Polynomial multiplication identity

Question

Let $R$ be a nontrivial ring with unity and consider the formal power series ring $R[X]$ over $R$. Let $X$ be the element of $R^{\mathbf{N}}$ for which $X_1=1$ and $X_n=0$ for all $n\ne 1$. Then for every polynomial $p\in R[X]$, there exists a unique $n\in\mathbf{N}$ such that $p_k=0$ for all $n<k$ and $$p=\sum_{k=0}^np_kX^k.$$

Let $p,q\in R[X]$ be polynomials. I want to prove that $$\left(\sum_{k=0}^np_kX^k\right)\left(\sum_{j=0}^mq_jX^j\right)=\sum_{k=0}^{n+m}\left(\sum_{j=0}^kp_jq_{k-j}\right)X^k.$$

Attempt:

$$\begin{align*} \left(\sum_{k=0}^np_kX^k\right)\left(\sum_{j=0}^mq_jX^j\right)&=\sum_{k=0}^n\sum_{j=0}^mp_kX^kq_jX^j\\ &=\sum_{k=0}^n\sum_{j=0}^mp_kq_jX^{j+k}. \end{align*}$$ I am unable to proceed further than this. I probably have to perform an index shift, but I can't see it. Any hints?

Edit:

Actually I think the identity, as it stands, is incorrect. What should the first summation interval be? $[0,n+m]$?

I think it's only reindexing and renaming your indices to get there.. — Berci, Aug 14 '20 at 21:27
The summand should be from $0$ to $n+m$ as you say, or just $0$ to $\infty$ (if you define $p_k = 0$ for $k \geq n$, and same for $q_j$.) — Jair Taylor, Aug 14 '20 at 21:35
This is usually true by definition. What's the definition of multiplication of power series you are using? — Jair Taylor, Aug 14 '20 at 21:35
@JairTaylor I am using the convolution definition: i.e. $(pq)n=\sum{j=0}^np_jq_{n-j}$ for $n\in\mathbf{N}$. — alf262, Aug 14 '20 at 21:39
In that case, there's nothing to prove that I can see, except I suppose the fact that you can limit the index to only go to $n+m$. — Jair Taylor, Aug 14 '20 at 21:50

score 1 · Answer 1 · answered Aug 19 '20 at 20:57

We use the coefficient of operator $[X^n]$ to denote the coefficient of $X^n$ of a polynomial. We see the polynomials, both at the left-hand and at the right-hand side have degree $n+m$ and consider therefore $q$ with $0\leq q\leq n+m$.

We obtain \begin{align*} [X^q]\left(\sum_{k=0}^np_kX^k\right)\left(\sum_{j=0}^mq_jX^j\right) &=\sum_{k=0}^{q}p_k[X^{k-q}]\sum_{j=0}^mq_jX^j\tag{1}\\ &=\sum_{k=0}^{q}p_kq_{k-q}\tag{2} \end{align*}

Comment:

In (1) we apply the rule $[X^r]X^sP(X) = [X^{r-s}]P(X)$ of the coefficient of operator. Note the upper index of the outer sum is set to $q$. This is admissible, since in case $q\leq n$ only coefficients less or equal to $q$ can provide a non-zero contribution. In case $q>n$ this is also fine, since then we have $p_k=0$ whenever $n<k\leq q$.
In (2) we select the coefficient of $X^{k-q}$, which is $0$ in case $k-q>m$ and the wanted sum on the other hand.

Since the degree of the polynomial from the left-hand side of (1) is $m+n$, we conclude from (1) and (2): \begin{align*} P(X)&=\left(\sum_{k=0}^np_kX^k\right)\left(\sum_{j=0}^mq_jX^j\right)\\ &=\sum_{q=0}^{m+n}\Big([Z^q]P(Z)\Big)X^q\\ &=\sum_{q=0}^{m+n}\left(\sum_{k=0}^{q}p_kq_{k-q}\right)X^q\\ \end{align*}

Sir, please help me with the problem at https://math.stackexchange.com/questions/3797044/distribution-of-the-number-of-trials-required-for-the-first-occurrence-of-the-ev — Junk Warrior, Sep 03 '20 at 00:29
@JunkWarrior: I will provide an answer at the weekend. Regards, — Markus Scheuer, Sep 04 '20 at 08:08

score 1 · Answer 2 · answered Aug 19 '20 at 21:25

Hint:

Unless you define the coefficients with an out-of-range index to be zero, you cannot express the product with a straight nested sum.

As the figure shows, the groups of coefficients for the terms with constant exponent form two triangular and one parallelogrammic patterns.

You will need a sum with a linearly increasing number of terms, one with a constant number, and one with a decreasing number. You can also arrange sums where the delimiting indexes are made of $\min/\max$ expressions.

Polynomial multiplication identity

2 Answers2