# How to sum numerator and denominator of a fraction?

I want to do sum over this. Can apply the summation to top and bottom separately?

$$\sum\limits_{i=1}^{n} \frac{-a(x_i-\mu)^2}{x_i}$$ $$=\frac{\sum\limits_{i=1}^{n}-a(x_i-\mu)^2}{\sum\limits_{i=1}^{n}x_i}$$

Is this correct?

Where can I find the rules to summations of a division, product, or addition?

Edit:
I want to solve for y, $$\frac{5}{y}=\sum\limits_{i=1}^{n} \frac{-a(x_i-\mu)^2}{x_i}$$ $$\frac{y}{5}=\sum\limits_{i=1}^{n} \frac{x_i}{-a(x_i-\mu)^2}$$

Is this correct?

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$\frac13+\frac23=\frac33$, not $\frac36$ – Mike May 18 '13 at 20:35
simple and elegant, thanks! – user13985 May 18 '13 at 20:39
If it looks like it's too simple to work, it probably doesn't. Could you cancel the $6$s in $\frac{16}{64} = \frac{1}{4}$? – Eric Jablow May 18 '13 at 21:22
Remember the rule: in mathematics, Nothing is True unless there’s a reason for it to be true. – Lubin May 18 '13 at 22:32
The part in the edit is incorrect. – egreg May 18 '13 at 23:36

No. For example $\frac{3 + 2}{1 + 1} = \frac{5}{2} \neq 5 = \frac{3}{1} + \frac{2}{1}$. There are no basic rules for division or product. Of cource, $\sum a_n + b_n = \sum a_n + \sum b_n$ which follows from definition.

One trick to use with products though is that $\ln \prod a_n = \sum \ln a_n$.

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$\ln \prod a_i = \sum \ln a_i$ where $a_i= \frac{1}{x_i}$ is ok? – user13985 May 18 '13 at 21:20
Yes, $a_i$ can be anything other than $0$. If there is some $a_i = 0$, then the product is $0$. – Jon Claus May 18 '13 at 21:24
Hi, I edit original question, is this true? – user13985 May 18 '13 at 23:33

As the example in the comment shows nicely, no - Another simple example shows why this can't work: $$\frac 12 + \frac 12 \neq \frac 24 = \frac 12$$

Next: You might want to visit Wikipedia's entry for Summation, which includes a nice list of valid manipulations that hold for sums.

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@Amzoti, lol, are you serious? – user13985 May 19 '13 at 13:35

One thing you can do which might be helpful is, assuming $x_i>0$ for all $i$, $$\frac{-a(x_i-\mu)^2}{x_i}=\frac{-a(x_i-\mu)^2}{(\sqrt{x_i})^2}=-a\left(\sqrt{x_i}-\frac{\mu}{\sqrt{x_i}}\right)^2$$

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I added "edit" section, how to solve for y? – user13985 May 18 '13 at 23:42