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MathSofiya

  • 3 years ago

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

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  1. TuringTest
    • 3 years ago
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    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
    • 3 years ago
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    at which point exactly do you feel lost?

  3. MathSofiya
    • 3 years ago
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    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
    • 3 years ago
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    but since it's (1-x)^2 I wont be able to do that as easily

  5. TuringTest
    • 3 years ago
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    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
    • 3 years ago
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    it does require an ability to recognize certain forms, which takes getting used to.

  7. MathSofiya
    • 3 years ago
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    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
    • 3 years ago
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    of *a function

  9. TuringTest
    • 3 years ago
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    exactly

  10. TuringTest
    • 3 years ago
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    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
    • 3 years ago
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    I see

  12. TuringTest
    • 3 years ago
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    or\[\frac a{1-r}\]...

  13. MathSofiya
    • 3 years ago
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    what about the R...radius of convergence

  14. TuringTest
    • 3 years ago
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    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
    • 3 years ago
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    so we just changed the names of our variables and do an index shift to get equation 1

  16. TuringTest
    • 3 years ago
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    ...and let a=1 like it says

  17. MathSofiya
    • 3 years ago
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    does this mean we assume that x is less than 1

  18. TuringTest
    • 3 years ago
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    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
    • 3 years ago
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    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
    • 3 years ago
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    makes sense.

  21. TuringTest
    • 3 years ago
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    maybe a good idea is to compare it with example 3 where the radius of convergence is not 1...

  22. TuringTest
    • 3 years ago
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    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
    • 3 years ago
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    *sorry I mean\[|r|<1\]

  24. TuringTest
    • 3 years ago
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    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
    • 3 years ago
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    typo above* should be a 1/5 in front of the series

  26. MathSofiya
    • 3 years ago
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    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
    • 3 years ago
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    I never wrote \[r=\frac15\]I substituted\[r=\frac x5\]

  28. TuringTest
    • 3 years ago
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    compare\[\frac 1{1-\frac x5}~~~~~~~~~~~\frac1{1-r}\]

  29. MathSofiya
    • 3 years ago
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    Oh I see r always equaled to \[\frac{x}{5}\] I didn't scroll up far enough

  30. TuringTest
    • 3 years ago
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    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
    • 3 years ago
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    I think it's making sense now

  32. TuringTest
    • 3 years ago
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    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
    • 3 years ago
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    written*

  34. MathSofiya
    • 3 years ago
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    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
    • 3 years ago
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    The same for me obviously. I just had to review it myself to get the hang of it :)

  36. MathSofiya
    • 3 years ago
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    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
    • 3 years ago
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    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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