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

This Question is Closed

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.3so 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
 4 years ago
Best ResponseYou've already chosen the best response.3at which point exactly do you feel lost?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I believe I do. I compare it to equation 2 \[\sum_{n=0}^\infty x^n=\frac{1}{1x}\] 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but since it's (1x)^2 I wont be able to do that as easily

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.3you 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 \[\frac1{(1x)^2}=\frac d{dx}\left(\frac1{1x}\right)\]

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.3it does require an ability to recognize certain forms, which takes getting used to.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so the whole idea behind the integration or finding the derivative of the function is to make the function into that simple form (equation 2)

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.3the goal is to somehow get your problem into the form\[\frac1{1r}\]so you can represent it as a power series

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.3or\[\frac a{1r}\]...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0what about the R...radius of convergence

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.3well you know (hopefully) that for a geometric series\[\sum_{n=1}^\infty ar^{n1}=\frac a{1r}~\textbf{ iff }~~\r<1\]

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.3so we just changed the names of our variables and do an index shift to get equation 1

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.3...and let a=1 like it says

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0does this mean we assume that x is less than 1

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.3if 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\[x<R\impliesR<x<R\]means the radius of convergence for that series is \(R\)

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.3in 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

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.3maybe a good idea is to compare it with example 3 where the radius of convergence is not 1...

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.3the trick of going from\[\frac1{5x}\to\frac15\frac1{1\frac x5}\]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\[\frac15\frac1{1r}=\sum_{n=0}^\infty r^n\]for which we know the radius of convergence is\[r<R\]

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.3*sorry I mean\[r<1\]

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.3so in this case in ex3 we get\[\frac x5<1\implies5<x<5\]so we find the radius of convergence to be 5

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.3typo above* should be a 1/5 in front of the series

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0got 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
 4 years ago
Best ResponseYou've already chosen the best response.3I never wrote \[r=\frac15\]I substituted\[r=\frac x5\]

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.3compare\[\frac 1{1\frac x5}~~~~~~~~~~~\frac1{1r}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh I see r always equaled to \[\frac{x}{5}\] I didn't scroll up far enough

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.3so we go straight to\[\frac15\frac1{1r}\text{converges}\impliesr<1\implies\frac x5<1\implies5<x<5\implies R=5\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I think it's making sense now

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.3should have writtenm\[\frac15\sum_{n=0}^\infty r^n\text{converges}\impliesr<1\implies\frac x5<1\]\[\implies5<x<5\implies R=5\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It'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
 4 years ago
Best ResponseYou've already chosen the best response.3The same for me obviously. I just had to review it myself to get the hang of it :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I'll do the problem \[\frac{3}{1x^4}\] 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
 4 years ago
Best ResponseYou've already chosen the best response.3Thank 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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