Cauchy-Hadamard Theorem

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Theorem

Complex Case

Let $\xi \in \C$ be a complex number.

Let $\ds \map S z = \sum_{n \mathop = 0}^\infty a_n \paren {z - \xi}^n$ be a (complex) power series about $\xi$.


Then the radius of convergence $R$ of $\map S z$ is given by:

$\ds \dfrac 1 R = \limsup_{n \mathop \to \infty} \cmod {a_n}^{1/n}$

If:

$\ds \limsup_{n \mathop \to \infty} \cmod {a_n}^{1/n} = 0$

then the radius of convergence is infinite, and $\map S z$ is absolutely convergent for all $z \in \C$.


Real Case

Let $\xi \in \R$ be a real number.

Let $\ds \map S x = \sum_{n \mathop = 0}^\infty a_n \paren {x - \xi}^n$ be a power series about $\xi$.


Then the radius of convergence $R$ of $S \paren x$ is given by:

$\ds \frac 1 R = \limsup_{n \mathop \to \infty} \size {a_n}^{1/n}$


If:

$\ds \frac 1 R = \limsup_{n \mathop \to \infty} \size {a_n}^{1/n} = 0$

then the radius of convergence is infinite and therefore the interval of convergence is $\R$.


Source of Name

This entry was named for Augustin Louis Cauchy and Jacques Salomon Hadamard.