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Find a power series representative of the function and determine it's interval of convergence. \[g(x)=\frac{1}{(1-x)^2}\]

Mathematics
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so do you understand how they get the power series representation of the function in that example? http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeriesandFunctions.aspx
at which point exactly do you feel lost?
I believe I do. I compare it to equation 2 \[\sum_{n=0}^\infty x^n=\frac{1}{1-x}\] so in this case it's \[\sum_{n=0}^\infty\] somehow I want to do this, make the equation into a similar form and substitute into the sum : |dw:1342721245413:dw|

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but since it's (1-x)^2 I wont be able to do that as easily
you have to think about how to reduce the power on the bottom term. hopefully integration comes to mind as a way to do that. Thinking along the lines of the fundamental theorem of calculus though here we don't actually integrate. Instead we just identify the fact that \[\frac1{(1-x)^2}=\frac d{dx}\left(\frac1{1-x}\right)\]
it does require an ability to recognize certain forms, which takes getting used to.
so the whole idea behind the integration or finding the derivative of the function is to make the function into that simple form (equation 2)
of *a function
exactly
the goal is to somehow get your problem into the form\[\frac1{1-r}\]so you can represent it as a power series
I see
or\[\frac a{1-r}\]...
what about the R...radius of convergence
well you know (hopefully) that for a geometric series\[\sum_{n=1}^\infty ar^{n-1}=\frac a{1-r}~\textbf{ iff }~~\|r|<1\]
so we just changed the names of our variables and do an index shift to get equation 1
...and let a=1 like it says
does this mean we assume that x is less than 1
if |x| is not less than 1 then the series is divergent, so the radius of convergence R is the limit as to what x can be\[|x|
in this case we know that for *any* geometric series we have from equation 1 that \[|x|<1\]and in the case of example 4 the radius of convergence is therefor 1
makes sense.
maybe a good idea is to compare it with example 3 where the radius of convergence is not 1...
the trick of going from\[\frac1{5-x}\to\frac15\frac1{1-\frac x5}\]is the same trick; we try to get this guy into the form in equation 1 in this case you ran say \(r=\frac x5\) and you get\[\frac15\frac1{1-r}=\sum_{n=0}^\infty r^n\]for which we know the radius of convergence is\[|r|
*sorry I mean\[|r|<1\]
so in this case in ex3 we get\[|\frac x5|<1\implies-5
typo above* should be a 1/5 in front of the series
got it. so \[r=\frac{1}{5}\] and \[\frac{1}{5} < R\] and R=1 so how did we get to \[\frac{x}{5}<1\] where did the x come from?
I never wrote \[r=\frac15\]I substituted\[r=\frac x5\]
compare\[\frac 1{1-\frac x5}~~~~~~~~~~~\frac1{1-r}\]
Oh I see r always equaled to \[\frac{x}{5}\] I didn't scroll up far enough
so we go straight to\[\frac15\frac1{1-r}\text{converges}\implies|r|<1\implies|\frac x5|<1\implies-5
I think it's making sense now
should have writtenm\[\frac15\sum_{n=0}^\infty r^n\text{converges}\implies|r|<1\implies|\frac x5|<1\]\[\implies-5
written*
It's a lot of info, and I *think* I get it now. I'll post a problem and do it, to test my understanding. I think I just have to do a problem like this once or twice for it to become second nature. But the above makes sense.
The same for me obviously. I just had to review it myself to get the hang of it :)
I'll do the problem \[\frac{3}{1-x^4}\] as far as I can and find the radius of convergence...and post it. Thanks again @TuringTest !!!!! I just went from clueless to finally getting a grip on Power series!
Thank you so much that wonderful comment. The real rewards go to the person who puts in the effort to learn though, so all credit to you I say :D

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