Summation Notation Question in McMillan's Theorem Proof

Question

Let me preface by saying that this question does not pertain as much to coding theory, as it does to mathematical notation. Every symbol in this question is a natural number.

Anyhow, I am currently reading through a proof of McMillan's Theorem. In the proof, there is the statement that \begin{equation}\tag{1} \left( \sum_{j=1}^m \dfrac{\alpha_j}{r^j} \right)^u = \left( \dfrac{\alpha_1}{r} + \cdots + \dfrac{\alpha_m}{r^m} \right)^u, \end{equation}

and so far, so good. However, the next line states that $(1)$ out can be multiplied out to yield

\begin{equation}\tag{2} \sum_{\substack{1 \leq i_j \leq m \\ i_1, i_2, \cdots, i_u}} \dfrac{\alpha_{i_1}}{r^{i_1}} \cdots \dfrac{\alpha_{i_u}}{r^{i_u}} = \sum_{\substack{1 \leq i_j \leq m \\ i_1, i_2, \cdots, i_u}} \dfrac{\alpha_{i_1} \alpha_{i_2} \cdots \alpha_{i_u}}{r^{i_1 + \cdots + i_u}}, \end{equation}

which confuses me. Can someone explain to me what the subscripts on the sigmas in equation $(2)$ are meant to indicate?

Also, I apologize if this question has already been answered somewhere else. I tried to look through the archives and didn't come across it.

Answer 1

To avoid clutter, let

$$f(i_1,i_2,\ldots,i_u)=\frac{\alpha_1}{r^{i_1}}\cdot\frac{\alpha_2}{r^{i_2}}\cdot\ldots\cdot\frac{\alpha_u}{r^{i_u}}\;.\tag{1}$$

(It could just as well be any function of $u$ variables.) Then

$$\sum_{\substack{1\le i_j\le m\\i_1,i_2,\ldots,i_u}}f(i_1,\ldots,i_u)\tag{2}$$

is the sum of the values $f(i_1,\ldots,i_u)$ as $\langle i_1,i_2,\ldots,i_u\rangle$ ranges over all $u$-tuples of integers satisfying the condition that $1\le i_j\le m$ for $j=1,\ldots,u$. For example, if $u=3$ and $m=2$, we have

$$\begin{align*}\sum_{\substack{1\le i_j\le 2\\i_1,i_2,i_3}}f(i_1,i_2,i_3)&=f(1,1,1)+f(1,1,2)+f(1,2,1)+f(1,2,2)\\ &\quad+f(2,1,1)+f(2,1,2)+f(2,2,1)+f(2,2,2)\;. \end{align*}$$

In general there will be $m^u$ terms in the sum, since each of the $u$ indices $i_j$ can take any of the $m$ values from $1$ through $m$.

When you multiply out

$$\left(\sum_{j=1}^m\frac{\alpha_j}{r^j}\right)^u=\underbrace{\left(\sum_{j=1}^m\frac{\alpha_j}{r^j}\right)\cdot\ldots\cdot\left(\sum_{j=1}^m\frac{\alpha_j}{r^j}\right)}_{u\text{ copies}}\;,\tag{3}$$

each term of the product (before you collect like terms) is the product of fractions $\frac{\alpha_j}{r^j}$, one from each of the $u$ copies, so it has exactly the form of our $f(i_1,\ldots,i_u)$ in $(1)$. $(2)$ is the sum of all of these terms, so it’s equal to $(3)$.

Answer 2

Would this help?

\begin{align} \left( \sum_{j=1}^2 \dfrac{\alpha_j}{r^j} \right)^3 &= \left( \dfrac{\alpha_1}{r}+\dfrac{\alpha_2}{r^2} \right)^3 \\ &= \left( \dfrac{\alpha_1}{r^1}+\dfrac{\alpha_2}{r^2} \right) \left( \dfrac{\alpha_1}{r^1}+\dfrac{\alpha_2}{r^2} \right) \left( \dfrac{\alpha_1}{r^1}+\dfrac{\alpha_2}{r^2} \right) \\ &= \dfrac{\alpha_1}{r^1}\dfrac{\alpha_1}{r^1}\dfrac{\alpha_1}{r^1}+\\ &\phantom{=} \left( \dfrac{\alpha_1}{r^1}\dfrac{\alpha_1}{r^1}\dfrac{\alpha_2}{r^2}+ \dfrac{\alpha_1}{r^1}\dfrac{\alpha_2}{r^2}\dfrac{\alpha_1}{r^1}+ \dfrac{\alpha_2}{r^2}\dfrac{\alpha_1}{r^1}\dfrac{\alpha_1}{r^1} \right) +\\ &\phantom{=} \left( \dfrac{\alpha_1}{r^1}\dfrac{\alpha_2}{r^2}\dfrac{\alpha_2}{r^2}+ \dfrac{\alpha_2}{r^2}\dfrac{\alpha_1}{r^1}\dfrac{\alpha_2}{r^2}+ \dfrac{\alpha_2}{r^2}\dfrac{\alpha_2}{r^2}\dfrac{\alpha_1}{r^1} \right) + \\ &\phantom{=} \dfrac{\alpha_2}{r^2}\dfrac{\alpha_2}{r^2}\dfrac{\alpha_2}{r^2}\\ &= \dfrac{\alpha_1\alpha_1\alpha_1}{r^{1+1+1}} + \\ &\phantom{=} \left( \dfrac{\alpha_1\alpha_1\alpha_2}{r^{1+1+2}}+ \dfrac{\alpha_1\alpha_2\alpha_1}{r^{1+2+1}}+ \dfrac{\alpha_2\alpha_1\alpha_1}{r^{2+1+1}} \right) +\\ &\phantom{=} \left( \dfrac{\alpha_1\alpha_2\alpha_2}{r^{1+2+2}}+ \dfrac{\alpha_2\alpha_1\alpha_2}{r^{r^2+1+2}}+ \dfrac{\alpha_2\alpha_2\alpha_1}{r^{r^2+2+1}} \right) + \\ &\phantom{=} \dfrac{\alpha_2\alpha_2\alpha_2}{r^{2+2+2}} \\ \end{align}