Summation of Products of n Numbers taken m at a time with Repetitions/Examples/Order 2

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Theorem

Let $a, b \in \Z$ be integers such that $b \ge a$.

Let $U$ be a set of $n = b - a + 1$ numbers $\set {x_a, x_{a + 1}, \dotsc, x_b}$.


Then:

$\ds \sum_{i \mathop = a}^b \sum_{j \mathop = a}^i x_i x_j = \dfrac 1 2 \paren {\paren {\sum_{i \mathop = a}^b x_i}^2 + \paren {\sum_{i \mathop = a}^b {x_i}^2} }$


Proof 1

Let:

\(\ds S_1\) \(:=\) \(\ds \sum_{i \mathop = a}^b \sum_{j \mathop = a}^i x_i x_j\)
\(\ds \) \(=\) \(\ds \sum_{j \mathop = a}^b \sum_{i \mathop = j}^b x_i x_j\) Summation of i from 1 to n of Summation of j from 1 to i
\(\ds \) \(=\) \(\ds \sum_{i \mathop = a}^b \sum_{j \mathop = i}^b x_i x_j\) Change of Index Variable of Summation
\(\ds \) \(=:\) \(\ds S_2\)


Then:

\(\ds 2 S_1\) \(=\) \(\ds S_1 + S_2\)
\(\ds \) \(=\) \(\ds \sum_{i \mathop = a}^b \paren {\sum_{j \mathop = a}^i x_i x_j + \sum_{j \mathop = i}^b x_i x_j}\)
\(\ds \) \(=\) \(\ds \sum_{i \mathop = a}^b \paren {\paren {\sum_{j \mathop = a}^b x_i x_j} + x_i x_i}\) Sum of Summations over Overlapping Domains: Example
\(\ds \) \(=\) \(\ds \sum_{i \mathop = a}^b \sum_{j \mathop = a}^b x_i x_j + \sum_{i \mathop = a}^b x_i x_i\) Sum of Summations equals Summation of Sum
\(\ds \) \(=\) \(\ds \paren {\sum_{i \mathop = a}^b x_i} \paren {\sum_{j \mathop = a}^b x_j} + \paren {\sum_{i \mathop = a}^b {x_i}^2}\) Change of Index Variable of Summation

The result follows on multiplying by $\dfrac 1 2$.

$\blacksquare$


Proof 2

We have that:

$\ds \sum_{i \mathop = a}^b \sum_{j \mathop = a}^i x_i x_j = \sum_{a \mathop \le j_1 \mathop \le j_2 \mathop \le b} x_{j_1} x_{j_2}$


From Summation of Products of n Numbers taken m at a time with Repetitions:

$\ds \sum_{a \mathop \le j_1 \mathop \le \cdots \mathop \le j_m \mathop \le b} x_{j_1} \cdots x_{j_m} = \sum_{\substack {k_1, k_2, \ldots, k_m \mathop \ge 0 \\ k_1 \mathop + 2 k_2 \mathop + \cdots \mathop + m k_m \mathop = m} } \dfrac { {S_1}^{k_1} } {1^{k_1} k_1 !} \dfrac { {S_2}^{k_2} } {2^{k_2} k_2 !} \cdots \dfrac { {S_m}^{k_m} } {m^{k_m} k_m !}$

where:

$S_r = \ds \sum_{k \mathop = a}^b {x_k}^r$ for $r \in \Z_{\ge 0}$.


Setting $m = 2$:

\(\ds \sum_{a \mathop \le j_1 \mathop \le j_2 \mathop \le b} x_{j_1} x_{j_2}\) \(=\) \(\ds \sum_{\substack {k_1, k_2 \mathop \ge 0 \\ k_1 \mathop + 2 k_2 \mathop = 2} } \dfrac { {S_1}^{k_1} } {1^{k_1} k_1 !} \dfrac { {S_2}^{k_2} } {2^{k_2} k_2 !}\)
\(\ds \) \(=\) \(\ds \dfrac { {S_1}^2 {S_2}^0} {\paren {1^2 \times 2!} \paren {2^0 \times 0!} } + \dfrac { {S_1}^0 {S_2}^1} {\paren {1^0 \times 0!} \paren {2^1 \times 1!} }\) as $k_1 = 2, k_2 = 0$ and $k_1 = 0, k_2 = 1$ are the only $k_1$ and $k_2$ to fulfil the criteria
\(\ds \) \(=\) \(\ds \frac 1 2 \paren { {S_1}^2 + S_2}\) simplifying

$\blacksquare$


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