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Version 4 |
| Without loss of generality, it may be assumed that $z_0 = 0$. |
Without loss of generality, it may be assumed that $z_0 = 0$. |
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| Let $c_n$ denote the coefficient of the $n$-th term in the Taylor series of $f$ about $0$. Let $r$ be a real number such that $0 < r < R$. Then $c_n$ may be expressed as an integral using Cauchy's derivative formula. |
Let $c_n$ denote the coefficient of the $n$-th term in the Taylor series of $f$ about $0$. Let $r$ be a real number such that $0 < r < R$. Then $c_n$ may be expressed as an integral using Cauchy's derivative formula. |
| $$c_n = {1 \over 2 \pi i} \oint_{|z| = r} {f(z) \over z^{n+1}} \, dz = {1 \over 2 \pi r^n} \int_{-\pi}^{+\pi} e^{-n \theta} f(r e^{i \theta}) \, d \theta$$ |
$$c_n = {1 \over 2 \pi i} \oint_{|z| = r} {f(z) \over z^{n+1}} \, dz = {1 \over 2 \pi r^n} \int_{-\pi}^{+\pi} e^{-n \theta} f(r e^{i \theta}) \, d \theta$$ |
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| Since $f$ is analytic, it is also continuous. Since a continuous function on a compact set is bounded, $|f| < B$ for some constant $B > 0$ on the circle $|z| = r$. Hence, we have |
Since $f$ is analytic, it is also continuous. Since a continuous function on a compact set is bounded, $|f| < B$ for some constant $B > 0$ on the circle $|z| = r$. Hence, we have |
| $$|c_n| = {1 \over 2 \pi r^n} \left| \int_{-\pi}^{+\pi} e^{-n \theta} f(r e^{i \theta}) \, d \theta \right| \le {1 \over 2 \pi r^n} \int_{-\pi}^{+\pi} | e^{-n \theta} f(r e^{i \theta}) | \, d \theta \le {1 \over 2 \pi r^n} \int_{-\pi}^{+\pi} B d \theta = {B \over r^n}$$ |
$$|c_n| = {1 \over 2 \pi r^n} \left| \int_{-\pi}^{+\pi} e^{-n \theta} f(r e^{i \theta}) \, d \theta \right| \le {1 \over 2 \pi r^n} \int_{-\pi}^{+\pi} | e^{-n \theta} f(r e^{i \theta}) | \, d \theta \le {1 \over 2 \pi r^n} \int_{-\pi}^{+\pi} B d \theta = {B \over r^n}$$ |
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Consequently, $\sqrt[n]{c_n} \le \sqrt[n]{B} / r$. Since $\lim_{n \to \infty} \sqrt[n]{B} = 1$, the radius of convergence must be greater than or equal to $r$. Since this is true for all $r < R$, it follows that the radius of convergence is greater than or equal to $R$.
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Consequently, $\sqrt[n]{c_n} \le \sqrt[n]{B} / r$. Since $\lim_{n \to \infty} \sqrt[n]{B} = 1$, the radius of convergence must be greater than or equal to $r$. Since this is true for all $r < R$, it follows that the radius of convergence is grater than or equal to $R$.
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