## MathSofiya 3 years ago Find a power series representative of the function and determine it's interval of convergence. $g(x)=\frac{1}{(1-x)^2}$

1. 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

2. TuringTest

at which point exactly do you feel lost?

3. MathSofiya

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|

4. MathSofiya

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

5. 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 $\frac1{(1-x)^2}=\frac d{dx}\left(\frac1{1-x}\right)$

6. TuringTest

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

7. MathSofiya

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)

8. MathSofiya

of *a function

9. TuringTest

exactly

10. TuringTest

the goal is to somehow get your problem into the form$\frac1{1-r}$so you can represent it as a power series

11. MathSofiya

I see

12. TuringTest

or$\frac a{1-r}$...

13. MathSofiya

14. TuringTest

well you know (hopefully) that for a geometric series$\sum_{n=1}^\infty ar^{n-1}=\frac a{1-r}~\textbf{ iff }~~\|r|<1$

15. TuringTest

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

16. TuringTest

...and let a=1 like it says

17. MathSofiya

does this mean we assume that x is less than 1

18. 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$|x|<R\implies-R<x<R$means the radius of convergence for that series is $$R$$

19. 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

20. MathSofiya

makes sense.

21. TuringTest

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

22. 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$|r|<R$

23. TuringTest

*sorry I mean$|r|<1$

24. TuringTest

so in this case in ex3 we get$|\frac x5|<1\implies-5<x<5$so we find the radius of convergence to be 5

25. TuringTest

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

26. MathSofiya

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?

27. TuringTest

I never wrote $r=\frac15$I substituted$r=\frac x5$

28. TuringTest

compare$\frac 1{1-\frac x5}~~~~~~~~~~~\frac1{1-r}$

29. MathSofiya

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

30. TuringTest

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

31. MathSofiya

I think it's making sense now

32. TuringTest

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

33. TuringTest

written*

34. MathSofiya

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.

35. TuringTest

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

36. MathSofiya

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!

37. 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