1. ## Summation 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 ?

2. ## Re: Summation division

$\frac{3i}{3i^2}\rightarrow 3i^-1\rightarrow \frac{3i}{3i^2}=\frac{1}{3i}$

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

?

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?

3. ## Re: 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?

4. ## Re: Summation division

Originally Posted by Paze
$\frac{3i}{3i^2}\rightarrow 3i^{-1}\rightarrow \frac{3i}{3i^2}=\frac{1}{3i}$
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?
[TEX]3i^{-1}[/TEX] gives $3i^{-1}$.

If an exponent has more that one character in it use {}.
It is a good habit to use {} for all exponents.

5. ## Re: Summation division

Paze, to get both $3i^{-1}$, put the entire exponent, -1, in braces: {-1}.

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

However, you seem to be talking about $\frac{\sum x_i}{\sum (x_i)^2}$ which is completely differenct. For example, if $x_i= i$, with i from 1 to 5, that becomes $\frac{1+ 2+ 3+ 4+ 5}{1+ 4+ 9+ 16+ 25}= \frac{15}{55}= \frac{3}{11}$ while if $x_1$, $x_2= 4$, $x_3= 5$, $x_4= 9$, $x_5= 12$ (chosen pretty much at random) then $\frac{\sum x_i}{\sum x_i^2}= \frac{1+ 4+ 5+ 9+ 12}{1+ 16+ 25+ 81+ 144}= \frac{31}{167}$.
There simply is no simple formula for a ratio of two sums. Addition and multiplicatio do not "play well together"!

6. ## Re: Summation division

Originally Posted by MN1987
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 ?
We assume that you mean finite sums.

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

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

then $\[\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?

7. ## Re: 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:
$\frac{{\sum\limits_{i = 1}^n {x_i } }}{{\sum\limits_{i = 1}^n {\left( {x_i } \right)^2 } }}$

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