I need some help with a summation problem i'm having.

dennis1590

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Aug 8, 2018
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[FONT=&quot]nΣxi2- (Σ[/FONT][FONT=&quot]x[/FONT]i)2 [FONT=&quot]= nΣ([/FONT][FONT=&quot]x[/FONT]i[FONT=&quot] - xavg[/FONT][FONT=&quot])2

So how is the left side converted to the right side. I tried everything. I only know that [/FONT]
[FONT=&quot]Σ[/FONT][FONT=&quot]x[/FONT]i =n​[FONT=&quot]x[/FONT]avg
 
Please show us one or two things that you tried.



When you have some time, please read the forum guidelines. Thank you! :cool:

Yes OK.Well. I did something like this. Sorry for the crooked english It's not my mothertongue.

nΣxi2- (Σxi)2 = nΣxi2- (nxavg)2 -> nΣxi2- n2xavg2 =n(Σxi2- xavgΣxi) = nΣ(xi2- xavgxi)

That's about the closest i got.
 
Seems like you didn't pay attention, during the registration process. Could that be why you're submitting all of these duplicate posts?

This notice
was written for you and your friends! :p

Yeah. This math forum is like rocket science compared to your averaged fora. And I was checking and hoping for a clue for my little problem. Turned out i'm ****ing with myself. This equation is from multivariate regressing BTW. Wanna make an equation to predict the alcohol content of my own brewed beer.
 
Yeah. This math forum is like rocket science compared to your averaged fora. And I was checking and hoping for a clue for my little problem. Turned out i'm ****ing with myself. This equation is from multivariate regressing BTW. Wanna make an equation to predict the alcohol content of my own brewed beer.
\(\displaystyle \displaystyle y = \left ( \sum_{j=1}^n (x_j - \bar x)^2 \right ), \text { where } \bar x = \dfrac{1}{n} * \left ( \sum_{j =1}^n x_j \right ).\)

\(\displaystyle \displaystyle \bar x = \dfrac{1}{n} * \left ( \sum_{j =1}^n x_j \right ) \implies n \bar x = \sum_{j=1}^n x_j.\)

\(\displaystyle \displaystyle y = \left ( \sum_{j=1}^n (x_j - \bar x)^2 \right ) = \left ( \sum_{j=1}^n x_j^2 - 2 x_j \bar x + (\bar x)^2 \right ) \implies\)

\(\displaystyle \displaystyle y = \left ( \sum_{j=1}^n x_j^2 \right )- 2 \bar x * \left ( \sum_{j=1}^n x_j \right ) + \left ( (\bar x)^2 * \sum_{j=1}^n 1 \right ) = \left ( \sum_{j=1}^n x_j^2 \right ) - 2 \bar x * (n * \bar x ) + ( \bar x)^2 * n \implies \)

\(\displaystyle \displaystyle y = \left ( \sum_{j=1}^n x_j^2 \right ) - 2n( \bar x)^2 + n( \bar x )^2 = \left ( \sum_{j=1}^n x_j^2 \right ) - n( \bar x)^2.\)

There you go.

If you divide through by n you get

\(\displaystyle \displaystyle \dfrac{y}{n} = \left ( \dfrac{1}{n} * \sum_{j=1}^n x_j^2 \right ) - ( \bar x )^2\) or

the mean of the squares minus the square of the mean.
 
… This math forum is like rocket science compared to your averaged fora …

… Wanna make an equation to predict the alcohol content of my own brewed beer …
You've conducted a study of average fora, have you? I'm curious to learn the percentage of misinformation you discovered. Some forums are way above average; freemathhelp is one of them. After 15+ years tutoring on-line, we know what works.

(I'm disheartened, when students suggest that reading and following instruction are rocket science.)

And, no, I'm not interested in developing an ETOH content-formula, for your beer. I'd rather drink it. :cool:
 
You've conducted a study of average fora, have you? I'm curious to learn the percentage of misinformation you discovered. Some forums are way above average; freemathhelp is one of them. After 15+ years tutoring on-line, we know what works.

(I'm disheartened, when students suggest that reading and following instruction are rocket science.)

And, no, I'm not interested in developing an ETOH content-formula, for your beer. I'd rather drink it. :cool:

Well at most fora you just post something and get moderated later on. I just assumed that it would be the same over here :D.

Thanks for helping me ;). The solution is always so simple when you see it.
 
Well at most fora you just post something and get moderated later on. I just assumed that it would be the same over here …
The last time vBulletin's moderation system stopped working, the forum was bombarded by what seemed like a thousand SPAM threads and posts interrupting others' conversations (providing a phone number in Asia, to the Love Doctor). That particular onslaught lasted several days, while the site owner was away, and was tantamount to a denial of service. Each board was clogged up; nobody could use the forum! There have been other issues (eg: pornography, illegal activities), in the past. Moderating new members prevents other members from being exposed to it all (when the system's working, heh).
 
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