Question

Quick question on sigma notation

Alright, so I'm given a series and I'm supposed to rewrite it in sigma notation. I'll attach the question and my answer to a post below.

I'm not sure if I've solved this correctly; my math course is very formal, even in teaching, which sometimes makes it hard to learn. Thanks in advance!

P.S. Am I allowed to factor out the (2/n) from the sum? I don't know if that's acceptable or not, but I can't see how it wouldn't be.

Answer 1

you can factor out the x's
don't know what to give as an example, but by integration by parts, of e^x/x dx
(differentiating the bottom x) I get:

e^x/x + e^x/x^2 + 2e^x/x^3 + 6e^x/x^4

where I write the sum as: \(\large\color{blueviolet}{ \displaystyle \sum_{ n=0 }^{ \infty } ~ (e^x~n!)/(xe^n)}\) and there I go ahead and factor of e^x.

Answer 2

I mean to factor out of e^x/x

Answer 3

your case is not like this

Answer 4

you have an "n" variable, the variable that represents a pattern, and by taking it out of the summation you will ruin the pattern and also the factored thing with an n, (be it n/2 as you did) itself would make no sense outside the sigma

Answer 5

So, no, I don't think you can take out n/2 like this

Answer 6

wait, is this pattern in your "question" coming from somewhere, or is it the original question?

Answer 7

In the picture, the expression labeled "the question" is a screenshot of the original question that I'm given

Answer 8

oh, i c

Answer 9

your expression is hard to understand.

Answer 10

you have n as number of terms, same as the n's inside the pattern.

Answer 11

My answer is hard to understand? Or are you saying the original expression is hard to understand

Answer 12

answer.

Answer 13

why did you put n on top?

Answer 14

I am actually lost reading this pattern-:(

Answer 15

I would imagine something like:

\(\large\color{blueviolet}{ [1-(\frac{\LARGE 2}{\LARGE x}-1)^2]\frac{\LARGE 2}{\LARGE x}}\)

\(\huge\color{blueviolet}{^._.}\)
\(\huge\color{blueviolet}{^._.}\)
\(\huge\color{blueviolet}{^._.}\)

\(\large\color{blueviolet}{ [1-(\frac{\LARGE 2n}{\LARGE x}-1)^2]\frac{\LARGE 2}{\LARGE x}}\)

Answer 16

saying that you multiply the top 2 of that inner fraction times 1, times 2, times 3, times, 4 and on till times n.

Answer 17

I assumed the original expression said the beginning of the pattern was [1 - ((2/n) - 1)^2] * (2/n) and the end of the pattern was [1 - ((2n/n) - 1)^2] * (2/n).

So if the sum starts at i = 1, then it would end at whatever n is

Answer 18

but you see what I am saying here?

Answer 19

I get what you mean by having 2/x instead of 2/n. But in the original expression they have n as the variable; not x. So wouldn't that mean that if n = 7, then the fraction would be like 2/7? Because that wouldn't be the same thing as 2/x

Answer 20

oh then yes, you can take n/2 out

Answer 21

your wouldn't be able any terms with "i" out, but n is just a variable, that is not the pattern variable (the pattern variable is i).

Answer 22

Here is a similar question (along with answer) that came from the same textbook

Answer 23

sorry for confusing you.
And yes, you are right.

Answer 24

yes, the second one is good as well

Answer 25

Alright, thanks Solomon :) and no problem, it's a really weird expression

Answer 26

yes, you bet, lol

Answer 27

what math course are you taking btw?

Answer 28

It's called AP Calculus BC but I just call it painful

Answer 29

well, AP courses are all harder....

Answer 30

they are preparing you well for the non-closed form answers

Answer 31

you will likely need sigma a lot to write your answers....

Answer 32

anyways, good luck with it!

Answer 33

Thanks, Solomon :)

Answer 34

yw