Assume two power series $\sum_{n\ge0}a_n x^n=f_a(x),\sum_{n\ge0}b_n x^n=f_b(x)$ with radii $r_a,r_b$ (respectively) and $r_a\lneqq r_b.$ Consider their Cauchy product $$\sum_{n\ge0}\left(\sum_{k=0}^n a_k b_{n-k}\right)x^n$$ and its radius of convergence $r$. We know that $r\ge r_a$. Now, could you give an example of functions such that $f_a(x)f_b(x)$ has a radius $r_a\lneqq r\lneqq r_b$? And is it possible to have $r_b\lneqq r\lneqq\infty$ (if so, could you give an example again)?


Consider the functions

$$ f(x) = \frac{1}{1-x} \frac{1}{1- x/2}, \qquad g(x) = (1-x),$$

without computing the power series representations it is clear that $f$ has radius of convergence $R_f = 1$, whilst $g$ has radius of convergence $R_g = \infty$.

However, the product

$$h(x) = f(x)g(x) = \frac{1}{1-x/2},$$

has radius of convergence $R_h = 2$, and in particular $R_f < R_h < R_g$.

Now if we consider instead

$$ f(x) = \frac{1}{\sqrt{1-x} } \frac{1}{1-x/2}, \qquad g(x) = \sqrt{1-x},$$

then we now have $R_f = R_g = 1$ and as before defining $h(x) = f(x)g(x)$ we have $R_h = 2$, so that now $R_f = R_g < R_h$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.