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

- anonymous

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- 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

at which point exactly do you feel lost?

- 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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## More answers

- anonymous

but since it's (1-x)^2 I wont be able to do that as easily

- 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

it does require an ability to recognize certain forms, which takes getting used to.

- 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

of *a function

- TuringTest

exactly

- 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

I see

- TuringTest

or\[\frac a{1-r}\]...

- anonymous

what about the R...radius of convergence

- 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

so we just changed the names of our variables and do an index shift to get equation 1

- TuringTest

...and let a=1 like it says

- anonymous

does this mean we assume that x is less than 1

- 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

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

makes sense.

- TuringTest

maybe a good idea is to compare it with example 3 where the radius of convergence is not 1...

- 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

*sorry I mean\[|r|<1\]

- TuringTest

so in this case in ex3 we get\[|\frac x5|<1\implies-5

- TuringTest

typo above*
should be a 1/5 in front of the series

- 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

I never wrote \[r=\frac15\]I substituted\[r=\frac x5\]

- TuringTest

compare\[\frac 1{1-\frac x5}~~~~~~~~~~~\frac1{1-r}\]

- anonymous

Oh I see r always equaled to \[\frac{x}{5}\] I didn't scroll up far enough

- TuringTest

so we go straight to\[\frac15\frac1{1-r}\text{converges}\implies|r|<1\implies|\frac x5|<1\implies-5

- anonymous

I think it's making sense now

- TuringTest

should have writtenm\[\frac15\sum_{n=0}^\infty r^n\text{converges}\implies|r|<1\implies|\frac x5|<1\]\[\implies-5

- TuringTest

written*

- 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

The same for me obviously. I just had to review it myself to get the hang of it :)

- 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

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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