# Summation notation

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• Aug 1st 2009, 06:58 PM
marie7
Summation 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 PM
AlephZero
Quote:

Originally Posted by marie7
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.

First of all, we need to be clear on what the problem is. Do you mean:

$\displaystyle \sum^4_{i=1}(xi-yi)^2$?
• Aug 1st 2009, 07:05 PM
marie7
yes that's correct sorry I don't know how to do the symbols
• Aug 1st 2009, 07:18 PM
AlephZero
Quote:

Originally Posted by marie7
yes that's correct sorry I don't know how to do the symbols

OK... well this is sort of an odd sum. The $\displaystyle x$ and $\displaystyle y$ aren't involved in the indexing, so we can take them out. We get

$\displaystyle (x-y)^2\sum_{i=1}^4 i^2.$

Now the sum should be very simple to solve.
• Aug 1st 2009, 07:21 PM
marie7
Quote:

Originally Posted by AlephZero
OK... well this is sort of an odd sum. The $\displaystyle x$ and $\displaystyle y$ aren't involved in the indexing, so we can take them out. We get

$\displaystyle (x-y)^2\sum_{i=1}^4 i^2.$

Now the sum should be very simple to solve.

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 PM
AlephZero
Quote:

Originally Posted by marie7
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.

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:

$\displaystyle \sum_{i=1}^4 (x_i - y_i)^2$?
• Aug 1st 2009, 07:38 PM
marie7
yeah it is.
• Aug 1st 2009, 07:44 PM
AlephZero
Well, then the easiest way to do it is just expand the sum, and substitute the values:

$\displaystyle \sum_{i=1}^4(x_i-y_i)^2 = (x_1-y_1)^2 + (x_2-y_2)^2+(x_3-y_3)^2+(x_4-y_4)^2.$

Simply plug in the values and do the arithmetic.
• Aug 1st 2009, 07:59 PM
marie7
I got -146 as my answer, is that correct?
• Aug 1st 2009, 08:07 PM
VonNemo19
No, Marie. 'fraid not.

Hint:

Spoiler:
Your answer cannot be negative because you're taking the same of "squares". Can A squared number be negative?
• Aug 1st 2009, 08:09 PM
marie7
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 PM
AlephZero
Quote:

Originally Posted by marie7
I got -146 as my answer, is that correct?

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 PM
marie7
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 PM
VonNemo19
Quote:

Originally Posted by marie7
I plugged the numbers in like this (2 -(-1))^2 + (-3-3)^2 + (10 - 5)^2 + (-5-7)^2

Am I right?

You plugged 'em in right, but unfortunately arrived at the wrong answer. What did you do after that?

Look at my last post for a hint.
• Aug 1st 2009, 08:14 PM
VonNemo19
Quote:

Originally Posted by marie7
yes but the upper limit is four so I only do from 1 to four not 1 to five, does that make sense?

Correct!
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