Sum of a Sum of a Squared Difference

Question

How did the author jump from the second equation to the third equation? I suspect there’s a rule I’m forgetting that allows for this, any help is appreciated.

I'm more interested in how they jumped from the first line to the second. Where do the 8s and 7s come from, what happened to good old $\mu$? — Vincent, Jul 19 '19 at 10:01

John Omielan · Accepted Answer · 2019-07-19T02:25:32.900

4

Note that

\begin{align} -\left(\sum_{i=1}^{160} (x_i - 8)^2 - \sum_{i=1}^{160} (x_i - 7)^2\right) & = -\left(\sum_{i=1}^{160} (x_i^2 - 16x_i + 64) - (x_i^2 - 14x_i + 49)\right) \\ & = -\left(\sum_{i=1}^{160} (-2x_i + 15)\right) \\ & = 2\sum_{i=1}^{160} x_i - 2400 \tag{1}\label{eq1} \end{align}

As you can see, the $15$ is a constant repeating $160$ times for a total of $15 \times 160 = 2400$. Also, the author used the minus sign in front to remove the first minus sign for the $2x_i$, moved the $2$ outside the summation and changed the plus to a minus for the sum of $2400$.

edited Jul 19 '19 at 02:25

answered Jul 19 '19 at 01:30

John Omielan

47,976

Brilliant, thanks for breaking that down for me. – Seraphim Jul 19 '19 at 01:36
@Seraphim You're welcome. I'm glad I could be of some help. – John Omielan Jul 19 '19 at 01:37

score 3 · Answer 2 · answered Jul 19 '19 at 01:56

3

A simpler way, using $a^2-b^2 = (a-b)(a+b) $, is

$\begin{array}\\ \sum_{i=1}^{160} (x_i - 8)^2 - \sum_{i=1}^{160} (x_i - 7)^2) &=\sum_{i=1}^{160} ((x_i - 8)^2 -(x_i - 7)^2)\\ &=\sum_{i=1}^{160} ((x_i - 8) -(x_i - 7))((x_i - 8) +(x_i - 7))\\ &=\sum_{i=1}^{160} (-1)(2x_i -15)\\ &=160\cdot 15-2\sum_{i=1}^{160} x_i\\ \end{array} $

answered Jul 19 '19 at 01:56

marty cohen

107,799

2

Simpler? Debatable. Cooler? Yes. – Hendrix Jul 19 '19 at 02:22
1

I think it's simple because you don't have to expand the squares. – marty cohen Jul 19 '19 at 02:30

Sum of a Sum of a Squared Difference

2 Answers2