2
$\begingroup$

I have the following product of two Taylor series:

$$f(x)g(x)=\frac{1}{z-1}\frac{1}{z-2}=\sum_{n=0}^{\infty} z^n \sum_{n=0}^{\infty} \frac{1}{2^{n+1}} z^n$$

I wanted to know 2 things:

1st. How can I simplify this product of series to get only one summation?
2nd. The radius of convergence of the each of the series is different (that of $f(x)$ is $1$ and that of $g(x)$ is $2$). When I have a product of series, like in this case, the radius of convergence is the smallest one? (in this case $1$)

Thanks.

$\endgroup$
  • 2
    $\begingroup$ The $x$ turned into $z$. Look at Cauchy product. $\endgroup$ – Git Gud Dec 30 '13 at 15:44
  • $\begingroup$ But Cauchy's product still leaves the final result in terms of a product of series. $\endgroup$ – Alejandro Dec 30 '13 at 15:45
  • $\begingroup$ But the inner most one is finite. $\endgroup$ – Git Gud Dec 30 '13 at 15:46
  • $\begingroup$ I see. So the final series will be: $\sum_{n=0}^{\infty} \left(1-\frac{1}{2^{n+1}}\right)z^n$. Thank you Git Gud. Can you help me with the 2nd question? $\endgroup$ – Alejandro Dec 30 '13 at 15:50
  • $\begingroup$ $\frac{1}{(x-1)(x-2)}=\frac{1}{x-2}-\frac{1}{x-1}$. $\endgroup$ – André Nicolas Dec 30 '13 at 15:52
4
$\begingroup$

You can try: $$ f(x)g(x)=\frac{1}{z-1}\frac{1}{z-2}=\frac{A}{z-1}+\frac{B}{z-2} $$ $A$ and $B$ are two variables you have to determine: $$ \frac{A(z-2) + B(z-1)}{(z-1)(z-2)} = \frac{1}{(z-1)(z-2)} $$ So $(A+B)z -2A -B = 1$, and this gives $A+B = 0$ and $2A+B = -1$. You can work out the rest. This way you will obtain a sum of the two series and not a product.

As for the region of convergence, it is always (at least) the intersection of the two original regions.

$\endgroup$
  • $\begingroup$ I see. So the region of convergence of a sum, product of quotient of two series is always the intersection of the two original regions? $\endgroup$ – Alejandro Dec 30 '13 at 15:56
  • $\begingroup$ Not exactly the intersection, but at least the intersection. Sometimes it's exactly the intersection, sometimes it is more. For example, if you sum $\frac{1}{1-z} + \frac{1}{z-1}$ the result is 0 and the region of convergence is all the complex plane. $\endgroup$ – gerd Dec 30 '13 at 15:58
  • $\begingroup$ Is then a general rule then? $\endgroup$ – Alejandro Dec 30 '13 at 16:00
  • 1
    $\begingroup$ Yes, get the final series expansion in the form $f(z) = \sum a_n z^n$ and apply the root test. en.wikipedia.org/wiki/Root_test $\endgroup$ – gerd Dec 30 '13 at 16:05
  • $\begingroup$ The radius of convergence is always the distance to the next singularity. So if your operation does not remove some of the poles and essential singularities, then the convergence radius of the result is the minimum of the radii of the operands. $\endgroup$ – LutzL Dec 30 '13 at 17:36
0
$\begingroup$

Convergance radius of $\sum_{n=0}^{\infty}\Big(1-\frac{1}{2^{n+1}}\Big)z^n$ is $1$ but this is purely accidental as you can clearly imagine $f(x) = 0$ and then the radius of $f(x)g(x)$ increases to $+\infty$.

$\endgroup$

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.