$A= a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+...$

$B= b_0+b_1x+b_2x^2+b_3x^3+b_4x^4+...$

$A\cdot B=a_0b_0+(a_0b_1+a_1b_0)x+(a_0b_2+a_1b_1+a_2b_0)x^2...$

So I can build a triangle by detailing the coefficients of the two above series:





I can keep writing, notice that the subscript for $a$ doesn't change. It is easy to write these coefficients in a triangle which I call the Cauchy triangle.

Recently, however, I encounter a problem that asks me to deal with the product of four power series instead of two. So I wish to build up a triangle that is similar to the case of two power series.

I write the four power series as followed:

$A= a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+...$

$B= b_0+b_1x+b_2x^2+b_3x^3+b_4x^4+...$

$C= c_0+c_1x+c_2x^2+c_3x^3+c_4x^4+...$

$D= d_0+d_1x+d_2x^2+d_3x^3+d_4x^4+...$

$A\cdot B\cdot C\cdot D=a_0b_0c_0d_0+(a_1b_0c_0d_0+a_0b_1c_0d_0+a_0b_0c_1d_0)x+(a_0b_2c_0d_0+a_2b_0c_0d_0+a_0b_0c_2d_0+a_0b_0c_0d_2 )x^2...$

This is the best I can try so far, is there a method to exhaustively list all these elements into a triangle like the case of two power series? I want a complete triangle to fully count the coefficients.

Please don't respond to this thread with full sigma notation, as I haven't mastered the procedure of manipulating these symbols yet.

  • $\begingroup$ For the $x^2$ term you've missed $6$ cases: $a_0b_0c_1d_1$ and so forth. For the $x$ term you missed $a_0b_0c_0d_1$ $\endgroup$ – saulspatz Nov 30 '19 at 3:36
  • $\begingroup$ @saulspatz, this is why I wish to make a triangle so that I don't miss counting any term. $\endgroup$ – James Warthington Nov 30 '19 at 3:45

When you multiply two power series, the general term is $$(a_0b_n+a_1b_{n-1}+\cdots+a_nb_0)x^n.$$ The subscripts sums to $n$. Note that there are always $n+1$ terms in the coefficient, since the subscript on $a$ must be $0\leq k\leq n$ and then there is only once choice for the subscript on $b$. The number of terms grows linearly with $n$, so they fit neatly in a triangle.

When you multiply $4$ power series a typical term in the coefficient of $x^n$ looks like $a_ib_jc_kd_l$ where $i+j+k+l=n$. There are $$\binom{n+3}{3}=\frac{(n+3)(n+2)(n+1)}{6}$$ such terms. (Google "stars and bars" if you don't know how I arrived at this expression.) Therefore, they won't fit it a triangle.

Obviously, knowing how many terms there should be will help you not miss any. What I would suggest is that in passing from $x^n$ to $x^{n+1}$ you add $1$ to each of the subscripts in the each term in the coefficient for $x^n$, taking care to avoid duplicates. For example, the coefficient of $x$ has $4$ and the coefficient for $x^2$ has $10$. If we add $1$ to each subscript in each term, we get $4$ times as many terms. This would give $16$ terms in the coefficient of $x^2$, not $10$. The reason is that there is more than one way to arrive at certain terms. For example $a_1b_1c_0d_0$ could arise from adding $1$ to the subscript of $a$ in $a_0b_1c_0d_0$ or from adding $1$ to the subscript of $b$ in $a_1b_0c_0d_0$.

That said, you will see that the numbers of terms quickly becomes too large to make listing them explicitly worthwhile. You really should learn how to use the sigma notation.

  • $\begingroup$ Thank you saulspatz! $\endgroup$ – James Warthington Nov 30 '19 at 4:20
  • $\begingroup$ @JamesWarthington It was my pleasure. $\endgroup$ – saulspatz Nov 30 '19 at 4:21

The coefficient of $x^n$ is the sum of all $a_ib_jc_kd_l$ with $i+j+k+l = n$.

You can see this by noting that the terms with such $a_ib_jc_kd_l$ is $a_ix^ib_jx^jc_kx^kd_lx^l$ $=a_ib_jc_kd_lx^{i+j+k+l}$.


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