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Let $f(z) = \sum a_n z^n$ be a power series with radius of convergence $R$. How do we show that $f$ is analytic in the circular region of radius $R$?

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    yeah I'm studying it ... I have proof in my book. I don't understand how it relates R so that |(f(z+h) - f(z))/h - f'(z))| = 0. Besides book is old ... I was looking for it somewhere online or ... on any ebook.2012-08-26

3 Answers 3

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Lemma Let $R$ be the radius of convergence of $f(z) = \sum_{n=0}^{\infty} a_nz^n$. Let $g(z) = \sum_{n=1}^{\infty} na_nz^{n-1}$.

Then the radius of convergence of $g(z)$ is $R$.

Proof: Let $R'$ be radius of convergence of $g(z)$. Since $|a_nz^n| \le |na_nz^n|$, $R' \le R$. So it suffices to prove $R \le R'$.

Let $z$ be such that $|z| < R$. Choose $r$ such that $|z| < r < R$. Let $\rho = \frac{|z|}{r}$.

Since $\sum_{n=1}^{\infty} a_nr^{n-1}$ converges, there exists $M > 0$ such that $|a_nr^{n-1}| \le M$ for all $n \ge 1$.

Then

$|na_nz^{n-1}| = n|a_n|r^{n-1}\rho^{n-1} \le nM\rho^{n-1}$

Since $0 \le \rho < 1$, $\sum_{n=1}^{\infty} nM \rho^{n-1}$ converges. Hence $\sum_{n=1}^{\infty} na_nz^{n-1}$ converges. Hence $R \le R'$. QED

Proposition Let $R$ be the radius of convergence of $f(z) = \sum_{n=0}^{\infty} a_nz^n$. Let $g(z) = \sum_{n=1}^{\infty} na_nz^{n-1}$.

Then $f'(z) = g(z)$ for every $|z| < R$.

Proof: Let $z$ be such that $|z| < R$. Choose $r$ such that $|z| < r < R$. Let $h$ be such that $0 < |h| \le r - |z|$

Consider $\frac{f(z+h) - f(z)}{h} = \sum_{n=0}^{\infty} a_n\frac{(z + h)^n - z^n}{h}$

By the formula $x^n - y^n = (x - y)(x^{n-1} + yx^{n-2} +\cdots+ y^{n-2}x + y^{n-1})$,

$|a_n\frac{(z + h)^n - z^n}{h}| = |a_n||(z + h)^{n-1} + z(z+h)^{n-2}+\cdots+ z^{n-2}(z+h) + z^{n-1}) \le n|a_n|r^{n-1}$

Define $\psi_n(h) = a_n\frac{(z + h)^n - z^n}{h}$ for $0 < |h| \le r - |z|$

Define $\psi_n(0) = n a_n z^{n-1}$.

Then $\psi_n(h)$ is continuous in $|h| \le r - |z|$. Since $\sum_{n=1}^{\infty} n|a_n|r^{n-1}$ converges by the lemma, $\Psi(h) = \sum_{n=1}^{\infty} \psi_n(h)$ converges uniformly in $|h| \le r - |z|$.

Hence $\Psi(h)$ is continuous in $|h| \le r - |z|$. In particular $\lim_{h \to 0} \Psi(h) = \Psi(0)$.

This implies $f'(z) = g(z)$.

QED

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I will show another proof.

Proposition Let $R$ be the radius of convergence of $f(z) = \sum_{n=0}^{\infty} a_nz^n$. Let $g(z) = \sum_{n=1}^{\infty} na_nz^{n-1}$.

Then $f'(z) = g(z)$ for every $|z| < R$.

Proof: By the lemma of my previous answer, the radius of convergence of $g(z)$ is $R$. Let $z$ be such that $|z| < R$. Since $\sum_{n=1}^{\infty} na_nz^{n-1}$ converges uniformly in every compact subset of $\{z; |z| < R\}$,

$\int_{0}^{z} g(\zeta) d\zeta = \sum_{n=1}^{\infty} \int_{0}^{z} na_n\zeta^{n-1} d\zeta = \sum_{n=1}^{\infty} a_nz^n = f(z) - a_0$,

where the integral path is any smooth curve inside the domain $\{z; |z| < R\}$ starting from $0$ to $z$.

Hence $f'(z) = g(z)$.

QED

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    @DonAntonio "It's neither polite nor used to write more than 1 answer to the same question" Why it's not polite to write more than one answer to the same question?2012-08-27
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I think all you need is

Theorem: A power series converges uniformly inside its convergence interval

Proof: Follows at once from Weierstrass M-test: let $\,r>0\,$ be the convergence radius of the power series $\sum_{n=1}^\infty a_n(z-a)^n$ and let $\,0<\rho, then: $|z-a|<\rho\Longrightarrow |a_n(z-a)^n|\leq |a_n|\rho^n=:M_n$ and since the series $\,\displaystyle{\sum_{m=1}^\infty M_n}\,$ converges (absolutely, of course) , we're done.

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    1) A power series converges *always* for values within its interval of convergence. 2) The series converges on the compact set $\,[\rho-\epsilon,\rho+\epsilon]\,$ , with $\,\epsilon\,$, small enough as to have \,\rho+\epsilon2012-08-26