1. Cauchy Product

Hi all,

Find an example of two power series $\sum a_{n}x^{n}$ and $\sum b_{n}x^{n}$ that have radii of convergence and respectively, such that the cauchy product $\sum c_{n}x^{n}$ ( where $c_{n}=\sum_{i+j=n} a_{i}b_{j}$) has a radius of convergence R:

(a) $R>min(R_{1},R_{2})$
(c) $max(R_{1},R_{2})

2. Originally Posted by BMWM5
Hi all,

Find an example of two power series $\sum a_{n}x^{n}$ and $\sum b_{n}x^{n}$ that have radii of convergence and respectively, such that the cauchy product $\sum c_{n}x^{n}$ ( where $c_{n}=\sum_{i+j=n} a_{i}b_{j}$) has a radius of convergence R:

(a) $R>min(R_{1},R_{2})$
(c) $max(R_{1},R_{2})

For (a), you could take the binomial series for $(1-x)^{1/2}$ and $(1-x)^{-1/2}$. They both have radius of convergence 1. But their product is the series 1+0+0+0+..., which obviously has infinite radius of convergence.

That won't be enough for (c), where you want the radius of convergence of the product to stay finite. But you could possibly adapt the solution to (a), for example by replacing $(1-x)^{1/2}$ by $(1-x)^{1/2}(2-x)^{1/2}$. The power series for that function will be the Cauchy product of the series for $(1-x)^{1/2}$ and $(2-x)^{1/2}$, and will still have radius of convergence 1. But when you multiply it by $(1-x)^{-1/2}$ the product is $(2-x)^{1/2}$, whose power series has radius of convergence 2.