Question

hi, for the expansion of power series (logarithmitic series) , ln(1+X) , why the condition for x is between -1 and 1 which x can be 1 but x cant be -1 ? 2. Relevant equations

Answer 1

Because log is not defined for negative numbers?

Answer 2

oh wait ..I read it wrong. 1 sec.

Answer 3

Can you revise the question?

Answer 4

`hi, for the expansion of power series (logarithmitic series) , ln(1+X)`
Do you mean ln(1-x) as is pictured?

Answer 5

yes

Answer 6

Mmm lemme just write some stuff out and see if I can make sense of this.

Answer 7

Since,
\[\Large\bf\sf \frac{d}{dx}\ln(1-x)=\frac{-1}{1-x}\]That implies that,\[\Large\bf\sf \ln(1-x)=\int\limits \frac{-1}{1-x}\;dx\quad=\quad -\int\frac{1}{1-x}\;dx\]And then we can write that as a geometric series,\[\Large\bf\sf =-\int\limits \sum_{n=0}^{\infty}x^n\;dx\]

Answer 8

thanks, but wat do u mean? can u try to type out using LA tex ?

Answer 9

\[\Large\bf\sf =-\sum_{n=0}^{\infty} \int\limits x^n\;dx\quad=\quad -\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}\]

Answer 10

So then those terms are giving us,\[\Large\bf\sf -x-\frac{x^2}{2}-\frac{x^3}{3}-...\]And the part we're confused on is the radius of convergence or something?

Answer 11

i can only see a bunch of equation... wat does large and bf , sf-x means?

Answer 12

Are you using Internet Explorer?
You probably should switch browsers.

It doesn't support the LaTeX plugin very well.
The equations aren't showing up correctly on your screen.

Answer 13

http://www.codecogs.com/latex/eqneditor.php i men this one...

Answer 14

i'm using chrome right now.....

Answer 15

We have the mathjax plugin on this site.
The equations should show up correctly on here.
I'm not sure why it's not working for you..

Answer 16

mathjax plug in? need to download? how to use it?

Answer 17

No it should work automatically :(
You're on chrome? Hmm that's weird..

Answer 18

can you suggest what should i do now to view the equation?

Answer 19

Is this your first time using the site?
Maybe you just need a reboot or something +_+
Blah I dunno!!

This is an old old thread but matt's response(tips for fixing latex display) near the top might still be relevant.
http://openstudy.com/study#/updates/4f4aa4c4e4b00c3c5d337cd2

Answer 20

thanks ! i can view it now after rebooting... i will view it later... currently busy with my other homework...

Answer 21

So I wrote it as a geometric series.
Recall that the geometric series only converges when that term is less than 1.\[\Large\bf\sf |x|\lt1\]Which we can write as,\[\Large\bf\sf -1\lt x\lt 1\]Hmm but that still doesn't explain the allowance for the lower bound.. hmmm

Answer 22

Maybe I misunderstand this question, but the way I read it all you're asking is to validate the regions of convergence, which is a trivial thing to do, but needs a careful analysis of the so called upper and lower bounds, or region endings. \[\large \log(1+x) = \sum_{n=1}^{+ \infty} \frac{(-1)^{n+1}}{n}x^n \]
so for \(x=1\) (only focus on the right hand side) we get \[ \large \sum_{n=1}^{+ \infty} \frac{(-1)^{n+1}}{n}\] which converges, it's the alternating harmonic series, and for \(x=-1\) we get \[\large \sum_{n=1}^{+ \infty} \frac{(-1)^{n+1}(-1)^n}{n}= - \sum_{n=1}^{+ \infty} \frac{1}{n} \] which diverges, because it's the harominc series.

Answer 23

If your question concerns the shifted \( \log(1-x)\) then you should already, from my example above, be able to see why it now works the other way around, meaning that it converges for \( -1 \leq x <1 \)

Answer 24

if x=1 then shoudnt the RHS =1^(n) , why you left it out?

Answer 25

\(1^n=1, \forall n \in \mathbb{Z}\), we sum over \(\mathbb{N}\) so you can say that:
\[1^1=1^2=1^3= \dots =1^n = 1 \]

Answer 26

You can't include -1 because \(\ln(-1+1)=\ln0\) is undefined.