Cauchy–Hadamard theorem

Template:Short description In mathematics, the Cauchy–Hadamard theorem is a result in complex analysis named after the French mathematicians Augustin Louis Cauchy and Jacques Hadamard, describing the radius of convergence of a power series. It was published in 1821 by Cauchy, but remained relatively unknown until Hadamard rediscovered it. Hadamard's first publication of this result was in 1888; he also included it as part of his 1892 Ph.D. thesis.

Theorem for one complex variable

Consider the formal power series in one complex variable z of the form $$f(z)=\sum _{n=0}^{\mathrm{\infty }}{c}_n(z-a{)}^n$$ where $a,c_n\in ℂ.$

Then the radius of convergence $R$ of f at the point a is given by $$\frac{1}{R}=\underset{n\to \mathrm{\infty }}{\mathrm{lim\; sup}}(|c_n{|}^{1/n})$$ where Template:Math denotes the limit superior, the limit as Template:Mvar approaches infinity of the supremum of the sequence values after the nth position. If the sequence values is unbounded so that the Template:Math is ∞, then the power series does not converge near Template:Math, while if the Template:Math is 0 then the radius of convergence is ∞, meaning that the series converges on the entire plane.

Proof

Without loss of generality assume that $a=0$. We will show first that the power series $\sum _n{c}_n{z}^n$ converges for $|z|<R$, and then that it diverges for $|z|>R$.

First suppose $|z|<R$. Let $t=1/R$ not be $0$ or $\pm \mathrm{\infty }.$ For any $\epsilon >0$, there exists only a finite number of $n$ such that $\sqrt[n]{|c_n|}\ge t+\epsilon$. Now $|c_n|\le (t+\epsilon {)}^n$ for all but a finite number of $c_n$, so the series $\sum _n{c}_n{z}^n$ converges if $|z|<1/(t+\epsilon )$. This proves the first part.

Conversely, for $\epsilon >0$, $|c_n|\ge (t-\epsilon {)}^n$ for infinitely many $c_n$, so if $|z|=1/(t-\epsilon )>R$, we see that the series cannot converge because its nth term does not tend to 0.

Theorem for several complex variables

Let $\alpha$ be an n-dimensional vector of natural numbers ($\alpha =({\alpha }_1,\cdots ,{\alpha }_n)\in {\mathbb{N}}^n$) with $\Vert \alpha \Vert :={\alpha }_1+\cdots +{\alpha }_n$, then $f(z)$ converges with radius of convergence $\rho =({\rho }_1,\cdots ,{\rho }_n)\in {ℝ}^n$, ${\rho }^{\alpha }={\rho }_1^{{\alpha }_1}\cdots {\rho }_n^{{\alpha }_n}$ if and only if $$\underset{\Vert \alpha \Vert \to \mathrm{\infty }}{\mathrm{lim\; sup}}\sqrt[\Vert \alpha \Vert ]{|c_{\alpha }|{\rho }^{\alpha }}=1$$ of the multidimensional power series $$f(z)=\sum _{\alpha \ge 0}{c}_{\alpha }(z-a{)}^{\alpha }:=\sum _{{\alpha }_1\ge 0,\dots ,{\alpha }_n\ge 0}{c}_{{\alpha }_1,\dots ,{\alpha }_n}(z_1-a_1{)}^{{\alpha }_1}\cdots (z_n-a_n{)}^{{\alpha }_n}.$$

Proof

From

Set $z=a+t\rho$ $(z_i=a_i+t{\rho }_i).$ Then

$\sum _{\alpha \ge 0}{c}_{\alpha }(z-a{)}^{\alpha }=\sum _{\alpha \ge 0}{c}_{\alpha }{\rho }^{\alpha }{t}^{\Vert \alpha \Vert }=\sum _{\mu \ge 0}(\sum _{\Vert \alpha \Vert =\mu }|c_{\alpha }|{\rho }^{\alpha })t^{\mu }.$

This is a power series in one variable $t$ which converges for $|t|<1$ and diverges for $|t|>1$. Therefore, by the Cauchy–Hadamard theorem for one variable

$\underset{\mu \to \mathrm{\infty }}{\mathrm{lim\; sup}}\sqrt[\mu ]{\sum _{\Vert \alpha \Vert =\mu }|c_{\alpha }|{\rho }^{\alpha }}=1.$

Setting $|c_m|{\rho }^m=\underset{\Vert \alpha \Vert =\mu }{\max }|c_{\alpha }|{\rho }^{\alpha }$ gives us an estimate

$|c_m|{\rho }^m\le \sum _{\Vert \alpha \Vert =\mu }|c_{\alpha }|{\rho }^{\alpha }\le (\mu +1{)}^n|c_m|{\rho }^m.$

Because $\sqrt[\mu ]{(\mu +1{)}^n}\to 1$ as $\mu \to \mathrm{\infty }$

$\sqrt[\mu ]{|c_m|{\rho }^m}\le \sqrt[\mu ]{\sum _{\Vert \alpha \Vert =\mu }|c_{\alpha }|{\rho }^{\alpha }}\le \sqrt[\mu ]{|c_m|{\rho }^m}⟹\sqrt[\mu ]{\sum _{\Vert \alpha \Vert =\mu }|c_{\alpha }|{\rho }^{\alpha }}=\sqrt[\mu ]{|c_m|{\rho }^m}(\mu \to \mathrm{\infty }).$

Therefore

$\underset{\Vert \alpha \Vert \to \mathrm{\infty }}{\mathrm{lim\; sup}}\sqrt[\Vert \alpha \Vert ]{|c_{\alpha }|{\rho }^{\alpha }}=\underset{\mu \to \mathrm{\infty }}{\mathrm{lim\; sup}}\sqrt[\mu ]{|c_m|{\rho }^m}=1.$

External links

• Template:MathWorld