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

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TuringTestBest ResponseYou've already chosen the best response.3
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
 one year ago

TuringTestBest ResponseYou've already chosen the best response.3
at which point exactly do you feel lost?
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
I 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
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
but since it's (1x)^2 I wont be able to do that as easily
 one year ago

TuringTestBest ResponseYou've already chosen the best response.3
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 \[\frac1{(1x)^2}=\frac d{dx}\left(\frac1{1x}\right)\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.3
it does require an ability to recognize certain forms, which takes getting used to.
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
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)
 one year ago

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

TuringTestBest ResponseYou've already chosen the best response.3
or\[\frac a{1r}\]...
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
what about the R...radius of convergence
 one year ago

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

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

TuringTestBest ResponseYou've already chosen the best response.3
...and let a=1 like it says
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
does this mean we assume that x is less than 1
 one year ago

TuringTestBest ResponseYou've already chosen the best response.3
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\[x<R\impliesR<x<R\]means the radius of convergence for that series is \(R\)
 one year ago

TuringTestBest ResponseYou've already chosen the best response.3
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
 one year ago

TuringTestBest ResponseYou've already chosen the best response.3
maybe a good idea is to compare it with example 3 where the radius of convergence is not 1...
 one year ago

TuringTestBest ResponseYou've already chosen the best response.3
the 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\[r<R\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.3
*sorry I mean\[r<1\]
 one year ago

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

TuringTestBest ResponseYou've already chosen the best response.3
typo above* should be a 1/5 in front of the series
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
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?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.3
I never wrote \[r=\frac15\]I substituted\[r=\frac x5\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.3
compare\[\frac 1{1\frac x5}~~~~~~~~~~~\frac1{1r}\]
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
Oh I see r always equaled to \[\frac{x}{5}\] I didn't scroll up far enough
 one year ago

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

MathSofiyaBest ResponseYou've already chosen the best response.1
I think it's making sense now
 one year ago

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

MathSofiyaBest ResponseYou've already chosen the best response.1
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.
 one year ago

TuringTestBest ResponseYou've already chosen the best response.3
The same for me obviously. I just had to review it myself to get the hang of it :)
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
I'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!
 one year ago

TuringTestBest ResponseYou've already chosen the best response.3
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
 one year ago
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