Question

Summation questions.

How do I square a summation when it has multiplication inside it. I know that you can take constants out of the sum, but, what about in this case, where the multiplication involves the "n" being summed over. such as:

Answer 1

\[(\sum_{i=1}^{N}x_i*\frac{ 1 }{ N })^2\]

Answer 2

If 1/N were a constant I know it could go outside the sum, in which case you would probably square it and multiply it times the sum, but since it involves the N of the sum, I am worried this isn't the case.

Answer 3

Wait, N is just the final number in the sum right!? So, I CAN pull it out, because 1/N isnt itterative, its just the size of the sample?

Answer 4

so I get \[(\sum_{i=1}^{N}x_i*\frac{ 1 }{ N })^2 = (\frac{ 1 }{ N } \sum_{i=1}^{N}x_i)^2\] which is then \[\frac{ 1 }{ N }^2 * (\sum_{i=1}^{N}x_i)^2\]\[= \frac{ 1 }{ N^2 } * (x_1+x_2+...x_n)^2\] right?

Answer 5

Ah yes, it IS a constant :) Good catch!

Answer 6

I should be more clear with the way I say that.
Constant with respect to the summation.*

Answer 7

Thanks for looking at this, It's part of a larger problem I've been working on for quite a while. Just to clarify, my final step looks good to you? except for I should have made the final x: x_(Capital N)

Answer 8

Here is a picture of the larger problem. I think the idea is proving an identity. The top of the page is my book definition of sigma^2 and mu, and I need to use them to show the alternate definition of sigma^2.

Answer 9

I think you are very close:
\[(1/N)\sum_{1}^{N}(Ci ^{2}-2C ^{i}\mu+\mu ^{2})\]
\[=\sum_{1}^{N}(C _{i}^{2}/N-2C _{i}/N+\mu ^{2}/N)\]
\[=\sum_{1}^{N}C _{i}^{2}/N-2\sum_{1}^{N}C _{i}/N+\sum_{1}^{N}\mu/N\]
\[=\sum_{1}^{N}C _{i}^{2}/N-2\sum_{1}^{N}C _{i}/N+\mu N/N\]
\[=\sum_{1}^{N}C _{i}^{2}/N-2\mu ^{2}+\mu ^{2}\]
\[=\sum_{1}^{N}C _{i}^{2}/N-\mu ^{2}\]

Answer 10

HTH

Answer 11

(forgot a mu next to the "2" in lines 2,3,4, but the result is the same)

Answer 12

Ok, I see what you are saying about there being a mu next to the 2 in line 2. In line 3, when you have substituted the summation in place of the mu however, should there be a c_i next to the 2 on the outside of the summation that was mu? @ybarrap

Answer 13

I'm a little confused by your question xar, is this the part you were confused by?
\[\large =\sum_{i=1}^N \frac{C^2_i}{N}-2\sum_{i=1}^N \frac{C_i \;\mu}{N}+\sum_{i=1}^N \frac{\mu^2}{N}\]

\[\large =\sum_{i=1}^N \frac{C^2_i}{N}-2\left(\color{royalblue}{\sum_{i=1}^N \frac{C_i \;}{N}}\right)\mu+\sum_{i=1}^N \frac{\mu^2}{N}\]

\[\large =\sum_{i=1}^N \frac{C^2_i}{N}-2\left(\color{royalblue}{\mu}\right)\mu+\sum_{i=1}^N \frac{\mu^2}{N}\]

Answer 14

Yes, the definition given for mu includes a c_i.. I guess I thought that it should be something like ... nevermind, I think I understand, the sum distributes to each term. Ug, thanks for your patience guys, I understand it now!

Answer 15

perfect!