0

$$S=(x_3-x_1)^2+(x_4-x_2)^2+(x_5-x_3)^2+(x_1-x_4)^2+(x_2-x_5)^2$$.

$$S'=(x_3-x_1)^2+(x_4+x_2)^2+(x_5-x_2-x_3)^2+(x_1+x_2-x_4)^2+(-x_2-x_5)^2$$

Find $S$–$S'$?

I thought that, I should manipulate the above equations on my own but I got stuck, How can I reach to the answer???

  • It's unclear what you're trying to do. Is this an equation to solve or is the right-hand side already the evaluated version of the left hand side? – Matti P. May 23 '19 at 12:44
  • What do you mean by 'the answer'? What are you trying to do with these terms? Usual quality requirements on Math.StackExchange are that you should try to provide at least a little bit of context for your question. – Jack Crawford May 23 '19 at 12:44
  • What are $x_1,x_2,x_3,x_4,x_5$? What are you trying to find? Please be more specific. – QuasarChaser May 23 '19 at 12:44
  • Ok wait let me edit my post –  May 23 '19 at 12:45
  • There is a whole lot of cancellation. Determining the difference term by term (square by square) is not difficult or much work. – Servaes May 23 '19 at 12:53

2 Answers2

1

Hint: Evaluate the difference term by term, i.e. square by square, and use the fact that $$a^2-b^2=(a-b)(a+b).$$ There is nothing difficult or painful about this.

In fact, it is quickly clear that in each case you get a multiple of $x_2$. The expression as a whole is quadratic, so it is a linear factor times $x_2$.

Servaes
  • 63,261
  • 7
  • 75
  • 163
0

If you don't want to perform the expansion manually, may be you could use Symbola or a similar software.

$$S-S'=(x_3-x_1)^2+(x_4-x_2)^2+(x_5-x_3)^2+(x_1-x_4)^2+(x_2-x_5)^2 -((x_3-x_1)^2+(x_4+x_2)^2+(x_5-x_2-x_3)^2+(x_1+x_2-x_4)^2+(-x_2-x_5)^2)$$

is

$$-2x_{3}x_{2}-2x_{2}^2-2x_{4}x_{2}-2x_{1}x_{2}-2x_{2}x_{5}$$

NoChance
  • 6,427