Cauchy-Hadamard Theorem/Real Case
Jump to navigation
Jump to search
Theorem
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 $\R$.
Proof
From the $n$th root test, $S \paren x$ is convergent if $\ds \limsup_{n \mathop \to \infty} \size {a_n \paren {x - \xi}^n}^{1/n} < 1$.
Thus:
\(\ds \size {a_n \paren {x - \xi}^n}^{1/n}\) | \(<\) | \(\ds 1\) | ||||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \size {a_n}^{1/n} \size {x - \xi}\) | \(<\) | \(\ds 1\) | |||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \size {a_n}^{1/n}\) | \(<\) | \(\ds \frac 1 {\size {x - \xi} }\) |
The result follows from the definition of radius of convergence.
$\blacksquare$
Source of Name
This entry was named for Augustin Louis Cauchy and Jacques Salomon Hadamard.
Sources
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 15.2$