Cauchy–Hadamard theorem

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
$f(z) = \sum_^ c_ (z-a)^$
where $a, c_n \in \Complex.$

Then the radius of convergence $R$ of ''f'' at the point ''a'' is given by
$\frac = \limsup_ \left( , c_ , ^ \right)$
where denotes the limit superior, the limit as approaches infinity of the supremum of the sequence values after the ''n''th position. If the sequence values is unbounded so that the is ∞, then the power series does not converge near , while if the 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, , and then that it diverges for $, z, >R$.

First suppose , z, . Let $t=1/R$ not be $0$ or $\pm\infty.$
For any $\varepsilon > 0$, there exists only a finite number of $n$ such that \sqrt \geq t+\varepsilon.
Now $, c_n, \leq (t+\varepsilon)^n$ for all but a finite number of $c_n$, so the series $\sum_n c_n z^n$ converges if $, z, < 1/(t+\varepsilon)$. This proves the first part.

Conversely, for $\varepsilon > 0$, $, c_n, \geq (t-\varepsilon)^n$ for infinitely many $c_n$, so if $, z, =1/(t-\varepsilon) > R$, we see that the series cannot converge because its ''n''th term does not tend to 0.

Theorem for several complex variables

Let $\alpha$ be an ''n''-dimensional vector of natural numbers ($\alpha = (\alpha_1, \cdots, \alpha_n) \in \N^n$) with $\, \alpha\, := \alpha_1 + \cdots + \alpha_n$, then $f(z)$ converges with radius of convergence $\rho = (\rho_1, \cdots, \rho_n) \in \R^n$, $\rho^\alpha = \rho_1^ \cdots \rho_n^$ if and only if
\limsup_ \sqrt \alpha\, 1
of the multidimensional power series
$f(z) = \sum_c_\alpha(z-a)^\alpha := \sum_c_(z_1-a_1)^\cdots(z_n-a_n)^.$

Proof

From

Set $z = a + t\rho$ $(z_i = a_i + t\rho_i).$ Then

:$\sum_ c_\alpha (z - a)^\alpha = \sum_ c_\alpha \rho^\alpha t^ = \sum_ \left( \sum_ , c_\alpha, \rho^\alpha \right) t^\mu.$

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

:\limsup_ \sqrt mu= 1.

Setting $, c_m, \rho^m = \max_ , c_\alpha, \rho^\alpha$ gives us an estimate

:$, c_m, \rho^m \leq \sum_ , c_\alpha, \rho^\alpha \leq (\mu + 1)^n , c_m, \rho^m.$

Because \sqrt mu\to 1 as $\mu \to \infty$

:\sqrt mu\leq \sqrt mu\leq \sqrt mu\implies \sqrt mu= \sqrt mu\qquad (\mu \to \infty).

Therefore

:\limsup_ \sqrt \alpha\, = \limsup_ \sqrt mu= 1.

Notes

External links

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{{DEFAULTSORT:Cauchy-Hadamard theorem
Augustin-Louis Cauchy
Series (mathematics)
Theorems in complex analysis