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Is there any demonstration or general formula of how the product between an infinite and a finite power series takes form?

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  • $\begingroup$ I guess "polynomial (function)" would be a more usual name for a finite power series, wouldn't it? And what do you mean "takes form"? You can multiply two infinite series, and thus also "finite ones", for example in the ring of power series over a ring, say. $\endgroup$
    – DonAntonio
    Commented Aug 10, 2016 at 18:08
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    $\begingroup$ Search Cauchys Product formula (en.wikipedia.org/wiki/Cauchy_product). For the finite power series the coefficients of this series will vanish for $n>N$. $\endgroup$
    – MrYouMath
    Commented Aug 10, 2016 at 18:09

1 Answer 1

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The product of power series has this form, where $a$ and $b$ are coefficient sequences. $a$ has infinitely many terms and $b$ has $n$ terms.

$$\sum_{i=0}^\infty a_ix^i\ \cdot\sum_{j=0}^nb_jx^j$$

The series can be distributed term by term.

$$=\sum_{i=0}^\infty\sum_{j=0}^na_ib_jx^{i+j}$$

Now we have to group the like powers of $x$, which is equivalent to grouping the terms with the same value of $i+j$. Let $k=i+j$. for a particular value of $k$, we can take all the terms starting at $i=k,j=0$ and iterating down to either $i=0,j=k$ or $i=k-n,j=n$, whichever comes first. The sum can then be written as:

$$\sum_{k=0}^\infty\sum_{j=0}^{min(n,k)}a_{k-j}b_jx^k$$

We can create a new coefficient sequence $c$ such that:

$$c_k=\sum_{j=0}^{min(n,k)}a_{k-j}b_j$$

The product can then be written as:

$$\sum_{i=0}^\infty a_ix^i\ \cdot\sum_{j=0}^nb_jx^j=\sum_{k=0}^\infty c_kx^k$$

With more information about $a$ and $b$, it may be possible to simplify the expression for $c$.

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