Cauchy product associativity proof

Question

Great. I need a proof that the Cauchy product is an associative operation. I can easily proof that it is a commutative operation, find identity series and find invertible and inverse series, BUT for some reason I fail to proof this damn associativity.

The proof should not use any fancy theorems or so... Rather, it should be a simple algebraic proof that $\forall_{n\ge n_0}\sum^n_{k=n_0}\sum^k_{l=n_0}a_lb_{k-l}c_{n-k}=\sum^n_{n=n_0}\sum^k_{l=n_0}a_{n-k}b_lc_{k-l}$.

Now I'm really sorry for this dumb question. Feel free to down-vote it as hard as you please, but could you kindly answer it? Thanks.

Answer 1

(Assuming you work with coefficients $\in\mathbb R$ or similar) The Cauchy product of the coefficients of polynomials $A,B$ gives us the coefficients of the unique(!) polynomial $P$ with $P(x)=A(x)B(x)$ for all $x$. Since $\bigl(A(x)B(x)\bigr)C(x) = A(x)\bigl(B(x)C(x)\bigr)$ for all $x$, both three-fold Cauchy products obtained must give the same result.

Answer 2

Here’s a very elementary approach, albeit one that’s not the kind of algebraic manipulation that you probably had in mind. Instead of just manipulating the expression, we identify the set of triples of indices that appear in terms of the two double summations. Let

$$I=\{\langle p,q,r\rangle\in\Bbb Z^3:p+q+r=n\text{ and }p\ge n_0\text{ and }q\ge 0\}\;.$$

Observe that if $n_0\le k\le n$ and $n_0\le\ell\le k$, then $\langle\ell,k-\ell,n-k\rangle\in I$ and $\langle n-k,\ell,k-\ell\rangle\in I$.

Conversely, if $\langle p,q,r\rangle\in I$, and we set $\ell=p$ and $k=p+q$, then $\langle p,q,r\rangle=\langle\ell,k-\ell,n-k\rangle$. Thus,

$$\sum_{k=n_0}^n\sum_{\ell=n_0}^ka_\ell b_{k-\ell}c_{n-k}=\sum_{\langle p,q,r\rangle\in I}a_pb_qc_r\;.$$

If instead we set $\ell=q$ and $k=q+r$, then $\langle p,q,r\rangle=\langle n-k,\ell,k-\ell\rangle$, so

$$\sum_{k=n_0}^n\sum_{\ell=n_0}^ka_{n-k}b_\ell c_{k-\ell}=\sum_{\langle p,q,r\rangle\in I}a_pb_qc_r\;.$$

Thus,

$$\sum_{k=n_0}^n\sum_{\ell=n_0}^ka_\ell b_{k-\ell}c_{n-k}=\sum_{\langle p,q,r\rangle\in I}a_pb_qc_r=\sum_{k=n_0}^n\sum_{\ell=n_0}^ka_{n-k}b_\ell c_{k-\ell}\;,$$

as desired.