Simple question. If you have a fraction where:

numerator = summation of Xi terms

denominator = summation of Xi^2 terms

Do standard rules of arithmetic apply? So that it equals the summation of 1/Xi ?

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- Feb 3rd 2013, 10:41 AMMN1987Summation division
Simple question. If you have a fraction where:

numerator = summation of Xi terms

denominator = summation of Xi^2 terms

Do standard rules of arithmetic apply? So that it equals the summation of 1/Xi ? - Feb 3rd 2013, 10:51 AMPazeRe: Summation division
$\displaystyle \frac{3i}{3i^2}\rightarrow 3i^-1\rightarrow \frac{3i}{3i^2}=\frac{1}{3i}$

Is this your question?

Or are you asking something in the lines of: $\displaystyle \frac{x+x^2+x^3}{x^2}=\frac{1}{x}+x+1$

?

P.S. I don't know how to make a negative exponent in latex. Only the - sign gets superscripted. Maybe someone can help me with that? - Feb 3rd 2013, 10:54 AMMN1987Re: Summation division
Replace the 3 in the first term with the sigma notation, and that's my question :) i.e., what does that simplify to?

- Feb 3rd 2013, 11:00 AMPlatoRe: Summation division
- Feb 3rd 2013, 11:03 AMHallsofIvyRe: Summation division
Paze, to get both $\displaystyle 3i^{-1}$, put the entire exponent, -1, in braces: {-1}.

MN1987, Paze is looking at the case of one term and I**think**he means $\displaystyle \frac{3i}{(3i)^2}= \frac{1}{3i}$

However, you seem to be talking about $\displaystyle \frac{\sum x_i}{\sum (x_i)^2}$ which is completely differenct. For example, if $\displaystyle x_i= i$, with i from 1 to 5, that becomes $\displaystyle \frac{1+ 2+ 3+ 4+ 5}{1+ 4+ 9+ 16+ 25}= \frac{15}{55}= \frac{3}{11}$ while if $\displaystyle x_1$, $\displaystyle x_2= 4$, $\displaystyle x_3= 5$, $\displaystyle x_4= 9$, $\displaystyle x_5= 12$ (chosen pretty much at random) then $\displaystyle \frac{\sum x_i}{\sum x_i^2}= \frac{1+ 4+ 5+ 9+ 12}{1+ 16+ 25+ 81+ 144}= \frac{31}{167}$.

There simply is no simple formula for a ratio of two sums. Addition and multiplicatio do not "play well together"! - Feb 3rd 2013, 11:08 AMPlatoRe: Summation division
We assume that you mean

**finite sums**.

$\displaystyle \frac{{\sum\limits_{k = 1}^n {x_k } }}{{\sum\limits_{k = 1}^n {\left( {x_k } \right)^2 } }}$

Let $\displaystyle S = \sum\limits_{k = 1}^n {x_k } \;\& \,T = \sum\limits_{k = 1}^n {\left( {x_k } \right)^2 } $

then $\displaystyle \[\frac{{\sum\limits_{k = 1}^n {x_k } }}{{\sum\limits_{k = 1}^n {\left( {x_k } \right)^2 } }} = \frac{{\sum\limits_{k = 1}^n {x_k } }}{T} = \sum\limits_{k = 1}^n {\frac{{x_k }}{T}} $

Is that what you mean? - Feb 3rd 2013, 11:22 AMMN1987Re: Summation division
Wow, I had no idea I had phrased my original question to be so ambiguous! I need to learn to use Latex notation.

Yes, I meant:

$\displaystyle \frac{{\sum\limits_{i = 1}^n {x_i } }}{{\sum\limits_{i = 1}^n {\left( {x_i } \right)^2 } }}$

I guess the consensus is that it does not simplify further than that, for all intents and purposes. It's obvious now looking at it... shame! It would've made my assignment problem alot easier to solve :)