I don't know how to write the symbol sigma on here but the lower limit is 1 and the upper limit is 4 and (xi - yi)^2. I need the method on how to do it.

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- Aug 1st 2009, 06:58 PMmarie7Summation notation
I don't know how to write the symbol sigma on here but the lower limit is 1 and the upper limit is 4 and (xi - yi)^2. I need the method on how to do it.

- Aug 1st 2009, 07:04 PMAlephZero
- Aug 1st 2009, 07:05 PMmarie7
yes that's correct sorry I don't know how to do the symbols

- Aug 1st 2009, 07:18 PMAlephZero
- Aug 1st 2009, 07:21 PMmarie7
Well the x and y have number values and I just wanted the method so I could do it myself but here are the numbers:

for x1 = 2, x2 = -3, x3 = 10 x4 = -5, x5 = -3

for y1 = -1, y2 = 3, y3 = 5, y4 = 7, y5 = -2.

i is supposed to represent for example 1 in x1, 2 in x2 etc. Its a little i. - Aug 1st 2009, 07:37 PMAlephZero
Alright, so when I asked you previously if I had typed up the problem correctly, and you said "yes," that wasn't true, evidently. And you are now providing, in addition, a hugely crucial portion of the problem that you also neglected to mention.

So now I ask you again. Think very hard. Is this the problem:

? - Aug 1st 2009, 07:38 PMmarie7
yeah it is.

- Aug 1st 2009, 07:44 PMAlephZero
Well, then the easiest way to do it is just expand the sum, and substitute the values:

Simply plug in the values and do the arithmetic. - Aug 1st 2009, 07:59 PMmarie7
I got -146 as my answer, is that correct?

- Aug 1st 2009, 08:07 PMVonNemo19
No, Marie. 'fraid not.

Hint:

__Spoiler__: - Aug 1st 2009, 08:09 PMmarie7
I plugged the numbers in like this (2 -(-1))^2 + (-3-3)^2 + (10 - 5)^2 + (-5-7)^2

Am I right? - Aug 1st 2009, 08:10 PMAlephZero
Ah, no, I'm afraid not, for at least 3 reasons. First, I have done the arithmetic myself based on the numbers you provided, and reached a different solution. Second, your answer is a negative number, and since the expression at hand involves the sum of four squares, a negative answer is impossible. Thirdly, you have provided a question involving a finite sum with 4 terms, and supplied data for 5 terms, so something is bound to be wrong with just about any answer you come up with.

- Aug 1st 2009, 08:12 PMmarie7
yes but the upper limit is four so I only do from 1 to four not 1 to five, does that make sense?

- Aug 1st 2009, 08:13 PMVonNemo19
- Aug 1st 2009, 08:14 PMVonNemo19