anonymous
  • anonymous
Find a power series representative of the function and determine it's interval of convergence. \[g(x)=\frac{1}{(1-x)^2}\]
Mathematics
chestercat
  • chestercat
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TuringTest
  • TuringTest
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
TuringTest
  • TuringTest
at which point exactly do you feel lost?
anonymous
  • anonymous
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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anonymous
  • anonymous
but since it's (1-x)^2 I wont be able to do that as easily
TuringTest
  • TuringTest
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 \[\frac1{(1-x)^2}=\frac d{dx}\left(\frac1{1-x}\right)\]
TuringTest
  • TuringTest
it does require an ability to recognize certain forms, which takes getting used to.
anonymous
  • anonymous
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)
anonymous
  • anonymous
of *a function
TuringTest
  • TuringTest
exactly
TuringTest
  • TuringTest
the goal is to somehow get your problem into the form\[\frac1{1-r}\]so you can represent it as a power series
anonymous
  • anonymous
I see
TuringTest
  • TuringTest
or\[\frac a{1-r}\]...
anonymous
  • anonymous
what about the R...radius of convergence
TuringTest
  • TuringTest
well you know (hopefully) that for a geometric series\[\sum_{n=1}^\infty ar^{n-1}=\frac a{1-r}~\textbf{ iff }~~\|r|<1\]
TuringTest
  • TuringTest
so we just changed the names of our variables and do an index shift to get equation 1
TuringTest
  • TuringTest
...and let a=1 like it says
anonymous
  • anonymous
does this mean we assume that x is less than 1
TuringTest
  • TuringTest
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|
TuringTest
  • TuringTest
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
anonymous
  • anonymous
makes sense.
TuringTest
  • TuringTest
maybe a good idea is to compare it with example 3 where the radius of convergence is not 1...
TuringTest
  • TuringTest
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|
TuringTest
  • TuringTest
*sorry I mean\[|r|<1\]
TuringTest
  • TuringTest
so in this case in ex3 we get\[|\frac x5|<1\implies-5
TuringTest
  • TuringTest
typo above* should be a 1/5 in front of the series
anonymous
  • anonymous
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?
TuringTest
  • TuringTest
I never wrote \[r=\frac15\]I substituted\[r=\frac x5\]
TuringTest
  • TuringTest
compare\[\frac 1{1-\frac x5}~~~~~~~~~~~\frac1{1-r}\]
anonymous
  • anonymous
Oh I see r always equaled to \[\frac{x}{5}\] I didn't scroll up far enough
TuringTest
  • TuringTest
so we go straight to\[\frac15\frac1{1-r}\text{converges}\implies|r|<1\implies|\frac x5|<1\implies-5
anonymous
  • anonymous
I think it's making sense now
TuringTest
  • TuringTest
should have writtenm\[\frac15\sum_{n=0}^\infty r^n\text{converges}\implies|r|<1\implies|\frac x5|<1\]\[\implies-5
TuringTest
  • TuringTest
written*
anonymous
  • anonymous
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.
TuringTest
  • TuringTest
The same for me obviously. I just had to review it myself to get the hang of it :)
anonymous
  • anonymous
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!
TuringTest
  • TuringTest
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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