# Cauchy products for 4 power series

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

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

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

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

$$a_0b_0$$

$$a_0b_1+a_1b_0$$

$$a_0b_2+a_1b_1+a_2b_0$$

$$a_0b_3+a_1b_2+a_2b_1+a_3b_0$$

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.

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

• Thank you saulspatz! – James Warthington Nov 30 '19 at 4:20
• @JamesWarthington It was my pleasure. – 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 $$a_ix^ib_jx^jc_kx^kd_lx^l$$ $$=a_ib_jc_kd_lx^{i+j+k+l}$$.