Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Say if I define a power series over some arbitrary field $F$ as

$$a = \sum^{ \infty }_{i = 0} a_{i} X^{i} $$

Then can I say:

$$ab = \sum^{ \infty }_{i = 0} \sum^{ \infty }_{j = 0} a_{i} b_{j} X^{i + j} $$

share|cite|improve this question
$i_1+j_1=i_2+j_2$ for many pairs. just use the cauchy product (gathering up terms with the same exponent) – yoyo Mar 28 '11 at 15:43
@user8603, @Bill Dubuque, @Mitch, @Jonas Kibelbek: My apologies for giving a wrong answer ("very misleading" was putting it mildly :-) -- I've deleted it. I wasn't aware of the notion of convergence for formal power series defined in Bill's answer -- thanks to all of you for educating me. Bill, your point about the "widespread confusion" about this is confirmed by the 7 upvotes I got for my "complete nonsense". :-) – joriki Mar 28 '11 at 22:27

Yes, using the natural notion of convergence for formal power series, the stated sum does indeed converge to the Cauchy product. One should beware - as exemplified in this thread - that there is widespread confusion about formal vs. functional power series - even by some experts (in other fields). Rota frequently told jokes in his lectures about certain distinguished mathematicians who published complete nonsense based on such confusion (Indiscrete Thoughts!)

In any case the basic ideas are quite simple if you merely take off your analyst hat and, instead, put on your algebraist or combinatorist hat. In particular, you should be able to find a correct discussion of convergence of formal power series in almost any good book on combinatorics or generating functions, e.g. here is an excerpt from Stanley's classic $\: $ Enumerative Combinatorics I.

enter image description here enter image description here enter image description here

share|cite|improve this answer

Not if you are talking about multiplying $a$ by a similarly defined formal power series $b$. The multiplication of formal power series can be written as: $ab = \sum_{i=0}^{\infty}(\sum_{j=0}^ia_jb_{i-j})X^i$

share|cite|improve this answer
It should be $X^i$; but also see my answer. – joriki Mar 28 '11 at 5:41
@joriki: Thanks, but I think that's obviously enough a type not to call it an "incorrect answer". – Alex Becker Mar 28 '11 at 5:43
I'm sorry, that didn't refer to the typo but to the fact that your answer basically says "no" when it should have been "yes and no" -- but in fact I weakened that to "incorrect or partial" even before I'd seen your comment -- sorry that the initial criticism was a bit too strong. – joriki Mar 28 '11 at 5:45
@joriki: That's why I said "not if... formal power series". But I know you didn't mean to blast me, and I agree that your answer is much fuller and more informative. – Alex Becker Mar 28 '11 at 5:52

Yes, you can: this is the definition of the product of the two formal power series $a$ and $b =\sum_{i=0}^\infty b_i X^i$.

share|cite|improve this answer
No, this is not the definition of the product of the formal power series; in fact it's not a power series. The product is $\sum_{k=0}^\infty(\sum_{i=0}^k a_ib_{k-i})X^k$; see and (Of course the two coincide if all the limits exist, but the point about formal power series is that you can manipulate them like this without worrying about whether the limits exist.) By the way, note that if you talk about "power series over some arbitrary field", these are actually just polynomials if the field is finite. – joriki Mar 28 '11 at 5:14
@Joriki: No, this is in fact correct (but far too terse). It appears that you are confusing formal power series with functional power series. In particular, formal power series over finite fields are not "just polynomials". – Bill Dubuque Mar 28 '11 at 17:41

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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