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I have $n$ power series. How can I find the power series of the product of these $n$ series? If there are two series $(a_m)$ and $(b_m)$ then the product series $(c_m)$ is given by the Cauchy product,

$$c_m = \sum_{k=0}^m a_k b_{m-k}$$

How does this generalize to more series?

Chris Taylor
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Ramesh
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    Please do a little research before asking a broad non specific question such as the one above. have you tried any multiplication? can you add some more detail please. – jimjim Jun 15 '11 at 09:11
  • The question needed a little cleaning up, but it was a long way from a 'broad non specific question' and it shows clear evidence of some research. I've edited to make it more obvious what's being asked. In general I think the first response to a question like this should be to edit for clarity rather than downvote. – Chris Taylor Jun 15 '11 at 09:20
  • @Chris, It is your cleaned upversion that deservesthe the +1 not the original question. – jimjim Jun 15 '11 at 10:00
  • @Arjang: I found the OP's original question to be clear and specific. – Pete L. Clark Jun 15 '11 at 12:21
  • @Pete @Chris, cool then my bad. I should have commented that I could'nt make sense of it and nothing towards the OP. – jimjim Jun 15 '11 at 21:24

2 Answers2

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$$\left(\sum_{n=0}^\infty a_{1,n} x^n\right)\left(\sum_{n=0}^\infty a_{2,n} x^n\right)\dots\left(\sum_{n=0}^\infty a_{l,n} x^n\right) = \sum_{n=0}^\infty \sum_{k_1 + k_2 + \dots + k_l = n } a_{1,k_1}a_{2,k_2}\dots a_{l,k_l} x^n $$

Alexander Thumm
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Presumably you are looking for an expression for the coefficients of $x^i$ in the product

$$\left(\sum_j a_j x^j \right) \left(\sum_k b_k x^k \right) \left(\sum_l c_l x^l \right) \ldots \left(\sum_m n_m x^m \right)$$

Generalising the Cauchy product, this is

$$\sum_i \left(\sum_{j=0}^{i} \sum_{k=0}^{i-j} \sum_{l=0}^{i-j-k} \cdots a_j b_k c_l \ldots n_{i-j-k-l-\cdots} \right) x^i$$

Henry
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