Multiplication of power series

Click For Summary
SUMMARY

The discussion centers on the multiplication of power series, specifically addressing the radius of convergence of the Cauchy product. It is established that the radius of convergence for the product of two power series, \(\Sigma a_n z^n\) and \(\Sigma b_n z^n\), is at least \(\min(R_1, R_2)\), where \(R_1\) and \(R_2\) are the radii of convergence of the individual series. An example using \(a_n = b_n = 2^{-n}\) demonstrates that the radius of convergence remains 2, contradicting the initial claim that it would be \(R_1 * R_2\). The discussion suggests that the question may have intended a different product definition.

PREREQUISITES
  • Understanding of power series and their convergence
  • Familiarity with the Cauchy product of series
  • Knowledge of limits and the \(\limsup\) concept
  • Basic algebraic manipulation of series
NEXT STEPS
  • Study the properties of the Cauchy product of power series
  • Learn about the implications of the \(\limsup\) in determining convergence
  • Explore different definitions of product for power series
  • Investigate examples of power series with varying radii of convergence
USEFUL FOR

Mathematicians, students studying complex analysis, and anyone interested in the properties of power series and their convergence behaviors.

sbashrawi
Messages
49
Reaction score
0

Homework Statement



Suppose that the power series [tex]\Sigma[/tex][a]_{n}[/tex]z^{n} and
\Sigma b_{n} z^{n} havr radii of convergence R! and R2 respictively. Prove that the radius
of convergence of the multiplication is at least R1 * R2

Homework Equations





The Attempt at a Solution



I found that the convergence radius is > min(R1, R2)
 
Physics news on Phys.org
Assuming you are talking about the Cauchy product of power series, namely [tex]\sum c_n z^n[/tex] where [tex]c_n = \sum_{0\leq j\leq n} a_j b_{n-j}[/tex], then you are correct (but the sign should be [tex]\geq[/tex] rather than [tex]>[/tex]), and the problem statement is wrong. For instance, if [tex]a_n = b_n = 2^{-n}[/tex] so that [tex]R_1 = R_2 = 2[/tex], then [tex]c_n = \frac{n + 1}{2^n}[/tex], and [tex]\limsup_{n\to\infty} \left( \frac{n + 1}{2^n} \right)^{1/n} = \frac12[/tex] so that the radius of convergence of [tex]\sum c_n z^n[/tex] is also [tex]2[/tex].

Perhaps the question intended to refer to a different notion of product, such as [tex]\sum d_n z^n[/tex] where [tex]d_n = a_n b_n[/tex]?
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 22 ·
Replies
22
Views
4K
Replies
6
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K