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Math Help - Power series question (ROC)

  1. #1
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    Power series question (ROC)

    hello,

    I see now that i did these power series by pattern and now I'm stuck big time

    Problem is ...

    \displaystyle \sum_{n=1}^{\infty} \frac {n}{3^nx^{2n}}

    so i found R as

    \displaystyle R = \lim_{n\to \infty} \big{|} \frac {a_n}{a_{n+1}}\big{|} = ... = 3

    and than i sow x is at power -2n and here is my problem...

    should I do it like this : (to see region of convergence)

    \displaystyle |x|< 3



    or like this ...

    \displaystyle |x^{-2}|< 3

    \displaystyle |x|< \frac{1}{\sqrt{3}}

    or I am way of with this ?

    can someone please explain to me what to do when i have "x" like i do here

    thanks in advance
    Last edited by sedam7; April 28th 2011 at 01:25 PM.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by sedam7 View Post
    hello,

    I see now that i did these power series by pattern and now I'm stuck big time

    Problem is ...

    \displaystyle \sum_{n=1}^{\infty} \frac {n}{3^nx^{2n}}

    so i found R as

    \displaystyle R = \lim_{n\to \infty} \big{|} \frac {a_n}{a_{n+1}}\big{|} = ... = 3

    and than i sow x is at power -2n and here is my problem...

    should I do it like this : (to see region of convergence)

    \displaystyle |x|< 3



    or like this ...

    \displaystyle |x^{-2}|< 3

    \displaystyle |x|< \frac{1}{\sqrt{3}}

    or I am way of with this ?

    can someone please explain to me what to do when i have "x" like i do here

    thanks in advance
    My first question to you (and I don't know the answer) is if you can use that test when the problem is not a power series.

    In any event you have your logic backward. |x| < \frac{1}{\sqrt{3}} is not the radius of convergence. The series converges on |x| > \frac{1}{\sqrt{3}}. So this is not a "radius" of convergence at all.

    -Dan
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  3. #3
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    hm... what do you mean ? can I use that test when series is not power series ? (I used D'Alambert criteria for testing of convergence )

    \displaystyle \sum_{n = 0}^\infty \frac{n}{3^n} x^{-2n}

    so i got

    a_n = \frac {n}{3^n}

    Than ...

    R=\lim_{x \to \infty} |\frac{a_n}{ a_{n+1}} | = \lim_{x \to \infty} \frac{\frac{n}{3^n}}{\frac{n+1}{3^n \cdot 3}} = 3

    then since in power series |x| <R is region of convergence .... i assumed that |x|<3 but... if I put x = 4 ... than .... using D'Alambert test i get ... .

    q= \lim_{x \to \infty} \frac{a_{n+1}}{a_n} = .... = \lim_{x \to \infty} \frac{n+1}{64n} = \frac {1}{64} < 1

    so series converge ? or I probably should go to sleep now and try it tomorow
    Last edited by sedam7; April 28th 2011 at 02:02 PM. Reason: Fixed some of the LaTeX
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by sedam7 View Post
    hm... what do you mean ? can I use that test when series is not power series ? (I used D'Alambert criteria for testing of convergence )

    \displaystyle \sum_{n = 0}^\infty \frac{n}{3^n} x^{-2n}

    so i got

    a_n = \frac {n}{3^n}

    Than ...

    R=\lim_{x \to \infty} |\frac{a_n}{ a_{n+1}} | = \lim_{x \to \infty} \frac{\frac{n}{3^n}}{\frac{n+1}{3^n \cdot 3}} = 3

    then since in power series |x| <R is region of convergence .... i assumed that |x|<3 but... if I put x = 4 ... than .... using D'Alambert test i get ... .

    q= \lim_{x \to \infty} \frac{a_{n+1}}{a_n} = .... = \lim_{x \to \infty} \frac{n+1}{64n} = \frac {1}{64} < 1

    so series converge ? or I probably should go to sleep now and try it tomorow
    Well as I said, I didn't know if the test were more general than power series.

    Look, if nothing else convinces you, try calculating x = 0.5. This is inside what you call your radius of convergence. It does not converge. Whereas, as you say, x = 4 does converge.

    -Dan
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    Thanks I was to tired and there's language issue ... and... Thank you very very much

    you have saved me



    |\frac{1}{x}|<3

    x\in (-\infty, -3] U [3, +\infty)


    I'm so stupid... sorry that i bother everybody with such stupidity ....
    Last edited by sedam7; April 28th 2011 at 02:22 PM.
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by sedam7 View Post
    Thanks I was to tired and there's language issue ... and... Thank you very very much

    you have saved me



    |\frac{1}{x}|<3

    x\in (-\infty, -3] U [3, +\infty)


    I'm so stupid... sorry that i bother everybody with such stupidity ....
    You aren't stupid. You made a mistake. Stupidity would have been not asking for help.

    -Dan
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  7. #7
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    Quote Originally Posted by sedam7 View Post
    |\frac{1}{x}|<3
    x\in (-\infty, -3] U [3, +\infty)
    You need to rethink that answer.
    The series converges for x=0.6 the sum is \frac{674}{4}.
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  8. #8
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    Quote Originally Posted by topsquark View Post
    This was bugging me and since I put in the work I thought I'd share. So the series converges for
    |x| > \frac{1}{\sqrt{3}}
    That is correct. I will point out that the absolute value of the root test works nicely.
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  9. #9
    Senior Member yeKciM's Avatar
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    yes, converge for |x| > \frac{1}{\sqrt{3}}

    and points  \displaystyle \frac {1}{\sqrt{3}} are not included
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  10. #10
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    okay i see what i have done sorry
    now I'm confused... in this problem i have to find radius of convergence ? what to say on this now ? heheheheh

    because these series converge in zero and for |x| > 1/ \sqrt{3} ?
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  11. #11
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    The difficulty is, as TopQuark said initially, that your series is NOT a power series. It does not have x to positive powers and so does NOT have a "radius of convergence". With x to a negative power, you get, as shown, that it converges outside some radius. Now, it is possible that the series is actually Sum (n/3^n) x^n?
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  12. #12
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    hmmmm....
    i was doing some problems from exams last year and i come to this one

    Find radius of convergence of the series:

    \displaystyle \sum_{n=1} ^{\infty} \frac{n}{3^n x^{2n}}

    then examine the convergence of the boundary points of intervals.
    so i have done what I wrote there and didn't know what to do else so i posted it here if someone could explain it ... actually as I look at it better ... three years ago was more or less the same problem ...

    Find radius of convergence of the series:

    \displaystyle \sum_{n=1} ^{\infty} \frac{n}{4^n x^{2n}}

    then examine the convergence of the boundary points of intervals.
    actually it's same just with 4^n ....


    now I'm very very confused soon I'll have exam and can't find any of them to see what to do with this or I'll just hope something like this I will not get
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  13. #13
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    Quote Originally Posted by sedam7 View Post
    hmmmm....
    i was doing some problems from exams last year and i come to this one so i have done what I wrote there and didn't know what to do else so i posted it here if someone could explain it ... actually as I look at it better ... three years ago was more or less the same problem ...
    actually it's same just with 4^n ....
    now I'm very very confused soon I'll have exam and can't find any of them to see what to do with this or I'll just hope something like this I will not get
    No one can blame you for being confused.
    I personally think that the instructor who set this question is badly confused.

    That ‘old’ series converges if |x|>\dfrac{1}{2} or in the set \left( { - \infty , - \frac{1}{2}} \right) \cup \left( {\frac{1}{2},\infty } \right).
    That is unbounded set. What is the radius of an unbounded set?
    You see why that instructor’s question is nonsense?
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