series $\\sum_{n \\ge 1} n q^{n-1}$ and $\\sum_{n \\ge 2} n (n-1) q^{n-2}$ converge. For $q = 0$ this is clear. For $q \u003e 0$ the ratio of consecutive terms tends to $q \u003c 1$, so the [[ratio test|0QNY6ZBXPN]] applies.\n2. Let $0 \\le s \u003c R$, where $R$ is the [[radius of convergence|045EWAXY9P]]. Choose $t$ with $s \u003c t \u003c R$. Since $R$ is a supremum, some $t' \u003e t$ has $(\\abs{a_n} t'^n)$ bounded, say by $M$. Then $\\abs{a_n} t^n \\le M$ for all $n$.\n3. Then $\\abs{a_n} s^n \\le M (s/t)^n$, $\\ n \\abs{a_n} s^{n-1} \\le \\frac{M}{t} n (s/t)^{n-1}$ and $n(n-1) \\abs{a_n} s^{n-2} \\le \\frac{M}{t^2} n(n-1) (s/t)^{n-2}$.\n4. By step 1 with $q = s/t$ and the [[comparison test|04Z5G0THN6]], the series $\\sum \\abs{a_n} s^n$, $\\sum n \\abs{a_n} s^{n-1}$ and $\\sum n(n-1) \\abs{a_n} s^{n-2}$ converge.\n5. Let $z \\in D$, and take $s = \\abs{z - c}$ in step 4. The series $\\sum a_n (z - c)^n$ and $\\sum n a_n (z - c)^{n-1}$ converge absolutely. Since $\\C$ is complete, [[they converge|082MGNWNA8]]. Put $g(z) = \\sum_{n \\ge 1} n a_n (z - c)^{n-1}$.\n6. The series for $g$ is the [[power series|0VARVQ9ZW1]] $\\sum_{m \\ge 0} b_m (z - c)^m$ with $b_m = (m+1) a_{m+1}$. Let $R'$ be its radius of convergence.\n7. If $0 \u003c s \u003c R$, the terms $\\abs{b_m} s^m = (m+1) \\abs{a_{m+1}} s^m$ are bounded, since their series converges by step 4. So $R' \\ge R$.\n8. If $s \u003e 0$ and $\\abs{b_m} s^m \\le K$ for all $m$, then $\\abs{a_{m+1}} s^{m+1} \\le (m+1) \\abs{a_{m+1}} s^{m+1} \\le K s$ for all $m$. So $(\\abs{a_n} s^n)$ is bounded and $s \\le R$. Hence $R' \\le R$, and $R' = R$.\n9. Let $A, B \\in \\C$ and $n \\ge 1$. Expanding the product and cancelling gives $A^n - B^n = (A - B) \\sum_{k=0}^{n-1} A^k B^{n-1-k}$, as for the [[finite geometric sum|05APWSX53S]].\n10. Assume $\\abs{A}, \\abs{B} \\le s$. By step 9 and the [[triangle inequality|0FJZMBX0ZZ]], $\\abs{A^k - B^k} \\le k s^{k-1} \\abs{A - B}$ for each $k \\ge 1$. This also holds for $k = 0$.\n11. Assume moreover $A \\ne B$ and $n \\ge 1$. By step 9, $\\frac{A^n - B^n}{A - B} - n B^{n-1} = \\sum_{k=0}^{n-1} (A^k - B^k) B^{n-1-k}$.\n12. By step 10, the modulus of this sum is at most $\\sum_{k=0}^{n-1} k s^{k-1} \\abs{A - B} s^{n-1-k} = \\frac{n(n-1)}{2} s^{n-2} \\abs{A - B}$. For $n = 1$ the sum is $0$.\n13. Fix $z \\in D$ and choose $s$ with $\\abs{z - c} \u003c s \u003c R$. Let $h \\ne 0$ with $\\abs{h} \u003c s - \\abs{z - c}$. Put $B = z - c$ and $A = z + h - c$. Then $\\abs{A}, \\abs{B} \u003c s$, $A - B = h$, and $z + h \\in D$.\n14. Subtracting the convergent series termwise, $\\frac{f(z + h) - f(z)}{h} - g(z) = \\sum_{n \\ge 1} a_n \\left( \\frac{A^n - B^n}{h} - n B^{n-1} \\right)$.\n15. By step 12, the modulus of this is at most $\\frac{\\abs{h}}{2} \\sum_{n \\ge 2} n (n-1) \\abs{a_n} s^{n-2} = \\frac{C \\abs{h}}{2}$, where $C$ is finite by step 4.\n16. So the difference quotient tends to $g(z)$ as $h \\to 0$. Thus $f$ is [[complex differentiable|0D1C9Y1EXA]] at $z$ with $f'(z) = \\sum_{n=1}^{\\infty} n a_n (z - c)^{n-1}$.\n17. This holds for each $z$ in the open disc $D$. So $f$ is [[holomorphic|0DG0Z72CQD]] on $D$, and by step 8 the series for $f'$ has radius $R$.\n\n## Idea\n\nInside the disc we can compare with a geometric series, which makes even the series with extra factors $n$ and $n(n-1)$ converge. The error in the difference quotient of $z^n$ is at most about $n^2 s^{n-2} \\abs{h}$, so the total error is a constant times $\\abs{h}$.\n",
  "commit": "68a8c3dc73fd55004b2796431c607c30960d1a85",
  "front": {
    "kind": "proof",
    "msc": [
      "30B10"
    ],
    "name": "Proof that power series are holomorphic in the disc of convergence",
    "proves": "0YMP521ZR5",
    "sources": [
      {
        "locator": "Differentiation and integration",
        "url": "https://en.wikipedia.org/wiki/Power_series"
      }
    ],
    "tag": "VQ7709JWXR",
    "uses": [
      "0VARVQ9ZW1",
      "045EWAXY9P",
      "0DG0Z72CQD",
      "0D1C9Y1EXA",
      "0QNY6ZBXPN",
      "04Z5G0THN6",
      "082MGNWNA8",
      "05APWSX53S",
      "0FJZMBX0ZZ"
    ]
  },
  "kind": "proof",
  "link": "https://math2.tamnd.com/node/VQ7709JWXR/",
  "links": [
    {
      "field": "proves",
      "target": "0YMP521ZR5"
    },
    {
      "field": "uses",
      "target": "0VARVQ9ZW1"
    },
    {
      "field": "uses",
      "target": "045EWAXY9P"
    },
    {
      "field": "uses",
      "target": "0DG0Z72CQD"
    },
    {
      "field": "uses",
      "target": "0D1C9Y1EXA"
    },
    {
      "field": "uses",
      "target": "0QNY6ZBXPN"
    },
    {
      "field": "uses",
      "target": "04Z5G0THN6"
    },
    {
      "field": "uses",
      "target": "082MGNWNA8"
    },
    {
      "field": "uses",
      "target": "05APWSX53S"
    },
    {
      "field": "uses",
      "target": "0FJZMBX0ZZ"
    }
  ],
  "markdown": "https://math2.tamnd.com/node/VQ7709JWXR.md",
  "msc": [
    "30B10"
  ],
  "name": "Proof that power series are holomorphic in the disc of convergence",
  "prov": null,
  "role": "",
  "source": "https://github.com/tamnd/math2/blob/main/nodes/VQ7/VQ7709JWXR.md",
  "status": null,
  "tag": "VQ7709JWXR",
  "uri": "m2://VQ7709JWXR",
  "url": "https://math2.tamnd.com/wiki/proof-that-power-series-are-holomorphic-in-the-disc-of-convergence/",
  "writing": {
    "tag": "VQ7709JWXR",
    "kind": "proof",
    "level": "W0",
    "next": [
      {
        "what": "an inventory row",
        "note": "the row of the inventory that the node accounts for",
        "have": 0,
        "want": 1
      }
    ]
  }
}