Nested operation notation convention for evaluation (particularly for Pi and Sigma)

Question

I have a question on the notation for two products.

For Pi:

Is this statement true?

$$\prod_{k=1}^{l}a_k\prod_{m=1}^{n}b_m = \prod_{k=1}^{l}\prod_{m=1}^{n}a_kb_m \neq a_1a_2\dots a_lb_1b_2\dots b_n = \left(\prod_{k=1}^{l}a_k\right)\left(\prod_{m=1}^{n}b_m\right)$$

For Sigma and Pi:

When dealing with double (or nested/multiple) operators, do we by convention always deal with the "rightmost" (or the innermost) operator first (this applies to any operator in general)?

$$\sum_{r=1}^{w}\prod_{u=1}^{w}g\left(r, u\right) = \sum_{r=1}^{w}g\left(r,1\right)g\left(r,2\right)\dots g\left(r,w\right) = \prod_{u=1}^{w}g\left(1, u\right) + \prod_{u=1}^{w}g\left(2, u\right) + \dots + \prod_{u=1}^{w}g\left(w, u\right)$$

$$\prod_{u=1}^{w}\sum_{r=1}^{w}g\left(r, u\right) = \prod_{u=1}^{w}\left[g\left(1,u\right) + g\left(2,u\right) + \dots + g\left(w,u\right)\right]$$

Or can we evaluate the outermost ("leftmost") first? Does the equivalence of operating from innermost and operating from the outermost depend on the types of operators (pi, sigma, integral, etc.)? Is operating from the innermost operator always the way to go?

Answer 1

First part:

The left-most equal sign in the first line is not correct.

We have per definition \begin{align*} \prod_{k=1}^la_k\prod_{m=1}^nb_m=\left(\prod_{k=1}^la_k\right)\left(\prod_{m=1}^nb_m\right) =\left(a_1\cdots a_l\right)\left(b_1\cdots b_m\right)\tag{1} \end{align*} On the other hand we have per definition \begin{align*} \prod_{k=1}^l\prod_{m=1}^na_kb_m &=\prod_{k=1}^l\left(\prod_{m=1}^na_k\right)b_m\tag{2}\\ &=\left(\prod_{k=1}^l\left (a_1\cdots a_l\right)\right)b_m\tag{3}\\ &=\left (a_1\cdots a_l\right)^lb_m\tag{4} \end{align*}

Comment:

• In (1) we note the product symbol $\prod$ and the multiplication operator $\cdot$ have the same precedence level. This means that without parentheses the $\prod$ symbol takes just one variable to the right of it and not more. This is the scope defined for the product symbol. This is indicated by the parentheses which enclose the product symbols.

• In (2) we apply precisely the same rules with a somewhat unexpected result. The scope of the inner product encloses $a_k$ but not $b_m$. If we want to have $b_m$ enclosed, we have to put it in parentheses resulting in \begin{align*} \prod_{m=1}^n\left(a_kb_m\right)=(a_kb_1)(a_kb_2)\cdots(a_kb_n)=a_k^nb_1\cdots b_n \end{align*}

• In (3) we have again a situation where $b_m$ is not within the scope of the product symbol, resulting in the representation (4). Note that $b_m$ is a variable with index $m$ which is not bound to a product symbol. See the concept of bound and free variables.

Second part:

When dealing with expressions like $\sum_{r=1}^{w}\prod_{u=1}^{w}g\left(r, u\right)$ we are thanks to the arithmetic rules and the precedence levels of multiplication and addition free to either start with the inner product or with the outer sum.

We obtain \begin{align*} &\sum_{r=1}^w\prod_{u=1}^wg(r,u)\\ &\qquad=\sum_{r=1}^wg(r,1)\cdots g(r,w)\\ &\qquad=g(1,1)\cdot\ldots\cdot g(1,w)+g(2,1)\cdot\ldots\cdot g(2,w)\\ &\qquad\qquad +\cdots+g(w,1)\cdot\ldots\cdot g(w,w)\tag{5} \end{align*} and \begin{align*} &\sum_{r=1}^w\prod_{u=1}^wg(r,u)\\ &\qquad=\prod_{u=1}^wg(1,u)+\prod_{u=1}^wg(2,u)+\cdots+\prod_{u=1}^wg(w,u)\\ &\qquad=g(1,1)\cdot\ldots\cdot g(1,w)+g(2,1)\cdot\ldots\cdot g(2,w)\\ &\qquad\qquad+\cdots+g(w,1)\cdot\ldots\cdot g(w,w)\tag{6} \end{align*} We observe (5) and (6) coincide.

The same holds for the other expression: $\prod_{u=1}^w\sum_{r=1}^wg(r,u)$.