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During high school I missed out on some of the math lessons due to health issues, one thing I missed out on was working with sums. I am now a second-year engineer student and I am still amazed how bad I am at this. I am reading a math course right now (statistics) and finally realized how big this issue is so here I am for help.

let's say we got a sample of x:

$x_1,x_2,...,x_9$

and we got these two equations:

$$\sum_{i=1}^9(ln(x_i)+ln(h))$$ $$\sum_{i=1}^9(ln(x_i)-ln(h))$$ h is some variable. How do we "break out" ln(h) of the sum so only $$\sum_{i=1}^9(ln(x_i))$$ is within the sum. I think it is possible to multiply ln(h) with 9 and put it infront like this: $$9ln(h)\sum_{i=1}^9(ln(x_i))$$ but I am not sure if it is correct and how the seccond equation would become? Is there any website with rules when working with this kind of summation?

J. Doe
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  • It's always useful to write out the sum like this: $$ \sum_{i=1}^9 \left( \ln{(x_i)} + \ln{(h)} \right) = \underbrace{\left( \ln{(x_1)} + \ln{(h)} \right)}{i =1} + \underbrace{\left( \ln{(x_2)} + \ln{(h)} \right)}{i = 2} + \ldots + \left( \ln{(x_9)} + \ln{(h)} \right) $$ From this you can probably already see where the $\ln{h}$ term should go. – Matti P. Oct 10 '18 at 12:07
  • $$\sum_{i=1}^9(ln(x_i)+ln(h))$$ is not an equation, it's an expression. – 5xum Oct 10 '18 at 12:10

3 Answers3

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$$\sum_{i=1}^9(\ln x_i+\ln h)$$ It is not permitted to multiply on the left by $\ln h$, but it is permitted to put the $\ln h$ terms in a separate sum: $$=\sum_{i=1}^9\ln x+\sum_{i=1}^9\ln h$$ Since this new sum's terms do not depend on the changing variable we can replace it by (the term) times (the number of indexes): $$=\sum_{i=1}^9\ln x+9\ln h$$ Similar reasoning applies for the sum with $-$ in it.

Parcly Taxel
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If you're confused about how to use shorthand notation like this, take a step back and consider what it is a shorthand for. In this case, $\sum_{i=1}^9(\ln(x_i)+\ln(h))$ is shorthand for $$ (\ln(x_1)+\ln(h)) + (\ln(x_2)+\ln(h)) + (\ln(x_3)+\ln(h)) + \\ (\ln(x_4)+\ln(h)) + (\ln(x_5)+\ln(h)) + (\ln(x_6)+\ln(h)) + \\ (\ln(x_7)+\ln(h)) + (\ln(x_8)+\ln(h)) + (\ln(x_9)+\ln(h)) $$ Once we remove the (now unnecessary) parentheses, we see that we have nine copies of $\ln(h)$ added together, and the standard shorthand for that is $9\cdot \ln(h)$. So using this and reorganising the terms a bit we get $$ 9\cdot \ln(h) + \ln(x_1) + \ln(x_2) + \ln(x_3) + \ln(x_4) + \\\ln(x_5) + \ln(x_6) + \ln(x_7) + \ln(x_8) + \ln(x_9) $$ And by now you can hopefully see that apart from the first term, all the other terms may be collected to become $\sum_{i = 1}^9\ln(x_i)$. So we get our answer $$ \sum_{i = 1}^9(\ln(x_i)+\ln(h)) = 9\cdot \ln(h) + \sum_{i = 1}^p\ln(x_i) $$

Arthur
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You should first convince yourself that you can "take out" $\ln(h)$ of both your numbers: $$\sum_{i=1}^9 (\ln(x_i) + \ln(h)) = 9\ln(h) + \sum_{i=1}^9 \ln(x_i)$$

You can also do it for the second number (be careful with the sign).

Maybe you could try to do some exercises from this here: https://www.math.ucdavis.edu/~kouba/CalcTwoDIRECTORY/summationdirectory/Summation.html

ngazagna
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